Lorentz transformation of time

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We have that

$\displaystyle {x}'=\gamma (x-vt)$

Since the laws of physics are the same in all inertial systems, that transformation must be possible to do in the opposite direction, but now with a plus for the velocity term because the relative motion is in the opposite direction.

$\displaystyle x=\gamma \left( {{x}'+v{t}'} \right)$

We now have two equations that can be used as a system of equations.

Try to derive the following from the above equations before looking at the rest of the page. This question is definitely beyond the DP course, but the questions further down the page are not.

$\displaystyle {t}'=\gamma \left( {t-\frac{{vx}}{{{{c}^{2}}}}} \right)$

We have

$\displaystyle x=\gamma \left( {{x}'+v{t}'} \right)\quad \quad \text{//divide by }\gamma$

$\displaystyle \frac{x}{\gamma }={x}'+v{t}'\quad \quad \quad \ \text{// do the substitution }\!\!~\!\!\text{ }{x}'=\gamma (x-vt)$

$\displaystyle \frac{x}{\gamma }=\gamma \left( {x-vt} \right)+v{t}' \quad \ \ \text{//divide by }\gamma$

$\displaystyle \frac{x}{{{{\gamma }^{2}}}}=x-vt+\frac{v}{\gamma }{t}'$

But

$\displaystyle \frac{1}{{{{\gamma }^{2}}}}=1-{{\left( {\frac{v}{c}} \right)}^{2}}=1-\frac{{{{v}^{2}}}}{{{{c}^{2}}}}$

$\displaystyle x\left( {1-\frac{{{{v}^{2}}}}{{{{c}^{2}}}}} \right)=x-vt+\frac{v}{\gamma }{t}'$

We can now expand the left-hand side. Then the x-terms will cancel to give us

$\displaystyle -\frac{v}{{{{c}^{2}}}}x=-t+\frac{{{t}'}}{\gamma }$

Dividing by v gives us

$\displaystyle -\frac{v}{{{{c}^{2}}}}x=-t+\frac{{{{t}' }}}{\gamma }$

We now add t and multiply by γ. This finally gives us

$\displaystyle {t}'=\gamma \left( {t-\frac{{vx}}{{{{c}^{2}}}}} \right)$

We now have the equations for Lorentz transformations for an object moving along the x-axis.

$\displaystyle \left\{ {\begin{array}{*{20}{c}} {{x}'=\gamma (x-vt)} \\ {{y}'=y} \\ {{z}'=z} \\ {t\prime =\gamma \left( {t-\frac{{vx}}{{{{c}^{2}}}}} \right)} \end{array}} \right.$

At low velocities, γ is very close to 1, and vx/c² is very close to 0. This will turn the Lorentz transformation into the non-relativistic Galilean transformations

$\displaystyle \left\{ {\begin{array}{*{20}{c}} {{x}'=x-vt} \\ {{y}'=y} \\ {{z}'=z} \\ {t\prime =t} \end{array}} \right.$

We continue with the spacecraft and asteroid from the previous page, where a spacecraft is flying past an asteroid at a relative speed of 0.80c. Two huge flashlights send out light flashes on each side of the asteroid as seen from the spacecraft. The flashes are sent at the same time, according to observers on the asteroid. The asteroid is 900 meters long as measured by the observers on the asteroid.
1. Find the time between the two light pulses in the reference frame of the spacecraft?
2. Which lamp flashed first in the reference frame of the spacecraft?

Link to key

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Last modified: Jun 10, 2025 @ 15:56