timo

you mean mp² instead of mc²? No, a simple check of the units will show you it cannot be mp².

I rather think your problem is that you think the energy for particles was E = hf. That´s only true for photons. The energy is E = sqrt[ (pc)² + (mc²)² ] just like you got it. Step-by-step derivation: - Start with a plain wave: [math] \psi = \exp \left( -i(Et - px) \right) [/math] - I´ll leave it 1D to save some time typing. Also I set hbar to one. It would cancel out in the next equation anyways. - Plug it into the Klein-Gordon wave-equation: [math] \left[ \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} + m^2 \right] \psi = 0 [/math] (note that due to my lazyness in typing I also set c=1). => [math] \left[ \dots \right] \psi = \left[ -E^2 \psi + p^2 \psi + m^2 \psi \right] = \left[ -E^2 + p^2 + m^2 \right] \psi = 0[/math] Since the wavefunction is nonzero the term in brackets must be zero => -E² + p² + m² = 0 => E² = p² + m² Putting back the c´s to get the SI units thus gives: E² = (pc)² + (mc²)² or E = +- sqrt[ (pc)² + (mc²)² ] Note that you also get negative total energies. If I had used the Dirac equation (which is a bit more complicated) in above those solutions with negative energies would be associated to anti-particles. To be honest I´m not completely sure if this can also be said in case of the Klein-Gordon equation.

The mathematical error is that (pc)² + (mc²)² does not equal (pc + mc²)² The correct result would not lead to any problems. Special relativistic quantum mechanics is actually very well established. Another note: mv²/2 is not the total energy - even in classical mechanics. It´s the kinetic energy.

That´s because you neglect the attraction the comet exerts on the sun because it´s so small. I also said this in my post you were referring to. Point two was added because Dror started to talk about "2-3 identical masses". I´d be surprised if the mass really canelled out when you consider both movement equation and field equation for both particles involved. The center of mass is a different one if you change the mass of the comet.

The paths will not be the same for two reasons: 1) Both comet and photon come in with a different velocity 2) The comet attracts the sun differently than a photon does. Point two becomes increasingly important as the mass of the object being bent on the sun is increased. This is however not because the movement of a comet is a gravitational field would be dependent on it´s mass. It´s because the comet due to it´s own gravtiational field does attract the sun and causes it to move. With the sun in a different position you have changed your gravitational field and thus the two gravitational field (the one that the comet moves in and the one the photon moves in) are not the same. For particles much smaller than the sun you can simply forget about their attraction on the sun because it´s so tiny. You certainly cannot neglect this attraction when you have identical masses. I´m not completely sure if this attraction on the sun must be considered for comets (I doubt it) but you can also take a neutron if you like. My statement that both paths are curved remains. btw.: If you completely ignore quantum mechanical effects and threat a photon as a point size particle it doesn´t even take much mass to trap light in an orbit. The (theoretical) radius of the orbit will just be so small that it would be inside the attracting object (it´s 3 km for our sun, for example). That´s because of the high velocity light has.

99p times 10 certainly isn´t 999p

Well, being a superconductor implies no resistance. As electron-phonon interaction is part of electrical resistance (if I remember that correctly) I´d suspect that there´s no interaction with the phonons in a superconductor. But maybe you could explain about cooper-pairs if you understood them. Because I haven´t understood how two electrons can form a bound state so far.

Actually the non-mathematical explanation was included and very simple: A particle´s movement in a gravitational field is independent of it´s mass. So when you negelct air resistance a feather falls down exactly as fast as a stone. This in fact is odd and contradicts everyday´s experience. I didn´t believe it myself until I saw the experiment demonstrating this. In Newtonian Gravity this is because the mass cancels out in the equation. In General Relativity the mass doesn´t even appear in the movement equation. Mass creates gravitational fields (in GR even photons do, but that´s a different story) but the movement of a particle that´s influenced by a gravitational field is independent of it´s mass. So if a comet´s path is curved by a sun so must be a photon´s.

The path of a (sufficiently small) particle that is affected by a gravitational field is independent of it´s mass. That´s true in both Relativity and Newtonian gravity. In Newtonian gravity: [math] m \vec a = \vec F = m \vec g \, \Rightarrow \, \vec a = \vec g[/math] g of course describes the gravitational field. Whether this equation is still true for m=0 is not completely certain from this (there are arguments for both views) but it´s not important as systems with a relevant light deviation are relativistic anyways. In General Relativity: [math] \forall i : \, a^i = \sum_{jk} -\Gamma^i_{jk} v^j v^k [/math] Even without fully understanding the relativistic movement equation my point should become clear: The Gamma ([math] \Gamma [/math]) is the analogy to g in the Newtonian equation. It describes the gravitational field. The mass of the particle affected does not appear in the equation. In short: Movement in a gravitational field is independent of mass.

Well, I can tell you the steps I´d take to get the answer: a) To deternime the diameter of the circle you use the height-sentence (or whatever it´s called in english): (PQ)² = XP * PY = XP * (PS + SY) = XP * (QR + XP). Solving for XP gives you the position of all points except T, U and S b) To determine VU and thus the position of the other points you can use the same method again: (UV)² = (XS + SV) * (XY - XS - SV) Note that UV = SV because SVUT is a square. Solve and be happy EDIT: Hehe, 12 hours without response and then two people answering within 10 minutes .

Didn´t understand question 1), sry. For q 2): As said phonons describe the crystal´s state (of oscillation). Assume a neutron is scattered by the crystal. An incoming neutron can interact with the crystal´s atoms and cause them to vibrate. The outgoing neutron can have a different momentum and energy than before. This difference is taken by the crystal (in form of the atom´s oscillating differently than before). In the phonon picture the same process reads as the neutron interactin with the phonons and destroying some of them (taking over the related momentum and energy) and/or creating some (giving energy/momentum away). The most basic (and also most probable, I think) effects are either absorbing or creating one single phonon. Same is true for other particles, of course. I choosed neutrons because they are actually most commonly used for measuring the dispersion relation E(p) of a crystal´s phonon spectrum. Sidenote to araise further confusion: I´d like to note that saying the phonons describe the crystal´s vibrations is an analogy to classical mechanics. From the QM perspective they simply describe it´s state. The classical picture of oscillating atoms would demand the atoms to be at a certain position with a certain velocity which they cannot be in QM. A similarity would be the harmonic oscillator. While in classical mechanics you´d have your particle bouncing aroung with a certain frequency, the energy-eigenfunctions of QM are not related to any movement at all (a superpoition of them is, though).

It´s the relativistic mass that´s nonzero (might also have other names). The relativistic mass is simply the energy divided by c² - or in other words: E = mc².

no, let´s just do the math: E² = (mc²)² + (pc)² => (mc²)² = E² - (pc)² with |p| = E/c => (mc²)² = E² - (E/c * c)² = E² - E² = 0 => m = 0

I am absolutely sure. Check the units if you don´t believe it. That´s also what my statement "1-c² doesn´t make much sense" meant: you can´t substract meters/second (with any nonzero exponent) from a number.

It´s getting warmer. If you now consider point a) in my 1st post and check your equation again (the c³ was a typo, wasn´t it?) you´ll come up with m=0. EDIT: Which is not much of a surprise since |p| = E/c implies m=0.

Indeed, that´s why they are only different in the dimensionless constant 2pi More helpfull: A photon´s momentum is not m*v (at least not if m is supposed to be the rest mass). The momentum of a photon is [math]p = \hbar k [/math] or [math] |p| = E/c = \hbar \omega / c [/math].

If h and hbar aren´t the same thing and related by a factor of 2pi what might that say for f and omega ?

a) E² = (mc²)² + (pc)², not (mc²)²+(pc²)² b) also it´s [math] E = \hbar \omega [/math] or E = hf c) plus: Your math is wrong. The minus sign must be in the numerator and I can´t see where the momentum went. 1-c² doens´t make much sense, either. No, there´s no upper limit on photon frequency. Extremely high energies will lead to some QFT effects like constant creation of particle-antiparticle effects, though.

The filename "helper101.dll" was not found when i wanted to look it up in the "DLL Help Database Search" on microsoft´s website (you better check that out for yourself - I just did a brief search to verify my assumption). Because of that it seems very probable to me that this particular dll is not part of windows but a spyware program (as allready suggested). From my experiences it is often sufficient to simply delete such files (you might want to rename it first or make a backup copy to be safe) and maybe search your windows registry for entries containing the file and also delete them.

Conservation of momentum is the same; total momentum is conserved. In fact -as it was allready noted- energy and momentum can be seen as very similar. In relativity they are both elements of the 4-momentum which is conserved. The 4-momentum is p = (E, px, py, pz).

it´s not exactly the kinetic energy only but the total energy. The vector containing energy and momentum is then called the 4-momentum.

- Phonons are bosons in the sense that any (natural) number of them can be in the same state. Same is true for the euro-cents on my bank account (ok, they can be negatve). Not sure if I´d call them particles because of that. - In the picture a wave is shown. The wave reusults from the displacement of the atoms off their rest position. As these exitations from the rest postions can be described by phonons (see last post about different bases of the movement equation´s solution) the wave can be described as a number of phonons being in the crystal.

I thought a lot and quite long about how to describe what and got lost in details so let´s try the short version: - The movement equation for an atom in a crystal is a differential equation F = m*a, where the force F depends on the atoms´relative positions. The movement of the atoms (the solution of the diffEq) are oscillations. - Differential equations have a vector space of functions as their solution. The particular solution for a given problem is a vector of this vector space that´s determined by boudary conditions. - As in real valued vector spaces like R² or R³ you can describe elements of the vector space of functions as a linear combination of a valid set of base functions (base vectors): [math] \vec v = a_1 \vec e_1 + a_2 \vec e_2 + ... [/math] - The basevectors (basefunctions) you chose are the phonons. Any oscillation can be described by a linear combination of phonon states. - Choosing the basefunctions is arbitrary in classical mechanics (though most people like to avoid headaches and chose the easiest one) but can become important in quantum mechanics (at least if you want to avoid serious headaches). The phonons are chosen to be states of definite energy and momentum. - If you scatter neutrons (a common way to experimental get the phonon distribution in crystals) on a crystal and the scattered neutron has energy and momentum different from the original one you can say that the crystal simply went to another state. Alternatively you can say that a phonon was produced (or destroyed). In this alternative view you give the phonons an actual existence as a particle. The same can be done with the electromagnetical field where it´s actually quite common to treat the basefunctions of the field as particles (photons, then). - Hope that helps.

No, a phonon doesn´t really have mass. In fact, I wouldn´t even say they are massless - they just don´t have an attribute called mass (what would be the mass of a wave in water?). Well, you could assign a mass to phonons like you apply the so-called "effective mass" to electrons in a crystal but this would be rather abstract (inverse of the 2nd derivative of the dispersion relation or something like that) and would not resemble the mass as you probably know it. I have not seen this being done so far. My guess would be that this is not nessecary because unlike electrons the number of phonons is not conserved. When you heat up your crystal, for example, the number of phonons is increased and more of the higher energetic phonons states are occupied (or occupied by more phonons as phonons are bosonic particles so a state can be occupied by more than one phonon). That´s a difference to the free electrons of a crystal. When the crystal's temperature is increased the number of electrons of course stays the same. The electrons just move to higher energetic states. Both phonons and free electrons contribute to thermal energy. Excitation of the bound electrons is usually not considered. About phonons being carriers of sound: Yes. Take a look at the "dispersion relation" in the wikipedia page. Sound waves have a long wavelenght in comparison with typical interatomic distances. This translates to very small wave-vectors k. The velocity of a wave packet is equal to (or porportional to, doesn´t really matter) the derivative of energy omega. As you can see, the curve in the plot is nearly a line for small k. So the slope of the line is the speed of sound in the crystal.

- A phonon is a state of oscillation of the crystal´s atoms. - As these states have definite momentum and energy and are quantized, they are usually threatened as particles and called quasi-particles (I´d think that´s because their existence is bound to a medium while similar particles like photons can exist without a medium). - As phonons describe the relative motion of the atoms (where relative means that the crystal´s movement in space as a whole is not considered) they are also used for calculating thermal properties like the heat capacity (although the free electrons also have to be considered for metals). - Phonons are not photons and not related to electromagetism.