# Relative 4-velocities

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1. Jan 14, 2015

### scottJH

1. The problem statement, all variables and given/known data
In a particular inertial frame of reference, a particle with 4-velocity V is observed by an
observer moving with 4-velocity U. Derive an expression for the speed of the particle
relative to the observer in terms of the invariant U · V

2. Relevant equations

$U.V=U'.V'$

3. The attempt at a solution

I used the relation the $U.V=U'.V'$ because U.V is invariant.

Using the rest frame of the observer I obtained

$U'.V' = -\gamma(u_R)c^2$

Then rearranging to find $u_R$ I obtained

$u_R = c*sqrt(1-(c^4/(U.V)^2))$

I'm just wondering if I used the correct method and got the correct result for $u_R$

Any insight is appreciated

2. Jan 14, 2015

### Orodruin

Staff Emeritus
Yes, your approach and result seem reasonable.

3. Jan 14, 2015

### Staff: Mentor

As a non-expert in relativity (far from it), I'm confused as to why a primed frame even needs to be introduced in this problem, since U and V are individually frame invariant. So the dot product of U and V can be calculated using the components of these vectors as reckoned with respect to any convenient reference frame. This is what scottH actually did. So why the need for the primes?

Chet

4. Jan 14, 2015

### Orodruin

Staff Emeritus
The point is that the rest frame of the observer is this convenient frame where you get an expression for the relative velocity in terms of the inner product. Naturally, once the derivation is done and the expression for the relative velocity in terms of the product is known, you can choose to evaluate $U\cdot V$ in any frame.

5. Jan 14, 2015

### Staff: Mentor

That's what I thought. So why the need for the primes?

Chet

6. Jan 14, 2015

### Orodruin

Staff Emeritus
Some people like to denote the same vector in different coordinate systems using primes for some reason ... I guess mostly for when writing in components instead of putting the prime on the indices.