Acceleartion and the laws of relativity

In summary, when a frame of reference undergoes acceleration/decelation (whatever,) the Lorentz equations and Special Relativity change.
  • #1
stevmg
696
3

Homework Statement



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.

Homework Equations


Lorentz transformations

The Attempt at a Solution



No clue
 
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  • #2
Special relativity doesn't deal with accelerating frames of reference. One must draw on General relativity.
 
  • #3
Eynstone said:
Special relativity doesn't deal with accelerating frames of reference. One must draw on General relativity.

That doesn't answer my question...

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

This is a myth.
stevmg said:
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.

For a treatment using special relativity, look at section 2.10 Hyperbolic motion in

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.118.2817&rep=rep1&type=pdf.

If you have questions about the treatment given there, fire away.
 
  • #5
George Jones said:
This is a myth.


For a treatment using special relativity, look at section 2.10 Hyperbolic motion in

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.118.2817&rep=rep1&type=pdf.

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

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 "[tex]\nabla[/tex]" kind of throw me.

Steve G



Steve G
 
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  • #6
stevmg said:
I know how to do this using SR.

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.
stevmg said:
I was asking how to do it in GR (the acceleration/deceleration part?)

For what particular spacetime. Without pinning down a spacetime, I can't give a quantitative answer.
stevmg said:
tht's a big book you sent me.

It's a draft version of

https://www.amazon.com/dp/144192406X/?tag=pfamazon01-20,

a copy of which I own.
stevmg said:
Is there a simpler explanation of what to do with the "accelerating frames?" as the weird operators, such as "[tex]\nabla[/tex]" kind of throw me.

But [itex]\nabla[/itex] isn't used in the section I referenced, section 2.10.
 
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  • #7
George Jones said:
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.

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

George Jones said:
For what particular spacetime. Without pinning down a spacetime, I can't give a quantitative answer.

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.

George Jones said:
It's a draft version of

https://www.amazon.com/dp/144192406X/?tag=pfamazon01-20,

a copy of which I own.

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

George Jones said:
But [itex]\nabla[/itex] isn't used in the section I referenced, section 2.10.

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
 
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Related to Acceleartion and the laws of relativity

1. What is acceleration?

Acceleration is the rate at which an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction.

2. How is acceleration related to the laws of relativity?

The laws of relativity, specifically Einstein's Theory of Special Relativity, state that the laws of physics are the same for all observers in uniform motion. This means that the concept of acceleration is relative, and the same acceleration may be perceived differently by different observers depending on their relative motion.

3. What is the equation for calculating acceleration?

The equation for acceleration is a = Δv / Δt, where a is acceleration, Δv is the change in velocity, and Δt is the change in time. This is also known as the average acceleration formula.

4. How does acceleration affect time and space?

Acceleration is one of the factors that can affect the perception of time and space according to the Theory of Special Relativity. As an object's acceleration increases, time slows down for that object and space contracts in the direction of motion. This effect becomes more significant as the object approaches the speed of light.

5. Can objects with mass reach the speed of light?

According to the Theory of Special Relativity, objects with mass cannot reach the speed of light because as they approach the speed of light, their mass increases infinitely. This means that an infinite amount of energy would be required to accelerate an object with mass to the speed of light, making it impossible to reach.

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