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| bio | website | motls.blogspot.com |
|---|---|---|
| location | Czech Republic | |
| age | 39 | |
| visits | member for | 2 years, 4 months |
| seen | 9 hours ago | |
| stats | profile views | 24,036 |
Hi, I am a string theorist and a publicist.
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May 14 |
comment |
Why doesn't one-photon-irreducible function have any pole at $q^2=0$? Hi, try to study section 1.4 here, mathematik.hu-berlin.de/~maphy/SkriptII11.pdf , Cutkosky rules - or elsewhere. It's a much stronger result than what you want. If there were a $1/q^2$ singularity, there would also be the imaginary part behaving as a delta-function, and those can be fully calculated via the rules. Those rules give you all the singularities and 1PI diagrams need to be cut by cutting several propagators at once and their singularity structure therefore depends on more momenta, as exactly dictated by the rules. |
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May 14 |
answered | Why doesn't one-photon-irreducible function have any pole at $q^2=0$? |
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May 13 |
comment |
Mass spectrum of Type I string theory 1) Yes, the gauginos and gauge bosons, respectively. 2) Orientifold projection, as described in any textbook. 3) It's just a notation, convention. 4) The representations are clear from the indices in the notation and described in any textbook. 5) Open strings aren't considered projections of closed strings here; they're an independent sector. Projected closed strings are unorientable closed string. 6) A Hagedorn tower described in any textbook. |
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May 13 |
answered | The status of $SU(3)_C$ symmetry in the Standard Model |
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May 12 |
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Some Dirac notation explanations Why don't you try to understand the notation correctly rather than trying to write down as many incorrect or marginally incorrect "identities" as you can? Also, I think it is not just notation you don't understand. The confusion in the question indicates that you don't understand the beef, the linear algebra, otherwise you wouldn't formulate the question in this way. Do you understand matrix multiplication? Ket vectors are just column thin matrices, bra vectors are the snigle rows - the hermitian conjugate to kets - and the operators are large matrices. What is so hard about it? |
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May 12 |
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Some Dirac notation explanations Notation is supposed to be useful and follow some restricted standards. Your "identities" apparently deliberately violate what a normal physicist using the Dirac notation would write down. In the 1st, it may be OK if the "dot" is a symbol for the inner product that assumes that one of the factors is hermitian conjugated. Physicists surely don't use such a notation. The same is true for 2nd. Physicists would normally write the 2nd without the dagger. the 3rd is nonsenical in any sense, operators such as $x$ shouldn't act on the "bracket side" of a vector. |
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May 11 |
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Supersymmetry and non-compact $R$-symmetry group? Dear @jancore, I meant that internal noncompact symmetries that produce states transforming as a linear finite-dimensional representation of the symmetry are unacceptable (for the simple "sign of thenorm" reason I fully described). The internal symmetry you mention (spacetime Lorentz group) doesn't lead to any linear representation. Similarly, SUGRA theories have noncompact groups like $E_{7(7)}$ but they're realized nonlinearly. |
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May 9 |
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Is conservation of statistics logically independent of spin? I think my answer speaks for itself and it's clear what the answers to your repeated questions are. It may be better to avoid repetitiveness; and to avoid writing things that you preemptively indicate to be irritated by. Otherwise canonical commutation relations always hold in any theory that may be derived by a quantization from its classical limit. Saying anything else means not to obey any rules or standards. Also, there can't be spin-1/2 tachyons because no Dirac equation can lead to $m^2\lt 0$. And yes, almost everything people said about tachyons in QFT in the 1960s was misguided. |
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May 9 |
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Is there any proof that the speed of gravity is limited? Dear Tomáši, it's not quite true that it takes an infinite time for the black hole to form. See e.g. motls.blogspot.com/2008/11/… for a related discussion... |
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May 9 |
answered | Is conservation of statistics logically independent of spin? |
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May 9 |
comment |
When a variation of a tensor is not a tensor? Exactly, @firtree ... The only difference if you consider the upper-index $\delta g$ relatively to the case of $\delta T$ is that you want to compute $\delta g^{ab}$ and the same object appears on the right hand side from the raising and Leibniz rule - so you get a sort of a self-referring equation (what you want to calculate is expressed in terms of the same thing). But this self-referring equation may still be solved - in my answer, it's solved in an easier way using the Kronecker delta. Such subtleties don't arise for $\delta T^{ab}$. |
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May 9 |
awarded | Guru |
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May 9 |
comment |
When a variation of a tensor is not a tensor? The convention to define the values of tensors with raised or lowered indices is a usual convention we apply to various tensors - we can't apply it to variations - but it has nothing to do with the question whether an object is a tensor! I wrote what the actual condition is above. Could you please reread my answer before you ask the same question again? |
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May 9 |
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When a variation of a tensor is not a tensor? No. Your naive – wrong – prescription for the variation of any tensor (even non-metric tensor) is always wrong as long as $\delta g_{\mu\nu}$ is nonzero. You must use the Leibniz rule $\delta(AB)=\delta A\cdot B+A\cdot\delta B$ to compute the variation. But the variation of a tensor, e.g. $\delta T^{\mu\nu}$, is always a tensor as well. There will be several terms on the right hand side and each of them is actually a tensor by itself! Again, you seem to misunderstand what the word "tensor" means. It doesn't mean that it comes with the naive rules to lower or raise indices. |
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May 9 |
revised |
When a variation of a tensor is not a tensor? added 989 characters in body |
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May 9 |
answered | When a variation of a tensor is not a tensor? |
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May 8 |
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If particles can find themselves spontaneously arranged, isn't entropy actually decreasing? Both major parts of the quoted claim above are completely wrong. There is no sharp open question about the identification of the statistical and thermodynamic entropy in the thermodynamic limit; and it is not true that entropy is only measurable for reversible processes - almost no processes in the real world are reversible so this would mean that entropy is almost never measurable which is just false. At any rate, if you followed your philosophy about the restriction, you should have honestly answered "my undestanding of the entropy doesn't allow me to discuss these matters", not what you did |
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May 8 |
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If particles can find themselves spontaneously arranged, isn't entropy actually decreasing? I am saying that this restriction is in no way necessary for anything and it conflicts with the discussion about what the entropy is doing in between which is a totally legitimate discussion. You may prevent yourself from talking about these matters by unjustified extra restrictions but you shouldn't and even if you do, it should prevent you from trying to answer similar questions as well. |
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May 8 |
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If particles can find themselves spontaneously arranged, isn't entropy actually decreasing? There are uncertainties in the definition of the entropy, in practice $\pm C\times k$ where $C$ is of order one and $k$ is the Boltzmann constant is the minimum error that is introduced by making conventions about ensembles etc. And in statistical physics, the entropy decreases, usually by a tiny amount, in a small percentage of the times. The larger decreases we consider, the less likely they become. But if we only look at long enough changes $\Delta t$ when the expected entropy change is macroscopic, the percentage of the "large steps" where entropy went down becomes zero. |
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May 8 |
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If particles can find themselves spontaneously arranged, isn't entropy actually decreasing? I don't think it's right, Christoph. What is true is that one needs some at least local equilibrium to define the temperature - because the temperature labels rather well-defined mixed states e.g. $\exp(-H/kT)$ in QM - but not the entropy. Entropy is well-defined for time-dependent processes. Indeed, it has to be well-defined because the second law of thermodynamics says how it changes during such processes. If we could only define the entropy at equilibrium, the second law about the "strict increase" would never hold because the increase would be incompatible with the equilibrium. |
