116,162 reputation
7180347
bio website motls.blogspot.com
location Czech Republic
age 41
visits member for 4 years, 7 months
seen 1 hour ago

Hi, I am a string theorist and a publicist.


Apr
19
comment fitting free QFTs into the Haag-Kastler algebraic formulation
Arnold, the exception you may have been thinking about is Coleman's proof that there are no Goldstone bosons in 2D, projecteuclid.org/…
Apr
19
comment fitting free QFTs into the Haag-Kastler algebraic formulation
Dear Urs, I agree that 2D CFT has been more successful as an example of the axiomatic treatment that may be reconciled with rigorous mathematicians. And it's ironic from a sociological viewpoint. But I disagree with your implicit suggestion which is written at least in between the lines that the AQFT school of thought itself represented a significant part of the 2D insights. See e.g. BPZ (4000 citations) www33.atwiki.jp/_pub/sakurazemi/reference/BPZ1984.pdf which is a paper that matters here. No AQFT references...
Apr
19
comment Reference paper to support information — energy relation ($kT \ln2 \rm\frac{J}{bit}$)
Dear Pygmalion, I changed one $E$ to $W$. Steffen, information surely "is not" energy. They're different quantities. But much like pretty much any pair of quantities in physics, they may have various relationships and there may exist important true sentences in which both of them appear. At a fixed temperature $T$, as you said, 1 bit allows to perform $kT\ln 2$ of work. But that doesn't mean that energy and information is the same thing, e.g. because $T$ isn't a universal constant. It's the temperature, another quantity. In a similar way, we don't say that voltage and current is the same thing
Apr
19
revised Reference paper to support information — energy relation ($kT \ln2 \rm\frac{J}{bit}$)
edited body
Apr
19
answered Reference paper to support information — energy relation ($kT \ln2 \rm\frac{J}{bit}$)
Apr
18
comment Commutating Annihilators with a beamsplitter
Hi, just use $[XY,Z]=XYZ-ZXY = XYZ-XZY+XZY-ZXY = X[Y,Z]+[X,Z]Y$ and the basic commutators $[a,a^\dagger]=1$ and similarly for $b$ while other commutators vanish. You will see that from the right hand side, only one term survives and it gives you what you need.
Apr
18
comment Why is quantum entanglement considered to be an active link between particles?
Yes, it's one of those hundreds of wrong popular books written by people who don't really understand quantum mechanics whom I was referring to.
Apr
18
answered Higgs mass and the hierarchy problem
Apr
18
comment How does the energy of a sound wave decrease with the distance
You meant square, not square root, right? Imagine the solution is something like $dE/dV=C \exp(-r/r_0)/r^2$.
Apr
18
comment Spin matrices in Dirac equation
Dear @lurscher, the spin-one equation is Maxwell's equation. The work needed to get from one to another is zero. Well, at most, you may change a basis or something like that. John: electron has $j=1/2$ because the eigenvalue of $J_z=J_{12}=\gamma_1\gamma_2/2$ acting on the electron state is $1/2$ times the original state. Photon has $J=1$ because when you act with $J_{12}$ on the photon state moving in the z-direction, you get $1$ times the original state. One-half is something else than one. For example, if people only eat 1/2 of a bread, 1 bread may feed 2 of them, not one.
Apr
17
comment Why are there Gravitons among the modes of oscillation in String Theory?
Before you get proper answers, you may read motls.blogspot.com/2007/05/…
Apr
17
answered Spin matrices in Dirac equation
Apr
16
comment Is it circular reasoning to derive Newton's laws from action minimization?
Dear @Nesp, a circle as a topological concept means that one gets from point A to B along one curve and then from B to A along another curve. For example, if you tried to say that some equation A is inevitable because of B and B is inevitable because of A, it would be circular. But here - and in most of science - there is nothing circular here. A implies B and/or B implies A but one still needs the agreement with observations to show that either A or B is true. On the other hand, such an agreement of theory with facts is a near-proof of both A and B, too.
Apr
14
comment Information conservation during quantum measurement
I will endorse genneth and post my answer as a comment only. ... Quantum measurements are fundamentally indeterministic and it's simply a law of physics that you get a random string when you do that. Liouville's theorem only applies to models of classical statistical physics, see en.wikipedia.org/wiki/Liouville%27s_theorem_(Hamiltonian) ... Our world isn't classical so it doesn't have to obey Liouville's theorem. The quantum extension only says that the evolution is unitary i.e. the total probability is 1 for any initial state but not that randomness is prohibited.
Apr
14
comment Walter Lewin Lecture 16 - Ball bouncing on wall?
Let me add a comment showing that the statements are not independent. It's useful to write the kinetic energy in terms of the momentum $P$, namely $E=(mv)^2/2m = p^2/2m$. Now, $p$ of the wall and the object are the same, up to the sign, due to the conservation of the momentum i.e. action and reaction. So you see that $p^2/2m$ is a decreasing function of $m$ for a fixed $p$, so the heavier object you have in the collision, the smaller kinetic energy it will carry.
Apr
14
comment Can relativistic kinetic energy be derived from Newtonian kinetic energy?
Thanks for the agreement, Ron. I was trying to fill in the gaps so that it would make more sense and resemble what Physiks lover wants - but it's so redundant. The resulting derivation is just a differential version of normal derivations; the beef has to be relativistic and it may be repulsive for the OP, anyway.
Apr
13
comment Autocorrelation Functions <---> Pair Correlation Functions
Vijay, typo: yes, thanks.
Apr
13
comment Autocorrelation Functions <---> Pair Correlation Functions
If your pairs are those in the "radial correlation function", there won't be any relationship, either. The autocorrelation is about the delaying in time while the radial correlation function is about shifting the objects in the radial direction. Those are different variables in which objects may be moved so the correlation functions won't be related, either.
Apr
13
answered Can relativistic kinetic energy be derived from Newtonian kinetic energy?
Apr
13
comment Calculating semi-major axis of binary stars from velocity, position and mass
I find this question very confusing, too. Are you calculating a 2-body problem or an N-body problem for a higher value of N? If it is the latter, the trajectories won't be ellipses, so it makes no sense to talk about the semi-major axes too accurately because these parameters only make sense for ellipses e.g. for a 2-body problem. In the 2-body case, the Wikipedia article clearly answers all your questions.