lorentz boost Archives - Time Travel, Quantum Entanglement and Quantum Computing https://stationarystates.com/tag/lorentz-boost/ Not only is the Universe stranger than we think, it is stranger than we can think...Hiesenberg Mon, 09 Sep 2024 18:28:42 +0000 en-US hourly 1 https://wordpress.org/?v=6.6.2 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. https://stationarystates.com/special-relativity/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/?utm_source=rss&utm_medium=rss&utm_campaign=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 Sun, 16 Jun 2024 04:40:17 +0000 https://stationarystates.com/?p=473 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 […]

The post 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. appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

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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 cc in one inertial frame SS also travels at the speed of light in a boosted frame S′S’, we need to consider the Lorentz transformation equations for a boost in the xx-direction. The Lorentz transformation equations for space and time coordinates are given by:

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

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

Assume an object is moving with speed cc in the xx-direction in the SS frame. The velocity components of the object in the SS frame are vx=cv_x = c, vy=0v_y = 0, and vz=0v_z = 0.

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

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

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

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

Substitute vx=cv_x = 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}}

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} = c

So, vx′=cv_x’ = c.

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

 

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