Photon momentum components [closed]

The question has now been amended to:

"Is there a systematic causal link between the direction of a photon and the direction the mass from which it emerges is moving in?

I'm not talking about the direction of the electron involved in radiating the photon, but the direction of the overall mass (atom or larger) from which the photon emerges, and asking whether such direction of such overall mass will systematically determine the path a photon takes when it flies off on its trip.

So let's say the mass is doing 0.5c along the x-axis, will this systematically determine the angle of trajectory which the photons which emerge from it take, or will they simply find their own trajectory independently according to other random factors? If it will systematically determine the trajectory of the photons, please state whether such determination will be in part or in full, and what the mechanism for such will be."

closed as unclear what you're asking by John Rennie, stafusa, Bill N, sammy gerbil, Jon CusterDec 26 '17 at 12:12

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• Momentum is a vector, and it is conserved. For a decay process, the total momentum of the system will remain constant, and that includes the direction of the sum of the momenta must be identical to the original particle velocity. – Bill N Dec 26 '17 at 0:10

Both momentum and energy have to be conserved (and all three directional components of momentum). So the original particle is deflected by the emission of the photon, and the photons direction and momentum must be such that everything is conserved.

Photons are elementary particles in the quantum mechanical standard model of particle physics.

An electron radiating a photon when accelerated does not lose any mass, its mass is invariant . Energy and momentum four vectors are conserved in whatever system one is examining them.

Here is what is happening at the photon electron and accelerating agent level.

The accelerating agent shown as Z provides the field on which an electron coming in from the left exchanges a virtual photon , the electron becomes virtual ( it is off mass shell) and then decays into an electron and a gamma/photon at the bottom.

Energy conservation means that the end products have the energy from the system eZ, (where Z can be a large magnet providing a magnetic field or just a nucleus'field ) to be taken up by the outgoing at the bottom eγ . The same with momentum conservation. As Z even as a nucleus has a much larger mass than the electron and is static on average, momentum conservation means that the incoming momentum of the electron is shared between the electron and the photon, according to the quantum mechanical scattering crossection which can be calculated.

The outgoing electron has the mass identifying it in the table. (Do not confuse relativistic mass with invariant mass. Relativistic mass is velocity dependent and cannot be used in four vector conservation laws.)

Energy conservation will be in the initial sums of the electron+field energy so that the electron increases its momentum at that instance, and is summed in energy conservation with the photon taking energy and the electron less than it would have if this basic feynman diagram did not have to hold, because of the laws of quantum mechanics.

• Anna ... you seem to want me to infer the answer rather than actually give it to me. You have made 2 statements which might be correct in themselves, but each leaves me wondering what I am to infer from them. Could you please take full cognisance of my question with regard to its intent, and answer it explicitly. – Analytical Dec 25 '17 at 23:20
• Your question is not answerable because you have a fundamental miss-conception. – anna v Dec 26 '17 at 5:24
• The photons emitted by radiation do not change the invariant mass (note the name) of the system they are leaving. They just take away energy and momentum of the system they are leaving, the energy provided by the accelerating source (an electron in its rest frame does not radiate), and momentum so that momentum will be conserved in the system . As the accelerating source has much larger mass than the electron then momentum changes seen involve only the electron and the photon. You are confusing invariant mass with relativistic mass which is not an invariant under lorentz trasformations. – anna v Dec 26 '17 at 5:31

As anna v pointed out, the scenario that you describe has a few issues.

Firstly, what is meant by the mass of the particle? If it the rest mass, then (for $c=1$) it is given by the dispersion relation: $$m_0 = \sqrt{E^2 - |{\bf p}|^2} .$$ Or is it the relativistic mass that one gets by boosting the particle to the velocity which gives it the momentum ${\bf p}$? If we are talking about the rest mass, then it would not change if the particle radiates a photon.

However, there is another consequence. After a particle that was on-shell (meaning that it obeyed the above dispersion relation) emitted a photon, which is also on-shell (obeying the dispersion relation $E_1^2 - |{\bf p}_1|^2=0$), the particle would in general not be on-shell. This is because of momentum and energy conservation. Hence, $$\sqrt{(E-E_1)^2 - |{\bf p}-{\bf p}_1|^2} \neq m_0 .$$ The resulting particle is therefore called off-shell, because it does not obey the dispersion relation (also called the mass shell condition).

What must happen is that the particle must emit another photon. Then it can again be on-shell. The combined result is given by the diagram that anna v shows in her answer.

• Thanks for your answers, however neither actually answers my question. Allow me to ask it another way: "Is there a systematic causal link between the direction of the speed of a photon and the direction of the speed of the mass from which it emerges?" So I'm not really talking about the direction of the speed of the electron involved in radiating it, but the direction of the speed of the overall mass (atom or larger), and asking whether such speed direction (velocity) of such overall mass will systematically determine the path a photon takes when it flies off on its trip. – Analytical Dec 26 '17 at 10:29
• So let's say the large mass is doing 0.5c along the x-axis, will this systematically determine the angle of trajectory which the photons which emerge from it take, or will they simply find their own trajectory independently according to other random factors? – Analytical Dec 26 '17 at 10:29
• The first photon can be emitted in any direction. It doesn't matter, because the single emission of a photon anyway does not satisfy momentum conservation. However, the emission of the second photon is then restricted by the conditions of the first photon, because it needs to satisfy the momentum conservation for the complete process. – flippiefanus Dec 29 '17 at 5:22