26,224 reputation
162135
bio website lightandmatter.com
location Fullerton, California
age
visits member for 3 years, 5 months
seen Dec 7 at 20:09

I teach physics at Fullerton College, a community college in Southern California. I have an undergrad degree in math and physics from Berkeley and a PhD in physics from Yale. Back when I was doing research, my field was experimental low-energy nuclear physics.


Nov
18
revised No-hair theorems for naked singularities?
added 92 characters in body
Nov
18
asked No-hair theorems for naked singularities?
Nov
18
revised Do all the spacelike curve terminate at the spatial infinity $i_0$ in the Penrose Diagram of a Schwarzchild black hole?
edited tags
Nov
18
comment Is chain reaction possible in stable isotopes?
U235 has a half-life of almost a billion years, so although it's not stable in the absolute sense, it's clearly right on the border, and in the practical sense it's "stable" enough to exist in the earth's crust. It's conceivable, although probably not likely, that there are nuclei in the transuranic island of stability that are stable. If so, then I'm sure they're fissile enough to undergo chain reactions.
Nov
18
comment Do all the spacelike curve terminate at the spatial infinity $i_0$ in the Penrose Diagram of a Schwarzchild black hole?
I would guess that the correct claim is that all spacelike geodesics terminate at $i^0$, in which case your counterexample might fail because it probably isn't a geodesic.
Nov
18
comment Do all the spacelike curve terminate at the spatial infinity $i_0$ in the Penrose Diagram of a Schwarzchild black hole?
Do you really mean all spacelike curves, not all spacelike geodesics, or all spacelike inextensible curves, or all complete spacelike geodesics? If the question is really about all spacelike curves, then there are easy counterexamples, including closed spacelike curves, the spacelike curve from Chicago to LA, and spacelike curves that terminate on the singularity. But all the books state the opposite. What books do you have in mind? Please quote exactly what they claim is true.
Nov
17
comment Does rotation increase mass?
related: physics.stackexchange.com/questions/94921/…
Nov
17
comment Does rotation increase mass?
The energy of a system in its own center of mass is exactly what the mass of a system is. If you don't believe that $\sqrt{E^2-p^2}$ is the correct relativistic definition of mass, then you should probably stop and consider what you think is the correct definition for a system of particles. You're also misunderstanding what invariance of mass means. Invariance of mass means that if a system has a certain mass in one frame, it has the same mass in other frames as well. It doesn't mean that the mass of a system of particles is independent of the motion of its constituents, which is not true.
Nov
17
comment $\beta^+$ decay question
[...] out of a Slater determinant of single-particle wavefunctions. It doesn't mean that "there aren't any individual particles." In the final paragraph, the discussion of pp fusion doesn't address the question, which is about beta decay, not fusion reactions that involve a weak-interaction process.
Nov
17
comment $\beta^+$ decay question
The first paragraph is pretty garbled, and it's not clear to me what you have in mind there. It seems as though you think that mass-energy equivalence, along with the fact that nuclear masses aren't equal to $Zm_p+Nm_n$, implies that neutron number and proton number aren't well defined. That's not true. If you have in mind something about the fact that protons and neutrons are composites of quarks, then that isn't coming through clearly here, and in any case isn't relevant. In the second paragraph, I think you're confused about what it means to put a many-body wavefunction together [...]
Nov
17
comment Is the charge distribution for an electric field unique?
Given the field in a region R, are you asking whether (1) the charge density is uniquely determined in R, or (2) whether it's uniquely determined everywhere? If 1, then Valter Moretti's comment answers your question. If 2, then TZDZ's answer.
Nov
17
comment $\beta^+$ decay question
In a nucleus (not just a bare proton), beta-decay of a proton can be allowed by conservation of mass-energy.
Nov
17
comment What's the escape velocity of Naked Singularities?
...probably can't form a naked singularity without exotic matter, so a negative mass wouldn't necessarily be surprising.)
Nov
17
comment What's the escape velocity of Naked Singularities?
And still two more issues. (1) Since you don't have a metric at the singularity, you can't define the velocity of a particle when it's at the singularity. By definition, the singularity is a termination point for geodesics that is reached over a finite affine parameter $\lambda$, so you could ask for $\lim_{\lambda\rightarrow0}v$. But this limit may not exist. For example, I find it plausible to imagine that $v$ could oscillate wildly in this limit and fail to approach any limiting value. (2) The ADM mass might be negative, in which case we might expect an escape velocity of zero. (You ...
Nov
17
comment What's the escape velocity of Naked Singularities?
Yet another possible definitional issue is that when you have a naked singularity, causality breaks down in a spectacular way: we no longer have existence and uniqueness for solutions of Cauchy problems, meaning that the laws of physics can't predict basic things like the motion of particles. If we can't necessarily predict the motion of particles, then it's not obvious to me that we can define whether a test particle is going to escape or not.
Nov
17
comment What's the escape velocity of Naked Singularities?
[...] unitless property. If we assume the spacetime is asymptotically flat (so that the notion of escape velocity makes sense), then the singularity should have some well-defined ADM or Bondi mass $m$. E.g., if you think you can make a naked singularity by overspinning or overcharging a black hole, then you might be able to form certain unitless ratios with $m$, such as $J/m^2$ for an overspun black hole. But there is no reason I can see to expect such a unitless ratio to be defined generically for all naked singularities.
Nov
17
comment What's the escape velocity of Naked Singularities?
I think this is potentially a very interesting question, but only if some of the issues of definitions and assumptions can be clarified. The reason that many physical objects have well defined escape velocities can be described in dimensional terms. In units where $c=1$ and $G=1$, velocity is unitless, so to have an escape velocity be a characteristic of the object, you need the object to have some intrinsic property that is unitless. In a case like the earth, the unitless ratio you can form is $m/r$. In the case of a naked singularity, it's not clear that there is any such [...]
Nov
17
comment How, in practice, could instantaneous signalling violate causality?
The philosophical issues are interesting, but not necessary in order to consider this problem. The relevant question is whether and how instantaneous signalling could be exploited in order to send a signal to the transmitter's own past world-line.
Nov
17
comment What's the escape velocity of Naked Singularities?
Also, There's no such thing as escape velocity for a black hole since the definition of a black hole singularity is that any line going to the singularity ends there. It's true that geodesic incompleteness is the definition of a singularity. However, a timelike singularity can cause geodesics to be past-incomplete, which means that stuff can emerge from them. The default expectation is given in a famous description by Earman: "all sorts of nasty things - TV sets showing Nixon's 'Checkers' speech, green slime, Japanese horror movie monsters, etc. - emerge helter-skelter from the singularity."
Nov
17
comment What's the escape velocity of Naked Singularities?
A naked singularity is created... This is not right. You're describing one hypothetical type of naked singularity, not a naked singularity in general. It may be possible to create a naked singularity by overspinning a black hole, but that's still being studied and there is not yet a conclusive answer. Overcharging may also be possible.