I Why do strong fields complicate our understanding of singularities?

[iuvalclejan]

Why doesn't this work if the field is strong? Or does it work as long as there are no singularities?

Mentor's Note: Original thread title was, "Calculating rest mass by integrating T_{00} over a 3 volume for static metric"

 

[PeterDonis]

Why doesn't this work

Because it's the wrong integral. Look up "Komar mass".

 

[pervect]

Why doesn't this work if the field is strong? Or does it work as long as there are no singularities?

Given a coordinate system (or, alternatively, a congruence of worldlines, if you are familiar with that concept), you'll get a number this way. But as $T_{00}$ is not a tensor, but just a piece of a tensor, there is no guarantee that you will get a number independent of your choice of coordinate, as the expression does not follow the tensor rules for generating an invariant quantity.

I don't have any more detailed specific illustrative worked out (such as specific coordinate choices giving different results) however.

I do have another relevant example, originating in MTW's text, "Gravitation". I don't have the specific page/section, it'd take some work to dig it out. My wording is probably not the same as MTW's and is probably less rigorous, be warned.

Consider a solid planet, that you cut up into smaller blocks. Then you pull all the blocks to infinity, dissassembling the planet. The result you find is that the mass of the dissassembled planet is different from the mass of the blocks. It takes work from an external source to dissassemble the planet - thus, the dissasembled planet has more mass than the assembled planet. The difference between these two masses is the Newtonian "gravitational binding energy" of the planet, the energy it would pull the planet apart and transport its pieces to infinity.

This illustrates that in the Newtonian limit, the idea of integrating $T_00$ is different from the idea of finding the system mass, because the system mass includes the gravitational binding energy, while the integral of $T^00$ does not. Again, one is motivated to find some tensor expression for the so-called "gravitational binding energy". Unfortunately, in general, there isn't one. I don't recall the details, but I do recall that MTW has a section on why you can't localize the gravitational binding energy as a tensor. MTW also disccusses the "pseudo-tensor" approach to mass in General Relativity.

Concepts of mass that do exist in GR require additional restrictions, such as the asymptotic flatness required for the ADM mass, and the stationary (or static) geometry for the Komar mass. As I recall, the pseudo-tensor approach also winds up requiring asymptotic flatness, but I could be wrong on this - my memory on this point is hazy.

Wald's text "General Relativity" has a detailed and highly mathematical workthrough for the Komar mass (though he doesn't use this name), and a brief discussion of the ADM mass. MTW covers the pseudo-tensor approach.

 

[iuvalclejan]

Because it's the wrong integral. Look up "Komar mass".

Thanks. Will study this tomorrow or the next day

 

[iuvalclejan]

Given a coordinate system (or, alternatively, a congruence of worldlines, if you are familiar with that concept), you'll get a number this way. But as $T_{00}$ is not a tensor, but just a piece of a tensor, there is no guarantee that you will get a number independent of your choice of coordinate, as the expression does not follow the tensor rules for generating an invariant quantity.

We don't need an invariant quantity, but the time component of a 4-vector, which that integral would be if we have the correct differential 1-form integrated over a 3-volume (with an appropriate Sqrt[-det[g]] in any coordinate system), and in a frame where there is no kinetic energy (all the T_{0j} are 0 for j=1-3, you're contracting T^{0j} j=0,3, with a 1-form, which includes the Levi-Civita tensor, maybe I should have T^{00} in the integral instead of T_{00}, minor detail).The problem seems to be with the singularity at r=0, see equation 23.20 on page 602 of my version of MTW.
As far as the issue with interpreting the resulting number as the rest-mass/energy in a 3-volume, yes, it includes some gravitational potential energy, as you point out. I am not so concerned about localizing that energy precisely (section20.4 of MTW), because, as they point out in their inimitable poetic style on page 603, box 23.1 , the argument of non-localizability "loses its bite" for a static spherically symmetric star.
*note: I edited/corrected some of my terminology. The Hodge duals of 3 forms are 1 forms.

 

[PeterDonis]

We don't need an invariant quantity

You do if you want the mass you calculate to be an invariant, which you should.

the time component of a 4-vector

Rest mass is not the time component of a 4-vector. Nor is it the integral of one.

equation 23.20 on page 602 of my version of MTW

Equation 23.20 is the exact mass integral for a spherically symmetric object, and assumes that you know the density $\rho$ as a function of areal radius $r$.

For a discussion of why this integral is correct, even though it apparently does not take into account either pressure or gravitational potential energy, see this Insights article:

https://www.physicsforums.com/insights/is-pressure-a-source-of-gravity/

That article also refers to other Insights articles where the Komar mass is discussed.

 

[PeterDonis]

The problem seems to be with the singularity at r=0

For a static, spherically symmetric object like a star, there is no singularity at $r = 0$. There is such a singularity in Schwarzschild spacetime, but Schwarzschild spacetime is a vacuum solution, with $T_{ab} = 0$ everywhere, and you cannot calculate its mass using the methods we are discussing in this thread.

 

[iuvalclejan]

As far as Komar mass, I'm not convinced it's something different from the T^{00} (not T_00) integral, except it does have hope for a non-zero finite value for a black hole where T00 can go to 0 at r=0, whereas Sqrt[-g_{00}] goes to infinity, so their product could be m.

 

[PeterDonis]

As far as Komar mass, I'm not convinced it's something different from the T^{00} (not T_00) integral

Have you got any actual math to back this up?

except it does have hope for a non-zero finite value for a black hole

Only if you rework the integral to be in terms of the Riemann tensor instead of the stress-energy tensor. Even then there are technical issues involved that I don't think you are grasping.

where T00 can go to 0 at r=0

The stress-energy tensor of a black hole is zero everywhere. It doesn't "go to 0 at r = 0".

whereas Sqrt[-g_{00}] goes to infinity

Calculating the Komar mass of Schwarzschild spacetime requires limiting the integral to the region outside the horizon, since that is the only region that is stationary (has a timelike Killing vector field), and the Komar mass is only defined for a stationary spacetime.

 

[PeterDonis]

I'm not convinced it's something different from the T^{00} (not T_00) integral

The Komar mass is an invariant. The $T^{00}$ (or $T_{00}$) integral is not.

 

[iuvalclejan]

Having trouble replying, it quotes the wrong message, so I'll do it by hand
"You do if you want the mass you calculate to be an invariant, which you should."
You're right. I'm interested in the mass energy inside a volume, which is more than just the rest mass/energy. So NOT invariant, but transforms as time component of a 4-vector.
"Equation 23.20 is the exact mass integral for a spherically symmetric object, and assumes that you know the density � as a function of areal radius �."
Or equivalently, it assumes you know T_00 or G_00. I'll look up the article, but I think you mean eqn (1) on the next page, not equation 23.20. It DOES take the internal energy (pressure) and gravitational energy into account, at least accrding to eqn (3) on the same page of MTW.
"Have you got any actual math to back this up?" No, because as it became clear in my next sentence, I was wrong. My integral is the total mass energy inside a volume, whereas Komar mass is the rest mass.
"Only if you rework the integral to be in terms of the Riemann tensor instead of the stress-energy tensor." Even then there are technical issues involved that I don't think you are grasping." I think you mean Einstein tensor, not Riemann. It shouldn't matter as they are related by the field equation.
"The stress-energy tensor of a black hole is zero everywhere. It doesn't "go to 0 at r = 0". Yes it must go to infinity, if you think of a collapsing star. At some point, just before all the mass has collapsed to r=0, the stress energy tensor is NOT 0 close to r, it is almost infinite, like a delta function.
"Calculating the Komar mass of Schwarzschild spacetime requires limiting the integral to the region outside the horizon, since that is the only region that is stationary (has a timelike Killing vector field), and the Komar mass is only defined for a stationary spacetime." I didn't realize that a timelike killing vector is necessary, just that none of the metric components have an explicit time dependence. I will have to think about it.
"For a static, spherically symmetric object like a star, there is no singularity at �=0. There is such a singularity in Schwarzschild spacetime, but Schwarzschild spacetime is a vacuum solution, with ���=0 everywhere, and you cannot calculate its mass using the methods we are discussing in this thread."
Unless you see the black hole as a limiting case of a collapsing star?

 

[iuvalclejan]

The Komar mass is an invariant. The $T^{00}$ (or $T_{00}$) integral is not.

OK, sorry I reposted that. Something is not working in my browser. Yes I agree. The T^00 integral is the time component of a 4 vector, not invariant. See my other reply

 

[PeterDonis]

I'm interested in the mass energy inside a volume, which is more than just the rest mass/energy.

No, it is the externally measured mass. The externally measured mass of the object already includes all the "corrections" due to other internal stress-energy components being nonzero, "gravitational potential energy", etc.

So NOT invariant, but transforms as time component of a 4-vector.

As I have already said, the externally measured mass is an invariant.

An isolated object can be described by a 4-vector (at least if you ignore its internal structure and assume its size to be negligible on whatever distance scale you are interested in), but the externally measured mass of the object will be the norm of the 4-vector, not its time component.

Regarding the section of MTW you quote, you appear to be confused about what the various equations represent.

Equation 23.19 represents the correct invariant integral for the externally measured mass of an isolated, spherically symmetric object.

Equation 23.20 is not an equation for the mass at all; it is just an explanation of why $m(0)$ in equation 23.19 has to be zero.

Equation (1) on the next page of MTW (inside Box 23.1) is the same as equation 23.19. Equation (3) in that box describes a possible split of the total mass, which is given by Equation (1), into three pieces. Equation (1) is not an equation for any of the three pieces by themselves; it is the equation for the total mass $m(r)$ on the LHS of Equation (3). The rest of Box 23.1 describes how the split described in Equation (3) is done. But none of that has any bearing at all on what we are discussing in this thread; we are discussing the total externally measured mass, $m(r)$, given by Equation (1) of Box 23.1, not any of the pieces into which it can be split.

Unless you see the black hole as a limiting case of a collapsing star?

No. A collapsing star is not stationary, and as I have already said, the Komar mass only applies to a stationary spacetime.

There are other mass integrals that you could apply to a collapsing star; the obvious one would be the ADM mass (which can also be applied to Schwarzschild spacetime). But those integrals do not involve the stress-energy tensor; they only look at the spacetime curvature as you approach infinity. So they don't give a way of calculating the mass by integrating over any stress-energy tensor.

 

[PeterDonis]

At some point, just before all the mass has collapsed to r=0, the stress energy tensor is NOT 0 close to r, it is almost infinite, like a delta function.

This is true of the interior of a collapsing star, but as I noted in post #13 just now, a collapsing star is not stationary, and the Komar mass only applies to a stationary spacetime.

 

[vanhees71]

We don't need an invariant quantity, but the time component of a 4-vector, which that integral would be if we have the correct differential 1-form integrated over a 3-volume (with an appropriate Sqrt[-det[g]] in any coordinate system), and in a frame where there is no kinetic energy (all the T_{0j} are 0 for j=1-3, you're contracting T^{0j} j=0,3, with a 1-form, which includes the Levi-Civita tensor, maybe I should have T^{00} in the integral instead of T_{00}, minor detail).The problem seems to be with the singularity at r=0, see equation 23.20 on page 602 of my version of MTW.
As far as the issue with interpreting the resulting number as the rest-mass/energy in a 3-volume, yes, it includes some gravitational potential energy, as you point out. I am not so concerned about localizing that energy precisely (section20.4 of MTW), because, as they point out in their inimitable poetic style on page 603, box 23.1 , the argument of non-localizability "loses its bite" for a static spherically symmetric star.
*note: I edited/corrected some of my terminology. The Hodge duals of 3 forms are 1 forms.

Lets go one step back and consider an analogous situation in classical electrodynamics within SRT (flat spacetime). There you have a Lagrangian for the electromagnetic field coupled to charged matter (for simplicity you can assume simply "dust matter" or an "ideal fluid" as a model). The you use Noether's theorem together with gauge invariance to get the energy-momentum tensor of this closed (!!!) system. Then this energy-momentum tensor fulfills in standard Lorentzian coordinates,
$$\partial_{\mu} T^{\mu \nu}=0,$$
which implies the equations of motion for the matter leading to the (relativistic version of) the Lorentz force. Then you can show that for total energy and momentum you can define the total energy-momentum vector by naively integrating over "entire space" at a fixed time coordinate $t=\text{const}$ and indeed get a conserved four-vector.

However, if this continuity equation is not fulfilled for some energy-momentum tensor, e.g., if you consider the electromagnetic part of the energy-momentum tensor only the corresponding integral does not define a four-vector. It's even not a priori clear, what you call the electromagnetic-field and the matter energy-momentum tensor, because you can shuffle parts back and forth between the two. E.g., in a dieelctric you can shuffle back and forth parts of the four-current to the fields in terms of polarization and vice versa. The physics, of course, always stays the same, and it indeed included in the total energy and momentum of the closed system.

In GR nothing of this holds, because there the continuity equation translates into
$$\nabla_{\mu} T^{\mu \nu}=0$$
in arbitrary coordinates, and the integration over "entire space" in some arbitrarily chosen "foliation of space time" does not define a total energy-momentum four-vector. So it's impossible to even define a generally covariant (i.e., gauge invariant!) total energy and momentum of the closed system of matter/radiation + gravitational field in any way. This was clear from the very beginning in 1915 when Einstein and the Göttingen group around Hilbert were discussing the theory, leading to the parallel discovery of the Einstein fielf equations in November 1915 (and the Einstein-Hilbert action) by Einstein and Hilbert. What's a bit less well known is that already in 1915 also Emmy Noether was involved in these debates, and that's how her famous paper about the relation between symmetries and conservation laws came out, ingeneously solving the problem completely, including the important difference between what we today call "global symmetries", which are true physical symmetries and "local gauge symmetries", which are merely a redundancy in description of physical entities rather than true physical symmetries. Unfortunately these subtleties have long been overlooked and had to be rediscovered much later again in the context of gauge theories in high-energy physics and (special) relativistic quantum field theory.

 

[iuvalclejan]

Peter Donnis: After reading the article you suggested, I am even more confused than before, because it claims that equation 4.19 (or (1) in box 1) of MTW for the mass energy inside a radius r IS the same as the Komar mass, so my original hunch was right? But this can't be, because m(r), at least according to the math I showed above (sorry, not latex formatted) should be the time component of a 4-vector, when T^{ij} can be diagonalized in some frame where T^{00} is just \rho, whereas the Komar mass should be an invariant, at least as far as I could tell from the Wikipedia article. Actually, the 3-volume integral (a surface of constant t in some coordinate system in 4 space) I have is not quite the same as equation 23.19, because it has the \Sqrt[-det[g]], which is missing in eqn 23.19, but is just r^2 Sin[theta] for a Schwarzchild metric, which gives the 4 Pi in equation 23.19 once the angular variables are integrated over. But in general there is a difference, and as you insist (and I'm almost convinced), we should not try to apply any of these to the S. metric.
As far as not integrating inside a horizon where the Killing vector in the time direction is spacelike instead of timelike, I didn't see where in the derivation of the Komar mass integral, the assumption that the Killing vector has to be timelike is used, but I will take it on faith for now (that it is not sufficient to have a Killing vector in the time direction, but that it also has to have a negative squared norm). Once a horizon forms, you are saying that the integral will still tell us the mass/energy, as long as the region of integration is only outside the horizon? It seems to me that if a horizon forms then the integral will not tell us what we want, even if we throw away the part inside the horizon (obvious in the case of a S. metric where T=0).

 

[PeterDonis]

equation 4.19 (or (1) in box 1) of MTW for the mass energy inside a radius r IS the same as the Komar mass

I think you mean equation 23.19. Yes, that integral is numerically the same as the Komar mass, although its form is different. The article discusses that.

m(r), at least according to the math I showed above (sorry, not latex formatted) should be the time component of a 4-vector

NO IT IS NOT. You have already been told repeatedly that this is wrong. If you are going to keep repeating wrong statements even after they have been corrected, discussion with you is pointless and this thread will simply be closed. The Komar mass, as you note, is an invariant, corresponding to the norm of the energy-momentum 4-vector describing an isolated object (when such a description is even meaningful). The function $m(r)$, or "mass inside areal radius $r$", only applies to a static, spherically symmetric object and is also an invariant; it can be thought of as what the Komar mass invariant would be if the object's surface were at areal radius $r$.

the 3-volume integral (a surface of constant t in some coordinate system in 4 space) I have

I don't know what integral you are referring to here, but if it's one you came up with yourself, I strongly suggest that you forget about it and focus on what is in actual textbooks.

I didn't see where in the derivation of the Komar mass integral, the assumption that the Killing vector has to be timelike is used

The definition of Komar mass requires that the spacetime is stationary. That means it has a timelike Killing vector field. I don't think MTW discusses the rationale for that in any detail; Wald might.

it is not sufficient to have a Killing vector in the time direction, but that it also has to have a negative squared norm

I don't know what you think the difference is between "a Killing vector in the time direction" and a timelike (i.e., negative squared norm) Killing vector. (Hint: there isn't one.)

Once a horizon forms

There is no "once a horizon forms". We are talking about a stationary spacetime; it's the same at all times. As you have already been told, a spacetime with a collapsing star in it is not stationary and the Komar mass is not even well-defined for it. Neither is the "mass function" $m(r)$, since, while a collapsing star can be spherically symmetric (as in the 1939 Oppenheimer Snyder model), it is not static (or stationary).

 

[iuvalclejan]

vanhees71, thanks for that reply. I admit I don't understand it, so I hope you will be patient in continuing to try to explain it to me.
1. Why must the (3-surface) integration of T^{ij} d\Sigma_j be over all of space, for fixed time, where \Sigma is the 3-surface? Why can't it be over a region of space enclosed by radius r and fixed time (in some coordinate system)?
2. To get a 4-vector, I am not integrating the covariant divergence of T^{ij} (which would give 0), but T{ij} itself, contracted with the one form that represents the 3-surface above. This gives a 4-vector valued one form, integrated over a 3-surface.
3. I don't understand why your comments about local gauge symmetry in GR are relevant

 

[PeterDonis]

To get a 4-vector, I am not integrating the covariant divergence of T^{ij} (which would give 0), but T{ij} itself, contracted with the one form that represents the 3-surface above. This gives a 4-vector valued one form, integrated over a 3-surface.

Properly defined, such an integral can give you the energy-momentum 4-vector for an isolated object. But then, as I have already pointed out, the object's rest mass will be the norm of this 4-vector, not its "time component".

 

[iuvalclejan]

Properly defined, such an integral can give you the energy-momentum 4-vector for an isolated object. But then, as I have already pointed out, the object's rest mass will be the norm of this 4-vector, not its "time component".

Right, we agree. So I am interested in the time component of this 4-vector, not its norm. It's different than the Komar mass (or equivalently the mass defined by 23.19) because of the \Sqrt[-det[g]] instead of just the Sqrt[gtt] factor, though it looks the same for a S. metric as eqn 23.19, but we agree we shouldn't use any of these formulas for a S. metric.
Please don't get upset and try to be patient, I am reading what your write, and trying to understand it, but I won't be intimidated to just accept something if I don't understand it. From your last post, it appears that at last YOU understand which integral I was talking about, and we can therefore hopefully forget about your previous statement:"If you are going to keep repeating wrong statements even after they have been corrected, discussion with you is pointless and this thread will simply be closed. " You can see now that it was a correct statement? If you get frustrated at my alleged stupidity, please don't close the post, maybe others can have more patience.
And anyway, I am not just trying to replicate old results, but come up with new ones (I hope to share these in due time). It would be nice if we did have a way to calculate the mass-energy inside a 3-volume (the time component of the mass-energy 4-vector), or the invariant mass-energy, even when we have singularities, and even when the Killing vector in the time direction for a static metric (meaning no explicit time dependence in metric) is no longer timelike (I mean that its squared norm is not negative anymore, but positive, to answer your question in your previous response. Isn't that what happens inside an event horizon?).

 

[PeterDonis]

I am interested in the time component of this 4-vector, not its norm

Why?

 

[PeterDonis]

I am not just trying to replicate old results, but come up with new ones

Please be aware that PF is not for doing personal research, which is what you describe here.

 

[iuvalclejan]

Why?

I can't tell you because PF is not for original research. So far we have only been talking about known physics.... If you want to know why, maybe we should take this offline?

 

[PeterDonis]

the Killing vector in the time direction

What do you mean by "the time direction"? (Hint: the only meaning I can assign to this is "in a timelike direction", meaning that there is no difference between "in the time direction" and "timelike".)

for a static metric (meaning no explicit time dependence in metric)

That's not quite what "static" means. "Static" means that there is a timelike Killing vector field that is hypersurface orthogonal. Given a static (or even just a stationary) spacetime, it is always possible to choose coordinates such that there is a "time" coordinate on which none of the metric coefficients depend; but it is also possible to have coordinates that do not have that property.

is no longer timelike (I mean that its squared norm is not negative anymore, but positive, to answer your question in your previous response. Isn't that what happens inside an event horizon?).

The Killing vector field in Schwarzschild spacetime that you refer to is timelike (negative squared norm) outside the horizon, null (zero squared norm) on the horizon, and spacelike (positive squared norm) inside the horizon. With the only valid meaning I can assign to the phrase "in the time direction", this Killing vector field would only be "in the time direction" outside the horizon, where it's timelike.

 

[PeterDonis]

I can't tell you because PF is not for original research.

Ah, ok. Then discussion of that aspect is off limits here.

 

[iuvalclejan]

That's not quite what "static" means. "Static" means that there is a timelike Killing vector field that is hypersurface orthogonal. Given a static (or even just a stationary) spacetime, it is always possible to choose coordinates such that there is a "time" coordinate on which none of the metric coefficients depend; but it is also possible to have coordinates that do not have that property.

OK, though I still have to take the timelike (in the sense of negative squared norm, not in the sense of in the direction of time in some coordinate system) part on faith. Maybe I need to get Wald. Which hypersurface are you referring to?

The Killing vector field in Schwarzschild spacetime that you refer to is timelike (negative squared norm) outside the horizon, null (zero squared norm) on the horizon, and spacelike (positive squared norm) inside the horizon. With the only valid meaning I can assign to the phrase "in the time direction", this Killing vector field would only be "in the time direction" outside the horizon, where it's timelike.

No disagreement on that. We were quibbling about the killing vector inside the horizon, where even though it points in the time direction, it is spacelike.

 

[PeterDonis]

I still have to take the timelike (in the sense of negative squared norm, not in the sense of in the direction of time in some coordinate system) part on faith

No, you don't. Calculating the norm of the Killing vector field in question in the three regions I described is straightforward as soon as you pick a coordinate chart that isn't singular at the horizon.

Which hypersurface are you referring to?

"Hypersurface orthogonal" means the spacetime (or at least the stationary region of the spacetime, which in this case is the region outside the horizon) can be foliated by spacelike hypersurfaces that are everywhere orthogonal to the integral curves of the Killing vector field. For the case in question, these hypersurfaces are the hypersurfaces of constant Schwarzschild $t$ coordinate.

the killing vector inside the horizon, where even though it points in the time direction

I have asked you several times now what you mean by "in the time direction". You can't just help yourself to that phrase without explaining what you mean by it.

 

[iuvalclejan]

No, you don't. Calculating the norm of the Killing vector field in question in the three regions I described is straightforward as soon as you pick a coordinate chart that isn't singular at the horizon.

So what happens if the norm is positive? What if there were no singularities but the space were of Euclidean signature (with no time dependence in metric, in some coordinate system)? I don't understand why the norm has to be negative. It's not about the ability to calculate the norm.

"Hypersurface orthogonal" means the spacetime (or at least the stationary region of the spacetime, which in this case is the region outside the horizon) can be foliated by spacelike hypersurfaces that are everywhere orthogonal to the integral curves of the Killing vector field. For the case in question, these hypersurfaces are the hypersurfaces of constant Schwarzschild $t$ coordinate.I have asked you several times now what you mean by "in the time direction". You can't just help yourself to that phrase without explaining what you mean by it.

It means orthogonal to a hypersurface of constant TIME (t in Schwarzchild coordinates).

 

[PeterDonis]

So what happens if the norm is positive?

Then the Killing vector field is spacelike, not timelike (assuming you mean the squared norm).

What if there were no singularities but the space were of Euclidean signature (with no time dependence in metric, in some coordinate system)?

Spacetime always has Lorentzian signature, not Euclidean signature. I don't know what you're talking about here.

I don't understand why the norm has to be negative.

The squared norm of a Killing vector field does not have to be negative. I have never claimed that it did. Indeed, I gave you regions of Schwarzschild spacetime in which the squared norm of the Killing vector field in question is zero or positive.

But in order for a spacetime to be stationary, it has to have a Killing vector field that is timelike, i.e., negative squared norm. And a spacetime has to be stationary for the Komar mass to be well-defined. That is by definition. Again, if you want details on the rationale for the definition, you're going to need to consult GR textbooks where the Komar mass and its rationale is discussed in detail. (Or you could look up Komar's original paper, I suppose.)

 

[PeterDonis]

It means orthogonal to a hypersurface of constant TIME (t in Schwarzchild coordinates).

The fact that a coordinate happens to be labeled $t$ does not mean it is a valid time coordinate. To be a time coordinate, the coordinate has to be timelike. The Schwarzschild $t$ is only timelike outside the horizon.

Also, there is no requirement that a time coordinate have integral curves that are hypersurface orthogonal. For example, in Kerr spacetime, the $t$ coordinate is timelike outside the ergosphere; but there is no set of hypersurfaces that are everywhere orthogonal to integral curves of $t$, as there are in Schwarzschild spacetime.

 

[iuvalclejan]

Spacetime always has Lorentzian signature, not Euclidean signature. I don't know what you're talking about here.

How do you know this for sure? Maybe there are regions that are Euclidean and small and haven't been discovered. Certainly Euclidean spacetime is used as a tool for path integration in QFT. The point of the question is to see why is the necessity for timelike Killing vectors in either the Komar mass, or the 4-vector gotten by integrating T over a 3-surface of constant time and radius less than some r. The Euclidean spacetime is a reductio ad absurdum to show that stationarity does not require a negative square norm Killing vector.

The squared norm of a Killing vector field does not have to be negative. I have never claimed that it did. Indeed, I gave you regions of Schwarzschild spacetime in which the squared norm of the Killing vector field in question is zero or positive.

But in order for a spacetime to be stationary, it has to have a Killing vector field that is timelike, i.e., negative squared norm. And a spacetime has to be stationary for the Komar mass to be well-defined. That is by definition. Again, if you want details on the rationale for the definition, you're going to need to consult GR textbooks where the Komar mass and its rationale is discussed in detail. (Or you could look up Komar's original paper, I suppose.)

OK, you say this quite dogmatically. I don't see it at all. Stationary seems like it just requires a Killing vector in the direction we call time. It doesn't have to have negative squared norm as far as I could tell. Or maybe it does for the Komar integral, but not for the other integral of T.

 

[PeterDonis]

How do you know this for sure?

Because spacetime includes spacelike, null, and timelike intervals. You need a Lorentzian signature for that.

Euclidean spacetime is used as a tool for path integration in QFT

As an intermediate mathematical tool, yes. But before extracting any physical observables, you always have to Wick rotate back to Lorentzian spacetime.

you say this quite dogmatically

No, I give the standard definition of a stationary spacetime that is in every GR textbook. If you don't like that standard definition, then you need to go convince all the textbook authors to change it. Good luck.

 

[PeterDonis]

Thread closed for moderation.

 

[Drakkith]

How do you know this for sure? Maybe there are regions that are Euclidean and small and haven't been discovered.

We don't know anything for sure. Everything Peter has said comes from a theory and various models that have proven extremely useful and accurate and have been verified with the best observations and experiments we can come up with. If there is some region of the universe that they don't apply, it isn't here and it isn't anywhere we've looked so far, barring perhaps some extreme conditions such as inside black holes and near the time of the big bang.

If you wish to go beyond standard GR and its associated models, then I wish you the best of luck.

Thread will remain closed since this conversation appears to be inherently about things which aren't supported by mainstream science.