Lorentz Transformation of relative velocity

  • #1
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Homework Statement


If two particles have velocities u and v in frame S, find their relative speed in frame S'.


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Homework Equations




The Attempt at a Solution


Isn't it strange that the relative speed doesn't depend on the velocity of the frame, ##\vec s##?

Since the two particles have velocities ##\vec u## and ##\vec v## in some reference frame S, I am to find the relative velocity and speed in frame S'.

Letting the relative velocity of the frame be ## \vec s##, the transformation of velocities are:

[tex]
\vec u' = \frac{1}{1 - \frac{\vec u \cdot \vec s}{c^2}}\left[ \frac{1}{\gamma_s} \vec u - \left( 1 - \frac{\vec u \cdot \vec s }{c^2}\frac{\gamma_s}{1 + \gamma_s} \right) \vec s \right] [/tex]

[tex]
\vec v' = \frac{1}{1 - \frac{\vec v \cdot \vec s}{c^2}}\left[ \frac{1}{\gamma_s} \vec v - \left( 1 - \frac{\vec v \cdot \vec s }{c^2}\frac{\gamma_s}{1 + \gamma_s} \right) \vec s \right] [/tex]

Taking ##\vec u' - \vec v'## only gives me components in ##\vec u, \vec v, \vec s ##. How do I extract the magnitude?
 

Answers and Replies

  • #2
TSny
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Note: the "speed of particle B relative to particle A" is the speed of particle B as measured in the reference frame moving with particle A.

So, what should you choose for primed frame, S'? Hence, what should be the velocity ##\vec{s}##?
 
  • #3
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Note: the "speed of particle B relative to particle A" is the speed of particle B as measured in the reference frame moving with particle A.

So, what should you choose for primed frame, S'? Hence, what should be the velocity ##\vec{s}##?
Yes, I misread the question. ##\vec s## is simply ##\vec v##. Then all that remains is to find the magnitude. But it looks very messy due to the ##\gamma_v##

[tex]
\vec w = \frac{1}{1 - \frac{\vec u \cdot \vec v}{c^2}}\left[ \frac{1}{\gamma_v} \vec u - \left( 1 - \frac{\vec u \cdot \vec v }{c^2}\frac{\gamma_v}{1 + \gamma_v} \right) \vec v \right] [/tex]

To find the magnitude I would take ##\sqrt{ \vec w \cdot \vec w}##
 
  • #4
TSny
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OK, that looks good. Yes, it is messy to deal with all the gamma factors. It will work out.

Another approach is to work out the x and y components of ##\vec{w}##, where the x direction is parallel to ##\vec{v}## and the y direction is perpendicular to ##\vec{v}## such that ##\vec{u}## is in the x-y plane.

There are well-known formulas for calculating these components. See below. These can be derived from your general formula.
 

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  • #5
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OK, that looks good. Yes, it is messy to deal with all the gamma factors. It will work out.

Another approach is to work out the x and y components of ##\vec{w}##, where the x direction is parallel to ##\vec{v}## and the y direction is perpendicular to ##\vec{v}## such that ##\vec{u}## is in the x-y plane.

There are well-known formulas for calculating these components. See below. These can be derived from your general formula.
I derived the expression ##
\vec w = \frac{1}{1 - \frac{\vec u \cdot \vec v}{c^2}}\left[ \frac{1}{\gamma_v} \vec u - \left( 1 - \frac{\vec u \cdot \vec v }{c^2}\frac{\gamma_v}{1 + \gamma_v} \right) \vec v \right] ## by using ## u_{||} = \frac{(\vec u \cdot \vec v) \vec v}{|\vec v|^2} ## and ## u_{\perp} = \vec u - \vec u_{||} ##.

I thought it would be easier, because finding the magnitudes of the horizontal and vertical components would involve angles and such.
 
  • #6
TSny
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I thought it would be easier, because finding the magnitudes of the horizontal and vertical components would involve angles and such.
If you want to work with the components, you can write ## u_{||} =u \cos \theta## and ## u_{\perp} =u \sin \theta##. Then the velocity-addition formulas for the two components will be fairly easy to work out.
 
  • #7
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If you want to work with the components, you can write ## u_{||} =u \cos \theta## and ## u_{\perp} =u \sin \theta##. Then the velocity-addition formulas for the two components will be fairly easy to work out.
So do you reckon it's easier to simply take ##\left[ \frac{u cos \theta - v}{1 - \frac{uv cos\theta}{c^2}} \right]^2 + \left[ \frac{u sin\theta}{\gamma (1- \frac{uv cos \theta}{c^2})} \right]^2 ##?
 
  • #8
TSny
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So do you reckon it's easier to simply take ##\left[ \frac{u cos \theta - v}{1 - \frac{uv cos\theta}{c^2}} \right]^2 + \left[ \frac{u sin\theta}{\gamma (1- \frac{uv cos \theta}{c^2})} \right]^2 ##?
Yes. Note that in the denominators you have ##1 - \frac{uv cos\theta}{c^2}## which is part of the answer you want.
 

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