# Homework Help: Acceleartion and the laws of relativity

1. Jul 13, 2010

### stevmg

1. The problem statement, all variables and given/known data

a) When a frame of reference undergoes acceleration/decelation (whatever,) what happens with regards to the Lorentz equations and Simple Relativity? How do the equations "change?"

b) It is said that when twins move apart (or clocks move apart) that this is a symmetrical situation and time would dilate in each frame with respect to the other. In the case where one holds one twin steady (inertial FOR) and moves the second twin - acceleration --> constant speed does this acceleration "force" one to use the reference frame in which there is no acceleration as inertial, or is arbitrary and only the amounts of time dilation is different in the two reference frames. Remember the reference frame in which there is acceleration has BOTH an initial period of acceleration which winds down to a "steady state" of constant speed.

c) With nothing else around, how would an observer know if he/she is accelerating/ decelerating or not? Remember, there is nothing else around to compare oneself to.

2. Relevant equations
Lorentz transformations

3. The attempt at a solution

No clue

2. Jul 16, 2010

### Eynstone

Special relativity doesn't deal with accelerating frames of reference. One must draw on General relativity.

3. Jul 16, 2010

### stevmg

FOR = frame of reference
If Terence ("stationary" twin) does not accelerate while Stella ("moving twin") accelerates, hits cruising constant speed, decelerates to "zero" (in Terence's FOR.) and then reacelerates the other way, again hits cruising speed and finally decelerates to zeo to meet Terence, then does Stella "feel" that she is the moving twin? How do you calculate her "proper time" during her periods of acceleration-deceleration. During steady stae of cruising speed, that's easy. Her proper time is
SQRT[(t'2 - t'1)2 - ((x'2/c) - (x'1/c))2]
where t'1 = time she entered steady state after acceleration, t'2 = time she decelerated
x'1 = point in space she entered "steady state" after acceleration, while x'2 = point in space she entered deceleration.
Doubling that gives the proper time she elapsed during steady state flights out and back..

How do you figure the proper times and distances for t'1, t'2, x'1 and x'2 for the acceleration decelaeration points while she was in acceleration/deceleration mode?

In othe words, how do I use GR to establish this?

I'll make it simpler:

Given the origin at (0, 0) in frame S. If I accelerate to the right at say g, then what ar e my time, distances after, say, time t1? How do I figure that out? In other words, work a simple problem to get me kickstarted.

Thanks,

Steve G

4. Jul 16, 2010

### George Jones

Staff Emeritus
This is a myth.
For a treatment using special relativity, look at section 2.10 Hyperbolic motion in

If you have questions about the treatment given there, fire away.

5. Jul 16, 2010

### stevmg

I know how to do this using SR. I was asking how to do it in GR (the acceleration/deceleration part?) tht's a big book you sent me. Is there a simpler explanation of what to do with the "accelerating frames?" as the weird operators, such as "$$\nabla$$" kind of throw me.

Steve G

Steve G

Last edited: Jul 16, 2010
6. Jul 16, 2010

### George Jones

Staff Emeritus
Now I'm confused. From your original post, it seemed to me that you were asking about acceleration/deceleration in special relativity, which is perfectly legitimate.
For what particular spacetime. Without pinning down a spacetime, I can't give a quantitative answer.
It's a draft version of

https://www.amazon.com/Einsteins-Ge...=sr_1_2?ie=UTF8&s=books&qid=1279291740&sr=8-2,

a copy of which I own.
But $\nabla$ isn't used in the section I referenced, section 2.10.

Last edited by a moderator: May 4, 2017
7. Jul 16, 2010

### stevmg

My mistake - didn't know that you could use acceleration/deceleration in SR. Is that explained in that chapter you referenced?

If that chapter covers it, I will work on it.

The problem I was thinking of was an acceleration of 7 g's until 0.6c then turning around and decelerating at 7 g's until you stop back where you started. I chose 7 gs because good pilots and reclining astronauts can take that for a while.

1 g on earth = 9.8m/sec2 or, say, 10 m/sec2 therefore (I don't know how to write that in Latex)
7 g = 70 m/sec2
v = 0.6c = 180,000,000 m/sec but I don't know if you can use the standard v = at and s= (1/2)70t2 = 35t2

Don't know where to get started. I've omitted the "crusing" speed of 0.6c to keep matters simple.

Is that one expensive little book! I looked it up online.

I know it has something to do with force fields at the site of what is being operated on but have no clue as to what it means.

Steve G

Last edited by a moderator: May 4, 2017