Annihilation and rest mass

[Mr Skeptic]

Imagine an electron and a positron (nearly) at rest a (nearly) infinite distance apart. Let the electron and positron fall directly toward each other (preferably without forming positronium) and annihilate. The energy of the system is [math]E = 2 m_e c^2[/math], and is conserved. But, as the particles fall, they gain kinetic energy, so that must mean that their rest mass/energy is decreasing. The rest mass decreases toward 0 as the speed of the particles increases toward c. Also, the dipole moment is decreasing toward 0. You could say that it looks more and more like light.

 

So my questions:

1. Do the various models that predict the rest mass of a particle account for such situations? That is, the rest mass seems to be variable.

2. How smoothly does the transformation from electron/positron to photons happen? Does their rest mass smoothly decrease as their velocity approaches c? What about the charge?

3. Would the electron and positron emit photons as they fall/accelerate?

4. If you put an electron on either side of a positron, what would be the movement of the extra electron when an electron/positron pair annihilate?

[NeonBlack]

Mass does not change.

 

Kinetic energy: Energy associated with motion.

 

Rest energy: Energy when an object is at rest. (Duh)

 

The two are completely independant.

 

Again, mass does not change!

[Spyman]

The rest mass is not variable, the relativistic mass increases as the speed approaches c.

 

The term mass in special relativity is commonly used by physicists to mean a quantity that does not depend on the observer or the inertial frame used to observe it. However, because the term relativistic mass is also used, this occasionally leads to confusion.

The invariant mass of an object (also known as the rest mass, intrinsic mass or proper mass) is an observer-independent quantity that is synonymous with mass. On the other hand, the relativistic mass of an object (also known as apparent mass) increases with its speed and therefore depends on one's frame of reference.

http://en.wikipedia.org/wiki/Mass_in_special_relativity

[swansont]

If the electron and positron form the positronium bound system, then they will emit photons as they "jump" down energy levels, eventually shedding 13.6 eV of energy, before annihilating.

 

If they don't form the bound state, they would emit some Bremsstrahlung. If they didn't, then they would have an extra 13.6 eV to emit during annihilation.

 

 

The annihilation photons are really short 13.6 eV if the annihilation happens from the ground state of positronium. But since they are 511 keV, that amount is negligible and doesn't show up in the calculations at this level of precision.

[D H]

Imagine an electron and a positron (nearly) at rest a (nearly) infinite distance apart. Let the electron and positron fall directly toward each other (preferably without forming positronium) and annihilate. The energy of the system is [math]E = 2 m_e c^2[/math], and is conserved. But, as the particles fall, they gain kinetic energy, so that must mean that their rest mass/energy is decreasing.

 

You forgot about electrical potential energy. The particles lose this potential energy as they fall toward each other.

[Mr Skeptic]

The rest mass is not variable, the relativistic mass increases as the speed approaches c.

 

So how do you account for conservation of energy then? The energy of the system is not increasing, because it is a closed system. At best, it would decrease due to radiation as swansont pointed out. Since the kinetic energy is increasing and the energy is constant, the rest energy must be decreasing.

[Farsight]

Mr Skeptic, here's an excerpt that may prove useful:

 

In terms of basic concepts however, we can understand the energy required to bring two electrons together as related to the extra energy required to separate the electron and the positron in pair production and send them flying apart. Without this, the electron and positron would immediately annihilate. In practice we do not require the 1/137th of 1022KeV which would give them a mutual escape velocity, because pair production does not occur in a total vacuum. It can be less, much less, as can be seen from binding energy. For example one binding energy level of positronium, an electron/positron “exotic atom” which usually lasts a microsecond before annihilating into two 511KeV gamma photons, is -0.0068 KeV. The binding energy is given as a negative figure because it’s a measure of how much energy might be obtained by letting an electron and a positron attract one another from infinity. The temporary positronium electron/positron configuration has less total energy than an electron and positron flying forcefully apart after pair production.

[swansont]

Oh, right. Positronium is a factor of 2 smaller than Hydrogen. 6.8 eV vs 13.6 eV

 

So how do you account for conservation of energy then? The energy of the system is not increasing, because it is a closed system. At best, it would decrease due to radiation as swansont pointed out. Since the kinetic energy is increasing and the energy is constant, the rest energy must be decreasing.

 

The potential energy is negative. If you start at rest, infinitely far apart, you will have KE + PE = 0

 

The mass of positronium is 6.8 eV/c² smaller than 2m_ec²

[Mr Skeptic]

The potential energy is negative. If you start at rest, infinitely far apart, you will have KE + PE = 0

 

Isn't that arbitrary? The only reason I have seen to define the potential at infinity as the zero point is because it removes the rather embarrassing infinity that inverse square laws tend to have at zero distance. Couldn't you just as well define the potential at infinity for a matter/antimatter system to be mc²?

 

Also, where does all the negative potential energy of the e+/e- pair go when they annihilate?

[Rocket Man]

does the acceleration of the particles over the arbirtarily large distance create any important magnetic effects?

 

the infinity at zero dissapears because a positron and an electron can only be so close together before they interact in some way violating the inverse square.

 

at infinite distance you have finite PE by the integration of the inverse square from the anihilation point to infinity.

if you define that as zero as swansont has done, the potential energy goes negative as the particles apporach eachother. that's accounted for by the kinetic energy on impact

[Spyman]

So how do you account for conservation of energy then?

I don't know, I only gave you the piece in the puzzle I did know...

 

You will have to follow the others explanation.

(swansont is the Expert.)

[swansont]

Isn't that arbitrary? The only reason I have seen to define the potential at infinity as the zero point is because it removes the rather embarrassing infinity that inverse square laws tend to have at zero distance. Couldn't you just as well define the potential at infinity for a matter/antimatter system to be mc²?

 

Also, where does all the negative potential energy of the e+/e- pair go when they annihilate?

 

The zeroes are arbitrary, but we only measure the change between two states, so that all gets subtracted out anyway. We choose some values to make the calculations easier, and choosing zero as the baseline is one way to do that.

 

I explained the last bit in post #8. The rest mass of positronium will be slightly smaller than the rest mass of the free constituents. Thus the photons would have slightly less energy.

[Farsight]

Isn't that arbitrary? The only reason I have seen to define the potential at infinity as the zero point is because it removes the rather embarrassing infinity that inverse square laws tend to have at zero distance. Couldn't you just as well define the potential at infinity for a matter/antimatter system to be mc²? Also, where does all the negative potential energy of the e+/e- pair go when they annihilate?

 

Skeptic, it's not as complicated as you maybe think. In pair production you need a 1022KeV gamma photon to make an electron and a positron. But if that's all you've got they don't fly apart. They just annihilate straight away. You need a bit more oomph so that they fly apart.

 

 

If you catch hold of the electron and positron and put them back close together to make some positronium, it will last a microsecond before annihilating to give you two 511KeV gamma photons. But in catching hold of them you nicked a bit of their fly-apart energy. The negative is just accounting convention. All the energy is conserved. It gets a little confusing because initially you're looking at things from the rest frame of the nucleus, and afterwards people tend to fuzz over the electron motion and the positron motion.

[Mr Skeptic]

The zeroes are arbitrary, but we only measure the change between two states, so that all gets subtracted out anyway. We choose some values to make the calculations easier, and choosing zero as the baseline is one way to do that.

 

Hm, I thought that the zero was arbitrarily chosen, which is slightly different than arbitrary. But if the only thing we can measure is the change in potential energy, then as you said it cancels out anyhow. There is a difference, though, if the total potential energy ever comes into play.

 

I explained the last bit in post #8. The rest mass of positronium will be slightly smaller than the rest mass of the free constituents. Thus the photons would have slightly less energy.

 

So its unavoidable for the e+/e- to radiate as they annihilate. I was hoping to avoid that, but I guess it is just entropy taking its toll. So if you stopped the e+/e- pair before they annihilated, they would have less rest energy than if they were far apart? And that would be due to having more negative potential energy?

 

I guess I was equating rest energy with potential energy to avoid negative energy. I don't see what the difference is, except that my way you don't have to believe in negative energy so it seems to me more Occam's Razor-friendly.

[swansont]

The negative term is bookkeeping, but it's handy, because you know a negative mechanical energy (KE + PE) indicates a bound state, which would require energy to separate them.