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| bio | website | motls.blogspot.com |
|---|---|---|
| location | Czech Republic | |
| age | 39 | |
| visits | member for | 2 years, 4 months |
| seen | 2 hours ago | |
| stats | profile views | 24,038 |
Hi, I am a string theorist and a publicist.
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Apr 22 |
comment |
Energy stored in fields Energy is a quantity; it is a number together with a unit. To talk about "quantities" is to talk "quantitatively" which is pretty much the same thing as "referring to mathematics". It's really the point of energy that it isn't a potato or bumblebee or anything with a specific shape; it "only" has the number which is what makes it important. In the international units, energy is counted in joules. You may imagine a joule is a bumblebee except that it looks like nothing of the sort. I wonder: what leads you to insist that you want to avoid mathematics? That's not a way to understand Nature. |
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Apr 18 |
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A four-dimensional integral in Peskin & Schroeder Dear Soliton, the $\exp(ik\cdot \epsilon)$ factor is the phase that is a part of the definition of the Fourier transform! It's nothing we have added to the function we're Fourier-transforming. |
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Apr 18 |
answered | A four-dimensional integral in Peskin & Schroeder |
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Apr 18 |
answered | How do I obtain a sense of scale when thinking about webers? |
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Apr 18 |
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When can a global symmetry be gauged? Third, you may always add $J\cdot A$ fields to an action so that you obtain a generating functional for correlation functions of $J$ in the original theory: the theory with the extra term doesn't have to have a gauge symmetry. The field $A$ is auxiliary. Fourth, treating it as auxiliary is less constraning because if it is dynamical, you have to impose the eqn of motion from varying $A$. Fifth, the equation of motion is $J=0$ unless you manage to write new "kinetic" terms for $A$ as well which is what makes the gauge symmetry physical and interesting but it's not guaranteed to exist. |
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Apr 18 |
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When can a global symmetry be gauged? Dear Tomáši, first, it's easy to add fields so that the new theory will have a gauge symmetry but in a typical case, one doesn't get an interesting theory because the new gauge symmetry just removes some degrees of freedom and it's more useful to erase them immediately with the symmetry, anyway. Second, I don't understand why you would consider extra fields to find out whether the original theory has conserved currents. If $S$ admits conserved currents, it does, otherwise it doesn't. |
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Apr 18 |
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Is it reasonable to interpret the Lamb shift as vacuum induced Stark shifts? An interesting interpretation! Do you know how to convert it to a calculation, at least an approximate one? |
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Apr 17 |
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What would happen to the Moon if Earth is turned into a black hole? The extremal Kerr $J=GM^2/c\sim R_{bh} Mc$. Now, the Earth-mass black hole has radius 9 mm or so so we get about $10^{30}$ Js. The Earth's spin, actual angular momentum now, is indeed over $10^{40}$, ten orders of magnitude too much. This discrepancy is of course linked to Earth's too low density that makes the collapse de facto impossible. |
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Apr 16 |
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When can a global symmetry be gauged? Thanks, Tomáš! Unfortunately, I can't help you with this because I don't understand your theory that is both gauge-symmetric as well as physically globally symmetric only. It's like the Princess Koloběžka (Scooter?) the First, right? ;-) youtube.com/watch?v=mBC9vr3nuiI What does it mean for a field to be called a "gauge field" if the normally associated with it gauge symmetry doesn't exist at all? |
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Apr 16 |
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How to put spin-1/2 Fermi sea into real space representation? The momentum space is much more appropriate to describe similar maths - the momentum space is a "part of the solution". It's important for the FS state that some states, those above kf, remain unoccupied. The "states above kf" is easy to be written in the energy eigenstate but they're complicated superpositions of position eigenstates. So you're effectively asking someone to write the solution without the most obvious step towards the solution. The result is guaranteed to look messy but why don't you just substitute the second formula to the first? It won't simplify. |
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Apr 16 |
answered | When can a global symmetry be gauged? |
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Apr 16 |
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What is kappa symmetry? Be warned, it's a technically very complex thing with limited physical implications. See e.g. intro to arxiv.org/abs/hep-th/9908045 for some background. Surprising that David McMahon chose this topic/formalism in a "demystified book". The kappa-symmetry is a local fermionic symmetry on the world sheet whose task is to remove the excessive number of spinor components of the Green-Schwarz "covariant" string down to 8 physical transverse fermions (8+8 on left/right). It may be done in some backgrounds - in others, the right known constructions don't start with a manifestly covariant start. |
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Apr 15 |
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What does it mean to integrate out fields from a theory? Hi @user15766, yes, at least when your rule (and the word "solved") is correctly extrapolated to higher orders. It is indeed a sort of a Born-Oppenheimer approximation in which the light (=slow) degrees of freedom are fixed and the heavy part of the theory is "solved". In practice, the calculation of the effective action leads to the evaluation of all the diagrams with at least one "heavy" propagator... In the context of the renormalization group, we also want to "integrate out" only a part of the field modes, those with energies between $E$ and $E+dE$. |
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Apr 15 |
revised |
What does it mean to integrate out fields from a theory? added 206 characters in body |
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Apr 15 |
answered | What does it mean to integrate out fields from a theory? |
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Apr 15 |
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Workdone in an equipotential surface is zero? Right, Richard. Moreover, it's also true that the work you have to do in this way may be made arbitrarily small by moving the body at a lower maximum speed. The maximum energy you borrow - and then repay - is given by the maximum kinetic energy the particle reaches during the transfer, $mv_{max}^2/2$, and $v_{max}$ may be chosen arbitrarily low. |
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Apr 15 |
answered | Physical Interpretation of the Integrand of the Feynman Path Integral |
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Apr 14 |
answered | Vector and Spinor Representation in Ramond-Neveu-Schwarz Superstring Theory |
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Apr 14 |
awarded | Enlightened |
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Apr 14 |
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If photons can be absorbed by electrons, wouldn't that mean light has a charge? It means exactly the opposite. It means that electron has a charge - the photon interacts with those particles that have a charge because the photon is the messenger of the electromagnetic force that is excited by this charge. But because the "gauge group" behind electromagnetism is Abelian, the messenger - photon - is neutral itself. |
