Relative 4-velocities

  • #1
6
0

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
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,933
6,739
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.
 
  • #3
20,974
4,604
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
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,933
6,739
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
20,974
4,604
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
 
  • #6
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,933
6,739
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.
 
  • Like
Likes Chestermiller

Related Threads on Relative 4-velocities

Replies
6
Views
1K
  • Last Post
Replies
1
Views
800
Replies
12
Views
5K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
949
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
0
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
1
Views
511
Top