# By photon's POV, would light travel infinity in no time?

1. Sep 17, 2006

### Skhandelwal

According to photon's standpoint, since light travels any distance in 0 time, then doesn't it mean that it can travel infinity in 0 time, but for my mind, that doesn't really made sense b/c I thought nothing could reach infinity, but apparently, light can.

2. Sep 17, 2006

### gnomedt

Photons don't experience time. It's the tradeoff for travelling at the speed of light. Therefore a "photon's standpoint" is meaningless.

Also, it is not true that light travels any distance in 0 time. In a vacuum, light travels 299792459 meters in one second.

Many of your questions are very elementary. Perhaps you should try Google first?

Last edited: Sep 17, 2006
3. Sep 17, 2006

### Mk

Light does not travel any distance in 0 time. It travels at the speed of light in whatever media.

4. Sep 17, 2006

### rcgldr

The limit of time passage approaches zero as an object appoaches the speed of light from a speed less than the speed of light, but I don't see any reason for an object that only exists at the speed of light not to experience time. The path of a photon changes due to gravitational fields. From the perpective of a photon, relative movement of other objects occurs, it's position relative to other objects changes with respect to time. A photon also has a frequency, a change of state versus time.

Ok, you have to deal with the fact that from the perspective of a photon, the speed of light isn't constant (another photon traveling in the same direction has zero velocity, while a photon not traveling in the same direction has observed velocity), but again, this is a special case for a frame of reference, with different set of rules than sub-light speed objects.

5. Sep 17, 2006

### Staff: Mentor

And many people (including me) insist that a photon has no meaningful "point of view" in the context of relativity, strictly speaking. The "point of view" of an object is the inertial reference frame in which the object is at rest. For a photon, there is no such inertial reference frame, because a photon must travel at the invariant speed c in all inertial reference frames.

At best, you can try to imagine a limiting case of an object with a very small mass, traveling very fast.

6. Sep 18, 2006

### michael879

you read his question wrong. He was kind of right in what he said. You can imagine a photons reference frame as just the limit as v->c. From our reference frame, a photon doesnt experience time and therefor can travel an infinite distance in what appears to in as 0 time. From a c reference frame however, all lengths are compacted to 0 so that it takes 0 time for the photon to get anywhere (in its reference frame).

7. Sep 18, 2006

### michael879

photons dont experience time, due to length contraction their paths are contracted to 0 so that they are emitted and absorbed simultaneously from their reference frame.

8. Sep 18, 2006

### JesseM

Photons do not have their own inertial rest frame in relativity! If you try to do the Lorentz transformation with v=c, you get nonsense (events having position or time coordinates of infinity). Plus, if they did have a rest frame this would violate Einstein's postulate that the laws of physics should look the same in all inertial frames, since a light wave cannot be at rest in the rest frames of any sublight object.

9. Sep 18, 2006

### DaveC426913

No. Inasmuch as we can pretend to "see" the photon's POV, think about what it would see:

It would see the entire universe be born, age and die in an instant. All the interactions the photon has with gravity etc, would happen simultaneously.

(In fact, if I think this through a little further, "all things happening simultaneously" is equivalent to - not a time dimension - but a spatial dimension!

Yeah. The couch's 'x position' in my living room, and the chair's 'x position' in my dining room are simultaneous! It is not as if the couch's distance from the wall happens, then the chair's position from the wall happens.

So, a photon experiences events in the universe like we experience a painting: it all happens at the same moment in time.)

10. Sep 18, 2006

### robphy

11. Sep 18, 2006

### michael879

I know everyone says this but its not really true. You dont get nonsense. And even in a reference frame traveling at c, light would still appear to have velocity c (one of the postulates of relativity). What laws of physics would be different?

12. Sep 18, 2006

### rcgldr

My point is they don't need an inertial rest frame in relativity.

Photons only exist at the speed of light. This is different than the limit of a sub-light object as it approaches the speed of light, so I believe that photons have their own special properties. Photons exist. They travel at a fixed speed (c). Their paths are affected by gravitational fields. They can be created and intercepted by electrons. They have a frequency.

My take on this is that from a photon pov or photon frame of reference (non-relativistic), time and distances are constant, but not necessarily zero. There is no relativity for photons. Assuming time and distances are non-zero and constant, then the speed of light is constant, but not relative, since all photons travel at the speed of light. The direction of one photon versus another determines it's "relative" speed in classic Newtonian fashion. Sub-light speed objects don't behave any different than light speed object when viewed from light speed frame of reference; there are no relativistic effects for a light speed frame of reference. A photon can't really "observe" another photon unless the photons pass through each other, but this doesn't matter for the case of a light speed frame of reference.

My point is that a frame of reference moving at the speed of light exists outside of relativity, and has it's own special properties.

Last edited: Sep 18, 2006
13. Sep 18, 2006

### pervect

Staff Emeritus
Everone says this because it is true. Photons do not have a "rest frame". There is no such thing as a frame of reference which "travels at the speed of light". One does get total mathematical nonsense if one tries to pretend otherwise.

One can use the intersecting worldlines of photons to create a coordinate system, but none of the coordinates in such a coordinate system represent either time or space. Such coordinates are abstract, they are not associated with "duration" or "distance". They are so-called "null coordinates", and are talked about in some of the threads robphy cited.

Note that the science advisors in this thread are unanimous on this point of the nonexistence of "frames" travelling at 'c'. One can read the same thing in the sci.physics.faq on the topic

and in most textbooks, too.

14. Sep 18, 2006

### JesseM

Are you familiar with the Lorentz transformation? You would get infinity for the x-coordinate and the t-coordinate of any event in a "frame of reference" with v=c!

You can of course come up with a coordinate system where a photon is at rest, but it won't be related to the inertial frames of relativity by a Lorentz transformation, and the laws of physics won't work the same way in this coordinate system as they do in SR.
If a frame is a photon's rest frame, then the photon must have a fixed position coordinate, by definition--that's just what "rest frame" means. There is no way to have a coordinate system that is both a "rest frame" for the photon and which also has the property that all light waves/photons move at c, the two notions are inherently contradictory.

15. Sep 18, 2006

### rcgldr

But photons are not abstract, they do exist, and they travel at the speed of light. Even if it requires a seperate set of rules, why can't something that actually exists have it's own frame of reference?

All the references I see mentioning that photons don't experience time seem to be based on the limits of what would happen if a sub-light object were accelerated to the speed of light, but this isn't possible. You have sub-light speed objects, and light speed objects. There's no reason that the same set of rules should apply to both.

Last edited: Sep 18, 2006
16. Sep 18, 2006

### JesseM

You can have a coordinate system where a photon is at rest, but the laws of physics won't work the same way as they do in inertial reference frames constructed according to Einstein's rules in special relativity. Along the same lines, if you have an object which is accelerating as seen in inertial frames, you can come up with a non-inertial coordinate system where the object is at rest, but you will also find that the laws of special relativity don't work in this coordinate system.

17. Sep 18, 2006

### michael879

I get that the math ends up being all 0s and infinities and I understand why everyone says you cant talk about a reference frame going at c. However, the infinities and 0s do make sense if you think about it. Youd basically be talking about the limits of an inertial frame as its speed goes to 0 relative to u. All distances would compact to 0, and all the clocks would go infinitly fast relative to yours. You would be born and die spontaneously. This all makes sense even though its a little weird.
Also, Im not talking about a photon's rest frame, Ive been talking about a reference frame going at c. Even in this reference frame photons would travel at c. ALL photons would still move at c, you wouldnt be following some specific photon.

18. Sep 18, 2006

### michael879

I get what your saying but comon, math teachers teach that 1/0 and 0/0 are undefined. I mean yes, 1/0 is undefined, but in nature it is infinity. saying 1/0 implies that your saying lim x->0 1/x.
As for 0/0 it solves to a set of all numbers: 0x = 0.

Im not saying your all wrong, Im just saying it is possible to imagine a reference frame going at c and you shouldnt just dismiss it as impossible. There is a group of properties that reference frames approach as they speed up to c.

Also, I wouldnt call infinity mathematical nonsense.

19. Sep 18, 2006

### clj4

Mathematically you are wrong on all accounts.

1. 1/0 is meaningless,
2. lim x->0 1/x is plus or minus infinity depending on approach (x>0 and x<0)

3. 0/0 is meaningless
4. lim x->0, y->0 y/x may or may not exist , depending on a complex set of rules (see l'Hospital) . It can be infinity, it can be a precise number, it can be non-existent as a limit

Any of the above has very little to do with whether one can attach a frame of reference to photons (you can but you shouldn't). Please try to listen to pervect, he really, really knows what he's talking about.

20. Sep 18, 2006

### michael879

I know what your both talking about but if you dont want to listen to me fine w/e. 0/0 is definatly NOT meaningless. I dont care what you say. 0 goes into 0 1 time, it goes into it 2 times, it goes into it 3 times etc. I never said anything about attaching a frame of reference to a photon, and the above has everything to do with why you shouldnt. If you think that 1/0 is meaningless then it follows that you think a reference frame with velocity c is meaningless. The reason it is meaningless is simply because time -> 1/0 (and maybe cause length goes to 0 I guess).

21. Sep 19, 2006

Staff Emeritus

Your statement is just wrong; you don't understand the math and are just trying to answer with everyday intutition a question that it never evolved to handle. The only productive thing you could do would be to drop your misconceptions and learn the math, but I'm not going to hold my breath.

22. Sep 19, 2006

### DaveC426913

No. Note you use the word "move". To move would require a measurement of distance and a measurement of time. Since there is no passage of time at c, it is meaningless to think you can measure how far anything has "moved".

No, it does not. If you insist on defining your own math, knock yourself out, but there's no need to convince anyone else, is there?

More accurately, 1/0 is undefined. i.e. it has no unique answer (which, I suppose, is what you mean when you say "0 goes into 0 1 time, it goes into it 2 times, it goes into it 3 times etc. ") A unique answer is required by the definition of multiplication/division.

Last edited: Sep 19, 2006
23. Sep 19, 2006

### lightarrow

About the initial question, which clearly has not an exact meaning, since nothing can travel exactly at the speed of light, let's investigate, instead, what would happen to a spaceship travelling to speed even closer to c: because c = 299,792,458 m/s and distances are Lorentz contracted, from his refer. frame, it would travel light years in even less time. At the limit v-->c, it would travel the entire universe in 0.0000000...seconds.

24. Sep 19, 2006

### rcgldr

Photons travel at the speed of light, and they are real.

25. Sep 19, 2006

### lightarrow

Of course. I was thinking to a spaceship or to any body with non zero mass.

Many people don't understand the fact such a body cannot reach or exceed the speed of light, because, maybe, they don't know that, in practice, light's speed is infinite:

The fact is that we have chosen a "bad" definition of speed, when we have chosen v = S/t, since space and time are not actually independent from each other; but we didn't know it! (Anyway, that one is the definition which we can deal better).

If we defined the speed of a body in a more appropriate way, that is, exactly in the way we define lenght, mass, time, that is, using a sample of it and adding n equal samples to make a sample n-times bigger, it's possible to show, mathematically, that the speed of light would become infinite.

For this reason, Skhandelwal initial statement is not completely wrong.

Last edited: Sep 19, 2006