Verify that a boost in the x-direction that any object traveling at speed c in an inertial frame S travels at speed c in the boosted frame.

Verify directly from the form of the Lorentz transformation representing a boost in the x-direction that any object traveling at speed c in an inertial frame S travels at speed c in the boosted frame.

 

o verify that an object traveling at the speed of light $c$ in one inertial frame $S$ also travels at the speed of light in a boosted frame S′S’S′, we need to consider the Lorentz transformation equations for a boost in the $x$-direction. The Lorentz transformation equations for space and time coordinates are given by:

x′=γ(x−vt)x’ = \gamma (x – vt)x′=γ(x−vt) y′=yy’ = yy′=y z′=zz’ = zz′=z t′=γ(t−vxc2)t’ = \gamma \left( t – \frac{vx}{c^2} \right)t′=γ(t−c2vx​)

where γ=11−v2c2\gamma = \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}γ=1−c2v2​​1​ is the Lorentz factor, $v$ is the relative velocity between the frames, and $c$ is the speed of light.

Assume an object is moving with speed $c$ in the $x$-direction in the $S$ frame. The velocity components of the object in the $S$ frame are vx=cv_x = cvx​=c, vy=0v_y = 0vy​=0, and vz=0v_z = 0vz​=0.

First, we need to find the velocity components of the object in the S′S’S′ frame. The velocity transformation equations for a boost in the $x$-direction are:

vx′=vx−v1−vvxc2v_x’ = \frac{v_x – v}{1 – \frac{v v_x}{c^2}}vx′​=1−c2vvx​​vx​−v​ vy′=vy1−v2c21−vvxc2v_y’ = \frac{v_y \sqrt{1 – \frac{v^2}{c^2}}}{1 – \frac{v v_x}{c^2}}vy′​=1−c2vvx​​vy​1−c2v2​​​ vz′=vz1−v2c21−vvxc2v_z’ = \frac{v_z \sqrt{1 – \frac{v^2}{c^2}}}{1 – \frac{v v_x}{c^2}}vz′​=1−c2vvx​​vz​1−c2v2​​​

Since vy=0v_y = 0vy​=0 and vz=0v_z = 0vz​=0, the transformations for vy′v_y’vy′​ and vz′v_z’vz′​ will both be zero. We are left with:

vx′=vx−v1−vvxc2v_x’ = \frac{v_x – v}{1 – \frac{v v_x}{c^2}}vx′​=1−c2vvx​​vx​−v​

Substitute vx=cv_x = cvx​=c into this equation:

vx′=c−v1−vcc2=c−v1−vcv_x’ = \frac{c – v}{1 – \frac{v c}{c^2}} = \frac{c – v}{1 – \frac{v}{c}}vx′​=1−c2vc​c−v​=1−cv​c−v​

Simplify the denominator:

vx′=c−vc−vc=c−v1−vc=c−v1−vc×cc=c(c−v)c−v=cv_x’ = \frac{c – v}{\frac{c – v}{c}} = \frac{c – v}{1 – \frac{v}{c}} = \frac{c – v}{1 – \frac{v}{c}} \times \frac{c}{c} = \frac{c(c – v)}{c – v} = cvx′​=cc−v​c−v​=1−cv​c−v​=1−cv​c−v​×cc​=c−vc(c−v)​=c

So, vx′=cv_x’ = cvx′​=c.

Therefore, in the boosted frame S′S’S′, the object still travels at the speed $c$. This verifies directly from the Lorentz transformation equations that any object traveling at the speed of light $c$ in one inertial frame travels at the speed of light $c$ in another inertial frame.