The speed of light is supposed to be the same for all observers in an inertial frame, no matter the observer's speed. As a result, time slows down for observers moving quickly, and this explains why light always seems to move away at the same speed. However, what if light moves toward a person in an inertial frame traveling at, say, half the speed of light? Without time dilation, the speed of light would appear to be be moving at a speed greater than the speed of light, in the observer's frame. The only way to fix this problem is for time dilation to occur; HOWEVER, to fix this problem, time should speed up for the observer, not slow down. And yet, for observers moving quickly, time slows down (which solves the problem of light moving away from an observer). Where is my misunderstanding in all of this/ what is going on here?

  • $\begingroup$ If you want to know what an inertial observer sees, and what times & distances they measure, you need to remember that an inertial observer is always at rest in their own frame. $\endgroup$
    – PM 2Ring
    Apr 18, 2019 at 2:54
  • $\begingroup$ Yea, so if the inertial frame is moving at half the speed of light, and light is moving toward the frame at the speed of light, wouldn't the observer see the light traveling at 1.5 times the speed of light w/o dilatation? Wouldn't we need to observer's time to speed up in order for them to really see the light move at the speed of light? $\endgroup$ Apr 18, 2019 at 3:38
  • 3
    $\begingroup$ You appear to have completely ignored the comment of @PM2Ring . Him: "An inertial observer is always at rest in his own frame." You: "So if the inertial frame is moving at half the speed of light..." ..... $\endgroup$
    – WillO
    Apr 18, 2019 at 3:54
  • $\begingroup$ Can't an inertial observer be at rest in his own frame even though his frame is moving? Ie if I'm on a train, isn't my inertial frame moving? I considered this problem as follows: I am on a non acceralerating spaceship (inertial frame) that moves toward some light at speed c/2. W/O S.R., I would see the light move at 3/2 C. With special relativity, time changes so that I always see light move at speed c. Thus, I reasoned that if my ship moves forward some distance in some time, and light moves toward me, time must change so that it seems as tho I the light moves at speed c. $\endgroup$ Apr 18, 2019 at 4:42
  • 2
    $\begingroup$ And here is your other misconception: " time should speed up for the observer, not slow down" . Time does not speed up or slow down; I don't even know what this would mean. If we are in motion relative to each other, then your clocks run normally in your frame and slow in my frame, while my clocks run normally in my frame and slow in your frame. There is no frame in which any clock runs faster than normal (unless the clock is broken!!). $\endgroup$
    – WillO
    Apr 18, 2019 at 5:18

4 Answers 4


A better way to state your scenario is to say an observer is stationary, looking straight ahead down the x-axis, and sees light emitted from a source which is located far from him in the +x direction, and traveling towards him at $c/2$ in the -x direction.

The main relativistic effects he sees are as follows:

– he measures the speed of this light (e.g. with mirrors and an interferometer, comparing it to known metersticks and a clock which are at rest with him) as $c$.

– if the light is emitted in pulses, he sees the time interval between each pulse as shorter (i.e. the pulses are in quicker succession) than are seen by an observer that is stationary with respect to the light source. This might be what you were trying to grasp at by saying "time speeds up"

– As a consequence of the previous point, the peaks and troughs of the light waves arrive more frequently to the observer than they were emitted in the source's frame, resulting in the frequency (and energy) of the light being increased compared to the frame of the light source. This is known as Doppler Blueshift.


What you are thinking about is that time dilation is relative.

Let's say that in your case there is an observer moving at the c/2. OK, but compared to what? There has to be a observer at rest. So the observer moving at c/2 is moving relative to the observer at rest.

Now the observer at rest has a clock, and this observer at rest will see its own clock tick normally. But if he compares his clock to the clock of the observer that is moving at c/2, he will see that the clock of the observer at rest is moving faster then the clock of the observer moving at c/2.

The observer moving at c/2 sees its own clock tick normally. But when he compares it to the other's clock, he will see that the clock of the observer who is moving at c/2 is ticking slower then the clock if the observer at rest.

This is due to time dilation.

Now as per SR, all observers will see light moving at speed c (when measured locally, in vacuum), regardless of the speed of the observers or their direction.

In your case, it does not matter whether the light is coming toward the observer or moving away from the observer, and it does not matter whether the observer is moving toward the light or away from the light. The observer moving at c/2 will always see light moving at speed c.

  • $\begingroup$ This helped, but I'm not sure it answered my question. Lets say theres a guy standing still in an inertial frame (as above). A spaceship moves away from him at c/2. Light moves toward the spaceship and the guy. (lets say this is all in one d). As you said, by SR, both observers will see light moving at speed c. However, by my reasoning, I think this would imply that the clock of the observer traveling speeds up, not that it slows down. Can you go through how this shows that the clock of the traveler slows down? $\endgroup$ Apr 18, 2019 at 4:31
  • $\begingroup$ I understand the case where light moves away from the traveling observer better. (When the light moves away, using the same speeds as above, without dilation the apparent speed would be c/2. If time is dilated s.t. it slows down by a factor of two, however, things work out, because the observer sees the light move at speed c. [Don't know if this is the right reasoning, but that's how I've been thinking about it] ). $\endgroup$ Apr 18, 2019 at 4:35
  • $\begingroup$ "Philosophy [i.e., physics] is written in this grand book - I mean the universe - which stands continually open to our gaze, but it cannot be understood unless one first learns to comprehend the language and interpret the characters in which it is written. It is written in the language of mathematics, and its characters are triangles, circles, and other geometrical figures, without which it is humanly impossible to understand a single word of it; without these, one is wandering around in a dark labyrinth." Galileo Galilei, Il Saggiatore (The Assayer, 1623). $\endgroup$
    – Elio Fabri
    Apr 18, 2019 at 12:30
  • $\begingroup$ @MarcelMazur ok so in your case the spaceship moves at c/2 away from the guy standing still. Light moves toward them. Both will see light coming at c. Why? You are right, time dilation causes the spaceship to slow down in time. In the time dimension, light moves at 0. The spaceship moves in the time dimension with c/2. The guy standing still moves in the time dimension with speed c. Now you need to read about the four vector. The four vector is built u so and the universre is built up so that its magnitude needs to be c always. $\endgroup$ Apr 18, 2019 at 15:47
  • $\begingroup$ @MarcelMazur So, light moves in the spatial dimensions woth speed c, and in the time dimension with speed 0. The spaceship moves in the spatial dimensions with speed c/2 and in the time dimension with speed c/2. The guy standing still moves in the spatial dimenaions with speed 0 and in the time dimension with speed c. All of them move in spacetime with speed c. Their four vector's magnitude is all c. If the spaceship moves at c/2 in space, it has to slow down in time to c/2. As it closes to the speed of light, it slows down in time more. $\endgroup$ Apr 18, 2019 at 15:50

What you are overlooking is the relativity of simultaneity. The cause of your confusion is that you think time slows down for the moving observer, which is the wrong way of looking at the effects of SR.

Suppose you are moving in a train. You are at the rear of a long carriage and a light flashes at the front. In your frame, the light reaches you after a certain time t'. To people on the platform, the light reaches you after a shorter time t, because you have been travelling towards the point in space at which the flash occurred. In this case t' is greater than t. Now if you repeat the experiment, but this time you sit at the front of the carriage and the light flashes at the rear, the arrangement is switched- people on the platform will think light took longer to reach you because you were travelling away from the point in space at which the light flashed in their frame

So in one case you have t' being greater than t, and in the other you have t' being less than t.

The way to understand this is that a plane of constant time in your train frame of reference is a tilted slice through time in the frame of the platform. Conversely a plane of constant time in the frame of the platform is a tilted slice through time in the frame of the train. That means clocks along the train are always out of synch with clocks along the platform, and it is the lack of synchronisation that allows the speed of light to be c in both frames.

The clocks on the train and on the platform all tick at the same rate, but the lack of synchronisation gives the effect of time dilation. To see this, imagine you are on the train passing a line of evenly spaced clocks on the platform. If those clocks are all out of synch to you, so that each clock you pass is set 1 second ahead of the last one you saw, then when you compare the time on your watch to the time on the passing clock, you will think your watch has lost another second (ie to be dilated) every time you pass another clock. In reality, your watch is ticking at the same rate as the clocks, but seems to be running slow because of the lack of synchronisation.

Now imagine someone on the platform watching a corresponding set of evenly spaced clocks on the passing train. To them, the clocks on the train are out of synch, each clock that passes being set 1s ahead of the previous one. To that person on the platform, their watch will seem to be losing 1s with every passing clock on the train, owing to the synchronisation effect.

So, both you and the person on the platform get the impression that your watches are running slow (ie you both think you are time dilated) when in fact your watches and all the clocks are ticking at the same rate, and the effect of dilation arises because the clocks in your respective frames are out of synch.

This is really a property of the geometry of flat spacetime. When two people are stationary relative to each other they have a common time axis and they exist on a common plane of simultaneity. But as soon as they move relative to each other their respective time axes point in different directions and their respective planes of simultaneity tilt, so that time in one frame becomes out of synch with time in the other.


Assuming that the light is emitted at point $A$ and the observer is at point $B$, the distance $AB = r =ct$ , if the emitter $A$ moves with a speed $v$ , we have:$ \vec{BA'}=\vec{BA}+ \vec{AA'}$, with: $\vec{BA'}=\vec{c}t'\;,\vec{BA}=\vec{c}_{0}t\;,\vec{AA'}=\vec{v}t'\;,|\vec{c}|=|\vec{c}_{0}|=c $ A simple calculation gives : $t'=\frac{t}{\frac{v}{c}\cos(\theta)+\sqrt{1-\frac{v^{2}}{c^{2}}\sin^{2}(\theta)} } $

for: $\theta=0,t'=\frac{t}{1+\frac{v}{c}}$


For $\theta=\pi/2$, we find : $E'=\sqrt{1-\frac{v^{2}}{c^{2}}}E$, this is the transverse Doppler effect.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.