# Light beam catching up to a driving truck

Suppose I have a truck which is $$\Delta x$$ meters long driving at $$v=c/5$$ and at $$t=0$$ a ligt beam hits the rear end of the truck.

I would now like to calculate the time it takes for the light to reach the front end of the truck for a resting observer.

Knowing that the truck gets Lorentz contracted for the resting observer I tried to calculate the time as follows: $$\Delta t=\frac{\Delta x}{\gamma (c-v)}$$

Sadly I'm not sure if calculating the relative speed by simply substracting the two speeds is allowed here because wer're dealing with relativistic speeds.

Or do I need to use the relativistic velocity addition formula in this case? Would be great if someone could help me out here!

• Hints: Do you think the truck is $\Delta x$ meters long, or does the driver think so? You know where the back of the truck is when the beam hits it. You know how fast light travels. You know how fast the truck travels. Where will the front of the truck be when light gets there? Mar 14, 2021 at 19:13
• For me (the resting observer) the truck should be $\frac{\Delta x}{\gamma}$ long, for the driver it's just normal $\Delta x$ long. Your question would lead me to the following equation: $$ct=\frac{\Delta x}{\gamma}+vt$$ Solving this for $t$ gives me again the formula from my post. So it should be correct I guess? Thanks also for taking the time to answer! Mar 14, 2021 at 19:22

Time dilation, length contraction and the velocity addition formula are all consequences of the Lorentz transformations. Using them won't get you anywhere you can't get from the transformations themselves, and only makes it easier to confuse yourself.

You have two events: Light hitting the back of the truck and light hitting the front of the truck. Write down the coordinates of these points in one frame, then transform to the other. Now you're done.

I find the easiest way to figure these kinds of things out is to assign explicit coordinates to events. In this case we have two events, A (the light hits the back of the truck) and B (the light hits the front of the truck). We have two coordinate systems, the Earth frame and the truck frame. For convenience lets put both origins at event A, so in both the Earth and truck frames the coordinates of A are (0,0). The truck is $$\Delta x$$ units long in its rest frame, and the speed of light is constant in all frames, so the coordinates of event B in the truck frame are (t, x) = ($$\Delta x/c$$, $$\Delta x$$). Now use the Lorentz transformation to convert those coordinates to the Earth frame:

$$\gamma=1 / \sqrt{1-{v^2}/{c^2}}$$

$$t' = \gamma(t - vx/{c^2}) = \gamma({\Delta x}/c - v{\Delta x}/c^2)$$

$$x' = \gamma(x - vt) = \gamma({\Delta x}/c - v{\Delta x}/c)$$

Since the time coordinate of event A is 0, the time for the light to hit the front of the truck in Earth's frame is the $$t'$$ above.

It's always helpful to use the full Lorentz transformation because it accounts for all relativistic effects (length contraction, time dilation, and relativity of simultaneity).

The short answer is, yes .

The formula you have used to calculate the time is correct.

Eric Smith has given a more mathematically rigorous derivation. Except , he has made a slight mistake of using Lorentz transformation. Since, he is transforming from truck frame to outside observer frame, he must use the inverse Lorentz transformation instead of the Lorentz transformation. So, the formula for t' he uses should have a +vx/c^2 instead of the -vx/c^2 that he uses. Once, you make that correction his formula will give the same result as the formula you have posited in your question. Just rearrange some of the terms, and both are the same formula