How Do Lorentz Transformations Affect Light Signal Velocities?

[JonathanT]

Homework Statement

Consider a light signal propagating in some arbitrary direction, with

v_x $\neq$ 0
v_y $\neq$ 0
v_z $\neq$ 0 and

v_x² + v_y² + v_z² = 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 = (v_x - V)/(1-v_xV/c²)

v'_y = (v_y√1-V²/c²))/(1-v_xV/c²)

v'_z = (v_z√1-V²/c²))/(1-v_xV/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.

 

[nlsherrill]

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.

 

[JonathanT]

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.

 

[nlsherrill]

no problem!

 

[paranormal]

Dear student,

The Lorentz velocity transformation equations are used to describe how velocities appear to change from one reference frame to another in special relativity. In this case, we are considering a light signal propagating in an arbitrary direction, meaning that it has nonzero velocities in all three spatial dimensions (vx, vy, vz). We also know from the given information that the sum of the squares of these velocities is equal to the speed of light squared (c2).

To show that the transformed velocities (v'x, v'y, v'z) also have a sum of squares equal to c2, we can use the Lorentz transformation equations for the components of velocity. These equations show how the transformed velocities (v'x, v'y, v'z) are related to the original velocities (vx, vy, vz) and the relative velocity between the two reference frames (V).

Starting with the transformed velocity in the x-direction (v'x), we can substitute in the Lorentz transformation equation for v'x and use the given information to simplify:

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

= (vx - V)(1+vxV/c2)/[(1-vxV/c2)(1+vxV/c2)] (multiplying by the conjugate)

= (vx - V + vxV2/c2)/(1 - vx2V2/c4) (expanding the denominator)

= (vx - V + vx(c2-vx2))/c2 (using vx2 + vy2 + vz2 = c2)

= (vx - V + vxc2 - vx3)/c2

= (vx + vxc2 - V - vx3)/c2 (rearranging terms)

= (vx√1-V2/c2 - V)/(1 - vxV/c2) (using the Lorentz transformation equation for vx)

= v'x (substituting back in the definition of v'x)

We can repeat this process for the transformed velocities in the y- and z-directions, and then use the Pythagorean theorem to show that the sum of their squares is equal to c2:

v'y = (vy√1-V2/c2)/(1-vxV/c2)

v'z = (vz√1-V2/c2)/(1-vxV/c2)

v'x2 + v'y