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.

9h
comment Temperature of a neutron star
@rob: it's hard math. Here's a paper (that I haven't read) where they try to: arxiv.org/abs/1211.2352
11h
comment Temperature of a neutron star
For a neutron star, I wouldn't be surprised if a significant portion of the entropy was accounted for by the spin alignment of the neutrons and the magnetic field, rather than mechanical kinetic energy.
17h
comment My book uses helium weight in reaction formula instead of weight alpha particle?
@AdamRedwine: the binding energy is of O(13.6 eV), while the rest mass of an electron is .511 MeV/c^{2}. So, the mass of the electrons swamps the binding energy of the electrons.
17h
comment Can an underdensity in space act as a negative matter density powering a warp drive?
@LuboŇ°Motl: he's trying to say that you get a "negative energy density" simulated by having positive matter + cosmological constant in region A, and then a void of just cosmological constant in region B, which you could then call the "negative relative energy region". This doesn't violate the energy conditions, though, which you need for these warp drive solutions, as you know.
19h
comment Can an underdensity in space act as a negative matter density powering a warp drive?
@Void: yes, but the fact that you need to violate a GR energy condition somewhere to "count" as negative energy density remains. These conditions are locally defined, so there is no cheating by creating "holes" in energy density. You actually need matter with negative mass.
1d
answered Explanation for $M{\ddot{r}}=-\nabla \phi$
1d
comment Does spin-0 or spin-2 describe massive or massless particles?
YOu can have both massive and massless particles at both spins. There are beleived to be stability problems with massive spin-2 theories, though.
1d
comment Can the magnetic fields of EM radiation be harnessed or measured?
Note that the magnetic field is generally only negligible because most ordinary objects travel with speeds $\ll c$. For objects moving relativistically, then since the magnetic force $\propto vB$, and since $E/B =c$ tells us that the magnitude of the electric field is greater than that of the magnetic field by a factor of $c$, we see that the forces are then roughly equal.
1d
comment F-mu-nu notation
@SmikGames: the latter one would be a field tensor that takes values over a Yang-Mills group
1d
comment Why did the Matrix simulate 1999 instead of a pre-computer year?
And also left the whole premise for the movie themodynamically ridiculous.
1d
comment The Alcubierre drive and closed timelike curves
Alcubierre's original published metric is free of CTC. You can get superluminal global travel without CTC, so long as there's no turnaround -- just think of the stars beyond the cosmological horizon. Their proper distance with respect to us is greater than the speed of light, but it just doesn't matter, because there's no way to send a signal back and forth, or for them to "turn around".
2d
answered The Alcubierre drive and closed timelike curves
2d
comment Why there does not exist any Gravitational Magnetic Field?
@Dvij: you can't predict normal magnetism using classical mechanics only. To see that you need magnetism if you have an electrical force, you have to make an appeal to Lorentz invariance.
2d
comment Charge Distribution in Reissner-Nordström Black Holes
Same thing. The Kerr solution is an electrovac solution (and reissner-nordstrom is a special case of the Kerr solution)
2d
answered Charge Distribution in Reissner-Nordström Black Holes
2d
answered The commutator of Killing vectors
2d
comment In a Big Crunch, would there be more mass than at the Big Bang?
One quibble: some non-asymptotically flat spacetimes do have definiable local energies over things like the cosmological horizon--you just need a null or timelike 3-surface with a null or timelike killing vector, and you can define the energy contained in that surface. This isn't the case for cosmological horizons, though.
Jul
28
comment How can space be euclidean when light bends?
Space can be made aribitrarily close to Minkowski(spatial part Euclidean) by choosing a small enough four-cube of spacetime. So, sufficiently small volumes are, in a sense, exactly Euclidean.
Jul
28
comment Maximum curvature in a black hole
Could you expand this question a little bit? It reads to me like "If a is true, then is it possible that a is valid?"
Jul
28
comment On the coordinate independence of general relativity
In particular, there are several solutions to Einstein's equation in vacuum that are NOT just Minkowski space, because they contain gravitational radiation, for example (other, more exotic contents are possible)
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