To add comments or start new threads please go to the full version of: Limits On Acceleration?
PhysForum Science, Physics and Technology Discussion Forums > Relativity, Quantum Mechanics and New Theories > Relativity, Quantum Mechanics, New Theories

Shanester1979
In essence it seems that there may be some types of limits on the acceleration of an object.

We would like to think that there are none.

But the argument is more obvious than day, and I don't know why this question only recently occured to me.

There are thought to be universal speed limits (i.e. the speed of light)

But, if there is a universal speed limit, then perhaps there must be some constraints on acceleration.

Quite obviously any given acceleration cannot be maintained indefinately. If an object were observed accelerating at 3.0*10^8m/s^2, then it the observation could only be for less than a second.

If acceleration were sufficiently great, the time over which such an acceleration could be observed would have to become increasingly small.

Are there practical limits on the 'smallness' of the time interval over which a measurement can be made?

Trout
QUOTE (Shanester1979+Jul 24 2008, 10:44 PM)

But, if there is a universal speed limit, then perhaps there must be some constraints on acceleration.


Not a correct reasoning, it is easy to prove that a constantly accelerated object will never attain the speed of light, no matter the value of the acceleration. You need to study a chapter called "hyperbolic motion" in SR. If you did that , you will learn that :

v(t)=at/sqrt(1+(at/c^2))

where
t=coordinate time
a=acceleration (can be any FINITE value)
c=speed of light

If you calculate the limit for t->infinity you get v=c.
In fact, v=c is the horizontal asymptote of the curve v=v(t)
Shanester1979
QUOTE (Trout+Jul 24 2008, 10:53 PM)
Not a correct reasoning, it is easy to prove that a constantly accelerated object will never attain the speed of light. You need to study a chapter called "hyperbolic motion" in SR. If you did that , you will learn that :

v(t)=at/sqrt(1+(at/c^2))

where
t=coordinate time
a=acceleration (can be any FINITE value)
c=speed of light

If you calculate the limit fot t->infinity you get v=c.
In fact, v=c is the horizontal asymptote of the curve v=v(t)

there may be some differences between 'proper acceleration', and acceleration here.

a given proper accerleration can be maintained indefinately. Put yourself as the accelerating object. You would agree that your acceleration could be constant indefinately. The proper acceleration would be the acceleration I observed of you from my inertial frame ( though yours in not) if I were to adapt my time to the time measured by your clock. And likewise, this is possible because proper speed may be any value.

I, however cannot observe by my measurements of time, and space, an indefinate increase in speed by you. my measurement of accleration is dv/dt. Proper acceleration , which I'll call 'alpha' is gamma*dv/dt.

I believe your formula should read

v(t)='alpha'*t/sqrt(1-(v(t)^2/c^2), with alpa equivalent to dv/d'tau'

the above collapse to
v= a*t, for a constant aceleration

in other words dv=a(t)*dt, if accleration varies with time.



Similarily, you may notice that your formula has a flaw, since the units in your sqrt term are not right.
Trout
QUOTE (Shanester1979+Jul 24 2008, 11:19 PM)
there may be some differences between 'proper acceleration', and  acceleration here.

a given proper accerleration can be maintained indefinately.  Put yourself as the accelerating object.  You would agree that your acceleration could be constant indefinately.  The proper acceleration would be the acceleration I observed of you from my inertial frame ( though yours in not) if I were to adapt my time to the time measured by your clock.  And likewise, this is possible because proper speed may be any value.

I, however cannot observe by my measurements of time, and space, an indefinate increase in speed by you.  my measurement of accleration is dv/dt.  Proper acceleration , which I'll call 'alpha' is gamma*dv/dt.

I believe your formula should read

v(t)='alpha'*t/sqrt(1-(v(t)^2/c^2), with alpa equivalent to dv/d'tau'

the above collapse to
v= a*t, for a constant aceleration

in other words dv=a(t)*dt, if accleration varies with time.



Similarily, you may notice that your formula has a flaw, since the units in your sqrt term are not right.

There was a misplaced parens:

v(t)=at/sqrt(1+(at/c)^2)

The rest of my post is correct. I suggest that you read on hyperbolic motion.
Shanester1979
QUOTE (Trout+Jul 25 2008, 12:23 AM)
There was a misplaced parens:

v(t)=at/sqrt(1+(at/c)^2)

The rest of my post is correct. I suggest that you read on hyperbolic motion.

Well, I'll certainly give it a look.

I'm still concerned that there may be some confusion between 'proper acceleration', and 'acceleration' here.

Actually in what I posted there was in careless error.


What I should have stated was that, I thought perhaps your formula should read

v(t)='alpha'*t/squrt(1+('alpha't/c)^2)^3, more correctly


with 'alpha' = 'gamma'^3*a

a=acceleration

'alpha'=proper acceleration=dv/d'tau' , with tau the proper time of the accelerating object.

you'll see ofcourse that

v(t)*squrt(1+('alpha'*t/c)^2)^3='alpha'*t, with 'alpha'=(gamma^3)*a

ofcourse squrt(1+(alpha*t/c)^2)=squrt(1-v(t)^2/c^2)=gamma, you can check this if you wish.

The end result is that

v=a*t, for constant motion

and dv(t)=a(t)*dt

This is exactly compatable with what you posted if you replace what you state as 'a' with what I believe should be 'alpha.'


Is that wrong? ...As I said I double check the items to which you refer, because I would like to see for sure that there is not a mistake in the above, but, I am pretty
sure it is generally correct. And if it is not, it won't be the first time I've been wrong.

However, what I refer to as 'a(t)' may certainly not be maintained indefineatly if there are limits on 'v(t)'.


Anyway thanks, I appreciate your comment.
Trout
QUOTE (Shanester1979+Jul 25 2008, 01:32 AM)
Well, I'll certainly give it a look.

I'm still concerned that there may be some confusion between 'proper acceleration', and  'acceleration' here.

Actually in what I posted there was in careless error.


What I should have stated was that, I thought perhaps your formula should read

v(t)='alpha'*t/squrt(1+('alpha't/c)^2)^3,

No , it should't. Do you want to learn or do you want to argue?
Shanester1979
QUOTE (Trout+Jul 25 2008, 02:30 AM)
No , it should't. Do you want to lear or do you want to argue?

well, I'd like to learn exactly what you are trying to say. I'd also like to argue because that can be a process of learning.


However, one will definately, in this reference fram never, see anything accelerating at any rate if it has already reached light's speed.


Your obviously incorret that the quantity dv/dt could be defined in such a way that it is other than zero for an object traveling at c.


What YOU are talking about is du/dt, an entirely different concept.
Trout
QUOTE (Shanester1979+Jul 25 2008, 02:38 AM)



However, one will definately, in this reference fram never,




1. Which "this frame" are you talking about?


QUOTE
see anything accelerating at any rate if it has already reached light's speed. 


2. Who has told you about any "object having already reached light's speed"?Didn't I show you that v(t) has c as a horizontal asymptote?
Don't you know that reaching c is impossible? (I gave a different proof elswhere in this forum, based on the fact that the mechanical work necessary to do so would be infinite, feel free to search for it).

QUOTE (->
QUOTE
see anything accelerating at any rate if it has already reached light's speed. 


2. Who has told you about any "object having already reached light's speed"?Didn't I show you that v(t) has c as a horizontal asymptote?
Don't you know that reaching c is impossible? (I gave a different proof elswhere in this forum, based on the fact that the mechanical work necessary to do so would be infinite, feel free to search for it).

Your obviously incorret that the quantity dv/dt could be defined in such a way that it is other than zero for an object traveling at c.


3. Again, who's talking about any "object traveling at c"? I thought that I just showed you that c is unattainable. Are you even reading what I post?

QUOTE


What YOU are talking about is du/dt, an entirely different concept.


4. What is u? You have not defined it and you are pulling it out of nowhere, like you pulled the "alpha" formula from your butt in the earlier post.
Shanester1979
QUOTE (Trout+Jul 25 2008, 02:30 AM)
No , it should't. Do you want to learn or do you want to argue?

And apparently, your lack of willinginss to acually THINK must prevent you from formulating an intelligent argument. I liked your other comments, some were even insightful, but what the HELL dude! Tell me why you think I'm wrong, (in your OWN words that is). Good lord! Are you capable!?

I actually portrayed you as an intelligent thinker before. LOL
Trout
QUOTE (Shanester1979+Jul 25 2008, 02:54 AM)
And apparently, your lack of willinginss to acually THINK must prevent you from formulating an intelligent argument. I liked your other comments, some were even insightful, but what the HELL dude!  Tell me why you think I'm wrong, (in your OWN words that is).  Good lord!  Are you capable!?

I actually portrayed you as an intelligent thinker before. LOL

I think that you wrote a whole bunch of nonsense, so go back and answer the questions I asked you above and we'll see......
Trout
QUOTE (Shanester1979+Jul 24 2008, 10:44 PM)


Quite obviously any given acceleration cannot be maintained indefinately.  If an object were observed accelerating at 3.0*10^8m/s^2, then it the observation could only be for less than a second.


Not obvious at all. Not correct either.
If a=c/sec

then, from the formula:

v=at/sqrt(1+(at/c)^2)

you get v=ct'/sqrt(1+t'^2)<c for any amount of elapsed coordinate time t.
Here t'=t/1sec is unitless.
This is because, expressed in unitless terms, t is always smaller than sqrt(1+t^2).

Most of the times one's "intuition" gets trumped by the proper application of math and physics. biggrin.gif

Now, there are some very interesting papers that talk about the effects of continously accelerating realistic objects (i.e. objects that aren't Born rigid). They will compress until they self-destruct, as also mentioned as an explanation in the Ehrenfest paradox. But somehow I don't think that this was what you were after in the OP.
Shanester1979
QUOTE (Trout+Jul 25 2008, 03:05 AM)
I think that you wrote a whole bunch of nonsense, so go back and answer the questions I asked you above and we'll see......

mu(tau) is the so-called 'proper-speed' of an object.

This quantity is essentially the speed that an 'inertial' frame would measure another object at, according to the time, that that other object would measure, according the first objects's frame of reference, whether 'they' were accelerating or not. That quantity is invariant.

The difference between 'acceration', and 'proper acceleration' is that 'acceleration' is defined as 'dv(t)/dt.' Proper acceleration is defined as du(v)/d'tau'


dt=d'tau'*gamma,... , we should all be aware that gamma=1/squrt(1-v^2/c^2) and that integration of the previous should yield delta(t)=delta(tau)*gamma.


Simply put,

mu(tau)=dx/d'tau'

x is the coordinate measured by the 'inertial' frame of reference ( not the accelerating object)

It is known and evident that dt=d'tau'*'gamma'




or,
d'tau'=dt/gamma


by the definition of proper speed,

mu(tau)=dx/d'tau', however d'tau'=dx/mu(tau) (simple algebra)

Concequently,

The proper acelleration is d(mu)/dtau=d(v(tau))/d(tau)

=d^2(x)/d(tau)^2, with d(tau)=dt/gamma


By definition, d(tau)^2=dt^2/gamma^2


Therefore

alpha=(d^2(x))/dt^2*gamma^2

=a(t)*gamma^3. That is the relation between 'proper acceleration' and 'accelerartion'.

Point and case. Maybe your confusing one's intuition with another's. Still I appreciate your comments, don't take it personally.
Confused2
QUOTE (Trout+)

Now, there are some very interesting papers that talk about the effects of continously accelerating realistic objects (i.e. objects that aren't Born rigid). They will compress until they self-destruct,..


Which is what makes me think Trout is a crank.
Trout
QUOTE (Shanester1979+Jul 25 2008, 05:02 AM)
mu(tau) is the so-called 'proper-speed' of an object.

This quantity is essentially the speed that an 'inertial' frame would measure another object at, according to the time, that that other object would measure, according the first objects's frame of reference, whether 'they' were accelerating or not. That quantity is invariant.








This is not a sentence , it appears that English is not your native language. Do you want to try again?

QUOTE
The difference between 'acceration', and 'proper acceleration' is that 'acceleration' is defined as 'dv(t)/dt.'  Proper acceleration is defined as du(v)/d'tau'


What is du(v) ? Heck, what is u(v)? You keep introducing variables and they make no sense.

I had to snip from there on since you aren't making sense. You didn't not even get to the point of addressing questions 2,3,4.


Let's try again:

v(t)=at/sqrt(1+(at/c)^2)

where
t=coordinate time (NOT "proper time")
a=coordinate acceleration (NOT "proper acceleration")
c=speed of light
v=coordinate speed

If you want to know how this is derived, I can give you a link.

Now, ALL of the above are measured from the frame of reference of one observer. At t=0 , v was 0 and the object was subjected to constant coordinate acceleration a. After ANY value of t, v<c.

If you calculate the limit for t->infinity you get v=c.
In fact, v=c is the horizontal asymptote of the curve v=v(t)

Now, if you want to introduce proper time,speed,acceleration in the problem (instead of coordinate values), the answer to your question can be found here.
Trout
QUOTE (Confused2+Jul 25 2008, 10:14 AM)
QUOTE (Trout+)

Now, there are some very interesting papers that talk about the effects of continously accelerating realistic objects (i.e. objects that aren't Born rigid). They will compress until they self-destruct,..


Which is what makes me think Trout is a crank.

It's ok that you didn't understand my post, no one expected you to.
As to being a crank, we all know what you are, no need to advertise.
SirShanson
Little bit of a sidetrack but please can I just ask, if or how does all these laws of acceleration apply to tachyons? Yes I do realize they are mostly theoretical. Is it simply the reverse as in, is it more of a limit of deceleration or what else? Any thoughts?

Many thanks,

SirS
Shanester1979
QUOTE (Trout+Jul 25 2008, 01:16 PM)
This is not a sentence , it appears that English is not your native language. Do you want to try again?



What is du(v) ? Heck, what is u(v)?  You keep introducing variables and they make no sense.

I had to snip from there on since you aren't making sense. You didn't not even get to the point of addressing questions 2,3,4.


Let's try again:

v(t)=at/sqrt(1+(at/c)^2)

where
t=coordinate time (NOT "proper time")
a=coordinate acceleration (NOT "proper acceleration")
c=speed of light
v=coordinate speed

If you want to know how this is derived, I can give you a link. 

Now, ALL of the above are measured from the frame of reference of one observer. At t=0 , v was 0 and the object was subjected to constant coordinate acceleration a. After ANY value of t, v<c.

If you calculate the limit for t->infinity you get v=c.
In fact, v=c is the horizontal asymptote of the curve v=v(t)

Now, if you want to introduce proper time,speed,acceleration in the problem (instead of coordinate values), the answer to your question can be found here.

Ugh!

If you had actually any significant background in relativity you shouldn't be so confused about the meaning of these variables. They should be pretty obvious to a person who is comfortable with the meaning of differential elements as well. Is this you? Or maybe its not. I don't know. Either way its nothing to feel too worried about.

I don't proclaim to know the level of YOUR understanding, but based on what you have written, I feel there may be some certain deficits in you true comprehension of some of the issues we've been discussing, and this topic is quickly becoming pretty overstated. All the answers are there there if you look for it.



If you don't think my variables make sense fine. Atleast you seem lto ike the link you gave me, so look it over yourself a few more times. I did, and the funny thing is they do , with a slightly different approach(not the most beautiful on in my opinion, a matter of aesthetics) about what I did in my derivation. However, as is not too surprising, the professor who wrote that presentation gave a couple of common mistake misunderstandings starting on page 8. Not a huge deal in this context, but maybe this is where part of your 'confusion' comes from.


In essence, what I refer to as 'mu' he refers to as 'Vapp.' So if you want to know what 'u' ( or mu) means just look at those slides you have, dont worry about my derivation if it plagues you.


In essence, however, what he refers to as 'a' is YOUR acceleration. This is NOT the same as dv/dt. This is why I don't prefer his approach to the matter. If you like it great! But if you should feel the need to quote other peoples equations, you should be aware of what the PHYSICAl INTERPRETATIONS of their variables are.

What I refer to as 'alpha' is what he refers to as 'Aapp'

What he refers to as 'a', I would typically call something like the clock acceration

The BOTTOM LINE is: However you decide to label your variables, 'dv/dt' may not increase indefinitely.

In essence your criticism of what I am saying, as I told you in the first place, is not valid, because the PHYSICAL quantity you are thinking of is 'your aceleration' not the 'acceleration' that I would observe of you. I hope this is clear to you by now, because I am not going to waste anymore time with this idiocy.


Let's get back to the relevant topic:

Are there realistic limits on 'observable acceleration', or dv/dt if you will?

TROUT you seem to be thinking about something else, but perhaps you were not clear on what I was talking about because you think 'english is not my first language.'

Perhaps there are some others that would lilke to comment on 'this topic' if Mr. Big Fish would be so kind as to let us talk about the observed accleration, 'dv/dt', in particular, for which I gave correctly that dv/dt='gamma'^3*du/dt.


And lets not forget the idea of: Is there a practical limit 'smallness' of the time interval that we can realistically make a measurement in? (maybe Quantum Mechanists might want to comment on this.)
Shanester1979
QUOTE (SirShanson+Jul 25 2008, 01:58 PM)
Little bit of a sidetrack but please can I just ask, if or how does all these laws of acceleration apply to tachyons? Yes I do realize they are mostly theoretical. Is it simply the reverse as in, is it more of a limit of deceleration or what else? Any thoughts?

Many thanks,

SirS

Not really my forte. But this is a discussion, so limits on acceleration is only an idea right now, not a fact, or a law, but a just concept to explore.
Confused2
Hi Shanester1979,

Welcome to the forum .. (IMO it's not going through one of it's best phases at the moment)

Trout and I have been through this a few times .. as far as I can make out he's never attempted to explain where 'the hand that crushes' comes from .. but that could be a result of my (very) limited ability.

-C2

Trout
QUOTE (Confused2+Jul 25 2008, 08:41 PM)
Hi Shanester1979,

Welcome to the forum .. (IMO it's not going through one of it's best phases at the moment)

Trout and I have been through this a few times .. as far as I can make out he's never attempted to explain where 'the hand that crushes' comes from .. but that could be a result of my (very) limited ability.

-C2

Not to crackpots, that's true. Never been able to get through you, that much is true. laugh.gif
MjolnirPants
QUOTE (Trout+Jul 24 2008, 10:53 PM (I corrected this equation, since there was an error in it; a misplaced parentheses))

v(t)=at/sqrt(1+(at/c)^2)

where
t=coordinate time
a=acceleration (can be any FINITE value)
c=speed of light

If you calculate the limit for t->infinity you get v=c.
In fact, v=c is the horizontal asymptote of the curve v=v(t)

Shanester,
can you show Trout's conclusion to be wrong by punchin in numbers fer the variables in that equation?
If not, can you show that this equation is not the proper one fer this purpose? (determinin the energy needed to accelerate an object to c.)

If so (to either one o those questions) please do so.
If not, then ya must admit that Trout is correct: While there is no limit to proper acceleration, as you claim, no amount o acceleration can get a massive object to c. Unless, o course, ya somehow eliminate time dilation. wink.gif
Confused2
QUOTE (Trout+)
Never been able to get through you, that much is true.


Yep .. I tend to keep the BS deflector set at MAX after the little problemo with gravitational potential.
Trout
QUOTE (Shanester1979+Jul 25 2008, 01:32 AM)



What I should have stated was that, I thought perhaps your formula should read

v(t)='alpha'*t/squrt(1+('alpha't/c)^2)^3, more correctly



No, it shouldn't. I wish you stopped making up hacky formulas. You either know hyperbolic motion or you don't.
I got tired of trying to wade through your fumbles and I wrote down the formulas for accelerated motion. You can download them from HERE. I fully expect you to continue to argue that you have written the correct thing and that I don't understand SR. I am used to this, I get this from the cranks in this forum on a daily basis.
As always, I wrote the above for the sane people who are truly interested in science. Which way will you lean? The jury is still out there. biggrin.gif
Trout
QUOTE (Confused2+Jul 25 2008, 09:44 PM)
QUOTE (Trout+)
Never been able to get through you, that much is true.


Yep .. I tend to keep the BS deflector set at MAX after the little problemo with gravitational potential.

You mean you are STILL Confused? wink.gif
inQZtive
It might be worth mentioning, that the "cranky self publishing format" that Trout has used to "publish" his "debunking material", has a "no downloads means we delete your file" policy.

"Just say NO to Crank".


He's always spamming these links, to try to keep himself "published".


He's turning out to be be a full-blooded crank.

biggrin.gif



J-
Trout
QUOTE (inQZtive+Jul 25 2008, 10:18 PM)
It might be worth mentioning, that the "cranky self publishing format" that Trout has used to "publish" his "debunking material", has a "no downloads means we delete your file" policy.



You are lying again, savefile.com deletes files if 30 days elapse without a download. I know that you can't do any science, why do you have to be a liar as well? Especially when you tell such stupid lies.

I said that the file I uploaded is for sane people, didn't I? laugh.gif
inQZtive
QUOTE
savefile.com deletes files if 30 days elapse without a download



Precisely.

Crank spammer.


Lying is normal for fraudulent debunkers like yourself. By definition.


Don't worry. Some people think "you're cool".


J-
gabba gabba hey
Trout, Shanester: With respect, I'm not sure either of you FULLY understand relativity, and I think it is best you avoid insulting each-other over misunderstandings and instead try to work together to further your understanding of the topic at hand.

To that end, it seems that you are analyzing the question from two different perspectives:

(1) Trout, you seem to be looking at the observed acceleration of a particle from any inertial reference frame, where the speed of the particle is defined as v(t)...this seems perfectly reasonable to me, although your analysis is sketchy at best.

(2) Shanester, you seem to be trying to analyze the problem in terms of the particles proper acceleration. This again, is an acceptable approach, but it is unnecessary to introduce the concepts of proper time and proper acceleration in this problem.

Also, both of you are using some confusing notation, which probably is not helping to clear up the misunderstandings between you. (i.e. your u, v, t, tau, alpha are not clearly defined)

Here is my attempt at answering the original question as posed by Shanester...

It is true that relativity imposes some constraints on the acceleration of a particle, but the constraint IS NOT in the form of a maximum possible acceleration. The acceleration of a particle, as viewed from any inertial reference frame, can in general, take on ANY finite instantaneous value. That is provided that whatever force (or pseudo-force) is responsible for this acceleration is strong enough to push the accelerating mass to the desired acceleration. However, as a massive object is accelerated, its relativistic mass increases and it requires a stronger force (i.e. more energy) to continue to maintain a specific value for the acceleration. As the particle approaches the speed of light, the relative mass, and subsequently the Energy required to maintain an acceleration, approaches infinity! In this way, the constraints on acceleration become clear: It is impossible to accelerate a particle up to the speed of light in any reference frame (unless you happen to have an infinite source of energy available). Mathematically, you can represent this constrain in many ways, but in the context of SR this is really the only constraint on acceleration.
Confused2
There is possible confusion between (say) a particle accelerated in a particle accelerator 'without limit' and a rocket with an inexhaustible supply of fuel accelerating 'without limit' - I am not sure Trout distinguishes between the two situations .. hence he gives the same 'crushing' reply to both.
Trout
QUOTE (gabba gabba hey+Jul 25 2008, 10:49 PM)
Trout, Shanester: With respect, I'm not sure either of you FULLY understand relativity, and I think it is best you avoid insulting each-other over misunderstandings and instead try to work together to further your understanding of the topic at hand.

To that end, it seems that you are analyzing the question from two different perspectives:

(1) Trout, you seem to be looking at the observed acceleration of a particle from any inertial reference frame, where the speed of the particle is defined as v(t)...this seems perfectly reasonable to me, although your analysis is sketchy at best.


Not really, I got tired of all the BS and I wrote up a complete analysis, from BOTH perspectives and with the correct equations,notations, etc. You can download it from here
If you checked, I also told Shane that he can search for an mechanical work-based analysis that I did earlier that proves that c is unattainable, since it would require an infinite amount of work. Feel free to search thru my earlier posts on the subject.
Trout
QUOTE (Confused2+Jul 25 2008, 10:59 PM)
There is possible confusion between (say) a particle accelerated in a particle accelerator 'without limit' and a rocket with an inexhaustible supply of fuel accelerating 'without limit' - I am not sure Trout distinguishes between the two situations .. hence he gives the same 'crushing' reply to both.

Nah, only a "Confused" would do such a dumb thing. wink.gif
Trout
QUOTE (inQZtive+Jul 25 2008, 10:49 PM)


Lying is normal for fraudulent debunkers like yourself. By definition.

You need to look in the mirror more often. laugh.gif
Confused2
QUOTE (Trout+)
Nah, only a "Confused" would do such a dumb thing.


Indeed? Oh really? You surprise me.
inQZtive
Trout Posted on Today at 11:10 PM
QUOTE
You need to look in the mirror more often.



ohmy.gif


You need to look at what you post (BS) more often.


The number of people calling you on that is growing daily.


laugh.gif
MjolnirPants
QUOTE (inQZtive+Jul 25 2008, 11:21 PM)
You need to look at what you post (BS) more often.


The number of people calling you on that is growing daily.

Ya'll need to start showin him to be full o BS before tellin him he's full o BS...
Trout
QUOTE (MjolnirPants+Jul 25 2008, 11:35 PM)
Ya'll need to start showin him to be full o BS before tellin him he's full o BS...

biggrin.gif
gabba gabba hey
QUOTE (Trout+Jul 25 2008, 11:01 PM)
Not really, I got tired of all the BS and I wrote up a complete analysis, from BOTH perspectives and with the correct equations,notations, etc. You can download it from here
If you checked, I also told Shane that he can search for an mechanical work-based analysis that I did earlier that proves that c is unattainable, since it would require an infinite amount of work. Feel free to search thru my earlier posts on the subject.

I just went over your "complete" analysis and I find it to be neither "complete" or correct....

In part (1) of your analysis, you implicitly define a_c=d v_c/dt.

Then, in part (2) you analyze hyperbolic motion starting with x^2-c^2t^2=d^2 where d=c^2/a_c, WITHOUT stating where this comes from OR how it is applicable to the problem at hand.

You then go on to derive the equation v_c= a_ct/sqrt(1+a_c^2t^2/c^2). This derivation is only correct if you assume (or define) a_c is constant (which you did not explicitly do). If a_c is constant than your derivation is fine, but then a_c cannot be dv_c/dt (as you implied it was in part 1) for dv_c/dt=a_ct*(1+a_c^2t^2/c^2)^-1.5=a_c is only true for a_c=0!!!

Since part (3) relies both on part (1) and part(2), part(3) is also incorrect!
Confused2
Deleted .. the post would have seemed sarcastic after the previous one .. there was no such intent.
magpies
Seems to me the amount of acceleration is hugely dependant on the amount of pressure or leway one can get. I'm just saying if you are to physicaly weak to light the match it will never get lit by you. Perhaps humans just dont have the pressure in us to go past the speed of light or mby we do but I think the interesting question would be did the universe itself ever have the pressure to pass the speed limit at any point. Be that when it was in big bang mode or in contraction mode or some other mode im not thinking of.
inQZtive
MjolnirPants Posted on Today at 11:35 PM
QUOTE
Ya'll need to start showin him to be full o BS before tellin him he's full o BS... 


You bet. I already did, in the "Is there a Medium in Space" thread.

Trials only happen once. They're duly recorded, just like threads. If you'd like to read about it (determine a proof for yourself), have at 'er!

That's my way of saying, "please don't ask me to prove it again". There were several "witnesses" (like a jury). They agreed that this was the case. There were others, who disagreed of course. Comments on this past case are welcome.

This is the same reason why the "Milo Woolf" thread has been trashed.. by people asking for another "trial". It doesn't work that way. It just has served to keep the focus off Physics, for a little while longer.

He is the one that was making claims. They were shown to be "premature" at best, downright deception, at worst. In the process, he was caught in 3 different lies. Those are heavy charges, I know. But they've been shown to be true..

IF Trout actually knows what he says he knows.

I tend to believe his level of education, so that leaves purposeful deception, aka "lies", as the best description of his actions, IMO (and several others). This is in order to appear "right", and "really smart", at all costs; again IMO.

IF he didn't know, this would be an entirely different ball game.


One last, minor point: I don't think he's FULL (100%) of BS. He does have some neurons firing in there, and I've seen him be right a couple of times.



J-


ps - let's not further trash this one as well!
Shanester1979
QUOTE (Trout+Jul 25 2008, 10:06 PM)
No, it shouldn't. I wish you stopped making up hacky formulas. You either know hyperbolic motion or you don't.
I got tired of trying to wade through your fumbles and I wrote down the formulas for accelerated motion. You can download them from HERE. I fully expect you to continue to argue that you have written the correct thing and that I don't understand SR. I am used to this, I get this from the cranks in this forum on a daily basis.
As always, I wrote the above for the sane people who are truly interested in science. Which way will you lean? The jury is still out there.  biggrin.gif

I think your in ability to comprehend the difference between the observed dv/dt, and the accleration 'felt' by an by the accelerating object speaks for itself, right now. Its a rather simple distinction.

OH, and by the way, there will never be any 'crushing' of an object.



A 'sane' person actually interested in science you wouldn't feel such a compulsion to dismiss everything that were not within their outlook, that they did not first try to comprehend.


This is why I read your arguments, digested them, showed you why what you were talking about was not what I am talking about, and asked you to look at your own slides, and digest the difference between, these topics:


Vapp and V, Aapp, and dV/dt.

If you did so you would get that dv/dt, even by your equations, cannot be continously maintained...No insults intended.

My formulas are not hacky. Try reading a graduate level mechanics, electrodynamics, or relativity book, as these are more rigourous sources on the topic, over the internet, or slides from a freshman or sophomore physics course that you yourself may not be have ever taken. If you have no such books as I suggested, I suggest finding a Library, or perhaps amazon to order, over bickering on the internet.

If you don't like what I have to say quit making a stink over it. I don't need your approval. Go find some more rigorous books you can get through, and try not to argue with the author too much. Those Weinberg's and Hawkings and such usually tend to know what they are speaking about more than Sophomore college course profs in general, but there are always acceptions to the rule.
MjolnirPants
QUOTE (inQZtive+Jul 26 2008, 12:03 AM)
You bet. I already did, in the "Is there a Medium in Space" thread.

Ya mean where C2 shows that Wollf's 'solutions' satisfy the equation in polar coordinates? Even though Wollf claims they do so spherically, an Trout jes keeps sayin "they're not the right solutions"? wink.gif
Looks to me more like ya didn't get where the problem was, an Trout didn't point it out to ya.

Gimme a link, if that ain't what yer talkin about.
Trout
QUOTE (gabba gabba hey+Jul 25 2008, 11:48 PM)
I just went over your "complete" analysis and I find it to be neither "complete" or correct....

In part (1) of your analysis, you implicitly define a_c=d v_c/dt.






Correct. Do you have a problem with the definition?

QUOTE
Then, in part (2) you analyze hyperbolic motion starting with x^2-c^2t^2=d^2 where d=c^2/a_c, WITHOUT stating where this comes from OR how it is applicable to the problem at hand.


Comes from here and also from here. Again, do you have a problem with the definition?

QUOTE (->
QUOTE
Then, in part (2) you analyze hyperbolic motion starting with x^2-c^2t^2=d^2 where d=c^2/a_c, WITHOUT stating where this comes from OR how it is applicable to the problem at hand.


Comes from here and also from here. Again, do you have a problem with the definition?

Since part (3) relies both on part (1) and part(2), part(3) is also incorrect!


Part 3 rhas NOTHING to do with parts 1 and 2.So, contrary to what you think, it doesn't rely on the other parts. Would you please drop the patronizing tone?
Shanester1979
QUOTE (MjolnirPants+Jul 25 2008, 09:42 PM)
Shanester,
can you show Trout's conclusion to be wrong by punchin in numbers fer the variables in that equation?
If not, can you show that this equation is not the proper one fer this purpose? (determinin the energy needed to accelerate an object to c.)

If so (to either one o those questions) please do so.
If not, then ya must admit that Trout is correct: While there is no limit to proper acceleration, as you claim, no amount o acceleration can get a massive object to c. Unless, o course, ya somehow eliminate time dilation. wink.gif

Well, there is an error in Trout's equation, but that's apparently due to a typo with parenthesis.


Despite all of the bickering about this, I believe what Trout is talking about is the acceleration 'felt' by an object. This is analogous to what an astronaut may feel when he is on a rocket ship taking off.


I am talking about the acceleration 'seen' by an observer. This is analogous to what you would see when a rocket ship takes off and speeds into the sky, if you were watching from the ground.

There are clearly limits on how much you can 'see' this rocket ship speed up, since the speed of light places a barrier on this. What you observe is dV/dt

Or the change in speed according to your frame of reference. This can only be oserved for a given dV/dt for a finite time interval.

I did not set out to say that TROUT is wrong persay. I set out initially to say that there was some difference between, dv/dt and proper acceleration.

We are not on the rockit we are observing it. That is why TROUTS equation, regardless of errors or no errors, is not the right equation to apply here. In truth, my only real goal was to show that you could derive trouts equation from the equations I was refering to. These are not incompatable ideas. But it didn't seem to get through the defense so to speak. I believe if TROUT were to think about this he would probably get what I am saying, as well as what I am not attempting to say.

In fact probably most anyone should get the difference.

In fact, maybe we can start a topic about what Trout is talking about just to show amiability, despite the various blows, no hard feelings. Actually of all the comments I've recieved on another topic, TROUT's was arguably the best one. This is part of the reason I was so surprised to see his stubborness here.

But to re-emphasize this topic is not about the force felt by the 'astronaughts' by intent. We are interested in the motion of the rockit ship according to your observations from the ground. Call it what you want, Ed, Rose, dV/dt, Big Cow, whatever. Hell I'll even start a topic about what TROUT is talking about so we try to be fair to everyone's interests, and stop confusing different concepts. I'll call it TROUTS Rocket Ship, fair enough?





Trout
QUOTE (Shanester1979+Jul 26 2008, 12:57 AM)


Despite all of the bickering about this, I believe what Trout is talking about is the acceleration 'felt' by an object. This is analogous to what an astronaut may feel when he is on a rocket ship taking off.









The definitions are quite clear. The math is quite clear. I am talking about both "proper" AND "coordinate" acceleration.

QUOTE
I am talking about the acceleration 'seen' by an observer.  This is analogous to what you  would see when a rocket ship takes off and speeds into the sky, if you were watching from the ground.


Yes, this is the "coordinate" acceleration.

QUOTE (->
QUOTE
I am talking about the acceleration 'seen' by an observer.  This is analogous to what you  would see when a rocket ship takes off and speeds into the sky, if you were watching from the ground.


Yes, this is the "coordinate" acceleration.


There are clearly limits on how much you can 'see' this rocket ship speed up, since the speed of light places a barrier on this.  What you observe is dV/dt


If you paid attention to the formulas for "ccordinate" and "proper" speed, you would have noticed that BOTH of them have a horyzontal asymptote at c. This is why math is a much superior tool to prose when it comes to physics.

QUOTE

I did not set out to say that TROUT is wrong persay.  I set out initially to say that there was some difference between, dv/dt and proper acceleration.


True, dv/dt is the coordinate acceleration, so it cannot be the same as proper acceleration.

QUOTE (->
QUOTE

I did not set out to say that TROUT is wrong persay.  I set out initially to say that there was some difference between, dv/dt and proper acceleration.


True, dv/dt is the coordinate acceleration, so it cannot be the same as proper acceleration.


We are not on the rockit we are observing it.  That is why TROUTS equation, regardless of errors or no errors, is  not the right equation to apply here.


I gave you BOTH formulas. In BOTH cases, the speed is limited to c. If you go back to my first post, this is the error I called you on.

QUOTE

But to re-emphasize this topic is not about the force felt by the 'astronaughts'  by intent.  We are interested in the motion of the rockit ship according to your observations from the ground. Call it what you want, Ed, Rose, dV/dt, Big Cow, whatever.  Hell I'll even start a topic about what TROUT is talking about so we try to be fair to everyone's interests, and stop confusing different concepts.  I'll call it TROUTS Rocket Ship, fair enough?


This is described by the equations in my paragraph 2. These are the equations of hyperbolic motion that I posted when the subject came up.



Shanester1979
QUOTE (Trout+Jul 26 2008, 01:07 AM)
.



True, dv/dt is the coordinate acceleration, so it cannot be the same as proper acceleration.








Indeed,

and 'coordinate accerlation' may not be maintained at a constant rate indefinatley. That is all I've been saying. I hope this clears the slate. Fair enough?

dv/dt=gamma^3*proper aceleration

=proper acceleration/(1+Vapp/c^2)^3/2

gabba gabba hey
QUOTE (Trout+Jul 26 2008, 12:54 AM)
Correct. Do you have a problem with the definition?



Comes from here and also from here. Again, do you have a problem with the definition?



Part 3 rhas NOTHING to do with parts 1 and 2.So, contrary to what you think, it doesn't rely on the other parts.  Would you please drop the patronizing tone?

You failed to address my most important point: the definition in part (1) is inconsistent with the v_c you derive in part two!

In part (1) you have a_c=dv_c/dt, but taking the derivative of the v_c you get in part(2), you get dv_c/dt=a_c/(1+a^2t^2/c^2)^3/2 which should be equal to a_c from part(1), but that can only be true if a_c=0!.

The reason for this error, is that the a_c used in the first reference you posted is the acceleration at the event (d,0) as measured by a commover (co-mover being the proper term) not(!) the acceleration in a reference frame where the particle is moving(!), while the a_c you use in part (1) is the acceleration in a frame where the instantaneous velocity is in general non-zero (i.e. not a co-moving frame). This is why I am calling your analysis incorrect!

As for my problem with your definitions; You didn't explicitly give these definitions in your "complete" analysis. You also didn't draw any meaningful conclusions in that analysis, and these are the reasons I called your analysis incomplete.

I was wrong about part(3) depending on (!) and (2). However, the equation for v_c derived in part 2 does not prove your point because you misapplied it.
Trout
QUOTE (gabba gabba hey+Jul 26 2008, 01:25 AM)
You failed to address my most important point: the definition in part (1) is inconsistent with the v_c you derive in part two!






No, I didn't.
Part 1 gives a definition for proper and coordinate acceleration. It is the pair of definitions that justifies the formula a_c=gamma^3*a_p

QUOTE
The reason for this error, is that the a_c used in the first reference you posted is the acceleration at the event (d,0) as measured by a commover (co-mover being the proper term) not(!) the acceleration in a reference frame where the particle is moving(!), while the a_c you use in part (1) is the acceleration in a frame where the instantaneous velocity is in general non-zero (i.e. not a co-moving frame). In part (3) you use the results of BOTH part (1) and part(2), but(!) these results are inconsistent! This is why I am calling your analysis incorrect!


All you need to do is drop the _c in part 2. Thats all.

QUOTE (->
QUOTE
The reason for this error, is that the a_c used in the first reference you posted is the acceleration at the event (d,0) as measured by a commover (co-mover being the proper term) not(!) the acceleration in a reference frame where the particle is moving(!), while the a_c you use in part (1) is the acceleration in a frame where the instantaneous velocity is in general non-zero (i.e. not a co-moving frame). In part (3) you use the results of BOTH part (1) and part(2), but(!) these results are inconsistent! This is why I am calling your analysis incorrect!


All you need to do is drop the _c in part 2. Thats all.


As for my problem with your definitions; You didn't explicitly give these definitions in  your "complete" analysis. You also didn't draw any meaningful conclusions in that analysis, and these are the reasons I called your analysis incomplete.


The definitions are quite clear. The math is quite clear. It is not my problem that it is insufficient to you.
gabba gabba hey
QUOTE (Trout+Jul 26 2008, 01:52 AM)
All you need to do is drop the _c in part 2. Thats all.


The definitions are quite clear. The math is quite clear. It is not my problem that it is insufficient to you.

Dropping the_c in part (2) is fine, but the a in part 2 is still the acceleration as measured in a co-moving frame. And your derivations in part (1) lead us to believe that you were going to show that the co-ordinate velocity in a non-co-moving frame. Which one did you actually intend?

The author of chapter 13 clearly defined his terms, you did not. The definitions you used only became clear AFTER reading your reference. It is like handing in paper to your physics professor with almost nothing but math and telling him after he's given you a c-, that the definitions and conclusions should have been clear to him because they were given in some textbook he's never read.

The equation for v arrived at in part 2, does show that v asymptotically approaches c for any constant acceleration a, but this only becomes clear after reading the reference you gave and discovering what you meant by v,a,d etc..
gabba gabba hey
Also, thanks for the feedback Trout...I wasn't aware that I could effectively portray a "patronizing tone" through my written posts.
MjolnirPants
QUOTE (Shanester1979+Jul 26 2008, 12:57 AM)
Well, there is an error in Trout's equation, but that's apparently due to a typo with parenthesis.

I did not set out to say that TROUT is wrong persay.   I set out initially to say that there was some difference between, dv/dt and proper acceleration.

We are not on the rockit we are observing it.  That is why TROUTS equation, regardless of errors or no errors, is  not the right equation to apply here.  In truth, my only real goal was to show that you could derive trouts equation from the equations I was refering to.  These are not incompatable ideas.  But it didn't seem to get through the defense so to speak.  I believe if TROUT were to think about this he would probably get what I am saying, as well as what I am not attempting to say.

In fact probably most anyone should get the difference.

In fact, maybe we can start a topic about what Trout is talking about just to show amiability, despite the various blows, no hard feelings.  Actually of all the comments I've recieved on another topic, TROUT's was arguably the best one.  This is part of the reason I was so surprised to see his stubborness here.

But to re-emphasize this topic is not about the force felt by the 'astronaughts'  by intent.  We are interested in the motion of the rockit ship according to your observations from the ground. Call it what you want, Ed, Rose, dV/dt, Big Cow, whatever.  Hell I'll even start a topic about what TROUT is talking about so we try to be fair to everyone's interests, and stop confusing different concepts.  I'll call it TROUTS Rocket Ship, fair enough?

Indeed.

QUOTE
Despite all of the bickering about this,  I believe what Trout is talking about is the acceleration 'felt' by an object.  This is analogous to what an astronaut may feel when he is on a rocket ship taking off. I am talking about the acceleration 'seen' by an observer.  This is analogous to what you  would see when a rocket ship takes off and speeds into the sky, if you were watching from the ground.

Indeed.

QUOTE (->
QUOTE
Despite all of the bickering about this,  I believe what Trout is talking about is the acceleration 'felt' by an object.  This is analogous to what an astronaut may feel when he is on a rocket ship taking off. I am talking about the acceleration 'seen' by an observer.  This is analogous to what you  would see when a rocket ship takes off and speeds into the sky, if you were watching from the ground.

Indeed.

There are clearly limits on how much you can 'see' this rocket ship speed up, since the speed of light places a barrier on this.  What you observe is dV/dt

As this hypothetical object approaches c, it's acceleration will be seen by an outside observer in an intertial reference frame to continue, although it will appear to approach zero as the object's velocity approaches c. Now here's the kicker:
The object's velocity never reaches c, relative to the observer. There is no point at which that acceleration actually reaches 0, relative to this outside observer, jes like there's no point at which the object will measure itself to be travelin at c.
The acceleration continues on forever, even to the outside observer, although this outside observer will see the acceleration shrinkin.
Yanno how you can define the set 1/n->0, even though there is no n that will produce 0? It's the same thing here. A more common analogy would be the guy who starts out at one goal post on a football field, hops halfway to the other goalpost, then half the remainin distance, then half the remainin distance...
Do ya get it now?
(There's even more to it, involvin the equivalence principle but fer the purposes o this discussion, that stuff seems extraneous...)

Jes fer giggles, I plugged that rate o acceleration ya mentioned earlier into Trout's equation.
QUOTE
v(t)=at/sqrt(1+(at/c)^2)
where
t=coordinate time
a=acceleration (can be any FINITE value)
c=speed of light

3*10^8m/s^2 fer one second gives us 212,095,329.9m/s.

3*10^8m/s^2 fer two seconds gives us 268,179,621.5m/s.
Trout
QUOTE (gabba gabba hey+Jul 26 2008, 02:08 AM)
Dropping the_c in part (2) is fine, but the a in part 2 is still the acceleration as measured in a co-moving frame.


Yes, a in part 2 is proper acceleration (acceleration as measure in the co-moving frame). So, the formula

v(t)=at/sqrt(1+(at/c)^2) is a hybrid formula (contains both coordinate and proper variable) but it is correct nevertheless. Can you concede that?

QUOTE
And your derivations in part (1) lead us to believe that you were going to show that the co-ordinate velocity in a non-co-moving frame. Which one did you actually intend?


There is no connection between part 1 and 2. I simply wanted to point out an interesting defintion for proper acceleration and the connection between a_c and a_p.

QUOTE (->
QUOTE
And your derivations in part (1) lead us to believe that you were going to show that the co-ordinate velocity in a non-co-moving frame. Which one did you actually intend?


There is no connection between part 1 and 2. I simply wanted to point out an interesting defintion for proper acceleration and the connection between a_c and a_p.

The author of chapter 13 clearly defined his terms, you did not. The definitions you used only became clear AFTER reading your reference. It is like handing in paper to your physics professor with almost nothing but math and telling him after he's given you a c-, that the definitions and conclusions should have been clear to him because they were given in some textbook he's never read.


It was not intended to be a paper , I just got tired about all that back and forth with Shane where there was very little math and a lot of "prose". I do not intend to "publish" the writeup, I only wanted to put up some math to replace the prose, ok?
\
QUOTE

The equation for v arrived at in part 2, does show that v asymptotically approaches c for any constant acceleration a, but this only becomes clear after reading the reference you gave and discovering what you meant by v,a,d etc..


Good, when and if I update the writeup I will add a "References" section.
biggrin.gif

I will also add an "Acknowledgement" section thanking you for your suggestions. biggrin.gif
I will not change one line of the math though since it is all correct (the only change is that I will replace a_c with a in section 2). I hope that you will not mind. biggrin.gif
MjolnirPants
QUOTE (gabba gabba hey+Jul 26 2008, 02:18 AM)
Also, thanks for the feedback Trout...I wasn't aware that I could effectively portray a "patronizing tone" through my written posts.

Ya bolded an capitalized key words an used exclaimation points, addin emphesis to yer post.
That gives the post a "tone" which is directly analogous to a person's "tone of voice" when speaking.
In this case, sentences like these:
QUOTE
The author of chapter 13 clearly defined his terms, you did not. The definitions you used only became clear AFTER reading your reference.

and these:
QUOTE (->
QUOTE
The author of chapter 13 clearly defined his terms, you did not. The definitions you used only became clear AFTER reading your reference.

and these:
I just went over your "complete" analysis and I find it to be neither "complete" or correct....

Then, in part (2) you analyze hyperbolic motion starting with x^2-c^2t^2=d^2 where d=c^2/a_c, WITHOUT stating where this comes from OR how it is applicable to the problem at hand.

If a_c is constant than your derivation is fine, but then a_c cannot be dv_c/dt (as you implied it was in part 1) for dv_c/dt=a_ct*(1+a_c^2t^2/c^2)^-1.5=a_c is only true for a_c=0!!!

Since part (3) relies both on part (1) and part(2), part(3) is also incorrect!

Try sayin those sentences outloud, with emphesis in place. Sounds... Confrontational, at best, don't it?
On top o that, ya made a ((rather large, IMO) mistake in that last sentence I quoted. Part 3 don't rely on parts 1 & 2 at all.
That ya would lecture him without botherin to read part 3 (an a confrontational tone often seems like a lecture to most folk) suggests condescension.

I know that that's a longer post than is needed jes to say "I agree with Trout" but I wanted to be clear as to why I agree with Trout.
Trout
An updated version can be downloaded here. I decided to show how each equation was derived.
PhysOrg scientific forums are totally dedicated to science, physics, and technology. Besides topical forums such as nanotechnology, quantum physics, silicon and III-V technology, applied physics, materials, space and others, you can also join our news and publications discussions. We also provide an off-topic forum category. If you need specific help on a scientific problem or have a question related to physics or technology, visit the PhysOrg Forums. Here you’ll find experts from various fields online every day.
To quit out of "lo-fi" mode and return to the regular forums, please click here.