The discussion centers around the validity of the Schwarzschild metric inside a black hole, contrasting it with its established evidence outside of massive bodies like the Sun. Participants also explore the implications of black holes formed from photons, including their mass properties and the conditions required for their formation.
Participants express a range of views on the applicability of the Schwarzschild metric inside black holes, with no consensus reached. The discussion remains unresolved regarding the implications of forming black holes from photons and the associated mass properties.
Limitations include the theoretical nature of the Schwarzschild metric's application inside black holes, the dependence on definitions of mass in the context of photons, and the unresolved status of experimental verification of these concepts.
Are you proposing an alternative?kurious said:
No.
kurious said:
kurious said:
There is no evidence of this. Nobody has ever been inside a event horizon to tell us if its correct. And if he was there then we'd have to be inside the event horizon too. While we might find out the answer, we'd pay for it with our lives. Frankly, I, personally am not that anxious to find out.kurious said:
Interesting question. If there were a system of photons which had a zero total momentum then it would be possible for a black hole to form. All that is required is that the photons be located within the black hole. Since the photons have mass then its possible for a black hole to form. This mass would then be the M in the Schwarzschild metric. But this M would be the E/c2 as measured in the zero momentum frame. Some people like to call that "rest mass".
At what fixed point in space? Why would a black hole form?kurious said:
You're speaking of commuinication faster than light using entangled states. There is some controversy about whether this is possible or not. People claim to have done it in fact. But I've seen this discussed in the physics literature. I have no opinion on this otherwise since I haven't studied it in detail.
kurious said:
Its not restmass, its just mass. Restmass is a bad word. And, it is the stress-energy tensor that acts as the source term in Einstein's field equations.marlon said:
Yes. Misner, Thorne and Wheeler call it mass-energy, or sometimes they call it just "mass", as Wald does at times (at least one place in his text that I know of) but its all the same thing. So long as you pay close attention to how I described it above.marlon said:
I don't recall ever saying that a black hole could move at the speed of light.
This might be relevant... no time to check now. No idea what they could be made of, but remember that GR is a classical theory.Ivan Seeking said:
marlon said:
Relativistic mass is not another name for energy. Proportionality of two physical quantities does not mean that the two physical quantities have the same physical meaning. They are defined differently. In fact in relativistic electrodynanics, the term energy is used to describe the sum E = Kinetic Energy + Rest Energy + Potential energy = K + E0 + V = mc2 + V. The quantity T = mc2 is inertial energy and the quantity E is "energy" or "total energy". It would be a mistake to confuse E with T. E is proportional to the time component of the generalized 4-momentum while T is proportional to the regular 4-momentum.pervect said:
No. There is no place for relativistic mass in modern relativity. It is the stress-energy tensor that is the source term in Einstein's field equation.marlon said:
This part is not an opinion, it is a incorrect statement. There is a huge difference between identity and equality. When you claim that it is an identitiy, as you are, you're claiming that the term "relativistic mass" is a name which was created to mean the exact same thing as "energy", I.e. a synonym. That is incorrect. Relativistic mass is defined in one way (as the m such that mv is a conserved quantity) and inertial energy, T (which is the notation used in some portions of Goldstein 3rd Ed.), is defined in another way, i.e. T = energy in the absense of potential energy, V. This is sometimes called free-particle energy (e.g. Jackson 2nd Ed.). It is then proven that T = mc2.pervect said:
That view confuses the physical quantities which generate the gravitational field with the mathematical quantity which describes them. But to each his own. But if that is your view then it would be inconsistent to say that charge generates an EM field. So you might want to avoid that in the future should the subject arise. To be consistent with your view then you'd have to say that 4-current generates a EM field. Note: There can be g-fields in the absesnce of matter (e.g. gravitational radiation) and there can be EM fields in the absence of charge (e.g. cosmic background radiation).
In GR when someone speaks of spacetime curvature they are speaking of the non-vanishing of the Riemann tensorHaelfix said:
That doesn't make much sense to me. What do you mean "everything is in the choice of connection"? Any geometry can be described by the metric as well as the affine connection. Affine geometry and metric geometry are two ways of detecting curvature. Curvature is a description of properties of a manifold. It is no more manmade than any other quantity in mathematics.
This is a common misconception. Gaussian curvature describes excess radius for a closed path in higher dimensional space just as Riemannian curvature describes the variation in a vector parrallel transported around a closed path in a higher dimensional spacetime. In both cases the path taken can be described as a displacement along a 2d surface, but in both cases the surface is in a higher dimensional imbedding. Your argument is just like claiming that Riemannian spacetime curvature only applies to a 2 dimensional spacetime which is rediculous.pmb_phy said: