# If a photon had a stopwatch

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1. May 5, 2015

I know that a photon traveling from the Sun takes around 8 minutes to get to Earth, but, how long does the photon "think" {if it had a stopwatch} it took to get from the Sun to Earth ?

2. May 5, 2015

### Chronos

zero. A particle traveling at c does not experience time irrespective of distance traveled.

3. May 5, 2015

### Staff: Mentor

I think a better way of putting this is that the concept of "elapsed time" does not apply to a photon. In other words, it's not that the answer to the question is "zero"; it's that the question itself is not well-defined, because a photon cannot have a "stopwatch" in the first place.

4. May 6, 2015

### haushofer

If you give us a Lorentz transformation which brings us to the rest frame of the photon, then we can bring a stopwatch and answer your question :P

5. May 6, 2015

OK, forget the photon, suppose a bloke in a "spaceship" {a bloke who owns an atomic stopwatch, or the like} zipped across 93 Million miles, how long would he think he'd traveled ?

6. May 6, 2015

### Staff: Mentor

That depends on how fast he is moving relative to us on earth. Remember that from his point of view, he is not moving at all; he's sitting in his spaceship while the earth and sun are moving towards him. So as he is sitting there, first the sun passes by his window, and then a little later the earth passes by. You're asking how long is "a little later" - just divide the distance between sun and earth by the speed to get the answer. The only trick is that distance is not 93 million miles - that's what is for us at rest in the solar system, but the solar system is moving relative to him so he'll see a smaller distance because of length contraction.

7. May 6, 2015

Thanks for the above Nugatory, I'm with you on the first part, how "he" would see the sun go by and the earth heading towards him, but, if I divide the distance by the speed of light I get around 8 minutes but that's from my point of view / my time scale, I'm not moving at the speed of light but the photon is, so the photon's time to cross that distance will be considerably much less than the 8 minutes that I perceive.

8. May 6, 2015

### Staff: Mentor

Well, again, the photon cannot measure time. But for astronaut "zippy" who is going at relativistic speeds the time can be much shorter than 8 minutes.

9. May 6, 2015

Appreciated, that the photons time will be different, but, "much" is a moveable feast, shouldn't it be calculable ?

10. May 6, 2015

### DaveC426913

'different' isn't the correct term, the correct term is 'does not apply'.

Yes. See posts 2 and 3.

If you're not satisfied with the correct answer, what kind of answer are you looking for?

11. May 6, 2015

### Bandersnatch

If you mean time elapsed on a spaceship travelling at some fraction of c with respect to the Earth rest frame, use Lorentz transformations to get the length contraction of the Sun-Earth distance in the rest frame of the ship:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html
Then divide the distance you get by the velocity of the ship to get the time elapsed.

As was mentioned before, V=c shouldn't be used in the calculations. Always use some fraction of c (it can be arbitrarily close to c, e.g. 0.9999999c).

12. May 6, 2015

### ZapperZ

Staff Emeritus
May I suggest that, before you get buried even deeper into this, you read one of our FAQ entry?

Please note that our physics, including relativity, are meant to be used, and were formulated, for frames of reference in which c is always a constant and having the same value in all reference frame. Your question implicitly require that one can transform into a photon's frame where one if moving at the same speed as c (i.e. the speed of light is zero in that frame), thus violating all the rules that we currently have. Whether you believe that it can be done or not is a different matter, but what we KNOW is that our known physics are not meant for such a situation. That is why you got the type of response you received from PeterDonis. The situation you are asking for is unrealistic and has no answer, because the concept of "elapsed time" in a photon's rest frame is meaningless and has no formulation.

This is such a complex issue, even if you don't realize it, and especially when from what I can tell, you still don't have a good understanding of the more established part of Special Relativity. Don't you think it is prudent that you try and learn that first before delving into the more exotic aspect of the theory?

Zz.

13. May 6, 2015

Thanks for the above, particularly for the link, appreciated, the calculator on there resolves my curiosity.

14. May 6, 2015

Hello Zapper, I guess the "problem" is, we aren't all on the same rung of the ladder, or even on the same ladder {of interest}. I'm a retired marine engineer with a curiosity about certain elements of the cosmos. I had hoped that my question, presented in layman's terms, might have solicited an opportunity for someone to enlighten me in similar terminology rather than present some egotist the opportunity to show how vastly superior they are to folks.

When folks have approached me requesting my expertise I have never looked on them with disgust, disregard or disrespect, just because they don't know what I know.
I don't have an appetite or a need to do that, or a frail enough ego to seek to put people down.

15. May 6, 2015

### Staff: Mentor

If, as the sun passes the observer in the ship, he sends a flash of that light in the direction of the earth that flash of light will reach the earth before the earth reaches the observer. The observer in the ship will find that the distance traveled by the light (which is considerably less than 93 million miles because of length contraction) divided by the time it takes for the light to reach the earth (which is considerably less than eight minutes) is equal to the speed of light (of course).

Someone on earth will interpret the same sequence of events as a flash of light being emitted from the ship as it passes the sun 93 million miles away; it takes the flash of light traveling at c eight minutes to cover the 93 million miles distance and reach the earth.

One interpretation uses clocks and distance measurements from a frame in which the earth and sun are at rest, and the other uses clocks and distance measurements from a frame in which the ship is at rest. Neither interpretation tells us anything about "the photon's time" - that concept isn't even meaningful because we'd need a clock from a frame in which the photon is at rest to give us the photon's time, and there is no such frame because the thing is moving at the speed of light in all frames. (This is a slightly longer way of saying what haushofer said above. It's also an FAQ in the STEM learning section of this forum https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170/).

16. May 6, 2015

### ZapperZ

Staff Emeritus
I'm not sure where this came from. You appear to either have misread, or misinterpret the intention of the post. But more importantly, you are missing the LESSONS that you could have learned from it.

There's nothing wrong with asking the question. But in response, you also are being given a clear example here where, in physics, what you ask is often as important as the answer you are seeking. The question you are asking is similar to asking "when did you stop beating your wife?". It is based on a premise that hasn't been established to be correct.

When dealing with physics, this is an extremely important issue, because the terms we all use have very clear, and specific underlying formalism. Even "space" and "time" have clear and definite description. It is why I spent time trying to impress upon you that there is a reason why you are not getting a clear, straight-forward answer.

Unfortunately, it appears that my effort in trying to make you understand all this seems to be taken the wrong way. So I will not waste my time any longer.

Zz.

17. May 6, 2015

Thanks again Nugatory for your time and consideration, it's valued and appreciated.

18. May 6, 2015

### Staff: Mentor

Sure. Just use the Lorentz transform or the spacetime interval or even the simplified time dilation calculation. Whatever threshold you set as "much shorter than 8 minutes", there is some velocity which can be calculated and, if astronaut Zippy goes that fast, his recorded time will be "much shorter than 8 minutes". For example, if you say that "1 minute is much shorter than 8 minutes" then as long as Zippy goes .9922 c or higher then the measured time will be less than 1 minute.

Here is a little calculator for this:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html#c2

EDIT: Oops, I see that Bandersnatch already posted the same calculator.

Last edited: May 6, 2015
19. May 6, 2015

### DaveC426913

Well, I for one am certainly sorry if anything I said led you to think I am an egotist, or feel superior, or feel disgust, disregard or disrespect.

20. May 7, 2015

### Filip Larsen

You could perhaps also ask what happens if you have a spaceship of mass M (at rest in some reference frame) and you impart it with kinetic energy E so that it passes through the two points A and B (also at rest). For an observer on the ship the passage from A to B will take a certain time T. Now, if you repeat the experiment so that everything is the same except that now the spaceship has less mass (so, specifically E and the distance between A and B are the same as before) then the observer would now measure a lower T.

If you repeat this argument and effectively let M go towards zero, then you would find that T also goes towards zero.