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ProfRob

 Discussion on question by Miss Unders…

Imported from a comment discussion on astronomy.stackexchange....
ProfRob
Nov 20, 2024 18:00
@MissUnderstand you have not understood what I'm saying. The stars are not "all the same size", either in absolute terms or in angular size terms. Anyway,, I've answered the question you could have asked.
ProfRob
Nov 20, 2024 18:00
The answer will vary between some maximum number for the star with the largest angular size to as many orders of magnitude smaller as you like for other stars. Which is why the question as asked is pointless and requires clarification.
ProfRob
Nov 20, 2024 18:00
Which star?.....
 

 Discussion on question by Ask Questio…

Imported from a comment discussion on astronomy.stackexchange....
ProfRob
Oct 22, 2024 21:57
No, the concept isn't new. But what do you mean by "old light" - light does not experience any proper time. If you mean "tired light", then that tired idea has been argued to death and there is oodles of evidence that show it is not a viable explanation for Hubble's law. What is the "force" that you think causes the redshift of light? I think there is a reasonable question in here, but I'm struggling to see what it is. Are you just asking whether there is any direct evidence for expansion? i.e. Seeing something get further away with time, rather than simply inferring a speed from redshift.
ProfRob
Oct 22, 2024 21:57
I think you have to come up with a specific scenario that can be discussed, or rather you have to say exactly what it is you want to vary and what you are holding constant.. All we can do is look for variations in dimensionless constants like the fine structure constant - the variations of which could be due to variations of several constants with units.
 

 Discussion on answer by safesphere: W…

Imported from a comment discussion on physics.stackexchange.co...
ProfRob
Dec 8, 2023 17:58
"Also, please feel free to post your own answer to express your views and opinions, as this site is not for extended conversations in comments". The biter, bit.
 

 Discussion on answer by ProfRob: Line…

Imported from a comment discussion on physics.stackexchange.co...
ProfRob
Nov 17, 2023 14:15
@CMarius Correct, it forms no part of my answer since rotation was not part of the original question. The reference was to show you that the rectilinear motion experiment has been done and there is no voltage produced.
ProfRob
Nov 17, 2023 14:15
@CMarius What edge of what magnet? The Lorentz force is $q(\vec{E} + \vec{v}\times\vec{B})$ and it is zero in both frames of reference, as I demonstrated.
ProfRob
Nov 17, 2023 14:15
"Does it matter "how included" the copper plate is in the magnetic filed? Does the size of the magnet w.r.t the size of the copper plate make any difference?" are NEW questions. @CMarius
ProfRob
Nov 17, 2023 14:15
It would be if you kept the magnets/magnetic field stationary in the laboratory frame of reference. In which case in the laboratory frame there is no electric field and a force $q\vec{v} \times \vec{B}$ on the charged particle. In the stationary frame of the charge particles there is an electric field $\gamma \vec{v}\times \vec{B}$ and a magnetic field of $\gamma \vec{v}\times \vec{B}$ that the particles are not moving with respect to. Hence $\vec{F}' = q\gamma \vec{v}\times \vec{B}$ (note that Lorentz force is not invariant).
 

 Discussion between ProfRob and Alfred

Imported from a comment discussion on physics.stackexchange.co...
ProfRob
Nov 11, 2023 14:05
Any discussion about exactly how big the Fermi energy is, is also irrelevant because it is always far bigger than required to ionise carbon -- "of order MeV" (which means 1-10 MeV) as I originally said.
ProfRob
Nov 11, 2023 14:05
You said "once degenerate it does not give up any heat". What is "it" if not the degenerate core? The degenerate core does give up heat - from the ions. That is what white dwarf cooling is and the question was solely about white dwarf cooling not all this stuff about what happened before it became a white dwarf.
ProfRob
Nov 9, 2023 17:47
In terms of heat getting out. Of course the ions are basically an ideal gas (until they solidify). It is they that contain all the thermal energy and it is they that cool, not the electrons.
ProfRob
Nov 9, 2023 17:45
geogebra.org/m/ACZt5h7a
ProfRob
Nov 9, 2023 17:44
Have aplay with this widget I made (a long time ago, not for this discussion!). You can set the T to 1E8K and the density to 10^10 kg/m^3 (about right for the centre of an average white dwarf. You can see the Fermi energy is about 1 MeV and the degeneracy parameter is 60 before it has even started on the WD cooling phase.
ProfRob
Nov 9, 2023 17:43
The white dwarf forms by shedding its envelope and it moves to the top left of the HR diagram. From there is cools at almost constant radius (and constant Fermi energy). The value of the Fermi energy is in the range 1-10 MeV, depending on the mass of the white dwarf ("of order MeV). Electrons have energies from zero up to the Fermi energy.
ProfRob
Nov 9, 2023 17:40
Let's both cool down, your answer has got plenty of upvotes. Yes, the material that becomes the AGB core and then the white dwarf must cool, but it becomes hotter - a natural consequence of the virial theorem. The reason the core becomes degenerate is not because its temperature gets lower it is because it becomes more dense.
ProfRob
Nov 8, 2023 19:21
I'm also really annoyed that you chose to directly criticise my answer, based on your own misreading of it, without having the courtesy of leaving a comment beneath my answer so that I know what you'd done.
ProfRob
Nov 8, 2023 19:19
This is misleading. The degenerate white dwarf is already sitting at the core of its giant progenitor. The main reason it decreases in radius is losing its envelope. Yes, there is some shrinkage after this, but that is of the non-degenerate atmosphere which accounts for about 1 percent of the white dwarf mass and much less than 1 percent of its thermal energy.
ProfRob
Nov 8, 2023 19:18
(iv) Then in your comments "Before becoming a white dwarf, a star is hot. Fuel exhausted, it goes through a path in the Hertzsprung-Russell diagram but eventually it radiates away and shrinks." This is misleading. The degenerate white dwarf is already sitting at the core of its giant progenitor.
ProfRob
Nov 8, 2023 19:15
(iii) "If it is too heavy, the energy reaches this value and keeps increasing. " ditto. The Fermi energy does not increase for a star of a given mass.
ProfRob
Nov 8, 2023 19:14
(ii) "it is as if gravitational energy is unable to give more "degeneracy" energy to the electrons. But this is not so." You may understand what you mean, but this reads like you think that gravitational contraction can increase the Fermi energy.
ProfRob
Nov 8, 2023 19:14
If you meant that paragraph to be talking about some progenitor object then you need to be clear. However, the carbon core of an asymptotic giant branch star, that will become a white dwarf, is already highly degenerate.
ProfRob
Nov 8, 2023 19:14
Where you give your misleading impression that the Fermi energy can change: (i) The entire paragraph that begins "When the star is hot...". The star we are discussing in this question is a white dwarf. Even a hot white dwarf is highly degenerate througout 99 percent of its interior. The Fermi energy of the electrons in the bulk of the star will not change (significantly) as it cools.
ProfRob
Nov 8, 2023 19:09
I don't mention a maximum Fermi energy at all. I don't use the word "maximum", or the phrase "Fermi energy". What I discuss is that the electrons in a white dwarf have a distribution of energies from zero up to some large value. That is because the electrons do have a distribution of energies from zero up to the Fermi energy and the Fermi energy is "of order MeV", right?
ProfRob
Nov 8, 2023 18:50
There is a maximum Fermi energy - it is arounfd 10 MeV (for a carbon white dwarf) as I said above. However, that is irrelevant to the question and I did not write it in my answer, so I'm not sure what you are on about. My downvote is because your answer and your comments give the impression that you do not understand that the Fermi energy of a white dwarf is more-or-less fixed as it cools.
ProfRob
Nov 8, 2023 18:50
The question is about white dwarfs and I don't understand your comment about maximum Fermi energies, as I said for a given WD it doesn;t change. The "Chandrasekhar mass" - as explained by you, is the mass at which the radius of the WD shrinks to zero and the Fermi energy does in fact become infnite. The actual upper limit, is set by General Relativity (not considered by Chandrasekhar) or by electron capture and both occur when the Fermi energy reaches around 10 MeV in the core of the white dwarf. None of which has anything to do with the OP's question. Your downvote has been reciprocated.
ProfRob
Nov 8, 2023 18:50
A typical white dwarf (of mass $0.6M_\odot$) has an average density of a few $10^{9}$ kg/m$^3$, is made of ionised carbon and has a typical electron Fermi (total) energy of 0.8 MeV, whatever its temperature. More massive white dwarfs have higher Fermi energies, less massive white dwarfs have lower Fermi energies. They are "of order MeV". Actually, the stuff you've written about the Chandrasekhar mass has nothing to do with the question at all. I've reversed my upvote for your needless edit.
ProfRob
Nov 8, 2023 18:50
The electrons in a white dwarf are highly degenerate with $E_F/k_BT$ of several hundred, even at temperatures of tens of thousands of K. The only way to increase the Fermi energy is to add mass to the white dwarf. White dwarrfs of constant mass do not get significantly smaller or denser as they cool and so the Fermi energy does not significantly increase with time.
 

 Discussion on answer by d_b: Is the m…

Imported from a comment discussion on physics.stackexchange.co...
ProfRob
Aug 8, 2023 03:59
@JohnDoty It depends entirely on the zeropoint. Today is 20-01-1445 in Iran. A zero magnetic field on the other hand has a well-defined meaning.
ProfRob
Aug 8, 2023 03:59
@JohnDoty we measure potential difference.
 

 Discussion on question by Cham: Numer…

Imported from a comment discussion on physics.stackexchange.co...
ProfRob
Aug 6, 2023 01:05
You have still not extended the x-axis of the logarithmic plot far enough for us to see what is happening as the density becomes small, even for the Newtonian case. Why is the "normalised" central pressure 0.5?
ProfRob
Aug 6, 2023 01:05
@Cham I don't know. Besides, that isn't your problem - because you've now checked that, right?
ProfRob
Aug 6, 2023 01:05
It is well known that a polytrope with $n>5$ $(\gamma = 1+1/n$) does not have a finite radius.
ProfRob
Aug 6, 2023 01:05
Also,please make sure you are showing us an example with $\gamma >1.2$ because the radius is infinite for smaller values
ProfRob
Aug 6, 2023 01:05
Can you plot log p on the y-axis. It's impossible to see what's going on when the density should vary over many orders of magnitude.
 

 Discussion on question by Ed_Gravy: W…

Imported from a comment discussion on astronomy.stackexchange....
ProfRob
Jul 18, 2023 18:27
@Sten more to the point, they are not "measured" to be moving faster than the speed of light either (what would the redshift of something be travelling at $>c$) ?
 

 Discussion on answer by anna v: What …

Imported from a comment discussion on physics.stackexchange.co...
ProfRob
Aug 4, 2022 15:40
@tobi_s The title of the paper is "The Newtonian gravitational deflection of light revisited". It could hardly be more relevant to the original question. The velocity does of course change, whilst the speed remains constant.
ProfRob
Aug 4, 2022 15:40
@annav you continue to repeat that. Yet there were predictions of the acceleration (a change in velocity) of light, using Newtonian mechanics, prior to General Relativity and Quantum Mechanics. An example of that treatment is here arxiv.org/pdf/physics/0508030.pdf and it uses neither Special Relativity or Quantum mechanics. The answer is wrong by a factor of 2, but it is not zero.
ProfRob
Aug 4, 2022 15:40
@tobi_s the question asks about interpreting the trajectory of light in terms of Newtonian physics. That can and has been done (and gets the wrong answer), but does not predict no acceleration.
ProfRob
Aug 4, 2022 15:40
So @tobi_s when we see light bend and we are using Newtonian physics, how is that unaccelerated (in Newtonian physics)?
ProfRob
Aug 4, 2022 15:40
Right. And velocity isn't the same as speed and contrary to your answer, the velocity of light can change, it doesn't involve quantum mechanics and Special Relativity isn't relevant to a situation involving gravity.
ProfRob
Aug 4, 2022 15:40
@Abbas when light is gravitationally lensed then obviously it does accelerate.
ProfRob
Aug 4, 2022 15:40
Light always moves with a speed of $c$ when measured locally. Photons can and do appear to be accelerated.
 

 Discussion on question by Ricky: Has …

Imported from a comment discussion on astronomy.stackexchange....
ProfRob
Jul 9, 2022 17:46
Does this answer your question? How many new stars visible to the naked eye have appeared since humanity started making records?
ProfRob
Jul 9, 2022 17:46
I can see why it may not be, but please clarify in the question why this is not a duplicate.
 

 Discussion on answer by SDV: Velocity…

Imported from a comment discussion on astronomy.stackexchange....
ProfRob
Mar 18, 2022 21:10
Show your calculation. Here is mine. The RMS velocity is the square root of 3kT/m, where m is the mass of a proton. This yields 600 km/s for a temperature of 15 million Kelvin. The RMS velocity for 5000 K is much lower,. That you can't do this very basic calculation, even when challenged on it, says much about the rest of your thesis.
ProfRob
Mar 18, 2022 14:20
@SDV $1.5\times 10^7$ K is of course the temperature in the centre of the Sun and the thermal velocities I quote are those in the centre. The fact that you don't understand that $1.5\times 10^7$ means 15 million makes me think you are wasting everybody's time, including your own.
ProfRob
Mar 18, 2022 14:20
Centre of star $T\simeq 1.5\times 10^7$K. Thermal RMS velocity of a proton $\simeq$ 600 km/s. Time dilation effect: a factor of 1.000002.
 

 Discussion on question by hm2020: On …

Imported from a comment discussion on astronomy.stackexchange....
ProfRob
Sep 20, 2021 18:13
The ascent stage was not ballistic, so reference to the escape velocity isn't relevant. It also didn't need to escape from the Moon it needed to match speeds with the command module. The reference (in comments) to aerodynamics applied to something travelling in near-vacuum is nonsense.