elfmotat

The Schwarzschild solution is (as a necessary assumption) t-symmetric, so it doesn't make any sense to talk about changes in the central mass. A changing mass distribution will produce an entirely different solution to the EFE's. I'm also not really sure how you arrived at that metric. It looks pretty nonsensical to me.

Spacetime is curved by energy and momentum. Basically, light follows Newton's first law through curved spacetime. It simply follows the straightest possible path.

The quarks that make up protons and neutrons are similar to the electron in the sense that they are point particles. Protons and neutrons themselves do occupy a certain amount of volume, though the concept of volume in quantum physics is not very clear-cut. It's difficult to say how much space something occupies due to inherent uncertainties. I also assure you that I'm not making anything up.

Why do you find this strange? Why do you think things must have volume to possess mass?

They are point particles, i.e. they have no volume.

Just like when going from Newtonian physics to special relativity with increasing speed, there is no clear-cut line where one "takes over." When speeds get faster and faster the effects of SR become more and more evident, until eventually Newtonian predictions are no longer useful. There's no distinct speed where this happens; it's a gradual change. Likewise, going to smaller and smaller scales will result in the inverse square law becoming less and less accurate until eventually it's nonsensical to even talk about, and you'd need quantum gravity to explain what's happening.

Square each speed, add them together, divide by 3, then take the square root. [math]v_{rms}=\sqrt{\frac{1}{n}\sum_{i=1}^n (v_i)^2}[/math] where vi is the speed of the i'th particle (i.e. i=1,2,3...) and n is the number of particles.

You're exceptionally good at dodging questions; have you considered politics? This is what I want you to answer: -What is "the standard four-velocity?" I've never heard of this before. How does this get around the fact that the four-velocity of a photon is undefined? -What do you mean by "the relation between and ?" What relation exists between a general affine parameter and dt? -What do you mean by "symmetries of the Schwarzschild metric," and why would symmetry in a particular vacuum solution imply constant speed in general? -What does the following mean (please expand it out, or show it using more conventional notation): -How do said the said relation between and and symmetries in the Schwarzschild solution imply the following: -How does ds2=0 imply the following: - (Most importantly) do you agree that, in my example, the person on the platform would measure the speed of light to differ from c? If so, how do you reconcile this with your belief that the speed of light in a vacuum is always c? If not, please point out the error(s) in my logic/calculation. This result was waited in pure physical terms because v, the standard velocity, is a gauge-invariant three-vector. I can't find either of those sections, nor that quote. Please provide page or section numbers.

Okay, so [math]\tau[/math] is just a random affine parameter in your case. Fair enough. Most people use [math]\tau[/math] to represent proper time. You still haven't addressed the fact that the four-velocity of a photon is undefined. You explained what every symbol is, but, like I said, I can't made heads or tails of your notation. I know what [math]\gamma_{ij}[/math] is, I know what [math]g_{ij}[/math] is, and I know what [math]\gamma_{ij}v^iv^j[/math] is, but I have absolutely no idea what [math]g_{ii}^2 \left( \frac{dx^i}{dt} \right)^2[/math] is. You also didn't explain what you mean by "the relation between [math]d\tau[/math] and [math]dt[/math]" or "symmetries of the Schwarzschild metric," and I don't see how ds2=0 implies: [math]|\mathbf{v}| = \sqrt{g_{ii}^2 \left( \frac{dx^i}{dt} \right)^2} = \sqrt{c^2}[/math] Yes, LOCALLY!!! He explained that that was what he meant: "In general relativity, the appropriate generalisation is that the speed of light is constant in any freely falling reference frame (in a region small enough that tidal effects can be neglected)." See above. Yes, everyone in this thread besides you. Your "computation" looks quite nonsensical to me. You also completely ignored the most important part of my last post. I'll state it again so that you'll be sure to see it: "That's the thing though - my example was completely physical! A man bouncing a light signal off a mirror in Schwarzschild spacetime would measure the speed of light to differ from c. If he measured the distance with a ruler and timed how long it took for the light to return and hit his detector, he would find a result that differs from c! Now, if he were freely falling and conducted the same experiment over a sufficiently small scale, then he would measure v=c. That's why we say the speed of light islocally c. Your "definition" (you haven't really given one yet) of the speed of light in GR seems to be the unphysical one. My definition is simply that which we can measure!"

Disagree with what?

The four-velocity of a photon is undefined! (Check any textbook.) Okay, except whose proper time are we differentiating by? Certainly not the photon's! I can't follow this notation at all. Please explain exactly what you're doing here. Once again, I don't follow. John Baez of UCR (among others - his link was just the first I found) disagrees: "The problem here comes from the fact that speed is a coordinate-dependent quantity, and is therefore somewhat ambiguous. To determine speed (distance moved/time taken) you must first choose some standards of distance and time, and different choices can give different answers. This is already true in special relativity: if you measure the speed of light in an accelerating reference frame, the answer will, in general, differ from c.In special relativity, the speed of light is constant when measured in any inertial frame. In general relativity, the appropriate generalisation is that the speed of light is constant in any freely falling reference frame (in a region small enough that tidal effects can be neglected). In this passage, Einstein is not talking about a freely falling frame, but rather about a frame at rest relative to a source of gravity. In such a frame, the speed of light can differ from c, basically because of the effect of gravity (spacetime curvature) on clocks and rulers." Einstein himself also disagrees: ". . . according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [. . .] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position." http://math.ucr.edu/...d_of_light.html EDIT: I just looked over the thread and realized that both of these have been posted already, though you've so far failed to address them. That's the thing though - my example was completely physical! A man bouncing a light signal off a mirror in Schwarzschild spacetime would measure the speed of light to differ from c. If he measured the distance with a ruler and timed how long it took for the light to return and hit his detector, he would find a result that differs from c! Now, if he were freely falling and conducted the same experiment over a sufficiently small scale, then he would measure v=c. That's why we say the speed of light is locally c. Your "definition" (you haven't really given one yet) of the speed of light in GR seems to be the unphysical one. My definition is simply that which we can measure!

I'm not sure what you're asking. Trying posing your question in a different way.

I'll work in units where c=G=1 for simplicity. Let someone standing on a platform at Schwarzschild coordinate r1 in Schwarzschild spacetime emit a light signal radially downwards at time t1. It reflects off a mirror at some height r2 and time t2, and returns to the initial height where it hits a detector. So, what does this person measure the speed of light to be? The proper distance (distance he would measure with a ruler), s, between r1 andr2 is: [math]s=\int_{r_1}^{r_2} \left (1-\frac{2M}{r} \right )^{-1/2}dr[/math] Since we'll assume he's at a height where r>>2M, we can simplify this with a binomial expansion and discard higher order terms. [math]s=\int_{r_1}^{r_2} \left (1+\frac{M}{r} \right )dr=\Delta r+Mln\left ( \frac{r_2}{r_1} \right )[/math] The proper time the person measures between t1 and t2 is simply a time-dilated coordinate time: [math]\tau =\sqrt{1-\frac{2M}{r_1}}\Delta t[/math] But using the same simplifying approximation as before gives: [math]\tau =\left (1-\frac{M}{r_1} \right )\Delta t[/math] Δt can be found in terms of r by using the fact that light travels on null geodesics (i.e. ds2=0): [math]ds^2=0~~\Rightarrow ~~dt=\left ( 1-\frac{2M}{r} \right )^{-1}dr\approx \left (1+\frac{2M}{r} \right )dr[/math] [math]\Rightarrow ~~ \Delta t=\Delta r+2Mln\left ( \frac{r_2}{r_1} \right )[/math] [math]\tau =\left (1-\frac{M}{r_1} \right )\left (\Delta r+2Mln\left ( \frac{r_2}{r_1}\right ) \right )[/math] Multiplying the times and distances by two gives the round trip time and distance. So the guy on the platform measures the speed of light to be: [math]v=s/\tau =\frac{\Delta r+Mln\left ( \frac{r_2}{r_1} \right )}{\left (1-\frac{M}{r_1} \right )\left ( \Delta r+2Mln\left ( \frac{r_2}{r_1}\right ) \right )}[/math] As you can see, even to first order as M gets larger the speed of light slows. As M goes to zero, we get that v=1, which is what we would expect. Even changing the separation Δr will yield different speeds. Conclusion: light doesn't globally travel at c. Only locally does it do so.

So what's the point, may I ask?

I can't make heads or tails of this.

Not really. LET, if anything, has more to do with the M-M experiment than previous models given that it was invented for the purpose of explaining the experiment's null result.

That's exactly why SR took over - basic Occam's razor. SR doesn't prove aether doesn't exist, it just makes it unnecessary.

Aether as originally conceived has indeed been disproven. Undetectable aether as described in LET has not been disproven , and is not falsifiable. The OP asked whether or not "aether" has been disproven, not whether "aether as originally conceived" has been disproven. Aether itself (as per the OP) has not been disproven.

First of all, you realize that Newtonian gravity is only a low-velocity weak-field approximation to general relativity, right? The main issue, however, is that you provide absolutely no quantitative resolution to the problem. It's not enough to say in words that you solved something, you need to provide numbers.

I do not dispute that theories may be falsifiable. The existence of an ill-defined concept like aether, however, is not. We have not disproven the existence of aether for the reason that one can always introduce new phenomenon to explain why motion with respect to it is undetectable. This is exactly what Lorentz did when he invented the concepts of "local time" and "length contraction." No, it doesn't. You seem to be confusing earlier ether models with LET. LET gives the Lorentz transformation as the way of transforming between reference frames for the explicit purpose of explaining the null result of the M&M experiment. In their predictions LET and SR are completely equivalent. The difference comes from the interpretation of the Lorentz transformation. Bear in mind, I definitely do not advocate LET as a better alternative to SR.

Has it been disproved? Of course not. You can't prove that invisible magic pixies don't permeate all of space either. You can never prove that something doesn't exist. What relativity does is make aether unnecessary, unlikely, and ugly. This doesn't mean it doesn't exist - for example, Lorentz ether theory is physically equivalent to special relativity. LET has been abandoned, not because it makes bad predictions (it makes the same predictions as SR), but because it's not nearly as elegant, it requires more assumptions, and it does not easily generalize to something that resembles general relativity.

I forgot to ask before, why do you think that energy flux would contribute to the energy density?

I don't see it. Is there something obvious I'm missing? [math]B=|\bold{B}|[/math] if that's what you're worried about.

You can't sum with [math]\delta_{ii}[/math] in a valid tensor equation. An index can appear no more than twice in any term, or else it becomes unclear which index you're summing over. Your notation is far more ambiguous. For example, how would you express the i0'th component squared, but not summed over (i is a free index)? The what convention? You sacrifice clarity for brevity. This topic has become sufficiently derailed at this point, so I won't be responding (in this thread) to any more points about notation. Feel free to PM me if you'd like to continue this conversation. Anyway, the SET for an electromagnetic field is: [math]T^{\mu\nu} = \frac{1}{\mu_0} \left[ F^{\mu \alpha}F^\nu_{\ \alpha} - \frac{1}{4} \eta^{\mu\nu}F_{\alpha\beta} F^{\alpha\beta}\right]=\left [\begin{matrix} \frac{1}{2}\left(\epsilon_0 E^2+\frac{1}{\mu_0}B^2\right) & S_x/c & S_y/c & S_z/c \\S_x/c & -\sigma_{xx} & -\sigma_{xy} & -\sigma_{xz} \\S_y/c & -\sigma_{yx} & -\sigma_{yy} & -\sigma_{yz} \\S_z/c & -\sigma_{zx} & -\sigma_{zy} & -\sigma_{zz} \end{matrix} \right ][/math] where F is the electromagnetic tensor, S is the Poynting vector, and: [math]\sigma_{ij} = \epsilon_0 E_i E_j + \frac{1}{{\mu _0 }}B_i B_j - \frac{1}{2} \left( \epsilon_0 E^2 + \frac{1}{\mu _0}B^2 \right)\delta _{ij}[/math] The energy density is given by T00 (no other components contribute to the energy density - they represent energy and momentum flux), which in this case is: [math]T^{00}=\frac{1}{2}\left(\epsilon_0 E^2+\frac{1}{\mu_0}B^2\right)[/math] Setting E=0 (it's a purely magnetic field) and dividing by c2 to get it in units of mass density gives: [math]\rho \equiv T^{00}/c^2=\frac{1}{2\mu_0 c^2}B^2=\frac{1}{2}\epsilon_0 B^2[/math] which is what I had in my original post. Of course, should one choose to boost to another frame then things would get a lot messier and other components of T (from the original frame) would contribute.

You seem to have a misunderstanding about what the summation convention is. The summation convention implies a sum over a repeated upper+lower index. There is no summation implied by [math](T^{i0})^2[/math]. For example: [math]A_\mu A^\mu [/math] implies a summation. [math]A_\mu A^\nu [/math] does not imply summation [math]A^\mu [/math] are the components of a vector. It clearly does not imply summation. Likewise, [math](T^{i0})^2[/math] does not imply summation.