Deriving particle speed from invariant U·V in the observer's frame

[scottJH]

Homework Statement

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

Homework Equations

$U.V=U'.V'$[/B]

The Attempt at a Solution

[/B]
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

 

[Orodruin]

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

Any insight is appreciated

Yes, your approach and result seem reasonable.

 

[Chestermiller]

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

 

[Orodruin]

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.

 

[Chestermiller]

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.

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

Chet

 

[Orodruin]

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.