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| bio | website | vladimirkalitvianski.wordpres… |
|---|---|---|
| location | Grenoble, France | |
| age | 54 | |
| visits | member for | 2 years, 4 months |
| seen | 4 hours ago | |
| stats | profile views | 3,669 |
Theoretical Physicist,
Research group: http://groups.google.com/group/qed-reformulation
Blog: http://vladimirkalitvianski.wordpress.com
Blog in Russian: http://fishers-in-the-snow.blogspot.com
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May 15 |
comment |
Many photons, one quantum field? @user15766: Yes, it is still possible to deal with one field, but very complicated one due to too many sources/absorbers. The boundary conditions I mentioned are approximate (simplified) solutions to the total field-matter equations. So we replace one total field with many similar ones acting each in different places of Universe (i.e., without interference). |
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May 15 |
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Many photons, one quantum field? @levitopher: You implicitly assume superpositions of photon states. How about mixtures which are not superpositions? |
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May 15 |
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Many photons, one quantum field? @levitopher: How about loss of interference due to retardation of photons? Loss of coherence must be described somehow. Randomness of wave phases is a mean to "label different fields". |
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May 14 |
answered | Many photons, one quantum field? |
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May 9 |
comment |
Julian Schwinger videos, is there any? @MarkWayne: Thanks a lot!!! |
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Apr 20 |
comment |
Bremsstrahlung: why is electron slowed/stopped by the positive nucleus? If two particles are point-like, they never collide, but pass by (miss each other due to zero sizes). The interaction force is a long-range one ($F\propto 1/r^2$), so for their deflections it is not necessary to collide face-to-face. |
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Apr 20 |
comment |
Bremsstrahlung: why is electron slowed/stopped by the positive nucleus? If the electron is deflected by a nucleus, the nucleus is also affected: it obtains kinetic energy from the electron. So the kinetic energy of the electron decreases in each collision. Bremsstrahlung is also a loss of energy due to creating photons. |
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Apr 20 |
comment |
Bremsstrahlung: why is electron slowed/stopped by the positive nucleus? When the electron is approaching, it is accelerated with attraction. When the electron goes away it is still attracted, but in the opposite direction, so it is decelerated. Half path it is accelerated, half path it is decelerated. The global potential difference is zero, so there is no net gain/loss (if the nucleus is not moving, a heavy one). |
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Apr 20 |
answered | Bremsstrahlung: why is electron slowed/stopped by the positive nucleus? |
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Apr 14 |
revised |
Does spatial coupling prohibit resonances due to an external source field? added 212 characters in body |
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Apr 13 |
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Does spatial coupling prohibit resonances due to an external source field? Introduce a Fourier transform for $\varphi(x,t)$ like $\int dk e^{ikx}\cdot\chi(k,t)$. Each k-th harmonics $\chi(k,t)$ will obey an ordinary equation like your driven oscillator, I guess. One of them will be resonant. |
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Apr 13 |
comment |
Does spatial coupling prohibit resonances due to an external source field? So make the integration with it and you will see the solution different from yours. If initially there is no free oscillations ($x_0(t)=0$), then the external force will give and take away energy to the oscillator. In a resonance case the energy (amplitude) will be growing. |
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Apr 13 |
revised |
Does spatial coupling prohibit resonances due to an external source field? added 97 characters in body |
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Apr 13 |
comment |
Does spatial coupling prohibit resonances due to an external source field? The solution in your post is not as general as one can imagine. The solution I speak about is $x(t) = x_0(t)+\int_0^tdt'F(t')\sin [(t-t')\omega_0]/\omega_0$. |
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Apr 13 |
revised |
Does spatial coupling prohibit resonances due to an external source field? added 94 characters in body |
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Apr 13 |
answered | Does spatial coupling prohibit resonances due to an external source field? |
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Apr 13 |
comment |
Does spatial coupling prohibit resonances due to an external source field? Your first SHO equation should have a full time derivative $d^2/dt^2$, not a partial one. Its solution, if the force starts acting at t = 0, is different - the amplitude grows with time. The Klein-Gordon equation should have either $\partial^2\varphi(x,t)/\partial t^2$ or $\omega_0^2 \varphi (x)$, not both of them. |
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Apr 13 |
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Where are exactly the charges of charged capacitor? @AlfredCentauri: You are, of course, right. |
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Apr 12 |
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Where are exactly the charges of charged capacitor? @AlfredCentauri: We can always charge one plate of capacitor and leave the other neutral or charge it differently ;-) (not with a battery, of course). |
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Apr 12 |
comment |
Why does Quantum Electrodynamics Allow a Photon to Exist Temporarily as a Positron and an Electron? @Lagerbaer: Yes, a collision may have many different channels (issues) of occurring. |
