Special relativity postulate says that speed of light is constant for all inertial observers, meaning that measures of time and distance must accommodate in a way to show that speed of light is the same for all.

I know about length contraction and time dilation. Under Lorentz transformations, the length measured by an observer in rest frame is shorter than length measured by an observer who's moving along with the thing he measures (for example, a rod of length $L$), moving relative to the resting observer. And the time between two physical events happening in a moving frame is greater for a resting observer than it is for the other one.

What bothers me are equations for these phenomena:

$$\Delta t = \gamma\Delta t'$$ $\Delta t'$ here means proper time, in other words time measured between two events happening in a moving frame of reference, measured by an observer in that frame. So that means, from this equation that $\Delta t'$ is smaller than $\Delta t$. For a guy on the ground, time between two events is longer.

$$\Delta L = \Delta L' / \gamma$$ In the same way, $\Delta L'$ is the length of, let's say a rod, measured by an observer for which the rod is at rest. And let's say they are moving with a velocity relative to the observer in a resting frame. The latter will measure, according to the equation, a shorter length.

Now I am finally getting to the problem that bothers me: If speed of light is the same for both observers, and speed is distance divided by time, meaning that

c = distance/time,

it means that for a moving observer $c = \Delta L'/ \Delta t'$ and for an observer at rest $c = \Delta L / \Delta t$

So if $\Delta L' > \Delta L$, shouldn't it be that also $\Delta t'> \Delta t$ since $\Delta L'/ \Delta t'$ must be equal to $\Delta L/ \Delta t$? From these two equations it follows that two observers measure different magnitudes for speed of light, and they should measure the same.

What am I doing wrong here?


4 Answers 4


One of the big pitfalls of physics equations is using them in situations where they do not apply. In this case, the time dilation formula only applies when the two events occur in the same location in one of the frames. That is the frame that defines $t’$. Here, if you are measuring the speed of light, then there is no frame where the two events are at the same place. So the time dilation formula simply does not apply.

Instead, you must use the full Lorentz transform. When you do that you will indeed find that the speed is c in both frames.

It is my recommendation that introductory students not use the time dilation or length contraction formulas at all. It is too likely to make the sort of mistake you have made. Instead, I recommend always using the Lorentz transform. It will automatically simplify whenever appropriate and will not allow this sort of mistake in situations where the simplified formulas don’t apply.

  • 2
    $\begingroup$ I heartily endorse the last paragraph. $\endgroup$ Commented May 2, 2022 at 17:58

Dale's answer is correct, but this might illuminate it with an example:

Say Gary is on the ground beside a pretty long train carriage - 100 metres long. He measures that the photons from a flash of light at one end of the carriage get to the other end after (100m)/c $\approx 3.3 * 10^{-7}$ seconds.

Sarah is in space going at c/2, so the relevant $\gamma$ is $1/\sqrt{1-(1/4)} = 2/\sqrt{3} \approx 1.15$. She sees the train carriage as being 100m/$\gamma$ $\approx 86.6$ metres long. However she also sees the train carriage moving at c/2! If she is moving in the opposite direction to the flash, she sees it take (86.6m)/c $\approx 2.89 * 10^{-7}$ seconds to cover the 86.6 metres but the other end of the carriage has moved in that time as far as Sarah is concerned! It has moved (c/2) $* 2.89 * 10^{-7} s \approx 43.3$ metres, and the end of the carriage is still moving according to Sarah. Summing up the series will get you to the result that Sarah sees the flash of light take longer to get from one end to the other, so her $\Delta t'$ is indeed $\gamma$ times Gary's $\Delta t$. Even though she measures the length of the train carriage to be shorter, she does not agree that the light has the same distance to cover to get from one end of the train carriage to the other.

  • $\begingroup$ I hope that you started to ask yourself about the case where the flash of light is moving from the end of the train furthest from Sarah back towards her, in the opposite direction of motion. Then Sarah sees the light travel a shorter distance, but as WillO succinctly put, it is a mistake to think of the speed being from one end of a thing to the other end, because where the ends are in space is not agreed on between observers. $\endgroup$ Commented May 2, 2022 at 18:56

Other answers are right but here is the quickest way to see where you went wrong:

The speed of light (or anything) is distance-traveled/time, not length-of-some-rod/time. Of course if the rod is stationary and the light travels from one end of the rod to the other, these are the same thing. But this rod is not stationary.


For a purely qualitative answer, the way that I often think about relativistic rods is to imagine that the rod changes color uniformly in its own rest frame. That is, at any given moment in its rest frame, it uniformly changes from red to orange, yellow, etc. through the color spectrum.

An observer moving relative to it, in contrast, will notice that the rod will appear to be rainbow-colored, akin to:

Rainbow Rod

(This is true even if one accounts for the differences in time(=distance/$c$) between different points of the rod relative to our eyes.)

Using this rainbow rod, consider the measurements made by both observers:

  • The observer who is stationary with the rod will measure its length while the entire rod is of uniform color (red).
  • The observer who is moving relative to it will measure its length from one end of one color (red) to the other end which is a different color (violet).

Now it is more evident that the the act of measuring a rod is finding the space-like distance between two spacetime events, which correspond to the two spacetime points at each end of the rod within the observer's reference frame. Moreover these two events can never be the same in both frames—the stationary observer measures the distance between two red points on the rod, but the moving observer measure it between a red and violet point—which occur at different times relative to the observer stationary with the rod.

What am I doing wrong here?

You're using formulas that assume that the measurement occurs between two spacetime points (events) that are the same in both reference frames. They are not.

The resolution, as other answers have detailed, is to identify what the actual events (points in spacetime) are and to apply the Lorentz transformation to these points to answer questions of how these are measured in different reference frames.

Image credit: https://commons.wikimedia.org/wiki/File:Spectrum4websiteEval.png

  • $\begingroup$ While this is somewhat useful as a mnemonic (for simultaneity) , it has some superficial overlap with the Doppler effect... but with important differences. Length-contraction is a speed-dependent effect involving the spatial-separation between two parallel timelike-lines (worldlines of the ends of the rod). Doppler, however, is a velocity-dependent effect (approaching yields blueshift -vs- receding yields redshift) involving the spatial-separation between two parallel lightlike-lines (successive wavefronts). In my opinion, care is needed not to confuse this mnemonic with the Doppler effect. $\endgroup$
    – robphy
    Commented May 2, 2022 at 17:04

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