# Lorentz Transformation of relative velocity

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1. Jan 9, 2015

### unscientific

1. The problem statement, all variables and given/known data
If two particles have velocities u and v in frame S, find their relative speed in frame S'.

2. Relevant equations

3. 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:

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

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

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

2. Jan 10, 2015

### TSny

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. Jan 10, 2015

### unscientific

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 $\gamma_v$

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

To find the magnitude I would take $\sqrt{ \vec w \cdot \vec w}$

4. Jan 10, 2015

### TSny

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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Last edited: Jan 10, 2015
5. Jan 10, 2015

### unscientific

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. Jan 10, 2015

### TSny

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. Jan 10, 2015

### unscientific

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. Jan 10, 2015

### TSny

Yes. Note that in the denominators you have $1 - \frac{uv cos\theta}{c^2}$ which is part of the answer you want.