YT2095 Posted January 19, 2008 Share Posted January 19, 2008 I`m trying to picture the idea behind this when it comes to particles and the energy released on collision. is this "Spin" literal? like an angle grinder meeting another angle grinder, if they spin in opposite directions there is a Massive amount of energy given off, but if one is stationary then there is only Half that energy, and if they are Both in the same direction then there is even less energy. is that anywhere Close to conceptualising this? for the up down, is that like 2 cars hitting when traveling in opposite directions? one car maybe at 60MPH the other also at 60MPH but combines are 120MPH. is that close to the idea? Link to comment Share on other sites More sharing options...
timo Posted January 19, 2008 Share Posted January 19, 2008 Energy released is purely determined by conservation of energy. The sum of the energies of the particles after collision will equal the sum of the energies of the particles before collision (same statement goes for momentum). What you call "energy released" is just some mass (to be considered something like condensed energy or potential energy for this case) of the initial particles converted into kinetic energy of the final particles. Spin does not play a role, except that in some cases spin orientation can contribute a term to energy (e.g. for a bound electron in a magnetic field). Link to comment Share on other sites More sharing options...
YT2095 Posted January 19, 2008 Author Share Posted January 19, 2008 Huh? I don`t understand Any of that. Link to comment Share on other sites More sharing options...
timo Posted January 19, 2008 Share Posted January 19, 2008 Example particle collision: A + B -> C + D. Let's say A and B are electron and positron at rest (meaning their energy is E=mc² = 511 keV for each) and C and D are photons. Due to conservation of energy, the sum of the energies of C and D must equal 2*511 keV. Due to conservation of momentum, they must have the same energy (it's not too hard to show, but the reason doesn't really matter here). Therefore, each of the photons must have an energy of 511 keV which -due to photons being massless- must be kinetic energy. So what happens is that an electron and a positron at rest annihilate into two photons with an energy of 511 keV that fly away in opposite directions (directions coming from conservation of momentum). You might claim that this process released 1022 keV of energy. Spin did not come into that calculation at any point. I might have misunderstood you, but you seemed to ask whether spin determines the energy balance of a reaction. The answer is: The energy balance is always E(after) = E(before). So the energy after the collision is completely determined by the energy before the collision (because it is the same). Whether the spins affect the inital energy depends on the situation (more precisely on the scenario you set up and the simplifications you use). In my example above, the energy was independent of the spin orientation. Link to comment Share on other sites More sharing options...
YT2095 Posted January 19, 2008 Author Share Posted January 19, 2008 no no no, I was thinking Up/Down and Spin as actual PHYSICAL movement and Direction. like a car moving or a grinder disc spinning. Link to comment Share on other sites More sharing options...
iNow Posted January 19, 2008 Share Posted January 19, 2008 I found this article pretty useful: http://www-hep.phys.unm.edu/ Many particles, such as electrons, protons, and neutrons, behave like spinning tops. Unlike classical tops, however, the spin of these particles is an intrinsic quantum mechanical phenomenon. This spin is responsible for many fundamental properties of matter, including the proton's magnetic moment, the different phases of matter in low-temperature physics, the properties of neutron stars, and the stability of the known universe. In recent experiments, a number of research groups have ... More at the link above. Link to comment Share on other sites More sharing options...
timo Posted January 19, 2008 Share Posted January 19, 2008 Hm, ok. No, spin is not a physical motion of something in space. Link to comment Share on other sites More sharing options...
YT2095 Posted January 19, 2008 Author Share Posted January 19, 2008 are you Sure that`s the correct link? I can`t find that quote anywhere on that page? Hm, ok. No, spin is not a physical motion of something in space. *sigh* oh well, it`s back to the drawing board for me again, I really thought it was but when it comes to Maths and Physics I`m the 1`st to admit I`m pretty stupid! Link to comment Share on other sites More sharing options...
timo Posted January 19, 2008 Share Posted January 19, 2008 The thought is not so bad. Spin is very similar to angular momentum (the momentum of rotation) both in its mathematical description and in some physical effects. And I think that historically, the first idea of spin actually was about the electrons rotating around some inner axis. It might possibly be sometimes helpful to actually think of spin like that as long as you know where that picture breaks down, for example when electrons are considered point-like and hence cannot rotate around some inner axis. Can't really say, I've never pictured spin like that and personally wouldn't advice doing so. Link to comment Share on other sites More sharing options...
YT2095 Posted January 19, 2008 Author Share Posted January 19, 2008 does it affect the Direction (Path) of the resultant products after a collision? in my mental model it does, a bit like dropping a spinning glass ball from a height, it will smash sure, but the most of the broken matter will travel in the direction of the spin. Link to comment Share on other sites More sharing options...
iNow Posted January 19, 2008 Share Posted January 19, 2008 are you Sure that`s the correct link? I can`t find that quote anywhere on that page? My bad. Time for some more coffee. Here's the link I intended to share: http://nmcpp.phys.unm.edu/ and another (perhaps more specific to your line of inquiry): http://www.newton.dep.anl.gov/askasci/phy00/phy00562.htm After that, I really need to bow out and let the experts address your questions. Cheers. Link to comment Share on other sites More sharing options...
YT2095 Posted January 19, 2008 Author Share Posted January 19, 2008 Nice, I understood almost everything said in that last link, Cheerz! Link to comment Share on other sites More sharing options...
timo Posted January 19, 2008 Share Posted January 19, 2008 does it affect the Direction (Path) of the resultant products after a collision? In billard, the motion direction of the balls after collision is completely described by how the balls hit each other. This is not the case in QM. However, there can be preferred directions, meaning if you repeat the collision several times, the resultant products (technical term: final state particles) will be more often end up being travelling in one direction than into some other. Spin can influence the preferred directions of the final state particles. This even has applications to current particle physics. Some foreword, because it will be necessary. The term "spin" can either refer to (a) a property of the particle class, e.g. "electrons have spin-1/2". It's a property of the particle class in the same sense as mass is a property of the particle class. (b) a property of some individual particle, e.g. "the electron has spin up". It's a property of the individual particle in the same sense as momentum is a property of the individual particle. This kind of spin is also referred to as "spin orientation", "polarisation" or "helicity". @a: If two particles merge into an intermediate particle which then decays, the spin of the intermediate particle can influence the preferred directions of the final state particles. This can come handy in searches for gravitons and Higgs-bosons which have a unique spin for elementary particles (2 and 0, respectively). @b: Here's a plot of mine: Forget about the blue dots, which explicitely show some incorrect results for comparison. The process is an incoming photon interacting with an electron at rest. Imagine the horizontal axis to be some preferred direction of the particles after collision. The value plotted is the preferrancy of this direction for a collision where the electron and the photon had a certain spin-orientation relative to the preferrancy for a collision where the electron had no definite spin orientation. If spin wouldn't affect the directions, then both the red and the green line would be =1 (parallel to the horizontal axis). Since this isn't the case, the spin orientation clearly has an effect on the preferred directions. This very process has some use for particle detectors, but I cannot remember the details (and possibly never even really knew them). Link to comment Share on other sites More sharing options...
YT2095 Posted January 19, 2008 Author Share Posted January 19, 2008 aha, that`s making a bit More sense to me now, Cheers! Link to comment Share on other sites More sharing options...
Severian Posted January 20, 2008 Share Posted January 20, 2008 The conservation of angular momentum is a direct consequence of the laws of nature being rotationally invariant. Now, when Dirac invented the Dirac equation to describe electrons, to his surprise he showed that it didn't conserve angular momentum unless the electrons themselves had an additional source of "intrinsic" angular momentum. Since the laws of physics are obviously rotationally invariant, and thus angular momentum is conserved, the electrons must have this extra angular momentum, called 'spin'. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now