2,412 reputation
1839
bio website none.seriously
location Switzerland
age 30
visits member for 4 years, 1 month
seen yesterday

Physicist.


Nov
28
awarded  Notable Question
Nov
27
comment Why is quantum entanglement considered to be an active link between particles?
While I agree that Luboš's answer is superior to this one, I do wonder why it was downvoted - any improvement suggestions?
Nov
18
comment How does interpreting negative energy electrons as positrons solve the negative energy problem?
@Dave you can use classical electrodynamics for that, when you swap the sign of all time derivatives and of all charge terms in the Maxwell equations, they remain the same.
Nov
18
answered How does interpreting negative energy electrons as positrons solve the negative energy problem?
Nov
12
awarded  Notable Question
Nov
2
awarded  Yearling
Oct
26
comment Imaginary time is to inverse temperature what imaginary entropy is to …?
That's some very interesting reading you got there (I should have known to check John Baez on something like this...), though as you mention he relates $\beta$ with $-i/\hbar$ and not with $it$ as the Wick-Rotation does, or did I mix something up there? I guess I need to read that paper more thoroughly, haven't done any QFT for years :/
Oct
25
comment Are there models/simulations of antigravitational antimatter-galaxies?
I've given this some more thought, and I think one must also consider the repulsion between the pair particles into consideration. Though I'm not entirely certain how. A first guess would be that by forcing them to stay together (which you propose would have them levitate upwards) they might remain in place - actually I think for a consistent theory (which it would have to be, obviously), the potential gravitational energy of antiparticles would also be negative (similarly to the Dirac sea)
Oct
25
asked Imaginary time is to inverse temperature what imaginary entropy is to …?
Oct
20
comment How do tachyons violate causality?
Reminds me of this answer, where I found an interesting paper according to which Tachyons would have to be Spin-0 particles and yet Fermions.
Oct
18
comment While finding escape velocity, does human mass counts as the mass of Earth?
@FranciscoPresencia Indeed, but there is also the other direction, energy being radiated away from Earth. Either way, it is likely that Earth's total mass is not constant.
Oct
16
comment Is a “third quantization” possible?
@DanielSank Precisely - basically one interpretation could be 1st quantization: individual modes per particle, 2nd quantization: the "modes" are the various ensemble populations, 3rd quantization: modes = which particle fields are excitated in a field-field (or "superfield" mayhaps)
Oct
16
comment Motion described by $m \frac{\mathrm{d}^2 x}{\mathrm{d}t^2}=-k\frac{\mathrm{d}^{\frac12 }x}{\mathrm{d}t^{\frac12}}$
6 7
Oct
16
comment Motion described by $m \frac{\mathrm{d}^2 x}{\mathrm{d}t^2}=-k\frac{\mathrm{d}^{\frac12 }x}{\mathrm{d}t^{\frac12}}$
note to self: 1 2 3a 3d 3 4 5
Oct
16
revised Motion described by $m \frac{\mathrm{d}^2 x}{\mathrm{d}t^2}=-k\frac{\mathrm{d}^{\frac12 }x}{\mathrm{d}t^{\frac12}}$
added 487 characters in body
Oct
15
comment Motion described by $m \frac{\mathrm{d}^2 x}{\mathrm{d}t^2}=-k\frac{\mathrm{d}^{\frac12 }x}{\mathrm{d}t^{\frac12}}$
@user121330 aww, that's a shame :( guess that renders my answer useless, I'll edit or delete it tomorrow...
Oct
15
comment Motion described by $m \frac{\mathrm{d}^2 x}{\mathrm{d}t^2}=-k\frac{\mathrm{d}^{\frac12 }x}{\mathrm{d}t^{\frac12}}$
@user148432 You basically have to insert the general solution (with four integration constants) of the 4th order ODE into the original equation (thereby also taking the half-derivative) and see which of them fulfil that as well - just like you insert the potential solutions to something like $\sqrt{2x}+x=-1$ obtained from $(x+1)^2=2x$ into the former equation to see which ones are actually valid (and which ones aren't because you basically discarded a sign when squaring). I just am not sure what kind of "sign" you discard when half-deriving the equation
Oct
15
comment Motion described by $m \frac{\mathrm{d}^2 x}{\mathrm{d}t^2}=-k\frac{\mathrm{d}^{\frac12 }x}{\mathrm{d}t^{\frac12}}$
@EmilioPisanty I honestly don't have much experience with fractional derivatives, but my guess is that similarly to squaring an equation (or taking derivatives of normal ODEs) you get more solutions than are valid for the original equation, i.e. two initial conditions should suffice. But as you state, the non-locality may yield curious effects. However, JamalS' answer (and others) clearly suggest two initial conditions just like for a usual 2nd degree ODE are required.
Oct
15
comment Which is more fundamental, Fields or Particles?
@DanielSank Since OP hasn't been here in a while and only asked two questions (no accepted answer in either) I guess they either moved on or lost their cookie and didn't register, thus cannot accept an answer anymore...
Oct
15
comment Which is more fundamental, Fields or Particles?
Related: How are forces “mediated”?