Inverting the metric coefficients in the Schwarzschild line element

[Bishal Banjara]

TL;DR Summary

I want to know if the metrics in Schwarzschild is inversed, then Einstein's tensor components get modified or not.

Assuming the line element $ds^s=e^{2\alpha}dt^2-e^{2\beta}dr^2-r^2{d\Omega}^2$as usual into the form $ds^s=e^{-2\alpha}dt^2-e^{-2\beta}dr^2-r^2{d\Omega}^2$, I found that the $G_{tt}$ tensor component of first expression do not reconcile with the second one though, it fits for $G_{rr}$component. I followed the way as it is guided in https://web.stanford.edu/~oas/SI/SRGR/notes/SchwarzschildSolution.pdf for all calculations of Christoffels, Ricci tensor components and Einstein's tensor components. But I applied the final result for both exterior and an interior solution treating full Einstein equation following Carroll's book at pp.231. I obtained $G_{tt}=\frac{1}{r^2}e^{2\beta-2\alpha}[2r\delta_r\beta+1-e^{-2\beta}]$ instead of $G_{tt}=\frac{1}{r^2}e^{-2\beta+2\alpha}[2r\delta_r\beta-1+e^{2\beta}].$I want to know whether I am doing wrong though metric is independent to the final result or it doesn't reconcile, naturally.

 

[PeroK]

Perhaps it's best to learn a bit of Latex:

https://www.physicsforums.com/help/latexhelp/

 

[Ibix]

As far as I can tell you've simply substituted $-\alpha$ and $-\beta$ for $\alpha$ and $\beta$ in your initial expression. That should simply feed through to the final expression as minus signs in front of the $\alpha$ and $\beta$. That you haven't got that implies to me that you've made an arithmetic slip somewhere.

 

[PeterDonis]

Assuming the line element $ds^s=e^{2\alpha}dt^2-e^{2\beta}dr^2-r^2{d\Omega}^2$as usual into the form $ds^s=e^{-2\alpha}dt^2-e^{-2\beta}dr^2-r^2{d\Omega}^2$

Why would you want to do that?

 

[PeterDonis]

Moderator's note: Changed thread level to "I" and edit thread title to be more descriptive of the OP question

 

[Bishal Banjara]

Why would you want to do that?

I am trying to derive the new metric coefficients for Schwarzschild line element.

 

[PeterDonis]

I am trying to derive the new metric coefficients for Schwarzschild line element.

What do you mean by "the new metric coefficients"? What are you trying to accomplish? You can't just arbitrarily change the metric coefficients to something else.

 

[Bishal Banjara]

What do you mean by "the new metric coefficients"? What are you trying to accomplish? You can't just arbitrarily change the metric coefficients to something else.

Nope, I am aware and following the proper method with all classical correspondences. If you could answer, tell me what is the difference between Einstein tensor $G^{\alpha\beta}$ and $G_{\alpha\beta}$, Reimann tensor $R^{\alpha\beta}$ and$R_{\alpha\beta}$, Stress-energy-momentum tensor$T^{\alpha\beta}$ and $T_{\alpha\beta}$? I found most of the textbooks use the contravariant type notations at first but use covariant type notations later for the case of Schwarzschild solution, for example, a first course in GR by Schutz.

 

[Bishal Banjara]

As far as I can tell you've simply substituted $-\alpha$ and $-\beta$ for $\alpha$ and $\beta$ in your initial expression. That should simply feed through to the final expression as minus signs in front of the $\alpha$ and $\beta$. That you haven't got that implies to me that you've made an arithmetic slip somewhere.

Yes, at apparent sense your logic is correct but I rechecked it enough, I still got the same expression. If you could answer, tell me what is the difference between Einstein tensor $G^{\alpha\beta}$ and $G_{\alpha\beta}$, Reimann tensor $R^{\alpha\beta}$ and$R_{\alpha\beta}$, Stress-emergy momentum tensor$T^{\alpha\beta}$ and $T_{\alpha\beta}$? I found most of the textbooks use the contravariant type notations at first but use covariant type notations later for the case of Schwarzschild solution, for example, a first course in GR by Schutz.

 

[PeterDonis]

I am aware and following the proper method with all classical correspondences

What "proper method" are you talking about?

If you could answer, tell me what is the difference between Einstein tensor $G^{\alpha\beta}$ and $G_{\alpha\beta}$, Reimann tensor $R^{\alpha\beta}$ and$R_{\alpha\beta}$, Stress-energy-momentum tensor$T^{\alpha\beta}$ and $T_{\alpha\beta}$?

Raising two indexes using the inverse metric. If the "proper method" you refer to is that you are trying to obtain the inverse metric $g^{\alpha \beta}$, you are doing it wrong; the result you have obtained is not the inverse metric $g^{\alpha \beta}$ corresponding to the Schwarzschild metric $g_{\alpha \beta}$.

 

[PeterDonis]

most of the textbooks use the contravariant type notations at first but use covariant type notations later for the case of Schwarzschild solution

If you are writing a metric as a line element $ds^2$, you have to use the covariant metric, i.e., the metric $g_{\alpha \beta}$ with two lower indexes. There is no such thing as a "line element" $ds^2$ using the inverse metric $g^{\alpha \beta}$ with two upper indexes.

 

[Ibix]

Yes, at apparent sense your logic is correct but I rechecked it enough, I still got the same expression.

Without seeing what you did, I can't comment.

If you could answer, tell me what is the difference between Einstein tensor $G^{\alpha\beta}$ and $G_{\alpha\beta}$, Reimann tensor $R^{\alpha\beta}$ and$R_{\alpha\beta}$, Stress-emergy momentum tensor$T^{\alpha\beta}$ and $T_{\alpha\beta}$? I found most of the textbooks use the contravariant type notations at first but use covariant type notations later for the case of Schwarzschild solution, for example, a first course in GR by Schutz.

As Peter says, you just raise and lower induces using the metric or inverse metric. That means that there isn't much difference - if an expression uses one version you can always sub in the expression for another. So you can use whichever form is most convenient for your particular application.

 

[Bishal Banjara]

What "proper method" are you talking about?
Raising two indexes using the inverse metric. If the "proper method" you refer to is that you are trying to obtain the inverse metric $g^{\alpha \beta}$, you are doing it wrong; the result you have obtained is not the inverse metric $g^{\alpha \beta}$ corresponding to the Schwarzschild metric $g_{\alpha \beta}$.

I think you miss understood me. I mentioned the problem relating the notations for Einstein's equation, not for the case of the metric tensor $g_{\alpha\beta}$ in Schwarzschild case. Ok, tell me what is the difference between $G^{\alpha\beta}=R^{\apha\beta}-1/2g^{\alpha\beta}R=kT^{\alpha\beta}$ and $G_{\alpha\beta}=R_{\apha\beta}-1/2g_{\alpha\beta}R=kT_{\alpha\beta}$? I would be very thankful if you again tell me whether $R^{\alpha\beta}$ is inverse of $R_{\alpha\beta}$? let this 'proper method' rely on me.

 

[PeterDonis]

I mentioned the problem relating the notations for Einstein's equation

I have no idea what problem you are talking about. Raising and lowering indexes on tensors is not a problem; it's a basic operation in tensor calculus.

Ok, tell me what is the difference between $G^{\alpha\beta}=R^{\alpha\beta}-1/2g^{\alpha\beta}R=kT^{\alpha\beta}$ and $G_{\alpha\beta}=R_{\apha\beta}-1/2g_{\alpha\beta}R=kT_{\alpha\beta}$?

Nothing except that the indexes are raised in one and lowered in the other. They are both perfectly valid expressions of the Einstein Field Equation; as @Ibix said, which one you use depends on your particular application.

I would be very thankful if you again tell me whether $R^{\alpha\beta}$ is inverse of $R_{\alpha\beta}$?

What do you mean by "inverse"?

let this 'proper method' rely on me.

I am not confident that you understand what you are doing, so I don't think it's a good idea to just rely on you.

 

[Ibix]

Also, just by the way, $R_{\alpha\beta}$ is the Ricci tensor. The Riemann tensor is $R^\alpha{}_{\beta\gamma\delta}$. $R_{\alpha\beta}=R^\mu{}_{\mu\alpha\beta}$.

 

[Bishal Banjara]

All these kinds of stuff I am discussing are all meant to reconcile the $G_{oo}$ (as we get, as usual without metric inversion in Schwarzschild solution) with the new form as I have mentioned in my post. I am intended to reconcile the calculation from both sides. Would you please give me some references on the issue I discussed just a moment ago? Also, tell me how to calculate the components of $R^{\alpha\beta}$. I am familiar with $R_{\alpha\beta}$.

 

[Bishal Banjara]

Nothing except that the indexes are raised in one and lowered in the other. They are both perfectly valid expressions of the Einstein Field Equation; as @Ibix said, which one you use depends on your particular application

Please write to me what it means by application? Don't they refer the same state?

 

[PeterDonis]

All these kinds of stuff I am discussing are all meant to reconcile the $G_{oo}$ (as we get, as usual without metric inversion in Schwarzschild solution) with the new form as I have mentioned in my post.

This makes no sense. If you cannot explain what you are trying to do, this thread will be closed.

I am intended to reconcile the calculation from both sides.

This makes no sense. Please explain what you are trying to do or this thread will be closed.

Would you please give me some references on the issue I discussed just a moment ago?

I do not understand what your issue is.

Also, tell me how to calculate the components of $R^{\alpha\beta}$. I am familiar with $R_{\alpha\beta}$.

If you do not already understand how to raise and lower indexes on tensors using the metric and inverse metric, then I am not surprised that nothing you are doing makes sense, since you lack a basic understanding of how tensors work. I suggest that you take the time to fix that first by learning basic tensor algebra from a textbook. Sean Carroll's online lecture notes on GR give a good introductory treatment:

https://arxiv.org/abs/gr-qc/9712019

 

[PeterDonis]

Please write to me what it means by application?

It just means that whether you use tensors with upper indexes or lower indexes will depend on what specific problem you are trying to solve. Both forms are physically equivalent.

 

[PeterDonis]

(as we get, as usual without metric inversion in Schwarzschild solution)

If you mean that what you are trying to do is to obtain the inverse metric for the Schwarzschild solution (in your chosen coordinates), then you are doing it wrong. I strongly suggest taking some time to learn basic differential geometry from a textbook. The lecture notes I linked to in an earlier post just now would be a good choice, since they give a good introductory treatment of differential geometry.

Note, btw, that the concept of "inverse" only makes sense for the metric tensor, not for tensors in general.

 

[pervect]

I second the recommendation to find a textbook. It appears you're struggling with the difference between vectors and covectors.

The metric tensor $g_{ab}$ is able to find the length of a vector $u^a$ by the tensor equation $g_{ab} u^a u^b$.

The metric tensor can lower the index of a vector to make it a covector as well. Co-vectors have a lower index, so while $u^a$ is a vector, $u_b$ is a covector.

The tensor equation to lower an index would be $g_b = g_{ab} u^a$.

vectors and covectors in tensor analysis are similar to column and row vectors in matrix algebra.

If you want to lower the index on a rank 2 tensor, you need to use the metric tensor twice. In tensor notation $R_{ab} = g_{ac} g_{bd} R^{cd}$

The inverse metric tensor satisfies the tensor equation $g^{ab} g_{bc} = \delta^a{}_c$

Graphically, vectors $u^a$ can be represented by little lines with arrows pointing in a given direction. co-vectors can be represented by stacks of plates. This approach may not be universal, but MTW, for instance, uses it. One needs two forms of vectors - row vectors and column vectors in matrix algebra, vectors and covectors in tensor notation - because they transform differently under a change of coordinates.

There is a natural paring between covectors and vectors that give the inner product.

In tensor notation $g_{ab} u^a v^b = (g_{ab} u^a v^b = u_b v^b$.

You'll also need the Einstein summation convetion for any of this to make sense.

Any rank 2 tensor of the form $T^{ab}$ can be regarded as a linear map from two covectors $u_a, u_b$ to a scalar, while any rank 2 tensor of the form $T_{ab}$ can be regarded as a lienar map from two vectors to a scalar.

The existence of the metric tensor is what allows one to raise and lower indices. Because indices can be raised and lowerd freely, no distinction is made between upper and lower indices. When you calculate a scalar quantity though, a tensor expression must have the same number of upper and lower indices.

 

[Bishal Banjara]

As we get Einstein's tensor component$G_{tt}=\frac{1}{r^2}e^{2(\alpha-\beta)}[2r\delta_{r}\beta-1+e^{2\beta}]$ for Schwarzschild solution $ds^2=-e^{2\alpha}dt^2+e^{2\beta}dr^2+r^2{d\Omega}^2$. Please, write me what we will get for $G^{oo}$ for the same Schwarzschild solution (if I am not wrong) with Einstein's equation as $G^{oo}=R^{oo}-1/2g^{oo}R=kT^{oo}$ in detail (it might be easier if you refer some references for proper calculations). If it solves, my problem will be solved.

 

[Ibix]

$G^{ab}=g^{ai}g^{bj}G_{ij}$. So $G^{00}=g^{0i}g^{0j}G_{ij}$. Since the metric is diagonal, this simplifies to $G^{00}=g^{00}g^{00}G_{00}$. Again because the metric is diagonal this is equal to $(g_{00})^{-2}G_{00}$.

 

[Bishal Banjara]

$G^{ab}=g^{ai}g^{bj}G_{ij}$. So $G^{00}=g^{0i}g^{0j}G_{ij}$. Since the metric is diagonal, this simplifies to $G^{00}=g^{00}g^{00}G_{00}$. Again because the metric is diagonal this is equal to $(g_{00})^{-2}G_{00}$.

Excellent, thank you!

 

[Bishal Banjara]

Excellent, thank you!

This means this property equally holds for $R_{oo}$, $T_{oo}$ as well, isn't it?

 

[PeterDonis]

This means this property equally holds for $R_{oo}$, $T_{oo}$ as well, isn't it?

What @Ibix did was raise the two indexes on the Einstein tensor. You can do the same operation with any tensor.

You gave this thread a thread level of "A" when you started it. That means you should have a graduate level background in the subject matter. Anyone with that background would already know how to raise and lower indexes on tensors. Even someone with an "I" level (undergraduate) background would know that, since tensor algebra is typically taught at the undergraduate level now.

I strongly suggest that you take the time to learn this subject from a textbook.

 

[Bishal Banjara]

What @Ibix did was raise the two indexes on the Einstein tensor. You can do the same operation with any tensor.

You gave this thread a thread level of "A" when you started it. That means you should have a graduate level background in the subject matter. Anyone with that background would already know how to raise and lower indexes on tensors. Even someone with an "I" level (undergraduate) background would know that, since tensor algebra is typically taught at the undergraduate level now.

I strongly suggest that you take the time to learn this subject from a textbook.

Thank you for your suggestion, it is not true that I am unaware of lowering and raising indexes but I was a bit confused with a similar appearance of both form of Einstein's equation without the extra terms from ${g_{oo}}^2$. Now, I got the point that these extra terms get canceled from both sides of full Einstein's equation for Schwarzschild's case. Hence, there is no difference in either form of covariant or contravariant tensor notation.

 

[PeterDonis]

it is not true that I am unaware of lowering and raising indexes

I am not sure you understand the implications of it, though. See below.

I was a bit confused with a similar appearance of both form of Einstein's equation without the extra terms from ${g_{oo}}^2$.

That's because such extra terms are never there; for any tensor equation, the "lower indexes" form and the "raised indexes" form will look the same. When you raise or lower an index on any tensor equation, the "extra terms" will always cancel.

there is no difference in either form of covariant or contravariant tensor notation.

As above, that is true for any tensor equation.

 

[Bishal Banjara]

I am not sure you understand the implications of it, though. See below.
That's because such extra terms are never there; for any tensor equation, the "lower indexes" form and the "raised indexes" form will look the same. When you raise or lower an index on any tensor equation, the "extra terms" will always cancel.
As above, that is true for any tensor equation.

Ok, I am convinced but still, there is no concluding remark on my original question. From my post, "I want to know whether I am doing wrong though metric is independent to the final result or it doesn't reconcile, naturally", what should we be concluded?

 

[Bishal Banjara]

Ok, I am convinced but still, there is no concluding remark on my original question. From my post, "I want to know whether I am doing wrong though metric is independent to the final result or it doesn't reconcile, naturally", what should we be concluded?

I believe, a single sentence is sufficient to reply.

 

[PeterDonis]

"I want to know whether I am doing wrong though metric is independent to the final result or it doesn't reconcile, naturally", what should we be concluded?

I'm not sure what you're asking.

If you're asking whether what you did in the OP of this thread is correct, I have already said that it's wrong, and explained why. See my posts #10 and #11.

If you are asking whether you can obtain the inverse metric $g^{\alpha \beta}$ by raising both indexes on the metric $g_{\alpha \beta}$, it should be obvious that you can't, since in order to raise indexes you need to already know the inverse metric $g^{\alpha \beta}$. You obtain the inverse metric by considering the metric as a matrix and obtaining its matrix inverse.

 

[Bishal Banjara]

I'm not sure what you're asking.

If you're asking whether what you did in the OP of this thread is correct, I have already said that it's wrong, and explained why. See my posts #10 and #11.

If you are asking whether you can obtain the inverse metric $g^{\alpha \beta}$ by raising both indexes on the metric $g_{\alpha \beta}$, it should be obvious that you can't, since in order to raise indexes you need to already know the inverse metric $g^{\alpha \beta}$. You obtain the inverse metric by considering the metric as a matrix and obtaining its matrix inverse.

The only way I could make my question very simple, be like, what if inverting the metric coefficients $g_{oo}$ and $g_{rr}$ of the usual Schwarzschild solution for the final result calculation of Einstein's tensor components $G_{oo}$ and $G_{rr}$? Does this final result after inverting the metrics coincide to the initial result of the original Schwarzschild solution?

 

[PeterDonis]

inverting the metric coefficients $g_{oo}$ and $g_{rr}$

Is a meaningless, wrong thing to do. It makes no sense.

Does this final result after inverting the metrics coincide to the initial result of the original Schwarzschild solution?

No. It is just nonsense. See above.