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TimRias

 Discussion on answer by John Rennie: …

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TimRias
May 22, 2024 19:32
Your argument fails exactly when $t$ is chosen to be a hyperboloidal time coordinate.
 

 Discussion on answer by Dan Romik: In…

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TimRias
Feb 20, 2024 03:41
@xLeitix Given that I have also seen a fair amount of institutions give exactly those types of updates for (more senior) searches, I'm not convinced that there is anything more to this rude behaviour than people accepting that this is just the way it is done. It deserves a bit more push back.
TimRias
Feb 20, 2024 03:41
@xLeitix Late letters are almost as bad as no letters. The polite thing to do is to send an e-mail when candidates have not made the short list, and when an offer has been made to a candidate. The effort to do is very low. The reason that employers don't is that they feel that this will make candidates less likely to accept in the case they do want get back to them (nobody likes being a "second" choice). This per definition is string the applicants along, which is rude. So, yes, sending letters only after the entire procedure has been concluded is also rude.
TimRias
Feb 20, 2024 03:41
As standard as the practice of not sending rejection letters is, does not take away from the fact how rude this practice is.
 

 Discussion on answer by safesphere: W…

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TimRias
Dec 8, 2023 17:58
The singularity theorem does not state that all geodesics need to hit the singularity. It states that there exist null geodesics with finite affine length that cannot be extended.
TimRias
Dec 8, 2023 17:58
"In a nutshell, the Penrose Singularity Theorem states that light inside the horizon cannot escape to infinity and therefore must end up in a singularity." This is not what the theorem states. The statement is that there are light rays with a finite affine length that cannot be extended. The example you give is a curve that actually has infinite affine length.
 

 Discussion on question by z.10.46: Ca…

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TimRias
Jun 13, 2023 23:04
Is this Homework?
 

 Discussion on question by Igarashi: T…

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TimRias
May 26, 2023 08:58
The Schwarzschild solution is somewhat pathological in that its Cauchy slices go through (at least) regions I and III. This is a silly consequence for taking an eternal black hole. It is physically more sensible look at a solution that includes formation of the black hole, like the (infalling) Vaidya metric. This has nice well defined Cauchy slices which provide initial conditions, that will form an horizon in their future.
TimRias
May 26, 2023 08:58
@Safesphere: GR is a well posed initial value problem in 3+1 dimensions. Given initial conditions specified on an appropriate (Cauchy) time slice the future spacetime is completely determined by the equations of motion, i.e. the Einstein equation. In particular, boundary conditions do not play a role.
 

 Discussion on question by Alice: How …

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TimRias
Oct 2, 2022 05:12
Due to the lack of tax treaty between Germany and Brazil, it is likely this person’s German income would be taxed as if she is a German (tax) resident. (While possibly being taxed again in Brazil)
TimRias
Oct 2, 2022 05:12
It is also not clear to me whether German labor laws would have anything to say about taking a second full time job outside of Germany, since this would be outside the jurisdiction of German law. At worst this would a contract violation of the contact this person has with the German university. If they don’t object…
 

 Discussion on answer by TimRias: Is t…

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TimRias
Jul 8, 2022 01:43
@nigel222 The Sun still is not dense enough, but Neutron Stars are (see answer, and LSAG report linked).
TimRias
Jul 8, 2022 01:43
@jmoreno The black hole would in principle be produced in the center of mass-frame of the collision. In the case of the LHC this means it can be produced with a very low velocity relative to the Earth. For Cosmic rays however, this means that the black hole would be produced at velocities well above the escape velocity.
TimRias
Jul 8, 2022 01:43
@computercarguy I would not put much faith in that 10^16 kg limit (from a poorly sourced paragraph in Wikipedia). It is based on a speculative extensition of general relativity (Einstein-Cartan theory) and the baseless assumption that the black hole would have to formed from fermions.
 

 Discussion on question by Oddthinking…

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TimRias
Jun 2, 2022 00:40
I would suggest changing the last part of the question to “Is this evidence that one or both are crisis actors?” Proving that the men are not actors would require an IMHO unethical amount of digging into their private lives prior to the Uvalde shooting. Let’s not invite people to do that.
 

 Discussion on answer by markvs: Is au…

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mmeent
May 1, 2022 03:11
@Issel One important difference is the use of physical samples, which are in limited supply.
 

 Discussion on answer by Dan Romik: Do…

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mmeent
Apr 29, 2022 08:04
@DRF No it does not mean that at all. It just means that if insurance companies want to take risk taking behavior into account, they have to do so without making sexist assumptions. E.g. look at number of traffic infractions.
mmeent
Apr 29, 2022 08:04
This answer is based on the assumption that it is rational for insurers to try to maximize their profit. This is a questionable assumption. Given the fact that quite a few insurers are technically "not-for-profit" organisations, this behaviour may be described is fairly irrational.
 

 Discussion on answer by Allure: Has N…

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mmeent
Mar 4, 2022 14:27
@Allure yes, it was and led to NATO operations such as en.wikipedia.org/wiki/Operation_Active_Endeavour, but Operation Enduring Freedom was not one of those operations.
mmeent
Mar 4, 2022 14:27
The invasion of Afghanistan was not a NATO operation. Although NATO later became officially involved, the initial invasion was the US and the UK acting on their own.
 

 Discussion on question by Coconut: Ca…

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mmeent
Aug 12, 2021 23:48
Does Maple cite the academic works its algorithms are based on?
 

 Discussion on question by Evan Carrol…

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mmeent
Feb 17, 2021 18:58
@fredsbend See e.g. texastribune.org/2021/02/16/natural-gas-power-storm, it seems a lot of the Texas gas infrastructure simply was not designed to operated in freezing conditions.
 

 Discussion on answer by Wolphram jonn…

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mmeent
Nov 17, 2020 15:31
Then what number do you get when you evaluate $ds^2$ for Schwarzschild at the point $t=1$, $r=10$ $\phi=2$, and $\theta=3$? @StephenG
mmeent
Nov 17, 2020 15:31
@StephenG If it were a scalar field as you claim you should be able to evaluate it at a point and get a number. $ds^2$ is a precisely defined mathematically object and that object is a symmetric rank 2 tensor field.
mmeent
Nov 17, 2020 15:31
@StephenG Wald is very explicit in his book (pg 23): " Sometimes the notation $ds^2$ is used in place of $g$ to represent the metric tensor, ..."
mmeent
Nov 17, 2020 15:31
Or try box 3.2 in MTW.
mmeent
Nov 17, 2020 15:31
No, I did not. Please go back to your introduction text differential geometry. You can come back and apologize after.
mmeent
Nov 17, 2020 15:31
If it had indices it would represent the components of a tensor. $ds^2$ is an abstract representation of a tensor.
mmeent
Nov 17, 2020 15:31
The $dx^\mu$ are not components of a vector, they are basis covectors.
mmeent
Nov 17, 2020 15:31
A symmetric rank-2 covariant tensor.
mmeent
Nov 17, 2020 15:31
$ds^2$ is not a scalar.
 

 Discussion on question by vsz: Is thi…

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mmeent
Aug 25, 2020 15:01
Have you tried googling the source listed on the image?
 

 Discussion on answer by henning -- re…

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mmeent
Jul 24, 2020 21:17
@m.raynal If you are unable to realize that such expressions are simply manners of speech that characterize the formal register in some cultures, than that would be a major red flag for your suitability for any sort of managerial task.
2
mmeent
Jul 24, 2020 21:17
The suggestion in @henning--reinstateMonica's comment that students regardless of their culture should have completely assimilated to Western academic professional standards during their undergrad, is pretty much the epitome of introducing a cultural bias.
 

 Discussion on answer by james turner:…

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mmeent
Jun 8, 2020 09:07
You would then have to first assume half of the fatalities involved another car. (As opposed to e.g. a tree)
mmeent
Jun 8, 2020 09:07
@tim I don't think that table distinguishes between occupants of the car driven by the drunk driver, or occupants of any other vehicle involved. The table is therefore not very useful for determine the "number of known victims".
mmeent
Jun 8, 2020 09:07
According to the link you posted that is only about 11k (30k is the total of traffic related fatalities). A significant portion of those deaths are either the driver theirself or occupants of their car. Both would know the driver.
 

 Discussion on question by Árpád Szend…

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mmeent
Feb 27, 2020 22:03
@safesphere There is absolutely no problem with construct a frame at the horizon. An easy very physical way of doing this is to start with a time slice with initial data representing a collapsing shell of matter. If you evolve this initial data in time, you will naturally end up with a timeslice that has an apparent horizon and therefore must extend inside the event horizon of the formed black hole.
 

 Discussion on answer by mmeent: Is th…

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mmeent
Feb 25, 2020 16:15
@A.V.S. I can see if I can find some time to do that. It would do very little to reduce the ambiguity on the force due to its gauge freedom. That freedom remains in the small mass-ratio limit.
mmeent
Feb 25, 2020 12:36
@Mitsuko An interesting observation about the GSF (due to your countryman Mino) is that one can always use the gauge freedom to choose coordinates such that the GSF on an orbiting body is constant. No matter how eccentric the orbit.
mmeent
Feb 25, 2020 12:36
@Mitsuko The average part of the GSF is well-defined, it is the oscillatory pieces that are extremely subtle. And no, fictious forces do not disappear in the classical limit. Think about the Coriolis force on Earth for example, or the fact that the gravitational acceleration on the equator is smaller due to the Earth's rotation.
mmeent
Feb 25, 2020 12:36
@Mitsuko The problem comes about due to the fact that we cannot distinguish gravitational forces from "fictious" forces due to the choice of coordinates. As a consequence you can always apply local coordinate transformations that change the local gravitational force due to the emission of GWs. When the accelerating force itself is gravitational this becomes even worse.
mmeent
Feb 25, 2020 12:36
@G.Smith Regarding the $JJ$ term. Each current multipole enters the expression for the radiation in the far field at a factor $1/c$ lower than the corresponding mass multipole. So as you suggested the first $JJ$ term comes in $1/c^2$ down from the leading order term. The leading $IJ$ term however comes in at the same order as the leading $II$ term because it starts at a lower multipole order.
mmeent
Feb 25, 2020 12:36
@G.Smith I don't think you should find equality, since there is also a torque being exerted by the radiation reaction.
mmeent
Feb 25, 2020 12:36
@G.Smith Yes the last $I$ should have been a $J$. I'' check on the coefficient tomorrow.
 

 Discussion on question by safesphere:…

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mmeent
Feb 16, 2020 21:57
@RobJeffries Axisymmetry does not guarantee that $\ddot{Q}=0$! (nor does it guarantee $\dddot{Q}=0$) This can easily be seen by considering the case of two masses accelerating towards each other on a head on collision. Such systems are well known to emit gravitational radiation.
 

 Discussion on answer by mmeent: Blink…

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mmeent
Aug 30, 2019 10:33
@safesphere Coordinate transformations only need to be well defined on the patch where they are applied. So that is not an issue. It is also irrelevant, since I've already explained to you why a timeslicing cannot be tangent to the event horizon.
mmeent
Aug 30, 2019 10:33
@safesphere That again uses statements in Schwarzschild coordinates, which are not defined at the horizon, and consequently can't be used to make sensible statements about things at the crossing.
mmeent
Aug 30, 2019 10:33
@safesphere There is no possible time slicing in which all objects would cross the event horizon at the same time. (That would imply that the time slices are null at the horizon, meaning that they are not time slices.)
mmeent
Aug 30, 2019 10:33
@Ivella Yes, you will still be able to see the flashes from the flashlight. In fact you will continue to see them after crossing the horizon (assuming the flashlight keeps flashing). The flashes that reach you before you cross the event horizon, will have been emitted before the event horizon.