Jerry Schirmer

Austin, TX

Age: 34

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.

39m
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.
2h
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.
3h
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.
1d
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.
1d
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.
1d
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.
2d
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.
2d
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.
2d
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.
2d
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.
2d
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.
2d
answered Does or should the metric expansion of space imply a locally observable increase in kinetic energy?
2d
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
15
comment How many joules of energy are required to convert 1 liter of water into hydrogen and oxygen (theoretically)?
this belongs on <a href="chemistry.stackexchange.com">chemistry</a>;
Aug
15
comment How does "warp drive" not violate Special Relativity causality constraints?
It definitely breaks causality, provided you have more than one drive, or the ability for the thing to turn around.
Aug
15
comment A common definition of a scalar
And you can have exotic systems with negative pressures, certainly, like dark energy
Aug
14
comment Rectifying incomplete popular notions in cosmology
@Andrew: general relativity is, conceptually, quite distinct from QFT and QM. A lot of progress has been made to reformulate GR in a more field-theoretic way, and to geometrize QFT in terms of fibre bundles, but they are still very conceptually distinct things.
Aug
14
answered Tidal forces in free fall
Aug
14
comment An atomic bomb explodes inside of an "unbreakable" container which is on a scale. Does the "weight" of the container change?
Is the box also perfectly insulated? if yes, then the weight doesn't change. If it eventually leaks out all of the heat of the explosion, it will definitely have less mass after it cools down to room temperature.
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