Thomas

@Cleonis There may be different solutions to the rotational shape given appropriate starting conditions, but for an ideally symmetric situation it would require a physical cause/disturbance to break this symmetry. And I am not aware that this would have been directly observed in experiments (which should not be too difficult to do). All the photos and illustrations regarding this effect are artists impressions, also the photo of the so-called 'dwarf planet' Haumea.

@Cleonis This alone can't be enough. You still need a physically preferred axis which determines the main axis of the ellipsoid. This could really only be due to the gravitational influence of another body.

If you consider equilibrium conditions in the theory of gas kinetics, it turns out that you can have an equilibrium between two constituents with different energies/temperatures. The velocity distribution function is not Maxwellian in this case. Maxwell himself was wrong here by claiming that an equilibrium would automatically result in a Maxwell distribution function. See my web page plasmaphysics.org.uk/maxwell.htm for more.

@Chris I was just giving c1=1.5c as an insider info. Of course, the observers would not know this. They just measure t1= 2/3/c and t2=2/c . But they can calculate c1=1/t1=c*3/2 and c2=1/t2=c*1/2 (assuming unit distance) so c1=c(1+0.5) and c2=c(1-0.5), or in general c1=c(1+q) , c2=c(1-q). Again, this involves one-way speeds (with clocks assumed synchronized), but unless q=0 you could not derive LT from this (this is why q=0 in Einstein's derivations).

@Chris As I said, it is not about round trip speed here (which does not come into it) but in determining c in contrast to c1 and c2

@Chris Now assume, hypothetically, the speed of light to be anisotropic (e.g. in case of a moving ether). If light would move in one direction with c1=1.5c and in the other with c2=0.5 c, two observers at unit distance would time this at t1= 2/3/c and t2=2/c respectively. They would therefore derive c1=1/t1=c(1+0.5) and c2=1/t2=c(1-0.5).

@Chris Again, you are missing the point: the derivation of the LT assumes/requires isotropic light speed (it is not just a convention as frequently claimed). Any round trip assumption is only made for clock synchronization purposes. If you assume the clocks to be synchronized you don't have to bother with round trips.

@Chris In any case, even with c/(1+q), c/(1-q) you can not get a generalized version of the LT without violating the relativity principle for the motion between the two frames (the unprimed frame would have to move with a different speed relatively to the primed frame than vice versa unless q=0). Just try it.

@Chris Sorry, I misread your notation. I though the last bit was the result of your calculation. Note though that your round trip assumption does not apply here. In bartleby.com/173/a1.html it is about two different light signals travelling in opposite directions, and any speed anisotropy added here should be of the form c(1+q), c(1-q) (analogously to the case of a swimmer swimming with and against the flow, the flow should have speed c).

@Chris "(L1+L2)/(L1/v1+L2/v2)=2v1v2/(v1+v2)=/=(v1+v2)/2." Which yields, as required, c for v1=c(1+q) ; v2=c(1-q) . For your suggestion v1=c/(1+q); v2=c/(1-q) it would yield c/(1-q^2) for the two way speed of light

@Chris ((1-q)c+(1+q)c)/2 = c

@Chris Have a look again at mu edited answer: I have shown there from first principles that it is not possible to derive a coordinate transformation for an anisotropic speed of light. Anyway, your very argument is itself based on the usual (i.e. two-way) time dilation formula. Otherwise it is without any basis.

@Chris I do not assume it is isotropic. I did not calculate the difference between the two time dilations quite correctly though. I have corrected my answer now in this respect. If the speed of light is 1.5c in one direction and 0.5 c in the other, moving both clocks with 1m/sec by 1m would introduce a time difference due to time dilation of 2*10^-17 sec, which would not be measurable.

@Chris "If the speed of light is non-isotropic, the clocks will drift out of synchronization exactly enough so that they read the same value when the light arrives." The de-synchronization due to time dilation would be completely negligible in case the light speed anisotropy is a substantial fraction of the speed of light itself. See my re-edited answer for more

@Chris ""if the speed of light is anisotropic in a theory, that theory's Lorentz transformation must have a different form" There is no transformation possible in case of an anisotropic speed of light whilst keeping up the (one-way) invariance of c and the principle of relativity (see me re-edited answer for more)

@Chris Two synchronized clocks would stay synchronized when moved with identical speed over the same distance if the speed of light is isotropic (and the latter is actually implied by the Lorentz transformation; you could not derive if the speed of light is different in the two directions).

@Felicia Sure the clocks would show different times if c would be different in the two directions. If the speed would be $0.5c$ in one direction and $1.5c$ in the other (two way speed =$c$), the light would take $3$ times as long to one detector than to the other. But anyway, the derivation of the Lorentz transformation implies the speed to be the same in the two directions. You could not derive it for different speeds (the factor $1-v^2/c^2$ comes from the fact that the speeds are the same ($(1-v/c)*(1+v/c)$))

@Felicia If the speed of light is the same in all directions, both clocks will stop at the same time. If they don't stop at the same time, you know therefore that the speed of light is different. In any case, note that you can ignore the relativistic time dilation effect on the timings here as this is only of second order in $v/c$ and won't even be measurable for pedestrian transport speeds $v$. Any time difference measured will directly reflect the anisotropy of the speed of light ( for an absolute value of the speed of light you would obviuously additionally need a separate two-way timing).

@Felicia "You forget a crucial thing" Such as?

@Felicia "You can't measure which of the two directions has biggest speed. That's why I asked the question." Read my answer again, and you'll realize that you can..

@pela ------This distance is constant if space is locally not expanding. If you have two other objects in a different region of space that is expanding, you will measure the distance between those two objects as increasing. If you then bring your measuring stick back to the objects in the non-expanding region, you will still see them at the initial (smaller) distance. The expansion of the other region is completely irrelevant for what you measure there.

@pela I don't understand your remark regarding the absolute size of space. Any length/scale you measure is always relative to some standard. If you have two objects (assuming negligible mass) at rest relatively to each other, the distance between them is what you measure with e.g. a measuring stick. ----

@pela According to GR, photons adjust to the local metric. Give me a good reason why photons should adjust locally to e g. the Schwarzschild metric (by being deflected by masses) but not to the local FLRW metric.

@pela Obviously, if you assume a global expansion, then the photon will never see the scale factor a(t) getting smaller again, unless the expansion turns into a contraction. However, if you have regions of the universe that don't take part in the global expansion, then the photon will see the scale factor getting smaller as soon as it enters such a region. The metric in this region will be the same metric as the one when the photon was emitted.

@pela Note also the quote from Steven Weinberg (taken from the Wikpedia redshift article): "The increase of wavelength from emission to absorption of light does not depend on the rate of change of a(t) at the times of emission or absorption, but on the increase of a(t) in the whole period from emission to absorption". In other words, the redshift should only be determined by the ratio of the local scale factors at the locations/moments of emission and absorption. If these are equal (in case space did locally not expand) there should not be any redshift (the period in between is irrelevant).

@pela There is no absolute scale involved with my argument, only the relative local scale factors at the moment of emission of the photon and some moment whilst in the Hubble flow on the one hand, and the relative scale factors between the Hubble flow and when it has entered our galaxy/ solar system on the other. The photon would effectively see space contracting when making the latter transition.

@pela It does contract by the same factor it was expanded by: if the photon is emitted at a scale factor a=1, then expands to a scale factor say a=2, it contracts back to a scale factor a=1 again if it enters a region of space that did not take part in the expansion.

@RobJeffries - Again, your equation, as used by you, does not describe a scattering problem. It describes continuous (true) absorption (i.e..photo-ionization) and thermal (black body) emission. With this you won't be getting any absorption lines. And if you include scattering in the total absorption coefficient you have to include the scattering term for your source function as well. I am not sure though why you would want to consider continuous absorption in the first place. It is quite evident that above the photosphere the opacity is negligible outside the absorption lines.

@Rob Jeffries-The sign for the equation is opposite in this reference as they define μ=-1 going into the plane parallel medium. See irina.eas.gatech.edu/EAS8803_Fall2017/petty_11.pdf ; Eqs.(11.9) to (11.11). for the general form. These equations also make clear that the mean specific intensity J (i.e. the local source function in the medium) can consist of two parts, a local thermal emission (related to the Planck function) and an emission due to scattering. You claim your equation treats scattering, then you have to consider the second term only and your temperature argument fails.

@called2voyage - The page I linked to is based on a chapter of my Ph.D. thesis. As this is not generally available, I re-worked it somewhat and put it on my website (the corresponding numerical code linked from that page as well)

@Rob Jeffries - The first equation in plasmaphysics.org.uk/radiative_transfer.htm is the radiative transfer equation for a scattering atmosphere in integral form. You can call it a formal solution of the differential form if you like, but it is not an actual solution yet (which can only be done numerically). If you have a look at Eq.(13.1) in irina.eas.gatech.edu/EAS8803_Fall2017/petty_13.pdf you can see the same equation in your differential form. The point is that the second term on the right depends again on I, not on some internal thermal source term J.

@Felis Super - If the reference in the above comment is too technical, try this one irina.eas.gatech.edu/EAS8803_Fall2017/petty_13.pdf It is quite technical as well in places but has good explanations and figures, although it does not specifically refer to absorption lines in the sun

@Rob Jeffries - The equation above is not appropriate for a purely scattering atmosphere.The specific intensity J is in this case an implicit function of I and has nothing to do with thermal excitation. See plasmaphysics.org.uk/radiative_transfer.htm for the correct form. The solution of this gives a result similar to (plasmaphysics.org.uk/imgs/scattering2.jpg). Continuum source is to the left here, main curve is local excitation source function (due to the external continuum source), inset curve scattered line intensity looking left (without continuum source).