KJW

What difficulties are you having with this?

Although the equations are symmetric, that "running movie backward" is a solution that is not symmetric. It illustrates the point that a symmetric equation need not have a symmetric solution, although the symmetry-transformation of a solution is a solution. However, auxiliary conditions that select a particular solution from the set of all solutions need not possess the same symmetries as the equations, and indeed must not possess the symmetries of the equation that are not possessed by the particular solution. You didn't respond to what I said about transition probabilities not being symmetric to time reversal. It should also be noted that covariance is not the same as invariance. For example, the rate equation mentioned above is covariant. But because the rate constant changes sign, along with rate of change, under time reversal, the arrow of time is maintained as the direction for which the rate constant is positive. Why are you often mentioning the Big Crunch? The current evidence is that the expansion of the universe is accelerating and that the Big Crunch is unlikely.

No, heavy water and normal water have the same number of electrons.

The equivalence principle is observed! In what way is the equivalence principle observed? So if there aren’t any electrons, for example if I have a sample consisting only of protons and neutrons (ie ionized hydrogen) in a stationary state, then this sample will exhibit no gravity? Also, you are forgetting that gravity is generated by things other than massive particles - for example electromagnetic fields have a gravitational influence, as do pressure, stress, strain etc. Perhaps you (@Evgen Ivashura) can explain why heavy water is heavier than normal water.

What about the equivalence principle?

But what does that actually mean? I've already said that symmetric equations do not necessarily lead to symmetric solutions. Symmetry in physics is actually quite subtle. No, there is a fundamental asymmetry in physics: Transition probabilities do not possess time-reversal symmetry. Suppose one has a microscopically reversible reaction: X transitions to Y with a given probability over a given interval of time, and Y transitions to X with the same probability over the same interval of time. Here, reversibility refers to either species transitioning to the other species. But both transitions are in the future direction. If one has more X than Y (or more Y than X) at a particular time and look at the composition a given interval of time later, it would be seen than the later composition obeys the statistics of the transition probability. But if one starts with the composition at the later time and considers the statistics of the transition to the earlier time, one would see that the earlier composition does not obey the statistics of the transition probability or any transition probability (seen by considering the time-reversed statistics at various times). And the reason why transition probabilities do not possess time-reversal symmetry is because probability is a positive number. But note that the rate equation of the reaction is covariant with respect to time reversal. That is because the rate constant, unlike the transition probability, has a negative value for the time-reversed reaction.

The equations of general relativity possess time-reversal symmetry, as do the general solutions of those equations, taken as sets of all solutions for each of the equations. But the individual solutions taken from a set of all solutions, perhaps specific solutions resulting from specifying auxiliary conditions, often do not share the symmetries of the equation, such as time-reversal symmetry. In other words, applying a symmetry transformation often transforms one solution to a different solution, not the same solution. It is my understanding that for a wave equation, the initial conditions contain all the information required for the retarded wave solution, rendering the advanced wave solution redundant. Also, the retarded wave solution is an expanding wave solution in the future direction while the advanced wave solution is an expanding wave solution in the past direction. But an expanding wave solution in the past direction is a contracting wave solution in the future direction and would require carefully arranged emitters to produce such a solution, noting that reality possesses an arrow of time which enforces the notion of causality and the second law of thermodynamics. This does not violate general relativity as covariance is maintained in spite of the asymmetry implied by the arrow of time. I suppose the simplest way to explain this is that the existence of symmetry in the laws of physics does not imply the absence of asymmetry in the laws of physics.

LOL. +1

[math]\text{Let:}\ \ \ g_{rs} = \eta_{rs} + \kappa_{rs}\ \ \ ;\ \ \ g^{rs} = \buildrel\rm{\small -1}\over{\eta}{\!}^{rs} + \buildrel\rm{\small -1}\over{\kappa}{\!}^{rs}[/math] [math]\buildrel\rm{\small -1}\over{\eta}{\!}^{ru} \eta_{us} = \delta^r_s\ \ \ ;\ \ \ (\buildrel\rm{\small -1}\over{\eta}{\!}^{ru} + \buildrel\rm{\small -1}\over{\kappa}{\!}^{ru})(\eta_{us} + \kappa_{us}) = \delta^r_s[/math]

@Anton Rize The point I was making about the distinction between observing from outside and observing from inside is that there isn't going to be a [math]\sqrt{3}[/math] factor difference in any observed quantity because in any non-relativistic system [math]\kappa^2 = \dfrac{r_s}{r} \ll 1[/math] and [math]\beta^2 = \dfrac{r_s}{2r} \ll 1[/math], and therefore [math]1 - (\kappa^2 + \beta^2) \approx 1[/math]. This is the factor that governs any difference between observations made from inside and observations made from outside. For example, at the surface of the sun, [math]\dfrac{r_s}{r} = 4.24 \times 10^{-6}[/math], requiring very accurate measurments to even detect the gravitational redshift of spectral lines.

I thought I had made it clear that it is NOT the distance from Earth that determines Io's measured orbital period but the rate at which Jupiter is moving away from Earth or towards Earth. What is being measured is essentially the Doppler effect (although the Doppler effect itself was not discovered until 166 years later by Christian Doppler). Of course the actual value of Io's orbital period is not changing. That's the whole point. The change in the measured value from the unchanging true value is the amount of time it takes light to travel the change in the distance from Earth that Jupiter has moved over the orbital period of Io.

Reference? Wikipedia said: "By timing the eclipses of Jupiter's moon Io, Rømer estimated that light would take about 22 minutes to travel a distance equal to the diameter of Earth's orbit around the Sun." and gave the free-to-read reference: https://babel.hathitrust.org/cgi/pt?id=uc1.b4375710&seq=1 (see in particular: https://babel.hathitrust.org/cgi/pt?id=uc1.b4375710&seq=12). Anyway, it doesn't really matter because neither value is particularly accurate. And it is quite absurd to regard Rømer's value as meaningful in a modern context. I already shown that the average deviation from the true value of the orbital period of Io is about 9 seconds over a period of about 43 hours. Do you really think Rømer's clock were sufficiently accurate to support your claim that 11 seconds is an accurate value? Even Isaac Newton gave a value that was much closer to the modern value in his 1704 book Opticks.

When Jupiter is closest to Earth (opposition), the orbital period of Io has the true value of about 42.5 hours. After this, Jupiter is moving away from Earth and therefore the measured orbital period of Io will be longer than the true value. Over the time it takes Jupiter to reach the furthest from Earth (conjunction), the deviations from the true orbital period of Io over the many orbits will have accumulated to about 16 minutes (measured as 22 minutes by Ole Rømer). And when Jupiter is furthest from Earth (conjunction), the orbital period of Io again has the true value of about 42.5 hours. After this, Jupiter is moving towards Earth and therefore the measured orbital period of Io will be shorter than the true value. Over the time it takes Jupiter to again reach the closest to Earth (opposition), the deviations from the true orbital period of Io over the many orbits will have accumulated to about –16 minutes (measured as –22 minutes by Ole Rømer). I did some calculations: Jupiter has a synodic period of 398.8 days from Earth, therefore the time from opposition to conjunction is 4785.6 hours. Over this period, Io will have orbited Jupiter 112.6 times. Therefore, on average, the deviation from the true orbital period of Io is 8.86 seconds (and –8.86 seconds from conjunction to opposition).

One thing I feel should be stated is that the greatest deviation from the true orbital period of Io occurs when Jupiter is moving away from Earth or towards Earth, not when Jupiter is furthest from Earth or closest to Earth. Indeed, when Jupiter is furthest from Earth or closest to Earth, the orbital period of Io will be close to the true value.

Did Ole Rømer measure the speed of light by looking at the Sun or at Jupiter's Io? If he measured it by looking at the Sun, it would be the radius. Or if he measured it by looking at Jupiter's Io, it would be the diameter. It's not about what Ole Rømer measured, it's about the value you are quoting. Rømer estimated that light would take about 22 minutes to travel a distance equal to the diameter of Earth's orbit around the Sun.

Ole Rømer's 11 minutes was for the time it takes light to travel the radius of Earth's orbit, not the diameter. Thus, Ole Rømer's speed of light was actually about 25% lower than the true value.

... that a topic in the Sandbox can be sent to the Trash Can. While the particular topic in the Sandbox was the same as the topic in other forums that were also set to the Trash Can, it struck me as a little odd that a topic in the Sandbox was sent to the Trash Can. That the topic was in the Sandbox was also odd, given the purpose of the Sandbox.

Living in a location with hot summer days, I often see mirages while driving along a road. Mirages are an example of total internal reflection. When light passes from a medium of low refractive index to a medium of high refractive index, the angle of refraction is less than the angle of incidence. But when light passes from a medium of high refractive index to a medium of low refractive index, the situation is reversed, and the angle of refraction is greater than the angle of incidence. That means there are angles of incidence less than 90° such that the angle of refraction would be greater than 90°, and at such high angles of incidents, the light is reflected instead. This is known as total internal reflection. When the surface of the ground is heated by the summer sun, the refractive index of the air just above the ground is less than the refractive index of the air higher up, forming the right conditions for total internal reflection to occur.

@jalaldn , did you not see this? You keep mentioning "the moment Io disappears behind Jupiter", whereas this is not the moment that interested Ole Rømer.

Yes, it's called "Solipsism". There is also "Solipsism syndrome".

Because the planet is in free-fall, it won't feel the "pull" of the black hole. Instead, the planet will experience the tidal force that tends to pull the poles apart and squeeze at the equator. If you are at the equator, you will feel this extra force towards the ground in addition to the force of gravity of the planet itself. If you are at either pole, you will feel less force than the force of gravity of the planet itself and may possibly float from the planet if the tidal force of the black hole is stronger than the force of gravity of the planet.

Which is still a classical solution. You are the one who brought up the observation that the classical picture is insufficient, and Quantum gravity may modify the situation. Why can't I do the same ? Fair enough. However, I mentioned the Reissner–Nordström metric to point out that a classical charged black hole is gravitationally repulsive at small distance, something that may be considered counter-intuitive.

But as @swansont pointed out, everybody else can say the same thing regardless of where they are. That is the nature of an expanding universe. Regardless of where you are in it, the universe will always appear to expand away from you.

The Newtonian solution has them both obeying the same 1/r^2 trend, so the repulsion is always bigger than the attraction by the same factor. How do you get enough of an attraction? You would need to propose some new physics. Also, charged black holes are gravitationally repulsive at sufficiently small distance. Note that the Reissner–Nordström metric (which describes a non-spinning charged black hole) has [math]g_{tt}[/math] of the form: [math]g_{tt} = c^2\left(1 - \dfrac{k_1 M}{r} + \dfrac{k_2 Q^2}{r^2}\right)[/math] Note that the [math]Q^2[/math] term is positive in contrast to the [math]M[/math] term. Also, at sufficiently small distance, the repulsive [math]Q^2[/math] will dominate, whereas at sufficiently large distance, the attractive [math]M[/math] will dominate.

I feel I need to point out that I don't necessarily think that fundamental particles are black holes, but rather that I disagree with the use of Hawking radiation to argue against the idea that fundamental particles are black holes. Or maybe it's neither. Perhaps we are dealing with "Quantum Gravity". Maybe the operative word in "black hole" is "hole" rather than "black". The point I was making is that just as quantum mechanics prevents an electron in an atomic orbital from radiating even though classical electrodynamics predicts that such an electron would radiate, a general relativistic quantum theory may prevent a subatomic black hole from radiating even though current quantum theory predicts such radiation. The same as what you get when you scatter a regular electron off another (???). Bear in mind that if an electron was a classical black hole, it would be a Kerr–Newman black hole, and according to https://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric#Dirac%E2%80%93Kerr%E2%80%93Newman_electron_model, an extremal solution with a naked singularity. Given that general relativity can be formulated in terms of spinors, half-integer spin particles would not be a problem. There is no suggestion that a subatomic particle would be a classical black hole. If a subatomic particle is a black hole, it would be a black hole from some form of general relativistic quantum theory and presumably would possess the same constraints as subatomic particles. But I need to reiterate that I am not making the claim that subatomic particle are black holes, and that the above is only to provide some plausibility to the idea.