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A particle's interaction (with anything it can interact with) can be thought of as it making a measurement of the physical quantity associated with the interaction, (e.g. electric field in case of the interaction between charged particles) and acting accordingly. To make a measurement one first needs a frame to make measurements "in" (a Lorentz frame in light of relativity). Assigning such frame to a photon appears to me, to be problematic in the sense that constant velocity of a photon in any inertial frame implies a photon in its own frame having a velocity $c$ (the speed of light). Now if we assume that no such frame exists for a photon, photons cannot interact with an other one.

Can we explain in this way that photons do not interact with each other, or more generally particles moving at velocity $c$ do not interact with similar particles?

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  • $\begingroup$ Welcome on SE-physics. Your question would benefit from a grammar and typos check, especially your punctuation. Also, choosing the correct tags will help people answering your question. $\endgroup$ – FraSchelle Dec 2 '13 at 20:26
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    $\begingroup$ @Oaoa In the first line I want to say that the particle itself is making the measurement and not any other observer.So I think the editing changed the main point!! $\endgroup$ – user35122 Dec 3 '13 at 5:07
  • $\begingroup$ Oups, sorry for that. Feel free to (un)edit the way you want. $\endgroup$ – FraSchelle Dec 3 '13 at 6:52
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    $\begingroup$ Seems to me you are re-thinking the wheel. Better acquire a background on interactions etc. This is a good source: hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html . One photon cannot have a rest frame, but two photons can: they have an invariant mass that can define a rest frame. $\endgroup$ – anna v Dec 3 '13 at 7:50
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    $\begingroup$ take the pio particle, it decays into two photons. The center of mass system of the two photons is the CMS of the pion too, i.e. where the pion has zero momentum. Photons interact very weakly with each other through higher order loop diagrams, but that is a long not simple story :en.wikipedia.org/wiki/Two-photon_physics $\endgroup$ – anna v Dec 3 '13 at 11:57
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A particle's interaction (with anything it can interact with) can be thought of as it making a measurement of the physical quantity associated with the interaction, (e.g. electric field in case of the interaction between charged particles) and acting accordingly.

When we use the term "particle" with "interaction" we are talking of elementary particles and we are in the framework of quantum mechanics. To make a measurement, yes,it is correct that an interaction between elementary particles has to take place. This is the table of elementary particles, and at the bottom line it is interactions between them that form the world we observe in the microworld, and collectively the world we observe in the macroworld we live and move in. The macro world emerges from innumerable interactions of the micro world constituents.

A field is not an elementary particle. A field in second quantization manifests elementary particles according the operators operating on the field, so it is not as simple. To measure the electric field a huge number of photons are involved and it is not a simple example as you think.

To make a measurement one first needs a frame to make measurements "in" (a Lorentz frame in light of relativity).

A lorenz frame can be any frame correctly defined. All interactions observed in that frame can be transformed to other moving frames, but we tend to work with lorenz invariant quantities so as not to worry about transformations,

Assigning such frame to a photon appears to me, to be problematic in the sense that constant velocity of a photon in any inertial frame implies a photon in its own frame having a velocity c (the speed of light).

A photon has no rest frame, is what you mean. There is no reason for it to have a rest frame other than the prejudices we carry from the macro world. The lorenz transformation assures that the mathematics of any frame are correct for photons too.

Now if we assume that no such frame exists for a photon, photons cannot interact with an other one.

Two photons define a rest frame , because two photons have an invariant mass which has a frame where all momenta are zero. Example the pi0 decay to two photons.

Can we explain in this way that photons do not interact with each other, or more generally particles moving at velocity c do not interact with similar particles?

No, we cannot, because they do interact through exchanges of virtual particles at higher orders in the mathematical expansions of the solutions of the specific problem, and thus with small probabilities. Gamma gamma scattering experiments exist, and gammas are high energy photons. They are even talking of gamma gamma colliders,

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A particle's interaction (with anything it can interact with) can be thought of as it making a measurement of the physical quantity associated with the interaction,

This isn't true. All measurements are interactions, but not all interactions are measurements.

When we talk about a Lorentz frame in special relativity, we're talking about a very sophisticated construct, which is sort of like filling the universe with clocks, rulers, and signaling devices, and then carrying out a complicated surveying process. A single photon doesn't do anything like this.

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A particle's interaction(with anything it can interact with) can be thought of as ,it making a measurement (of the physical quantity associated with the interaction-eg electric field in case of charged particles interaction)and acting accordingly

This isn't the correct way to think about interactions, and as you later point out this (kind of) violates causality. When two particles make a measurement of each other they have already interacted!

Firstly, you should expand the framework that you're trying to solve this problem in. In order to fully describe particle interactions you have to move from describing individual relativistic particles to talking about relativistic interacting fields whose quanta are particles, or a 'quantum field theory'. The reason for this is that the photon itself is a quantum of the the electromagnetic field which itself is fundamentally relativistic. In this framework, interacting fields are constantly interacting, so your problem does not arise.

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  • $\begingroup$ Here interaction between two photons has not already happened. Absence of a frame to associate with a photon has made it not able to participate in any interaction with another photon. Are photons of light and photons which are quanta of em interaction physically same? $\endgroup$ – user35122 Dec 3 '13 at 5:18
  • $\begingroup$ Yes :) photons of light and quanta of em are physically the same. This is why the different wavelengths of light are described as the electromagnetic spectrum. You should not think about individual photon interactions but instead you have to think about photon fields (the EM field). As such you should not think about the frame of reference of individual photons, but of the photon field - which does not have to be travelling at the speed of light. $\endgroup$ – kd88 Dec 3 '13 at 15:08

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