Center of momentum and mass energy equivalence

[Dale]

the symmetric emission of photons

It is only symmetric if the positron and electron have no net momentum. If they have momentum then the photons are emitted asymmetrically in the detector frame, and the decay does not occur on the line between the two detectors. This leads to blurring in the images, and is one of the fundamental limits to the resolution for some PET tracers.

This is precisely the case where energy and mass are different

 

[Ibix]

It is VERY cool that we see the massive positron decay as two photons moving in opposing directions...AT c!~ lol bye bye mass..oh wait. Ima call it a system so its still massive.

Mass is conserved in a closed system, so the two photons must have mass. You can treat the two photons separately in which case you don't have a closed system. I don't really understand why you find this a problem.

 

[vanhees71]

No, each of the photons is massless. Both photons together have a total invariant mass of $\sqrt{s}$, i.e., the center-mass energy of the colliding electron-positron pair. That's just a definition of the kinematics of scattering theory.

 

[nitsuj]

Mass is conserved in a closed system, so the two photons must have mass. You can treat the two photons separately in which case you don't have a closed system. I don't really understand why you find this a problem.

Because you're calling two photons moving in opposite directions a closed system. I cannot understand how two thing moving at c in opposite directions could thought of as a "closed system"; causally they couldn't be farther apart.

 

[SiennaTheGr8]

Because you're calling two photons moving in opposite directions a closed system. I cannot understand how two thing moving at c in opposite directions could thought of as a "closed system"; causally they couldn't be farther apart.

A "closed" system (aka an "isolated" system) is simply one that doesn't interact in any way with its environment. It's an idealization—no system is truly closed in the real world.

In some contexts—thermodynamics especially—it's common to distinguish between a "closed" system (no mass leaves or enters, though energy might) and an "isolated" system (no mass or energy leaves or enters; i.e., the system doesn't interact with its environment at all). But in special relativity this distinction becomes less useful ($E_0 = mc^2$), and so the terms are often used interchangeably to mean the latter.

Note that whether a system is closed has nothing at all to do with how far apart its constituents are. For many purposes our galaxy is well-approximated as a closed system.

 

[Mister T]

Because you're calling two photons moving in opposite directions a closed system. I cannot understand how two thing moving at c in opposite directions could thought of as a "closed system"; causally they couldn't be farther apart.

An imaginary boundary separates a system from its environment.

The system is anything within the boundary. If you choose it to be an electron-positron pair about to collide then that's what it is. It doesn't cease to be a system after the interaction. Moreover, if nothing crosses the boundary then the system is said to be closed. The laws of physics describe this interaction and in one of those descriptions the mass of the system doesn't change during the interaction as long as nothing crosses the boundary.

 

[PAllen]

Because you're calling two photons moving in opposite directions a closed system. I cannot understand how two thing moving at c in opposite directions could thought of as a "closed system"; causally they couldn't be farther apart.

Consider two equal energy photons moving towards each other. If high enough energy, they can (very rarely) interact producing a positron+electron pair. Or, they can both be absorbed by a body. In either case, the invariant mass of the system remains constant. The e/p pair would have the same invariant mass the photons did before. The absorbing body would increase in mass by the invariant mass of the two photons. If it makes sense to attribute mass to the photon pair moving towards each other, do you really want to say this mass disappears if they pass each other without interacting?

 

[Dale]

I cannot understand how two thing moving at c in opposite directions could thought of as a "closed system"; causally they couldn't be farther apart.

As @Mister T mentioned above, they can clearly be called a system simply by arbitrarily choosing your system boundaries to include them. Whether they are a closed system depends on if momentum or energy are crossing the system boundary. It does not matter if the interior of the system is interacting with itself.

Suppose that you start with the initial electron and positron, and suppose further that you draw your system boundaries very far away. Suppose further that there is nothing else inside the boundaries and that no energy or momentum cross the boundary.

Then, it should be clear that the system has some well-defined energy and momentum and therefore some mass. Now, since those are conserved then they must remain constant whether or not the electron and positron anhilate. If they do not anhilate then the mass is attributed to the positron and electron system since that is the only thing inside the boundary. If they do anhilate then the mass is attributed to the two photon system since that is the only thing inside the boundary.

 

[Mister T]

@nitsuj Perhaps you are thinking of a bound system, such as a neutral Hydrogen atom, where the nucleus and the orbital electron are bound by an attractive electrical force. There's an interaction bonding the constituents. But a system need not be bound. The constituents of a system need not be interacting. If you ionize that neutral Hydrogen atom by removing the electron so that it's very far from the nucleus you can still have the same system. Of course, you would need to transfer enough energy to the system to perform that ionization, but when you do that you increase the mass of the system by the amount of energy that you transfer.

The fact that a pair of oppositely moving photons are not interacting, and are indeed causally disconnected as you say, has no bearing on the claim that they constitute a system. If it did then the ionized atom and the free electron wouldn't constitute a system. Think about that nucleus and that electron moving apart even though they are so far from each other that their interaction is negligible. The kinetic energy that each has relative to their center of momentum contributes to the mass of the system, if it didn't energy, momentum, and mass wouldn't be conserved! Note that $m^2=E^2-p^2$; therefore if any two of those three are conserved, so is the third.

 

[Ibix]

Because you're calling two photons moving in opposite directions a closed system. I cannot understand how two thing moving at c in opposite directions could thought of as a "closed system"; causally they couldn't be farther apart.

In addition to all the comments that have come before, it's also worth noting that the photons do not need to be traveling in opposite directions. Merely not traveling exactly parallel is enough.

Edit: Of course, two photons traveling in opposite directions is two photons just traveling not quite parallel, as described in some other frame.

 

[nitsuj]

Ironic that I just bought a tiny veil of tritium to light the inside of my watch...I went to read about the type of radiation and came here to ask will it be firing off these symmetric photons..I see there's no love for my "it's things moving apart at c, those two things meaningless to each other physically.." lol

 

[Dale]

I see there's no love for my "it's things moving apart at c, those two things meaningless to each other physically.."

Even if they are meaningless to each other they are not meaningless to us

 

[SiennaTheGr8]

Even if they are meaningless to each other they are not meaningless to us

Deep.

 

[nitsuj]

Even if they are meaningless to each other they are not meaningless to us

'nuff said :)

So read more and it doesn't look like tritium makes such photon systems...by does have electron and electron neutrino "pairs" ...very interested to read about the angles they part but have yet to find something on that