2,321 reputation
1539
bio website none.seriously
location Switzerland
age 30
visits member for 3 years, 11 months
seen 5 hours ago

Physicist.


7h
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.
1d
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”?
Oct
15
answered Motion described by $m \frac{\mathrm{d}^2 x}{\mathrm{d}t^2}=-k\frac{\mathrm{d}^{\frac12 }x}{\mathrm{d}t^{\frac12}}$
Oct
15
comment Motion described by $m \frac{\mathrm{d}^2 x}{\mathrm{d}t^2}=-k\frac{\mathrm{d}^{\frac12 }x}{\mathrm{d}t^{\frac12}}$
It is good practice to state what you have done so far instead of letting answerers start from scratch. At its current state, your (interesting) question does unfortunately sound quite like a "hey, anyone please work this out for me"...
Oct
15
comment Which is more fundamental, Fields or Particles?
Great answer - though for completeness you could also mention the density matrix and the relationship to Statistical Physics / Thermodynamics. Then again, you might end up writing an entire book :D
Oct
15
comment Is a “third quantization” possible?
...Your point does however inspire me to claim that the "obvious" generalization to "third" quantization would be from "single particle in distinguishable states" via "indistinguishable particles in states" to "various particles in distinguishable particle-states", i.e. a quantization of the particle kind, which would basically be String Theory...
Oct
15
comment Is a “third quantization” possible?
@DanielSank Great answer you posted there :) However, in your explanation you're assuming sufficiently distinguishable (though not necessarily non-interacting) states the (fixed amount of) particles can take, states which strongly depend on the system considered. What I mean is the very general idea that just like $|1\rangle$ can be represented as $\int\,d^3x\,|x\rangle\underbrace{\langle x|1\rangle}_{=\psi_1(x)}$, the multi-particle state $|112\rangle$ is something like $\int\mathcal D\psi\,|\psi(x)\rangle\underbrace{\langle\psi(x)|112\rangle}_{=:\Phi(\psi)}$...
Oct
9
comment What challenges needed to be overcome to create (blue) LEDs?
That doesn't explain why the Nobel Prize wasn't awarded to the red-LED creators instead or as well
Oct
8
revised Is the Gibbs-Duhem equation always valid?
added 607 characters in body
Oct
8
comment Is the Gibbs-Duhem equation always valid?
(or, apparently equivalent, when is $G\neq\sum_i\mu_i N_i$?)
Oct
8
asked Is the Gibbs-Duhem equation always valid?