Acceleration/Deceleration in SR

[stevmg]

I just found out that you can do acceleration/deceleration problems in SR. I didn't know that.

The problem I was thinking of was the classic Terence/Stella problem of recent fame on this Forum. See this post by Jesse M who solves this for constant velocities:
https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

Basically, Terence and Stella are on Earth. Terence stays put while Stella accelerates, to the right, say, at 7 g (about 70 m/sec² s until achieving a velocity of 0.6c (or 180,000,000 m/sec) to the right and then turns around and decelerates at 7 g's until she reaches or catches up with Terence on Earth. I chose 7 g's because good pilots and reclining astronauts can take that for a while.

I don't know where to get started. I've omitted the "crusing" speed of 0.6c to keep matters simple. In other words, Stella's rocket goes out and immediately turns around to come back.

I don't know if you can use the standard v = at and s= (1/2)70t² = 35t² to figure alloted time and distance in Terence's frame and Stella's accelerating, then decelerating frame. I assume you have to bust her travels into two frames - one out and one in.

Please give me a kickstart. I know how to do it at "steady state" (constant v = 0.6 c) for out and back.

Do you calculate first t for the given parameters, then s by using the t already calculated, and using the formula:
f X s = energy expended. Then obtain the v for the energy by KE = (1/2)mv² and then with the v from the energy equation (not the v = 0.6 c) and the s apply the Lorentz transforms? I don't think so, although that would be a first good "guess."


Steve G

 

[yuiop]

Most of the equations you need to handle acceleration in SR are given in simple form here:

http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/rocket.html

That might help.

 

[starthaus]

I just found out that you can do acceleration/deceleration problems in SR. I didn't know that.

The problem I was thinking of was the classic Terence/Stella problem of recent fame on this Forum. See this post by Jesse M who solves this for constant velocities:
https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

Basically, Terence and Stella are on Earth. Terence stays put while Stella accelerates, to the right, say, at 7 g (about 70 m/sec² s until achieving a velocity of 0.6c (or 180,000,000 m/sec) to the right and then turns around and decelerates at 7 g's until she reaches or catches up with Terence on Earth. I chose 7 g's because good pilots and reclining astronauts can take that for a while.

I don't know where to get started. I've omitted the "crusing" speed of 0.6c to keep matters simple. In other words, Stella's rocket goes out and immediately turns around to come back.

I don't know if you can use the standard v = at and s= (1/2)70t² = 35t² to figure alloted time and distance in Terence's frame and Stella's accelerating, then decelerating frame. I assume you have to bust her travels into two frames - one out and one in.

Please give me a kickstart. I know how to do it at "steady state" (constant v = 0.6 c) for out and back.

Do you calculate first t for the given parameters, then s by using the t already calculated, and using the formula:
f X s = energy expended. Then obtain the v for the energy by KE = (1/2)mv² and then with the v from the energy equation (not the v = 0.6 c) and the s apply the Lorentz transforms? I don't think so, although that would be a first good "guess."


Steve G

You can start by reading https://www.physicsforums.com/blog.php?b=1911 . I wrote a few files on the subject.

 

[stevmg]

You can start by reading https://www.physicsforums.com/blog.php?b=1911 . I wrote a few files on the subject.

Is there an "Acceleration in SR part I? All I see is part II.

Steve G

 

[stevmg]

Hold on - I got to your blog and retrieved the other three .pdf files you wrote on this subject.

Thanks,
SG

 

[starthaus]

Is there an "Acceleration in SR part I? All I see is part II.

Steve G

Yes but it is a https://www.physicsforums.com/blog.php?b=1893 that part II (don't ask). So, it is a good idea to start with part II :-)

 

[stevmg]

Ok...

 

[Mike_Fontenot]

Years ago, I derived a simple equation (called the "CADO" equation) that explicitly gives the ageing of the home twin during accelerations by the traveler (according to the traveler). The equation is especially easy to use for idealized traveling twin problems with instantaneous speed changes. But it also works for finite accelerations. I've got a detailed example with +-1g accelerations on my webpage:

http://home.comcast.net/~mlfasf

And I've published a paper giving the derivation of the CADO equation:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

 

[stevmg]

Copied it to a .pdf file.

Thanks,

SMG

WRT: "Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

How about a reprint or a site where I can see it?

 

[Dale]

I also like this arxiv article.
http://arxiv.org/abs/gr-qc/0104077

It explicitly works on the twin paradox for a finite acceleration and gives a reasonable coordinate system to use for the traveling twin at all points.

 

[Mike_Fontenot]

WRT: "Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

How about a reprint or a site where I can see it?

As far as I know, it's not online anywhere (and I don't have it in any kind of "emailable" form. And the journal didn't give me any reprints, like some journals (at least used to) do.

I also don't know if you can get a copy of just the single article from "Physics Essays" (they DO have a webpage), or perhaps an entire back issue for a reasonable price.

But most university libraries should either have it, or else be able to get it for you via inter-library loan, so that should work if you're anywhere near a university.

BTW, you were not alone in thinking that special relativity can't handle accelerations...that's a common misconception, even among physicists who should know better.

 

[Mike_Fontenot]

I also like this arxiv article.
http://arxiv.org/abs/gr-qc/0104077

It explicitly works on the twin paradox for a finite acceleration and gives a reasonable coordinate system to use for the traveling twin at all points.

From just a quick look at that link, it appears to be a definition of simultaneity such that ALL observers will agree about the simultaneity of two given separated events. If so, that's a big mistake (in my opinion).

The ONLY definition of simultaneity that doesn't contradict the observer's own elementary measurements and elementary calculations, is the one I have given in my previously referenced paper. And with that definition, observers in relative motion with respect to one another WON'T agree about the simultaneity of any two given separated events.

If an observer has to disregard his own elementary measurements and calculations, that involve only first-principles, he simply CAN'T do any physics.

Mike Fontenot

 

[Fredrik]

From just a quick look at that link, it appears to be a definition of simultaneity such that ALL observers will agree about the simultaneity of two given separated events.

It isn't. It's just a standard way to associate a coordinate system with the motion of an object. (I think this is what MTW calls the object's "proper reference frame"). Take the world line to be the t axis and assign coordinates to other events by generalizing this idea: If light is emitted at x=0 at t=-T, then reflected somewhere, and returned to x=0 at t=T, we assign t=0 (and x=T) to the reflection event. (If we apply this procedure to a timelike geodesic, we get a global inertial frame. If we apply it to the world line of an object doing constant proper acceleration, we get Rindler coordinates. These authors are just applying the same idea to the astronaut twin's world line).

 

[Dale]

From just a quick look at that link, it appears to be a definition of simultaneity such that ALL observers will agree about the simultaneity of two given separated events. If so, that's a big mistake (in my opinion).

You are mis-reading it. The bulk of the paper describes how the different observers disagree, including plots of the stay-at-home twin's worldline in the traveling twin's frame. It is just a definition of simultaneity that works everywhere in a non-inertial frame.

 

[Mike_Fontenot]

You are mis-reading it.

That is certainly possible.

But the important question is, for an accelerating traveler, does the value they compute for the current age of the home twin, at any given age of the traverer, ACCORDING TO THE TRAVELER, agree with my result or not?

If it does, then it is just an alternative way of arriving at my result.

If it does not, then their result will contradict the traveler's own elementary measurements, combined with his own elementary calculations. And the traveler CAN'T do any physics, if he is forced to disregard his own measurements.

Mike Fontenot

 

[Dale]

does the value they compute for the current age of the home twin, at any given age of the traverer, ACCORDING TO THE TRAVELER, agree with my result or not?

I don't know your result enough to answer that.

If it does not, then their result will contradict the traveler's own elementary measurements, combined with his own elementary calculations. And the traveler CAN'T do any physics, if he is forced to disregard his own measurements.

The idea that physics can only be done using some form of your simultaneity convention is just silly. As long as you know the metric you can do physics using any coordinates and any simultaneity convention.

 

[starthaus]

I don't know your result enough to answer that.

The idea that physics can only be done using some form of your simultaneity convention is just silly. As long as you know the metric you can do physics using any coordinates and any simultaneity convention.

PE is a journal known for publishing fringe or outright incorrect (and/or anti-mainstream) articles. Since Mike is unwilling to provide a copy of the paper, it is impossible to figure out to what extent his paper is correct.

 

[Mike_Fontenot]

I don't know your result enough to answer that.

OK, here's a question that's trivial to answer with my equation:

Suppose two people (say, Tom and Sue) are stationary with respect to one another, when they are both 30 years old, and that they are 40 lightyears apart.

Then suppose that Tom instantaneously changes his speed so that he is moving away from Sue at 0.866c.

Tom is still 30 years old. How old is Sue, according to Tom?

What's the answer, according to the link you posted?

Mike Fontenot

 

[starthaus]

OK, here's a question that's trivial to answer with my equation:

Suppose two people (say, Tom and Sue) are stationary with respect to one another, when they are both 30 years old, and that they are 40 lightyears apart.

Then suppose that Tom instantaneously changes his speed so that he is moving away from Sue at 0.866c.

Tom is still 30 years old.

Not forever

How old is Sue, according to Tom?

Older. How much older depends on the amount of time elapsed on Tom's clock.
For a complete solution, see here

 

[Fredrik]

Older. How much older depends on the amount of time elapsed on Tom's clock.

I think you read the question wrong, or maybe I did. I'd say that the answer is "younger" regardless of whether we use the comoving inertial frame or the radar time notion of simultaneity as in the Dolby & Gull article. Also, I think Mike meant that 0 time has elapsed on Tom's clock at the event we're supposed to consider.

OK, here's a question that's trivial to answer with my equation:

Suppose two people (say, Tom and Sue) are stationary with respect to one another, when they are both 30 years old, and that they are 40 lightyears apart.

Then suppose that Tom instantaneously changes his speed so that he is moving away from Sue at 0.866c.

Tom is still 30 years old. How old is Sue, according to Tom?

What's the answer, according to the link you posted?

I don't feel like doing any calculations right now, but in the diagram I'm drawing in my head, I can see that Sue would be much younger than 30 in Tom's comoving inertial frame (after the boost), because its simultaneity lines have slope v in the diagram, so the boost event is simultaneous with an "early" event on Sue's world line.

In the coordinate system that Dolby & Gull are using, things are much more complicated. Simultaneity starts getting messed up 40 years before the boost event, because that's how long it takes for light to go from Sue to Tom. I don't seem to be able to work out the details in my head. I think her aging rate just keeps getting slower in (D&G's version of) Tom's frame, for a long time starting 40 years before the boost event, and ending...uh...at the event on Tom's world line that's simultaneous in his comoving inertial frame with Sue age 70. After that, her aging rate in (D&G's version of) Tom's frame is the same as in the comoving inertial frame, i.e. slow by a factor of gamma.

 

[starthaus]

I think you read the question wrong, or maybe I did. I'd say that the answer is "younger" regardless of whether we use the comoving inertial frame or the radar time notion of simultaneity as in the Dolby & Gull article. Also, I think Mike meant that 0 time has elapsed on Tom's clock at the event we're supposed to consider.

I read it as in "Sue is the stay-at-home" twin while Tom accelerates away. Therefore Tom is younger , so Sue is older.

 

[Dale]

OK, here's a question that's trivial to answer with my equation:

Suppose two people (say, Tom and Sue) are stationary with respect to one another, when they are both 30 years old, and that they are 40 lightyears apart.

Then suppose that Tom instantaneously changes his speed so that he is moving away from Sue at 0.866c.

Tom is still 30 years old. How old is Sue, according to Tom?

What's the answer, according to the link you posted?

Using Tom's radar coordinates Sue is about 6.9 years old, both immediately before and immediately after Tom's acceleration. In radar coordinates her age increases monotonically and continuously.

Of course, that is assuming that Tom remains inertial after the acceleration and that by "when they are both 30 years old" you are referring to simultaneity in the reference frame where they were both initially at rest.

 

[Austin0]

OK, here's a question that's trivial to answer with my equation:

Suppose two people (say, Tom and Sue) are stationary with respect to one another, when they are both 30 years old, and that they are 40 lightyears apart.

Then suppose that Tom instantaneously changes his speed so that he is moving away from Sue at 0.866c.

Tom is still 30 years old. How old is Sue, according to Tom?

What's the answer, according to the link you posted?

Mike Fontenot

She hasnt been born yet. Just a quick mental appraisal without figuring out the actual gamma factor

 

[Mike_Fontenot]

She hasn't been born yet. Just a quick mental appraisal without figuring out the actual gamma factor

You are right. Immediately after Tom's instantaneous speed change, Sue is -4.6 years old (according to Tom)...i.e., Sue's mother-to-be will age another 4.6 years before she gives birth to Sue.

Any other answer will contradict Tom's own elementary measurements and calculations.

(It takes about 15 seconds to calculate that result, using my CADO equation).

Mike Fontenot

 

[Mike_Fontenot]

Also, I think Mike meant that 0 time has elapsed on Tom's clock at the event we're supposed to consider.

Yes, you got it right. The whole question is just about what happens to Sue's age (according to Tom) during the instantaneous speed change that Tom undergoes.

 

[Dale]

Any other answer will contradict Tom's own elementary measurements and calculations.

I am going to call BS on this. Did you even read the Dolby and Gull paper? It shows exactly how their convention is compatible with Tom's "elementary measurements and calculations". If you believe otherwise then it is up to you to prove it.

In any case, simultaneity is purely a matter of convention and in non-inertial frames there is certainly no need to use any specific convention. Physics will work fine whatever convention is adopted. The advantage of the D&G approach is that it leaves you with an invertible coordinate system, which is always good if you want to do any physics calculations.

 

[Mike_Fontenot]

[...]
In the coordinate system that Dolby & Gull are using, things are much more complicated. Simultaneity starts getting messed up 40 years before the boost event, because that's how long it takes for light to go from Sue to Tom. [...]

Thanks for your description of Dolby & Gull simultaneity, Fredrik.

If Sue, and her mother, never accelerate at all, consider the following two alternatives for Tom (and his mother):

In both alternatives, from birth until Tom is 30 years old, he and his mother don't accelerate. And, until Tom is 30 years old, he and Sue are 40 lightyears apart (as in my original question). (And each of the childrens' mothers is always co-located with her child).

But then, the two alternatives differ. The alteratives are:

1) Tom (and his mother) never accelerate.

or

2) At age 30, Tom and his mother do the instantaneous speed change described in my original question...they go from zero velocity (relative to Sue) to +0.866c (in the direction away from Sue), in essentially zero time on their watches.

If I understood your description correctly, Fredrik, it is impossible for Tom's mother, when Tom is one year old, to know how old Sue's mother is then (assuming of course that Sue's mother is still alive), unless Tom's mother already knows whether she and Tom will accelerate or not, 29 years later in their lives.

It sounds like, in Dolby & Gull, that no one can make realtime assertions about simultaneity, unless they are mystics. If you don't believe in mystics, then I guess, for Dolby and Gull, that simultaneity is for historians, only, to discuss.

Mike Fontenot

 

[Dale]

It sounds like, in Dolby & Gull, that no one can make realtime assertions about simultaneity, ... that simultaneity is for historians, only, to discuss.

That is true in general. Realtime assertions about simultaneity would require superluminal communication.

If your mysterious approach does that then it either violates relativity or causality.

 

[Austin0]

Immediately after Tom's instantaneous speed change, Sue is -4.6 years old (according to Tom)...i.e., Sue's mother-to-be will age another 4.6 years before she gives birth to Sue.


(It takes about 15 seconds to calculate that result, using my CADO equation).

Mike Fontenot

I have had a second take on the scenario.

Prior to Tom's acceleration his frame mate Bob is vacationing at rest 26 ly towards earth.

As Tom initiates acceleration Bob is rudely awaken only to get a last fleeting glimpse of the Earth rushing towards him due to the Earth frame's contraction.
Not only terminating Bob but incidently nailing Sue's mom on the way to the library.

My question is this : ...Does Tom simultaneously lose all memory of Sue or does he wait until the information regardiung her prenatal demise (i.e. history) reachs him at c ?

What does CADO predict?? :-)

 

[Fredrik]

It sounds like, in Dolby & Gull, that no one can make realtime assertions about simultaneity, unless they are mystics. If you don't believe in mystics, then I guess, for Dolby and Gull, that simultaneity is for historians, only, to discuss.

DaleSpam told you the most essential point here:

That is true in general. Realtime assertions about simultaneity would require superluminal communication.

I'd like to add that the D&G approach has a feature that at first sounds even weirder than that, because what events are simultaneous with you right now depends on how you will move in the future. When I first realized this (about 15 years ago), I thought this makes the "radar" notion of simultaneity pretty dumb. But I eventually realized that simultaneity in inertial frame is no less weird. An inertial frame is just what you end up with if you apply this idea to a world line that's straight now and forever. So the only reason why simultaneity in inertial frames doesn't depend on "how you will move in the future" is that you are simply assuming that your velocity always has been and always will be the same as it is now.