elfmotat

Daniel, it has become quite obvious that you'd rather argue your crackpot ideas than actually learn something. We've all been trying very hard to clear up your incredibly numerous misconceptions, but your overwhelming response has been to either treat facts like opinions or ignore them completely when it suits your cause. I took the time to write you a detailed post, and your response was to ignore large chunks and provide me with a link to a crackpot's forum post which you probably didn't even read yourself. If you had, you would have realized that the OP was torn to shreds by responses to the topic. I don't see what good it would do to continue trying to make you see reason. You're wasting my and everyone else's time which could be better spent on people who are actually willing to learn.

No it doesn't. What exactly is your intention for this thread?

No. They called it a black hole because a black hole is the only known object that could be so small and yet that massive. Anything that dense has to be a black hole. Buchdahl's theorem proves that for a stable, uncollapsed spherical object of mass M and radius R, [math]GM/c^2R \leq 4/9[/math]. What this means is that if you were to construct a star of radius [math]R=9GM/4c^2[/math] and you gave it a little inward push, it would have no choice but to collapse into a black hole. Resisting collapse would require infinite internal pressure. Yes, as I just explained. It depends on what you mean by "density." It's more customary to compare the mass and the radius. A black hole of mass M has a radius [math]R=2GM/c^2[/math]. What do you mean by "limited information?" We know a great deal about the centers of galaxies. We can determine the central black hole's mass from the way nearby stars orbit it. No. People assume they exist because they do exist! If you think otherwise then you're going to have to explain away mountains of observational evidence. Scientists don't "want" anything but the truth. You're the one who seems to be letting bias get in the way of understanding how nature works. You're on the internet. How could you possibly claim that any information is inaccessible? These are called "experiments." We humans have also conducted many experiments and compiled tons of observational evidence that General Relativity is a reliable theory as well. You might not have been personally involved with the experiments, but you can easily find the details and data online. I really don't know what you mean by this. An example would help. That's called being closed-minded. Around 240 BC Eratosthenes knew the Earth must be round, and he even estimated its circumference which he calculated using shadows from the Sun at various locations (Greece and Egypt). He was correct to a good margin of error. Where would we be if we only accepted the evidence directly in front of our eyes? We'd probably still think the Earth was flat. The world of "practical science" already has discovered black holes. The evidence is free and readily available to you. You don't accept it because you have some strange bias against them.

You use the Stokes and Gauss theorems to switch back and forth between the integral and differential forms of Maxwell's equations. Stokes also pops up a lot (the 4-d version of it) when varying Lagrange densities, so you can use it to eliminate boundary terms. The usual 3-d Stokes and Gauss theorems actually fall out of the generalized Stokes theorem, i.e. they're both consequences of a more general theorem.

OP, do you have a question? Or a point? It certainly appears so.

Exactly! Unless you have some reference point to compare your motion to you won't be able to tell that you're "accelerating" due to gravity. If you were placed in a room where you were weightless, you wouldn't be able to tell whether you were free-falling towards a massive body or just floating around in empty space. There's no local test you can perform which can distinguish the two! It's not until you start resisting gravity that you can actually measure your "acceleration." I.e. if you're standing on the Earth an accelerometer will read "9.8 m/s2." If you jump off a cliff, the accelerometer will read zero during your fall.

Your crackpot ideas have no place in the science section.

You don't seem to understand how a scientific theory works. Using Special and General Relativity, we can predict to high accuracy exactly how much time dilation should occur, and experiments agree wonderfully with our predictions. We're not simply looking at a clock that was moving, seeing that it read off a slightly different amount of time than a stationary clock and then calling the difference "time dilation." We're looking at a clock that was moving and seeing that the time it read off is the time difference predicted by relativity. Do you think this is just a happy coincidence? You're going to have a tough time explaining away over a century of experimental data that confirms relativity's predictions. Time dilation is real, whether it sits well with you or not. You're trying to change the data to fit your theory instead of the reverse. You should learn to accept that nature doesn't always work the way you want it to.

Maybe not kind, but certainly appropriate. The post I was responding to was: You literally just made up nonsense.

You just defined [math]L_{\mu \nu}\equiv G_{\mu \nu}[/math]. Now you're saying they're equal only when [math]\Omega=0[/math]. Which is it?

They're the same because all you did was change the symbol for the Einstein tensor to L and the symbol for the CC to omega. That doesn't change anything.

You can't use GR to describe quantum phenomena nor to predict the value of the cosmological constant. Since the Ricci scalar is just a contraction of the Ricci tensor, if the Ricci tensor is dynamic then so is the Ricci scalar: [math]\frac{dR}{dt}=\frac{d}{dt}[g_{\mu \nu}R^{\mu \nu}][/math] It doesn't. The Einstein tensor is just [math]R^{\mu \nu}-\frac{1}{2}g_{\mu \nu }R[/math]. The CC comes from considering the following action for gravity: [math]S=\int \left [ \frac{1}{2\kappa}(R-2\Lambda) +\mathcal{L}_{M} \right ]\sqrt{-g}d^4x[/math] where [math]\mathcal{L}_{M}[/math] is the Lagrangian density for the matter distribution in spacetime, and [math]g=det(g_{\mu \nu})[/math]. Varying this action W.R.T the metric yields the field equations. But that's exactly the same thing as general relativity. Those are exactly the same field equations; all you did was change the symbols. If you want a new classical metric theory of gravity, you need to consider: Scalar-tensor or scalar-vector-tensor theories Additional spacial dimensions Higher-order terms in the action Examples of the first would be Brans-Dicke Theory and MOG. Brans-Dicke in particular is a popular competitor with GR. In these theories, additional scalar and vector fields contribute to gravitational interaction. Another type of variation on gravity is the addition of extra spacial dimensions. Theories of this kind are called Kaluza-Klein theories, and have some interesting consequences. For example, (4+1)-D KK theory unifies classical electromagnetism with GR - its field equations separate into the Einstein field equations and Maxwell's equations. If you consider powers, derivatives, different contractions, etc. of the curvature in the action then you can come up with infinitely many different theories, each with different field equations. Einstein-Hilbert action (which GR is derived from) is the simplest possible metric gravity theory because the action is just: [math]S=\int \alpha R \sqrt{-g}d^4x[/math] where [math]\alpha[/math] is some constant. A simple example of a higher-order gravity theory would be: [math]S=\int \left [ \alpha_{1} R+\alpha_{2} R^2 \right ]\sqrt{-g}d^4x[/math] Indeed, you can come up with infinitely many theories this way: where [math].~.~.[/math] represents the infinitely many other scalars we can form from the curvature tensor. This type of GR alternative is interesting because with good choice of higher-order terms, you can render gravity renormalizable which is good for quantization. Unfortunately, these theories also generally feature ghosts.

What is "condensed space?" I've never heard of anything remotely like this before.

Ah, I see. Well, the properties of the cosmological constant are derived mostly from the FLRW metric which is an exact solution to the field equations which describes the expanding universe: [math]ds^2=-dt^2+a(t)^2d\Sigma^2[/math] where [math]d\Sigma^2=\frac{dr^2}{1-kr^2}+r^2(d\theta ^2+sin^2\theta d\phi ^2)[/math], a(t) is a function called the "scale factor" of space, and k is a measure of the overall curvature (shape) of the universe. From observations, we know that the universe is probably flat (k=0). If we assume the stress-energy tensor is approximately a perfect fluid, from the field equations we get: [math]{\dot \rho} = - 3 \frac{\dot a}{a}\left(\rho+\frac{p}{c^{2}}\right)[/math] [math]\frac{\ddot a}{a} = - \frac{4\pi G}{3}\left(\rho + \frac{3p}{c^{2}}\right) + \frac{\Lambda c^{2}}{3}[/math] As you can see from the second equation, the pressure and energy density cause a deceleration in the expansion of the universe whereas the cosmological constant causes an acceleration in the expansion of the universe. We see that if we assume the cosmological constant term to be part of the energy density and pressure, we can eliminate the CC term from the above equations: [math]\rho_{\Lambda}=\frac{\Lambda }{8\pi G}[/math] [math]p_{\Lambda}=-\frac{\Lambda }{8\pi G}[/math] [math]\rho =\rho_T+\rho_{\Lambda }[/math] [math]p =p_T+p_{\Lambda }[/math] where [math]\rho_T[/math] and [math]p_T[/math] are the energy density and pressure contributed by the stress-energy tensor. So, we can see that the cosmological constant can be treated as a form of energy density (what you could call "dark energy") with negative pressure: [math]p_\Lambda =-\rho _\Lambda [/math]

I said "time is what's measured by clocks," not "clocks cause time to exist." You're being silly. Why would your state of consciousness have any affect on the rate at which clocks run?

Because "relative velocity" is defined as the velocity of one object in the rest frame of another, and there's no such thing as a rest frame for a photon. That's just the way it is, despite your objections. As timo was explaining, the separation velocity between two parallel photons is 2c. That's the only kind of velocity you can define between photons. "QA - No. QB - No." ???

Though your image is confusing I know exactly what you're talking about, and I don't see the problem. Antiderivatives, like differential equations, yield a family of solutions. The specific solution that fits the physical situation is determined by boundary conditions. I'm not quite sure what about this you find problematic. I'm not sure what the Poisson equation has to do with scalar/vector field definitions. First of all, I don't see what your point is. Second, k1 is a component, not a magnitude. Magnitudes are invariant under diffeomorphisms because the components change in a precise way.

Time is what's measured by clocks as far as physics is concerned. Questions like "is time real or a manmade concept" are untestable philosophical questions. There's no way to test which is correct.

Yes, I'm well aware that photons don't have a rest frame. That's why everyone has been saying that that relative velocity of one photon with respect to another is undefined.

You're using a confusing definition of "relative velocity." What you're describing is separation velocity. "Relative velocity" is traditionally defined as the the speed of one object from the rest frame of another.

I can't make heads nor tails of any of your pictures. I have no idea what they're supposed to represent. Again, from what I could tell, you seem to have a problem with gauge invariance. It's a very simple concept really: the derivative of a constant is zero. So taking the derivative of a function is equivalent to taking the derivative of a function plus a constant. In other words, the function f'(x) is invariant under the gauge transformation f(x)→f(x)+C. There's nothing magical, inconsistent, or otherwise here. That's simply not true. A scalar field can have infinite degrees of freedom. It can have a different value at every point in space. The key idea is that if you plug in a point xu the scalar function (we'll call f) returns a single number f(xu)=k1. If you were to plug in some other point, say yu, then you would get another number f(yu)=k2. It is not necessarily the case that k1=k2. An example of a scalar field is a temperature field, where each point in space has a different temperature. In order to have a rank-1 tensor (vector or one-form) field, when you plug a point xu into a vector av, you get n numbers back where n is the dimension of the space. In three dimensions for example: av(xu)=(k1,k2,k3). In order to have a rank-2 tensor field, when you plug xu into the field you get back n2 numbers. I.e. in 3D: bav(xu) = ((k11,k12,k13), (k21,k22,k23), (k31,k32,k33)). A rank-2 tensor field can't magically turn into a scalar field; that makes literally no sense. What I mean is that, without changing the form of the equation to something that resembles Gauss' Law, the Poisson equation is not a vector equation. It states that the Laplacian of some scalar function is equal to some other function. No vectors here.

No, you really aren't. You're being extremely vague in your mission to write out your "theory" without using math. You realize that physics is a quantitative science, right? We need numerical predictions to test whether or not your "theory" agrees with experiment. That means numbers! And where are we going to get those numbers? If your "theory" has no equations then we can't get them from anywhere! So your "theory" isn't really a theory at all - it's just a jumble of ill-defined concepts.

A vector is defined as something that transforms like a vector: [math]a^{A}=\frac{\partial x^A}{\partial x^a}a^a[/math] Of course adding or subtracting a constant vector is going to make a difference. What gave you the idea that it didn't? For example: They are both constant vectors, but the decision of whether to add or subtract them makes a huge difference. In what way could you possibly consider the above vectors equivalent? There are no vectors in the Poisson equation to begin with, so I'm not sure what you're talking about there. As far a Gauss' theorem goes, that's just a gauge invariance. Are you equally troubled by the fact that in classical mechanics the potential [math]V[/math] yields the the same equations of motion as the potential [math]V+C[/math]? What about in classical electrodynamics, where the magnetic vector potential and electric scalar potential yield the same equations of motion under [math]\mathbf{A}\rightarrow \mathbf{A}+\nabla f[/math] and [math]\phi \rightarrow \phi -\frac{\partial f}{\partial t}[/math] ? Or the fact a Lagrangian [math]L[/math] produces the same equations of motion as a function of that Lagrangian: [math]f(L)[/math]? I could go on, but I think you get the idea. If you really have a problem with gauge invariance then you have a lot of work ahead of you because you're going to have to rewrite the entirety of mathematics and physics. As long as the cosmological constant is set to zero, yes. No. All of the components are zero, not just the diagonal ones: [math]G_{\mu \nu}=0[/math] I'm starting to think you have no idea what the field equations are, nor how they work. By making the metric flat you already assumed that the cosmological constant was zero! As I already showed you in a previous post, if the CC is nonzero then the vacuum field equations reduce to [math]R_{\mu \nu}=g_{\mu \nu}\Lambda [/math]. Even in a vacuum the spacetime is curved, so the metric is obviously not Minkowski. You can't have it both ways: either the metric is flat and the CC is zero, or the CC is nonzero and you have a curved metric. Those are the field equations, not a definition of the Einstein tensor. I don't even know what that means. How can a rank-2 tensor "become" a scalar without contractions? That Einstein quote has absolutely nothing to do with the cosmological constant. His footnote refers to: "It must be pointed out that there is only a minimum of arbitrariness in the choice of these equations. For besides Guv there is no tensor of second rank which is formed from the guv and its derivatives, contains no derivations higher than second, and is linear in these derivatives." What his footnote means is that his above description is technically valid for any tensor [math]R_{\mu \nu}+\lambda Rg_{\mu \nu}[/math] for arbitrary constant [math]\lambda[/math]. The reason [math]\lambda=-1/2[/math] is chosen is because of the Bianchi identities. If any other value for [math]\lambda[/math] is used then the equation [math]\nabla_{\nu}G^{\mu \nu}[/math] wouldn't be valid, and so neither would [math]\nabla_{\nu}T^{\mu \nu}[/math] and then we wouldn't have energy/momentum conservation. You did, when you made the metric Minkowski. It does by definition. The Einstein tensor is a measure of curvature. If spacetime is flat then there is, by definition, no curvature. Again, I have no idea what that means. I'm not really sure what your responding to here. I was simply saying that the field equations when the metric is defined as Minkowski are not useful or interesting.

That was a good read. It's unfortunate A-wal did't bother to look at it.

How could it possibly change the physics? The CC term is simply absorbed into a new tensor. [math]"G_{\mu \nu}"=G_{\mu \nu }+g_{\mu \nu}\Lambda =\kappa T_{\mu \nu}[/math] There's nothing special here, it's just simple algebra. What are you talking about? The definition of a vector in the context of differential geometry is standard. There's absolutely nothing wrong with the derivation of GR - all you have to do is vary the Hilbert action with respect to the metric. Obviously GR is incomplete because it doesn't get along with QFT, but that doesn't mean there's anything "wrong" with it. It's a mathematically consistent theory. Can you show me in detail how it does? Because I have no idea what you're talking about. How could a tensor "violate the summation properties of tensors?" I don't even know what that means. Again, this is just basic algebra. The Einstein tensor is defined as [math]G_{\mu \nu }\equiv R_{\mu \nu}-\frac{1}{2}g_{\mu \nu}R[/math]. So we see that the only way for your equation could possibly be correct is if: The cosmological constant is set to zero. [math]G_{\mu \nu }[/math] is something other than the Einstein tensor. The "G" tensor defined by MTW would work here, so long as the sign of the CC is reversed. It wasn't, and I don't know why it would need to be. This is just math here; there's nothing to refute. I'm also unsure how it's relevant. Um, no. It's not an opinion, you're needlessly restricting the equations to the point where they're useless. Since you made the metric flat, you've already defined all of the information you could possibly extract from the field equations. With a flat metric, the curvature (Riemann, Ricci, and Einstein tensors) is identically zero, the stress-energy tensor is zero, and the cosmological constant has to be zero (which is physically wrong anyway). Nothing about this is interesting or useful.