# B Connected events at opposite ends of a photon's travel

1. May 31, 2016

### KenJackson

Someone started a recent thread with a statement that included "light exists outside of time". That excerpt was flatly rejected, and the post generated such a lively response that it was closed, having "wandered off into philosophy".

It made me wonder if "light exists outside of time" describes the thought I've had for some time.

Suppose, on a planet around a star a billion light years away, some event in an atom results in a photon traveling toward Earth. Much later, the photon strikes an atom on earth causing an event witch absorbs the photon's energy.

Now both Earth and the remote planet have experienced about a billion years as the photon made its way from one to the other. But the photon itself did not experience any time at all (right?).

So can we say that from the photon's point of view, the events on both planets happened together and are connected, as if there was really only one event?

2. May 31, 2016

### Stephanus

Yes, I think so. Considering both Earth and the remote planet are comoving. That they are in the same frame. But it might not be that way because the earth revolves around the sun about ...108 thousands km/hour or 30km/sec. I don't know about the orbital speed of the planet. But providing the planet has the same distance as the earth's to the sun, we can say that for all practical purpose they are both comoving in relativity point of view.
Because their speed is negligible in relativity point of view.
I think so. Because time stops for every object that moves at c. Object that moves at c will not experience time. Come on, even the great Albert Einstein wondered how the universe would look like if he rode in a photon.
And one more thing that I learn in PF.
Because of this velocity addition formula. $w = \frac{u+v}{1+uv}$, so every objects that moves at c wrt to something always moves at c wrt everything.
You can say that. "The events on both planets happened together wrt, with respect to the photon.
NOW has no significant meaning in relativity.

3. May 31, 2016

### Vitro

The short answer is: there is no such thing as "photon's point of view" or "experience". There are many discussions around detailing why that is, but I can't think of specific ones to link here.

There's a one paragraph answer in the FAQ: https://www.physicsforums.com/threads/rest-frame-of-a-photon.511170. Rest frame of a photon and photon's point of view are equivalent concepts.

4. May 31, 2016

### Stephanus

Thanks Vitro for the answer (although it's not for me.) So the rest frame of an object is the object point of view? Is this the term in SR?
And I knew that photon has no rest frame, but I didn't know that photon does not have point of view.
It means that we can't draw photon worldline horizontally in Minkowski diagram. at least mathematically?

5. May 31, 2016

### Ibix

@Stephanus - you are always "here" from your own point of view. So your point of view is a frame in which you are at rest - always at the same place with the world moving around you. But light cannot have a rest frame (if it's not accelerating then any such frame would be inertial - so light would have to move at 3x108m/s and be at rest at the same time) so it cannot have a point of view.

6. May 31, 2016

### robphy

Given a light ray from event A to event B,
B is to the future of A. That is, A happens then B happens. They are distinct events.

7. May 31, 2016

### KenJackson

We seem to have a singularity, or at least a conundrum. Many things in math that don't make sense can be solved by taking the limit as something approaches something. So let's replace the photon with an unknown particle traveling a tiny bit slower than light.

It has a point of view and it experienced much less than a billion years during its trip. Now take the limit as its speed approaches the speed of light. Doesn't the limit of the time experienced approach zero?

8. May 31, 2016

### Mister T

Loosely speaking, photons do not experience time. But let's formulate the situation in a way that allows us to speak rigorously. Let's say Earth and the remote planet are indeed $10^9$ light years apart and are at rest relative to each other. A particle travels from the planet to Earth so fast that it experiences one second of time. So, a billion years will have elapsed for people on Earth, but only one second of time will have elapsed for the particle. If it was, for example, a particle that lives for only two seconds in its own rest frame, it will indeed live to make the journey.

9. May 31, 2016

### Mister T

For the particle, yes. For people on Earth, no.

10. May 31, 2016

### KenJackson

(Wow. You and I posted a similar theme at about the same instant from our distance. That seems to be coincidental cubed--timing, theme, and vaguely illustrative.)

You seem to confirm my thought. I'm still trying to connect the two events on the two planets that happened a billion years apart as observed from either planet. They seem to me to have happened at the same instant with respect to light and are therefore two parts of one event.

I'm not sure what significance this has, but its utterly fascinating.

11. Jun 1, 2016

### Ibix

You can make the time the particle experiences in the crossing arbitrarily small by making the particle move arbitrarily close to the speed of light, yes. However, to the particle, it is at rest and your clocks are running slow. The reason for the short duration of the crossing (actually, the stars are moving in this frame, not the particle, so crossing isn't quite the right word) is that the distance between the planets is length contracted.

Try to carry this reasoning on to something travelling at light speed, though, and you run into nothing but contradictions. If you want to say it experiences no time then, in its experience, you experience no time because you are moving at light speed relative to it. Yet your clocks advance. If you want to say that it experiences no time between emission and reception because the distance between the two events is zero, it can say the same about you. Yet your rulers make useful measurements. If everything is length contracted to the same place, what would you mean by "first" in the phrase "the photon is absorbed by the first thing it strikes"?

There isn't a coherent way to describe a photon's "point of view". It's a contradiction even talking about it, for the reason I gave Stephanus. Mathematically, the problem is that you are assigning the same coordinate to multiple events in spacetime and hoping to reason from there. It's kind of like asserting that all points on the surface of the Earth lie on the equator and hoping you will be able to navigate anyway.

12. Jun 1, 2016

### Laurie K

KenJackson, I can remember reading recently that cosmological red shift occurs between a photons emission point and its subsequent absorption point.

If the emission point continues to emit photons during the photons traveling time and the emitted 'in transit' photons are not blocked or distorted along the way then there should exist a continuous stream of discrete photons between the source and the observer at the time of the observation i.e. now, unless you regard the entire photon stream as one photon (a billion light year long quanta of light?).

13. Jun 4, 2016

### ExecNight

When the photon leaves the planet, it actually instantaneously strikes the Earth therefore the event is completed. But, it takes one billion years for the universe to process this information. And therefore we can not observe something when the information about the event has not reached us yet.

Although this sounds very ridiculous, this is actually what time dilation equation says in my understanding. When v = c, t becomes indefinite is pretty much what the above can be interpreted as, don't you think?

*Ducks*

14. Jun 4, 2016

### Ibix

When v=c you are violating the assumptions upon which the time dilation equation rests. You can put v=c in to it, but the answer is meaningless.

15. Jun 4, 2016

### Mister T

No, it doesn't. The amount of time that elapses between the two events depends on the frame of reference in which that time is measured. There is no frame of reference in which that elapsed time is zero.

An event has no extension is space and no duration in time. It therefore makes no sense to say that an event is completed or not completed. It's simply an assignment of spacetime coordinates.

That's true in one frame of reference, but not true in many many others.

Last edited: Jun 4, 2016
16. Jun 5, 2016

### ExecNight

Well, we don't know that because like it was said here, current equations doesn't give an answer to the situation. But we also know there are two frames in the universe that move wrt each other faster than c, galaxies far enough experience this because of the expansion of space. So, this is actually a natural thing that happens. Yet, not modeled how.

Like i said, this is unknown since current equations explaining this phenomenon are not answering the situation.

Exactly, actually we should be able to experience this phenomenon. If we can launch a telescope into our solar system and accelerate it to say 100.000 km/s. Our observable universe, should differ from its observable universe. Wish i had a few billion dollars to try this out.

17. Jun 5, 2016

### m4r35n357

This is not an equation, it is a definition, and you don't get to challenge it ;)

As for your "instantaneous" light pulse, what you have done is calculated the spacetime interval, then interpreted it as if it were a (time) coordinate interval. Schoolboy error.

18. Jun 5, 2016

### Staff: Mentor

This situation is completely and correctly modeled by general relativity. Although the recession velocity of two sufficiently distant galaxies will exceed $c$, this recession velocity is not a speed in the sense that you're trying to use it - you can't multiply it by a time to get a distance travelled and it has nothing to do with the $v$ that appears in the time dilation and length contraction equations of special relativity. You certainly cannot use it to define two inertial reference frames that are moving at speeds greater than $c$ relative to one another.

19. Jun 5, 2016

### Staff: Mentor

Only if you assume a certain simultaneity convention. That is, you have to pick some event on Earth that is simultaneous with the event of the remote planet emitting the photon; and you have to pick some event on the remote planet that is simultaneous with the event of the Earth receiving the photon. Only if you make certain choices for both of those things will the statement quoted above be true. And it is easy to make alternate choices for those things that make that statement false.

What this is really telling you is that the statement quoted above is not a statement about physics; it's a statement about human conventions. So you have to be very careful trying to draw physical inferences from it.

No, because there wasn't only one event, there were two. And there were also an infinite number of events in between them, all on the photon's worldline. The fact that the "arc length" of the photon's worldline is zero does not mean there are no events on it. It just means those events are not distinguished by arc length; they are distinguished by some other parameter.

Yes, but that doesn't mean what you appear to think it means. See above.

20. Jun 5, 2016

### KenJackson

The intention was that the cause and timing of the event on the remote planet are unknown and unimportant--only that it resulted in a photon being emitted.

And the event on earth would be caused when the photon struck an atom, that is, the photon caused the event, so it wouldn't be picked.

Also, I didn't state it, but I assumed the planets would be moving much, much slower than the speed of light with respect to each other.

I suspect all the points have been touched on this one.