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From what i know, photons are moving at the speed of light and that means the time that photons are experiencing is zero, but here is the thing... If the time is zero how can they travel astronomical distances? From the relation of distance with speed and time D=C×t Then, D=(3×10^8)×0=0m How we can see (not physically) photons moving (or coming) from another galaxy?

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    $\begingroup$ D=C×t is not relativistically invarant. $\endgroup$ – Lewis Miller Dec 23 '18 at 2:39
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    $\begingroup$ physics.stackexchange.com/q/168091/37364 $\endgroup$ – mmesser314 Dec 23 '18 at 4:55
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    $\begingroup$ 1- There is no inertial frame of reference the moves at the speed of light with respect to another inertial frame, so it does not make sense to say $\Delta t=0$ for photons, there is no way to compare the passage of time for a photon. 2- Even if you could go to the frame of reference of the photon and $\Delta t=0$ there, the speed of the photon there would be 0, it wouldn’t make sense to say $v=c$ in that frame. $\endgroup$ – Hugo V Dec 23 '18 at 5:20
  • $\begingroup$ The distance a photon moves is $d=c \times t$ where t is the time elapsed in you frame of reference, which is expected, since if you say a photon is moving at the speed of light, it has to change its position exactly at the speed of light $\endgroup$ – Hugo V Dec 23 '18 at 5:28
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    $\begingroup$ They change in position because that's what it means to be moving. $\endgroup$ – WillO Dec 24 '18 at 0:51
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I think the pitfall here is that you talk about "the time" when there is actually more than one time to consider. There's the time experienced by the photon, and a time experienced by anyone moving at sub-$c$ velocity.

Minkowski Spacetime

This diagram from Wikipedia is a nice illustration. Imagine you have two friends, F1 and F2, that move at velocities $v'$ and $v''$. A photon is whizzing past all of you.That is the yellow line going through the 1st and 3rd quadrants. Now, the representation of the Lorentz transformation between different reference frames is that the axes of the frames that aren't yours seem to close around the photon world line like scissors. How you perceive a moving particle - the photon or your friends - is seen as their projections onto the space and time axes. Your friend F2 at $v''$ will have different projections onto your $ct$ and $x$ axes than onto the axes of your friend F1 moving at $v'$. Their space will seem shorter, and their time slower, to you, and you will measure $v''$ as higher than will F1. All of you will experience $c$ as the same, though.

But - and here comes the crucial point - At the speed of light, the spatial and time axes are the same axis. The scissors have closed completely. The projection onto the space axis along the direction of the time axis is not defined. Both time and distance are zero. That, however is only in ther reference frame of the photon. Meanwhile, in the other (black and blue) reference frames, the world line of the photon still has nice and well defined projections onto your time and space axes, and all is good.

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    $\begingroup$ Photons do not have a reference frame, by definition. A reference frame is a set of one timelike and three spacelike orthonormal vectors. $\endgroup$ – Dale Dec 24 '18 at 1:11
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Remember that length contraction also takes place: If you were to run along side the photon at close to the speed of light, the distance between the galaxies would be contracted by the Lorentz Factor γ, which 'cancels' with the time dilation.

You could imagine that at the speed of light, the distance between the galaxies becomes zero, but you have do be careful with diving by zeroes.

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Light moves at 3x10^8 m/s not an individual photon.

It is meaningless to talk about the velocity of an individual photon. It is a quantum particle. When it is detected at a distance x from the source, it has probability for travelling along any possible path from the source to the detector. (See Feynman's QED) The time at which it was emitted by the source is unknown. So neither the distance travelled nor time taken are experimentally determinable.

When the velocity of light is measured it is the average for a large number of photons. Or if you prefer, a light wave

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The first fact that should be corrected is that photons are not created equal. Which means that they all don’t travel at the speed of light. They travel near the speed of light. And we are talking about traveling 99.9 percent of the speed of light.

Some special photons that are still under research actually slow down exponentially over distance.

So why are we able to see?

Well due to the principal of reflection. The mass of material that photons bounce and are absorbed by actually redirects some of the light. This is why when you take a flashlight, turn it on and point it at the ground it will not be as bright as if it were to be the same distance from you and the ground but pointing towards your eyes.

So basically only half of the light that we see most of the time is actually reflected.

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