B Electron-positron bremsstrahlung

[metastable]

If I have an electron and positron in a vacuum, initially at rest with respect to each other and in close proximity, will they produce any bremsstrahlung radiation?

 

[Meir Achuz]

Yes, but this can be treated classically only if they are far apart.

 

[Vanadium 50]

They will certainly radiate but I am not sure I would call it "bremsstrahlung".

 

[metastable]

What happens as initial rest distance decreases?

 

[Vanadium 50]

Where are you going with this?

 

[metastable]

I am still just wondering about the original question which stated "initially at rest with respect to each other and in close proximity."

 

[Vanadium 50]

Which was answered.

 

[metastable]

So the answer is yes? I didn’t understand what was meant exactly by “but this can be treated classically only if they are far apart.”

 

[king vitamin]

Eventually it cannot be treated classically. Quantum electrodynamics (QED) must be used to describe their radiation/interaction (1) if the total mass-energy of the system approaches that of m_e c^2, or (2) when the distance between the two is comparable to their Compton wavelengths. (There are probably other limits I am not thinking of.) Eventually they annihilate, which is a textbook QED process.

 

[metastable]

Case A
Initial electron-positron rest distance = 5 femtometer

Case B
Initial electron-positron rest distance = 10 femtometer

Do we see any difference in the total energy radiated (bremsstrahlung or otherwise) during the subsequent e- + e+ -> γ + γ reactions?

 

[king vitamin]

Sure, those two scenarios have a different initial potential energy and will therefore have a different amount of total radiated energy.

 

[mfb]

Particles with a position so well-constrained will have a huge uncertainty in their momentum, by the way. You'll get a large range of results just from repeating the experiment many times.

Concerning the total energy: Total energy is always conserved. That part is easy. There can be more than two photons in the final state.

 

[metastable]

If I repeat the experiment many thousands of times, each time measuring the total radiation emitted after placing the particles at the lowest experimentally achievable initial rest distance (with some uncertainty as to their actual initial positions and momentums in each repetition of the experiment), will I ever measure the subsequent e- + e+ -> γ + γ radiation emitted totaling less than 2 *
0.5109989461(13) MeV (electron rest mass)?

 

[mfb]

If you prepare the initial state as positronium it is possible. Note that it doesn't have a well-defined distance between the two particles.

 

[metastable]

The positronium article states:

"Para-positronium can decay into any even number of photons (2, 4, 6, ...), but the probability quickly decreases with the number: the branching ratio for decay into 4 photons is 1.439(2)×10−6.[1]"

In the case of 4-photon-emission para-positronium decay... are the 4 photons emitted sequentially in pairs or simultaneously? Are the 4 photons expected to have roughly the same energy or does each pair have a different value? If different, do we expect the higher energy photons to be emitted prior or subsequently to the lower energy photons?

 

[mfb]

are the 4 photons emitted sequentially in pairs or simultaneously?

Within the problems to define the emission time: Simultaneously. It doesn't make sense to pair them either. They have four different energy values.

You seem to avoid your main question. What do you actually want to know?

 

[Vanadium 50]

will I ever measure the subsequent e- + e+ -> γ + γ radiation emitted totaling less than 2 *
0.5109989461(13) MeV (electron rest mass)?

If you constrain the system to start at 5 fm apart, you don't have a well defined energy.

Again, where are you going with this? If you have some bigger picture in mind, why won't you tell us what it is? Or are you just flinging random questions at us?

 

[metastable]

The positronium article states:

"The mass of positronium is 1.022 MeV, which is twice the electron mass minus the binding energy of a few eV."

Does the positronium system have less mass/energy than its constituent fundamental particles? Can I write an equation relating the electron mass with its antiparticle charge-separation distance?

 

[mfb]

Does the positronium system have less mass/energy than its constituent fundamental particles?

Yes, that's what the article told you already.
This is a general property of bound states (with exceptions that are irrelevant here). As another example a hydrogen atom has an energy slightly below the sum of energies of proton and electron.

Can I write an equation relating the electron mass with its antiparticle charge-separation distance?

A well-defined distance doesn't correspond to a well-defined energy and small distances cannot be treated classically, as everyone in the thread told you already.

 

[metastable]

Within the problems to define the emission time: Simultaneously. It doesn't make sense to pair them either. They have four different energy values.

If I’m looking at a large quantity of separate para-positronium -> 4 photon annihilations, and I’m using a photon detector apparatus which is substantially in the same rest frame as the center of mass of the positronium systems, and from one such decay I observe 4 photons of 4 different wavelengths whose combined energies total to a few eV less than double the electron rest mass (on account of the positronium binding energy), how am I to reasonably conclude they were emitted simultaneously? Is the emission duration for each photon independent of the 4 different wavelengths?

 

[mfb]

how am I to reasonably conclude they were emitted simultaneously?

You can measure their arrival time.

Is the emission duration for each photon independent of the 4 different wavelengths?

"Emission duration" is not a meaningful concept.

 

[metastable]

If I look at the 4 discrete energy values of each of the 4 detected photons from one of the para-positronium annihilations, and then I divide the wavelength of each of the emitted photons (in meters) by the speed of light in meters per second (photon wavelength meters / meter per second)-- or if I divide 1 second by the frequency of each of the emitted photons in cycles per second (1 second / photon frequency hz) -- how do I reasonably conclude I haven't just calculated the varying emission durations for each separate photon which had been detected?

 

[Vanadium 50]

how do I reasonably conclude I haven't just calculated the varying emission durations

"Emission duration" is not a meaningful concept.

Again, where are you going with this? If you have some bigger picture in mind, why won't you tell us what it is? Or are you just flinging random questions at us?

 

[metastable]

If the emission durations of the 4 photons vary, and the detection of the photons is simultaneous, and detectors are at substantially the same rest distance from the center of mass of the para-positronium collisions, how can the emissions be "simultaneous?"

 

[Vanadium 50]

"Emission duration" is not a meaningful concept.

 

[metastable]

if i divide the wavelength of each of the (4) emitted photons (in meters) by the speed of light in meters per second (photon wavelength meters / meter per second), what then have I calculated (if not the emission duration)?

 

[mfb]

You have calculated the inverse of the frequency of the photons.

 

[Vanadium 50]

The inverse of the frequency of the photons.

 

[metastable]

& is the inverse of frequency not the duration of emission?

 

[mfb]

"Emission duration" is not a meaningful concept.

This thread is going in circles. If you keep asking the same question over and over again I'll close it. There is simply no point in repeating the answer for the 10th time.

 

[metastable]

i take the photon length (wavelength) in cycles per second (hz)... next i take the inverse... seconds per cycle... the inverse of cycles per second (seconds per cycle) is not the duration of emission?

 

[mfb]

As you have been told many times, there is no such thing, repeating this many more times wouldn't help. I closed this thread.