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

[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.

 

[PeterDonis]

@Bishal Banjara I am closing this thread since your original question has been addressed.