# Paradox Regarding Time Dilation at the Speed of Light

Consider the following scenario: On earth, I pulse a laser focused at a far away mirror such that the time it takes for the light to reflect off the mirror and arrive back at me is at least a few seconds. Immediately after the laser pulse, I place a second mirror where the reflected beam will arrive. In the photon's frame, it would appear that the universe is moving around passed the photon at the speed of light. Furthermore, time on the moving universe would be infinitely dilated. Thus, the photon would never see the second mirror being raised. So how does it reflect off of it?

EDIT: To be clear, I don't think moving frame velocity being equal to the speed of light here is necessary. I've written a similar example to my question below:

Consider an electron accelerated to 0.9c and then released in some arbitrary direction. After some amount of time, let the electron encounter an infinitely strong potential such that it reflects back to its release point. Meanwhile, an experimentalist calculates when they expect the electron to arrive and turns on an electric field right before it arrives that will curve the electron's trajectory so that it collides into a detector. In the electron's frame, the universe's time is dilated so that it never sees the electric field being turned on. Thus, how does it interact with the field and get detected?

• In this case as pointed out in the answers, there is no photon frame as its mass is 0. But also please keep in mind when working with thought experiments that paradoxes arising from supposed infinities indicate that one reaches the limit of the equations/assumptions/postulates in the framework displaying infinities. Example: the electron is a point particle, the field grows like 1/r^2, the field is infinite (approaches infinity) at the electron point, tells us that a new theoretical framework than classical electrodynamics is necessary. – anna v Jan 3 '14 at 5:42
• Please refer to my edit. Also, in what limit are you claiming that special relativity fails? The only case I can think of in my example is perhaps during the period where the electron is reflected off of the potential barrier. However, I don't think that will change the problem as usually such cases can still be worked out using special relativity by integrating over the instantaneous changes in velocity. – mcFreid Jan 3 '14 at 5:47
• in your second example it is not special relativity but the "infinite strong potential" that fails and calls in quantum mechanics, as in my example in the first comment. There are no infinitely strong potentials because nature is quantized. Special relativity works within quantum field theory bu the concept of classical potential fails. – anna v Jan 3 '14 at 6:36
• I added an edit addressing your edit. – user12029 Jan 3 '14 at 7:32

## 2 Answers

Leaving aside the troublesome photon and considering a highly boosted particle, this is just another version of the twin paradox. My favourite resolution for this is to point out that the particle is in different inertial frames when leaving and returning to the Earth, so it must have accelerated in between. It's during this period that the electron sees the electric field being turned on.

To see this start with the electron leaving the Earth and use the usual Lorentz transformations to treat it's outward journey. Now start at the point the electron returns to the Earth and evolve time backwards again using the Lorentz contractions (this is obviously symmetric to the outward journey). At the spatial point where the two trajectories meet the times on the Earth clock don't match. The acceleration fills in the bit in between them.

Calculating the effect of the acceleration is straightforward, and the equations are given in John Baez's article on the relativistic rocket. You could with enough effort (it's somewhat messy) calculate what time on the electron's clock corresponds to the time on Earth that the field is switched on. I'll leave this as an exercise for the reader :-)

• I wonder though, couldn't the same experiment be performed without the need for reflection. Consider two people on earth, person A and person B, that synchronize their clocks. person B tells person A to expect an electron in exactly 1 hour. person B then slowly travels to an electron accelerator and fires off an electron at person B such that person A will see it arrive in 1 hour. Knowing this, person B turns on an electric field 1 second (or some arbitrarily short time) before the electron is to arrive. Again, I don't see how the electron could see the electric field being turned on. – mcFreid Jan 3 '14 at 11:10
• The only remedy I can think of is that it has to do with time dilation during the acceleration process of the electron, but it seems to me that if you made the distance between person A and B long enough you could neglect the effects of this part of the acceleration process (i.e. consider the electron to be accelerated far enough in the past such that in it's frame, it would not observe the universe's time to speed up sufficiently to the point where it is expected to arrive at person A) – mcFreid Jan 3 '14 at 11:12
• If the electron is not created (e.g. by pair production) during the expt then its clock is initially synchronised with A and B, and its time is continuous so there must be a mapping between when B turns on the field and the electron time. If the electron is created during the experiment then different observers will disagree about the time the electron was created. I think you need to come up with a more concrete expt if you want a detailed analysis of what's going on. – John Rennie Jan 3 '14 at 11:17
• Well, person A know's how fast he is planning to walk away from person B and know's for how long (i.e. how far) he is planning to walk. Thus if B agrees to turn on the electric field detector at, say, 1pm January 30th 2014, then person A can know exactly when to send the electron towards person B by calculating and accounting for the difference in his clock due to travelling. So, I don't see why there would be any synchronization issues. Once person A arrives at the accelerator, he runs the experiment that both creates an electron and accelerates it to near $c$. – mcFreid Jan 3 '14 at 12:00
• He then fires the electron out of the accelerator towards person B such that it arrives precisely at the time agreed upon (according to person A's clock). Furthermore, person B walks far enough away from person A such that the amount of time the electron spends accelerating compared to the amount of time it is a free electron moving constant velocity towards person A is negligibly small. – mcFreid Jan 3 '14 at 12:00

"X's frame" usually refers to the frame in which X is stationary. With this definition, the photon has no frame, because the photon must always move at the speed of light. You start to err as soon as you say "In the photon's frame."

I'd try to help you further by discussing length contraction and frequency with the relativistic doppler effect (because as $v\to c$ frequency going to $0$ or $\infty$ tells us whether the wave is essentially spread out over all of space or is essentially a point), but it's difficult for me to set up a limiting example because it's hard to figure out what the question is. You might want to try it by finding explicit formulas. Draw your Minkowski diagram with whatever origin you decide, write your electromagnetic field strength as a function of space and time (ignoring all the fancy electromagnetic effects; I am referring to something like $\text{magnitude}(t)=\sin(\omega (x-ct))$), write your spacetime events of the photon being emitted or reflected, apply your Lorentz transformation, and look at what happens in your other frame as the frame velocity goes to $c$ or $-c$. You might get an argument saying, "This happens in an infinite amount of time, but the photon is so red it takes an infinite amount of time to reflect", where "infinite" is taken in the sense of limits. But what you see may depend on direction and your choice of origin! (that is physically meaningless because of course what you see depends on your frame of reference)

Regardless, you will always see the photon moving towards or away from you at the speed of light, and you will always see that events happen in the correct order. There is no "photon's frame".

Edit to address the new question

If we're dealing with an electron in space we might as well talk about a spaceship arriving somewhere and turning around, because the case is analogous and is more commonly discussed. It sounds precisely like you're asking about the twin paradox. The answer is that in the electron's instantaneous rest frame at laboratory time $t$, during the short period of time the electron is accelerating its plane of simultaneity changes dramatically. If you drew your coordinate system in the electron's frame it would look like you're travelling forwards in time on the detector and emitter's side, and backwards in time on the side opposite from the detector and emitter. This is not actual time travel because the information the electron receives is the information on its light cone, which does stay constant through all frames of reference. (Lorentz transformations take light cones to light cones.) This is illustrated as the difference between the red and blue lines in this twin paradox Wikipedia diagram.

So this is a meaningful question, but don't mistake the word "paradox" for an actual paradox. Everything is consistent.

One way to look at it is the frequency of a light emitted at the origin. Since light is a wave it can represent the frequency of information arriving at the spaceship (or electron). The formula derived on the relativistic doppler effect page gives the formula for the frequency detected at the spaceship/electron as $\sqrt{\frac{1-\beta}{1+\beta}}f_s$, where $\beta$ has the standard meaning and $f_s$ is the frequency of the source. ($\beta$ is negative if the observer [spaceship/electron] and source [detector/emitter] are moving towards each other).

At first as the electron moves away, $\beta$ is positive and the detected frequency is lower/redder. As $\beta$ is negative the spaceship/electron moves back towards the detector and the frequency is higher/bluer. So while the time of the detector is dilated by $1/\gamma$ in the frame of the electron in both cases with constant velocities, the acceleration means that the rate at which information reaches the spaceship is different in both cases. This could be interpreted comically: As the spaceship flies away people around the detector seem to move in slow motion, and as the spaceship flies back people seem to be sped up, despite the fact that in your coordinate system the detector and people are slowed down by a factor of $1/\gamma$. This is simply because the earth and light are both heading towards you, and so light "piles up" in front of the earth and reaches you in one big burst.

• I've added an edit to include an example for a near-speed-of-light moving electron. – mcFreid Jan 3 '14 at 4:58