## Why don't photons experience time?

 Quote by PeterDonis We can't. It's physically impossible. That means there's no point in even considering such hypotheses; it is like asking what would happen if 2 were an odd number. Because there's no such thing. Objects that move on null worldlines are fundamentally different, physically, from objects that move on timelike worldlines. Trying to understand photons by making analogies with objects that move almost at the speed of light is not a good strategy; it focuses attention on the wrong things.
Then pardon the analogy, It just attempt to extent the OP logic,
 Quote by la6ki If yes, then why can't we extend the same logic to a photon?
What I'm trying to say is, particle moving at 0.9999c will see it's clock ticking slower (after it return to earth) than 0.99c. If the travel is 0.999999999999...9c it will see its clock nearly stop ticking when it return. By that analogy only then when it 1c the ticking will be no longer nearly stop, but stand still.

The same way $$\lim_{x\to a^-} f(x)\ = 0$$ when right left limit in not same, we still draw empty circle at the end of the line, that is at y=0

But then off course the saying is different when v=c. I get it now.

Btw, saying "Time stand still" is better than saying "There is no time" or "Time do not apply", imho.

Can you answer my previous question, how to write down the event of emitted and absorb of photon and why it's null interval is zero, is because dt=0 and dl=0, or because c.dt=dl

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 Quote by SysAdmin What I'm trying to say is, particle moving at 0.9999c will see it's clock ticking slower (after it return to earth) than 0.99c.
But for this to happen, the moving particle has to turn around. If it just keeps moving at .99c or .9999c, it will never come back to Earth. It's the turning around and coming back, so that it meets up again with a clock that stayed on Earth, that produces the difference in elapsed time.

My point is that, by focusing on the speed the particle is traveling, instead of the fact that it goes out, then turns around and comes back, you're focusing on the wrong thing.

 Quote by SysAdmin Btw, saying "Time stand still" is much more acceptable than saying "There is no time" or "Time do not apply", imho.
Not if saying "time stands still" gives the wrong impression and leads to incorrect inferences.

 Quote by SysAdmin Can you answer my previous question, how to write down the event of emitted and absorb of photon and why it's null interval is zero, is because dt=0 and dl=0, or because c.dt=dl
Because c dt = dl. Look at the two events I wrote down:

(0, 0, 0, 0) and (1000, 1000, 0, 0)

I used units where c = 1, so we have dt = 1000 and dl = 1000; the interval is null because the two are equal.

 Quote by PeterDonis Not if saying "time stands still" gives the wrong impression and leads to incorrect inferences.
The "time stands still", is after I read the text book. As reader it less confusing for me.

 Quote by PeterDonis But for this to happen, the moving particle has to turn around. If it just keeps moving at .99c or .9999c, it will never come back to Earth. It's the turning around and coming back, so that it meets up again with a clock that stayed on Earth, that produces the difference in elapsed time. My point is that, by focusing on the speed the particle is traveling, instead of the fact that it goes out, then turns around and comes back, you're focusing on the wrong thing.
I'm trying to see the OP point of view, that is why I edit the comment and add the limit notation, you know, if the limit goes to zero, than it will no surprise if you draw the blank circle in the end of the line, that is y=0

 just to play Lucifer's Lawyer in the photon cause: I submit that there is ample reason to consider the passage of photon time. Consider: Our concept and measure of time is based on change, Periodic fluctuation. In this regard photons are intrinsically endowed. SO the photon proper time interval between two points is simply the number of cycles of EM phase transition. $\Delta$t=(D/c)f The time difference between two photons of varying frequency is obviously relative Time Dilation Lite. I rest my case

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PeterDonis

 Because c dt = dl. Look at the two events I wrote down: (0, 0, 0, 0) and (1000, 1000, 0, 0) I used units where c = 1, so we have dt = 1000 and dl = 1000; the interval is null because the two are equal.

very helpful insight....I had forgotten the distinction between a null interval in space versus a null interval in space-time.......

a nice brief discussion here for others who may be learning:

http://en.wikipedia.org/wiki/Spaceti...time_intervals

and also 'null events of a photon' trace out a lightcone....illustration here,
in flat space-time

http://en.wikipedia.org/wiki/Light-cone

This kind of light-cone has some characteristics of null surfaces in cosmological,
also accelerated, horizons, right??

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