I know physicists don't often like to hear about "photon's frame of reference", so I apologize in advance :)

If the time dilation at the speed on light is infinite, it should mean that if Bob is observing a photon traveling from the sun to the earth, then from the photon's frame of reference (sorry again:), no time should have passed for Bob from the moment of its emittance to the moment of it reaching earth.

How come then that in reality, roughly 8 minutes pass for Bob? How does the fact that light has a finite speed settle with the theory of infinite time dilation?

And one more thing: if the events of emittance and reaching the earth are simultaneous for the photon (sorry:), does that imply that the photon exists in a 4D space, where all events are simultaneous?

  • $\begingroup$ Well, in the "photon's frame of reference", the photon gets to a point it sees instantaneously (in terms of special relativity intervals, interval is 'lightlike'). However, in Bob's frame of reference, he sees a photon travelling $3 \times 10^8 $ meters in a second (because that's how the speed of light remains constant, the interval is 'timelike'), giving the 8 minutes. $\endgroup$ May 11, 2016 at 6:36
  • $\begingroup$ The simple fact is that the relevant "distance" between any two points of the universe seems to be (nearly) zero when we imagine that we go at (nearly) the speed of light. That is, indeed, true. The travel time that it takes to get from here to anywhere in the universe can be made as small as we wish, if we can afford the energy or, if we can reduce ourselves to a light wave. You should, by the way, stop thinking about photons as material objects. They are merely measurements of the state of a field. If there is no measurement, it makes little sense to talk about them, at all. $\endgroup$
    – CuriousOne
    May 11, 2016 at 6:47

1 Answer 1


You have slightly misunderstood time dilation. Suppose you are moving at some speed $v$ relative to me, then in my coordinates the time $t$ I measure for you to move a distance $d$ is:

$$ t = \frac{d}{v} $$

Andd even if you are a ray of light this still applies so the time I measure for a ray of light to move a distance $d$ is:

$$ t = \frac{d}{c} $$

So the time I measure for a ray of light to travel the eight light-minutes from the Sun is just eight minutes.

Time dilation means that the time $t'$ you measure for you to move from the Sun to me is not equal to the time $t$ that I measure. The time you measure is:

$$ t' = t\sqrt{1 - \frac{v^2}{c^2}} $$

And for any $v \gt 0$ this means $t' \lt t$ i.e. you measure less time than I do. In limit of $v \rightarrow c$ your time $t' \rightarrow 0$.

You're final question cannot be answered because anything moving at $c$ has no rest frame. You simply can't ask how spacetime appears to a photon because the question is meaningless.

  • $\begingroup$ You could ask the question for a neutrino though. $\endgroup$
    – csiz
    May 11, 2016 at 11:13
  • $\begingroup$ So if I'm a neutrino and I stare at Bob's clock I would see it moving 8 minutes during my approach (which takes t' very little for me). Why do I see Bob moving in fast forward, while Bob sees me in slowmotion? $\endgroup$
    – csiz
    May 11, 2016 at 11:26
  • $\begingroup$ I've just realized that Bob would also see me in fast forward (well if neutrinos interacted with light; lets switch to spaceship) . For he would first see my image when I leave the sun right before I pass him because I'm moving at nearly c. $\endgroup$
    – csiz
    May 11, 2016 at 11:45

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