I How to obtain other inverse metrics than that of Schwarzschild?

[Bishal Banjara]

TL;DR Summary

Deriving $$ds^2=-(1-2GM/r)^-1dt^2+(1-2GM/r)dr^2+r^2d\omega^2$$

The Schwarzschild solution could simply be expressed as $$ds^2=-(1-2GM/r)dt^2+(1-2GM/r)^-1dr^2+r^2d\omega^2$$. Is it possible that we could obtained a new metric into the form as
$$ds^2=-(1-2GM/r)^-1dt^2+(1-2GM/r)dr^2+r^2d\omega^2$$? If possible, what are the steps and procedures that should be done to derive it in formal way?

 

[Ibix]

Use two $ signs, not one, for paragraph LaTeX here, and two # signs for inline maths.

You've just written down the metric you want. Plug it into the Einstein Field Equations and see what stress-energy tensor you need to get it. The End. (Except that the required stress-energy will likely be physically implausible.)

 

[Bishal Banjara]

Use two $ signs, not one, for paragraph LaTeX here, and two # signs for inline maths.

You've just written down the metric you want. Plug it into the Einstein Field Equations and see what stress-energy tensor you need to get it. The End. (Except that the required stress-energy will likely be physically implausible.)

This metric have significant impact. Let me freed up with this point you just mentioned. I mean let not go through stress-energy tensor. I am in need to derive that metric only. The consequences will be tackled/cleared later by myself.

 

[Ibix]

Normally you start with physical constraints (like "I want two stars orbiting each other"), write down some ansatz for the stress-energy tensor and try to derive the metric from that. That's how you derive a metric.

But you've said what metric you want. So there's nothing to derive except the stress energy tensor. Generate that, and you will see what (probably unlikely) physical circumstances lead to the metric you specified.

 

[Bishal Banjara]

Normally you start with physical constraints (like "I want two stars orbiting each other"), write down some ansatz for the stress-energy tensor and try to derive the metric from that. That's how you derive a metric.

But you've said what metric you want. So there's nothing to derive except the stress energy tensor. Generate that, and you will see what (probably unlikely) physical circumstances lead to the metric you specified.

Of course, you are right. In this case, I simply want the procedure to obtain the line element that I mentioned above in formal way. I tried it on conventional basis but the result always accords to Schwarzschild.

 

[PeterDonis]

I simply want the procedure to obtain the line element that I mentioned above in formal way.

What do you mean by "the formal way"?

 

[Bishal Banjara]

What do you mean by "the formal way"?

It means, not by coordinate transformation. I want it by the use of christoffels, ricci tensor components and so on..

 

[Ibix]

Are you under the impression that the new line element you have written down also represents Schwarzschild spacetime but in different coordinates? If so, you are mistaken. It manifestly has the same symmetries, but it is not a vacuum spacetime.

In order to derive it in a similar way to textbook derivations of the Schwarzschild metric you would need to know the stress-energy tensor. But unless you have some physical circumstances in mind the only way to get that is to put the metric through the Einstein Field Equations.

 

[Bishal Banjara]

Are you under the impression that the new line element you have written down also represents Schwarzschild spacetime but in different coordinates? If so, you are mistaken. It manifestly has the same symmetries, but it is not a vacuum spacetime.

In order to derive it in a similar way to textbook derivations of the Schwarzschild metric you would need to know the stress-energy tensor. But unless you have some physical circumstances in mind the only way to get that is to put the metric through the Einstein Field Equations.

No. I am expecting a new line element.

 

[Ibix]

This is making no sense.

Let's go back to basics. To derive anything you need to specify some starting assumptions. For example, to derive the Schwarzschild metric the usual route is to specify spherical symmetry, time independence, and a vacuum. What assumptions are you making?

 

[Bishal Banjara]

This is making no sense.

Let's go back to basics. To derive anything you need to specify some starting assumptions. For example, to derive the Schwarzschild metric the usual route is to specify spherical symmetry, time independence, and a vacuum. What assumptions are you making?

All those are expected excepted. If I am not wrong, could we start with $$ds^2=g^{\mu\nu} dx_\mu dx_\nu$$?

 

[Ibix]

All those are expected excepted.

As I've already pointed out, the spacetime described by the metric is not a vacuum. You cannot get to it if you initially demand a vacuum.

If I am not wrong, could we start with $$ds^2=g^{\mu\nu} dx_\mu dx_\nu$$?

That is very little better than saying "I intend to use maths to solve this problem".

You need to specify some constraints to be able to find a solution. You seem happy to assume spherical symmetry and time invariance. What do you want to assume about the stress energy tensor?

 

[Bishal Banjara]

As I've already pointed out, the spacetime described by the metric is not a vacuum. You cannot get to it if you initially demand a vacuum.

That is very little better than saying "I intend to use maths to solve this problem".

You need to specify some constraints to be able to find a solution. You seem happy to assume spherical symmetry and time invariance. What do you want to assume about the stress energy tensor?

ok, let it be non-vacuum. I want that line element to be derived only, right now. It can't be solved without stress energy tensor?

 

[PeterDonis]

I want it by the use of christoffels, ricci tensor components and so on..

You have to already know the metric to calculate those. You can't find a metric by using those.

It can't be solved without stress energy tensor?

As @Ibix has said, if you already know what metric you want, you dont have to "solve" anything. You just compute the Einstein tensor of your metric, multiply it by $8 \pi$, and that's the stress-energy tensor that produces that metric. However, there is no guarantee that this stress-energy tensor will represent anything that's physically possible or reasonable.

 

[Ibix]

ok, let it be non-vacuum. I want that line element to be derived only, right now. It can't be solved without stress energy tensor?

Correct.

Take a step back and think about what you are asking. Specifying the metric is specifying the gravitational field. If you have chosen a gravitational field the only question left is "what matter/energy distribution could produce this". Saying "I've got this gravitational field, how do I derive it" doesn't make sense - you can only ask "where does matter have to be to give me this gravitational field".

As Peter and I have both said, there's no guarantee that the stress-energy distribution you get from feeding an arbitrary metric through the Einstein Field Equations corresponds to any realistic scenario.

 

[Bishal Banjara]

You have to already know the metric to calculate those. You can't find a metric by using those.As @Ibix has said, if you already know what metric you want, you dont have to "solve" anything. You just compute the Einstein tensor of your metric, multiply it by $8 \pi$, and that's the stress-energy tensor that produces that metric. However, there is no guarantee that this stress-energy tensor will represent anything that's physically possible or reasonable.

"You just compute the Einstein tensor of your metric, multiply it by $8 \pi$, and that's the stress-energy tensor that produces that metric."

So far I have understood, we need Ricci tensors and curvature scalar inorder to compute the Einstein tensor. For Ricci tensors to be calculated we need Riemann tensors. And Riemann tensors are evaluated by Christoffel's. I am confused what actually did you mean. Lets take some simple reference. At pg. 231 Sean Carroll's book, the equations 5.139 and 5.140 could be obtained by following your instructions because, we already have calculated $G_tt$ and $G_rr$ there at equation 5.135. And what I am targeting is to obtain the $G_tt$ and $G_rr$ itself. (my expectation is obtaining $G_tt$ and $G_rr$ just in reverse way than that of equation 5.135)

 

[PeterDonis]

So far I have understood, we need Ricci tensors and curvature scalar inorder to compute the Einstein tensor.

Yes, and you compute the Ricci tensor and the Ricci scalar from the metric.

For Ricci tensors to be calculated we need Riemann tensors.

Yes.

And Riemann tensors are evaluated by Christoffel's.

Yes, and to calculate the Christoffel symbols, you need to know the metric.

what I am targeting is to obtain the $G_tt$ and $G_rr$ itself.

And to do that for the metric you gave in your OP, you would just calculate them: calculate the Christoffel symbols from the metric, calculate the Riemann tensor from the Christoffel symbols, calculate the Ricci tensor and Ricci scalar and Einstein tensor from the Riemann tensor. Then you know all the components of $G$ for the metric you gave in your OP. Then, as I said, you multiply the Einstein tensor by $8 \pi$ to get the stress-energy tensor that would produce the metric you gave.

At pg. 231 Sean Carroll's book, the equations 5.139 and 5.140 could be obtained by following your instructions because, we already have calculated $G_tt$ and $G_rr$ there at equation 5.135.

Yes, and equation 5.135 was calculated from the metric given in equation 5.133. But that that metric is not the Schwarzschild metric you gave in the OP of this thread; it is a general metric for any static, spherically symmetric spacetime, and has two unknown functions in it, $\alpha$ and $\beta$. So the Einstein tensor computed from it gives differential equations for those unknown functions. Then Carroll assumes that the stress-energy tensor is a perfect fluid, and writes down the two differential equations to be solved for the two unknown functions, equations 5.139 and 5.140. But he could only assume a form for the stress-energy tensor because the metric was not completely specified; there were two unknown functions in it. So his solution is not unique; it only applies to the case where the stress-energy tensor is in fact a perfect fluid. There could be other solutions if we made some other assumption about the stress-energy tensor.

But in the other (not Schwarzschild) metric you wrote down in the OP of this thread, the one you are asking questions about, there are no unknown functions; all of the metric coefficients are completely specified. So there is no freedom to make any assumptions about the stress-energy tensor; the Einstein tensor computed from this metric will not be a set of differential equations for unknown functions, it will just be a set of fixed expressions in terms of the constants that appear in the metric coefficients, and the coordinates. So the stress-energy tensor components can only be whatever those fixed expressions are, multiplied by $8 \pi$. There is no freedom of choice at all.

 

[Bishal Banjara]

Yes, and you compute the Ricci tensor and the Ricci scalar from the metric.Yes.Yes, and to calculate the Christoffel symbols, you need to know the metric.And to do that for the metric you gave in your OP, you would just calculate them: calculate the Christoffel symbols from the metric, calculate the Riemann tensor from the Christoffel symbols, calculate the Ricci tensor and Ricci scalar and Einstein tensor from the Riemann tensor. Then you know all the components of $G$ for the metric you gave in your OP. Then, as I said, you multiply the Einstein tensor by $8 \pi$ to get the stress-energy tensor that would produce the metric you gave.Yes, and equation 5.135 was calculated from the metric given in equation 5.133. But that that metric is not the Schwarzschild metric you gave in the OP of this thread; it is a general metric for any static, spherically symmetric spacetime, and has two unknown functions in it, $\alpha$ and $\beta$. So the Einstein tensor computed from it gives differential equations for those unknown functions. Then Carroll assumes that the stress-energy tensor is a perfect fluid, and writes down the two differential equations to be solved for the two unknown functions, equations 5.139 and 5.140. But he could only assume a form for the stress-energy tensor because the metric was not completely specified; there were two unknown functions in it. So his solution is not unique; it only applies to the case where the stress-energy tensor is in fact a perfect fluid. There could be other solutions if we made some other assumption about the stress-energy tensor.

But in the other (not Schwarzschild) metric you wrote down in the OP of this thread, the one you are asking questions about, there are no unknown functions; all of the metric coefficients are completely specified. So there is no freedom to make any assumptions about the stress-energy tensor; the Einstein tensor computed from this metric will not be a set of differential equations for unknown functions, it will just be a set of fixed expressions in terms of the constants that appear in the metric coefficients, and the coordinates. So the stress-energy tensor components can only be whatever those fixed expressions are, multiplied by $8 \pi$. There is no freedom of choice at all.

ok I will do the exercise.

 

[Bishal Banjara]

Then, as I said, you multiply the Einstein tensor by 8π to get the stress-energy tensor that would produce the metric you gave.

As Einstein tensor component be $G_oo=8\pi T_oo$ then I have to multiply $8\pi$$ to stress energy tensor component $T_oo$ to get Einstein tensor $G_oo$ or multiply $8\pi$ to $G_oo$ to get Stress enenrgy tensor component $T_oo$?

 

[Ibix]

PeterDonis made a typo - $G_{ab}=8\pi T_{ab}$ is correct.

 

[PeterDonis]

PeterDonis made a typo - $G_{ab}=8\pi T_{ab}$ is correct.

Ah, yes, sorry, I put the factor of $8 \pi$ in the wrong place. If you know the Einstein tensor, you would divide by $8 \pi$ to get the stress-energy tensor. Sorry for the mixup on my part.

 

[Bishal Banjara]

I got an idea from the quora( from my private message with V. T. Tooth) that the Schwarzschild solution itself could be indexed up and could be written down into the new line element directly as what I am targeting. And he said that is not new line element, the other form Schwarzschild's itself. What about it?

 

[PeterDonis]

I got an idea from the quora( from my private message with V. T. Tooth) that the Schwarzschild solution itself could be indexed up and could be written down into the new line element directly as what I am targeting. And he said that is not new line element, the other form Schwarzschild's itself. What about it?

You could just exchange the $t$ and $r$ labels in the Schwarzschild metric (which some people suggest doing inside the horizon to emphasize the fact that the "radial" coordinate, which is called $r$ normally but which would be called $t$ after the switching) is timelike there. However, that would make the factor in the metric coefficients $(1 - 2M / t)$, not $(1 - 2M / r)$. But your other metric in the OP, even though it switches the $dt^2$ and $dr^2$ around, still has $(1 - 2M / r)$, so it's not the same as the re-indexed Schwarzschild metric.

 

[Bishal Banjara]

I asked...."But the calculation of Einstein tensor components upon direct substitution of new metrics by myself yield some finite value while it should be 0 inorder to incorporate the schwarzschild solution. It is because the new solution actually seems as the non vacuum solution. Am I wrong somewhere?"

He replies...."Get it please already: raising or lowering indices does *not* create a new metric. It is just a different representation of the same entity. If you were to write down a *covariant* metric with the components that are the inverses of the (covariant) Schwarzschild metric, then you are indeed writing down a new, weird metric that is not a vacuum solution of Einstein’s field equations. But this is entirely different from raising and lowering indices."
It means he is taking this as just raising and lowering index case.

 

[Bishal Banjara]

You could just exchange the $t$ and $r$ labels in the Schwarzschild metric (which some people suggest doing inside the horizon to emphasize the fact that the "radial" coordinate, which is called $r$ normally but which would be called $t$ after the switching) is timelike there. However, that would make the factor in the metric coefficients $(1 - 2M / t)$, not $(1 - 2M / r)$. But your other metric in the OP, even though it switches the $dt^2$ and $dr^2$ around, still has $(1 - 2M / r)$, so it's not the same as the re-indexed Schwarzschild metric.

I am aware of this.

 

[Ibix]

It means he is taking this as just raising and lowering index case.

You can certainly write the metric tensor with contravariant indices, $g^{ab}$ instead of $g_{ab}$. In the case of a diagonal metric the on-diagonal components of the contravariant version are $g^{aa}=1/g_{aa}$, where summation over repeated indices is not implied. But it would be wrong to simply substitute those raised-index terms into the line element, if that's what you did in the OP. That is a different metric and that is how we have been interpreting you.

So you need to decide what it is you want. If you just want to know the upper-index form of the metric then you simply write the lower-index metric as a matrix and find the inverse as described in linear algebra textbooks. If you want a completely different metric then you need to proceed as advised in this thread.

 

[vanhees71]

Let's recapitulate the basics: The covariant components of the metric are $g_{ab}$. The contravariant components by definition are given by the inverse of the corresponding matrix, i.e., you have
$$g_{ab} g^{bc}=\delta_a^c.$$
Since for the Schwarzschild metric
$$g_{ab}=\mathrm{diag}(1-r_S/r,-(1-r_S/r)^{-1},-r^2,-r^2 \sin^2 \vartheta))$$
here the inverse is very easy to find,
$$g^{ab}=\mathrm{diag}((1-r_S/r)^{-1},-(1-r_S/r),-1/r^2,-1/(r^2 \sin^2 \vartheta)).$$

 

[PeterDonis]

It means he is taking this as just raising and lowering index case.

But the inverse metric $g^{ab}$ is not a line element; you can't write it as $ds^2$. You wrote your alternate metric in the OP as $ds^2$. That indicates that it was intended to be an ordinary metric, not just the inverse metric corresponding to the Schwarzschild metric.

Also, the last two coefficients you wrote in your alternate metric in the OP are not the inverses of the Schwarzschild metric coefficients.

 

[vanhees71]

The line element is, of course, to be formulated with the covariant metric components,
$$\mathrm{d} s^2=g_{ab} \mathrm{d} q^a \mathrm{d} q^b.$$

 

[Ibix]

Worth noting that all line elements in the OP have an $r^2d\omega^2$ term, which is not consistent with the idea that the terms are $g^{ab}\mathrm{d}x^a\mathrm{d}x^b$ terms, even if that were a legal statement.

 

[PeterDonis]

Worth noting that all line elements in the OP have an $r^2d\omega^2$ term, which is not consistent with the idea that the terms are $g^{ab}\mathrm{d}x^a\mathrm{d}x^b$ terms, even if that were a legal statement.

I'm assuming that is just shorthand for the angular coordinate terms in a spherically symmetric spacetime in Schwarzschild coordinates (i.e., $r$ is the areal radius). It's more common to see the Omega capitalized in this usage, i.e., $r^2 d \Omega^2$.

 

[Ibix]

Sure, but if the OP were mistakenly writing $g^{ab}\mathrm{d}x^a\mathrm{d}x^b$ then the relevant term would have $1/r^2$. So it's either what we initially understood (a different metric) or it's a mess that includes both inverse metric and metric components in the line element.

 

[vanhees71]

I think, it's a mess ;-)).

 

[Bishal Banjara]

But the inverse metric $g^{ab}$ is not a line element; you can't write it as $ds^2$. You wrote your alternate metric in the OP as $ds^2$. That indicates that it was intended to be an ordinary metric, not just the inverse metric corresponding to the Schwarzschild metric.

Also, the last two coefficients you wrote in your alternate metric in the OP are not the inverses of the Schwarzschild metric coefficients.

He wrote previously...."By lowering the indices of dt and dr. So instead of $g_00dx^0dx^0$, how about $g^00dx_0dx_0=g^00(g_00dx^0)g_00dx^0=g^00g_00_g00dx^0dx^0=g_00dx^0dx^0? See how easy it is?
In the case of Schwarzschild, $g_00=1–2GM/c^2r$ and of course $g^00=1/g_00$ for a diagonal metric."

As well wrote..." Indexed notation helps. We can either write ds^2=g_μνdx^μdx^ν or we can write ds^2=g^μνdx_μdx_ν, but then we realize that dx^α=g^αβdx_β and of course g^αβg_βγ=δ^α_γ, so g^μνdx_μdx_ν=gμνdx^μdx^ν identically."

 

[PeterDonis]

He wrote previously....

Who wrote this?

 

[Bishal Banjara]

Who wrote this?

https://www.quora.com/profile/Viktor-T-Toth-1 see link....

 

[PeterDonis]

https://www.quora.com/profile/Viktor-T-Toth-1 see link....

Where does he write the things you quoted?

 

[Bishal Banjara]

Where does he write the things you quoted?

He wrote this in private message with me.
Also he writes" In other words, if you raise the indices of the metric, you must lower the indices of the coordinates. The summation indices must always balance upstairs and downstairs. Many introductory relativity textbooks skip over this and “simplify” the expressions, e.g., by not distinguishing between covariant and contravariant indices; by doing so, they do a disservice to readers/students."

 

[vanhees71]

I don't know any serious textbook, which does not carefully use co- (lower) and contravariant (upper) indices. The only exception is that in Euclidean field theory in special relativity or in old-fashioned books, where they use the "$\mathrm{i} c t$ convention" for real-time physics, where you have $\delta_{\mu \nu}$ instead of $\eta_{\mu \nu}$ anyway, and you don't need to distinguish between upper and lower indices (although using the Cartesian Ricci calculus the mathematical distinction between vector components/basis vectors and dual-vector components/co-basis vectors is lost).

 

[PeterDonis]

He wrote this in private message with me.

That's not a valid reference for a PF discussion.

if you raise the indices of the metric, you must lower the indices of the coordinates

This doesn't make sense. The coordinates themselves don't have indices.

The $dx^\mu$ that appear in the line element $ds^2$ are coordinate differentials. It doesn't really make sense to lower the indices on those either.

The metric can also be used to give the inner product of two vectors. In that case it makes sense to lower the indices on the vectors (to get the corresponding covectors), as long as you also raise the indices on the metric (to get the inverse metric). The inner product remains invariant under these operations. But in the OP of this thread you didn't write the inner product of vectors, you wrote line elements. Those have coordinate differentials in them, not vectors.

 

[Bishal Banjara]

The inner product remains invariant under these operations. But in the OP of this thread you didn't write the inner product of vectors, you wrote line elements. Those have coordinate differentials in them, not vectors

If I want to write the line element with all its metric coefficients just inverse than Schwarzschild line element, then is he right? I mean $$g_{\mu\nu} dx^\mu dx^\nu=g^{\mu\nu} dx_\mu dx_\nu$$.

 

[vanhees71]

Formally yes, but it doesn't make too much sense.

 

[PeterDonis]

If I want to write the line element with all its metric coefficients just inverse than Schwarzschild line element, then is he right? I mean $$g_{\mu\nu} dx^\mu dx^\nu=g^{\mu\nu} dx_\mu dx_\nu$$.

As @vanhees71 says, formally this is correct, but what do $dx_\mu$ and $dx_\nu$ even mean? As I said before, those are coordinate differentials, and it doesn't make sense to lower indexes on those.

 

[vanhees71]

It's simply defined by
$$\mathrm{d} x_{\mu} =g_{\mu \nu} \mathrm{d} x^{\nu}.$$
The $\mathrm{d} x^{\mu}$ are vector components with respect to the holonomous basis $\partial_{\mu}$, and the $\mathrm{d} x_{\mu}$ are co-vector components with respect to the corresponding dual basis, which is $\mathrm{d} x^{\mu}$ itself.

 

[PeterDonis]

It's simply defined by
$$\mathrm{d} x_{\mu} =g_{\mu \nu} \mathrm{d} x^{\nu}.$$
The $\mathrm{d} x^{\mu}$ are vector components with respect to the holonomous basis $\partial_{\mu}$, and the $\mathrm{d} x_{\mu}$ are co-vector components with respect to the corresponding dual basis, which is $\mathrm{d} x^{\mu}$ itself.

I understand how it's defined mathematically. But you yourself said it doesn't make too much sense. Are you now saying it does?

 

[vanhees71]

Of course, you know this, and I still don't see, why I should rewrite it with contravariant components. Some quantities are "naturally described" by contravariant components, as the $\mathrm{d} x^{\mu}$, others by covariant components. E.g., the metric is "naturally described" by covariant components $g_{\mu \nu}$, because after all it maps to vectors bilinearly to a scalar and as such is a 2nd-rank tensor (field).

 

[PeterDonis]

I still don't see, why I should rewrite it with contravariant components.

Neither do I. I'm not the one asking about that, the OP is.

 

[PeterDonis]

If I want to write the line element with all its metric coefficients just inverse than Schwarzschild line element

As we've said, you can do this mathematically but it doesn't make much sense. However, as we have already pointed out, this is not what you did in the OP. The alternate metric you wrote in the OP is not what you get if you take the inverse metric of the standard Schwarzschild metric and pretend you can write it as a line element.

 

[Bishal Banjara]

He writes.....In terms of geometry, $dx^μ$ is an infinitesimal vector in the $x^μ$ direction. I think of $dx^μ$ as a corresponding line, surface, volume, etc. element (depending on the dimensionality of the manifold) to which $dx^μ$ is normal. The inner product, $dx_μdx^μ=g_μνdx^νdx^μ=ds^2$ is just the invariant line element. (Indeed, that is the formal definition of a covector like $dx_μ$: a covector is what maps a vector into a scalar.) This should tell you that $dx^μ$ and $dx_μ$ carry the same information in different form, just as I can specify a direction and a magnitude in 3D space using either an arrow (a vector) or a surface element of a given size facing a given direction. Two different representations, but the same information content, same degrees of freedom.

 

[PeterDonis]

He writes

Unless you can give a published, valid source for these quotes, they are out of bounds here.