Lorentz Velocity Transformation

  • Thread starter JonathanT
  • Start date
  • #1
18
0

Homework Statement


Consider a light signal propagating in some arbitrary direction, with

vx [itex]\neq[/itex] 0
vy [itex]\neq[/itex] 0
vz [itex]\neq[/itex] 0 and

vx2 + vy2 + vz2 = c2


Use the Lorentz transformation equations for the components of velocity to show that


v'x2 + v'y2 + v'z2 = c2

Homework Equations



Combination of Velocities

v'x = (vx - V)/(1-vxV/c2)

v'y = (vy√1-V2/c2))/(1-vxV/c2)

v'z = (vz√1-V2/c2))/(1-vxV/c2)



The Attempt at a Solution



I know this is just a simple algebra proof but for some reason I'm getting stuck on it. Maybe I'm using the wrong equations?

I would really appreciate being shown where to start for this proof. Thanks in advance for the help and/or time.
 

Answers and Replies

  • #2
323
1
You should get the transformation equations for vx, vy, vz to the prime variables and insert them into vx^2 +vy^2 +vz^2 = c^2 to show this reduces to vx'^2 +vy'^2 +vz'^2 = c^2. Since the speed of light is constant in all frames, the change in frames should be invariant under the lorentz transformation. Just be careful with your algebra...if it starts to get too messy then you probably did a simplification that led you in the wrong direction.
 
  • #3
18
0
Thank you. I apparently was making a stupid algebra mistake and over complicating it. Just knowing I was heading in the right direction helped. Made me find my mistake. Thanks again.
 
  • #4
323
1
no problem!
 

Related Threads on Lorentz Velocity Transformation

  • Last Post
Replies
1
Views
14K
Replies
2
Views
1K
Replies
7
Views
971
  • Last Post
Replies
3
Views
2K
Replies
3
Views
6K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
2K
Top