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Nov
5
comment What is the significance of the branch cut in renormalization group logarithms?
Well, a good question. I would like to see some more complete answer as well. The branch cut from $\ln(E)$ surely means that singular physics we understand takes place at $E=0$ as well as $E=\infty$ and you're not allowed to circumvent these points. It's somewhat questionable whether all the functions with $\ln(E)$ have natural extrapolations to the complex plane. Also, note that a log may be written as an integral of $1/(E-E_0)$ so the log branch cut comes from a continuous union of many ordinary first-order poles. In $N=2$ SUSY QFTs (Seiberg-Witten etc.), there are many sharper facts on logs
Nov
5
answered Global symmetry in string theory
Nov
4
comment What is the minimum wavelength of electromagnetic radiation?
Well, @Ron, it's surely possible to distinguish them in principle, even though it may be hard. The simplest way is when you accelerate a whole laboratory to nearly the same velocity as the object so the relative momentum becomes manageable, and then you use e.g. the LHC calorimeters etc. to figure out whether it was a photon or something else.
Nov
4
answered Lorentz transformation in light cone coordinates in string theory
Nov
3
answered What does Gribov's last paper tell about coloured states?
Nov
3
comment Gravitational waves as dark energy?
However, if you gauge-fix the diffeomorphisms and/or otherwise define, locally and approximately, the stress-energy tensor of the gravitational waves, this stress-energy tensor will scale in the same wave as for electromagnetic waves.
Nov
3
comment Gravitational waves as dark energy?
Dear @mtrencseni, as WIMP says, your comment that the waves don't really get reflected is valid. A material that would be able to reflect them nearly perfectly cannot exist, in fact. Still, I need some well-defined action of them to calculate the pressure using naive mechanics. Of course, if you don't need such visual aid, you may just use the formulae for general relativity where the pressure is given by the stress-energy tensor. Of course, the (matter) stress-energy tensor for pure gravity waves is really zero and one deals with many issues about the "triviality" of energy in GR.
Nov
3
answered Why is there no double counting of $s$- and $t$-channels in string theory?
Nov
3
comment Gravitational waves as dark energy?
Alternatively, you may argue that in 4 dimensions, the stress-energy tensor of radiation has to be traceless because the classical theory describing the radiation has no dimensionful constants (conformal symmetry). That means that $p_{xx}=p_{yy}=p_{zz}$ by rotational symmetry and all of them have to be $\rho/3$ to get zero for $\rho-3 \times \rho/3$.
Nov
3
comment Gravitational waves as dark energy?
Hi! The same derivation holds for any particles or waves moving by the speed of light. Take a graviton of momentum $\vec p$ in a box $L^3$. It takes $L/v_x$ of time to go from the left boundary to the right one; in each collision, the momentum given to the walls is $2 p_x$. That's $2 p_x\cdot v_x/Lc$ of momentum per unit time. Sum over $x,y,z$ to get momentum per time $p\cdot v/Lc$. Divide by the area of the cube, $6L^2$, to get $pressure=Force/Area = p\cdot v/3L^3 = E/3L^3 = \rho/3$ for any particles/waves moving by the speed $c$.
Nov
3
revised Gravitational waves as dark energy?
added 56 characters in body
Nov
3
reviewed Approve building-physics tag wiki excerpt
Nov
3
comment How broad is the term “String Theory”?
Dear @Harry, any relativistic model with extended objects that has at most a finite number of adjustable parameters and avoids short-distance inconsistencies may be called "string theory" and there can't be any other theory matching these constraints besides those that are considered "string theory" by "string theorists". I assure you that if you find another "qualitatively new" solution that matches the simple criteria, and it will be a real working one, not just a crackpot "proposal of a concept", you will become very famous among HEP-TH physicists. String theory is the only game in town.
Nov
3
comment Why don't experimental physics groups have statisticians in it?
Exactly, @Anna.
Nov
3
comment Why don't experimental physics groups have statisticians in it?
+1, Ron... ;-) Well, because physicists and especially experimenters must simply know statistics (at least the relevant one) well, usually better than the people who are "just" statisticians.
Nov
3
comment Accelerometer Orientation Conversion
yes, it's just a rotation of a vector by 45 degrees. Note that $\cos 45$ degrees is equal to $1/\sqrt{2}$.
Nov
3
answered Gravitational waves as dark energy?
Nov
3
comment What is the minimum wavelength of electromagnetic radiation?
Dear @BCS, it is completely irrelevant how a photon was generated. Whatever the circumstances of the birth were, you may always boost the whole arrangement and get a photon of a shorter wavelength than before. There is no lower bound on the allowed wavelength. If you are uncertain what is allowed with "large energy densities", "short wavelengths" etc. and if it looks like a conflict of many infinities, try to find a reference frame in which the situation doesn't look too extreme. That's what relativity allows you to do. You will see that a single photon is never "less or more extreme".
Nov
3
comment What is the minimum wavelength of electromagnetic radiation?
Dear @Ron, I agree that the momentum conservation prohibits the emission of just one photon from a black hole; otherwise, nope. A photon is always distinguishable from a black hole. Its rest mass is zero; the rest mass of any black hole is always greater than the Planck mass (times a numerical constant of order one).
Nov
3
comment Does the rotation of the earth dramatically affect airplane flight time?
Dear @Kris, the centrifugal force won't be separately manifested because in all these contexts, only the total of the centrifugal plus the gravitational acceleration will be seen. The Coriolis force won't add energy extra work because it's perpendicular to the direction of the motion - much like in the case of the magnetic force acting on an electric charge. So both of these forces are really inconsequential for the timing in an airplane - they don't speed up or slow down the aircraft - and one could argue that none of them is related to the "big" effect proposed by the OP.