# What a photon experiences

1. Jun 13, 2015

### $H3K Hi there, I have some simple questions concerning a thought experiment about photons. Set-up: There are two molecules in the universe and in between them there is empty space. A photon of light is emitted from molecule A and is absorbed by molecule B. From the perspective of the photon, does any time pass and has any distance been traversed? And if not, then does the photon experience "teleportation" between molecule A and molecule B because there is no time nor space to cross it in (according to the photon)? And (this is a very big IF, if the other two are answered "no" and "yes" respectively) - is it possible that the photon does not actually exist at all between massive particles? I appreciate any help! Thanks,$H3K

Last edited: Jun 13, 2015
2. Jun 13, 2015

### Staff: Mentor

There is no such thing as "the perspective of a photon" in the sense you mean; there is no such thing as an inertial frame in which the photon is at rest. So the concepts of "time passed" and "distance traversed" don't make sense from the "perspective" of photons. They only make sense in an inertial frame; in any such frame, the photon is moving at $c$, and the time it takes to cover a given distance is determined accordingly.

3. Jun 14, 2015

### Ibix

One of the postulates of special relativity is that the speed of light is constant in all inertial reference frames. If there were a reference frame in which a photon is at rest, then light would be travelling at zero and 3x108m/s at the same time. The notion of a "perspective of light" is self-contradictory, unfortunately.

To put it another way, Einstein defined time as "what clocks measure". Since you can't accelerate a clock to the speed of light, time isn't defined at the speed of light.

4. Jun 14, 2015

### $H3K Hi PeterDonis and Ibix, Thanks for your responses. These questions were part of a thought-experiement in response to astrophysicist Dr. Pamela Gay's claim that photons in fact do not experience time: http://www.universetoday.com/111603/does-light-experience-time/ Apparently, there are a few physicists out there who think that time and distance don't exist for photons. But, I also understand your point of view. Thanks for responding. I suppose I am asking for validity for the claim from the article linked above: "Photons can take hundreds of thousands of years to travel from the core of the Sun until they reach the surface and fly off into space. And yet, that final journey, that could take it billions of light years across space, was no different from jumping from atom to atom."$H3K

Last edited: Jun 14, 2015
5. Jun 14, 2015

### Ibix

That article says that Dr Gay says that "[p]hotons do not experience time". It then goes on to say that photons experience zero time. The first version is certainly consistent with what Peter and I said; the second is not. It's not clear exactly what Dr Gay said and what is the author's (mis-)interpretation of what she said - so I wouldn't put too much faith in it.

There certainly are scientists who write about photons experiencing zero time. Brian Greene is the most famous of them, as far as I know. Peter is fond of pointing out that he never uses the concept in his research work, but only in his pop-sci presentations. The obvious conclusion is that he doesn't think the idea is good for much beyond selling books.

The article's author is also waxing poetical in the bit you quoted. When it arrives at Earth, the photon has climbed part way out of the Sun's gravity well, so will be red-shifted compared to a similar measurement made on the surface of the Sun. If the photon heads out into inter-galactic space it might land on the telescope of an alien astronomer many billions of light years away, who will see it red-shifted due to the expansion of space as it travelled. Those outcomes are different for photons that were identical when emitted.

In short, there can be changes to a photon in flight. That strongly suggests that "experiences zero time" is wrong. "Time for a photon" being nonsense (oddly enough) makes more sense.

6. Jun 14, 2015

### $H3K Thank you Ibix, That was a very helpful explanation. It had occurred to me earlier that should amount of space traveled over time not really effect photons because they can't experience it, then it should be the sheer number of electrons they have hopped to/from that determines any change in its composition. However, I hadn't thought of red-shift due to gravitational or expansionary spacial warping, and surely if photons passed directly over the space to their next electrons, then we would hardly be able to have any red-shifting or blue-shifting, for a photon hopping between some arbitrary number of atoms towards the center of the sun (blueshift) would yield the same properties as a photon having crossed the same number of atoms over many light years through expanding space (red-shift). Thanks a bunch. Good explanation, I have one more question, though. Although you seem convinced, and I am becoming moreso, that photons do not experience time (not zero-time) then what about distance? If they do not experience distance, nor time, for those concepts are alien to the photon, then, distance and time being the only definitive indicators of dimension, it seems we'll have to drop all dimensions altogether and conclude that dimensions do not exist (or are a null point) for photons or anything else traveling at the speed of light?$H3K

Last edited: Jun 14, 2015
7. Jun 14, 2015

### Ibix

Short on time, so will be brief.

The key point is that asking what a photon experiences is incoherent in relativity. There's simply no coherent way to "look from a photon's perspective". So there is no way to answer what distance it experiences any more than what time it experiences (or what the flowers smell like along the way, or anything else).

8. Jun 14, 2015

### Stephanus

Can someone please tell me how we should calculate in SR?
Do we have to consider that we are NOT moving, and everything else comes to us (or away from us)?
Except photon?
Do photons have rest frame? I think from PeterDonis answer, they don't. But I need confirmation.
Thanks.

9. Jun 14, 2015

### Stephanus

Yep, yesterday I watched a youtube video about light not photon. Not the video that you link above, a different one. Light doesn't experience time because
$\gamma = \frac{1}{\sqrt{1-v^2}}$ But if V >= 1 (can V > 1??), then $\gamma = \infty$
So if $\text {length } = \frac{1}{\infty}$, length should be zero. Time should be instant. Might right, migh be wrong. Who knows?
But this already too much for me. Still struggling with basic SR, here.

10. Jun 14, 2015

### PeroK

You're not the first on this forum to be trying to understand SR, but actually to be struggling with the concept of reference frames. You might benefit from a better knowledge of classical kinematics. Reference frames are not particular to SR. Take an example from classical physics:

You're in a car, moving towards a road sign. In your reference frame you are at rest and the sign is moving towards you. But, you also know that the sign is not moving with respect to the Earth/road and you are. There is no mystery to this. You can do calculations from your reference frame or the the reference frame in which the Earth (road and road sign) are at rest.

And, in fact, when doing classical problems, choosing the best reference frame often makes the problems easier to solve. For example, if you have two objects moving towards each other, you could study this from the point of view (rest frame) of either object; or, the reference frame where their centre of mass is at rest - which is often very useful. Or, of course, from your reference frame as a "stationary" observer.

SR adds a twist in that lengths and elapsed times are not the same when you move between reference frames. And, clocks synchonised in one frame will not be synchronised in another. This is why you need Lorentz.

Now, if we go back to the classical case: what is your absolute velocity? Well, let's say the car is moving at 100km/h (relative to the road). But, the Earth is rotating and orbiting the sun. And, the sun is orbiting the centre of the galaxy. And the galaxy is on a collision course with the Andromeda galaxy. The point is that even in classical physics, there is no absolute velocity. You have to choose your reference frame.

You're at risk of making a mystery out of a plain matter, in my opinion.

Last edited: Jun 14, 2015
11. Jun 14, 2015

### Stephanus

Thanks!

12. Jun 14, 2015

### 1977ub

I recently read through this online book which gave me a lot of food for thought.
http://www.mathpages.com/rr/rrtoc.htm
Reflections on Relativity

>>... once a photon is emitted, its phase does not advance. In a sense, the ancients who conceived of sight as something like a blind man's incompressible cane, feeling distant objects, were correct, because our retinas actually are in "direct" contact, via null intervals, with the sources of light. The null interval plays the role of the incompressible cane, and the wavelike properties we "feel" are really the advancing quantum phases of the source.<<

>>This image of a photon as a single unified event with a coordinated emission and absorption seems unsatisfactory to many people, partly because it doesn't allow for the concept of a "free photon", i.e., a photon that was never emitted and is never absorbed. However, it's worth remembering that we have no direct experience of "free photons", nor of any "free particles", because ultimately all our experience is comprised of completed interactions.<<
http://mathpages.com/rr/s9-09/9-09.htm

13. Jun 14, 2015

### Staff: Mentor

That article is a pop science article. Pop science articles, even when written by scientists, are not good sources to use if you actually want to understand the science.

This is a good illustration of what I just said above. The quote contradicts itself, because it's using vague ordinary language instead of precise mathematics.

14. Jun 14, 2015

### Staff: Mentor

They don't.

15. Jun 18, 2015

### gliss

One thing I find myself thinking about is that because a photon is traveling at the speed of light is has zero length in its direction of travel. So would it be accurate to say that a photon only exists in two dimensions?

16. Jun 18, 2015

### Staff: Mentor

That is not correct. The formula that might lead you to believe that (the length contraction formula) does not apply to photons.

17. Jun 30, 2015

### 1977ub

Is a photon thought to have size in any dimension?

18. Jun 30, 2015

### Staff: Mentor

A photon is not an "object" in the ordinary sense, and does not have a meaningful "size". The term "photon", when used in the context of SR and GR, is really a shorthand for "a very short pulse of EM radiation", which is modeled as an object with zero invariant mass traveling on a null worldline. This model is only an approximation, which is sufficient for most relativity problems but is certainly not intended to make any claims about the fundamental nature of photons. For that you need quantum mechanics, which will tell you what I said in the first sentence above.

19. Jul 28, 2015

### $H3K I know that this is an old post, but Leonard Susskind of Stanford University cites Paul Dirac's work that internally, any particle (photon or otherwise) traveling at the speed of light can not experience any change because time is 0. At 43m 48s 20. Jul 28, 2015 ### PeterDonis ### Staff: Mentor This is not wrong, exactly, but it can be misleading, because it invites the further inferences you make in your OP in this thread, all of which are wrong. Everything that I and the other responders to you in this thread have said still stands. The fundamental problem here, as I said before, is that you are trying to use ordinary language to analyze something that can only be properly analyzed with math. Susskind is similarly trying to use ordinary language to express something that can only be properly expressed with math. 21. Jul 29, 2015 ### Finny As the prior post says, its easy to to misread what Susskind means. He gives great insights in many of his lectures, but they cannot be taken too far. Photons are not so well suited to 'simple' answers. Instead, if you think of the observed time of an object slowing as the speed of the object increases, and approaches zero as speed approaches c, you'll be closer. Yet here, at speeds less than c, local time, that is proper time in the frame of the object, remains unchanged. That's relativity. But photons are quantum particles; described by another theory, Quantum Electrodynamics. Reconciling exactly with Einstein's relativity is not so simple. https://en.wikipedia.org/wiki/Quantum_electrodynamics {QED} yes. We detect a photon at the receiver, your molecule B. We do not know for sure what is between the emitter and receiver. The photon seems to be a point like particle since it is absorbed or emitted by arbitrarily small systems, systems much smaller than its wavelength, such as an atomic nucleus. That's what we detect. The wiki article gives a description of Richard Feynman's view in QED: ".... photons and electrons do, somehow, move from point to point and electrons, somehow, emit and absorb photons. We do not know how these things happen, but the theory tells us about the probabilities of these things happening. 22. Jul 29, 2015 ### PeterDonis ### Staff: Mentor This is not quite correct. We cannot assign the photon a definite trajectory between the emitter and the receiver. But that does not mean the photon does not exist at all between the emitter and the receiver. The correct way to describe it (at least, as well as it can be described without math) would be to say that a quantum field exists between the emitter and the receiver; "photon" is the name we give to the state of this field that gives rise to the observed emission and reception. 23. Jul 30, 2015 ### Finny C'mon, Cut me some slack here! [LOL] It's not quite 'incorrect" either. But I like the rest of that post. " We do not know for sure what is between the emitter and receiver. " pretty well sums up a lot od discussions in these forums....We don't know the photon 'size', if it has a single frequency it must be infinite in extent [which to some may mean it can't even be between emitter and receiver!!], we have a model without an ability to test 'what is there', when we do test all we get a point like perturbation, and I think the photon may not even have a well defined particle state until detected. See Weinberg text below on whether a quantum field even 'exists'. OTOH, PeterDonis has a good point that we cannot describe a photon in classical terms.....it does not fit our everyday 'logic' which, to paraphrase Susskind, Evolution has not prepared us for the quantum world.... For the OP,$H3K:

There are some great discussions about particles and fields in these forums.....

One is "what is a particle" which dissects particle definitions in general, and more specific to photons,this:

From that discussion:
via Weinberg's "The quantum theory of fields" vol.1.....

"For realistic systems with varying numbers of particles we build the Fock space as a direct sum of products of irreducible representations spaces. Then the sole purpose of quantum fields (=certain linear combinations of particle creation and annihilation operators) is to provide "building blocks" for interacting generators of the Poincare group in the Fock space. In this logic quantum fields are no more than mathematical tools.

.... strictly speaking there are two distinct notions of particles in QFT. Local particle states correspond to the real objects observed by finite size detectors. .... On the other hand, global particle states....can be defined only under certain conditions...... uniquely-defined particle states do not exist in general, in QFT on a curved spacetime."

24. Jul 30, 2015

### Staff: Mentor

It has a state--more precisely, the underlying quantum field has a state--but that state is not a "particle" state until the photon is detected. The interaction of the photon field with the detector puts the field in a state that we call a "particle" state (because it corresponds to a localized detection event).

But note carefully how he defines the term "quantum field" in that passage: "(=certain linear combinations of particle creation and annihilation operators) ". Here he isn't describing the photon or its "field" in the sense we've been using the term here. He's describing a particular mathematical model and labeling a particular part of that model as the "quantum field".

When we say there is a quantum field called the "photon field" in between the emitter and the detector, we are using "quantum field" in a different sense: to mean "whatever thing it is that propagates information about the photon from the emission event to the detection event". We know there must be something that does that because we can put a detector anywhere; if we put a new detector halfway between the emitter and the original detector, we will detect something. There is never a "gap" where we can put as many detectors as we like and never detect anything; so we conclude that there is something "real" in between the emitter and any given detector, and we call that real something the "quantum field" of the photon. It doesn't matter how we model the situation mathematically; nothing requires us to use linear combinations of particle creation and annihilation operators, and the real something in between the emitter and the detector doesn't magically appear when we adopt that particular mathematical model.

It's also worth commenting on this that you quoted from Weinberg:

Here he is not saying that no quantum field states exist globally, or in curved spacetime. He is only saying that the field states cannot be globally described as "particle" states, or uniquely defined as "particle" states in curved spacetime. The states are still there; they just don't fit the requirements we impose in order to apply the label "particle states" to them. But there are many, many quantum field states that are not "particle states", but are still perfectly good states and are just as "real".

25. Jul 30, 2015

### Finny

First, You'll note I liked your post. What you posted makes sense and is in general accord with what I think I know. [You know what Feynman said about THAT!] . And added a bit to what I know, too, so thanks.

But I think all that supports my earlier comments: Those fields and states are wonderful handy dandy models, and without question are marvels in themselves because of their highly accurate predictions, but we can't detect those 'real' fields and states. Additional credit to you for putting real in parenthesis!

I'll remain a bit skeptical they are just as 'real' as particle detections. From that other more general particle discussion I referenced:

Anyway, great discussion.....I hope the OP had as much fun as I.