I hope I'll be forgiven if what I'm about to write is not entirely clear.

I find it very difficult to find an explanation of SR that is coherent when stated in English, or diagrammatically, rather than in mathematical terms - particularly in relation to effects like time dilation. It would appear many wrestle unsuccessfully in translating between the two forms of description.

To draw a picture of a situation in classical physics, imagine that a person A is throwing a football to person B - both are stationary, and the football is thrown at the speed of light. The time the football takes to travel between the two (the "travel time") is a purely a function of how far apart they are (the speed of the football's travel already being a given).

Again, staying in the realm of the classical model of physics, if the person B is moving backwards away from A, then the travel time is a function of (a) the distance between A and B when the football was thrown, plus (b) the amount of extra distance between A and B, which B has gained during the time elapsed between the instant when the football was thrown and the instant when it was caught.

If A were throwing a succession of balls to B (for example, throwing a ball once every second), whilst B was moving constantly backwards away from A, then B would receive them at a slower rate than A were throwing them (the implication being that, as B gains distance from A, an increasing number of balls would be seen to be mid-air, so that eventually the distance between them would be such that A would be throwing the next ball before B had received the previous one).

This much I assume is uncontroversial.

Now in special relativity, as B's motion away from A approaches relativistic speeds, the "travel time" of the football is not just a function of (a) the distance between A and B when the football is thrown, and (b) the extra distance gained by B until the football is received, but also (c) some additional amount that is not accounted for by the classical model of physics, but by a function of B's speed and the Lorentz factor.

Have I understood the situation correctly?

If so, I will shortly come to my actual question, but first I will just make explicit some more of my thinking. In this football analogy, the football itself is a discrete thing that is both thrown and subsequently caught in a single instant. That is, the throwing of the ball is something that occurs in an instant, and it's catching occurs in an instant - the ball is either in-hand or else it is in mid-air, but never both.

But light waves actually have a cycle-time (or alternatively, light conceived as a particle has a non-zero diameter, and it's full extent is not present purely at a single point in space, but in a volume of space). Properly understood then, light is therefore not "thrown" in an instant, or "caught" in an instant, but rather it takes a period of time for each to fully occur (to emit a full wave, and for a full wave to be received).

It is not easy to capture this behaviour with the football-throwing analogy (or any analogy that involves the use of a single familiar physical object). The only way I can sensibly modify the analogy is by introducing and analysing two footballs as a single transaction, one thrown first to represent the leading/rising edge of the light wave, and another thrown subsequently to represent the tail/falling edge of the light wave.

Using this modified analogy of light as being a single transaction involving two footballs, would I be right (if those reading this have followed me so far) that the Lorentz factor describes the additional period of time (which I referred to as component "c" earlier) that accrues between B catching the first football, and catching the second football? That is, the amount of time that it takes him to fully catch the two-ball transaction, when the implication (mentioned earlier) of him moving away from A at speed is that the time between catching each single football increases?

The key insight I'm employing here is that the throwing and the catching (the transmission and reception) is not an event that takes place in an instant, but a process that takes time to complete (and if B is moving away from A, then it takes place across more than a single point in space, so that the trailing edge of the light wave must travel farther than the leading edge in order to reach and be absorbed by B). Have I got this right or am I quite wrong?

  • $\begingroup$ Yes what you are describing is the relativistic doppler effect. If A throws the footballs with a frequency $f$, B will receive them at some modified frequency $f'$. Even classically there is a doppler effect just due to the changing distance, but time dilation also needs to be taken into account for relativistic velocities. $\endgroup$
    – octonion
    Commented Dec 23, 2017 at 10:45

1 Answer 1


There are two directly measured proper times: $T_A$ which is the time A waits between throwing two footballs, and $T_B$ which is the time between B catching two footballs. We want to derive a formula relating them, so we will first choose a reference frame. If we choose the reference frame where A is at rest, then B is moving and A calculates a dilated time interval $T_B'$. $T_A$ will be related to $T_B'$ through the classical Doppler effect. And then we can use time dilation to relate $T_A$ and $T_B$ directly. We can also do the same in B's frame and I will show it gives the same result so there is no paradox.

$B$ is moving with velocity $v$ away from A, and the speed of the football is $c$, the speed of light.

A's frame:

When B catches the first football, the distance of the second football from B is $c T_A$. But B is moving away so it actually takes $$T_B'=\frac{c}{c-v}T_A$$ for that second football to catch up to B. If we stop right here this is the classical Doppler effect with a moving observer and stationary source.

Since B is moving away and thus $v$ is positive, $T$ is bigger than $T_0$ so you might think that's the explanation of time dilation. But that's not the case, really we haven't even used relativity yet. Notice by the way that if B was moving towards A, $v$ is negative and then $T$ would be shorter than $T_0$, so this clearly isn't the same thing as time dilation.

Now to relate $T_B'$ to the proper time $T_B$ we just use the time dilation factor $$T_B'=\frac{c}{\sqrt{c^2-v^2}}T_B$$ So now combining the two equations we get the relativistic Doppler effect for the time $T_B$ that B actually observes between catches compared to the time $T_A$ that A actually observes between throws $$\boxed{T_B=\sqrt{\frac{c+v}{c-v}}T_A}$$

B's frame

To show there is no paradox, I'll derive the same formula starting in B's frame. In B's frame, A is moving and throwing the football every time interval $T_A'$.

The length between the two traveling footballs is $T_A'(c+v)$ since A moves away during that time. So the time interval B catches them is $$T_B=\frac{c+v}{c}T_A'$$ This is the classical Doppler effect with a moving source and stationary observer.

Now if you just substitute in our expression relating $T_A$ and $T_A'$ through time dilation $$T_A'=\frac{c}{\sqrt{c^2-v^2}}T_A,$$ we get the exact same formula relating $T_A, T_B$ as before. So there is no paradox.

  • $\begingroup$ just to confirm my understanding, is the time dilation factor present equally and identically when B is approaching A as much as when he is receding from A? That is, as you say, when B is approaching A, the frequency of each catch increases according to classical Doppler, and yet relativistically there will still be less of an increase in the frequency than would be expected classically. Or to put it another way, time dilates (the frequency of catches reduces) according to the speed of B, regardless of whether B is moving toward or away from A at that speed? $\endgroup$
    – Steve
    Commented Dec 24, 2017 at 21:44
  • $\begingroup$ Yeah that's right, since the dilation factor only depends on $v^2$ it's the same approaching or receding. But actually there will be more of an increase in frequency than classically since the dilated time interval is T in A's frame where B is moving, not T' in B's frame where B is at rest. It $\endgroup$
    – octonion
    Commented Dec 25, 2017 at 5:52
  • $\begingroup$ I'm still not sure that I've understood which way the relativistic component bears on the frequency that B receives. Does it's effect correspond to the "directionality" of the classical behaviour but in extra amounts (i.e. moving away decreases frequency, moving towards increases), or does it operate purely as a one-sided bias in the received frequency (related to speed, but independent of direction)? If it is the latter, is it a one-way bias that decreases the received frequency in both directions (as I originally understood), or a one-way increase (as you appear lastly to say)? $\endgroup$
    – Steve
    Commented Dec 25, 2017 at 7:06
  • $\begingroup$ I tried to explain this in my answer but maybe it would help to put into different words. The classical frequency $1/T$ is the frequency calculated in A's frame of football catches which are happening at a moving object (observer B). The idea of time dilation is that A is observing time to be slowed down for the moving object, so in B's frame, where B is not moving, the football catches must be happening quicker. That is why the frequency is increasing compared to the classical answer. $\endgroup$
    – octonion
    Commented Dec 25, 2017 at 16:59
  • $\begingroup$ That has only confused me further, but I hope you'll bear with me, because I'm missing something. I am interested in the frequency of catches that B perceives (in B's frame, measured according to the clock that B has with him) as a result of balls thrown by A. If time dilation occurs in both directions of movement (i.e. if the dilatory effect experienced as A and B recede from one another is not recaptured as time contraction on the inbound stretch if A and B reconverge, as it would be in purely classical Doppler), that seems to break the symmetry and relativity of the situation. $\endgroup$
    – Steve
    Commented Dec 26, 2017 at 9:16

Your Answer

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

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