A Is the variation of the metric ##\delta g_{\mu\nu}## a tensor?

[PeterDonis]

physicists with their sloppy use of language would

Yes, and this is the relativity forum, not one of the math forums, so physics sloppiness wins out here.

 

[haushofer]

Thanks. Even with this general caution, may we take it obvious that in tensor division of
$$ \bar{g}_{\mu\nu} := g_{\mu\nu}+\delta g_{\mu\nu} $$
all $\bar{g}_{\mu\nu}$ , $g_{\mu\nu}$ and $\delta g_{\mu\nu}$ are tensors in the world of metric undertaking variation ?

Yes.

 

[anuttarasammyak]

I will summarize what I have written.

The index-raising and -lowering relation for the metric tensor

$$ g_{\mu\nu}=g_{\mu\alpha}g_{\nu\beta}g^{\alpha\beta} ,$$

holds for a varied metric tensor as

$$\bar{g}_{\mu\nu}=\bar{g}_{\mu\alpha}\bar{g}_{\nu\beta}\bar{g}^{\alpha\beta} $$

where

$$ \bar{g}_{\mu\nu} := g_{\mu\nu}+\delta g_{\mu\nu} .$$
$$ \bar{g}^{\mu\nu} := g^{\mu\nu}+\delta g^{\mu\nu} .$$

Equation (*) in the OP follows from this.

I agree with the OP that if $\delta g_{\mu\nu}$ were a tensor, then equation (**) in the OP would hold.

A tensor satisfying (*) remains a mystery to me.

 

[JimWhoKnew]

I will summarize what I have written.

The index-raising and -lowering relation for the metric tensor

$$ g_{\mu\nu}=g_{\mu\alpha}g_{\nu\beta}g^{\alpha\beta} ,$$

holds for a varied metric tensor as

$$\bar{g}_{\mu\nu}=\bar{g}_{\mu\alpha}\bar{g}_{\nu\beta}\bar{g}^{\alpha\beta} $$

where

$$ \bar{g}_{\mu\nu} := g_{\mu\nu}+\delta g_{\mu\nu} .$$
$$ \bar{g}^{\mu\nu} := g^{\mu\nu}+\delta g^{\mu\nu} .$$

Equation (*) in the OP follows from this.

I agree with the OP that if $\delta g_{\mu\nu}$ were a tensor, then equation (**) in the OP would hold.

A tensor satisfying (*) remains a mystery to me.

As @haushofer remarked in #18, it is the behavior under coordinate transformations that determines whether something is a tensor. If Weinberg says $~\delta g_{\mu\nu}~$ qualifies, I'd take it seriously.

I think the notation is misleading. I'll repeat my argument from #25 in a slightly modified form: consider a regular spacetime with a (single!) metric $~g_{\mu\nu}~$ . Suppose we have a smooth symmetric tensor field $~A_{\mu\nu}~$ which has the same signature as the metric at every point. Suppose further that there is also a symmetric tensor field $~B^{\mu\nu}~$ satisfying the same signature condition, such that $~B^{\mu\rho}A_{\rho\nu}=\delta^\mu{}_\nu~$ . Would you necessarily expect $~B^{\mu\nu}=A^{\mu\nu}~~$?

Edit: added "symmetric"

 

[anuttarasammyak]

As @haushofer remarked in #18, it is the behavior under coordinate transformations that determines whether something is a tensor.

Going back to basics, Dirac's textbook defines a tensor as follows:

$$ T^{\alpha'\beta'}_{\gamma'}=x^{\alpha'}_{,\lambda}x^{\beta'}_{,\mu}x^{\nu}_{,\gamma'}T^{\lambda\mu}_{\nu}
\ \ \ (3.6)$$
How exactly should I perform coordinate transformations to investigate the quantity I'm considering? Or is there a more obvious way to check?

 

[anuttarasammyak]

consider a regular spacetime with a (single!) metric gμν . Suppose we have a smooth symmetric tensor field Aμν which has the same signature as the metric at every point. Suppose further that there is also a symmetric tensor field Bμν satisfying the same signature condition, such that BμρAρν=δμν . Would you necessarily expect Bμν=Aμν ?

I am sorry I don't understand "the same signature". If A is g itself, I think that B = A = g.
A=mg, B=1/m g where m is a real number $\neq 0$ satisfy the relation.
I am sorry not to be smart enough to relate A $\neq$ B with the discussion.

 

[JimWhoKnew]

Going back to basics, Dirac's textbook defines a tensor as follows:

$$ T^{\alpha'\beta'}_{\gamma'}=x^{\alpha'}_{,\lambda}x^{\beta'}_{,\mu}x^{\nu}_{,\gamma'}T^{\lambda\mu}_{\nu}
\ \ \ (3.6)$$

So what @haushofer and I said above is in agreement with Dirac's definition. A little nitpicking: if you write $~T^{\lambda\mu}_\nu~$ , then after raising the $\nu$ index, you don't know the order of $~\{\nu,\lambda,\mu\}~$ from left to right.

How exactly should I perform coordinate transformations to investigate the quantity I'm considering? Or is there a more obvious way to check?

In general, you'll need more information. For example: we know that the connections ($\Gamma$'s) are not tensors, because we know how they should be transformed. In the example of #34 we assume that $~A_{\mu\nu}~$ is a tensor field, so it follows that $~A_{\mu\nu}\pm g_{\mu\nu}~$ are also tensor fields.

I am sorry I don't understand "the same signature".

Given a specific inertial frame in Minkowski spacetime, $~C(x)_{\mu\nu}:=\delta_{\mu\nu}~$ is a symmetric invertible smooth tensor field whose signature is different from that of $~\eta_{\mu\nu}~$. So the signature assumption is needed in the example of #34.

If A is g itself, I think that B = A = g.

That's correct, but I want to consider the general case, not only trivial cases. As I said in #34, I find that the notation in OP (following Weinberg and others) obscures the property of "sameness" as pointed in #25. That's why I use A and B instead.

A=mg, B=1/m g where m is a real number $\neq0$ satisfy the relation. I am sorry not to be smart enough to relate A$\neq$B with the discussion.

No need to get personal.

If $~A_{\mu\nu}=m~g_{\mu\nu}~$ then $~A^{\mu\nu}=m~g^{\mu\nu}\neq g^{\mu\nu}/m=B^{\mu\nu} ~~$.
What does this example imply when we write $~\bar{g}_{\mu\nu}:=m~g_{\mu\nu}~$ (and still use g to raise and lower indices)?

Hint: equation (**) in OP is wrong, while (*) is correct to first order in $\delta g~$.

I hoe this study is on a right track.

I don't understand your example. If your manifold is 1D (the ring), it should be charted by a single coordinate. If it is a 2D Euclidean as in the drawing, the line element should contain terms proportional to $~drdr, drd\theta, d\theta d\theta~~$. In 2D Euclidean space, you can easily see that a short arrow tangent to the constant r ring (in your lower drawing) is not necessarily tangent to the constant r' curve (meaning it may have a radial component in the primed coordinates).

 

[anuttarasammyak]

First of all, thanks @JimWhoKnew

In general, you'll need more information.

So in general cases which could include $\delta g^{\mu\nu}$, do we need further information or condition to say it is a tensor or not, though the answer of #32 was affirmative ?

In the example of #34 we assume that Aμν is a tensor field, so it follows that Aμν±gμν are also tensor fields.

I agree that the sum of tensors is a tensor. And sum of no tensors could be sometimes a tensor.

What does this example imply when we write g¯μν:=m gμν (and still use g to raise and lower indices)?

Well, the metric under variation $$\bar{g}^{\mu\nu}=g^{\mu\nu}+(m-1)g^{\mu\nu}=g^{\mu\nu}+\delta g^{\mu\nu}$$
$$ \delta g^{\mu\nu} = (m-1)g^{\mu\nu} $$
from (*)
$$ \delta g_{\mu\nu} = -(m-1)g_{\mu\nu} $$
I may be wrong and/or missing the point. Your further coaching would be appreciated.

 

[JimWhoKnew]

So in general cases which could include $\delta g^{\mu\nu}$, do we need further information or condition to say it is a tensor or not, though the answer of #32 was affirmative ?

@haushofer has already addressed this question in #11. I don't want to get into it here. You'd better consult a textbook (d'Inverno has a nice treatment, despite a little sign inconsistency). Very sloppily: under a coordinate transformation $~T(x)\rightarrow T'(x')~~$. If $~T(x)~$ is a tensor field, so are $~T'(x')~$ , $~T'(x)~$ and $~T'(x)-T(x)~~$. Now, we know that the metric is a tensor field, therefore $~\delta g^{\mu\nu}~$ is a tensor field too (we have to be careful with this economic notation).

Well, the metric under variation $$\bar{g}^{\mu\nu}=g^{\mu\nu}+(m-1)g^{\mu\nu}=g^{\mu\nu}+\delta g^{\mu\nu}$$
$$ \delta g^{\mu\nu} = (m-1)g^{\mu\nu} $$
from (*)
$$ \delta g_{\mu\nu} = -(m-1)g_{\mu\nu} $$
I may be wrong and/or missing the point. Your further coaching would be appreciated.

With $~A_{\mu\nu}=m~g_{\mu\nu}~$ and $~B^{\mu\nu}=g^{\mu\nu}/m~$
1) Find $~\hat{\delta}_{\mu\nu}:=A_{\mu\nu}-g_{\mu\nu}~$ (as a function of g and m).
2) Find $~\tilde{\delta}^{\mu\nu}:=B^{\mu\nu}-g^{\mu\nu}~$.
3) Find $~\hat{\delta}^{\mu\nu}~$.
4) Assume $m=1+\epsilon~$ where $|\epsilon|\ll 1~$. Expand the results of (2) and (3) above to first order in $\epsilon$. Compare with equation (*) in OP.

The important point in my posts: $~A^{\mu\nu}\neq B^{\mu\nu}~$ in general (and therefore equation (**) in OP is wrong).

 

[anuttarasammyak]

With Aμν=m gμν and Bμν=gμν/m
1) Find δ^μν:=Aμν−gμν (as a function of g and m).
2) Find δ~μν:=Bμν−gμν .
3) Find δ^μν .
4) Assume m=1+ϵ where |ϵ|≪1 . Expand the results of (2) and (3) above to first order in ϵ. Compare with equation (*) in OP.

Thanks. I will restate it to confirm my understanding. Here I try not to refer controversial (at least to me) tensor and concentrate on what I found in making use of this example.

The metric
$$ \bar{g}_{\mu\nu}=mg_{\mu\nu} $$
With
$$ \bar{g}_{\mu\alpha} \bar{g}^{\alpha\nu} =\delta^{\nu}_{\mu}$$
$$ \bar{g}^{\mu\nu}=\frac{1}{m}g^{\mu\nu} $$
Say A which satisfies the equation
$$ A_{\mu\nu}=\bar{g}_{\mu\alpha} \bar{g}_{\nu\beta}A^{\alpha\beta} $$,
$\bar{g}_{\mu\nu}$ is A as easily confirmed in this example. The old metric $g_{\mu\nu}$ is not A, actually,
$$ mg_{\mu\nu}=\bar{g}_{\mu\alpha} \bar{g}_{\nu\beta}\frac{1}{m}g^{\alpha\beta} $$
Let us rewrite this equation as
$$ g_{\mu\nu}+\delta g_{\mu\nu}=\bar{g}_{\mu\alpha} \bar{g}_{\nu\beta}(g^{\alpha\beta}+\delta g^{\alpha\beta} )$$
where
$$\delta g_{\mu\nu}=(m-1)g_{\mu\nu}$$
$$\delta g^{\mu\nu}=(\frac{1}{m}-1)g^{\mu\nu}$$
Expecting the variation to be small
$$m=1+\epsilon,\ \ |\epsilon|<<1$$
$$\delta g_{\mu\nu}=\epsilon g_{\mu\nu}$$
$$\delta g^{\mu\nu}=-\epsilon g^{\mu\nu}+0(\epsilon^2)$$

Both $g_{\mu\nu}$ and $\delta g_{\mu\nu}$ are not A. Their sum $g_{\mu\nu}+\delta g_{\mu\nu}$ is A. $\delta g_{\mu\nu}$ and $\delta g^{\mu\nu}$ is introduced to be so. We should not be surprised to see such a conventional quantity satisfies the peculiar equation (*).
Equation (*) holds by changing $g$ to $\bar{g}$. We can regard one metric original and the other varied and vice versa.

The important point in my posts: $~A^{\mu\nu}\neq B^{\mu\nu}~$ in general (and therefore equation (**) in OP is wrong).

In case that A and B are metric, A=B. I used it.

 

[JimWhoKnew]

The metric
$$ \bar{g}_{\mu\nu}=mg_{\mu\nu} $$

I intentionally defined $\hat{\delta}$ and $\tilde{\delta}~$. g is the only metric in this example, and should be used to raise and lower indices. I expected answers to the 4 simple steps as stated.

Edit:
If later on you want to rename $g$ as $\bar{g}$ and regard it as the "new" metric, it doesn't matter, as long as you are consistent and unambiguous. In such a case, the "old" metric will take the form of A and B, which will not represent the same tensor.

 

[anuttarasammyak]

g is the only metric in this example

To me the point is two metrics: original and varied. In varied metric world what kind of things or relations remain and what does not ?

 

[JimWhoKnew]

To me the point is two metrics: original and varied. In varied metric world what kind of things or relations remain and what does not ?

Let's look at it step by step. Point out where you disagree.
We have the original metric $g$ and the varied $\bar{g}~$. Is it legitimate to take the difference$$g_{\mu\nu}=\bar{g}_{\mu\nu}-\delta g_{\mu\nu}\quad ? \tag{1}$$If the answer is "no", the discussion in OP is irrelevant. So let's explore the other option. Now we ask whether there is a meaning to raising the indices of $~\delta g_{\mu\nu}~$. Again, if the answer is "no", the discussion in OP is irrelevant. If "yes", then we are faced with the question how should it be done? We may attempt to preserve symmetry between the metrics by raising one index by $g$ and the other by $\bar{g}~$. But now we have to choose which one raises the left index, which breaks the symmetry. Moreover, the contravariant form will not be symmetric, which is an undesired feature. The other (and more logical) option is to raise both indices with the same metric. Without loss of generality, we may choose $\bar{g}~$ ("the varied world"). But this choice breaks the symmetry between the metrics, making $~\bar{g}~$ the preferred one. With this choice, the RHS of (1) is raised by $~\bar{g}^{\mu\nu}~$ , so to maintain consistency, the LHS must also be raised by $~\bar{g}^{\mu\nu}~$. But now we get the inevitable result that the contravariant form $~\bar{g}^{\mu\rho}g_{\rho\sigma}\bar{g}^{\sigma\nu}~$ is not the inverse of the covariant form $~g_{\mu\nu}~$ (assuming $~\delta g_{\mu\nu}\neq0~~$), and this was my point all along.

Edit:
The OP (following Weinberg) defines $~\delta g^{\mu\nu}:=\bar{g}^{\mu\nu}-g^{\mu\nu}~$ , where $~g^{\mu\nu}~$ is the inverse of $~g_{\mu\nu}~$ . In the framework of the above discussion, where $~\bar{g}~$ was chosen as prefered, this $~\delta g^{\mu\nu}~$ is not the contravariant form of $~\delta g_{\mu\nu}~$ (so equation (**) in OP is inconsistent under the above assumptions, while (*) is satisfied).

 

[anuttarasammyak]

I don't understand your example.

My bad. I withdraw it.

 

[anuttarasammyak]

We may attempt to preserve symmetry between the metrics by raising one index by g and the other by g¯ . But now we have to choose which one raises the left index, which breaks the symmetry. Moreover, the contravariant form will not be symmetric, which is an undesired feature. The other (and more logical) option is to raise both indices with the same metric.

I agree that index raise- lower operation should be done using one same, old or new, metric for all the indexes of all the entities in equations.

 

[anuttarasammyak]

The OP (following Weinberg) defines δgμν:=g¯μν−gμν , where gμν is the inverse of gμν . In the framework of the above discussion, where g¯ was chosen as prefered, this δgμν is not the contravariant form of δgμν (so equation (**) in OP is inconsistent under the above assumptions, while (*) is satisfied).

$$\bar{g}_{\mu\nu} - g_{\mu\nu} = \delta g _{\mu\nu} = - \delta \bar{g}_{\mu\nu} $$
$\bar{g}_{\mu\nu}$ is a tensor in the world where metric is $\bar{g}_{\mu\nu}$ itself.
$g_{\mu\nu}$ is a tensor in the world where metric is $g_{\mu\nu}$ itself.
Their difference $\delta g _{\mu\nu} = - \delta \bar{g}_{\mu\nu}$ belong to neither or both the worlds. I observe that minus sign in (*) , which is shared in both the worlds, comes from this ambivalence.

 

[JimWhoKnew]

$$\bar{g}_{\mu\nu} - g_{\mu\nu} = \delta g _{\mu\nu} = - \delta \bar{g}_{\mu\nu} $$
$\bar{g}_{\mu\nu}$ is a tensor in the world where metric is $\bar{g}_{\mu\nu}$ itself.
$g_{\mu\nu}$ is a tensor in the world where metric is $g_{\mu\nu}$ itself.
Their difference $\delta g _{\mu\nu} = - \delta \bar{g}_{\mu\nu}$ belong to neither or both the worlds. I observe that minus sign in (*) , which is shared in both the worlds, comes from this ambivalence.

The minus sign in (*) follows from the definitions which are used, and from the requirement that $~g^{\mu\nu}~$ and $~\bar{g}^{\mu\nu}~$ are the inverse of $~g_{\mu\nu}~$ and $~\bar{g}_{\mu\nu}~$ respectively (in both worlds, assuming its legitimacy as addressed in #43). Since only terms up to first order in $~\delta g~$ (same as your $~-\delta \bar{g}~~$) are maintained, it is trivial to note that (*) is invariant under "barring".

 

[anuttarasammyak]

The minus sign in (*) follows from the definitions which are used, and from the requirement that $~g^{\mu\nu}~$ and $~\bar{g}^{\mu\nu}~$ are the inverse of $~g_{\mu\nu}~$ and $~\bar{g}_{\mu\nu}~$ respectively (in both worlds, assuming its legitimacy as addressed in #43). Since only terms up to first order in $~\delta g~$ (same as your $~-\delta \bar{g}~~$) are maintained, it is trivial to note that (*) is invariant under "barring".

Yes. For
$$g_{\mu\nu}= g_{\mu\alpha} g_{\beta\nu} g^{\alpha\beta},\ g^{\mu\alpha} g_{\alpha\nu}=\delta^\mu_\nu $$
symbolically I write
$$ 1=1*1*1,\ 1*1=1 $$
and for all barred things
$$ 1+\epsilon=(1+\epsilon)(1+\epsilon)(1-\epsilon),\ (1+\epsilon)*(1-\epsilon)=1 $$
in first order of $\epsilon$. When one becomes fat, the other becomes thin.

Only up to first order is the rule on variation though I do not want to argue it is trivial or not.

 

[anuttarasammyak]

To me the point is two metrics: original and varied. In varied metric world what kind of things or relations remain and what does not ?

An example of my point.
Say light is passing on element $dx^\mu$, i.e.,
$$ g_{\mu\nu}dx^\mu dx^\nu = 0 $$
After variation of metric
$$ \bar{g}_{\mu\nu}dx^\mu dx^\nu \neq 0 $$
The light changes its course and $dx^\mu$ means something different from the light path.
$dx^\mu$ is a 4-vector in the sense of invariance under coordinate transformation in the old metric world. It is also a 4-vector in the sense of invariance under coordinate transformation in the new metric world. May we say it that the metric change has nothing to do with 4-vector property of $dx^\mu$ ?

 

[JimWhoKnew]

Yes. For
$$g_{\mu\nu}= g_{\mu\alpha} g_{\beta\nu} g^{\alpha\beta},\ g^{\mu\alpha} g_{\alpha\nu}=\delta^\mu_\nu $$
symbolically I write
$$ 1=1*1*1,\ 1*1=1 $$
and for all barred things
$$ 1+\epsilon=(1+\epsilon)(1+\epsilon)(1-\epsilon),\ (1+\epsilon)*(1-\epsilon)=1 $$
in first order of $\epsilon$. When one becomes fat, the other becomes thin.

Only up to first order is the rule on variation though I do not want to argue it is trivial or not.

With "the requirement that $~g^{\mu\nu}~$ and $~\bar{g}^{\mu\nu}~$ are the inverse of $~g_{\mu\nu}~$ and $~\bar{g}_{\mu\nu}~$ respectively", expand the RHS of$$0=\bar{g}^{\mu\rho}\bar{g}_{\rho\nu}-g^{\mu\rho}g_{\rho\nu}=\bar{g}^{\mu\rho}\bar{g}_{\rho\nu}-\left(\bar{g}^{\mu\rho}-\delta\bar{g}^{\mu\rho}\right)\left(\bar{g}_{\rho\nu}-\delta\bar{g}_{\rho\nu}\right)\quad.$$Since the result is already first order in $~\delta\bar{g}~$ , we can now get rid of the $~\bar{g}^{\mu\rho}~$ by using either $~\bar{g}_{\sigma\mu}~$ or $~g_{\sigma\mu}~$.
Equally, we could have expressed the middle term by the un-barred quantities to derive (*). Not trivial?