Hi guys,
I need help solving this system of differential equations. Thank you !
Actually, I managed to make some progress, see here. I still need some help.
At a first glance, for the general initial value problem: multiply the first equation by v_x, the second by v_y - add them together, then you can integrate up once. This will give you v_x in terms of v_y, which you can then use in the first. The resulting ODE is horrendous.
If you're just looking for any solution, then it's easy to find the static one. A common tactic when dealing with Lorentz invariant equations is to look for a static solution then apply a boost to get the more general solution.
If you're just looking for any solution, then it's easy to find the static one. A common tactic when dealing with Lorentz invariant equations is to look for a static solution then apply a boost to get the more general solution.
QUOTE (Euler+Mar 24 2008, 08:50 AM)
At a first glance, for the general initial value problem: multiply the first equation by v_x, the second by v_y - add them together, then you can integrate up once. This will give you v_x in terms of v_y, which you can then use in the first. The resulting ODE is horrendous.
Thank you, Euler.
This was my first idea but I had to discard it immediately. You can't do that because the
\gamma term involves BOTH v_x^2 and v_y^2 .
And yes, I did get the "horrendous non-linear ODE in v_x, (dv_x)/dt and (d^2v_x)/dt^2
If you're just looking for any solution, then it's easy to find the static one. A common tactic when dealing with Lorentz invariant equations is to look for a static solution then apply a boost to get the more general solution.
How do I do that? Can you show me?
Thank you, Euler.
This was my first idea but I had to discard it immediately. You can't do that because the
\gamma term involves BOTH v_x^2 and v_y^2 .
And yes, I did get the "horrendous non-linear ODE in v_x, (dv_x)/dt and (d^2v_x)/dt^2
QUOTE
If you're just looking for any solution, then it's easy to find the static one. A common tactic when dealing with Lorentz invariant equations is to look for a static solution then apply a boost to get the more general solution.
How do I do that? Can you show me?
QUOTE (Trout+Mar 24 2008, 01:44 PM)
This was my first idea but I had to discard it immediately. You can't do that because the \gamma term involves BOTH v_x^2 and v_y^2 .
You can... the resulting ODE is:
\gamma^2 \frac{d}{dt}( v_x^2 + v_y^2 ) - (2E/B) \frac{d v_x}{dt} = 0
Now use the chain rule on the first term. If it isn't clear, set X=v_x^2 + v_y^2 so that the first term in the above is:
c^2 (1- X/c^2)^{-1} \frac{dX}{dt} = -c^2 \frac{d}{dt} ln(1-X/c^2)
Unless my scribbles have betrayed me. With regards my second comment: if a PDE is Lorentz invariant, my suggestion is akin to choosing a convenient frame of reference to solve the problem, then applying a Lorentz transformation to obtain the solution from a general reference frame. For an exposition of the mathematics of this technique, I would recommend taking a look at Olver's book.
You can... the resulting ODE is:
\gamma^2 \frac{d}{dt}( v_x^2 + v_y^2 ) - (2E/B) \frac{d v_x}{dt} = 0
Now use the chain rule on the first term. If it isn't clear, set X=v_x^2 + v_y^2 so that the first term in the above is:
c^2 (1- X/c^2)^{-1} \frac{dX}{dt} = -c^2 \frac{d}{dt} ln(1-X/c^2)
Unless my scribbles have betrayed me. With regards my second comment: if a PDE is Lorentz invariant, my suggestion is akin to choosing a convenient frame of reference to solve the problem, then applying a Lorentz transformation to obtain the solution from a general reference frame. For an exposition of the mathematics of this technique, I would recommend taking a look at Olver's book.
QUOTE (Euler+Mar 24 2008, 01:57 PM)
You can... the resulting ODE is:
\gamma^2 \frac{d}{dt}( v_x^2 + v_y^2 ) - (2E/B) \frac{d v_x}{dt} = 0
Now use the chain rule on the first term. If it isn't clear, set X=v_x^2 + v_y^2 so that the first term in the above is:
c^2 (1- X/c^2)^{-1} \frac{dX}{dt} = -c^2 \frac{d}{dt} ln(1-X/c^2)
Unless my scribbles have betrayed me. With regards my second comment: if a PDE is Lorentz invariant, my suggestion is akin to choosing a convenient frame of reference to solve the problem, then applying a Lorentz transformation to obtain the solution from a general reference frame. For an exposition of the mathematics of this technique, I would recommend taking a look at Olver's book.
This is not what I'm getting:
\gamma^3*(v_x*dv_x/dt+v_y*dv_y/dt)-qEv_y=0
Now, with X=v_x^2+v_y^2
the above reduces to :
m_0/sqrt(1-X)^3*dX/dt=qEv_y
I could integrate both sides wrt t EXCEPT X is really a function of v_y.
\gamma^2 \frac{d}{dt}( v_x^2 + v_y^2 ) - (2E/B) \frac{d v_x}{dt} = 0
Now use the chain rule on the first term. If it isn't clear, set X=v_x^2 + v_y^2 so that the first term in the above is:
c^2 (1- X/c^2)^{-1} \frac{dX}{dt} = -c^2 \frac{d}{dt} ln(1-X/c^2)
Unless my scribbles have betrayed me. With regards my second comment: if a PDE is Lorentz invariant, my suggestion is akin to choosing a convenient frame of reference to solve the problem, then applying a Lorentz transformation to obtain the solution from a general reference frame. For an exposition of the mathematics of this technique, I would recommend taking a look at Olver's book.
This is not what I'm getting:
\gamma^3*(v_x*dv_x/dt+v_y*dv_y/dt)-qEv_y=0
Now, with X=v_x^2+v_y^2
the above reduces to :
m_0/sqrt(1-X)^3*dX/dt=qEv_y
I could integrate both sides wrt t EXCEPT X is really a function of v_y.
QUOTE (Trout+Mar 24 2008, 02:09 PM)
This is not what I'm getting:
\gamma^3*(v_x*dv_x/dt+v_y*dv_y/dt)-qEv_y=0
But your first equation gives:
qEv_y = frac{E\gamma m_0}{B} \frac{dv_x}{dt}
...
\gamma^3*(v_x*dv_x/dt+v_y*dv_y/dt)-qEv_y=0
But your first equation gives:
qEv_y = frac{E\gamma m_0}{B} \frac{dv_x}{dt}
...
QUOTE (Euler+Mar 24 2008, 02:11 PM)
But your first equation gives:
qEv_y = frac{E\gamma m_0}{B} \frac{dv_x}{dt}
...
Please PM me your email, I'll send you a .doc file with everything I did.
qEv_y = frac{E\gamma m_0}{B} \frac{dv_x}{dt}
...
Please PM me your email, I'll send you a .doc file with everything I did.
I'll go through it in more detail: multiplying your first equation by v_x and your second by v_y gives:
\frac{\gamma^3 m_0}{2} \frac{d}{dt} (v_x^2+v_y^2) - qE v_y = 0
Now use your first equation, to express v_y in terms of dv_x/dt. This gives:
\gamma^2 \frac{d}{dt}( v_x^2 + v_y^2 ) - (2E/B) \frac{d v_x}{dt} = 0 (*)
Which is the original expression I gave you. Using the chain rule on the first term, you can express (*) as a total t-derivative, integrating up then gives you v_y in terms of v_x. The best thing to do, if you don't follow the steps above, is to use your given equations to show that the LHS of really does vanish.
\frac{\gamma^3 m_0}{2} \frac{d}{dt} (v_x^2+v_y^2) - qE v_y = 0
Now use your first equation, to express v_y in terms of dv_x/dt. This gives:
\gamma^2 \frac{d}{dt}( v_x^2 + v_y^2 ) - (2E/B) \frac{d v_x}{dt} = 0 (*)
Which is the original expression I gave you. Using the chain rule on the first term, you can express (*) as a total t-derivative, integrating up then gives you v_y in terms of v_x. The best thing to do, if you don't follow the steps above, is to use your given equations to show that the LHS of really does vanish.
QUOTE (Euler+Mar 24 2008, 02:11 PM)
But your first equation gives:
qEv_y = frac{E\gamma m_0}{B} \frac{dv_x}{dt}
...
No, my first equation (see the uploaded file) is:
\gamma*m0*dv_x/dt=qBv_y
so, after multiplication with v_x, I got:
\gamma*m0*v_xdv_x/dt=qBv_x*v_y
The second equation is:
\gamma*m0*dv_y/dt-qE*\gamma^-2=-qBv_x
so, after multiplication with v_y:
\gamma*m0*v_y*dv_y/dt-qE*v_y\gamma^-2=-qBv_x*v_y
Add them together and you get:
\gamma*m0*(v_xdv_x/dt+v_y*dv_y/dt)=qE*v_y\gamma^-2
or:
\gamma^3*m0*(v_xdv_x/dt+v_y*dv_y/dt)=qE*v_y
The induction B disappears completely.
qEv_y = frac{E\gamma m_0}{B} \frac{dv_x}{dt}
...
No, my first equation (see the uploaded file) is:
\gamma*m0*dv_x/dt=qBv_y
so, after multiplication with v_x, I got:
\gamma*m0*v_xdv_x/dt=qBv_x*v_y
The second equation is:
\gamma*m0*dv_y/dt-qE*\gamma^-2=-qBv_x
so, after multiplication with v_y:
\gamma*m0*v_y*dv_y/dt-qE*v_y\gamma^-2=-qBv_x*v_y
Add them together and you get:
\gamma*m0*(v_xdv_x/dt+v_y*dv_y/dt)=qE*v_y\gamma^-2
or:
\gamma^3*m0*(v_xdv_x/dt+v_y*dv_y/dt)=qE*v_y
The induction B disappears completely.
QUOTE (Trout+Mar 24 2008, 02:26 PM)
(1) \gamma*m0*dv_x/dt=qBv_y
(2) \gamma^3*m0*(v_xdv_x/dt+v_y*dv_y/dt)=qE*v_y
Use (1) to express v_y in terms of dv_x/dt, then put this into (2). Use also the fact that vdv/dt = (1/2) d(v^2)/dt.
As I said, if you don't follow the steps the best thing to do is show that the LHS of my expression in (*) vanishes if the equations you've given are satisfied.
(2) \gamma^3*m0*(v_xdv_x/dt+v_y*dv_y/dt)=qE*v_y
Use (1) to express v_y in terms of dv_x/dt, then put this into (2). Use also the fact that vdv/dt = (1/2) d(v^2)/dt.
As I said, if you don't follow the steps the best thing to do is show that the LHS of my expression in (*) vanishes if the equations you've given are satisfied.
QUOTE (Euler+Mar 24 2008, 02:23 PM)
I'll go through it in more detail: multiplying your first equation by v_x and your second by v_y gives:
\frac{\gamma^3 m_0}{2} \frac{d}{dt} (v_x^2+v_y^2) - qE v_y = 0
The best thing to do, if you don't follow the steps above, is to use your given equations to show that the LHS of really does vanish.
Yes, this is what I got.
Now use your first equation, to express v_y in terms of dv_x/dt. This gives:
\gamma^2 \frac{d}{dt}( v_x^2 + v_y^2 ) - (2E/B) \frac{d v_x}{dt} = 0 (*)
Which is the original expression I gave you.
Yes, I agree.
Yes, I agree.
Using the chain rule on the first term, you can express (*) as a total t-derivative, integrating up then gives you v_y in terms of v_x.
OK, this is now the part where I get stuck:
X=v_x^2+v_y^2
so (*) becomes
1/(1-X)*dX/dt= (2E/B)*dv_x/dt
BUT! X is a function of v_x, so I cannot really integrate, can I?
Besides, if you look at the upload, I already got v_y as a function of v_x and dv_x/dt.
The best thing to do, if you don't follow the steps above, is to use your given equations to show that the LHS of really does vanish.
I don't get this part.
Look where I got so far, please feel free to download and edit the file.If you could return it to me, this would be much easier than strugglung with posting here. Thank you very much for your help.
\frac{\gamma^3 m_0}{2} \frac{d}{dt} (v_x^2+v_y^2) - qE v_y = 0
The best thing to do, if you don't follow the steps above, is to use your given equations to show that the LHS of really does vanish.
Yes, this is what I got.
QUOTE
Now use your first equation, to express v_y in terms of dv_x/dt. This gives:
\gamma^2 \frac{d}{dt}( v_x^2 + v_y^2 ) - (2E/B) \frac{d v_x}{dt} = 0 (*)
Which is the original expression I gave you.
Yes, I agree.
QUOTE (->
| QUOTE |
Now use your first equation, to express v_y in terms of dv_x/dt. This gives: \gamma^2 \frac{d}{dt}( v_x^2 + v_y^2 ) - (2E/B) \frac{d v_x}{dt} = 0 (*) Which is the original expression I gave you. |
Yes, I agree.
Using the chain rule on the first term, you can express (*) as a total t-derivative, integrating up then gives you v_y in terms of v_x.
OK, this is now the part where I get stuck:
X=v_x^2+v_y^2
so (*) becomes
1/(1-X)*dX/dt= (2E/B)*dv_x/dt
BUT! X is a function of v_x, so I cannot really integrate, can I?
Besides, if you look at the upload, I already got v_y as a function of v_x and dv_x/dt.
QUOTE
The best thing to do, if you don't follow the steps above, is to use your given equations to show that the LHS of really does vanish.
I don't get this part.
Look where I got so far, please feel free to download and edit the file.If you could return it to me, this would be much easier than strugglung with posting here. Thank you very much for your help.
QUOTE (Trout+Mar 24 2008, 02:36 PM)
1/(1-X)*dX/dt= (2E/B)*dv_x/dt
1/(1-X) dX/dt = -d/dt ln(1-X)
1/(1-X) dX/dt = -d/dt ln(1-X)
QUOTE (Euler+Mar 24 2008, 02:39 PM)
1/(1-X) dX/dt = -d/dt ln(1-X)
Yes, I understand but X is a function of v_x, so can I really integrate by variable separation?
Look here where I got so far , I already got v_y as a function of v_x and dv_x/dt.
Yes, I understand but X is a function of v_x, so can I really integrate by variable separation?
Look here where I got so far , I already got v_y as a function of v_x and dv_x/dt.
QUOTE (Trout+Mar 24 2008, 02:46 PM)
Yes, I understand but X is a function of v_x, so can I really integrate by variable separation?
Using what I've just given, you essentially have:
d/dt f(t) = 0
Which you can integrate up with respect to t to get f(t)=const. In this case, we have;
f(t) = -c^2 ln(1-X/c^2) - (2E/B)v_x
where X = v_x^2 + v_y^2. Aside from all this, if you're just looking for any solution, just look for an easy one - i.e static (so v_x, v_y are constant). This will give you easy equations for solve for v_x and v_y.
Using what I've just given, you essentially have:
d/dt f(t) = 0
Which you can integrate up with respect to t to get f(t)=const. In this case, we have;
f(t) = -c^2 ln(1-X/c^2) - (2E/B)v_x
where X = v_x^2 + v_y^2. Aside from all this, if you're just looking for any solution, just look for an easy one - i.e static (so v_x, v_y are constant). This will give you easy equations for solve for v_x and v_y.
QUOTE (Euler+Mar 24 2008, 02:52 PM)
Using what I've just given, you essentially have:
d/dt f(t) = 0
Which you can integrate up with respect to t to get f(t)=const. In this case, we have;
f(t) = -c^2 ln(1-X/c^2) - (2E/B)v_x
where X = v_x^2 + v_y^2.
Yes, I see. Thank you.
Can't do that. The motion is definitely accelerated so v_x and v_y are definitely not constant.
Can you look at what I did so far?
d/dt f(t) = 0
Which you can integrate up with respect to t to get f(t)=const. In this case, we have;
f(t) = -c^2 ln(1-X/c^2) - (2E/B)v_x
where X = v_x^2 + v_y^2.
Yes, I see. Thank you.
QUOTE
Aside from all this, if you're just looking for any solution, just look for an easy one - i.e static (so v_x, v_y are constant). This will give you easy equations for solve for v_x and v_y.
Can't do that. The motion is definitely accelerated so v_x and v_y are definitely not constant.
Can you look at what I did so far?
QUOTE (Trout+Mar 24 2008, 02:57 PM)
Yes, I see. Thank you.
No problem.
No problem.
QUOTE (Trout+Mar 24 2008, 02:57 PM)
Can't do that. The motion is definitely accelerated so v_x and v_y are definitely not constant. Can you look at what I did so far?
Well, now you can use your new expression for v_y in terms of v_x in your first equation. This gives you a separable ODE in (v_x, t), so in principle you are done. Whether or not the resulting integral has a closed form expression in terms of elementary functions is a question for differential Galois theory, and I imagine pretty far from what you want!
The main point is that our work so far give us v_y = F(v_x). Using this in your first equation gives:
dv_x/dt = G(v_x),
i.e a seperable ODE in (v_x, t) (for appropriate functions F,G). From my POV the problem solved! From your POV, you need to worry about the resulting integral. I'd recommend pouring a big mug of coffee and finding the G(v_x) explicitly - you can proceed from there.
Well, now you can use your new expression for v_y in terms of v_x in your first equation. This gives you a separable ODE in (v_x, t), so in principle you are done. Whether or not the resulting integral has a closed form expression in terms of elementary functions is a question for differential Galois theory, and I imagine pretty far from what you want!
The main point is that our work so far give us v_y = F(v_x). Using this in your first equation gives:
dv_x/dt = G(v_x),
i.e a seperable ODE in (v_x, t) (for appropriate functions F,G). From my POV the problem solved! From your POV, you need to worry about the resulting integral. I'd recommend pouring a big mug of coffee and finding the G(v_x) explicitly - you can proceed from there.
QUOTE (Euler+Mar 24 2008, 03:14 PM)
No problem.
Well, now you can use your new expression for v_y in terms of v_x in your first equation. This gives you a separable ODE in (v_x, t), so in principle you are done. Whether or not the resulting integral has a closed form expression in terms of elementary functions is a question for differential Galois theory, and I imagine pretty far from what you want!
The main point is that our work so far give us v_y = F(v_x). Using this in your first equation gives:
dv_x/dt = G(v_x),
i.e a seperable ODE in (v_x, t) (for appropriate functions F,G). From my POV the problem solved! From your POV, you need to worry about the resulting integral. I'd recommend pouring a big mug of coffee and finding the G(v_x) explicitly - you can proceed from there.
Thank you for all your help, this is what I got, see here. Nasty.
Well, now you can use your new expression for v_y in terms of v_x in your first equation. This gives you a separable ODE in (v_x, t), so in principle you are done. Whether or not the resulting integral has a closed form expression in terms of elementary functions is a question for differential Galois theory, and I imagine pretty far from what you want!
The main point is that our work so far give us v_y = F(v_x). Using this in your first equation gives:
dv_x/dt = G(v_x),
i.e a seperable ODE in (v_x, t) (for appropriate functions F,G). From my POV the problem solved! From your POV, you need to worry about the resulting integral. I'd recommend pouring a big mug of coffee and finding the G(v_x) explicitly - you can proceed from there.
Thank you for all your help, this is what I got, see here. Nasty.
QUOTE (Trout+Mar 24 2008, 03:57 PM)
Thank you for all your help, this is what I got, see here. Nasty.
I'd agree with your conclusion: I doubt that integral is expressible in terms of elementary functions.
I'd agree with your conclusion: I doubt that integral is expressible in terms of elementary functions.
Hi Trout and Euler,
It seems that in the time I took to get around to posting, you two have worked out all the difficulties! I still would like to mention two things:
First, Trout, you are missing a factor of 2 in the help_final file, just after the point where you define the variable X. It comes form differentiating the squares of the velocities. This does not affect the character of the final separated equation, which I agree looks like it's not going to have a closed form expression in terms of elementary functions. I would suggest a flow line plot on a v_x v_y plane as a good way of visualizing the dynamics.
Second, I have some doubts about the initial equations. Certainly this strongly resembles the physical situation of electromagnetic forces acting on a particle in motion. Specifically, it looks like the synchrotron equation. What I am seeing here is the vector equation F = ma, where there are constant uniform E and B fields in the y and z directions respectively. A particle with rest mass m0 is traveling along a trajectory which would be a circle if there were no E, but which instead is pushed progressively further in the y direction, so that the particle moves through a series of loops like a person doodling spirals with a pen. The particle is moving fast enough that its mass must be treated as gamma*m0. So far so good. But I don't think that the factor of gamma^-2 on the E term is appropriate. In a rest frame in which E and B have their constant uniform values, the classical laws of electromagnetic interaction should apply, with the only relativistic effect coming from the mass increase. Therefore, it seems to me that you should have just a plain E, not E*gamma^-2. This of course drastically affects the solvability of the equations!
If you are SURE that this term is correct, then you have the solution you want via Euler's hints, although it is not mathematically in a "nice" form. On the other hand, if the gamma^-2 factor does not belong there, then the equations beome much easier. The 1/(1-X) factor will disappear, the time integral of both sides will now give a linear, not a logarithmic, relation between X and vx, and the final integration May be closed form after all.
Hope that helps!
--Stuart Anderson
It seems that in the time I took to get around to posting, you two have worked out all the difficulties! I still would like to mention two things:
First, Trout, you are missing a factor of 2 in the help_final file, just after the point where you define the variable X. It comes form differentiating the squares of the velocities. This does not affect the character of the final separated equation, which I agree looks like it's not going to have a closed form expression in terms of elementary functions. I would suggest a flow line plot on a v_x v_y plane as a good way of visualizing the dynamics.
Second, I have some doubts about the initial equations. Certainly this strongly resembles the physical situation of electromagnetic forces acting on a particle in motion. Specifically, it looks like the synchrotron equation. What I am seeing here is the vector equation F = ma, where there are constant uniform E and B fields in the y and z directions respectively. A particle with rest mass m0 is traveling along a trajectory which would be a circle if there were no E, but which instead is pushed progressively further in the y direction, so that the particle moves through a series of loops like a person doodling spirals with a pen. The particle is moving fast enough that its mass must be treated as gamma*m0. So far so good. But I don't think that the factor of gamma^-2 on the E term is appropriate. In a rest frame in which E and B have their constant uniform values, the classical laws of electromagnetic interaction should apply, with the only relativistic effect coming from the mass increase. Therefore, it seems to me that you should have just a plain E, not E*gamma^-2. This of course drastically affects the solvability of the equations!
If you are SURE that this term is correct, then you have the solution you want via Euler's hints, although it is not mathematically in a "nice" form. On the other hand, if the gamma^-2 factor does not belong there, then the equations beome much easier. The 1/(1-X) factor will disappear, the time integral of both sides will now give a linear, not a logarithmic, relation between X and vx, and the final integration May be closed form after all.
Hope that helps!
--Stuart Anderson
I was never able to judge if gamma or if "y" was meant in the E term.
QUOTE (mr_homm+Mar 24 2008, 07:10 PM)
It seems that in the time I took to get around to posting, you two have worked out all the difficulties!
What difficulties!?! The problem was solved in my first post! I might have been a little brief with the details I guess...
What difficulties!?! The problem was solved in my first post! I might have been a little brief with the details I guess...
QUOTE (mr_homm+Mar 24 2008, 07:10 PM)
Second, I have some doubts about the initial equations. Certainly this strongly resembles the physical situation of electromagnetic forces acting on a particle in motion.
As a (trainee) mathematician, I am not required to question the physical relevance of a set of equations! It would be nice to find out if the associated physical problem is such that the equations admit a tractable solution.
As a (trainee) mathematician, I am not required to question the physical relevance of a set of equations! It would be nice to find out if the associated physical problem is such that the equations admit a tractable solution.
I just want to say I read the new help_doc file, and it rubbed my ego something fierce.
That's not to say I can work through it (by any means) but that I actually understood what all those math symbols meant. Hehe, I'm one step closer to my goal of true, unadulterated scientific nerdiness.
(Now that I've learned some of the alphabet, I need to start learning to spell...)
That's not to say I can work through it (by any means) but that I actually understood what all those math symbols meant. Hehe, I'm one step closer to my goal of true, unadulterated scientific nerdiness.
(Now that I've learned some of the alphabet, I need to start learning to spell...)
QUOTE (mr_homm+Mar 24 2008, 07:10 PM)
Hi Trout and Euler,
It seems that in the time I took to get around to posting, you two have worked out all the difficulties! I still would like to mention two things:
First, Trout, you are missing a factor of 2 in the help_final file, just after the point where you define the variable X. It comes form differentiating the squares of the velocities. This does not affect the character of the final separated equation, which I agree looks like it's not going to have a closed form expression in terms of elementary functions.
Yes, I know.
Correct again.
Correct again.
Specifically, it looks like the synchrotron equation. What I am seeing here is the vector equation F = ma, where there are constant uniform E and B fields in the y and z directions respectively. A particle with rest mass m0 is traveling along a trajectory which would be a circle if there were no E, but which instead is pushed progressively further in the y direction, so that the particle moves through a series of loops like a person doodling spirals with a pen. The particle is moving fast enough that its mass must be treated as gamma*m0. So far so good. But I don't think that the factor of gamma^-2 on the E term is appropriate. In a rest frame in which E and B have their constant uniform values, the classical laws of electromagnetic interaction should apply, with the only relativistic effect coming from the mass increase. Therefore, it seems to me that you should have just a plain E, not E*gamma^-2. This of course drastically affects the solvability of the equations!
I can vouch for the fact that the equations are correct. I can send you the derivation that arrived to the equations in the uploaded file.
If you are SURE that this term is correct, then you have the solution you want via Euler's hints, although it is not mathematically in a "nice" form.
Yes, we all seem to agree on this.
Yes, we all seem to agree on this.
On the other hand, if the gamma^-2 factor does not belong there, then the equations beome much easier. The 1/(1-X) factor will disappear, the time integral of both sides will now give a linear, not a logarithmic, relation between X and vx, and the final integration May be closed form after all.
Hope that helps!
--Stuart Anderson
The "gamma" factor needs to be there , unfortunately.
Yeah, that's too bad Trout. They should make it easier for you.
Or, maybe you should switch majors!
Welcome to therapy, punk.
Yeah, that's too bad Trout. They should make it easier for you.
So, you have two choices as hobbies: stalker or clown.
d2v/dt2+2v/c^2(1+(v/c)^2)(dv/dt)^2+g/L*(1-(v/c)^2)*v=0 (6)
I need to check if the above has a closed solution. We know that v varies from 0 (the highest point of the pendulum) to v_max (at the lowest point) , so we should be looking for a periodic function as a solution.
Continued:
turns out that I can reduce eq.6 to a first order ODE by using the substitution w(v)=(dv/dt)^2
Then d2v/dt2=d(sqrt(w)/dt=.5/sqrt(w)*dw/dt=.5/sqrt(w)*dw/dv*dv/dt=.5*dw/dv
So, the equation becomes:
dw/dv+2f(v)*w+2h(v)=0 (7)
where f(v)=2v/c^2*gamma^2 and h(v)=g/L*gamma^-2
We don't even need to use the approximation for gamma^2 anymore, we can use the exact formula.
Now, (7) is a first order ODE. We are in good shape. More tomorrow.
It seems that in the time I took to get around to posting, you two have worked out all the difficulties! I still would like to mention two things:
First, Trout, you are missing a factor of 2 in the help_final file, just after the point where you define the variable X. It comes form differentiating the squares of the velocities. This does not affect the character of the final separated equation, which I agree looks like it's not going to have a closed form expression in terms of elementary functions.
Yes, I know.
QUOTE
Second, I have some doubts about the initial equations. Certainly this strongly resembles the physical situation of electromagnetic forces acting on a particle in motion.
Correct again.
QUOTE (->
| QUOTE |
| Second, I have some doubts about the initial equations. Certainly this strongly resembles the physical situation of electromagnetic forces acting on a particle in motion. |
Correct again.
Specifically, it looks like the synchrotron equation. What I am seeing here is the vector equation F = ma, where there are constant uniform E and B fields in the y and z directions respectively. A particle with rest mass m0 is traveling along a trajectory which would be a circle if there were no E, but which instead is pushed progressively further in the y direction, so that the particle moves through a series of loops like a person doodling spirals with a pen. The particle is moving fast enough that its mass must be treated as gamma*m0. So far so good. But I don't think that the factor of gamma^-2 on the E term is appropriate. In a rest frame in which E and B have their constant uniform values, the classical laws of electromagnetic interaction should apply, with the only relativistic effect coming from the mass increase. Therefore, it seems to me that you should have just a plain E, not E*gamma^-2. This of course drastically affects the solvability of the equations!
I can vouch for the fact that the equations are correct. I can send you the derivation that arrived to the equations in the uploaded file.
QUOTE
If you are SURE that this term is correct, then you have the solution you want via Euler's hints, although it is not mathematically in a "nice" form.
Yes, we all seem to agree on this.
QUOTE (->
| QUOTE |
If you are SURE that this term is correct, then you have the solution you want via Euler's hints, although it is not mathematically in a "nice" form. |
Yes, we all seem to agree on this.
On the other hand, if the gamma^-2 factor does not belong there, then the equations beome much easier. The 1/(1-X) factor will disappear, the time integral of both sides will now give a linear, not a logarithmic, relation between X and vx, and the final integration May be closed form after all.
Hope that helps!
--Stuart Anderson
The "gamma" factor needs to be there , unfortunately.
@Euler
I said there were difficulties, I didn't say they were yours
Reminds me of a friend I had in school. I was doing a crossword with a pencil when this other gent came by and said, rather haughtily, "Oh, I never use pencil. I do crosswords in ink." My friend said witheringly, "you mean you need to write them down?"
As to the origin of the differential equations, of course from a purely mathematical standpoint the equations themselves are the starting point. However, it is sometimes useful (even to a mathematician) to have an idea of what the equations are modeling, because insights derived from that perspective may aid in solving the equations. For instance, known conservation laws often simply hand you the first integral. Having got that hint, you are of course free to be as rigorous as you wish in pursuing the solution. But of course I fully agree that the physical interpretation of the equation is not your department.
@Trout
Yes, please,I would like to see the derivation, just because I'm now curious as to how the gamma^-2 factor arises. I can't see it intuitively. Thanks.
--Stuart Anderson
I said there were difficulties, I didn't say they were yours
Reminds me of a friend I had in school. I was doing a crossword with a pencil when this other gent came by and said, rather haughtily, "Oh, I never use pencil. I do crosswords in ink." My friend said witheringly, "you mean you need to write them down?"
As to the origin of the differential equations, of course from a purely mathematical standpoint the equations themselves are the starting point. However, it is sometimes useful (even to a mathematician) to have an idea of what the equations are modeling, because insights derived from that perspective may aid in solving the equations. For instance, known conservation laws often simply hand you the first integral. Having got that hint, you are of course free to be as rigorous as you wish in pursuing the solution. But of course I fully agree that the physical interpretation of the equation is not your department.
@Trout
Yes, please,I would like to see the derivation, just because I'm now curious as to how the gamma^-2 factor arises. I can't see it intuitively. Thanks.
--Stuart Anderson
My post, was of course, tongue in cheek! I'd also like to express my interest in the derivation - assuming my rusty physics brain can follow! 
QUOTE (mr_homm+Mar 24 2008, 09:30 PM)
@Euler
I said there were difficulties, I didn't say they were yours
Reminds me of a friend I had in school. I was doing a crossword with a pencil when this other gent came by and said, rather haughtily, "Oh, I never use pencil. I do crosswords in ink." My friend said witheringly, "you mean you need to write them down?"
As to the origin of the differential equations, of course from a purely mathematical standpoint the equations themselves are the starting point. However, it is sometimes useful (even to a mathematician) to have an idea of what the equations are modeling, because insights derived from that perspective may aid in solving the equations. For instance, known conservation laws often simply hand you the first integral. Having got that hint, you are of course free to be as rigorous as you wish in pursuing the solution. But of course I fully agree that the physical interpretation of the equation is not your department.
@Trout
Yes, please,I would like to see the derivation, just because I'm now curious as to how the gamma^-2 factor arises. I can't see it intuitively. Thanks.
--Stuart Anderson
If the initial equations are right (I think they are) then we have a very ugly problem on our hands:
1. I verified that the integral representing v_x doesn't have a closed form
2. This means that we can't get a closed form for v_y either
3. The above mean that we can't integrate the equations of motion x=x(t) and y=y(t)
I said there were difficulties, I didn't say they were yours
Reminds me of a friend I had in school. I was doing a crossword with a pencil when this other gent came by and said, rather haughtily, "Oh, I never use pencil. I do crosswords in ink." My friend said witheringly, "you mean you need to write them down?"
As to the origin of the differential equations, of course from a purely mathematical standpoint the equations themselves are the starting point. However, it is sometimes useful (even to a mathematician) to have an idea of what the equations are modeling, because insights derived from that perspective may aid in solving the equations. For instance, known conservation laws often simply hand you the first integral. Having got that hint, you are of course free to be as rigorous as you wish in pursuing the solution. But of course I fully agree that the physical interpretation of the equation is not your department.
@Trout
Yes, please,I would like to see the derivation, just because I'm now curious as to how the gamma^-2 factor arises. I can't see it intuitively. Thanks.
--Stuart Anderson
If the initial equations are right (I think they are) then we have a very ugly problem on our hands:
1. I verified that the integral representing v_x doesn't have a closed form
2. This means that we can't get a closed form for v_y either
3. The above mean that we can't integrate the equations of motion x=x(t) and y=y(t)
QUOTE (Trout+Mar 25 2008, 01:39 AM)
3. The above mean that we can't integrate the equations of motion x=x(t) and y=y(t)
Except numerically -- which for applications like engineering is good enough.
Except numerically -- which for applications like engineering is good enough.
QUOTE (rpenner+Mar 25 2008, 06:12 AM)
Except numerically -- which for applications like engineering is good enough.
Yes, but this is disappointing. It appeared to be such a nice problem to solve. It turned ugly rather quickly. Two different attacks and they both fizzled.
Actually, since we cannot get closed expressions for the speeds, the equation of motion would be a pain even in the case of using numerical integration. There must be a way to solve it, between you , Euler and mr_homm. If necessary, bring others in, we need more mathematicians. Where the heck is AlphaNumeric when we need him?
On the positive side, we are having a sane thread , with no crackpots' "participation" and "contributions". This is a "first", makes the place look like a real science forum.
Yes, but this is disappointing. It appeared to be such a nice problem to solve. It turned ugly rather quickly. Two different attacks and they both fizzled.
Actually, since we cannot get closed expressions for the speeds, the equation of motion would be a pain even in the case of using numerical integration. There must be a way to solve it, between you , Euler and mr_homm. If necessary, bring others in, we need more mathematicians. Where the heck is AlphaNumeric when we need him?
On the positive side, we are having a sane thread , with no crackpots' "participation" and "contributions". This is a "first", makes the place look like a real science forum.
QUOTE (Trout+Mar 25 2008, 06:19 AM)
Yes, but this is disappointing. It appeared to be such a nice problem to solve. It turned ugly rather quickly. Two different attacks and they both fizzled.
Actually, since we cannot get closed expressions for the speeds, the equation of motion would be a pain even in the case of using numerical integration. There must be a way to solve it, between you , Euler and mr_homm. If necessary, bring others in, we need more mathematicians. Where the heck is AlphaNumeric when we need him?
On the positive side, we are having a sane thread , with no crackpots' "participation" and "contributions". This is a "first", makes the place look like a real science forum.
OK,
It is solved, the problem was much tougher than anticipated. See the solution here. mr_homm gave me a great idea on the attack.
Actually, since we cannot get closed expressions for the speeds, the equation of motion would be a pain even in the case of using numerical integration. There must be a way to solve it, between you , Euler and mr_homm. If necessary, bring others in, we need more mathematicians. Where the heck is AlphaNumeric when we need him?
On the positive side, we are having a sane thread , with no crackpots' "participation" and "contributions". This is a "first", makes the place look like a real science forum.
OK,
It is solved, the problem was much tougher than anticipated. See the solution here. mr_homm gave me a great idea on the attack.
On a different forum a crank managed to stump the moderators pretty good with the following problem (closely related to the one that we've solved in this thread):
This one is a toughie. Some of the moderators wrote some prose but none of them rolled up their sleeves to put down any equations to back up their point. And yes, SR works with accelerated motion just fine.
QUOTE
A pendulum of mass m and length L is executing oscillations with a small angle "theta" in a frame attached to the Earth (gravitational acceleration can be considered constant and equal to g). Find out the oscillation period T as measured from a frame moving at a constant speed v , where v is a significant fraction of c (you cannot ignore relativistic effects). The frame is moving in a direction perpendicular to the plane of oscillation, such that there is no length contraction for L
This one is a toughie. Some of the moderators wrote some prose but none of them rolled up their sleeves to put down any equations to back up their point. And yes, SR works with accelerated motion just fine.
Hi Trout,
This one doesn't look too bad to me, unless I'm missing something. Let the pendulum swing through the origin of coordinates, so that the lowest point on its swing passes directly through the origin. Since the pendulum executes periodic motion, it is only necessary to consider time at which it passes through the origin on each swing. This will define two events with coordinates (x,y,z,ct) = (0,0,0,0) and (0,0,0,cT), corresponding to passages through the origin one period apart. Even though the pendulum is moving and accelerating, these two events contain all the information necessary to define the period of the pendulum in the earth frame. For the assumptions in the problem, T = sqrt(l/g).
Now in another frame, it is only necessary to transform the spacetime interval between these two events in order to find the new period. Let the second frame have velocity v relative to the earth, so that gamma = 1/sqrt(1-(v/c)^2). Then by the Lorentz transformation, (0,0,0,cT) transforms to gamma*(-vT,0,0,cT), so that in the new frame, the period is simply gamma*T, during which time the pendulum and the earth frame move a distance -gamma*vT relative to the second frame. That's all there is to the calculation.
The pendulum therefore appears to swing slower by a factor of gamma as seen from the second frame.
I think that's all there is to the problem. Am I missing something?
--Stuart Anderson
This one doesn't look too bad to me, unless I'm missing something. Let the pendulum swing through the origin of coordinates, so that the lowest point on its swing passes directly through the origin. Since the pendulum executes periodic motion, it is only necessary to consider time at which it passes through the origin on each swing. This will define two events with coordinates (x,y,z,ct) = (0,0,0,0) and (0,0,0,cT), corresponding to passages through the origin one period apart. Even though the pendulum is moving and accelerating, these two events contain all the information necessary to define the period of the pendulum in the earth frame. For the assumptions in the problem, T = sqrt(l/g).
Now in another frame, it is only necessary to transform the spacetime interval between these two events in order to find the new period. Let the second frame have velocity v relative to the earth, so that gamma = 1/sqrt(1-(v/c)^2). Then by the Lorentz transformation, (0,0,0,cT) transforms to gamma*(-vT,0,0,cT), so that in the new frame, the period is simply gamma*T, during which time the pendulum and the earth frame move a distance -gamma*vT relative to the second frame. That's all there is to the calculation.
The pendulum therefore appears to swing slower by a factor of gamma as seen from the second frame.
I think that's all there is to the problem. Am I missing something?
--Stuart Anderson
QUOTE (mr_homm+Jul 1 2008, 12:27 AM)
Hi Trout,
This one doesn't look too bad to me, unless I'm missing something. Let the pendulum swing through the origin of coordinates, so that the lowest point on its swing passes directly through the origin. Since the pendulum executes periodic motion, it is only necessary to consider time at which it passes through the origin on each swing. This will define two events with coordinates (x,y,z,ct) = (0,0,0,0) and (0,0,0,cT), corresponding to passages through the origin one period apart. Even though the pendulum is moving and accelerating, these two events contain all the information necessary to define the period of the pendulum in the earth frame. For the assumptions in the problem, T = sqrt(l/g).
Now in another frame, it is only necessary to transform the spacetime interval between these two events in order to find the new period. Let the second frame have velocity v relative to the earth, so that gamma = 1/sqrt(1-(v/c)^2). Then by the Lorentz transformation, (0,0,0,cT) transforms to gamma*(-vT,0,0,cT), so that in the new frame, the period is simply gamma*T, during which time the pendulum and the earth frame move a distance -gamma*vT relative to the second frame. That's all there is to the calculation.
The pendulum therefore appears to swing slower by a factor of gamma as seen from the second frame.
I think that's all there is to the problem. Am I missing something?
--Stuart Anderson
Yes, you seem to be missing the fact that the period of the pendulum (T=sqrt(L/g)) has been derived not only by using Newtonian mechanics but also for the case of observer at rest wrt pendulum.
So, the equation used was :
m*d2x/dt2=-mg sin(theta) (1)
The same equation, for SR is quite different since F is no longer ma but dp/dt=d(gamma*m_0*v)/dt
So, in SR, eq(1) becomes a much nastier "animal":
d2(gamma*v)/dt2=-gama*g*sin(theta) (2)
In both cases (1) and (2), for small theta you can approximate sin(theta)=theta=x/L
so (1) simplifies to:
d2x/dt2=-gx/L (1')
which gives the well known T=sqrt(L/g)
Unfortunately, (2) doesn't get any simpler:
d2(gamma*v)/dt2=-gama*g*x/L (2')
...a very nasty one because gamma=1/sqrt(1-(v/c)^2)....
This one doesn't look too bad to me, unless I'm missing something. Let the pendulum swing through the origin of coordinates, so that the lowest point on its swing passes directly through the origin. Since the pendulum executes periodic motion, it is only necessary to consider time at which it passes through the origin on each swing. This will define two events with coordinates (x,y,z,ct) = (0,0,0,0) and (0,0,0,cT), corresponding to passages through the origin one period apart. Even though the pendulum is moving and accelerating, these two events contain all the information necessary to define the period of the pendulum in the earth frame. For the assumptions in the problem, T = sqrt(l/g).
Now in another frame, it is only necessary to transform the spacetime interval between these two events in order to find the new period. Let the second frame have velocity v relative to the earth, so that gamma = 1/sqrt(1-(v/c)^2). Then by the Lorentz transformation, (0,0,0,cT) transforms to gamma*(-vT,0,0,cT), so that in the new frame, the period is simply gamma*T, during which time the pendulum and the earth frame move a distance -gamma*vT relative to the second frame. That's all there is to the calculation.
The pendulum therefore appears to swing slower by a factor of gamma as seen from the second frame.
I think that's all there is to the problem. Am I missing something?
--Stuart Anderson
Yes, you seem to be missing the fact that the period of the pendulum (T=sqrt(L/g)) has been derived not only by using Newtonian mechanics but also for the case of observer at rest wrt pendulum.
So, the equation used was :
m*d2x/dt2=-mg sin(theta) (1)
The same equation, for SR is quite different since F is no longer ma but dp/dt=d(gamma*m_0*v)/dt
So, in SR, eq(1) becomes a much nastier "animal":
d2(gamma*v)/dt2=-gama*g*sin(theta) (2)
In both cases (1) and (2), for small theta you can approximate sin(theta)=theta=x/L
so (1) simplifies to:
d2x/dt2=-gx/L (1')
which gives the well known T=sqrt(L/g)
Unfortunately, (2) doesn't get any simpler:
d2(gamma*v)/dt2=-gama*g*x/L (2')
...a very nasty one because gamma=1/sqrt(1-(v/c)^2)....
QUOTE
Unfortunately, (2) doesn't get any simpler:
Yeah, that's too bad Trout. They should make it easier for you.
Or, maybe you should switch majors!
Welcome to therapy, punk.
QUOTE (inQZtive+Jul 1 2008, 03:57 AM)
Yeah, that's too bad Trout. They should make it easier for you.
So, you have two choices as hobbies: stalker or clown.
How do I know you're not lying again?
QUOTE (mr_homm+Jul 1 2008, 12:27 AM)
Hi Trout,
This one doesn't look too bad to me, unless I'm missing something. Let the pendulum swing through the origin of coordinates, so that the lowest point on its swing passes directly through the origin. Since the pendulum executes periodic motion, it is only necessary to consider time at which it passes through the origin on each swing. This will define two events with coordinates (x,y,z,ct) = (0,0,0,0) and (0,0,0,cT), corresponding to passages through the origin one period apart. Even though the pendulum is moving and accelerating, these two events contain all the information necessary to define the period of the pendulum in the earth frame. For the assumptions in the problem, T = sqrt(l/g).
Now in another frame, it is only necessary to transform the spacetime interval between these two events in order to find the new period. Let the second frame have velocity v relative to the earth, so that gamma = 1/sqrt(1-(v/c)^2). Then by the Lorentz transformation, (0,0,0,cT) transforms to gamma*(-vT,0,0,cT), so that in the new frame, the period is simply gamma*T, during which time the pendulum and the earth frame move a distance -gamma*vT relative to the second frame. That's all there is to the calculation.
The pendulum therefore appears to swing slower by a factor of gamma as seen from the second frame.
I think that's all there is to the problem. Am I missing something?
--Stuart Anderson
The problem with the above is that it leads to a paradox:
1. On one hand, you are absolutely correct, T'=gamma*T , where gamma=1/sqrt(1-(V/c)^2) and V is the relative speed between frames.
2. On the other hand, if we use the Newtonian expression for the period (T=sqrt(L/g)), then it would APPEAR that T'=T. (turns out that this is not correct, since g does not transform into g, accelerations transform differently in SR)
This is how the crank managed to (almost) stump the moderators on the other forum. As we've seen, T is NOT equal to sqrt(L/g) in SR, so we need to be able to solve the nasty differential equation that describes the pendulum motion in SR. I predict that the period will depend on something like a higher power of g.
This one doesn't look too bad to me, unless I'm missing something. Let the pendulum swing through the origin of coordinates, so that the lowest point on its swing passes directly through the origin. Since the pendulum executes periodic motion, it is only necessary to consider time at which it passes through the origin on each swing. This will define two events with coordinates (x,y,z,ct) = (0,0,0,0) and (0,0,0,cT), corresponding to passages through the origin one period apart. Even though the pendulum is moving and accelerating, these two events contain all the information necessary to define the period of the pendulum in the earth frame. For the assumptions in the problem, T = sqrt(l/g).
Now in another frame, it is only necessary to transform the spacetime interval between these two events in order to find the new period. Let the second frame have velocity v relative to the earth, so that gamma = 1/sqrt(1-(v/c)^2). Then by the Lorentz transformation, (0,0,0,cT) transforms to gamma*(-vT,0,0,cT), so that in the new frame, the period is simply gamma*T, during which time the pendulum and the earth frame move a distance -gamma*vT relative to the second frame. That's all there is to the calculation.
The pendulum therefore appears to swing slower by a factor of gamma as seen from the second frame.
I think that's all there is to the problem. Am I missing something?
--Stuart Anderson
The problem with the above is that it leads to a paradox:
1. On one hand, you are absolutely correct, T'=gamma*T , where gamma=1/sqrt(1-(V/c)^2) and V is the relative speed between frames.
2. On the other hand, if we use the Newtonian expression for the period (T=sqrt(L/g)), then it would APPEAR that T'=T. (turns out that this is not correct, since g does not transform into g, accelerations transform differently in SR)
This is how the crank managed to (almost) stump the moderators on the other forum. As we've seen, T is NOT equal to sqrt(L/g) in SR, so we need to be able to solve the nasty differential equation that describes the pendulum motion in SR. I predict that the period will depend on something like a higher power of g.
We can make some progress observing that:
d(gamma*v)/dt = -g/L*gamma*x (1)
becomes:
gamma^2*dv/dt=-g/L*x (2)
by observing that
d(gamma*v)/dt=gamma^3*dv/dt (3)
Now, taking the time derivative of (2) we get:
d(gamma^2*dv/dt)dt=-g/L*v (4)
This leads to:
d2v/dt2+2v/c^2*gamma^2(dv/dt)^2+g/L*gamma^(-2)*v=0 (5)
Now, we know that there is no closed solution to the above (we have seen that from the previous problem of moving a charged particle in an EM field). Nevertheless, if we assume v<<c (the pendulum should move at speeds much smaller than c, we can make the following approximation:
gamma^2=1+(v/c)^2
so we end up with:
d2v/dt2+2v/c^2(1+(v/c)^2)(dv/dt)^2+g/L*(1-(v/c)^2)*v=0 (6)
I need to check if the above has a closed solution. We know that v varies from 0 (the highest point of the pendulum) to v_max (at the lowest point) , so we should be looking for a periodic function as a solution.
d(gamma*v)/dt = -g/L*gamma*x (1)
becomes:
gamma^2*dv/dt=-g/L*x (2)
by observing that
d(gamma*v)/dt=gamma^3*dv/dt (3)
Now, taking the time derivative of (2) we get:
d(gamma^2*dv/dt)dt=-g/L*v (4)
This leads to:
d2v/dt2+2v/c^2*gamma^2(dv/dt)^2+g/L*gamma^(-2)*v=0 (5)
Now, we know that there is no closed solution to the above (we have seen that from the previous problem of moving a charged particle in an EM field). Nevertheless, if we assume v<<c (the pendulum should move at speeds much smaller than c, we can make the following approximation:
gamma^2=1+(v/c)^2
so we end up with:
d2v/dt2+2v/c^2(1+(v/c)^2)(dv/dt)^2+g/L*(1-(v/c)^2)*v=0 (6)
I need to check if the above has a closed solution. We know that v varies from 0 (the highest point of the pendulum) to v_max (at the lowest point) , so we should be looking for a periodic function as a solution.
QUOTE (Trout+Jul 2 2008, 09:58 PM)
d2v/dt2+2v/c^2(1+(v/c)^2)(dv/dt)^2+g/L*(1-(v/c)^2)*v=0 (6)
I need to check if the above has a closed solution. We know that v varies from 0 (the highest point of the pendulum) to v_max (at the lowest point) , so we should be looking for a periodic function as a solution.
Continued:
turns out that I can reduce eq.6 to a first order ODE by using the substitution w(v)=(dv/dt)^2
Then d2v/dt2=d(sqrt(w)/dt=.5/sqrt(w)*dw/dt=.5/sqrt(w)*dw/dv*dv/dt=.5*dw/dv
So, the equation becomes:
dw/dv+2f(v)*w+2h(v)=0 (7)
where f(v)=2v/c^2*gamma^2 and h(v)=g/L*gamma^-2
We don't even need to use the approximation for gamma^2 anymore, we can use the exact formula.
Now, (7) is a first order ODE. We are in good shape. More tomorrow.
QUOTE (Trout+Jul 3 2008, 02:33 AM)
Continued:
turns out that I can reduce eq.6 to a first order ODE by using the substitution w(v)=(dv/dt)^2
Then d2v/dt2=d(sqrt(w)/dt=.5/sqrt(w)*dw/dt=.5/sqrt(w)*dw/dv*dv/dt=.5*dw/dv
So, the equation becomes:
dw/dv+2f(v)*w+2h(v)=0 (7)
where f(v)=2v/c^2*gamma^2 and h(v)=g/L*gamma^-2
We don't even need to use the approximation for gamma^2 anymore, we can use the exact formula.
Now, (7) is a first order ODE. We are in good shape. More tomorrow.
The solution for (7) is:
w=exp(F)(C-2*Integral{exp(-F)h(v)dv}
where F=-2*Integral[f(v)dv]=4sqrt(1-(v/c)^2)
Unfortunately,
Integral{exp(-F)h(v)dv}=Integral{exp(-4*sqrt(1-(v/c)^2)*(1-(v/c)^2)dv}
does not have a closed form.
turns out that I can reduce eq.6 to a first order ODE by using the substitution w(v)=(dv/dt)^2
Then d2v/dt2=d(sqrt(w)/dt=.5/sqrt(w)*dw/dt=.5/sqrt(w)*dw/dv*dv/dt=.5*dw/dv
So, the equation becomes:
dw/dv+2f(v)*w+2h(v)=0 (7)
where f(v)=2v/c^2*gamma^2 and h(v)=g/L*gamma^-2
We don't even need to use the approximation for gamma^2 anymore, we can use the exact formula.
Now, (7) is a first order ODE. We are in good shape. More tomorrow.
The solution for (7) is:
w=exp(F)(C-2*Integral{exp(-F)h(v)dv}
where F=-2*Integral[f(v)dv]=4sqrt(1-(v/c)^2)
Unfortunately,
Integral{exp(-F)h(v)dv}=Integral{exp(-4*sqrt(1-(v/c)^2)*(1-(v/c)^2)dv}
does not have a closed form.
Sorry, but if T = √(l/g) and l -> l and T -> γT then what's wrong with g -> γ^(-2)g ?
At least this is my reasoning based on slowed down motion picture analysis.
At least this is my reasoning based on slowed down motion picture analysis.
QUOTE (rpenner+Jul 3 2008, 07:05 PM)
Sorry, but if T = √(l/g) and l -> l and T -> γT then what's wrong with g -> γ^(-2)g ?
At least this is my reasoning based on slowed down motion picture analysis.
1. Because this is not how acceleration transverse to the direction of motion transforms in SR.
2. Because the formula g'=g*gamma^-2 appears ad-hoc
3. Because we need to work things from base principles. Can you help?
At least this is my reasoning based on slowed down motion picture analysis.
1. Because this is not how acceleration transverse to the direction of motion transforms in SR.
2. Because the formula g'=g*gamma^-2 appears ad-hoc
3. Because we need to work things from base principles. Can you help?
I disagree, at least in the limiting case of small motions.
Setting four-acceleration equal to four-acceleration, we have:
\gamma_u^2\mathbf{a'} + \frac{\mathbf{a' \cdot u}}{c^2} \gamma_u^4\mathbf{u} = \mathbf{a}
γ^2 a' + (a' · u / c²) γ^4 u = a
Thus, when we can neglect the down-component of the pendulum's motion, we have:
\gamma_u^2\mathbf{a'} = \mathbf{a}
γ^2 a' = a
or
\mathbf{a'} = \gamma_u^{-2} \mathbf{a}
a' = γ^(-2) a
Which is absolutely non-paradoxical as my above post should make clear.
Setting four-acceleration equal to four-acceleration, we have:
\gamma_u^2\mathbf{a'} + \frac{\mathbf{a' \cdot u}}{c^2} \gamma_u^4\mathbf{u} = \mathbf{a}
γ^2 a' + (a' · u / c²) γ^4 u = a
Thus, when we can neglect the down-component of the pendulum's motion, we have:
\gamma_u^2\mathbf{a'} = \mathbf{a}
γ^2 a' = a
or
\mathbf{a'} = \gamma_u^{-2} \mathbf{a}
a' = γ^(-2) a
Which is absolutely non-paradoxical as my above post should make clear.
QUOTE (rpenner+Jul 3 2008, 07:58 PM)
I disagree, at least in the limiting case of small motions.
Setting four-acceleration equal to four-acceleration, we have:
\gamma_u^2\mathbf{a'} + \frac{\mathbf{a' \cdot u}}{c^2} \gamma_u^4\mathbf{u} = \mathbf{a}
γ^2 a' + (a' · u / c²) γ^4 u = a
Thus, when we can neglect the down-component of the pendulum's motion, we have:
\gamma_u^2\mathbf{a'} = \mathbf{a}
γ^2 a' = a
or
\mathbf{a'} = \gamma_u^{-2} \mathbf{a}
a' = γ^(-2) a
Which is absolutely non-paradoxical as my above post should make clear.
Precisely my point 2, what allows you to neglect the term (a' · u / c²) γ^4 u? The division by c^2?
Aside from this, I would still want to solve the equation of motion in SR. This seems to reduce to an ODE that has no closed form. Any thoughts?
Setting four-acceleration equal to four-acceleration, we have:
\gamma_u^2\mathbf{a'} + \frac{\mathbf{a' \cdot u}}{c^2} \gamma_u^4\mathbf{u} = \mathbf{a}
γ^2 a' + (a' · u / c²) γ^4 u = a
Thus, when we can neglect the down-component of the pendulum's motion, we have:
\gamma_u^2\mathbf{a'} = \mathbf{a}
γ^2 a' = a
or
\mathbf{a'} = \gamma_u^{-2} \mathbf{a}
a' = γ^(-2) a
Which is absolutely non-paradoxical as my above post should make clear.
Precisely my point 2, what allows you to neglect the term (a' · u / c²) γ^4 u? The division by c^2?
Aside from this, I would still want to solve the equation of motion in SR. This seems to reduce to an ODE that has no closed form. Any thoughts?
The dot-product between the acceleration and the frame velocity means that you can neglect this term to the extent in which they are perpendicular.
If the pendulum is at rest, then u is perpendicular to a and a', and the result is exact.
If the pendulum is of length L, and under acceleration g, and in the small-angle approximation, sin θ << 1, where the period is T=2π√(L/g), then the speed, v, is small relative to √(2gL) which would be the maximum speed if the pendulum were executing the arc of a semi-circle. And so the second term is negligible when the speed of the observer is less than c - sin²θgL/2c, i.e. when the gamma related to the frame is less than 1/√(sin²θgL/c²). Which is to say there is no gamma so large that the small angle approximation doesn't save us, even for largish pendulums.
If the pendulum is at rest, then u is perpendicular to a and a', and the result is exact.
If the pendulum is of length L, and under acceleration g, and in the small-angle approximation, sin θ << 1, where the period is T=2π√(L/g), then the speed, v, is small relative to √(2gL) which would be the maximum speed if the pendulum were executing the arc of a semi-circle. And so the second term is negligible when the speed of the observer is less than c - sin²θgL/2c, i.e. when the gamma related to the frame is less than 1/√(sin²θgL/c²). Which is to say there is no gamma so large that the small angle approximation doesn't save us, even for largish pendulums.
QUOTE (rpenner+Jul 3 2008, 10:56 PM)
The dot-product between the acceleration and the frame velocity means that you can neglect this term to the extent in which they are perpendicular.
If the pendulum is at rest, then u is perpendicular to a and a', and the result is exact.
If the pendulum is of length L, and under acceleration g, and in the small-angle approximation, sin θ << 1, where the period is T=2π√(L/g), then the speed, v, is small relative to √(2gL) which would be the maximum speed if the pendulum were executing the arc of a semi-circle. And so the second term is negligible when the speed of the observer is less than c - sin²θgL/2c, i.e. when the gamma related to the frame is less than 1/√(sin²θgL/c²). Which is to say there is no gamma so large that the small angle approximation doesn't save us, even for largish pendulums.
OK, sounds like a good approximation.
I think that we can get an even better approximation by using the equations for 3-acceleration transformation:
a'_y=gamma^-2[(1+v_xV/c^2)^-2*a_y-v_yV/c^2(1+v_xV/c^2)^-3*a_x]
In the particular case we are talking about (tranforming g=g_y) a_x is exacly 0 so:
a'_y=gamma^-2*(1+v_xV/c^2)^-2*a_y (no approximation needed)
or, since a_y=g:
g'=gamma^-2*(1+v_xV/c^2)^-2*g
So, the only approximation that we need to do is , for small v_x (which is the case):
1+v_xV/c^2=1
Rpenner,
I would still like some help in solving the equation of the pendulum in SR. I arrived (again) to an ODE that seems not to have a closed solution. Actually, we don't need the solution, we only need to show that it is peridical and that the period depends linearly on gamma. Any ideas?
If the pendulum is at rest, then u is perpendicular to a and a', and the result is exact.
If the pendulum is of length L, and under acceleration g, and in the small-angle approximation, sin θ << 1, where the period is T=2π√(L/g), then the speed, v, is small relative to √(2gL) which would be the maximum speed if the pendulum were executing the arc of a semi-circle. And so the second term is negligible when the speed of the observer is less than c - sin²θgL/2c, i.e. when the gamma related to the frame is less than 1/√(sin²θgL/c²). Which is to say there is no gamma so large that the small angle approximation doesn't save us, even for largish pendulums.
OK, sounds like a good approximation.
I think that we can get an even better approximation by using the equations for 3-acceleration transformation:
a'_y=gamma^-2[(1+v_xV/c^2)^-2*a_y-v_yV/c^2(1+v_xV/c^2)^-3*a_x]
In the particular case we are talking about (tranforming g=g_y) a_x is exacly 0 so:
a'_y=gamma^-2*(1+v_xV/c^2)^-2*a_y (no approximation needed)
or, since a_y=g:
g'=gamma^-2*(1+v_xV/c^2)^-2*g
So, the only approximation that we need to do is , for small v_x (which is the case):
1+v_xV/c^2=1
Rpenner,
I would still like some help in solving the equation of the pendulum in SR. I arrived (again) to an ODE that seems not to have a closed solution. Actually, we don't need the solution, we only need to show that it is peridical and that the period depends linearly on gamma. Any ideas?
QUOTE (Trout+Jul 4 2008, 12:27 AM)
OK, sounds like a good approximation.
I think that we can get an even better approximation by using the equations for 3-acceleration transformation:
a'_y=gamma^-2[(1+v_xV/c^2)^-2*a_y-v_yV/c^2(1+v_xV/c^2)^-3*a_x]
In the particular case we are talking about (tranforming g=g_y) a_x is exacly 0 so:
a'_y=gamma^-2*(1+v_xV/c^2)^-2*a_y (no approximation needed)
or, since a_y=g:
g'=gamma^-2*(1+v_xV/c^2)^-2*g
So, the only approximation that we need to do is , for small v_x (which is the case):
1+v_xV/c^2=1
Rpenner,
I would still like some help in solving the equation of the pendulum in SR. I arrived (again) to an ODE that seems not to have a closed solution. Actually, we don't need the solution, we only need to show that it is peridical and that the period depends linearly on gamma. Any ideas?
OK, here is the deal with this problem:
Since m_0 *d^2x/dt^2=k (constant) is Galilei invariant but not
Lorentz invariant, we have to redefine the impulse in SR as p=
\gamma(v)*m_0*v instead of the Newtoniam m_0*v.
1. We can prove that in the proper frame of the object F :
d(\gamma(v)*v)/dt=\gamma(v)^3*dv/dt
2. We can also prove (after some computations) that , in a frame F'
moving with constant speed V wrt the frame F:
d(\gamma(v')*v')/dt'=\gamma(v')^3*dv'/dt'
3. We can further prove that
\gamma(v')^3*dv'/dt'=\gamma(v)^3*dv/dt=k/m_0
This means that by redefining the relativistic impulse as
\gamma(v)*m_0*v , the equations of motion under constant force (F=k)
are Lorentz invariant. So far, so good.
The situation when F is NOT constant is much more complicated and I know of no solution
Here are several examples:
1. F= - q*x (common spring)
2. F= -q *sin(theta) (common pendulum)
3. An even nastier case is the case of the torsion pendulum where we
need the relativistic equivalent for the Newtonian p=I*d(theta)/dt
where I is the momentum of inertia . It is not obvious what that
formula would be.
In these particular cases, the fact that the left term of the equation of motion
is invariant (i.e. \gamma(v')^3*dv'/dt'=\gamma(v)^3*dv/dt) is of no good,
since the right term is obviously not Lorentz invariant since neither
x, nor theta are Lorentz invariants.
Of course, Hooke law and the pendulum law are laws derived
empirically, so the obvious approach would be to redefine them in
such a fashion that they become Lorentz invariant. Did any of you see any
literature on this?
I think that we can get an even better approximation by using the equations for 3-acceleration transformation:
a'_y=gamma^-2[(1+v_xV/c^2)^-2*a_y-v_yV/c^2(1+v_xV/c^2)^-3*a_x]
In the particular case we are talking about (tranforming g=g_y) a_x is exacly 0 so:
a'_y=gamma^-2*(1+v_xV/c^2)^-2*a_y (no approximation needed)
or, since a_y=g:
g'=gamma^-2*(1+v_xV/c^2)^-2*g
So, the only approximation that we need to do is , for small v_x (which is the case):
1+v_xV/c^2=1
Rpenner,
I would still like some help in solving the equation of the pendulum in SR. I arrived (again) to an ODE that seems not to have a closed solution. Actually, we don't need the solution, we only need to show that it is peridical and that the period depends linearly on gamma. Any ideas?
OK, here is the deal with this problem:
Since m_0 *d^2x/dt^2=k (constant) is Galilei invariant but not
Lorentz invariant, we have to redefine the impulse in SR as p=
\gamma(v)*m_0*v instead of the Newtoniam m_0*v.
1. We can prove that in the proper frame of the object F :
d(\gamma(v)*v)/dt=\gamma(v)^3*dv/dt
2. We can also prove (after some computations) that , in a frame F'
moving with constant speed V wrt the frame F:
d(\gamma(v')*v')/dt'=\gamma(v')^3*dv'/dt'
3. We can further prove that
\gamma(v')^3*dv'/dt'=\gamma(v)^3*dv/dt=k/m_0
This means that by redefining the relativistic impulse as
\gamma(v)*m_0*v , the equations of motion under constant force (F=k)
are Lorentz invariant. So far, so good.
The situation when F is NOT constant is much more complicated and I know of no solution
Here are several examples:
1. F= - q*x (common spring)
2. F= -q *sin(theta) (common pendulum)
3. An even nastier case is the case of the torsion pendulum where we
need the relativistic equivalent for the Newtonian p=I*d(theta)/dt
where I is the momentum of inertia . It is not obvious what that
formula would be.
In these particular cases, the fact that the left term of the equation of motion
is invariant (i.e. \gamma(v')^3*dv'/dt'=\gamma(v)^3*dv/dt) is of no good,
since the right term is obviously not Lorentz invariant since neither
x, nor theta are Lorentz invariants.
Of course, Hooke law and the pendulum law are laws derived
empirically, so the obvious approach would be to redefine them in
such a fashion that they become Lorentz invariant. Did any of you see any
literature on this?
Hi Trout and rpenner,
QUOTE (Trout+)
The problem with the above is that it leads to a paradox:
1. On one hand, you are absolutely correct, T'=gamma*T , where gamma=1/sqrt(1-(V/c)^2) and V is the relative speed between frames.
2. On the other hand, if we use the Newtonian expression for the period (T=sqrt(L/g)), then it would APPEAR that T'=T. (turns out that this is not correct, since g does not transform into g, accelerations transform differently in SR)
This is how the crank managed to (almost) stump the moderators on the other forum. As we've seen, T is NOT equal to sqrt(L/g) in SR, so we need to be able to solve the nasty differential equation that describes the pendulum motion in SR. I predict that the period will depend on something like a higher power of g.
Trout's comments
Rpenner's first answer ( t' = gamma t ) is absolutely correct provided that the mass of the observer is small relative to the system and is in free fall. You do not transform the acceleration of the pendulum and "earth" system to the moving particle, it is in "free fall" so is currently "unaccelerated". In free fall the paths of all co-moving particles are "all" on straight line trajectories even in a gravity field(as measured from any other freely falling particle) when in close proximity to the pendulum origin (forget about possibly hitting the "earth object"). Trout is right about SR being applicable but he probably read that from the original source. A clock is a clock regardless of the state of motion.... A pendulum constitutes a clock in any man's language. This "relative" motion in SR does not affect the physics as observed in the other frame (or for that matter any other relatively moving frame). The event is the same and is "approximately" unaffected by any external motion of test particles (the other observers). The only difference in the frames is how it is viewed in relative motion. The "accelerated clocks" run slow when all optical effects are removed (... remember Trout?). Not the acceleration due to "gravity" but SR Twin's Paradox due to motion. There is "no paradox" if you choose the "accelerated" frame, less time elapses there relative to the "observed" pendulum frame. All measurements must be made "close" to the pendulum "observation" being made for this condition to hold. You must ignore optical effects (time delays due to proper motion and propagation times) to gain any real answers... that is timings must be made in the same frame.... That is for the Einstein condition for "synchronization" to apply.
These timings are actually only influenced by "time dilation" as noted by rpenner. You will not be able to make a valid observation over several periods of oscillation (if traveling near the speed of light) and a "local observation" is required and extrapolated for the full cycle. It does not matter for how long the moving frame is close to the pendulum since it has no influence on the physics and distant observations need correction for propagation times and optical effects of stellar aberration. In general this is "difficult" but it is not the main question. What you need to know is only the period of the pendulum.
In the rest frame of the pendulum... it motion is unaffected by any external "observers". Distant observations are "invalid" due to long distance changes in spacetime curvature resulting in significant optical time delays along paths. In the region around this pendulum a thousand particles traveling in different directions will all be executing approximately linear paths (relative to each other... check it out that is correct and it occurs even at low velocity) if all are in free fall even if velocities are different. In order to remove the effects of propagation delay you need to measure the systems in a rest frame relative to the two systems. Instantaneous rest frames may be "proposed" to solve the problem these could be placed at small incremental distances along the "moving frame's" linear path (assuming the only "non-moving frame" is the pendulum and the planet to simplify the problem). Arbitrarily "pick" the moving frames to be brought to a halt instantaneously and to "suffer" the time dilation relative to the "stay at home clock... pendulum thingy". An "intellectually satisfying" way to do this is to propose that this "pendulum" is indeed a "clock" and we observe the dial on that clock. We observe that dial a moment later at some "short" distance away and the difference in times is a dilated time difference which neglects the propagation times from the first position to the next position. A simultaneous measurement is made with the moving clock. The difference will be related to "gamma" when corrected for the propagation time of light to the new position.
Starting with the point of closest approach to the pendulum as t = 0. Clocks are synchronized at the start of the path s = 0 and synchronized again at the end of the path t = t1, s = s1. At the end of the path correction for time dilation is made to correct between frames v•t1 (since the pendulum is taken as being in a "stationary" frame for the sake of simplifying this problem conceptually) and for the propagation time for light... -s1/C is a correction for this displacement (deduct this from the time read on the dial when at position s1). The corrected period of time measured is related through gamma. Significance must be given to the relative signs of the motion. If you do not like that then think of the problem as "time reversed"... the answer for a pendulum will still be the same. It will always be the same "fraction" of the motion observed up close which depends only on the relative velocity and that displacement. Different relative velocities result in different gammas. The "period" becomes gamma•period. Every way you look at it... it is gamma•t. Optical phenomena do not count as relativistic time dilation.... end of story.
Anyway you said that this "almost" stumped the moderators... so you already know the answer. What is the point? If you think this is going to help us you better "spill the beans" now. You have my take on it so what is yours?
Cheers
1. On one hand, you are absolutely correct, T'=gamma*T , where gamma=1/sqrt(1-(V/c)^2) and V is the relative speed between frames.
2. On the other hand, if we use the Newtonian expression for the period (T=sqrt(L/g)), then it would APPEAR that T'=T. (turns out that this is not correct, since g does not transform into g, accelerations transform differently in SR)
This is how the crank managed to (almost) stump the moderators on the other forum. As we've seen, T is NOT equal to sqrt(L/g) in SR, so we need to be able to solve the nasty differential equation that describes the pendulum motion in SR. I predict that the period will depend on something like a higher power of g.
Trout's comments
Rpenner's first answer ( t' = gamma t ) is absolutely correct provided that the mass of the observer is small relative to the system and is in free fall. You do not transform the acceleration of the pendulum and "earth" system to the moving particle, it is in "free fall" so is currently "unaccelerated". In free fall the paths of all co-moving particles are "all" on straight line trajectories even in a gravity field(as measured from any other freely falling particle) when in close proximity to the pendulum origin (forget about possibly hitting the "earth object"). Trout is right about SR being applicable but he probably read that from the original source. A clock is a clock regardless of the state of motion.... A pendulum constitutes a clock in any man's language. This "relative" motion in SR does not affect the physics as observed in the other frame (or for that matter any other relatively moving frame). The event is the same and is "approximately" unaffected by any external motion of test particles (the other observers). The only difference in the frames is how it is viewed in relative motion. The "accelerated clocks" run slow when all optical effects are removed (... remember Trout?). Not the acceleration due to "gravity" but SR Twin's Paradox due to motion. There is "no paradox" if you choose the "accelerated" frame, less time elapses there relative to the "observed" pendulum frame. All measurements must be made "close" to the pendulum "observation" being made for this condition to hold. You must ignore optical effects (time delays due to proper motion and propagation times) to gain any real answers... that is timings must be made in the same frame.... That is for the Einstein condition for "synchronization" to apply.
These timings are actually only influenced by "time dilation" as noted by rpenner. You will not be able to make a valid observation over several periods of oscillation (if traveling near the speed of light) and a "local observation" is required and extrapolated for the full cycle. It does not matter for how long the moving frame is close to the pendulum since it has no influence on the physics and distant observations need correction for propagation times and optical effects of stellar aberration. In general this is "difficult" but it is not the main question. What you need to know is only the period of the pendulum.
In the rest frame of the pendulum... it motion is unaffected by any external "observers". Distant observations are "invalid" due to long distance changes in spacetime curvature resulting in significant optical time delays along paths. In the region around this pendulum a thousand particles traveling in different directions will all be executing approximately linear paths (relative to each other... check it out that is correct and it occurs even at low velocity) if all are in free fall even if velocities are different. In order to remove the effects of propagation delay you need to measure the systems in a rest frame relative to the two systems. Instantaneous rest frames may be "proposed" to solve the problem these could be placed at small incremental distances along the "moving frame's" linear path (assuming the only "non-moving frame" is the pendulum and the planet to simplify the problem). Arbitrarily "pick" the moving frames to be brought to a halt instantaneously and to "suffer" the time dilation relative to the "stay at home clock... pendulum thingy". An "intellectually satisfying" way to do this is to propose that this "pendulum" is indeed a "clock" and we observe the dial on that clock. We observe that dial a moment later at some "short" distance away and the difference in times is a dilated time difference which neglects the propagation times from the first position to the next position. A simultaneous measurement is made with the moving clock. The difference will be related to "gamma" when corrected for the propagation time of light to the new position.
Starting with the point of closest approach to the pendulum as t = 0. Clocks are synchronized at the start of the path s = 0 and synchronized again at the end of the path t = t1, s = s1. At the end of the path correction for time dilation is made to correct between frames v•t1 (since the pendulum is taken as being in a "stationary" frame for the sake of simplifying this problem conceptually) and for the propagation time for light... -s1/C is a correction for this displacement (deduct this from the time read on the dial when at position s1). The corrected period of time measured is related through gamma. Significance must be given to the relative signs of the motion. If you do not like that then think of the problem as "time reversed"... the answer for a pendulum will still be the same. It will always be the same "fraction" of the motion observed up close which depends only on the relative velocity and that displacement. Different relative velocities result in different gammas. The "period" becomes gamma•period. Every way you look at it... it is gamma•t. Optical phenomena do not count as relativistic time dilation.... end of story.
Anyway you said that this "almost" stumped the moderators... so you already know the answer. What is the point? If you think this is going to help us you better "spill the beans" now. You have my take on it so what is yours?
Cheers
QUOTE (Good Elf+Jul 13 2008, 07:21 AM)
Hi Trout and rpenner,
Rpenner's first answer ( t' = gamma t ) is absolutely correct provided that the mass of the observer is small relative to the system and is in free fall.
This is obvious but , as I explained, this is not the issue.
You do not transform the acceleration of the pendulum and "earth" system to the moving particle,
No one did that, so this is irrelevant.
No one did that, so this is irrelevant.
Trout is right about SR being applicable but he probably read that from the original source.
What are you talking about?
A clock is a clock regardless of the state of motion.... A pendulum constitutes a clock in any man's language. This "relative" motion in SR does not affect the physics as observed in the other frame (or for that matter any other relatively moving frame).
Again, you are stating the obvious while you are completely missing the point of the thread.
Again, you are stating the obvious while you are completely missing the point of the thread.
The event is the same and is "approximately" unaffected by any external motion of test particles (the other observers).
<I had to snip the rest of your babbling>
Anyway you said that this "almost" stumped the moderators... so you already know the answer. What is the point? If you think this is going to help us you better "spill the beans" now. You have my take on it so what is yours?
Cheers
The moderators couldn't answer the poster's question. I f you took some time to really understand the posts, you'd have understood that.
No one could answer. At least I am trying to work out a mathematical solution.
And you posted again a very long and irrelevant answer to a question no one asked in the form of a story.
Well known SR transformation laws tell us that ... g'=β^-2 g
You don't know that, I just showed you that this may be an acceptable approximation.
Anyway, I am way past the point of showing that the period is frame-invariant, I am at the point at trying to find a covariant form of the different empiracal laws shown in my previous post. Thank you for the attempt, though.
The above is obviously incorrect since you left out the term dβ/dt'm_0 u'. (You gave a justification for this but it is not acceptable).
The reason is that β in p=βmu is a function of u. β=1/sqrt(1-(u/c)^2). I can see that you came up with a series of approximations in order to justify your point but this defeats the whole point, you are bringing the problem back in the classical domain u<<c. This is not interesting.
If you redo the computations correctly you end up with a much nastier ODE, the one I was asking help with. You say correctly that the resulting ODE has no symbolic solution, we already know that since it showed up in the EM problem that started this thread. Nevertheless, mr_homm and I defeated that problem (without resorting to hacks). I am asking for help in order to solve this new problem.
F'_net=dp'/dt'=(d/dt')(γ' m_0 u')=m_0 [u'dγ'/dt' + γ'du'/dt']
Correct
No, this is exactly what I called you on, the whole problem is that you can't assume u to be so small as to u_x=0. You are subconciously reducing the problem to its Newtonian limits.
No, this is exactly what I called you on, the whole problem is that you can't assume u to be so small as to u_x=0. You are subconciously reducing the problem to its Newtonian limits.
Therefore γ≈1 and γ'≈β and the time derivatives of both are negligible
No, again. Please look at the sample solution for the EM problem. You can't reduce the problem to γ≈1 and γ'≈β because in effect this reduces it to the Newtonian limit.
Once again, I got the solution to the point where there are no approximations, unfortunately the resulting ODE has no symbolic solution.
I am past this point, I ma looking for a covariant formulation of the pendulum law.
Now, if you want to examine the regime where u_y and u_z are not small compared to c, the approximation γ'≈β is of course no longer valid, however, neither is the original solution T=2π(l/g)^0.5 (!!!), for even in Frame S where u_x=0, you end up with an ugly ODE! T'=βT will still be correct but you will need to solve either the ODE that arises in S or the ODE that arises in S' to find T or T' I am sure you can find a few papers that deal with the solution to the ODE in S, but I suspect that all solutions are numerical and not analytical. If I get a chance later, I will try to find a paper on the relativistic pendulum and see if there are any analytical solutions.
Thank you, this is exactly what we are looking for.
More generally, a covariant reformulation of the pendulum law, Hooke law, torsion pendulum law (see my earlier post here).
We cannot use the current formulations because they produce frame-variant results.
Rpenner's first answer ( t' = gamma t ) is absolutely correct provided that the mass of the observer is small relative to the system and is in free fall.
This is obvious but , as I explained, this is not the issue.
QUOTE
You do not transform the acceleration of the pendulum and "earth" system to the moving particle,
No one did that, so this is irrelevant.
QUOTE (->
| QUOTE |
You do not transform the acceleration of the pendulum and "earth" system to the moving particle, |
No one did that, so this is irrelevant.
Trout is right about SR being applicable but he probably read that from the original source.
What are you talking about?
QUOTE
A clock is a clock regardless of the state of motion.... A pendulum constitutes a clock in any man's language. This "relative" motion in SR does not affect the physics as observed in the other frame (or for that matter any other relatively moving frame).
Again, you are stating the obvious while you are completely missing the point of the thread.
QUOTE (->
| QUOTE |
A clock is a clock regardless of the state of motion.... A pendulum constitutes a clock in any man's language. This "relative" motion in SR does not affect the physics as observed in the other frame (or for that matter any other relatively moving frame). |
Again, you are stating the obvious while you are completely missing the point of the thread.
The event is the same and is "approximately" unaffected by any external motion of test particles (the other observers).
<I had to snip the rest of your babbling>
Anyway you said that this "almost" stumped the moderators... so you already know the answer. What is the point? If you think this is going to help us you better "spill the beans" now. You have my take on it so what is yours?
Cheers
The moderators couldn't answer the poster's question. I f you took some time to really understand the posts, you'd have understood that.
No one could answer. At least I am trying to work out a mathematical solution.
And you posted again a very long and irrelevant answer to a question no one asked in the form of a story.
QUOTE (mr_homm+Jul 1 2008, 12:27 AM)
Hi Trout,
This one doesn't look too bad to me, unless I'm missing something. Let the pendulum swing through the origin of coordinates, so that the lowest point on its swing passes directly through the origin. Since the pendulum executes periodic motion, it is only necessary to consider time at which it passes through the origin on each swing. This will define two events with coordinates (x,y,z,ct) = (0,0,0,0) and (0,0,0,cT), corresponding to passages through the origin one period apart. Even though the pendulum is moving and accelerating, these two events contain all the information necessary to define the period of the pendulum in the earth frame. For the assumptions in the problem, T = sqrt(l/g).
Now in another frame, it is only necessary to transform the spacetime interval between these two events in order to find the new period. Let the second frame have velocity v relative to the earth, so that gamma = 1/sqrt(1-(v/c)^2). Then by the Lorentz transformation, (0,0,0,cT) transforms to gamma*(-vT,0,0,cT), so that in the new frame, the period is simply gamma*T, during which time the pendulum and the earth frame move a distance -gamma*vT relative to the second frame. That's all there is to the calculation.
The pendulum therefore appears to swing slower by a factor of gamma as seen from the second frame.
I think that's all there is to the problem. Am I missing something?
--Stuart Anderson
This solution is absolutely correct (aside from a missing factor of 2Pi), and is probably the simplest way of solving this problem.
This one doesn't look too bad to me, unless I'm missing something. Let the pendulum swing through the origin of coordinates, so that the lowest point on its swing passes directly through the origin. Since the pendulum executes periodic motion, it is only necessary to consider time at which it passes through the origin on each swing. This will define two events with coordinates (x,y,z,ct) = (0,0,0,0) and (0,0,0,cT), corresponding to passages through the origin one period apart. Even though the pendulum is moving and accelerating, these two events contain all the information necessary to define the period of the pendulum in the earth frame. For the assumptions in the problem, T = sqrt(l/g).
Now in another frame, it is only necessary to transform the spacetime interval between these two events in order to find the new period. Let the second frame have velocity v relative to the earth, so that gamma = 1/sqrt(1-(v/c)^2). Then by the Lorentz transformation, (0,0,0,cT) transforms to gamma*(-vT,0,0,cT), so that in the new frame, the period is simply gamma*T, during which time the pendulum and the earth frame move a distance -gamma*vT relative to the second frame. That's all there is to the calculation.
The pendulum therefore appears to swing slower by a factor of gamma as seen from the second frame.
I think that's all there is to the problem. Am I missing something?
--Stuart Anderson
This solution is absolutely correct (aside from a missing factor of 2Pi), and is probably the simplest way of solving this problem.
QUOTE (Trout+Jul 4 2008, 12:27 AM)
OK, sounds like a good approximation.
I think that we can get an even better approximation by using the equations for 3-acceleration transformation:
a'_y=gamma^-2[(1+v_xV/c^2)^-2*a_y-v_yV/c^2(1+v_xV/c^2)^-3*a_x]
In the particular case we are talking about (tranforming g=g_y) a_x is exacly 0 so:
a'_y=gamma^-2*(1+v_xV/c^2)^-2*a_y (no approximation needed)
or, since a_y=g:
g'=gamma^-2*(1+v_xV/c^2)^-2*g
So, the only approximation that we need to do is , for small v_x (which is the case):
1+v_xV/c^2=1
Rpenner,
I would still like some help in solving the equation of the pendulum in SR. I arrived (again) to an ODE that seems not to have a closed solution. Actually, we don't need the solution, we only need to show that it is peridical and that the period depends linearly on gamma. Any ideas¿
I think in order to avoid confusion you first need to properly define your terms. I suggest you use v for the speed of the observer relative to the plane of the pendulum and u for the velocity of the mass at the end of the pendulum. Also (!), SR is easiest to work with in Cartesian coordinates and so I recommend you use them! In the ODE you came up with, you use x to represent the arc length which is not Cartesian! and so there is bound to be some confusion when computing the solution. Instead, I recommend you use (x,y,z) to represent the coordinates of the mass at the end of the pendulum and {i,j,k} to represent the corresponding unit vectors.
Here is my solution:
Let gravity point in the negative z-direction, with the pendulum hanging from the origin and swinging in the y-z plane. And let the observer be travelling in the positive x-direction with speed v. And let us denote the angle that the pendulum makes with the negative z-axis by θ.
(1) Compute the net force on the pendulum's mass in this moving frame:
F'_net= F'_gravity + F'-string= [-m'g'k] + [m'g'cos(θ')k-m'g'sin(θ')j]
= [-m'g'k] + [m'g'(-z'/L')k-m'g'(y'/L)'j]
= -m'g'[(y'/L')j + (1+(z'/L'))k]
Where the (')s denote that the corresponding value is computed in the moving frame.
Since the observer is travelling perpendicular to the plane of the pendulum, there will be no length contraction for the string and so L'=L.
Well known SR transformation laws tell us that m'=γm_0 and g'=β^-2 g where γ'=(1-u'^2/c^2)^-0.5 and β=(1-v^2/c^2)^-0.5
And so;
F'_net= -(γ'/β^2)m_0 g[(y'/L)j + (1+(z'/L))k]
(2) Apply Newton's second law to obtain the equation of motion:
F'_net=dp'/dt'=(d/dt')(γ' m_0 u')=m_0 [u'dγ'/dt' + γ'du'/dt']
Now, for small θ (and subsequently small θ') and small m_0, u_y and u_z (and u'_y and u'_z) are negligible compared with v (u'_x=v and u_x=0) and c. Therefore γ≈1 and γ'≈β and the time derivatives of both are negligible (A better approximation would result in an ODE that cannot be solved analytically, but if solved numerically would lead to the same result).
And so we have:
F'_net=dp'/dt'=(d/dt')(γ' m_0 u')=βm_0 du'/dt']
=βm_0[ du'_x/dt'i +du'_y/dt'j +du'_z/dt'k]
SR transformation laws for accelerations give du'_x/dt'=β^-3 (1-u_x v/c^2)^-3 du_x/dt= 0 since u_x=0 in the rest frame (frame at rest relative to the pendulum's anchor point)
=> F'_net=βm_0[ du'_y/dt'j +du'_z/dt'k]
=βm_0[ d^2y'/dt'^2j +d^2z'/dt'^2k]= -(γ'/β^2)m_0 g[(y'/L)j + (1+(z'/L))k]
Concentrating on the y'-component:
βm_0 d^2y'/dt'^2 = (γ'/β^2)m_0 g[(y'/L)=(1/β)m_0 g[(y'/L)
=> d^2y'/dt'^2 = (g/Lβ^2) y'
=> y'= y'_max cos[(Lβ^2/g)^0.5 t'] where y'=y'_max at t'=0
=> T' =2πβ (L/g)^0.5 which agrees with Mr. homm's solution
I think that we can get an even better approximation by using the equations for 3-acceleration transformation:
a'_y=gamma^-2[(1+v_xV/c^2)^-2*a_y-v_yV/c^2(1+v_xV/c^2)^-3*a_x]
In the particular case we are talking about (tranforming g=g_y) a_x is exacly 0 so:
a'_y=gamma^-2*(1+v_xV/c^2)^-2*a_y (no approximation needed)
or, since a_y=g:
g'=gamma^-2*(1+v_xV/c^2)^-2*g
So, the only approximation that we need to do is , for small v_x (which is the case):
1+v_xV/c^2=1
Rpenner,
I would still like some help in solving the equation of the pendulum in SR. I arrived (again) to an ODE that seems not to have a closed solution. Actually, we don't need the solution, we only need to show that it is peridical and that the period depends linearly on gamma. Any ideas¿
I think in order to avoid confusion you first need to properly define your terms. I suggest you use v for the speed of the observer relative to the plane of the pendulum and u for the velocity of the mass at the end of the pendulum. Also (!), SR is easiest to work with in Cartesian coordinates and so I recommend you use them! In the ODE you came up with, you use x to represent the arc length which is not Cartesian! and so there is bound to be some confusion when computing the solution. Instead, I recommend you use (x,y,z) to represent the coordinates of the mass at the end of the pendulum and {i,j,k} to represent the corresponding unit vectors.
Here is my solution:
Let gravity point in the negative z-direction, with the pendulum hanging from the origin and swinging in the y-z plane. And let the observer be travelling in the positive x-direction with speed v. And let us denote the angle that the pendulum makes with the negative z-axis by θ.
(1) Compute the net force on the pendulum's mass in this moving frame:
F'_net= F'_gravity + F'-string= [-m'g'k] + [m'g'cos(θ')k-m'g'sin(θ')j]
= [-m'g'k] + [m'g'(-z'/L')k-m'g'(y'/L)'j]
= -m'g'[(y'/L')j + (1+(z'/L'))k]
Where the (')s denote that the corresponding value is computed in the moving frame.
Since the observer is travelling perpendicular to the plane of the pendulum, there will be no length contraction for the string and so L'=L.
Well known SR transformation laws tell us that m'=γm_0 and g'=β^-2 g where γ'=(1-u'^2/c^2)^-0.5 and β=(1-v^2/c^2)^-0.5
And so;
F'_net= -(γ'/β^2)m_0 g[(y'/L)j + (1+(z'/L))k]
(2) Apply Newton's second law to obtain the equation of motion:
F'_net=dp'/dt'=(d/dt')(γ' m_0 u')=m_0 [u'dγ'/dt' + γ'du'/dt']
Now, for small θ (and subsequently small θ') and small m_0, u_y and u_z (and u'_y and u'_z) are negligible compared with v (u'_x=v and u_x=0) and c. Therefore γ≈1 and γ'≈β and the time derivatives of both are negligible (A better approximation would result in an ODE that cannot be solved analytically, but if solved numerically would lead to the same result).
And so we have:
F'_net=dp'/dt'=(d/dt')(γ' m_0 u')=βm_0 du'/dt']
=βm_0[ du'_x/dt'i +du'_y/dt'j +du'_z/dt'k]
SR transformation laws for accelerations give du'_x/dt'=β^-3 (1-u_x v/c^2)^-3 du_x/dt= 0 since u_x=0 in the rest frame (frame at rest relative to the pendulum's anchor point)
=> F'_net=βm_0[ du'_y/dt'j +du'_z/dt'k]
=βm_0[ d^2y'/dt'^2j +d^2z'/dt'^2k]= -(γ'/β^2)m_0 g[(y'/L)j + (1+(z'/L))k]
Concentrating on the y'-component:
βm_0 d^2y'/dt'^2 = (γ'/β^2)m_0 g[(y'/L)=(1/β)m_0 g[(y'/L)
=> d^2y'/dt'^2 = (g/Lβ^2) y'
=> y'= y'_max cos[(Lβ^2/g)^0.5 t'] where y'=y'_max at t'=0
=> T' =2πβ (L/g)^0.5 which agrees with Mr. homm's solution
QUOTE (gabba gabba hey+Jul 13 2008, 07:53 PM)
Well known SR transformation laws tell us that ... g'=β^-2 g
You don't know that, I just showed you that this may be an acceptable approximation.
Anyway, I am way past the point of showing that the period is frame-invariant, I am at the point at trying to find a covariant form of the different empiracal laws shown in my previous post. Thank you for the attempt, though.
QUOTE
F'_net=dp'/dt'=(d/dt')(γ' m_0 u')=βm_0 du'/dt'
The above is obviously incorrect since you left out the term dβ/dt'm_0 u'. (You gave a justification for this but it is not acceptable).
The reason is that β in p=βmu is a function of u. β=1/sqrt(1-(u/c)^2). I can see that you came up with a series of approximations in order to justify your point but this defeats the whole point, you are bringing the problem back in the classical domain u<<c. This is not interesting.
If you redo the computations correctly you end up with a much nastier ODE, the one I was asking help with. You say correctly that the resulting ODE has no symbolic solution, we already know that since it showed up in the EM problem that started this thread. Nevertheless, mr_homm and I defeated that problem (without resorting to hacks). I am asking for help in order to solve this new problem.
QUOTE (Trout+Jul 14 2008, 03:59 AM)
You don't know that, I just showed you that this may be an acceptable approximation.
Anyway, I am way past the point of showing that the period is frame-invariant, I am at the point at trying to find a covariant form of the different empiracal laws shown in my previous post. Thank you for the attempt, though.
The above is obviously incorrect since you left out the term dβ/dt'm_0 u'. (You gave a justification for this but it is not acceptable).
The reason is that β in p=βmu is a function of u. β=1/sqrt(1-(u/c)^2). I can see that you came up with a series of approximations in order to justify your point but this defeats the whole point, you are bringing the problem back in the classical domain u<<c. This is not interesting.
If you redo the computations correctly you end up with a much nastier ODE, the one I was asking help with. You say correctly that the resulting ODE has no symbolic solution, we already know that since it showed up in the EM problem that started this thread. Nevertheless, mr_homm and I defeated that problem (without resorting to hacks). I am asking for help in order to solve this new problem.
β is NOT a function of u! β is a function of v, which is constant and so dβ/dt'= dβ/dt = 0! The only approximation I used for my solution was that γ'=(1-u'^2/c^2)^-0.5 ≈(1-v^2/c^2)^-0.5=β. In most cases, this is a valid approximation since u'^2=(u'_x)^2 + (u'_y)^2 +(u'_z)^2 =v^2 + (u'_y)^2 +(u'_z)^2. When g is close to its value on earth and m_0 is significantly less than the mass of a star or black-hole, u_y and u_z are much smaller than v. Since the moving frame S' is in motion perpendicular to the plane of the pendulum, the transformation laws for u_y and u_z are simple: u'_y= β^-1 u_y and u'_z = β^-1 u_x, and therefore if u_x and u_y are much smaller than c, so are u'_y and u'_z. And so; u^2/c^2= v^2/c^2+(u'_y)^2/c^2+(u'_z)^2/c^2≈v^2/c^2 because vis presumably close to c and each of the other two terms are <<<<1.
Also, I do know that g'=β^-2g (!). This is not an approximation, it is exact (at least in this case where g is assumed to be constant)! To calculate it I used the following transformation law for accelerations for a body moving with velocity u in frame S and velocity u' in S' where S' is moving at speed v along the positive x-axis relative to S: d(u'_z)/dt'= β^-2 (1-u_x v/c^2)^-2 d(u_z)/dt - vu_z c^-2 β^-2 (1-u_x v/c^2)^-2 d(u_x)/dt. In order to use this transformation law to find g', we need only examine what happens to a particle in free-fall in S:
d(u_z)/dt= g and d(u_x)/dt=d(u_y)/dt=0 for a free-falling particle in S
:. g'=d(u'_z)/dt'= β^-2 (1-u_x v/c^2)^-2 *(g) - vu_z c^-2 β^-2 (1-u_x v/c^2)^-2 *(0)
= β^-2 (1-u_x v/c^2)^-2 *(g) = β^-2 (1-(0)*v/c^2)^-2 *(g)=β^-2 g
Now, if you want to examine the regime where u_y and u_z are not small compared to c, the approximation γ'≈β is of course no longer valid, however, neither is the original solution T=2π(l/g)^0.5 (!!!), for even in Frame S where u_x=0, you end up with an ugly ODE! T'=βT will still be correct but you will need to solve either the ODE that arises in S or the ODE that arises in S' to find T or T' I am sure you can find a few papers that deal with the solution to the ODE in S, but I suspect that all solutions are numerical and not analytical. If I get a chance later, I will try to find a paper on the relativistic pendulum and see if there are any analytical solutions.
Anyway, I am way past the point of showing that the period is frame-invariant, I am at the point at trying to find a covariant form of the different empiracal laws shown in my previous post. Thank you for the attempt, though.
The above is obviously incorrect since you left out the term dβ/dt'm_0 u'. (You gave a justification for this but it is not acceptable).
The reason is that β in p=βmu is a function of u. β=1/sqrt(1-(u/c)^2). I can see that you came up with a series of approximations in order to justify your point but this defeats the whole point, you are bringing the problem back in the classical domain u<<c. This is not interesting.
If you redo the computations correctly you end up with a much nastier ODE, the one I was asking help with. You say correctly that the resulting ODE has no symbolic solution, we already know that since it showed up in the EM problem that started this thread. Nevertheless, mr_homm and I defeated that problem (without resorting to hacks). I am asking for help in order to solve this new problem.
β is NOT a function of u! β is a function of v, which is constant and so dβ/dt'= dβ/dt = 0! The only approximation I used for my solution was that γ'=(1-u'^2/c^2)^-0.5 ≈(1-v^2/c^2)^-0.5=β. In most cases, this is a valid approximation since u'^2=(u'_x)^2 + (u'_y)^2 +(u'_z)^2 =v^2 + (u'_y)^2 +(u'_z)^2. When g is close to its value on earth and m_0 is significantly less than the mass of a star or black-hole, u_y and u_z are much smaller than v. Since the moving frame S' is in motion perpendicular to the plane of the pendulum, the transformation laws for u_y and u_z are simple: u'_y= β^-1 u_y and u'_z = β^-1 u_x, and therefore if u_x and u_y are much smaller than c, so are u'_y and u'_z. And so; u^2/c^2= v^2/c^2+(u'_y)^2/c^2+(u'_z)^2/c^2≈v^2/c^2 because vis presumably close to c and each of the other two terms are <<<<1.
Also, I do know that g'=β^-2g (!). This is not an approximation, it is exact (at least in this case where g is assumed to be constant)! To calculate it I used the following transformation law for accelerations for a body moving with velocity u in frame S and velocity u' in S' where S' is moving at speed v along the positive x-axis relative to S: d(u'_z)/dt'= β^-2 (1-u_x v/c^2)^-2 d(u_z)/dt - vu_z c^-2 β^-2 (1-u_x v/c^2)^-2 d(u_x)/dt. In order to use this transformation law to find g', we need only examine what happens to a particle in free-fall in S:
d(u_z)/dt= g and d(u_x)/dt=d(u_y)/dt=0 for a free-falling particle in S
:. g'=d(u'_z)/dt'= β^-2 (1-u_x v/c^2)^-2 *(g) - vu_z c^-2 β^-2 (1-u_x v/c^2)^-2 *(0)
= β^-2 (1-u_x v/c^2)^-2 *(g) = β^-2 (1-(0)*v/c^2)^-2 *(g)=β^-2 g
Now, if you want to examine the regime where u_y and u_z are not small compared to c, the approximation γ'≈β is of course no longer valid, however, neither is the original solution T=2π(l/g)^0.5 (!!!), for even in Frame S where u_x=0, you end up with an ugly ODE! T'=βT will still be correct but you will need to solve either the ODE that arises in S or the ODE that arises in S' to find T or T' I am sure you can find a few papers that deal with the solution to the ODE in S, but I suspect that all solutions are numerical and not analytical. If I get a chance later, I will try to find a paper on the relativistic pendulum and see if there are any analytical solutions.
QUOTE (gabba gabba hey+Jul 13 2008, 07:53 PM)
F'_net=dp'/dt'=(d/dt')(γ' m_0 u')=m_0 [u'dγ'/dt' + γ'du'/dt']
Correct
QUOTE
Now, for small θ (and subsequently small θ') and small m_0, u_y and u_z (and u'_y and u'_z) are negligible compared with v (u'_x=v and u_x=0) and c.
No, this is exactly what I called you on, the whole problem is that you can't assume u to be so small as to u_x=0. You are subconciously reducing the problem to its Newtonian limits.
QUOTE (->
| QUOTE |
| Now, for small θ (and subsequently small θ') and small m_0, u_y and u_z (and u'_y and u'_z) are negligible compared with v (u'_x=v and u_x=0) and c. |
No, this is exactly what I called you on, the whole problem is that you can't assume u to be so small as to u_x=0. You are subconciously reducing the problem to its Newtonian limits.
Therefore γ≈1 and γ'≈β and the time derivatives of both are negligible
No, again. Please look at the sample solution for the EM problem. You can't reduce the problem to γ≈1 and γ'≈β because in effect this reduces it to the Newtonian limit.
Once again, I got the solution to the point where there are no approximations, unfortunately the resulting ODE has no symbolic solution.
I am past this point, I ma looking for a covariant formulation of the pendulum law.
QUOTE (gabba gabba hey+Jul 14 2008, 09:02 AM)
Now, if you want to examine the regime where u_y and u_z are not small compared to c, the approximation γ'≈β is of course no longer valid, however, neither is the original solution T=2π(l/g)^0.5 (!!!), for even in Frame S where u_x=0, you end up with an ugly ODE! T'=βT will still be correct but you will need to solve either the ODE that arises in S or the ODE that arises in S' to find T or T' I am sure you can find a few papers that deal with the solution to the ODE in S, but I suspect that all solutions are numerical and not analytical. If I get a chance later, I will try to find a paper on the relativistic pendulum and see if there are any analytical solutions.
Thank you, this is exactly what we are looking for.
More generally, a covariant reformulation of the pendulum law, Hooke law, torsion pendulum law (see my earlier post here).
We cannot use the current formulations because they produce frame-variant results.
QUOTE (Trout+Jul 14 2008, 03:04 PM)
Thank you, this is exactly what we are looking for.
More generally, a covariant reformulation of the pendulum law, Hooke law, torsion pendulum law (see my earlier post here).
We cannot use the current formulations because they produce frame-variant results.
Okay, I'll work on this a little later and see if I can find a solution for you. My effort will utilize the same notation as my previous posts, so I think we should try to clear up some obvious misunderstandings you have with my notation.....
More generally, a covariant reformulation of the pendulum law, Hooke law, torsion pendulum law (see my earlier post here).
We cannot use the current formulations because they produce frame-variant results.
Okay, I'll work on this a little later and see if I can find a solution for you. My effort will utilize the same notation as my previous posts, so I think we should try to clear up some obvious misunderstandings you have with my notation.....
QUOTE (gabba gabba hey+Jul 14 2008, 03:16 PM)
Okay, I'll work on this a little later and see if I can find a solution for you. My effort will utilize the same notation as my previous posts, so I think we should try to clear up some obvious misunderstandings you have with my notation.....
Thank you.
I don't think I have any misunderstanding with your notation, I think that you needed to understand the problem statement before you plunged into the approximations that reduce the problem to the Newtonian limit. Try to stay clear of u<<c and we''ll be on the same page. Don't get me wrong, I appreciate interacting with you, there are very few non-cranks, true physicists in this forum. You clearly know what you are doing.
As an aside, here is the class of problem that got us started on this quest.
Thank you.
I don't think I have any misunderstanding with your notation, I think that you needed to understand the problem statement before you plunged into the approximations that reduce the problem to the Newtonian limit. Try to stay clear of u<<c and we''ll be on the same page. Don't get me wrong, I appreciate interacting with you, there are very few non-cranks, true physicists in this forum. You clearly know what you are doing.
As an aside, here is the class of problem that got us started on this quest.
QUOTE (Trout+Jul 14 2008, 02:58 PM)
Correct
No, this is exactly what I called you on, the whole problem is that you can't assume u to be so small as to u_x=0. You are subconciously reducing the problem to its Newtonian limits.
No, again. Please look at the sample solution for the EM problem. You can't reduce the problem to γ≈1 and γ'≈β because in effect this reduces it to the Newtonian limit.
Once again, I got the solution to the point where there are no approximations, unfortunately the resulting ODE has no symbolic solution.
I am past this point, I ma looking for a covariant formulation of the pendulum law.
These are the misunderstandings my last post referred to:
First, I am not ASSUMING u_x is small, I am DEFINING u_x to be zero! u_x is defined as the x-component of the velocity of the mass at in the end of the pendulum IN FRAME S. Where frame S is defined to be at rest relative to the anchor point of the pendulum. The pendulum swings in the y-z plane and so u_y and u_z are in general, non-zero. But! the pendulum is not moving inthe x-direction in this frame, and so u_x=0 by definition.
My, ODE however, was computed in the S' frame, which is moving at constant speed v along the positive(shared) x-axis (this is the frame of reference of the observer moving at speed v towards/away from the plane of the pendulum). For this reason, u'_x= -v , again by definition.
The only assumptions I made were regarding the y- and z-components of the velocity in each frame. For small theta, these velocity components are small (this is the newtonian limit) compared to c.
I will now work on the relativistic case, where u_y and u_z are NOT small relative to c. The solution for the period will still be frame variant according to T'=βT, but T and T' will likely be more complicated than the Newtonian solutions I calculated previously. This is as it should be though, because any time period computed in two inertial reference frames are in general, frame variant.
While I am working on the non-Newtonian solution, feel free to post any problems you might still have with my Newtonian solution.
No, this is exactly what I called you on, the whole problem is that you can't assume u to be so small as to u_x=0. You are subconciously reducing the problem to its Newtonian limits.
No, again. Please look at the sample solution for the EM problem. You can't reduce the problem to γ≈1 and γ'≈β because in effect this reduces it to the Newtonian limit.
Once again, I got the solution to the point where there are no approximations, unfortunately the resulting ODE has no symbolic solution.
I am past this point, I ma looking for a covariant formulation of the pendulum law.
These are the misunderstandings my last post referred to:
First, I am not ASSUMING u_x is small, I am DEFINING u_x to be zero! u_x is defined as the x-component of the velocity of the mass at in the end of the pendulum IN FRAME S. Where frame S is defined to be at rest relative to the anchor point of the pendulum. The pendulum swings in the y-z plane and so u_y and u_z are in general, non-zero. But! the pendulum is not moving inthe x-direction in this frame, and so u_x=0 by definition.
My, ODE however, was computed in the S' frame, which is moving at constant speed v along the positive(shared) x-axis (this is the frame of reference of the observer moving at speed v towards/away from the plane of the pendulum). For this reason, u'_x= -v , again by definition.
The only assumptions I made were regarding the y- and z-components of the velocity in each frame. For small theta, these velocity components are small (this is the newtonian limit) compared to c.
I will now work on the relativistic case, where u_y and u_z are NOT small relative to c. The solution for the period will still be frame variant according to T'=βT, but T and T' will likely be more complicated than the Newtonian solutions I calculated previously. This is as it should be though, because any time period computed in two inertial reference frames are in general, frame variant.
While I am working on the non-Newtonian solution, feel free to post any problems you might still have with my Newtonian solution.
QUOTE (Trout+Jul 14 2008, 03:28 PM)
Thank you.
I don't think I have any misunderstanding with your notation, I think that you needed to understand the problem statement before you plunged into the approximations that reduce the problem to the Newtonian limit. Try to stay clear of u<<c and we''ll be on the same page. Don't get me wrong, I appreciate interacting with you, there are very few non-cranks, true physicists in this forum. You clearly know what you are doing.
As an aside, here is the class of problem that got us started on this quest.
When I first tried downloading the E-M problem last week, it was no longer hosted on the website. Now that you have provided a working link, I see what the original question was. The solution you came up with is indeed analytical and a nice demonstration of how choosing the right reference frame can make a problem easier to solve. Well done to you and Mr.Homm on that one!
As, for the pendulum problem, I thought that you were simply looking for T' in the case where T is given by the Newtonian solution, now that We've cleared up that misunderstanding I can begin to answer the question you had intended....
I don't think I have any misunderstanding with your notation, I think that you needed to understand the problem statement before you plunged into the approximations that reduce the problem to the Newtonian limit. Try to stay clear of u<<c and we''ll be on the same page. Don't get me wrong, I appreciate interacting with you, there are very few non-cranks, true physicists in this forum. You clearly know what you are doing.
As an aside, here is the class of problem that got us started on this quest.
When I first tried downloading the E-M problem last week, it was no longer hosted on the website. Now that you have provided a working link, I see what the original question was. The solution you came up with is indeed analytical and a nice demonstration of how choosing the right reference frame can make a problem easier to solve. Well done to you and Mr.Homm on that one!
As, for the pendulum problem, I thought that you were simply looking for T' in the case where T is given by the Newtonian solution, now that We've cleared up that misunderstanding I can begin to answer the question you had intended....
I found a numerical solution here: relativistiic pendulum
I don't think there is an analytical solution, but I will post my version of the ODE in a little while.
I don't think there is an analytical solution, but I will post my version of the ODE in a little while.
QUOTE (gabba gabba hey+Jul 14 2008, 04:14 PM)
I found a numerical solution here: relativistiic pendulum
I don't think there is an analytical solution, but I will post my version of the ODE in a little while.
This is a good paper. Look at his eq(9) . The left hand member for this type of problems is always \gamma(u)*du/dt.This is because they all start with d(\gamma*u)/dt. I have already shown that this expression is frame invariant. So far, so good.
Unfortunately, for all the examples I gave (Hooke, standard and torsion pendulum) the right hand side is NOT frame invariant. This means that we can't get covariant relativistic formulation of these empirical "laws" by starting from their Newtonian expression. So, the "program" of this thread has been finding a covariant relativistic formulation to replace the original Newtonian one.
In a sense, this is the same as the reformulation of the impulse as \gamma*m_0*v and the Lagrangian as -m_0c^2/gamma instead of -mv^2/2.
Thank you for your help, I truly appreciate your participation.
I don't think there is an analytical solution, but I will post my version of the ODE in a little while.
This is a good paper. Look at his eq(9) . The left hand member for this type of problems is always \gamma(u)*du/dt.This is because they all start with d(\gamma*u)/dt. I have already shown that this expression is frame invariant. So far, so good.
Unfortunately, for all the examples I gave (Hooke, standard and torsion pendulum) the right hand side is NOT frame invariant. This means that we can't get covariant relativistic formulation of these empirical "laws" by starting from their Newtonian expression. So, the "program" of this thread has been finding a covariant relativistic formulation to replace the original Newtonian one.
In a sense, this is the same as the reformulation of the impulse as \gamma*m_0*v and the Lagrangian as -m_0c^2/gamma instead of -mv^2/2.
Thank you for your help, I truly appreciate your participation.
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