This question already has an answer here:

My understanding is that light does not experience time. In attempting to understand what the universe would be like from the perspective of a photon, the answer I get is that the universe would be essentially 'frozen'. Answers I've seen describe the photon in a universe where it can move through space, but with events around it not moving at all in time.

So...from my frame of reference, it takes about 4 hours for light to go from Neptune to where I am on Earth. Conceivably I could move somewhere during that time to dodge the photons. But if from the photons perspective I don't have time to move out of the way, then there is a paradox of sorts.

Therefore I must be missing something here. Can someone explain how the motion of other things is possible from the perspective of a photon?


marked as duplicate by ACuriousMind, Kyle Kanos, Danu, Colin McFaul, Jim Nov 25 '14 at 21:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The answer is that a photon has no perspective. "A photon experiences no time", while a popular explanation, is wrong, because special relativity says that light moves at $c$ in every frame of reference, so you can never say that a photon is at rest. $\endgroup$ – Javier Nov 25 '14 at 20:14
  • 1
    $\begingroup$ There is no photon frame of reference, so the "perspective of a photon" is a meaningless statement. $\endgroup$ – Kyle Kanos Nov 25 '14 at 20:14
  • $\begingroup$ Oddly, the answer to 'Would time freeze if you could travel at the speed of light?' states that 'you could infer that time is 'frozen' to an object traveling at the speed of light.' Fortunately some of the answers here explain why that is inaccurate. $\endgroup$ – user2092608 Nov 25 '14 at 22:41

A photon cannot be said to have its own inertial reference frame, because inertial reference are defined to be a family of coordinate systems that satisfy the two fundamental postulates of SR, one of which is that light moves at c in all frames. You could construct a coordinate system where the photon was at rest, but since this coordinate system wouldn't be an inertial frame there'd be no reason to expect equations that are specific to inertial frames, like the time dilation equation, would still give correct predictions in this frame.

One thing you might try is considering the limit as an inertial object's velocity approaches c relative to the rest frame of the galaxy, and it's true that in this limit, clocks with low velocities relative to the galaxy would approach being completely stopped. Another point is that if I'm moving at some very large fraction of c relative to the galaxy, not only do I measure the galaxy to be highly compressed in the direction I'm going, but I also see clocks on either end of the galaxy as wildly out-of-sync due to the relativity of simultaneity...if the galaxy is 100,000 light years long in its own frame, and I'm moving at speed v relative to it, then two clocks at either end of the galaxy which are synchronized in the galaxy's frame will be out of sync by (100,000 ly)*(v)/c^2 in my frame. So, in the limit as v approaches c, clocks on either end of the galaxy are out-of-sync by 100,000 years, so at the same moment that it's 2014 A.D. on the leading edge, it's 102,014 A.D. on the trailing edge. And yet in the limit as v approaches c, the distance between clocks along the direction of motion is compressed to zero. So in the limit, perhaps you could say that the photon's entire history is traversed instantly, since it's going zero distance and all the different clock-readings it passes are squashed together on this zero-length path.

However, I think there would be aspects of this limit that wouldn't really be well-defined, in that they would depend on the details of what are the series of cases that you are using to construct the limit. For example, consider a particle A moving at some v < c, and a photon B moving at c. In the limit as v approaches c, both A and B are moving at c, and B is still moving at c in the frame of A. On the other hand, consider two particles moving at the same speed v, at rest relative to one another. In the limit as v approaches c, both A and B are moving at c, and B is at rest in the frame of A. So, the question of what velocity A would measure B to have "in the limit" would depend on which type of limit you used, even though in both cases they both have a velocity of c in the limit.

  • 1
    $\begingroup$ This answer hits upon the facet I like: That a photon never moves. From its perspective there was no spatial distance along its path, and it took no time to get from one end to the other. Like it is a point in a higher dimension. $\endgroup$ – kitsu.eb Jul 10 '15 at 0:03
  • $\begingroup$ @kitsu.eb - As I said though, there are problems with defining a photon's "perspective". And remember that when physicists talk about an object's "perspective" this is usually just shorthand for some type of coordinate system where it is at rest, which is human-defined rather than something forced on us by nature; although sometimes it can also refer to when various light signals cross its worldline, measured by its own proper time (but the proper time is always 0 between events on a photon worldline). $\endgroup$ – Hypnosifl Jul 10 '15 at 0:40

Two questions and two errors:

The (hypothetical) point of view of the photon is not "frozen" because you need time to perceive something frozen. But the photon has proper time zero, everything is reduced to one instant, thus nothing can be frozen.

"Dodge" a photon: Information is transmitted with light speed. As the photon is moving with light speed it is impossible for you to see the photon before it arrives on your eye, thus it is impossible to dodge a photon.

  • $\begingroup$ I appreciate your answer to the first question. It very simply explains things in a meaningful way. As far a dodging something goes, I don't think I need to see something to get out of its way. Nonetheless I will try and make things clearer in future questions. $\endgroup$ – user2092608 Nov 25 '14 at 22:45

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