Twin Spheres Paradox
Assume we have two magic spheres that can pass through each other and we can get inside them. The inertial frame logic of light claims if either of us flash our light, all points on the inside of the sphere will be struck at the same time by the light regardless of our actual motion.
In other words, light expands spherically stationary to the frame from the point of emission in the frame. Now, somehow we put these spheres at the same point in space and I flash the light. Both of us then would see the light strike our sphere points all at the same time.
This time I am going to move very fast relative to the other sphere. However, your sphere looks like it is moving very fast relative to me and I am at rest. At the instant we occupy the same exact space, I flash the light. Light speed is not infinite so it does take time to strike each point on the sphere. Well, since light expands stationary to me it hits my points all at the same time.
But what about you? It cannot hit your points all at the same time and mine also because I am moving relative to you and we no longer occupy the same point in space. But, light from a moving light source is supposed to proceed outward spherically from the point of emission in your frame which is the center of your sphere according to the SR light postulate.
Since one light sphere cannot intersect the two rigid body spheres at all points in each sphere simultaneously, we have a contradiction. In particular, since the center of the one light sphere is centered in each frame, then the center of the one light sphere would have to occupy two different points in space.
Try drawing a picture of the event even using length contraction, it is impossible. Make the figures anyway you like and you still cannot express a geometry/picture that works for the two spheres and one light sphere.
More specifically, SR encourages the use of disjoint frame to frame logic when deducing relative motion. However, given one light sphere and two rigid body frame spheres in relative motion, both frames are forced together to solve the problem and the problem must be solved with all known information. As such the following three conditions must be met with one light sphere.
1) Light must proceed spherically from the center of the "rest" sphere.
2) Light must proceed spherically from the center of the "moving" sphere.
3) Because there exists relative motion, the two centers of the rigid body spheres do not occupy the same point in space.
In fact, by looking at just the x-axis of the spheres, the at rest observer would claim both equal marks at a distance r were struck at the same time, and in addition the moving observer would make the same claim.
However, the relativity of simultaneity suggests if the at rest observer views the event as simultaneous, then an observer moving in collinear relative motion would not judge the same event as simultaneous it its frame. In fact, both observers would agree on the ordinality of events in the moving frame. In particular, in Einstein's train experiment, both observer M and M' judged M' did not see the front and back of the train struck at the same time and both observers agreed the front was struck first. They just disagreed on the timing, not the ordinality.
Yet, both observers view the event above from the x-axis in a simultaneous fashion which contradicts the relativity of simultaneity.
Basically, since SR is unable to produce a set of equations that satisfies all the 3 conditions listed above, then SR's understanding of light is false and it is that simple.
QUOTE (qwtyu+Nov 5 2009, 01:37 AM)
Twin Spheres Paradox
Assume we have two magic spheres that can pass through each other and we can get inside them. The inertial frame logic of light claims if either of us flash our light, all points on the inside of the sphere will be struck at the same time by the light regardless of our actual motion.
In other words, light expands spherically stationary to the frame from the point of emission in the frame. Now, somehow we put these spheres at the same point in space and I flash the light. Both of us then would see the light strike our sphere points all at the same time.
This time I am going to move very fast relative to the other sphere. However, your sphere looks like it is moving very fast relative to me and I am at rest. At the instant we occupy the same exact space, I flash the light. Light speed is not infinite so it does take time to strike each point on the sphere. Well, since light expands stationary to me it hits my points all at the same time.
But what about you? It cannot hit your points all at the same time and mine also because I am moving relative to you and we no longer occupy the same point in space. But, light from a moving light source is supposed to proceed outward spherically from the point of emission in your frame which is the center of your sphere according to the SR light postulate.
Since one light sphere cannot intersect the two rigid body spheres at all points in each sphere simultaneously, we have a contradiction. In particular, since the center of the one light sphere is centered in each frame, then the center of the one light sphere would have to occupy two different points in space.
Try drawing a picture of the event even using length contraction, it is impossible. Make the figures anyway you like and you still cannot express a geometry/picture that works for the two spheres and one light sphere.
More specifically, SR encourages the use of disjoint frame to frame logic when deducing relative motion. However, given one light sphere and two rigid body frame spheres in relative motion, both frames are forced together to solve the problem and the problem must be solved with all known information. As such the following three conditions must be met with one light sphere.
1) Light must proceed spherically from the center of the "rest" sphere.
2) Light must proceed spherically from the center of the "moving" sphere.
3) Because there exists relative motion, the two centers of the rigid body spheres do not occupy the same point in space.
In fact, by looking at just the x-axis of the spheres, the at rest observer would claim both equal marks at a distance r were struck at the same time, and in addition the moving observer would make the same claim.
However, the relativity of simultaneity suggests if the at rest observer views the event as simultaneous, then an observer moving in collinear relative motion would not judge the same event as simultaneous it its frame. In fact, both observers would agree on the ordinality of events in the moving frame. In particular, in Einstein's train experiment, both observer M and M' judged M' did not see the front and back of the train struck at the same time and both observers agreed the front was struck first. They just disagreed on the timing, not the ordinality.
Yet, both observers view the event above from the x-axis in a simultaneous fashion which contradicts the relativity of simultaneity.
Basically, since SR is unable to produce a set of equations that satisfies all the 3 conditions listed above, then SR's understanding of light is false and it is that simple.
It's fascinating how many folks claim to disprove something they have zero understanding of. Especially when it pertains to science.
Assume we have two magic spheres that can pass through each other and we can get inside them. The inertial frame logic of light claims if either of us flash our light, all points on the inside of the sphere will be struck at the same time by the light regardless of our actual motion.
In other words, light expands spherically stationary to the frame from the point of emission in the frame. Now, somehow we put these spheres at the same point in space and I flash the light. Both of us then would see the light strike our sphere points all at the same time.
This time I am going to move very fast relative to the other sphere. However, your sphere looks like it is moving very fast relative to me and I am at rest. At the instant we occupy the same exact space, I flash the light. Light speed is not infinite so it does take time to strike each point on the sphere. Well, since light expands stationary to me it hits my points all at the same time.
But what about you? It cannot hit your points all at the same time and mine also because I am moving relative to you and we no longer occupy the same point in space. But, light from a moving light source is supposed to proceed outward spherically from the point of emission in your frame which is the center of your sphere according to the SR light postulate.
Since one light sphere cannot intersect the two rigid body spheres at all points in each sphere simultaneously, we have a contradiction. In particular, since the center of the one light sphere is centered in each frame, then the center of the one light sphere would have to occupy two different points in space.
Try drawing a picture of the event even using length contraction, it is impossible. Make the figures anyway you like and you still cannot express a geometry/picture that works for the two spheres and one light sphere.
More specifically, SR encourages the use of disjoint frame to frame logic when deducing relative motion. However, given one light sphere and two rigid body frame spheres in relative motion, both frames are forced together to solve the problem and the problem must be solved with all known information. As such the following three conditions must be met with one light sphere.
1) Light must proceed spherically from the center of the "rest" sphere.
2) Light must proceed spherically from the center of the "moving" sphere.
3) Because there exists relative motion, the two centers of the rigid body spheres do not occupy the same point in space.
In fact, by looking at just the x-axis of the spheres, the at rest observer would claim both equal marks at a distance r were struck at the same time, and in addition the moving observer would make the same claim.
However, the relativity of simultaneity suggests if the at rest observer views the event as simultaneous, then an observer moving in collinear relative motion would not judge the same event as simultaneous it its frame. In fact, both observers would agree on the ordinality of events in the moving frame. In particular, in Einstein's train experiment, both observer M and M' judged M' did not see the front and back of the train struck at the same time and both observers agreed the front was struck first. They just disagreed on the timing, not the ordinality.
Yet, both observers view the event above from the x-axis in a simultaneous fashion which contradicts the relativity of simultaneity.
Basically, since SR is unable to produce a set of equations that satisfies all the 3 conditions listed above, then SR's understanding of light is false and it is that simple.
It's fascinating how many folks claim to disprove something they have zero understanding of. Especially when it pertains to science.
QUOTE (brucep+Nov 4 2009, 10:27 PM)
It's fascinating how many folks claim to disprove something they have zero understanding of. Especially when it pertains to science.
Yeah, it's amazing that someone would make such an incredibly bold claim when I can defeat their logic with 2 simple phrases...
"Frame invariance of c," and "time dilation."
Yeah, it's amazing that someone would make such an incredibly bold claim when I can defeat their logic with 2 simple phrases...
"Frame invariance of c," and "time dilation."
QUOTE (qwtyu+Nov 5 2009, 01:37 AM)
1) Light must proceed spherically from the center of the "rest" sphere.
2) Light must proceed spherically from the center of the "moving" sphere.
It does, dumbass.
QUOTE (MjolnirPants+Nov 5 2009, 04:46 AM)
Yeah, it's amazing that someone would make such an incredibly bold claim when I can defeat their logic with 2 simple phrases...
"Frame invariance of c," and "time dilation."
Fine, let's see the equations.
When someone claims to understand something in physics, they can produce the equations since the language of physics is math.
Above, the center of the light sphere must occupy two different points in space to satisfy the necessary conditions.
You are not able to work that out.
"Frame invariance of c," and "time dilation."
Fine, let's see the equations.
When someone claims to understand something in physics, they can produce the equations since the language of physics is math.
Above, the center of the light sphere must occupy two different points in space to satisfy the necessary conditions.
You are not able to work that out.
Looks Like A PHYSICAL CHALLENGE!
QUOTE (MjolnirPants+Nov 5 2009, 04:46 AM)
Yeah, it's amazing that someone would make such an incredibly bold claim when I can defeat their logic with 2 simple phrases...
"Frame invariance of c," and "time dilation."
The problem is that you must make the center of the light sphere exist at two different positions in space.
You cannot.
This would be like claiming a person can be at two different places at the same time.
"Frame invariance of c," and "time dilation."
The problem is that you must make the center of the light sphere exist at two different positions in space.
You cannot.
This would be like claiming a person can be at two different places at the same time.
it IS a PHYSICAL CHALLENGE!
QUOTE (qwtyu+Nov 5 2009, 12:07 PM)
Fine, let's see the equations.
When someone claims to understand something in physics, they can produce the equations since the language of physics is math.
Above, the center of the light sphere must occupy two different points in space to satisfy the necessary conditions.
You are not able to work that out.
Here , dumbass:
In the frame attached to the source, the wavefront is a sphere:
x^2+y^2+z^2=(ct)^2
Apply the Lorentz transforms in order to get the wavefront shape in the arbitrary frame where the light source is moving:
x'=\gamma(x-vt)
y'=y
z'=z
t'=\gamma(t-vx/c^2)
and you will get:
(x')^2+(y')^2+(z')^2=(ct')^2
When someone claims to understand something in physics, they can produce the equations since the language of physics is math.
Above, the center of the light sphere must occupy two different points in space to satisfy the necessary conditions.
You are not able to work that out.
Here , dumbass:
In the frame attached to the source, the wavefront is a sphere:
x^2+y^2+z^2=(ct)^2
Apply the Lorentz transforms in order to get the wavefront shape in the arbitrary frame where the light source is moving:
x'=\gamma(x-vt)
y'=y
z'=z
t'=\gamma(t-vx/c^2)
and you will get:
(x')^2+(y')^2+(z')^2=(ct')^2
QUOTE (Trout+Nov 5 2009, 02:48 PM)
Here , dumbass:
In the frame attached to the source, the wavefront is a sphere:
x^2+y^2+z^2=(ct)^2
Apply the Lorentz transforms in order to get the wavefront shape in the arbitrary frame where the light source is moving:
x'=\gamma(x-vt)
y'=y
z'=z
t'=\gamma(t-vx/c^2)
and you will get:
(x')^2+(y')^2+(z')^2=(ct')^2
Comical, I have seen this before exactly before.
You have confused using LT transformations when everything is being done in the rest frame.
So, you have successfully transfered to the Minkowsky space of the moving frame, but all calculations must exist within the rest frame to describe the event.
Let D be the diameter of both spheres when at rest with one another.
You will find the center of the moving sphere vt + D/(2λ) arfter time t in the corrds of the rest frame.
Thus, the center of the sphere is located at the origin (0,0,0) of the rest frame while at the same time is emerging from (vt/λ, 0, 0) after time t.
in the coords of the rest frame for the moving sphere.
In the frame attached to the source, the wavefront is a sphere:
x^2+y^2+z^2=(ct)^2
Apply the Lorentz transforms in order to get the wavefront shape in the arbitrary frame where the light source is moving:
x'=\gamma(x-vt)
y'=y
z'=z
t'=\gamma(t-vx/c^2)
and you will get:
(x')^2+(y')^2+(z')^2=(ct')^2
Comical, I have seen this before exactly before.
You have confused using LT transformations when everything is being done in the rest frame.
So, you have successfully transfered to the Minkowsky space of the moving frame, but all calculations must exist within the rest frame to describe the event.
Let D be the diameter of both spheres when at rest with one another.
You will find the center of the moving sphere vt + D/(2λ) arfter time t in the corrds of the rest frame.
Thus, the center of the sphere is located at the origin (0,0,0) of the rest frame while at the same time is emerging from (vt/λ, 0, 0) after time t.
in the coords of the rest frame for the moving sphere.
QUOTE (qwtyu+Nov 5 2009, 03:05 PM)
Comical, I have seen this exact thng before.
Means that you aren't only a dumbass, you are a stubborn dumbass who can't learn.
Means that you aren't only a dumbass, you are a stubborn dumbass who can't learn.
QUOTE (Trout+Nov 5 2009, 03:09 PM)
Means that you aren't only a dumbass, you are a stubborn dumbass who can't learn.
Curious, you seem like someone I have debated.
Anyway, that was an intresting error you made transfering to the space of the moving sphere to describe the event which was supposed to be described in the context of the rest frame.
Curious, you seem like someone I have debated.
Anyway, that was an intresting error you made transfering to the space of the moving sphere to describe the event which was supposed to be described in the context of the rest frame.
QUOTE (qwtyu+Nov 5 2009, 03:14 PM)
Curious, you seem like someone I have debated.
Anyway, that was an intresting error you made transfering to the space of the moving sphere to describe the event which was supposed to be described in the context of the rest frame.
What do you think x'=\gamma(x-vt) means?
Still a dumbass, I see.....
Anyway, that was an intresting error you made transfering to the space of the moving sphere to describe the event which was supposed to be described in the context of the rest frame.
What do you think x'=\gamma(x-vt) means?
Still a dumbass, I see.....
QUOTE (qwtyu+Nov 5 2009, 03:05 PM)
You will find the center of the moving sphere vt + D/(2λ) arfter time t in the corrds of the rest frame.
Thus, the center of the sphere is located at the origin (0,0,0) of the rest frame while at the same time is emerging from (vt + D/(2λ), y, z) in the coords of the rest frame for the moving sphere.
What do you think x'=\gamma(x-vt) means?
Still a dumbass, I see.
QUOTE (Trout+Nov 5 2009, 03:18 PM)
What do you think x'=\gamma(x-vt) means?
Still a dumbass, I see.....
Where is the origin of the light sphere when x'=\gamma(x-vt) ?
It is at (0,0,0) which is not the center of the moving sphere.
Since the ligfht flashed when both rigid body spheres were origined, then the center of the moving sphere is located at (vt/λ, 0, 0) after time t.
QUOTE (Trout+Nov 5 2009, 03:20 PM)
What do you think x'=\gamma(x-vt) means?
Still a dumbass, I see.
You see, the light sphere is located at -x' and x' on the x-axis origined in the moving frame.
That is what I am trying to tell you. You are in the coords of the moving frame and should be in the coords of the "at rest" frame.
Still a dumbass, I see.
You see, the light sphere is located at -x' and x' on the x-axis origined in the moving frame.
That is what I am trying to tell you. You are in the coords of the moving frame and should be in the coords of the "at rest" frame.
QUOTE (qwtyu+Nov 5 2009, 03:38 PM)
You see, the light sphere is located at -x' and x' on the x-axis origined in the moving frame.
That is what I am trying to tell you. You are in the coords of the moving frame and should be in the coords of the "at rest" frame.
So, you don't know what x'=\gamma(x-vt) means. Figures.
That is what I am trying to tell you. You are in the coords of the moving frame and should be in the coords of the "at rest" frame.
So, you don't know what x'=\gamma(x-vt) means. Figures.
QUOTE (Trout+Nov 5 2009, 03:46 PM)
So, you don't know what x'=\gamma(x-vt) means. Figures.
Of course I do.
You are not listening. You are in the coordinate system of the moving frame when you apply that.
Of course I do.
You are not listening. You are in the coordinate system of the moving frame when you apply that.
QUOTE (qwtyu+Nov 5 2009, 03:49 PM)
Of course I do.
You are not listening. You are in the coordinate system of the moving frame when you apply that.
Ummm, no.
Your affliction (natural dumbass) is not curable.
You are not listening. You are in the coordinate system of the moving frame when you apply that.
Ummm, no.
Your affliction (natural dumbass) is not curable.
QUOTE (Trout+Nov 5 2009, 03:46 PM)
So, you don't know what x'=\gamma(x-vt) means. Figures.
Here is the transformation in Galilean logic which will show you.
To origin O' with x, it is x' in the origin of O' which is x' = x - vt.
The origin of O' is located at vt in the coords of O.
That means x' must be less than x when vt becomes the origin for O' or (0,0) in O'.
| |
| |
| |
| |
| |
+----------------------------------
O ->vt O' x
|--------vt-------|-------x'------|
Here is the transformation in Galilean logic which will show you.
To origin O' with x, it is x' in the origin of O' which is x' = x - vt.
The origin of O' is located at vt in the coords of O.
That means x' must be less than x when vt becomes the origin for O' or (0,0) in O'.
CODE
| |
| |
| |
| |
| |
+----------------------------------
O ->vt O' x
|--------vt-------|-------x'------|
QUOTE (Trout+Nov 5 2009, 03:51 PM)
Ummm, no.
Your affliction (natural dumbass) is not curable.
Now, after learning what LT means with origins and studying the diagram I made, you are confessing that you have the origin at (0,0,0) and at (vt/λ, 0, 0) for the one light sphere when translated back to the coords of the at rest observers which was my whole point in the first place.
Your affliction (natural dumbass) is not curable.
Now, after learning what LT means with origins and studying the diagram I made, you are confessing that you have the origin at (0,0,0) and at (vt/λ, 0, 0) for the one light sphere when translated back to the coords of the at rest observers which was my whole point in the first place.
QUOTE (qwtyu+Nov 5 2009, 04:05 PM)
Now, after learning what LT means with origins and studying the diagram I made, you are confessing that you have the origin at (0,0,0) and at (vt/λ, 0, 0) for the one light sphere when translated back to the coords of the at rest observers which was my whole point in the first place.
"Dumbassness" isn't curable, dumbass. Take a hint, go away.
"Dumbassness" isn't curable, dumbass. Take a hint, go away.
QUOTE (Trout+Nov 5 2009, 04:06 PM)
"Dumbassness" isn't curable, dumbass. Take a hint, go away.
You are an interesting way of confessing you are wrong.
Did you learn from the diagram I made?
You are an interesting way of confessing you are wrong.
Did you learn from the diagram I made?
QUOTE (qwtyu+Nov 5 2009, 04:09 PM)
Did you learn from the diagram I made?
That you are a hopeless idiot.
That you are a hopeless idiot.
QUOTE (Trout+Nov 5 2009, 04:11 PM)
That you are a hopeless idiot.
Well, I am glad you at least learned when applying LT, you change origins.
Further, science is written in the language of math.
So, you attempted to do that and only proved my point once we figured out the origins.
Math is a good language for physics, don't you agree?
Well, I am glad you at least learned when applying LT, you change origins.
Further, science is written in the language of math.
So, you attempted to do that and only proved my point once we figured out the origins.
Math is a good language for physics, don't you agree?
QUOTE (qwtyu+Nov 5 2009, 04:15 PM)
Math is a good language for physics, don't you agree?
I agree that you are an attention seeking idiot who doesn't know what he's talking about.
I agree that you are an attention seeking idiot who doesn't know what he's talking about.
QUOTE (Trout+Nov 5 2009, 04:17 PM)
I agree that you are an attention seeking idiot who doesn't know what he's talking about.
Call me what you like I do not care.
But, the math you posted proved the one light sphere has two different origins one at (0,0,0) and one at (vt/λ, 0, 0)
Thanks, but I did not need your help proving the contradiction.
Call me what you like I do not care.
But, the math you posted proved the one light sphere has two different origins one at (0,0,0) and one at (vt/λ, 0, 0)
Thanks, but I did not need your help proving the contradiction.
QUOTE (qwtyu+Nov 5 2009, 04:22 PM)
But, the math you posted proved the one light sphere has two different origins one at (0,0,0) and one at (vt/λ, 0, 0)
No, dumbass, it is the same sphere, viewed from two different frames. How stooopid are you?
QUOTE (Trout+Nov 5 2009, 04:25 PM)
No, dumbass, it is the same sphere, viewed from two different frames. How stooopid are you?
You have lost track of the argument.
The one light sphere is origined at (0,0,0) in the rest frame according toi the light postulate.
But, the light source is in the moving frame.
Thus, it must be proceeding in that frame sphereically from the emission point in the frame.
In the coords of the rest frame, that origin is located at (vt/λ, 0, 0)
Thus, the light sphere must be emerging from the origin as well.
That is interesting, the origin of that light sphere is moving with the frame.
You have lost track of the argument.
The one light sphere is origined at (0,0,0) in the rest frame according toi the light postulate.
But, the light source is in the moving frame.
Thus, it must be proceeding in that frame sphereically from the emission point in the frame.
In the coords of the rest frame, that origin is located at (vt/λ, 0, 0)
Thus, the light sphere must be emerging from the origin as well.
That is interesting, the origin of that light sphere is moving with the frame.
QUOTE (qwtyu+Nov 5 2009, 04:30 PM)
The one light sphere is origined at (0,0,0) in the rest frame according toi the light postulate.
But, the light source is in the moving frame.
You are still an idiot, motion is relative in physics.
QUOTE (Trout+Nov 5 2009, 04:42 PM)
You are still an idiot, motion is relative in physics.
I am assuming the axioms of SR with this.
In that case, motion is relative, fine.
But, again, to satisfy both conditions of the light postulate for both spheres,
the light sphere must be origined at (0,0,0) and at (vt/λ,0,0).
You already took a stab at the math which proved my case, care to try another?
[Moderator: 50% likelihood I will ban this user and delete all posts. The poster is failing at communicating justified, factual and useful information.]
I am assuming the axioms of SR with this.
In that case, motion is relative, fine.
But, again, to satisfy both conditions of the light postulate for both spheres,
the light sphere must be origined at (0,0,0) and at (vt/λ,0,0).
You already took a stab at the math which proved my case, care to try another?
[Moderator: 50% likelihood I will ban this user and delete all posts. The poster is failing at communicating justified, factual and useful information.]
QUOTE (qwtyu+Nov 5 2009, 04:47 PM)
I am assuming the axioms of SR with this.
No, idiot, you are assuming absolute motion. You don't even know what you are doing.
No, idiot, you are assuming absolute motion. You don't even know what you are doing.
QUOTE (Trout+Nov 5 2009, 05:04 PM)
No, idiot, you are assuming absolute motion. You don't even know what you are doing.
I am not assuming any form of absolute motion. The rigid body spheres are in relative motion and the light sphere proceeds spherically from the emission point in the frame. That is what I am assuming.
Here is another way to put it.
According to the Relativity of Simultaneity,
If the "at rest" observer sees its front and back of the sphere struck at the same time, then the moving sphere will not see its front and back struck at the same time. Also both observers will agree om the results.
Front the perspective of the at rest frame Einstein said:
he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A
www.
bartleby
.com/173/9.html
From the perspective of the moving observer, Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:
www.
bartleby
.com/173/9.html
Thus, both observers should agree the moving sphere should see it front and back of the sphere struck at different times as in the relativity of Simultaneity.
However, the moving observer contains the light source and so claims its front and back should be struck simultaneously. But, this is inconsistent with the conclusions of the relativity of Simultaneity.
I am not assuming any form of absolute motion. The rigid body spheres are in relative motion and the light sphere proceeds spherically from the emission point in the frame. That is what I am assuming.
Here is another way to put it.
According to the Relativity of Simultaneity,
If the "at rest" observer sees its front and back of the sphere struck at the same time, then the moving sphere will not see its front and back struck at the same time. Also both observers will agree om the results.
Front the perspective of the at rest frame Einstein said:
he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A
www.
bartleby
.com/173/9.html
From the perspective of the moving observer, Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:
www.
bartleby
.com/173/9.html
Thus, both observers should agree the moving sphere should see it front and back of the sphere struck at different times as in the relativity of Simultaneity.
However, the moving observer contains the light source and so claims its front and back should be struck simultaneously. But, this is inconsistent with the conclusions of the relativity of Simultaneity.
QUOTE (qwtyu+Nov 5 2009, 10:05 AM)
Comical, I have seen this before exactly before.
I shal endeavor to translate this into English for those of us who don't speak Crankese.
"I don't know enough math to make my argument using it, and I don't want anyone to know this, so I'm simply going to pretend that I understand this, and see some error with it, when in reality the best I can do is copy and paste an equation from Wikipedia and hope nobody notices my complete lack of expertise."
I shal endeavor to translate this into English for those of us who don't speak Crankese.
"I don't know enough math to make my argument using it, and I don't want anyone to know this, so I'm simply going to pretend that I understand this, and see some error with it, when in reality the best I can do is copy and paste an equation from Wikipedia and hope nobody notices my complete lack of expertise."
......
QUOTE (MjolnirPants+Nov 5 2009, 05:50 PM)
I shal endeavor to translate this into English for those of us who don't speak Crankese.
"I don't know enough math to make my argument using it, and I don't want anyone to know this, so I'm simply going to pretend that I understand this, and see some error with it, when in reality the best I can do is copy and paste an equation from Wikipedia and hope nobody notices my complete lack of expertise."
Here are the equations posted.
These come straight from section 3 from Einstein's paper.
These describe the light sphere from the rest frame, the top equation, and the botton equations refers to the light sphere in an arbitrary frame with relative motion v.
"I don't know enough math to make my argument using it, and I don't want anyone to know this, so I'm simply going to pretend that I understand this, and see some error with it, when in reality the best I can do is copy and paste an equation from Wikipedia and hope nobody notices my complete lack of expertise."
QUOTE
Here , dumbass:
In the frame attached to the source, the wavefront is a sphere:
x^2+y^2+z^2=(ct)^2
Apply the Lorentz transforms in order to get the wavefront shape in the arbitrary frame where the light source is moving:
x'=\gamma(x-vt)
y'=y
z'=z
t'=\gamma(t-vx/c^2)
and you will get:
(x')^2+(y')^2+(z')^2=(ct')^2
In the frame attached to the source, the wavefront is a sphere:
x^2+y^2+z^2=(ct)^2
Apply the Lorentz transforms in order to get the wavefront shape in the arbitrary frame where the light source is moving:
x'=\gamma(x-vt)
y'=y
z'=z
t'=\gamma(t-vx/c^2)
and you will get:
(x')^2+(y')^2+(z')^2=(ct')^2
Here are the equations posted.
These come straight from section 3 from Einstein's paper.
These describe the light sphere from the rest frame, the top equation, and the botton equations refers to the light sphere in an arbitrary frame with relative motion v.
QUOTE (qwtyu+Nov 5 2009, 05:25 PM)
I am not assuming any form of absolute motion. The rigid body spheres are in relative motion and the light sphere proceeds spherically from the emission point in the frame. That is what I am assuming.
Here is another way to put it.
According to the Relativity of Simultaneity,
If the "at rest" observer sees its front and back of the sphere struck at the same time, then the moving sphere will not see its front and back struck at the same time. Also both observers will agree om the results.
Front the perspective of the at rest frame Einstein said:
he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A
www.
bartleby
.com/173/9.html
From the perspective of the moving observer, Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:
www.
bartleby
.com/173/9.html
Thus, both observers should agree the moving sphere should see it front and back of the sphere struck at different times as in the relativity of Simultaneity.
However, the moving observer contains the light source and so claims its front and back should be struck simultaneously. But, this is inconsistent with the conclusions of the relativity of Simultaneity.
OK, so you don't understand relativity of simultaneity either. It is very simple, really:
1. In frame F, the wavefront of the light is spherical : x^2+y^2+z^2=(ct)^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S simultaneously in frame F
2. In frame F', the wavefront of the light is also spherical : x'^2+y'^2+z'^2=(ct')^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S' simultaneously in frame F'
3. Now, because frames F and F' move wrt each other, the light fronts will not strike simultaneously:
- the two ends of the metal sphere S as viewed from F'
-the two ends of the metal sphere S' as viewed from F
This does not mean that the two wavefronts aren't spherical and that they don't touch all the points of the surface of the metal sphere S at the same time in F and all the points of the metal sphere S' at the same time in F' They do but you are too dumb to comprehend. There is no contradiction in SR, the only contradiction is in your head.
Here is another way to put it.
According to the Relativity of Simultaneity,
If the "at rest" observer sees its front and back of the sphere struck at the same time, then the moving sphere will not see its front and back struck at the same time. Also both observers will agree om the results.
Front the perspective of the at rest frame Einstein said:
he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A
www.
bartleby
.com/173/9.html
From the perspective of the moving observer, Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:
www.
bartleby
.com/173/9.html
Thus, both observers should agree the moving sphere should see it front and back of the sphere struck at different times as in the relativity of Simultaneity.
However, the moving observer contains the light source and so claims its front and back should be struck simultaneously. But, this is inconsistent with the conclusions of the relativity of Simultaneity.
OK, so you don't understand relativity of simultaneity either. It is very simple, really:
1. In frame F, the wavefront of the light is spherical : x^2+y^2+z^2=(ct)^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S simultaneously in frame F
2. In frame F', the wavefront of the light is also spherical : x'^2+y'^2+z'^2=(ct')^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S' simultaneously in frame F'
3. Now, because frames F and F' move wrt each other, the light fronts will not strike simultaneously:
- the two ends of the metal sphere S as viewed from F'
-the two ends of the metal sphere S' as viewed from F
This does not mean that the two wavefronts aren't spherical and that they don't touch all the points of the surface of the metal sphere S at the same time in F and all the points of the metal sphere S' at the same time in F' They do but you are too dumb to comprehend. There is no contradiction in SR, the only contradiction is in your head.
QUOTE (Trout+Nov 5 2009, 06:26 PM)
OK, so you don't understand relativity of simultaneity either. It is very simple, really:
1. In frame F, the wavefront of the light is spherical : x^2+y^2+z^2=(ct)^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S simultaneously in frame F
2. In frame F', the wavefront of the light is also spherical : x'^2+y'^2+z'^2=(ct')^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S' simultaneously in frame F'
3. Now, because frames F and F' move wrt each other, the light fronts will not strike simultaneously:
- the two ends of the metal sphere S as viewed from F'
-the two ends of the metal sphere S' as viewed from F
This does not mean that the two wavefronts aren't spherical and that they don't touch all the points of the surface of the metal sphere S at the same time in F and all the points of the metal sphere S' at the same time in F' They do but you are too dumb to comprehend. There is no contradiction in SR, the only contradiction is in your head.
The only problem with your analysis in part 2 is that Einstein was very clear,
Assuming, both spheres are exactly the same,
If the front and back are struck simultaneously for the at rest observer, then the moving observer will not see its front and back struck at the same time.
From the perspective of the moving observer, Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:
www.
bartleby
.com/173/9.html
Thus, the moving observer will conclude its front and back will not be struck at the same time.
The reason for that Einstein explains,
he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A
www.
bartleby
.com/173/9.html
Thus, when the light is emiitted at the origin of the rest frame, the moving sphere's front has moved past the at rest sphere's front after any time t. Hence, light will have further to travel to strike the front of the moving sphere as Einstein said,
whilst he is riding on ahead of the beam of light coming from A
www.
bartleby
.com/173/9.html
In addition, Einstein specifically said the moving observer would conclude the light strikes the front and back at different time. He also said the at rest observer would agree with the moving observer's assessment.
Finally, Einstein said,
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A.
www.
bartleby
.com/173/9.html
This confirms, the moving sphere would reach a different conclusion from the at rest observer regarding the striking of the front and back of their spheres with the one light sphere.
In fact, this would be much easier to see if the light source were contained in the at rest observer's sphere.
1. In frame F, the wavefront of the light is spherical : x^2+y^2+z^2=(ct)^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S simultaneously in frame F
2. In frame F', the wavefront of the light is also spherical : x'^2+y'^2+z'^2=(ct')^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S' simultaneously in frame F'
3. Now, because frames F and F' move wrt each other, the light fronts will not strike simultaneously:
- the two ends of the metal sphere S as viewed from F'
-the two ends of the metal sphere S' as viewed from F
This does not mean that the two wavefronts aren't spherical and that they don't touch all the points of the surface of the metal sphere S at the same time in F and all the points of the metal sphere S' at the same time in F' They do but you are too dumb to comprehend. There is no contradiction in SR, the only contradiction is in your head.
QUOTE
1. In frame F, the wavefront of the light is spherical : x^2+y^2+z^2=(ct)^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S simultaneously in frame F
2. In frame F', the wavefront of the light is also spherical : x'^2+y'^2+z'^2=(ct')^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S' simultaneously in frame F'
2. In frame F', the wavefront of the light is also spherical : x'^2+y'^2+z'^2=(ct')^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S' simultaneously in frame F'
The only problem with your analysis in part 2 is that Einstein was very clear,
Assuming, both spheres are exactly the same,
If the front and back are struck simultaneously for the at rest observer, then the moving observer will not see its front and back struck at the same time.
From the perspective of the moving observer, Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:
www.
bartleby
.com/173/9.html
Thus, the moving observer will conclude its front and back will not be struck at the same time.
The reason for that Einstein explains,
he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A
www.
bartleby
.com/173/9.html
Thus, when the light is emiitted at the origin of the rest frame, the moving sphere's front has moved past the at rest sphere's front after any time t. Hence, light will have further to travel to strike the front of the moving sphere as Einstein said,
whilst he is riding on ahead of the beam of light coming from A
www.
bartleby
.com/173/9.html
In addition, Einstein specifically said the moving observer would conclude the light strikes the front and back at different time. He also said the at rest observer would agree with the moving observer's assessment.
Finally, Einstein said,
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A.
www.
bartleby
.com/173/9.html
This confirms, the moving sphere would reach a different conclusion from the at rest observer regarding the striking of the front and back of their spheres with the one light sphere.
In fact, this would be much easier to see if the light source were contained in the at rest observer's sphere.
QUOTE (qwtyu+Nov 5 2009, 06:55 PM)
The only problem with your analysis in part 2 is that Einstein was very clear,
The only problem is that you are an idiot. This is not fixable since you were born this way.
QUOTE
If the front and back are struck simultaneously for the at rest observer, then the moving observer will not see its front and back struck at the same time.
The moving observer will see it's front and back struck at the same time. What the stationary observer and the moving observer will NOT agree on is WHEN the strikes take place.
QUOTE (qwtyu+Nov 5 2009, 06:55 PM)
The only problem with your analysis in part 2 is that Einstein was very clear,
Assuming, both spheres are exactly the same,
If the front and back are struck simultaneously for the at rest observer, then the moving observer will not see its front and back struck at the same time.
From the perspective of the moving observer, Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:
www.
bartleby
.com/173/9.html
Thus, the moving observer will conclude its front and back will not be struck at the same time.
The reason for that Einstein explains,
he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A
www.
bartleby
.com/173/9.html
Thus, when the light is emiitted at the origin of the rest frame, the moving sphere's front has moved past the at rest sphere's front after any time t. Hence, light will have further to travel to strike the front of the moving sphere as Einstein said,
whilst he is riding on ahead of the beam of light coming from A
www.
bartleby
.com/173/9.html
In addition, Einstein specifically said the moving observer would conclude the light strikes the front and back at different time. He also said the at rest observer would agree with the moving observer's assessment.
Finally, Einstein said,
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A.
www.
bartleby
.com/173/9.html
This confirms, the moving sphere would reach a different conclusion from the at rest observer regarding the striking of the front and back of their spheres with the one light sphere.
In fact, this would be much easier to see if the light source were contained in the at rest observer's sphere.
Trout laid it out perfectly for you. If you can't understand what he showed you then most likely you're not interested in understanding as much as' being right'. I hope the moderator doesn't remove all your comments because Trout's explanation is worth archiving. I don't think you're and idiot but you need to make an honest enquiry into what your trying to debunk before you start. Your time would have been better spent. IE you would have a better understanding of relativity theory and not made a fool of yourself in this science forum.
QUOTE (AlexG+Nov 5 2009, 07:39 PM)
The moving observer will see it's front and back struck at the same time. What the stationary observer and the moving observer will NOT agree on is WHEN the strikes take place.
OK, may we assume the light source is in the rest observer's frame?
O is the origin of the at rest observer and O' is the origin of the moving observer.
The light is emitted when O and O' are coincident and proceeds in all direction c from the source of light in the rest frame.
Below is a picture.
|-------<-----------O---------->--------|
vt-> |-------------------O'------------------|
This consistent with R of S that light will stike the back of O' before the front.
Further, both observers will agree. They will disagree on the timing though because of time dilation. This is a reverse of the original thought experiment in terms of the direction both light strikes travel.
From the perspective of the moving observer, Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:
www.
bartleby
.com/173/9.html
Therefore, O' will conclude the back direction of light at O emitted before the front.
QUOTE (qwtyu+Nov 5 2009, 08:03 PM)
OK, may we assume the light source is in the rest observer's frame?
O is the origin of the at rest observer and O' is the origin of the moving observer.
The light is emitted when O and O' are coincident and proceeds in all direction c from the source of light in the rest frame.
Below is a picture.
|-------<-----------O---------->--------|
vt-> |-------------------O'------------------|
This consistent with R of S that light will stike the back of O' before the front.
Further, both observers will agree. They will disagree on the timing though because of time dilation. This is a reverse of the original thought experiment in terms of the direction both light strikes travel.
From the perspective of the moving observer, Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:
www.
bartleby
.com/173/9.html
Therefore, O' will conclude the back direction of light at O emitted before the front.
You need to actually find out how Einstein's model works. Events are frame invariant. Until you figure out what that means you should put a cork in it.
O is the origin of the at rest observer and O' is the origin of the moving observer.
The light is emitted when O and O' are coincident and proceeds in all direction c from the source of light in the rest frame.
Below is a picture.
|-------<-----------O---------->--------|
vt-> |-------------------O'------------------|
This consistent with R of S that light will stike the back of O' before the front.
Further, both observers will agree. They will disagree on the timing though because of time dilation. This is a reverse of the original thought experiment in terms of the direction both light strikes travel.
From the perspective of the moving observer, Einstein said:
Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:
www.
bartleby
.com/173/9.html
Therefore, O' will conclude the back direction of light at O emitted before the front.
You need to actually find out how Einstein's model works. Events are frame invariant. Until you figure out what that means you should put a cork in it.
If he did, there'd be no illumination from his butt-candle. No shadows, no substance. Just gas in a cavern...
QUOTE (brucep+Nov 5 2009, 08:54 PM)
You need to actually find out how Einstein's model works. Events are frame invariant. Until you figure out what that means you should put a cork in it.
I agree a spacetime interval between two events is frame invariant.
But, that has nothing to do with the ordinality of the events to a frame.
For example, in the original R of S train embankment thought experiment, the ordinality of the lighting flash events was the front flash occured before the back flash for the moving train.
For the embankment observer, the flashes were simultaneous.
Had a train been going the other direction, that train would have concluded the back flash occured before the front flash.
Thus, the above picture demonstrates this logic.
I agree a spacetime interval between two events is frame invariant.
But, that has nothing to do with the ordinality of the events to a frame.
For example, in the original R of S train embankment thought experiment, the ordinality of the lighting flash events was the front flash occured before the back flash for the moving train.
For the embankment observer, the flashes were simultaneous.
Had a train been going the other direction, that train would have concluded the back flash occured before the front flash.
Thus, the above picture demonstrates this logic.
QUOTE (qwtyu+Nov 5 2009, 09:15 PM)
I agree a spacetime interval between two events is frame invariant.
But, that has nothing to do with the ordinality of the events to a frame.
For example, in the original R of S train embankment thought experiment, the ordinality of the lighting flash events was the front flash occured before the back flash for the moving train.
For the embankment observer, the flashes were simultaneous.
Had a train been going the other direction, that train would have concluded the back flash occured before the front flash.
Thus, the above picture demonstrates this logic.
'Events' are FRAME INVARIANT so it does have something to say about the order of events in any frame. If you actually learned something about this subject you might quit making ignorant comments
But, that has nothing to do with the ordinality of the events to a frame.
For example, in the original R of S train embankment thought experiment, the ordinality of the lighting flash events was the front flash occured before the back flash for the moving train.
For the embankment observer, the flashes were simultaneous.
Had a train been going the other direction, that train would have concluded the back flash occured before the front flash.
Thus, the above picture demonstrates this logic.
'Events' are FRAME INVARIANT so it does have something to say about the order of events in any frame. If you actually learned something about this subject you might quit making ignorant comments
QUOTE (brucep+Nov 5 2009, 09:37 PM)
'Events' are FRAME INVARIANT so it does have something to say about the order of events in any frame. If you actually learned something about this subject you might quit making ignorant comments
If you mean events are invariant to frames in terms of the light cone, then I am in agreement.
A Light cone is the path that a flash of light would take through spacetime. As time progresses, the light from the flash spreads out in a circle, and the result is a cone. In reality, there are three space dimensions, so the light actually forms a sphere in space, and the light cone is actually a fourth dimensional shape, but it's much easier to visualize it as a cone.
The above classifications hold true in any frame of reference; that is, an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers, no matter their frame of reference. This is why the concept is so powerful.
http:
//en.wikipedia.
org/wiki/Light_cone
Once an object is contained in the light sphere, then all observers will agree that object is contained in the light sphere and thus, that event is in the absolute past for all frames.
Thus, when these statements were posted,
Assuming Statement 3 above, the two ends of the metal sphere S as viewed from F' will not strike simultaneously contradicts statement 1 in terms of the light cone.
Specifically, F sees both (-r,0,0) and (+r,0,0) enter the light cone at the same time whereas F' sees those points enter at different times. Therefore, there will exist an instant for F where F claims both are in the light cone at the same time and F' in its own proper time claims one is in but one is not.
Thus, they disagree on the absolute past for the light cone for the events of light striking the front and back of the sphere.
But,
an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers
http:
//en.wikipedia
.org/wiki/Light_cone
Therefore, there cannot be any disagreement between frames of the order of objects being overtaken by the light sphere and so statements 1-3 above contradict the necessary conditions of the light cone.
If you mean events are invariant to frames in terms of the light cone, then I am in agreement.
A Light cone is the path that a flash of light would take through spacetime. As time progresses, the light from the flash spreads out in a circle, and the result is a cone. In reality, there are three space dimensions, so the light actually forms a sphere in space, and the light cone is actually a fourth dimensional shape, but it's much easier to visualize it as a cone.
The above classifications hold true in any frame of reference; that is, an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers, no matter their frame of reference. This is why the concept is so powerful.
http:
//en.wikipedia.
org/wiki/Light_cone
Once an object is contained in the light sphere, then all observers will agree that object is contained in the light sphere and thus, that event is in the absolute past for all frames.
Thus, when these statements were posted,
QUOTE
1. In frame F, the wavefront of the light is spherical : x^2+y^2+z^2=(ct)^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S simultaneously in frame F
2. In frame F', the wavefront of the light is also spherical : x'^2+y'^2+z'^2=(ct')^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S' simultaneously in frame F'
3. Now, because frames F and F' move wrt each other, the light fronts will not strike simultaneously:
- the two ends of the metal sphere S as viewed from F'
-the two ends of the metal sphere S' as viewed from F
2. In frame F', the wavefront of the light is also spherical : x'^2+y'^2+z'^2=(ct')^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S' simultaneously in frame F'
3. Now, because frames F and F' move wrt each other, the light fronts will not strike simultaneously:
- the two ends of the metal sphere S as viewed from F'
-the two ends of the metal sphere S' as viewed from F
Assuming Statement 3 above, the two ends of the metal sphere S as viewed from F' will not strike simultaneously contradicts statement 1 in terms of the light cone.
Specifically, F sees both (-r,0,0) and (+r,0,0) enter the light cone at the same time whereas F' sees those points enter at different times. Therefore, there will exist an instant for F where F claims both are in the light cone at the same time and F' in its own proper time claims one is in but one is not.
Thus, they disagree on the absolute past for the light cone for the events of light striking the front and back of the sphere.
But,
an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers
http:
//en.wikipedia
.org/wiki/Light_cone
Therefore, there cannot be any disagreement between frames of the order of objects being overtaken by the light sphere and so statements 1-3 above contradict the necessary conditions of the light cone.
QUOTE (qwtyu+Nov 5 2009, 11:59 PM)
Once an object is contained in the light sphere, then all observers will agree that object is contained in the light sphere and thus, that event is in the absolute past for all frames.
Bollocks, all points in 4D space have their own very unique frames.
Simply cannot wait for your next squirming idiocy.
Bollocks, all points in 4D space have their own very unique frames.
Simply cannot wait for your next squirming idiocy.
QUOTE (qwtyu+Nov 5 2009, 11:59 PM)
Assuming Statement 3 above, the two ends of the metal sphere S as viewed from F' will not strike simultaneously contradicts statement 1 in terms of the light cone.
No, it doesn't. Your idiocy is incurable.
QUOTE (Geoff Mollusc+Nov 6 2009, 01:21 AM)
Bollocks, all points in 4D space have their own very unique frames.
Simply cannot wait for your next squirming idiocy.
Bollocks, all points in 4D space have their own very unique frames.
This statement violates the definition of a math function.
All points in 4D space are mapped according to the "rest" frame to a specific relative motion frame.
Moreover, you have claimed a "prefered 4D space" which violates relativity.
What cannot be denied is when the light sphere overtakes an object, no frame in the universe can disagree.
Therefore, it is impossible for frame M and M' to see their respective front and back of their spheres struck simultaneously because there is one light sphere and the front and back of each right body sphere are not coincident.
Simply cannot wait for your next squirming idiocy.
Bollocks, all points in 4D space have their own very unique frames.
This statement violates the definition of a math function.
All points in 4D space are mapped according to the "rest" frame to a specific relative motion frame.
Moreover, you have claimed a "prefered 4D space" which violates relativity.
What cannot be denied is when the light sphere overtakes an object, no frame in the universe can disagree.
Therefore, it is impossible for frame M and M' to see their respective front and back of their spheres struck simultaneously because there is one light sphere and the front and back of each right body sphere are not coincident.
QUOTE (Trout+Nov 6 2009, 01:47 AM)
No, it doesn't. Your idiocy is incurable.
Not true.
You have both spheres being struck and the same time and yet not struck at the same time.
The light cone implements the absolute past such that no observer can disagree and does not allow such contradictions as you stated.
All observes must agree on the ordinality of events regarding the light cone and the absolute past.
The above classifications hold true in any frame of reference; that is, an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers, no matter their frame of reference. This is why the concept is so powerful.
http
://en.wikipedia
.org/wiki/Light_cone
Not true.
You have both spheres being struck and the same time and yet not struck at the same time.
The light cone implements the absolute past such that no observer can disagree and does not allow such contradictions as you stated.
All observes must agree on the ordinality of events regarding the light cone and the absolute past.
The above classifications hold true in any frame of reference; that is, an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers, no matter their frame of reference. This is why the concept is so powerful.
http
://en.wikipedia
.org/wiki/Light_cone
QUOTE (qwtyu+Nov 6 2009, 02:11 AM)
Not true.
You have both spheres being struck and the same time and yet not struck at the same time.
Like I said, you are an idiot. Incurable.
You have both spheres being struck and the same time and yet not struck at the same time.
Like I said, you are an idiot. Incurable.
QUOTE (qwtyu+Nov 6 2009, 01:59 AM)
Bollocks, all points in 4D space have their own very unique frames.
This statement violates the definition of a math function.
All points in 4D space are mapped according to the "rest" frame to a specific relative motion frame.
Moreover, you have claimed a "prefered 4D space" which violates relativity.
What cannot be denied is when the light sphere overtakes an object, no frame in the universe can disagree.
Therefore, it is impossible for frame M and M' to see their respective front and back of their spheres struck simultaneously because there is one light sphere and the front and back of each right body sphere are not coincident.
... excellent nonsense. Ever considered a career as understudy to a disturbingly retarded village idiot?.
[Moderator: Your continuing hostility towards unilluminating posts is unenlightened. Suspended 30 days.]
This statement violates the definition of a math function.
All points in 4D space are mapped according to the "rest" frame to a specific relative motion frame.
Moreover, you have claimed a "prefered 4D space" which violates relativity.
What cannot be denied is when the light sphere overtakes an object, no frame in the universe can disagree.
Therefore, it is impossible for frame M and M' to see their respective front and back of their spheres struck simultaneously because there is one light sphere and the front and back of each right body sphere are not coincident.
... excellent nonsense. Ever considered a career as understudy to a disturbingly retarded village idiot?. [Moderator: Your continuing hostility towards unilluminating posts is unenlightened. Suspended 30 days.]
QUOTE (Trout+Nov 6 2009, 02:14 AM)
Like I said, you are an idiot. Incurable.
Perhaps I misread your post.
In 1) you say F concludes the metal sphere S simultaneously
In 3) the two ends of the metal sphere S as viewed from F' the light fronts will not strike simultaneously.
Thus, two observers disagree on the ordinality of events implemented by the light cone.
Yet, no two observers can disagree on the ordinality of events with the light cone.
The above classifications hold true in any frame of reference; that is, an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers, no matter their frame of reference. This is why the concept is so powerful.
http
://en.wikipedia
.org/wiki/Light_cone
Perhaps I misread your post.
QUOTE
1. In frame F, the wavefront of the light is spherical : x^2+y^2+z^2=(ct)^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S simultaneously in frame F
2. In frame F', the wavefront of the light is also spherical : x'^2+y'^2+z'^2=(ct')^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S' simultaneously in frame F'
3. Now, because frames F and F' move wrt each other, the light fronts will not strike simultaneously:
- the two ends of the metal sphere S as viewed from F'
-the two ends of the metal sphere S' as viewed from F
2. In frame F', the wavefront of the light is also spherical : x'^2+y'^2+z'^2=(ct')^2, so the light will hit the points (-r,0,0) and (+r,0,0) of the metal sphere S' simultaneously in frame F'
3. Now, because frames F and F' move wrt each other, the light fronts will not strike simultaneously:
- the two ends of the metal sphere S as viewed from F'
-the two ends of the metal sphere S' as viewed from F
In 1) you say F concludes the metal sphere S simultaneously
In 3) the two ends of the metal sphere S as viewed from F' the light fronts will not strike simultaneously.
Thus, two observers disagree on the ordinality of events implemented by the light cone.
Yet, no two observers can disagree on the ordinality of events with the light cone.
The above classifications hold true in any frame of reference; that is, an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers, no matter their frame of reference. This is why the concept is so powerful.
http
://en.wikipedia
.org/wiki/Light_cone
QUOTE (qwtyu+Nov 5 2009, 11:59 PM)
If you mean events are invariant to frames in terms of the light cone, then I am in agreement.
A Light cone is the path that a flash of light would take through spacetime. As time progresses, the light from the flash spreads out in a circle, and the result is a cone. In reality, there are three space dimensions, so the light actually forms a sphere in space, and the light cone is actually a fourth dimensional shape, but it's much easier to visualize it as a cone.
The above classifications hold true in any frame of reference; that is, an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers, no matter their frame of reference. This is why the concept is so powerful.
http:
//en.wikipedia.
org/wiki/Light_cone
Once an object is contained in the light sphere, then all observers will agree that object is contained in the light sphere and thus, that event is in the absolute past for all frames.
Thus, when these statements were posted,
Assuming Statement 3 above, the two ends of the metal sphere S as viewed from F' will not strike simultaneously contradicts statement 1 in terms of the light cone.
Specifically, F sees both (-r,0,0) and (+r,0,0) enter the light cone at the same time whereas F' sees those points enter at different times. Therefore, there will exist an instant for F where F claims both are in the light cone at the same time and F' in its own proper time claims one is in but one is not.
Thus, they disagree on the absolute past for the light cone for the events of light striking the front and back of the sphere.
But,
an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers
http:
//en.wikipedia
.org/wiki/Light_cone
Therefore, there cannot be any disagreement between frames of the order of objects being overtaken by the light sphere and so statements 1-3 above contradict the necessary conditions of the light cone.
You'll probably never realize just how incompetent you are. You fit the profile referred to in the following article
http://www.sfgate.com/cgi-bin/article.cgi?.../18/MN73840.DTL
It doesn't matter whether an event is inside or outside the future or past light-cone of another event. THE LAWS OF PHYSICS are frame invariant.
A Light cone is the path that a flash of light would take through spacetime. As time progresses, the light from the flash spreads out in a circle, and the result is a cone. In reality, there are three space dimensions, so the light actually forms a sphere in space, and the light cone is actually a fourth dimensional shape, but it's much easier to visualize it as a cone.
The above classifications hold true in any frame of reference; that is, an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers, no matter their frame of reference. This is why the concept is so powerful.
http:
//en.wikipedia.
org/wiki/Light_cone
Once an object is contained in the light sphere, then all observers will agree that object is contained in the light sphere and thus, that event is in the absolute past for all frames.
Thus, when these statements were posted,
Assuming Statement 3 above, the two ends of the metal sphere S as viewed from F' will not strike simultaneously contradicts statement 1 in terms of the light cone.
Specifically, F sees both (-r,0,0) and (+r,0,0) enter the light cone at the same time whereas F' sees those points enter at different times. Therefore, there will exist an instant for F where F claims both are in the light cone at the same time and F' in its own proper time claims one is in but one is not.
Thus, they disagree on the absolute past for the light cone for the events of light striking the front and back of the sphere.
But,
an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers
http:
//en.wikipedia
.org/wiki/Light_cone
Therefore, there cannot be any disagreement between frames of the order of objects being overtaken by the light sphere and so statements 1-3 above contradict the necessary conditions of the light cone.
You'll probably never realize just how incompetent you are. You fit the profile referred to in the following article
http://www.sfgate.com/cgi-bin/article.cgi?.../18/MN73840.DTL
It doesn't matter whether an event is inside or outside the future or past light-cone of another event. THE LAWS OF PHYSICS are frame invariant.
QUOTE (brucep+Nov 6 2009, 02:33 AM)
You'll probably never realize just how incompetent you are. You fit the profile referred to in the following article
It doesn't matter whether an event is inside or outside the future or past light-cone of another event. THE LAWS OF PHYSICS are frame invariant.
It doesn't matter whether an event is inside or outside the future or past light-cone of another event. THE LAWS OF PHYSICS are frame invariant.
In way way you are correct. In one way you are wrong.
This thread is not trying to claim the laws are different frame to frame.
This thread is trying to examine the light cone that implements the absolute past.
Further, the light cone orders events by shining on them such that no observer can disagree on the ordinality of these events. They may disagree on the timing but not the ordinality.
Thus, the light sphere cannot meet the points of one sphere simultaneously and also meet the points of another sphere simultaneously with collinear relative motion since the center of the light sphere must exist in two different positions in space.
It doesn't matter whether an event is inside or outside the future or past light-cone of another event. THE LAWS OF PHYSICS are frame invariant.
It doesn't matter whether an event is inside or outside the future or past light-cone of another event. THE LAWS OF PHYSICS are frame invariant.
In way way you are correct. In one way you are wrong.
This thread is not trying to claim the laws are different frame to frame.
This thread is trying to examine the light cone that implements the absolute past.
Further, the light cone orders events by shining on them such that no observer can disagree on the ordinality of these events. They may disagree on the timing but not the ordinality.
Thus, the light sphere cannot meet the points of one sphere simultaneously and also meet the points of another sphere simultaneously with collinear relative motion since the center of the light sphere must exist in two different positions in space.
QUOTE (qwtyu+Nov 6 2009, 02:45 AM)
It doesn't matter whether an event is inside or outside the future or past light-cone of another event. THE LAWS OF PHYSICS are frame invariant.
In way way you are correct. In one way you are wrong.
This thread is not trying to claim the laws are different frame to frame.
This thread is trying to examine the light cone that implements the absolute past.
Further, the light cone orders events by shining on them such that no observer can disagree on the ordinality of these events. They may disagree on the timing but not the ordinality.
Thus, the light sphere cannot meet the points of one sphere simultaneously and also meet the points of another sphere simultaneously with collinear relative motion since the center of the light sphere must exist in two different positions in space.
You're the only one who thinks you're right about any of the nonsense you've been posting at this site. Trout's right you are a complete idiot.
In way way you are correct. In one way you are wrong.
This thread is not trying to claim the laws are different frame to frame.
This thread is trying to examine the light cone that implements the absolute past.
Further, the light cone orders events by shining on them such that no observer can disagree on the ordinality of these events. They may disagree on the timing but not the ordinality.
Thus, the light sphere cannot meet the points of one sphere simultaneously and also meet the points of another sphere simultaneously with collinear relative motion since the center of the light sphere must exist in two different positions in space.
You're the only one who thinks you're right about any of the nonsense you've been posting at this site. Trout's right you are a complete idiot.
QUOTE (qwtyu+Nov 6 2009, 02:24 AM)
In 1) you say F concludes the metal sphere S simultaneously
In 3) the two ends of the metal sphere S as viewed from F' the light fronts will not strike simultaneously.
Yes, this is correct. This is known as relativity of simultaneity.
You are an idiot. Don't worry, it isn't curable. You'll stay this way for the rest of your life.
Here's how to break this down via events.
Let r be the proper radius of the massive shell. Proper length is defined as the length as measured in the frame where a massive object travelling inertially is motionless, so r is well-defined.
Let o be the center of the shell. o is a place defined relative to the shell. If the shell is meausred to be moving in the x direction, then o also moves in the x direction.
Let h be another place, defined relative to the inertial observer as the place where the light came from. h never moves relative to the observer. h and o are the same for at least the instant when the light flashes. But if the massive shell is moving relative to the observer, if v is not zero, then h is not equal to o at any other time.
A is an event -- the event which happens at h at a certain time, and which also happens at o at a certain time, because of our definition of o.
F is an event -- the event which happens when the light hits one part of the shell.
R is an event -- the event which happens when the light hits another part of the shell. F and R (forward and reverse) happen at certain places and certain times.
Let's place our observer in a frame where the shell is motionless -- in which case everything is easy. Let's define A to be the origin of our space-time co-ordinate system. A = (0,0,0,0). F is forward in the x direction, and obviously happens at a distance of r and after a time of r/c, so its coordinates are (r/c, r,0,0). R is likewise found as (r/c, -r,0,0). Indeed these are just special points on the shell and there is an infinite of events where the light hits the shell, given by E(θ,φ) = (r/c, r cos φ,r sin φ cos θ,r sin φ sin θ), but we are just dealing with F = E(0,0), and R=E(π,0).
Finally, each of these events, W, has an associated time, tW, and place, pW, in this coordinate frame. And given two (different) events, we can compute the speed of a particle which would connect them.
Summarizing:
o = (--, 0, 0, 0)
h = (--, 0, 0, 0)
A = (0, 0, 0, 0)
tA = (0, --, --, --)
pA = (--, 0, 0, 0) = o = h
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
pR = (--, -r, 0, 0) = - pF
sAF = | pF - pA | / (tF - tA) = |r|/(r/c) = c
sAR = | pR - pA | / (tR - tA) = |r|/(r/c) = c
But since there is no relative motion between massive objects and our observer, there is no application of the special relativity yet.
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
Which are the famous Einstein velocity addition formulas.
Let r be the proper radius of the massive shell. Proper length is defined as the length as measured in the frame where a massive object travelling inertially is motionless, so r is well-defined.
Let o be the center of the shell. o is a place defined relative to the shell. If the shell is meausred to be moving in the x direction, then o also moves in the x direction.
Let h be another place, defined relative to the inertial observer as the place where the light came from. h never moves relative to the observer. h and o are the same for at least the instant when the light flashes. But if the massive shell is moving relative to the observer, if v is not zero, then h is not equal to o at any other time.
A is an event -- the event which happens at h at a certain time, and which also happens at o at a certain time, because of our definition of o.
F is an event -- the event which happens when the light hits one part of the shell.
R is an event -- the event which happens when the light hits another part of the shell. F and R (forward and reverse) happen at certain places and certain times.
Let's place our observer in a frame where the shell is motionless -- in which case everything is easy. Let's define A to be the origin of our space-time co-ordinate system. A = (0,0,0,0). F is forward in the x direction, and obviously happens at a distance of r and after a time of r/c, so its coordinates are (r/c, r,0,0). R is likewise found as (r/c, -r,0,0). Indeed these are just special points on the shell and there is an infinite of events where the light hits the shell, given by E(θ,φ) = (r/c, r cos φ,r sin φ cos θ,r sin φ sin θ), but we are just dealing with F = E(0,0), and R=E(π,0).
Finally, each of these events, W, has an associated time, tW, and place, pW, in this coordinate frame. And given two (different) events, we can compute the speed of a particle which would connect them.
Summarizing:
o = (--, 0, 0, 0)
h = (--, 0, 0, 0)
A = (0, 0, 0, 0)
tA = (0, --, --, --)
pA = (--, 0, 0, 0) = o = h
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
pR = (--, -r, 0, 0) = - pF
sAF = | pF - pA | / (tF - tA) = |r|/(r/c) = c
sAR = | pR - pA | / (tR - tA) = |r|/(r/c) = c
But since there is no relative motion between massive objects and our observer, there is no application of the special relativity yet.
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
Which are the famous Einstein velocity addition formulas.
QUOTE (rpenner+Nov 6 2009, 04:13 AM)
a.
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
b.
F' = ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
tF'=(r/c)√((c+v)/(c-v))
tR'=(r/c)√((c-v)/(c+v))
Very nice.
So:
a. in the frame of the sphere, the observer sees the events R(ear) and F(orward) happening simultaneously at tR=tF=r/c
b. in the moving frame, the observer sees the forward strike of the light front on the surface of the sphere located in the other frame happening later than the rear strike. This is expected since light is "chasing" the front end.
What is missing is the stuff that was shown earlier, each observer sees himself in the center of a spherical wavefront.
Now, watch the new idiocy coming out of the typewriter nutcase.
QUOTE (qwtyu+)
since the center of the light sphere must exist in two different positions in space.
I think you need to take into account some extra property of space (spacetime) .. which is what rpenner et al are trying to say.
-C2.
I think you need to take into account some extra property of space (spacetime) .. which is what rpenner et al are trying to say.
-C2.
QUOTE (Trout+Nov 6 2009, 04:25 AM)
Very nice.
So:
a. in the frame of the sphere, the observer sees the events R(ear) and F(orward) happening simultaneously at tR=tF=r/c
b. in the moving frame, the observer sees the forward strike of the light front on the surface of the sphere located in the other frame happening later than the rear strike. This is expected since light is "chasing" the front end.
What is missing is the stuff that was shown earlier, each observer sees himself in the center of a spherical wavefront.
Now, watch the new idiocy coming out of the typewriter nutcase.
Can you please tell me where I find information to do subscripts and superscripts?
That way, at least my answer will be more readable.
Thanks
So:
a. in the frame of the sphere, the observer sees the events R(ear) and F(orward) happening simultaneously at tR=tF=r/c
b. in the moving frame, the observer sees the forward strike of the light front on the surface of the sphere located in the other frame happening later than the rear strike. This is expected since light is "chasing" the front end.
What is missing is the stuff that was shown earlier, each observer sees himself in the center of a spherical wavefront.
Now, watch the new idiocy coming out of the typewriter nutcase.
Can you please tell me where I find information to do subscripts and superscripts?
That way, at least my answer will be more readable.
Thanks
QUOTE (rpenner+Nov 6 2009, 04:13 AM)
Here's how to break this down via events.
Let r be the proper radius of the massive shell. Proper length is defined as the length as measured in the frame where a massive object travelling inertially is motionless, so r is well-defined.
Let o be the center of the shell. o is a place defined relative to the shell. If the shell is meausred to be moving in the x direction, then o also moves in the x direction.
Let h be another place, defined relative to the inertial observer as the place where the light came from. h never moves relative to the observer. h and o are the same for at least the instant when the light flashes. But if the massive shell is moving relative to the observer, if v is not zero, then h is not equal to o at any other time.
A is an event -- the event which happens at h at a certain time, and which also happens at o at a certain time, because of our definition of o.
F is an event -- the event which happens when the light hits one part of the shell.
R is an event -- the event which happens when the light hits another part of the shell. F and R (forward and reverse) happen at certain places and certain times.
Let's place our observer in a frame where the shell is motionless -- in which case everything is easy. Let's define A to be the origin of our space-time co-ordinate system. A = (0,0,0,0). F is forward in the x direction, and obviously happens at a distance of r and after a time of r/c, so its coordinates are (r/c, r,0,0). R is likewise found as (r/c, -r,0,0). Indeed these are just special points on the shell and there is an infinite of events where the light hits the shell, given by E(θ,φ) = (r/c, r cos φ,r sin φ cos θ,r sin φ sin θ), but we are just dealing with F = E(0,0), and R=E(π,0).
Finally, each of these events, W, has an associated time, tW, and place, pW, in this coordinate frame. And given two (different) events, we can compute the speed of a particle which would connect them.
Summarizing:
o = (--, 0, 0, 0)
h = (--, 0, 0, 0)
A = (0, 0, 0, 0)
tA = (0, --, --, --)
pA = (--, 0, 0, 0) = o = h
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
pR = (--, -r, 0, 0) = - pF
sAF = | pF - pA | / (tF - tA) = |r|/(r/c) = c
sAR = | pR - pA | / (tR - tA) = |r|/(r/c) = c
But since there is no relative motion between massive objects and our observer, there is no application of the special relativity yet.
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
Which are the famous Einstein velocity addition formulas.
Sorry, I meant to quote you.
Can you tell how to do subscripts and superscripts?
Thanks
Let r be the proper radius of the massive shell. Proper length is defined as the length as measured in the frame where a massive object travelling inertially is motionless, so r is well-defined.
Let o be the center of the shell. o is a place defined relative to the shell. If the shell is meausred to be moving in the x direction, then o also moves in the x direction.
Let h be another place, defined relative to the inertial observer as the place where the light came from. h never moves relative to the observer. h and o are the same for at least the instant when the light flashes. But if the massive shell is moving relative to the observer, if v is not zero, then h is not equal to o at any other time.
A is an event -- the event which happens at h at a certain time, and which also happens at o at a certain time, because of our definition of o.
F is an event -- the event which happens when the light hits one part of the shell.
R is an event -- the event which happens when the light hits another part of the shell. F and R (forward and reverse) happen at certain places and certain times.
Let's place our observer in a frame where the shell is motionless -- in which case everything is easy. Let's define A to be the origin of our space-time co-ordinate system. A = (0,0,0,0). F is forward in the x direction, and obviously happens at a distance of r and after a time of r/c, so its coordinates are (r/c, r,0,0). R is likewise found as (r/c, -r,0,0). Indeed these are just special points on the shell and there is an infinite of events where the light hits the shell, given by E(θ,φ) = (r/c, r cos φ,r sin φ cos θ,r sin φ sin θ), but we are just dealing with F = E(0,0), and R=E(π,0).
Finally, each of these events, W, has an associated time, tW, and place, pW, in this coordinate frame. And given two (different) events, we can compute the speed of a particle which would connect them.
Summarizing:
o = (--, 0, 0, 0)
h = (--, 0, 0, 0)
A = (0, 0, 0, 0)
tA = (0, --, --, --)
pA = (--, 0, 0, 0) = o = h
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
pR = (--, -r, 0, 0) = - pF
sAF = | pF - pA | / (tF - tA) = |r|/(r/c) = c
sAR = | pR - pA | / (tR - tA) = |r|/(r/c) = c
But since there is no relative motion between massive objects and our observer, there is no application of the special relativity yet.
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
Which are the famous Einstein velocity addition formulas.
Sorry, I meant to quote you.
Can you tell how to do subscripts and superscripts?
Thanks
QUOTE (rpenner+Nov 6 2009, 04:13 AM)
Here's how to break this down via events.
Let r be the proper radius of the massive shell. Proper length is defined as the length as measured in the frame where a massive object travelling inertially is motionless, so r is well-defined.
Let o be the center of the shell. o is a place defined relative to the shell. If the shell is meausred to be moving in the x direction, then o also moves in the x direction.
Let h be another place, defined relative to the inertial observer as the place where the light came from. h never moves relative to the observer. h and o are the same for at least the instant when the light flashes. But if the massive shell is moving relative to the observer, if v is not zero, then h is not equal to o at any other time.
A is an event -- the event which happens at h at a certain time, and which also happens at o at a certain time, because of our definition of o.
F is an event -- the event which happens when the light hits one part of the shell.
R is an event -- the event which happens when the light hits another part of the shell. F and R (forward and reverse) happen at certain places and certain times.
Let's place our observer in a frame where the shell is motionless -- in which case everything is easy. Let's define A to be the origin of our space-time co-ordinate system. A = (0,0,0,0). F is forward in the x direction, and obviously happens at a distance of r and after a time of r/c, so its coordinates are (r/c, r,0,0). R is likewise found as (r/c, -r,0,0). Indeed these are just special points on the shell and there is an infinite of events where the light hits the shell, given by E(θ,φ) = (r/c, r cos φ,r sin φ cos θ,r sin φ sin θ), but we are just dealing with F = E(0,0), and R=E(π,0).
Finally, each of these events, W, has an associated time, tW, and place, pW, in this coordinate frame. And given two (different) events, we can compute the speed of a particle which would connect them.
Summarizing:
o = (--, 0, 0, 0)
h = (--, 0, 0, 0)
A = (0, 0, 0, 0)
tA = (0, --, --, --)
pA = (--, 0, 0, 0) = o = h
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
pR = (--, -r, 0, 0) = - pF
sAF = | pF - pA | / (tF - tA) = |r|/(r/c) = c
sAR = | pR - pA | / (tR - tA) = |r|/(r/c) = c
But since there is no relative motion between massive objects and our observer, there is no application of the special relativity yet.
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
Which are the famous Einstein velocity addition formulas.
I have a question on the above.
As I read it, you have o moving relative to o'
Then o is also moving at speed v in the x direction.
Then you also have o' moving it seems, origined at vt' at some t'
o' = (--,vt',0,0). o' is now a "place" than moves
I assume you mean the observer is o'
but h' is a place defined to be motionless in the frame of the observer
If h' is at rest with o', then I do not understand how you say the below.
h' = (--, 0, 0, 0)
Thus, if o' and h' are at rest, how can both origin conditions above be satisfied after any time t'?
Perhaps I misread it.
Let r be the proper radius of the massive shell. Proper length is defined as the length as measured in the frame where a massive object travelling inertially is motionless, so r is well-defined.
Let o be the center of the shell. o is a place defined relative to the shell. If the shell is meausred to be moving in the x direction, then o also moves in the x direction.
Let h be another place, defined relative to the inertial observer as the place where the light came from. h never moves relative to the observer. h and o are the same for at least the instant when the light flashes. But if the massive shell is moving relative to the observer, if v is not zero, then h is not equal to o at any other time.
A is an event -- the event which happens at h at a certain time, and which also happens at o at a certain time, because of our definition of o.
F is an event -- the event which happens when the light hits one part of the shell.
R is an event -- the event which happens when the light hits another part of the shell. F and R (forward and reverse) happen at certain places and certain times.
Let's place our observer in a frame where the shell is motionless -- in which case everything is easy. Let's define A to be the origin of our space-time co-ordinate system. A = (0,0,0,0). F is forward in the x direction, and obviously happens at a distance of r and after a time of r/c, so its coordinates are (r/c, r,0,0). R is likewise found as (r/c, -r,0,0). Indeed these are just special points on the shell and there is an infinite of events where the light hits the shell, given by E(θ,φ) = (r/c, r cos φ,r sin φ cos θ,r sin φ sin θ), but we are just dealing with F = E(0,0), and R=E(π,0).
Finally, each of these events, W, has an associated time, tW, and place, pW, in this coordinate frame. And given two (different) events, we can compute the speed of a particle which would connect them.
Summarizing:
o = (--, 0, 0, 0)
h = (--, 0, 0, 0)
A = (0, 0, 0, 0)
tA = (0, --, --, --)
pA = (--, 0, 0, 0) = o = h
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
pR = (--, -r, 0, 0) = - pF
sAF = | pF - pA | / (tF - tA) = |r|/(r/c) = c
sAR = | pR - pA | / (tR - tA) = |r|/(r/c) = c
But since there is no relative motion between massive objects and our observer, there is no application of the special relativity yet.
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
Which are the famous Einstein velocity addition formulas.
QUOTE
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
I have a question on the above.
As I read it, you have o moving relative to o'
Then o is also moving at speed v in the x direction.
Then you also have o' moving it seems, origined at vt' at some t'
o' = (--,vt',0,0). o' is now a "place" than moves
I assume you mean the observer is o'
but h' is a place defined to be motionless in the frame of the observer
If h' is at rest with o', then I do not understand how you say the below.
h' = (--, 0, 0, 0)
Thus, if o' and h' are at rest, how can both origin conditions above be satisfied after any time t'?
Perhaps I misread it.
o can't move relative to o' -- They are the same place.
On the other hand, h and h' are different places since they are defined relative to the observer.
But because the observer is not moving, o is written as a non-moving place, and because the observer' is moving, o' is seen as moving relative to observer'. That's true even in Galilean relativity.
On the other hand, h and h' are different places since they are defined relative to the observer.
But because the observer is not moving, o is written as a non-moving place, and because the observer' is moving, o' is seen as moving relative to observer'. That's true even in Galilean relativity.
QUOTE (rpenner+Nov 6 2009, 04:13 AM)
Here's how to break this down via events.
Let r be the proper radius of the massive shell. Proper length is defined as the length as measured in the frame where a massive object travelling inertially is motionless, so r is well-defined.
Let o be the center of the shell. o is a place defined relative to the shell. If the shell is meausred to be moving in the x direction, then o also moves in the x direction.
Let h be another place, defined relative to the inertial observer as the place where the light came from. h never moves relative to the observer. h and o are the same for at least the instant when the light flashes. But if the massive shell is moving relative to the observer, if v is not zero, then h is not equal to o at any other time.
A is an event -- the event which happens at h at a certain time, and which also happens at o at a certain time, because of our definition of o.
F is an event -- the event which happens when the light hits one part of the shell.
R is an event -- the event which happens when the light hits another part of the shell. F and R (forward and reverse) happen at certain places and certain times.
Let's place our observer in a frame where the shell is motionless -- in which case everything is easy. Let's define A to be the origin of our space-time co-ordinate system. A = (0,0,0,0). F is forward in the x direction, and obviously happens at a distance of r and after a time of r/c, so its coordinates are (r/c, r,0,0). R is likewise found as (r/c, -r,0,0). Indeed these are just special points on the shell and there is an infinite of events where the light hits the shell, given by E(θ,φ) = (r/c, r cos φ,r sin φ cos θ,r sin φ sin θ), but we are just dealing with F = E(0,0), and R=E(π,0).
Finally, each of these events, W, has an associated time, tW, and place, pW, in this coordinate frame. And given two (different) events, we can compute the speed of a particle which would connect them.
Summarizing:
o = (--, 0, 0, 0)
h = (--, 0, 0, 0)
A = (0, 0, 0, 0)
tA = (0, --, --, --)
pA = (--, 0, 0, 0) = o = h
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
pR = (--, -r, 0, 0) = - pF
sAF = | pF - pA | / (tF - tA) = |r|/(r/c) = c
sAR = | pR - pA | / (tR - tA) = |r|/(r/c) = c
But since there is no relative motion between massive objects and our observer, there is no application of the special relativity yet.
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
Which are the famous Einstein velocity addition formulas.
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
I have an additional question.
I do not understand this for F'.
Einstein wrote the elapsed time in the rest frame tB - tA for light to travel from mirror A to mirror B is rAB/( c - v ) where rAB is the distance between the mirrors as determined in the rest rest frame.
The light was emitted when A and the origin of the rest frame were coincident.
Thus, the mirror at B is exactly the front part of the moving sphere.
This logic is at the end of Chapter 2.
http
://www.fourmilab.ch/etexts/einstein/specrel/www/
Let r be the proper radius of the massive shell. Proper length is defined as the length as measured in the frame where a massive object travelling inertially is motionless, so r is well-defined.
Let o be the center of the shell. o is a place defined relative to the shell. If the shell is meausred to be moving in the x direction, then o also moves in the x direction.
Let h be another place, defined relative to the inertial observer as the place where the light came from. h never moves relative to the observer. h and o are the same for at least the instant when the light flashes. But if the massive shell is moving relative to the observer, if v is not zero, then h is not equal to o at any other time.
A is an event -- the event which happens at h at a certain time, and which also happens at o at a certain time, because of our definition of o.
F is an event -- the event which happens when the light hits one part of the shell.
R is an event -- the event which happens when the light hits another part of the shell. F and R (forward and reverse) happen at certain places and certain times.
Let's place our observer in a frame where the shell is motionless -- in which case everything is easy. Let's define A to be the origin of our space-time co-ordinate system. A = (0,0,0,0). F is forward in the x direction, and obviously happens at a distance of r and after a time of r/c, so its coordinates are (r/c, r,0,0). R is likewise found as (r/c, -r,0,0). Indeed these are just special points on the shell and there is an infinite of events where the light hits the shell, given by E(θ,φ) = (r/c, r cos φ,r sin φ cos θ,r sin φ sin θ), but we are just dealing with F = E(0,0), and R=E(π,0).
Finally, each of these events, W, has an associated time, tW, and place, pW, in this coordinate frame. And given two (different) events, we can compute the speed of a particle which would connect them.
Summarizing:
o = (--, 0, 0, 0)
h = (--, 0, 0, 0)
A = (0, 0, 0, 0)
tA = (0, --, --, --)
pA = (--, 0, 0, 0) = o = h
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
pR = (--, -r, 0, 0) = - pF
sAF = | pF - pA | / (tF - tA) = |r|/(r/c) = c
sAR = | pR - pA | / (tR - tA) = |r|/(r/c) = c
But since there is no relative motion between massive objects and our observer, there is no application of the special relativity yet.
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
Which are the famous Einstein velocity addition formulas.
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
I have an additional question.
I do not understand this for F'.
Einstein wrote the elapsed time in the rest frame tB - tA for light to travel from mirror A to mirror B is rAB/( c - v ) where rAB is the distance between the mirrors as determined in the rest rest frame.
The light was emitted when A and the origin of the rest frame were coincident.
Thus, the mirror at B is exactly the front part of the moving sphere.
This logic is at the end of Chapter 2.
http
://www.fourmilab.ch/etexts/einstein/specrel/www/
Proof of the logical invalidity of SR's version of the light postulate using the twin spheres.
Light Cone
A Light cone is the path that a flash of light would take through spacetime. As time progresses, the light from the flash spreads out in a circle, and the result is a cone. In reality, there are three space dimensions, so the light actually forms a sphere in space, and the light cone is actually a fourth dimensional shape, but it's much easier to visualize it as a cone.
The above classifications hold true in any frame of reference; that is, an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers, no matter their frame of reference. This is why the concept is so powerful.
http
://en.wikipedia.org/wiki/Light_cone
Causality
In classical physics a cause should always precede its effect. In relativity theory this requirement is strengthened so as to limit causes to the back (past) light cone of the event to be explained (the "effect"); nor can an event be a cause of any event outside the former event's front (future) light cone. These restrictions are consistent with the grounded belief that causal influences cannot travel faster than the speed of light.
http
://en.wikipedia.org/wiki/Causality_(physics)
Proof
This proof below will proceed by reductio ad absurdum and will depend on the facts above about the light cone.
First an event is defined as light striking an object.
Let O and O' be two objects and let there be one light sphere. Let E(O) mean object O was struck by the light sphere.
According to the logic of the light cone, one and only one of the following tricotomy holds:
1) Object O is struck by the light sphere before object O' written as E(O) < E(O')
2) Object O' is struck by the light sphere before object O written as E(O') < E(O)
3) Both O and O' were struck by the light sphere but neither condition 1 or 2 were ever true.
This tricotomy is just a restatement of causality defined above as implemented by the light cone. Also, as stated above, no observer in the universe can disagree on the ordinality of events as determined by one light sphere. This would be a violation of causality.
Assume there are two rigid body spheres S and S'. Let c be the center of S and c' be the center of S'.
Let Observer O be at rest with rigid body sphere S.
Let Observer O' be at rest with rigid body sphere S'.
S' has a light source at its center c'
Both spheres have a diameter D when they are at rest with one another.
Let L be the point where the sphere intersects the negative x-axis and R be the point where the sphere intersects the positive x-axis for S.
Let L' be the point where the sphere intersects the negative x-axis and R' be the point where the sphere intersects the positive x-axis for S' in the coordinates of observer O.
Assume S' is in relative motion v > 0 along the positive x-axis of observer O.
When the centers of the spheres happen to be coincident, the light source emits from c' of S'.
By the light postulate, the motion of c' does not affect the speed of light in any direction so O concludes L and R are struck by the light at the same time.
Thus, t_L = t_R.
Further, since the light source is at rest with O', O' concludes t_L' = t_R'. to satisfy the light postulate.
Finally, observer O, the stationary observer, calculates the time light strikes L' and R' as follows.
As in the relativity of simultaneity, O concludes the front of the sphere of S', named R', is moving away from the constant speed of light.
Thus, light must travel the length of the radius of S' in O which is D/(2λ) plus the distance S' moves in time t which is vt.This is consistent with Einstein's assertions at the end of chapter 2 which he writes:
tB - tA = rAB/(c-v) and t'A - tB = rAB/(c+v)
http
://www.fourmilab.ch/etexts/einstein/specrel/www/
Thus,
ct'_R' = D/(2λ) + vt'_R' .
ct'_R' - vt'_R' = D/(2λ)
t'_R' (c - v) = D/(2λ)
t'_R' = D/(2λ(c - v))
In the same way, for the constant light to strike the L', light must move D/(2λ) minus the distance S' moves in time t which is vt.
Thus,
ct'_L' = D/(2λ) - vt'_L'.
ct'_L' + vt'_L' = D/(2λ)
t'_L'(c + v) = D/(2λ)
t'_L' = D/(2λ(c + v))
Now, it is the case that O' concludes t_L' = t_R' by the light postulate as stated above.
But, it is also the case that no two observers in the universe can disagree on events as defined above based on the tricotomy relationship between two events as shown above.
Since O' concludes t_L' = t_R'., O has no choice but to conclude, t'_L' = t'_R', in order to satisfy the necessary condition of the light cone that no two observers can disagree on the ordinality of events as implemented by one light sphere. They can disagree on the timing of the ordinality of events, but not the ordinality.
Thus,
t'_R' = t'_L' = D/(2λ(c - v)) = D/(2λ(c + v))
1/(c - v) = 1/(c + v)
(c + v) = (c - v)
2v = 0
v = 0.
This is a contradiction since it was assumed we have relative motion v > 0.
Therefore, it is impossible to meet the conditions
1) Observer O concludes light struck the L' prior to R'
and
2) Observer O' concludes t_L' = t_R',
because of the tricotomy imposed by the light cone.
However, the relativity of simultaneity requires that condition 1 is met and the light postulate requires that condition 2 is met.
Hence, SR's version if the light postulate is not logically viable since the theory requires that both conditions are met.
Light Cone
A Light cone is the path that a flash of light would take through spacetime. As time progresses, the light from the flash spreads out in a circle, and the result is a cone. In reality, there are three space dimensions, so the light actually forms a sphere in space, and the light cone is actually a fourth dimensional shape, but it's much easier to visualize it as a cone.
The above classifications hold true in any frame of reference; that is, an event judged to be in the light cone by one observer, will also be judged to be in the same light cone by all other observers, no matter their frame of reference. This is why the concept is so powerful.
http
://en.wikipedia.org/wiki/Light_cone
Causality
In classical physics a cause should always precede its effect. In relativity theory this requirement is strengthened so as to limit causes to the back (past) light cone of the event to be explained (the "effect"); nor can an event be a cause of any event outside the former event's front (future) light cone. These restrictions are consistent with the grounded belief that causal influences cannot travel faster than the speed of light.
http
://en.wikipedia.org/wiki/Causality_(physics)
Proof
This proof below will proceed by reductio ad absurdum and will depend on the facts above about the light cone.
First an event is defined as light striking an object.
Let O and O' be two objects and let there be one light sphere. Let E(O) mean object O was struck by the light sphere.
According to the logic of the light cone, one and only one of the following tricotomy holds:
1) Object O is struck by the light sphere before object O' written as E(O) < E(O')
2) Object O' is struck by the light sphere before object O written as E(O') < E(O)
3) Both O and O' were struck by the light sphere but neither condition 1 or 2 were ever true.
This tricotomy is just a restatement of causality defined above as implemented by the light cone. Also, as stated above, no observer in the universe can disagree on the ordinality of events as determined by one light sphere. This would be a violation of causality.
Assume there are two rigid body spheres S and S'. Let c be the center of S and c' be the center of S'.
Let Observer O be at rest with rigid body sphere S.
Let Observer O' be at rest with rigid body sphere S'.
S' has a light source at its center c'
Both spheres have a diameter D when they are at rest with one another.
Let L be the point where the sphere intersects the negative x-axis and R be the point where the sphere intersects the positive x-axis for S.
Let L' be the point where the sphere intersects the negative x-axis and R' be the point where the sphere intersects the positive x-axis for S' in the coordinates of observer O.
Assume S' is in relative motion v > 0 along the positive x-axis of observer O.
When the centers of the spheres happen to be coincident, the light source emits from c' of S'.
By the light postulate, the motion of c' does not affect the speed of light in any direction so O concludes L and R are struck by the light at the same time.
Thus, t_L = t_R.
Further, since the light source is at rest with O', O' concludes t_L' = t_R'. to satisfy the light postulate.
Finally, observer O, the stationary observer, calculates the time light strikes L' and R' as follows.
As in the relativity of simultaneity, O concludes the front of the sphere of S', named R', is moving away from the constant speed of light.
Thus, light must travel the length of the radius of S' in O which is D/(2λ) plus the distance S' moves in time t which is vt.This is consistent with Einstein's assertions at the end of chapter 2 which he writes:
tB - tA = rAB/(c-v) and t'A - tB = rAB/(c+v)
http
://www.fourmilab.ch/etexts/einstein/specrel/www/
Thus,
ct'_R' = D/(2λ) + vt'_R' .
ct'_R' - vt'_R' = D/(2λ)
t'_R' (c - v) = D/(2λ)
t'_R' = D/(2λ(c - v))
In the same way, for the constant light to strike the L', light must move D/(2λ) minus the distance S' moves in time t which is vt.
Thus,
ct'_L' = D/(2λ) - vt'_L'.
ct'_L' + vt'_L' = D/(2λ)
t'_L'(c + v) = D/(2λ)
t'_L' = D/(2λ(c + v))
Now, it is the case that O' concludes t_L' = t_R' by the light postulate as stated above.
But, it is also the case that no two observers in the universe can disagree on events as defined above based on the tricotomy relationship between two events as shown above.
Since O' concludes t_L' = t_R'., O has no choice but to conclude, t'_L' = t'_R', in order to satisfy the necessary condition of the light cone that no two observers can disagree on the ordinality of events as implemented by one light sphere. They can disagree on the timing of the ordinality of events, but not the ordinality.
Thus,
t'_R' = t'_L' = D/(2λ(c - v)) = D/(2λ(c + v))
1/(c - v) = 1/(c + v)
(c + v) = (c - v)
2v = 0
v = 0.
This is a contradiction since it was assumed we have relative motion v > 0.
Therefore, it is impossible to meet the conditions
1) Observer O concludes light struck the L' prior to R'
and
2) Observer O' concludes t_L' = t_R',
because of the tricotomy imposed by the light cone.
However, the relativity of simultaneity requires that condition 1 is met and the light postulate requires that condition 2 is met.
Hence, SR's version if the light postulate is not logically viable since the theory requires that both conditions are met.
QUOTE (rpenner+Nov 6 2009, 04:13 AM)
Here's how to break this down via events.
Let r be the proper radius of the massive shell. Proper length is defined as the length as measured in the frame where a massive object travelling inertially is motionless, so r is well-defined.
Let o be the center of the shell. o is a place defined relative to the shell. If the shell is meausred to be moving in the x direction, then o also moves in the x direction.
Let h be another place, defined relative to the inertial observer as the place where the light came from. h never moves relative to the observer. h and o are the same for at least the instant when the light flashes. But if the massive shell is moving relative to the observer, if v is not zero, then h is not equal to o at any other time.
A is an event -- the event which happens at h at a certain time, and which also happens at o at a certain time, because of our definition of o.
F is an event -- the event which happens when the light hits one part of the shell.
R is an event -- the event which happens when the light hits another part of the shell. F and R (forward and reverse) happen at certain places and certain times.
Let's place our observer in a frame where the shell is motionless -- in which case everything is easy. Let's define A to be the origin of our space-time co-ordinate system. A = (0,0,0,0). F is forward in the x direction, and obviously happens at a distance of r and after a time of r/c, so its coordinates are (r/c, r,0,0). R is likewise found as (r/c, -r,0,0). Indeed these are just special points on the shell and there is an infinite of events where the light hits the shell, given by E(θ,φ) = (r/c, r cos φ,r sin φ cos θ,r sin φ sin θ), but we are just dealing with F = E(0,0), and R=E(π,0).
Finally, each of these events, W, has an associated time, tW, and place, pW, in this coordinate frame. And given two (different) events, we can compute the speed of a particle which would connect them.
Summarizing:
o = (--, 0, 0, 0)
h = (--, 0, 0, 0)
A = (0, 0, 0, 0)
tA = (0, --, --, --)
pA = (--, 0, 0, 0) = o = h
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
pR = (--, -r, 0, 0) = - pF
sAF = | pF - pA | / (tF - tA) = |r|/(r/c) = c
sAR = | pR - pA | / (tR - tA) = |r|/(r/c) = c
But since there is no relative motion between massive objects and our observer, there is no application of the special relativity yet.
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
Which are the famous Einstein velocity addition formulas.
OK, nevermind I can see
((r/c + vr/c²)/√(1-v²/c²) = r/(λ(c-v))
OK, nevermind I can see
((r/c + vr/c²)/√(1-v²/c²) = r/(λ(c-v))
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
OK
Then you wrote
This is to be expected since it is assumed light travels c in the rest frame and these calculations were done from the rest frame where that is assumed as c. I am OK here.
This is to be expected since it is assumed light travels c in the rest frame and these calculations were done from the rest frame where that is assumed as c. I am OK here.
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
This part appears to prove light is not moving at c in the moving frame but that is moving at u and that is consistent.
Did I understand this correctly?
This part appears to prove light is not moving at c in the moving frame but that is moving at u and that is consistent.
Did I understand this correctly?
Dumbass. Persistent.
Thus,
ct'_R' = D/(2λ) + vt'_R' .
ct'_R' - vt'_R' = D/(2λ)
t'_R' (c - v) = D/(2λ)
t'_R' = D/(2λ(c - v))
In the same way, for the constant light to strike the L', light must move D/(2λ) minus the distance S' moves in time t which is vt.
Thus,
ct'_L' = D/(2λ) - vt'_L'.
ct'_L' + vt'_L' = D/(2λ)
t'_L'(c + v) = D/(2λ)
t'_L' = D/(2λ(c + v))
Now, it is the case that O' concludes t_L' = t_R' by the light postulate as stated above.
But, it is also the case that no two observers in the universe can disagree on events as defined above based on the tricotomy relationship between two events as shown above.
Dumbass, this is the calculation the way the observer travelling with sphere S' sees the strikes on sphere S. The strikes are NOT simultaneous for him.
Only a dumbass like you could conclude such an idiocy. t'_R' is not equal to t'_L'. You don't understand the relativity of simultaneity.
It is a bit tricky.
The equations operate from the rest observer S.
These are the correct equations for S viewing S'
It is a bit tricky.
The equations operate from the rest observer S.
These are the correct equations for S viewing S'
Thus,
ct'_R' = D/(2λ) + vt'_R' .
ct'_R' - vt'_R' = D/(2λ)
t'_R' (c - v) = D/(2λ)
t'_R' = D/(2λ(c - v))
In the same way, for the constant light to strike the L', light must move D/(2λ) minus the distance S' moves in time t which is vt.
Thus,
ct'_L' = D/(2λ) - vt'_L'.
ct'_L' + vt'_L' = D/(2λ)
t'_L'(c + v) = D/(2λ)
t'_L' = D/(2λ(c + v))
Now, it is the case that O' concludes t_L' = t_R' by the light postulate as stated above.
But, it is also the case that no two observers in the universe can disagree on events as defined above based on the tricotomy relationship between two events as shown above.
As such, these are the correct equations for the interpretation of S' from S.
Given that this proof calculates the frame of S' from S, t'_R' > t'_L'. as should be concluded by the relativity of simultaneity.
However, the light postulate also demands that t_R' = t_L'.
Finally, the logic of the light cone demands that S and S' cannot disagree on the ordinality of E(R') and E(L') because there is one light sphere/cone.
But, R of S and the light postulate disagree on the ordinality E(R') and E(L').
This is the contradiction.
It is a bit tricky.
No, it isn't tricky, it is elementary.
Don't worry, your stupidity isn't curable.
Let r be the proper radius of the massive shell. Proper length is defined as the length as measured in the frame where a massive object travelling inertially is motionless, so r is well-defined.
Let o be the center of the shell. o is a place defined relative to the shell. If the shell is meausred to be moving in the x direction, then o also moves in the x direction.
Let h be another place, defined relative to the inertial observer as the place where the light came from. h never moves relative to the observer. h and o are the same for at least the instant when the light flashes. But if the massive shell is moving relative to the observer, if v is not zero, then h is not equal to o at any other time.
A is an event -- the event which happens at h at a certain time, and which also happens at o at a certain time, because of our definition of o.
F is an event -- the event which happens when the light hits one part of the shell.
R is an event -- the event which happens when the light hits another part of the shell. F and R (forward and reverse) happen at certain places and certain times.
Let's place our observer in a frame where the shell is motionless -- in which case everything is easy. Let's define A to be the origin of our space-time co-ordinate system. A = (0,0,0,0). F is forward in the x direction, and obviously happens at a distance of r and after a time of r/c, so its coordinates are (r/c, r,0,0). R is likewise found as (r/c, -r,0,0). Indeed these are just special points on the shell and there is an infinite of events where the light hits the shell, given by E(θ,φ) = (r/c, r cos φ,r sin φ cos θ,r sin φ sin θ), but we are just dealing with F = E(0,0), and R=E(π,0).
Finally, each of these events, W, has an associated time, tW, and place, pW, in this coordinate frame. And given two (different) events, we can compute the speed of a particle which would connect them.
Summarizing:
o = (--, 0, 0, 0)
h = (--, 0, 0, 0)
A = (0, 0, 0, 0)
tA = (0, --, --, --)
pA = (--, 0, 0, 0) = o = h
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
pR = (--, -r, 0, 0) = - pF
sAF = | pF - pA | / (tF - tA) = |r|/(r/c) = c
sAR = | pR - pA | / (tR - tA) = |r|/(r/c) = c
But since there is no relative motion between massive objects and our observer, there is no application of the special relativity yet.
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
Which are the famous Einstein velocity addition formulas.
QUOTE
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
OK, nevermind I can see
((r/c + vr/c²)/√(1-v²/c²) = r/(λ(c-v))
QUOTE (->
| QUOTE |
| So now we apply the Lorentz transforms to the three events. A' = (0, 0, 0, 0) F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0) = ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0) R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0) = ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0) |
OK, nevermind I can see
((r/c + vr/c²)/√(1-v²/c²) = r/(λ(c-v))
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
OK
Then you wrote
QUOTE
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
This is to be expected since it is assumed light travels c in the rest frame and these calculations were done from the rest frame where that is assumed as c. I am OK here.
QUOTE (->
| QUOTE |
| sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c |
This is to be expected since it is assumed light travels c in the rest frame and these calculations were done from the rest frame where that is assumed as c. I am OK here.
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
This part appears to prove light is not moving at c in the moving frame but that is moving at u and that is consistent.
Did I understand this correctly?
QUOTE (qwtyu+Nov 5 2009, 04:47 PM)
[Moderator: 50% likelihood I will ban this user and delete all posts. The poster is failing at communicating justified, factual and useful information.]
Probability levels at 75% and rising.
Probability levels at 75% and rising.
QUOTE (rpenner+Nov 6 2009, 04:13 AM)
Here's how to break this down via events.
Let r be the proper radius of the massive shell. Proper length is defined as the length as measured in the frame where a massive object travelling inertially is motionless, so r is well-defined.
Let o be the center of the shell. o is a place defined relative to the shell. If the shell is meausred to be moving in the x direction, then o also moves in the x direction.
Let h be another place, defined relative to the inertial observer as the place where the light came from. h never moves relative to the observer. h and o are the same for at least the instant when the light flashes. But if the massive shell is moving relative to the observer, if v is not zero, then h is not equal to o at any other time.
A is an event -- the event which happens at h at a certain time, and which also happens at o at a certain time, because of our definition of o.
F is an event -- the event which happens when the light hits one part of the shell.
R is an event -- the event which happens when the light hits another part of the shell. F and R (forward and reverse) happen at certain places and certain times.
Let's place our observer in a frame where the shell is motionless -- in which case everything is easy. Let's define A to be the origin of our space-time co-ordinate system. A = (0,0,0,0). F is forward in the x direction, and obviously happens at a distance of r and after a time of r/c, so its coordinates are (r/c, r,0,0). R is likewise found as (r/c, -r,0,0). Indeed these are just special points on the shell and there is an infinite of events where the light hits the shell, given by E(θ,φ) = (r/c, r cos φ,r sin φ cos θ,r sin φ sin θ), but we are just dealing with F = E(0,0), and R=E(π,0).
Finally, each of these events, W, has an associated time, tW, and place, pW, in this coordinate frame. And given two (different) events, we can compute the speed of a particle which would connect them.
Summarizing:
o = (--, 0, 0, 0)
h = (--, 0, 0, 0)
A = (0, 0, 0, 0)
tA = (0, --, --, --)
pA = (--, 0, 0, 0) = o = h
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
pR = (--, -r, 0, 0) = - pF
sAF = | pF - pA | / (tF - tA) = |r|/(r/c) = c
sAR = | pR - pA | / (tR - tA) = |r|/(r/c) = c
But since there is no relative motion between massive objects and our observer, there is no application of the special relativity yet.
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
Which are the famous Einstein velocity addition formulas.
That's nice, thanks. The only way I've done this [as an interested amateur] is solving the transformation equations dx=...., and dt=.... for dx/dt and defining u as the velocity in the laboratory frame and u' for the velocity in the rocket frame.
Let r be the proper radius of the massive shell. Proper length is defined as the length as measured in the frame where a massive object travelling inertially is motionless, so r is well-defined.
Let o be the center of the shell. o is a place defined relative to the shell. If the shell is meausred to be moving in the x direction, then o also moves in the x direction.
Let h be another place, defined relative to the inertial observer as the place where the light came from. h never moves relative to the observer. h and o are the same for at least the instant when the light flashes. But if the massive shell is moving relative to the observer, if v is not zero, then h is not equal to o at any other time.
A is an event -- the event which happens at h at a certain time, and which also happens at o at a certain time, because of our definition of o.
F is an event -- the event which happens when the light hits one part of the shell.
R is an event -- the event which happens when the light hits another part of the shell. F and R (forward and reverse) happen at certain places and certain times.
Let's place our observer in a frame where the shell is motionless -- in which case everything is easy. Let's define A to be the origin of our space-time co-ordinate system. A = (0,0,0,0). F is forward in the x direction, and obviously happens at a distance of r and after a time of r/c, so its coordinates are (r/c, r,0,0). R is likewise found as (r/c, -r,0,0). Indeed these are just special points on the shell and there is an infinite of events where the light hits the shell, given by E(θ,φ) = (r/c, r cos φ,r sin φ cos θ,r sin φ sin θ), but we are just dealing with F = E(0,0), and R=E(π,0).
Finally, each of these events, W, has an associated time, tW, and place, pW, in this coordinate frame. And given two (different) events, we can compute the speed of a particle which would connect them.
Summarizing:
o = (--, 0, 0, 0)
h = (--, 0, 0, 0)
A = (0, 0, 0, 0)
tA = (0, --, --, --)
pA = (--, 0, 0, 0) = o = h
F = (r/c, r, 0, 0)
tF = (r/c, --, --, --)
pF = (--, r, 0, 0)
R = (r/c, -r, 0, 0)
tR = (r/c, --, --, --) = tF
pR = (--, -r, 0, 0) = - pF
sAF = | pF - pA | / (tF - tA) = |r|/(r/c) = c
sAR = | pR - pA | / (tR - tA) = |r|/(r/c) = c
But since there is no relative motion between massive objects and our observer, there is no application of the special relativity yet.
What we need now is the assumption that a new observer (primed coordinates) sees the same massive sphere as moving at speed v. Then o is also moving at speed v in the x direction.
o' = (--,vt',0,0). o' is now a "place" than moves!
but h' is a place defined to be motionless in the frame of the observer, and at the time of the flash o' and h' are the same. So let's call that time, t' = 0.
h' = (--, 0, 0, 0)
So now we apply the Lorentz transforms to the three events.
A' = (0, 0, 0, 0)
F' = ((r/c + vr/c²)/√(1-v²/c²), (r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c+v)/(c-v)), r√((c+v)/(c-v)), 0, 0)
R' = ((r/c - vr/c²)/√(1-v²/c²), (-r + vr/c)/√(1-v²/c²), 0, 0)
= ((r/c)√((c-v)/(c+v)), -r√((c-v)/(c+v)), 0, 0)
That is the proper application of the Lorentz transforms. So now let's see if Einstein's assumption about the speed of light holds for the observer who sees the sphere move.
sA'F' = | pF' - pA' | / (tF' - tA') = |r|/(r/c) = c
sA'R' = | pR' - pA' | / (tR' - tA') = |r|/(r/c) = c
But what if the speed of the "light" from A to F and R wasn't moving at c but another speed, 0 < u < c.
Then
A = (0,0,0,0)
G = (r/u, r, 0, 0)
S = (r/u, -r, 0, 0)
and
A' = (0,0,0,0)
G' = ((r/u + vr/c²)/√(1-v²/c²), (r + vr/u)/√(1-v²/c²), 0, 0)
S' = ((r/u - vr/c²)/√(1-v²/c²), (-r + vr/u)/√(1-v²/c²), 0, 0)
sA'G' = | pG' - pA' | / (tG' - tA') = | r + vr/u | / (r/u + vr/c²) = (u + v)/(1 + uv/c²)
sA'S' = | pS' - pA' | / (tS' - tA') = | r - vr/u | / (r/u - vr/c²) = (u - v)/(1 - uv/c²)
Which are the famous Einstein velocity addition formulas.
That's nice, thanks. The only way I've done this [as an interested amateur] is solving the transformation equations dx=...., and dt=.... for dx/dt and defining u as the velocity in the laboratory frame and u' for the velocity in the rocket frame.
QUOTE (qwtyu+Nov 6 2009, 10:37 PM)
This part appears to prove light is not moving at c in the moving frame but that is moving at u and that is consistent.
Did I understand this correctly?
Dumbass. Persistent.
QUOTE (qwtyu+Nov 6 2009, 10:08 PM)
Thus,
ct'_R' = D/(2λ) + vt'_R' .
ct'_R' - vt'_R' = D/(2λ)
t'_R' (c - v) = D/(2λ)
t'_R' = D/(2λ(c - v))
In the same way, for the constant light to strike the L', light must move D/(2λ) minus the distance S' moves in time t which is vt.
Thus,
ct'_L' = D/(2λ) - vt'_L'.
ct'_L' + vt'_L' = D/(2λ)
t'_L'(c + v) = D/(2λ)
t'_L' = D/(2λ(c + v))
Now, it is the case that O' concludes t_L' = t_R' by the light postulate as stated above.
But, it is also the case that no two observers in the universe can disagree on events as defined above based on the tricotomy relationship between two events as shown above.
Dumbass, this is the calculation the way the observer travelling with sphere S' sees the strikes on sphere S. The strikes are NOT simultaneous for him.
QUOTE
Since O' concludes t_L' = t_R'., O has no choice but to conclude, t'_L' = t'_R', in order to satisfy the necessary condition of the light cone that no two observers can disagree on the ordinality of events as implemented by one light sphere. They can disagree on the timing of the ordinality of events, but not the ordinality.
Thus,
t'_R' = t'_L' = D/(2λ(c - v)) = D/(2λ(c + v))
1/(c - v) = 1/(c + v)
(c + v) = (c - v)
2v = 0
v = 0.
Thus,
t'_R' = t'_L' = D/(2λ(c - v)) = D/(2λ(c + v))
1/(c - v) = 1/(c + v)
(c + v) = (c - v)
2v = 0
v = 0.
Only a dumbass like you could conclude such an idiocy. t'_R' is not equal to t'_L'. You don't understand the relativity of simultaneity.
QUOTE
Dumbass, this is the calculation the way the observer travelling with sphere S' sees the strikes on sphere S. The strikes are NOT simultaneous for him.
It is a bit tricky.
The equations operate from the rest observer S.
These are the correct equations for S viewing S'
QUOTE (->
| QUOTE |
| Dumbass, this is the calculation the way the observer travelling with sphere S' sees the strikes on sphere S. The strikes are NOT simultaneous for him. |
It is a bit tricky.
The equations operate from the rest observer S.
These are the correct equations for S viewing S'
Thus,
ct'_R' = D/(2λ) + vt'_R' .
ct'_R' - vt'_R' = D/(2λ)
t'_R' (c - v) = D/(2λ)
t'_R' = D/(2λ(c - v))
In the same way, for the constant light to strike the L', light must move D/(2λ) minus the distance S' moves in time t which is vt.
Thus,
ct'_L' = D/(2λ) - vt'_L'.
ct'_L' + vt'_L' = D/(2λ)
t'_L'(c + v) = D/(2λ)
t'_L' = D/(2λ(c + v))
Now, it is the case that O' concludes t_L' = t_R' by the light postulate as stated above.
But, it is also the case that no two observers in the universe can disagree on events as defined above based on the tricotomy relationship between two events as shown above.
As such, these are the correct equations for the interpretation of S' from S.
QUOTE
Only a dumbass like you could conclude such an idiocy. t'_R' is not equal to t'_L'. You don't understand the relativity of simultaneity.
Given that this proof calculates the frame of S' from S, t'_R' > t'_L'. as should be concluded by the relativity of simultaneity.
However, the light postulate also demands that t_R' = t_L'.
Finally, the logic of the light cone demands that S and S' cannot disagree on the ordinality of E(R') and E(L') because there is one light sphere/cone.
But, R of S and the light postulate disagree on the ordinality E(R') and E(L').
This is the contradiction.
QUOTE (qwtyu+Nov 10 2009, 01:45 AM)
It is a bit tricky.
No, it isn't tricky, it is elementary.
Don't worry, your stupidity isn't curable.
QUOTE (Trout+Nov 10 2009, 02:04 AM)
No, it isn't tricky, it is elementary.
Don't worry, your stupidity isn't curable.
Well, at least you realize if you want to debate me, it will be here and not at BAUT.
So, here is the deal.
The light postulate forces S' to see its left and right sphere points hit at the same time.
R of S forces S to see S' s left and right sphere points hit at the different times.
The logic of the light cone forces neither to disagree.
This is pretty simple.
Further, this is all under the theory of special relativity.
There is no logical resolution.
Don't worry, your stupidity isn't curable.
Well, at least you realize if you want to debate me, it will be here and not at BAUT.
So, here is the deal.
The light postulate forces S' to see its left and right sphere points hit at the same time.
R of S forces S to see S' s left and right sphere points hit at the different times.
The logic of the light cone forces neither to disagree.
This is pretty simple.
Further, this is all under the theory of special relativity.
There is no logical resolution.
QUOTE (Trout+Nov 10 2009, 02:04 AM)
No, it isn't tricky, it is elementary.
Don't worry, your stupidity isn't curable.
OK, I must admit, I laughed for days over that twins thread you were in with those "experts".
I recall you saying you are not doing integration you are talking about it.
I must admit it was funny. They were too timid once you came around to actually perform an integration for fear of getting it all messed up.
Thus, you were able to make them cower just on the thought you might request them to perform integral calculus to support their calculations.
It was all very amusing.
[Moderator: Suspended 30 days for misappropriating the language but not the methods of mathematics. Would anyone care to take a stab at setting up the problem correctly and identifying specifically where poster went wrong?]
Don't worry, your stupidity isn't curable.
OK, I must admit, I laughed for days over that twins thread you were in with those "experts".
I recall you saying you are not doing integration you are talking about it.
I must admit it was funny. They were too timid once you came around to actually perform an integration for fear of getting it all messed up.
Thus, you were able to make them cower just on the thought you might request them to perform integral calculus to support their calculations.
It was all very amusing.
[Moderator: Suspended 30 days for misappropriating the language but not the methods of mathematics. Would anyone care to take a stab at setting up the problem correctly and identifying specifically where poster went wrong?]
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