Lorentz transformation of 4-acceleration

  • #1
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Homework Statement:
Problem 11, chapter IV of Rindler's "Introduction to Special Relativity": Use the fact that ##A## as given in (23.5) is a four-vector to give an alternative solution to the first part of Exercise II(12).
That exercise reads: In ##S^\prime## a particle is momentarily at rest and has acceleration ##\alpha## in the ##y^\prime##-direction. What is the magnitude and direction of its acceleration in ##S##?
Relevant Equations:
The expression for the four-vector acceleration (23.5) is $$A = \gamma \left(\frac{d\gamma}{dt}c, \frac{d\gamma}{dt}\mathbf{u} + \gamma\mathbf{a} \right).$$

Time dilation: ##\frac{dt}{dt^\prime} = \gamma.##

Velocity transformations: $$u_1 = \frac{u_1^\prime +v}{1 + u_1^\prime v/c^2},\, u_2 = \frac{u_2^\prime}{\gamma(1+u_1^\prime v/c^2)},\, u_3 = \frac{u_3^\prime}{\gamma(1+u_1^\prime v/c^2)}.$$

It should also be noted that the frame ##S^\prime## is moving in the positive ##x##-direction with speed ##v## as seen from the frame ##S##.
I have been getting back to studying physics after a long break and decided to go through the problems in Rindler. But there is something I don't quite understand in this problem.

To first answer the second part, Exercise II(12), I wrote $$\frac{du_2}{dt} = \frac{du_2}{du_2^\prime} \frac{du_2^\prime}{dt} = \frac{du_2}{du_2^\prime} \frac{du_2^\prime}{dt^\prime}\frac{dt^\prime}{dt}.$$ The first and the last factors are both ##\frac{1}{\gamma}## as can be seen from the time-dilation equation and the velocity transformation with ##u_1^\prime = 0##. The middle factor is the proper acceleration ##\alpha##. Thus, I get ##\alpha/\gamma^2##. For the ##x##- and ##z##-components, the acceleration is still zero since ##\frac{d u_1^\prime}{dt^\prime} = \frac{d u_3^\prime}{dt^\prime} =0##.

In order to do this with the four-vector, note that in the lab frame ##S##, we have the velocity ##\mathbf{u} = (v, 0, 0)##. Starting at rest, we get (I think) ##\frac{d\gamma}{dt} = 0##. Using the expression for the four-acceleration, I get $$A = (0,0,\gamma^2 \alpha, 0),$$ which is different from the previous result by a factor of ##\gamma^4##. Where did I go wrong?
 

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  • #2
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Homework Statement: Problem 11, chapter IV of Rindler's "Introduction to Special Relativity": Use the fact that ##A## as given in (23.5) is a four-vector to give an alternative solution to the first part of Exercise II(12).
That exercise reads: In ##S^\prime## a particle is momentarily at rest and has acceleration ##\alpha## in the ##y^\prime##-direction. What is the magnitude and direction of its acceleration in ##S##?
Homework Equations: The expression for the four-vector acceleration (23.5) is $$A = \gamma \left(\frac{d\gamma}{dt}c, \frac{d\gamma}{dt}\mathbf{u} + \gamma\mathbf{a} \right).$$

Time dilation: ##\frac{dt}{dt^\prime} = \gamma.##

Velocity transformations: $$u_1 = \frac{u_1^\prime +v}{1 + u_1^\prime v/c^2},\, u_2 = \frac{u_2^\prime}{\gamma(1+u_1^\prime v/c^2)},\, u_3 = \frac{u_3^\prime}{\gamma(1+u_1^\prime v/c^2)}.$$

It should also be noted that the frame ##S^\prime## is moving in the positive ##x##-direction with speed ##v## as seen from the frame ##S##.

I have been getting back to studying physics after a long break and decided to go through the problems in Rindler. But there is something I don't quite understand in this problem.

To first answer the second part, Exercise II(12), I wrote $$\frac{du_2}{dt} = \frac{du_2}{du_2^\prime} \frac{du_2^\prime}{dt} = \frac{du_2}{du_2^\prime} \frac{du_2^\prime}{dt^\prime}\frac{dt^\prime}{dt}.$$ The first and the last factors are both ##\frac{1}{\gamma}## as can be seen from the time-dilation equation and the velocity transformation with ##u_1^\prime = 0##. The middle factor is the proper acceleration ##\alpha##. Thus, I get ##\alpha/\gamma^2##. For the ##x##- and ##z##-components, the acceleration is still zero since ##\frac{d u_1^\prime}{dt^\prime} = \frac{d u_3^\prime}{dt^\prime} =0##.

In order to do this with the four-vector, note that in the lab frame ##S##, we have the velocity ##\mathbf{u} = (v, 0, 0)##. Starting at rest, we get (I think) ##\frac{d\gamma}{dt} = 0##. Using the expression for the four-acceleration, I get $$A = (0,0,\gamma^2 \alpha, 0),$$ which is different from the previous result by a factor of ##\gamma^4##. Where did I go wrong?

That might all be correct. Using the transformation of 4-vectors into frame S we get:

##A = (0,0, \alpha, 0)##

But, also, in frame S we have

##A^y = \gamma^2 a_y##

Where ##a_y## is the 3-acceleration in frame S.

This gives ##a_y = \alpha/ \gamma^2##.

Which is what you got by transforming the 3-acceleration.

All good!
 
  • #3
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. Using the expression for the four-acceleration, I get $$A = (0,0,\gamma^2 \alpha, 0),$$

On reflection this is not correct. You should either get ##\alpha## there, by simple transformation of a 4-vector under a boost in the x-direction:

##A^y = A'^y##

Or ##\gamma^2 a_y## where ##a_y## is the y-component of three-acceleration. By using the formula for 4-acceleration.
 
  • #4
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On reflection this is not correct. You should either get ##\alpha## there, by simple transformation of a 4-vector under a boost in the x-direction:

##A^y = A'^y##

Or ##\gamma^2 a_y## where ##a_y## is the y-component of three-acceleration. By using the formula for 4-acceleration.

Thanks for your reply. I completely agree with you, but do you know where in my reasoning I went wrong?
I guess a better solution is to write down the 4-acceleration in the primed frame and do a Lorentz transformation, but I'm curious as to what went wrong in the above.
 
  • #5
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Thanks for your reply. I completely agree with you, but do you know where in my reasoning I went wrong?
I guess a better solution is to write down the 4-acceleration in the primed frame and do a Lorentz transformation, but I'm curious as to what went wrong in the above.
You took ##a_y = \alpha## as explained above.

In fact, ##\alpha = a'_y## is the three acceleration in the primed frame.
 
  • #6
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You took ##a_y = \alpha## as explained above.

In fact, ##\alpha = a'_y## is the three acceleration in the primed frame.

Sorry, but I still don't quite understand.

Taking the three-acceleration in the primed frame as ##\alpha=a_y^\prime##, the 4-acceleration in the primed frame is $$A^\prime = (0,0,\gamma^2 \alpha, 0).$$

Applying a Lorentz transformation in the ##x##-direction then gives in the unprimed frame $$A = (0,0,\gamma^2 \alpha, 0).$$ Hence the three-acceleration in the unprimed frame is ##a_y = \alpha##, same as the primed frame. But this is not what I got when transforming the three-acceleration. Which one is wrong and why? This is really bugging me and I would like to understand this.
 
  • #7
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Sorry, but I still don't quite understand.

Taking the three-acceleration in the primed frame as ##\alpha=a_y^\prime##, the 4-acceleration in the primed frame is $$A^\prime = (0,0,\gamma^2 \alpha, 0).$$

In general, you need to develop a notation to distinguish between different gamma factors. In any case, ##\gamma' = 1##, where ##\gamma'## is the gamma-factor of the particle in the primed frame. You can take that out of the above equation.

In this case, as the particle is at rest in the primed frame, you can use ##\gamma## as the gamma factor of the Lorentz transformation between the frames and the gamma factor of the particle in the unprimed frame.

In general, you might want to number your gamma factors.
 
  • #8
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In general, you need to develop a notation to distinguish between different gamma factors. In any case, ##\gamma' = 1##, where ##\gamma'## is the gamma-factor of the particle in the primed frame. You can take that out of the above equation.

In this case, as the particle is at rest in the primed frame, you can use ##\gamma## as the gamma factor of the Lorentz transformation between the frames and the gamma factor of the particle in the unprimed frame.

In general, you might want to number your gamma factors.
I see, that makes a lot of sense. So in the primed frame, ##\gamma = 1## since the particle is initially at rest. Thus ##A^\prime = (0,0,\alpha,0)##, and transforming this gives ##A=(0,0,\alpha,0)##, which means that you can read off the three-acceleration as ##a_y = \alpha/\gamma^2##. (Obvious in hindsight...)

Thanks so much for your help!
 
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