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I am a Ph.D. general relativist working as a software engineer. I like to still go and do physics as a hobby, and to keep up my skill and knowledge.


16h
comment Bekenstein bound for electron?
@StudentT: he's dropping it into a black hole's horizon. If you take general relativity to be true all the way down to an electron's scale, there is no horizon, so none of Bekenstein's equations make any sense, since they are all based on crossing the horizon.
23h
answered Bekenstein bound for electron?
1d
comment Time reversal in Maxwell's electromagnetism
You can also arrive at this conclusion about the magnetic field by realizing that $F = q v \times B$ has to be invariant, so the minus sign in v has to be cancelled by a minus sign in B.
2d
comment Does or should the metric expansion of space imply a locally observable increase in kinetic energy?
@benrg the more I think of this,the more I think this is actually an open research question,mostly because you're appealing to the fact that the cosmology isn't locally important.But at that point,you have to consider the local cosmological field to be some sort of mean field that will be dependent on the particular way that matter from local galaxies is lumpy and inhomogenous, which breaks the local spherical symmetry on which the shell theorem depends. You can either model the cosmology as smooth all the way down, or it becomes lumpy.In the latter case, you can't ignore the lumpiness.
2d
comment Does or should the metric expansion of space imply a locally observable increase in kinetic energy?
the space is maximally symmetric, and in that patch, has a manifest timelike killing vector. By the global symmetries, you can rotate that patch wherever you want it. There might be caustics, but every point has a finite neighborhood with a timelike killing vector.
2d
comment Does or should the metric expansion of space imply a locally observable increase in kinetic energy?
de Sitter can be written down with the line element $ds^{s} = -\left(1- \lambda r^{2}\right)dt^{2} + \frac{dr^{2}}{1 - \lambda r^{2}} + r^{2}d\Omega^{2}$, which has a manifest timelike killing vector.
Aug
19
comment Does or should the metric expansion of space imply a locally observable increase in kinetic energy?
The cosmological constant is not a gravitational force between two objects. In fact, pure de Sitter space has no expansion at all -- it has a global timelike killing vector.
Aug
19
comment Does or should the metric expansion of space imply a locally observable increase in kinetic energy?
@benrg: but that's not true, and can't be true. $\Lambda$ can easily be hidden inside of $a(t)$, after all. And there is a new force on orbits (actually all orbits are ultimately unstable) in any model where you have a nonzero cosmology. This is manifest if you write the robertson-walker metric in terms of physical coordinates, rather than comoving coordinates. It requires a net force to deviate a particle from constant comoving distance.
Aug
19
comment Does or should the metric expansion of space imply a locally observable increase in kinetic energy?
@benrg: No. You're wrong. That statement contradicts an analysis of orbits in asymptotically Robertson-walker spacetimes. Orbits ARE static (but definitely altered) in Kerr-de Sitter space. And if you force a particle to move on something other than a geodesic, you definitely can extract work. Requiring macroscopically seperated objects to keep a constant proper distance is definitely forcing them not to move on a geodesic.
Aug
18
comment Does or should the metric expansion of space imply a locally observable increase in kinetic energy?
@BenCrowell: well, it has a well-defined ground state. Technically, the fact that it is in a cosmological background introduces a perturbation Hamiltonian which would change the energy levels, etc, etc. The effect would be ludicrously small, of course.
Aug
18
comment Is there a difference in handwritten nabla $\vec{\nabla}$ with an overset arrow and typeset nabla $\nabla$?
Just to add to the complications here, in General Relativity, the convention is to write the red symbol you give as $\nabla_{a}$, using abstract index notation.
Aug
18
comment Does or should the metric expansion of space imply a locally observable increase in kinetic energy?
To the best of our measurments, you can completely ignore gravity when doing any atomic or particle physics.
Aug
18
comment Does or should the metric expansion of space imply a locally observable increase in kinetic energy?
@JoshuaHonig: that effect would be yet smaller. The force you get will be proportional to the distance between the objects -- you're basically doing work against the velocity given by Hubble's law, after all.
Aug
18
comment What makes General Relativity conformal variant?
@Amirpouyan: this is a classical field theory question. Any answer that involves particles is going to be wrong. General relativity is married the coordinate system in a way that Maxwell theory is not. In GR, a coordinate tranformation is a rescaling of $g_{ab}$, which takes the same role that $A_{a}$ does in Maxwell theory. That a change of scale makes $A_{a}$ transform differently than the way that $g_{ab}$ changes is really all there is to this.
Aug
18
answered Does or should the metric expansion of space imply a locally observable increase in kinetic energy?
Aug
18
comment How can I derive the Hamiltonian of simple harmonic oscillator from this Lagrangian?
RAther than say "I made it this far:", could you walk us through the steps you took to get to the formula below that line? I say this because it's definitely wrong -- just look at the units your Hamiltonian should have, and the units that your terms have.
Aug
18
comment Would a three wheeled vehicle be faster than a four wheeled vehicle of the same weight?
Also, you have the additional problem that static friction is actually what makes the car go. You WANT friction.
Aug
17
comment Definition of derivative operator on a manifold
Wald's notation is a bit goofy in the second clause. He means that $t(f)$ is the directional derivative of f along the direction of $t^{a}$.
Aug
17
revised What makes General Relativity conformal variant?
added 437 characters in body
Aug
17
answered What makes General Relativity conformal variant?