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Obviously you are more interested in persisting in your errors than learning the answer.

First off, it is called "dilation", not "dialation". Second off, GR gives you all the components, including the effects due to both linear and rotational motion. Third off, both your t(1-Rs/r)^(-1/2) and t(1-v^2/c^2)^0.5 formulas are incorrect. If you do not understand the subject, just ask and I will explain.

You do not need to "combine both GR and SR effects", GR gives the COMPLETE prediction , all by itself. It falls out straight from the Schwarzschild solution.

See here

Detailed solution is given in R.C. Tolman "Relativity, Thermodynamics and Cosmology", p. 245.

I think that you and I are saying the same exact thing. I never said that the solutions cover the object itself, I said that exactly the opposite is true: the interior solutions are NOT vacuum solutions, the EXTERIOR ones , by contrast, ARE vacuum solutions. They all describe the gravitational field in the vacuum AROUND the respective OBJECT, OUTSIDE the object proper. One doesn't "cut out" the object lest the effects of the object charge, mass, momentum are nullified, one SEPARATES the external solution from the internal one. Maybe it is just an issue of terminology but one needs to be precise. The problem is one of semantics, I am trying to explain to ajb that one MUST have different sources PRESENT (charged, uncharged, rotating and non-rotating) in order to have interesting solutions to the EFEs. In ADDITION, contrary to his claims, they are ALL called.... VACUUM solutions. This is because they all describe thee vacuum surrounding the object. The vacuum solutions do NOT reduce to the ONLY ONE, the MOST TRIVIAL ONE, the one that has total absence of sources. This reduces to the flat spacetime solution, the Minkowski metric, totally uninteresting (and not the only vacuum solution known to physicists).

I have taken in what you wrote, you don't know what you are talking about. You may want to sleep on it. Alternatively, you may want to click on the links I provided you with. BTW, as credentials go, I have many more papers published than you do. I wasn't attacking your credentials, I was just pointing out that you have serious misconceptions about GR, that's all <shrug>.

Sure they do, otherwise you are stuck with your totally trivial solution, no gravitational sources, no gravitational field : Schwarzschild vacuum (which describes the spacetime geometry around a non-charged, non-rotating spherical mass), Kerr vacuum (which describes the geometry around a rotating spherical mass) Reissner-Nordstrom vacuum (which describes the spacetime geometry around a charged spherical mass), Kerr-Newman vacuum (which describes the spacetime geometry around a rotating charged spherical mass) You seem very unfamiliar with GR, have you taken any classes on this subject?

If you "cut out" the object, then you get no gravitational field. I take it that you are a mathematician, not a physicist. That also explains your confusion about the interior solutions: of course that the interior solutions cannot be vacuum solutions, the interior is obviously not a vacuum. Anyways, you are splitting hairs and you are splitting them badly. You must have missed my answer to timo, check out : Schwarzschild vacuum Kerr vacuum etc, etc.

I already explained : Schwarzschild vacuum solution Kerr vacuum solution Kerr-Newman vacuum solution Riesner-Nordstrom vacuum solution They are all vacuum solutions, they describe the gravitational field in the vacuum neighboring an interesting object, like an uncharged, non-rotating sphere (Schwarzschild), charged non-rotating sphere (Riessner), etc. This is basic GR, I do not understand why you have so much difficulty understanding. Huh? None of the so-called "interior solutions" claims to be a "vacuum solution", so your objection totally misses the point.

I understand the confusion: I cited the interesting cases, you two are citing the trivial case. The term "vacuum solution" has multiple connotations, I prefer the interesting one.

Not very interesting. Actually not interesting at all.

Let me illustrate with some examples: Schwarzschild vacuum (which describes the spacetime geometry around a spherical mass), Kerr vacuum (which describes the geometry around a rotating spherical mass) Reissner-Nordstrom vacuum (which describes the spacetime geometry around a charged spherical mass), Kerr-Newman vacuum (which describes the spacetime geometry around a rotating charged spherical mass) So, there are objects, more precisely there is one very specific object in the region, which is exactly what I said.

This is pretty much a rehash of what you posted here

Yep, there is a professor that puts forward the fuel problem to his students EVERY year and, like clockwork, the answer comes back: "Nope. Not possible". It is an issue of fuel, indeed. Actually, starting and stopping to take fuel is a lot worse. Absent any resistance, once you achieve cruising speed you let inertia take over, the ship should maintain speed w/o using any more fuel. The problem is that ramping up to 0.999...c takes too much fuel by itself.

absence of any gravitational bodies bar one.

Nonsense, you are confusing a matrix with its determinant.

E=total energy, p=momentum, you got the other 3 right.

Yes, I mentioned earlier that there are multiple valid explanations. I gave a kinematic one earlier in the thread. Here is your explanation, in mathematical terms: [math]E^2+(pc)^2=(mc^2)^2[/math] [math]p=\frac{mv}{\sqrt{1-(v/c)^2}}[/math] From the above two, one gets: [math]E^2+(\frac{mvc}{\sqrt{1-(v/c)^2}})^2=(mc^2)^2[/math] Explicitate [math]v[/math]: [math]v=c \sqrt{1-\frac{m^2c^4}{E^2}}<c[/math]

There are many proofs, here is one of them: In special relativity, the speed for accelerated motion is related to acceleration via the expression: [math]v=\frac{at}{\sqrt{1+(at/c)^2}}[/math] Simple algebra, shows that for any acceleration [math]a[/math] and for any [math]t[/math], no matter how large, [math]v<c[/math].

No, nothing can be "accelerated to the speed of light".

This is Zwicky's theory of "Tired Light". It has been proven incorrect.

Thank you for catching the error in the first proof.

Bah, I made two mistakes, not one, here is the correct derivation: [math](mc^2)^2=E^2-(pc)^2[/math] where : [math]E[/math] is the total energy [math]p=\frac{mv}{\sqrt{1-(v/c)^2}}[/math] is the impulse [math]v[/math] is the speed So: [math](mc^2)^2=E^2-(\frac{mv}{\sqrt{1-(v/c)^2}})^2[/math] This means: [math]v=c \sqrt{1-\frac{m^2c^4}{E^2}}<c[/math] Yes, the second derivation is much easier. Here is a third one: [math]p=\frac{mv}{\sqrt{1-(v/c)^2}}[/math] [math]v=\frac{c}{\sqrt{1+(m/p)^2}}<c[/math] This derivation has the added advantage to show that [math]v=c[/math] if and only if [math]m=0[/math], i.e. only massless particles can travel at c.

It is impossible for any object/particle having mass [math]m[/math] to travel at the light speed, let alone FTL. The explanation is simple: [math](mc^2)^2=E^2+(pc)^2[/math] where : [math]E[/math] is the total energy [math]p=mv[/math] is the impulse [math]v[/math] is the speed So: [math](mc^2)^2=E^2+(mvc)^2[/math] This means: [math]v=c \sqrt{1-\frac{m^2c^4}{E^2}}<c[/math] Another way to prove the above is by starting from: [math]E=\frac{mc^2}{\sqrt{1-(v/c)^2}}[/math] resulting into: [math]v=c \sqrt{1-\frac{m^2c^4}{E^2}}<c[/math]