Jump to content

George Dowell

Members
  • Posts

    5
  • Joined

  • Last visited

Everything posted by George Dowell

  1. No sir, sorry I'm not clear, it is only e-e+ is what I'm asking about. Now I see you said "in colliders" in your answer. That is the high energy mode I'm referring to, but want to know where between Thermalized and a Collider does e-e+. For example at 542 KE, it is not allowed. Before it can, either Bremsstrahlung or collisions happen, either to the positron alone or to a positron-electron atom (positronium), until the pair are at rest, then annihilation happens with the common dual but opposed 511 keV from the pair's rest energy. Outside of a collider environment, are there commonly known e-p annihilation's that produce, say 1000 keV pairs photons, or anything besides 511 keV pairs? Thank you. Geo
  2. Thanks. Could you point out a reference where these other energies are mentioned please. And if possible, where does the low energy ("thermalized" or "at rest" they"are calling it) stop and the high energy mode become possible? I'm only concerned with electrons and positrons, not the other positron/particle annihilations. Thanks for your time. Geo
  3. If I may add a practical example to verify the quoted above: As a Ham Radio operator, one of the many aspects that fascinated me was E.M.E.- That is a radio path which originates on the earth (me) travels to the moon, is reflected, a portion of that comes back to the earth. This happens on 144 MHz. When I listed to my own radio signal coming back, 2.7 +/- seconds later, it is on a different frequency at my location than the original frequency. The difference frequency is constantly changing due to the constant orbital motion, and will be above or below the transmitted frequency depending is the moon is advancing or receding relative to my position. We use this path to 2-way communicate with other radio stations scattered all over the world. This is Doppler shift and shows both + and - , or in light would be blue and red shift. The other earthly example is Police Speed Radar- they used continuous transmission 10 GHz when I worked on them, the signal hit a car and bounced back to the duplex antenna at the transmitter, where a tiny portion of the transmitting signal was mixed with the received signal, yielding a difference frequency equal to the Doppler shift caused by the vehicles velocity. This was in the audio range and could be accurately counted, the frequency being displayed in MPH for the operator to see. Both these are relativistic examples, and I believe the formulas some mentors mentioned were the same ones we applied to radio frequency Doppler shift. Geo>K0FF K0EME
  4. It's my understanding the electron-positron annihilation can only take place when both particles have no remaining kinetic energy, and my experiments bear that out - always two 180 degree opposed 511 keV photons. When positrons are created they do have kinetic energy, but before annihilation they loose it through collisions or radiative effects. What I'm trying to figure out and have no direct knowledge of, is the high energy mode (collider- like LEP), when they are forced together with very high kinetic energy naturally that energy is used up, transformed into particles and waves, conserving the original energy. Does this happen, before, during or after annihilation? Thanks Geo
  5. Electron-Positron Annihilation, low energy vs high energy mode- do both end in 511 keV Annihilation? Normal room temperature electron-positron annihilation is a well documented process, used in industrial, medical and metroloogy fields, among others. There are two modes of e-p annihilation, the room temperature or so-called low energy model, and the high energy model (think storage Rings, accelerators, colliders). In the latter category, do the e-p particles annihilate and produce 511 keV opposed photons like in a normal annihilation, or is there a threshold where e=p annihilation can be forced by kinetic energy alone? Citations/ references if you please, not personal opinions, I need to prove this to someone, one way or the other. Thank you for your time. George Dowell
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.