Dear Thierry

 A photon can't count as an observer. Time slows down for objects that move at close to the speed of light. 

regards 

Two objects which are at rest, one wth respect to the other, in one frame, are similarly at rest in any other frame. But the distance between the objects varies from frame to frame. Since the photons seem to as as moving together, so they move in any frame.

However, as Saeed Al Rashid said, a photon itself cannot be an observer. I will jst put the problem differently, there doesn't exist a frame in which a photon is at rest.

Oh, Remi,

==> "You are trying to anthropomorphise photons, how can they "see"?"

You understand in full that the question is what would be the state of motion of one photon, in a frame moving together with the other photon. We know that a photon is never at rest.

The Lorentz transformations, they say that two objects at rest with respect to one another, are so in whatever frame. Only the length of the distance between them differs from frame to frame. If a car travels with the forward and back lights on, the length of the car changes, but the forward lights remain at rest withe respect to the back lights.

As to interaction, no! The photons in the question are non-interacting.

If two any objects, including photons, move at the same physical speed  in the same direction with respect to a reference frame, the relative speed between the two objects is zero. It is a result of elementary kinematic.

If a photon cannot count as an observer, what are the requirements to "count" as an observer and why?

I remark that Daniele Sasso has been downvoted twice. This means that they think it is the contrary.

However, they don't explain their reasoning. How strange...

Remi, my friend,

I don't understand you. Prof. Saeed Al Rashid put the things so simply!

I suggest you to convince yourself by taking the Lorentz transformations! Consider two photons that pass through the same point in space at an interval ∆t by the lab clock. Let's now see what happens in the frame of one of them.

(1) ∆t' = [1 - (V/c)2]-1/2 [∆t - (V/c) (∆x/c)]

I denote by  ∆t '  the time interval in the frame of the photon, while ∆t and ∆x are in the lab frame. Since the photons pass through the same point in the lab frame, ∆x = 0. There remains

(2) ∆t ' = [1 - (V/c)2]-1/2∆t 

Since ∆t  is finite, if  V → c  one gets

(3) ∆t ' → ∞.

I am only detailing what Prof. Al Rashid. You know such things, they are simple for you. As to interaction between photons, no such problem is posed in the question.

With kind regards!

I want to add only a few comments in response to this question and above mentioned answers without agreeing or disagreeing with anybody else here:

1. The photons do not obey the Lorentz transformations. These transformations are meant for material objects in motion only.

2. Two coherent laser photons will appear as travelling in phase in the laboratory frame of reference i.e. to a 3rd observer.

3. As a wave, the two photons will add up (i.e. interfere) constructively or destructively depending upon their relative phase. During interference, the two photons observe each other's phase.

4. If there is a three level atom simultaneously interacting with two photons, in the lambda, V or ladder configurations, the resulting atomic coherence effects like EIT, CPT etc.  depend upon the relative phase of the two photons in addition to their detuning.

Muhammad Anwar,

Why photons don't obey Lorentz transformations? In the Lorentz transformations you can take the limit V --> c. Can you give an example that photons contradict Lorentz transformations?

Issues of interference in lab frame do not belong to this question. Note that interference patterns that we see in one frame don't look the same in another frame.

(By the way, about the statement that two photons add up, I am sorry, they don't add up. Simple interference, as you describe - 1st order correlations - does not involve two or more photons. To remind you Dirac's dictum, "a photon interferes only with itself". As an example, thermal light also produces interference and the phases there are incorrelated. It's second order correlations that imply two photons, but that is something else, two-particle interferometry, e.g. the HB&T effect.)

Sorry to disagree,

Sofia

Sofia D. Wechsler

We should not try to treat a photon like a Newton's particle simply moving with velocity c and apply the Lorentz transformations and expect the consequences of the Special Relativity. However, we can treat photons as light waves travelling with velocity c but then we have to consider that the velocity of light cannot be added or subtracted to itself.

Coming back to the original question where the theory of coherence might help us.

We want to tell something about the photons co-propagating in a laser beam. What is their "relative velocity" ?

If Lc is the coherence length of the laser, then the laser photons will remain coherent over a distance Lc or for a time tc = Lc / c, where c is the speed of light. The coherence length of the laser comes from its linewidth (Frequency uncertainty) via Heisenberg's uncertainty principle: (Energy uncertainty).(Time uncertainty) ~ h and (Energy uncertainty) = h.(Frequency uncertainty).

Now the photons in a laser beam separated beyond the coherence length lc become incoherent and therefore cannot interfere. Since interference is a manifestation of correlations, the photons separated beyond the coherence length are uncorrelated.

Looking at this problem in a relativistic way. I would like to make the following assertion without any reference.

"While the uncorrelated photons travel in their individual frames of reference, the correlated photons share the same frame of reference."

Therefore, the coherent photons share the same frame and the incoherent photons have different frames so the coherent photons appear to be at rest and the incoherent photons “might” appear as moving relative to each other. But, suppose if v is their relative velocity, then according to special relativity v + c = c and v – c = c i.e. v = 0. There is no relative velocity between the two photons even if they are incoherent.

The correlation between the photons means that there is an "exchange of something" between each pair of photons and the uncorrelated photons lack this "exchange of something". Therefore, by analogizing this "exchange" with "seeing", the photons can "see" each other only over their coherence length. The photons are 'unaware' of the other photons beyond their coherence length.

I agree with the preceding comment that examines in details the question.

Muhammad Anwar

==> "We should not try to treat a photon like a Newton's particle simply moving with velocity c and apply the Lorentz transformations and expect the consequences of the Special Relativity."

What are you talking about? We do not HAVE TO treat the photon as a wave. The treatment of the photon as a wave, or as a particle, depends on the case. The fact that the photon has no rest-mass doesn't impose one treatment or another. It's the particular experiment that imposes the treatment, how big is the wavelength of the photon vis-a-vis the dimensions of the particles with which it interacts - see examples below. 

The Compton effect showed the the photon behaves a particle with linear momentum ħҟ and energy ħω. It is NOT a Newtionian particle in the sense that it doesn't have a rest-mass. But we write the laws of energy and linear momentum conservation as for a collision between particles. Gamma rays also pass through matter as particles. In the decay of a photon into a electron-positron pair, we treat the photon as a particle.

The treatment as a wave is fit for cases in which the wavelength of the photon exceeds by far the dimensions of the particles of the involved material. For instance, you can record interference on a photographic plate if the wavelength of the photons are much greater than a spot obtained on the plate. But if the wavelength is much smaller than the diameter of the molecule, how can you see fringes? The photon would kick away some electrons and go out.

Now, if the theory of coherence is so important to you, I won't disturb you. You may develop on this issue as much as you desire. But the question was about the photon as a particle.

By the way, not all the light in the world is correlated, not all the beams of light are laser beams. THERMAL light is NON-COHERENT. In spontaneous down-conversion (DC) we get pairs of photons and there is an interval of typically 100 femtoseconds between the detection of the first photon in the pair and the second one. 

Best regards!

Sofia D. Wechsler

I like your spirit for physics but there is a 'gap' between our understanding of physics as our previous discussions. I see the photon concept mainly as the 'quantum of energy' and not as a quantum of mass otherwise it is just a 'wave-packet' which is a sum of many a waves inside it.

By the way, at the end of my comments above, I have used the special relativity where v+c=c, v-c=c & v=0. So if photon does not obey the special relativity (which is not the case), then this conclusion is probably not correct. But the fields too can be 'relativistic'. Here I am supporting you.

But I do not want to stick to one particular picture. I only thought logically and tried to answer the question to the best of my knowledge. People might agree or disagree with my arguments. This is what is PHYSICS. I don't mind it.

Thanks for your discussion. I really enjoyed it.

Best Regards

Dear Muhammad Anwar,

Thanks for your comment and for the good words.

By the way, there is nothing wrong in disagreeing with the views of another person (of course, as long as it is done with scientific arguments - unfortunately, there are cases, rare, thanks God - when the people are not polite). As you can see in this site, we are all of us colleagues. Polemics improve our knowledge, give us new information, or make us look at things that we know, from a new perspective.

By the way, of course the photon is a wave-packet, a superposition of Fourier components, but this wave-packet has a group linear momentum, h/λ.

The quantum field theory, which is relativistic, contains as a special branch the quantum optics. There, we add photons by number. For instance, a Fock state may have 1 photon, or two, or 3, etc. There even exist states of two photons named biphoton.

With kind regards,

Sofia

Sofia D. Wechsler

Thanks for your counseling. Mostly I agree with your point of view. But I again have a slight difference with you.

The Fock or number states |n1,n2,n3, ....> are meant to represent quantum mechanically the field intensity in terms of the number of photons n, say, in a laser cavity. I do not know at least if these refer to the addition of two or more photons to produce photons of higher energy or frequency.

I also like quantum optics and wish to implement it experimentally.

Best Regards

Anwar

Is it right that when an observer is traveling at v=0.9c, he allegedly sees light travel at c ? The same for even closer speeds to c?  So, also for an entire light beam? 

Thierry De Mees

Very smart!

It is true that if we treat photon as a Newtonian particle simply travelling with velocity c then Special Relativity predicts that its velocity is also observed to be c by an observer which is moving at velocity c.

But please read a couple of my comments above.

Thanks.

Anwar

Dear Muhammad Anwar, I know you consider light as a quantum of energy rather than a quantum of mass. However, does this change anything fundamental to my latest question? If it would, that would mean that there is a physical difference of being an observer containing mass, or a virtual observer containing energy.

What is justifiying such a difference w.r.t. the theory of relativity and how would that be described? Or is it just by "intuition", without a fundamental, physical justification?

Dear Thierry De Mees

What do you think about this equality  " c - c = c + c = c ", where c is the velocity of light? This is what special relativity predicts for photons assuming it to like a Newtonian particle.

Anwar, according to the use of the conventional mathematical symbols, which you use, in that case c=0.

But indeed, you mean that, physically speaking, in SRT, there is a flaw in conceiving the concept of velocity, as well for objects with a mass as for light, which has only acquired a mass by the grace of its velocity, defined by physical constants, hence physical properties of the medium in which it propagates.

This is precisely what my question is about.

Part of the issue is that light is often seen as a massive particle, while instead the rest mass is zero. The impulsion that any wave acquires with speed is given by the kinetic energy (motion) of the medium. So, it *appears* that the wave has a mass, while it is only the impulsion (and kinetic energy) that is transmitted at its wave speed by the medium.

Take the example of any wave (a ondulating blanket, sheet, rope, water wave, etc...): the propagation speed is the square root of the medium's specific energy (tension, energy density), devided by its mass density. That is the same as saying E=mc². However, it says that the energy that is propagated in a wave (initiated, say, by a process in particles) is involving a mass "m" of the medium.

Relativity followers don't like this representation, because it require more insights than pure mathematics, but I am convinced that only such an approach will make progress science.