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| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 4 months |
| seen | May 19 at 21:24 | |
| stats | profile views | 89 |
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Mar 21 |
awarded | Enlightened |
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Mar 21 |
awarded | Nice Answer |
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Feb 6 |
accepted | Low-energy gluodynamics as a string |
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Jan 26 |
awarded | Yearling |
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Oct 5 |
revised |
Scale invariance symmetry as a simple argument in an electrostatics problem deleted 17 characters in body |
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Oct 5 |
answered | Scale invariance symmetry as a simple argument in an electrostatics problem |
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Aug 28 |
answered | What happens when you shake a can of soda? |
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Jul 1 |
awarded | Mortarboard |
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May 16 |
comment |
Massless charged particles To be precise, you need to turn your partial derivatives into covariant derivatives to minimally couple the scalar the field to the photon field: $\mathcal{L} = D_\mu\phi^\ast D^\mu\phi$ for $D_\mu = \partial_\mu + ie\hat{Q}A_\mu$. From here, note that the photon-loop diagram would give a mass renormalization. Unless there is some symmetry which protects/prevents the renormalization of the $\phi$ field's mass, there is no reason to assume that this bare Lagrangian should give physically massless particles! |
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Mar 23 |
answered | Is a charged particle at rest affected by magnetic field? |
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Mar 21 |
revised |
How exactly is the propagator a Green's function for the Schrodinger equation added 1 characters in body |
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Mar 21 |
answered | Why aren't the spin-3/2 fields in the (3/2,0)+(0,3/2) representation? |
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Mar 20 |
comment |
How exactly is the propagator a Green's function for the Schrodinger equation You won't solve it this way. You will get the residual term from integrating over the $\delta$ functions. To solve the Schroedinger equation, you would not put the time ordering condition. Then convoling the initial state $\psi(x^\prime,t)$ with $K$ gives the solution at later or earlier times. |
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Mar 20 |
answered | How exactly is the propagator a Green's function for the Schrodinger equation |
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Mar 20 |
comment |
Low-energy gluodynamics as a string Yes I am familiar with the excellent work of Luescher and Weisz. This still isn't quite what I have in mind, though. I'll clarify above. |
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Mar 20 |
answered | Gauge invariance and the form of the Rarita-Schwinger action |
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Mar 19 |
asked | Low-energy gluodynamics as a string |
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Mar 8 |
answered | Another question about Shankar's notation |
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Mar 5 |
revised |
Decomposition of SU(N) adjoint representation under SU(2) added 155 characters in body |
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Mar 5 |
comment |
Gravitation in a space that is topologically toroidal You could stop summing after a fixed distance away, e.g. Using the Euclidean metric is right. The formalism of potentials is the non-relativistic limit. You're summing over purely spatial distances in the denominator, as usual for potentials. |
