Why is the metric tensor symmetric?

Question

I was reading Schutz, A First Course in General Relativity. On page 9, he argued that the metric tensor is symmetric:

$$ ds^2~=~\sum_{\alpha,\beta}\eta_{\alpha\beta} ~dx^{\alpha}~dx^{\beta} $$ $\text{Note that we can suppose}$ $\eta_{\alpha\beta}=\eta_{\beta\alpha}$ $\text{for all}$ $\alpha$ $\text{and}$ $\beta$ $\text{since only the sum}$ $\eta_{\alpha\beta}+\eta_{\beta\alpha}$ $\text{ever appears in the above equation when}$ $\alpha\neq\beta$.

I don't understand his argument. If someone can explain why, I would really appreciate it.

Answer 1

Assume $\eta_{\alpha\beta} \neq \eta_{\beta\alpha}$. Because it's irrelevant what letter we use for our indices, $$\eta_{\alpha\beta}dx^{\alpha}dx^{\beta}=\eta_{\beta\alpha}dx^{\beta}dx^{\alpha}.$$ Then $$\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2}(\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}$$ so only the symmetric part of $\eta_{\alpha\beta}$ would survive the sum. As such we may as well take $\eta_{\alpha\beta}$ to be symmetric in its definition.

Answer 2

The metric tensor is created from the spacetime interval equation. On top of that, $[dx^\alpha,dx^\beta]=0$. Suppose we have a 1+1 dimensional spacetime, if you are given an interval equation resembling:

$$ds^2=-a\,dx_0^2+b\,dx_1^2+c\,dx_0dx_1$$

Obviously, $\eta_{0\,0}=-a$ and $\eta_{1\,1}=b$. Now, you can assume that, say, $\eta_{0\,1}=c/3$ and $\eta_{1\,0}=2c/3$ such that you still get the $c\,dx_0dx_1$ term, but why would you? You know $c=\eta_{1\,0}+\eta_{0\,1}$ and that's the only requirement you have. Since $dx_0$ and $dx_1$ commute, it is easier for everybody if you just say the tensor is symmetric and set $\eta_{1\,0}=\eta_{0\,1}=c/2$. Nothing changes except that you have saved yourself some trouble later on.

Answer 3

I don't understand Schutz's argument either. But to explain why the metric is symmetric we should examine where the metric comes from.

The source of the metric is the Euclidean inner product. This measures lengths and angles in flat Euclidean space. And to do this it must be symmetric. This is because distance itself is symmetric: to measure the distance from a to b is the same as measuring the distance from b to a (this insight is due to @Andrew).

Now the Euclidean inner product is also positive definite and we can weaken this to be indefinite. In this case it is called the semi-Euclidean inner product.

It remains symmetric.

This is the kind of metric we have in General Relativity if spacetime was flat. In the general case where spacetime is smoothly curved we allow the semi-Euclidean inner product to smoothly vary from point to point. This is called a semi-Riemannian manifold and is the manifold used to describe spacetime in General Relativity.

Answer 4

Any metric space comes endowed with a function called the metric which measures distances between points, and by definition the metric is symmetric. The reason is simple: if you have two points $A$ and $B$, the distance between them shouldn't depend on your direction of travel. If it does, you aren't dealing with an ordinary sense of the word "distance", and need to use some more general concept.

The tangent space of a manifold is a metric space and the metric tensor measures distances between the endpoints of infinitesimal displacements in the tangent space (at least that's an intuitive if not-fully-rigorous way to say it). In order for the distance function in the tangent space $d(U, V) = g_{\mu\nu} U^\mu V^\mu$ to be symmetric $d(U,V)=d(V,U)$, the metric tensor $g_{\mu\nu}$ must be symmetric as well.

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The above is a math answer since it uses a mathematical definition. The physics answer is that we use the mathematical abstraction of a metric space because it models the real world well. The distance measure we're interested in special and general relativity is the spacetime interval. It is worth pointing out (as others have) that the spacetime interval does violate one of the requirements for a definition of distance function defined above, namely that it is non-negative, and indeed we aren't interested in normal metric spaces leading to Riemannian geometry, but pseudo-Riemannian geometry. However, the spacetime interval reduces to more ordinary notions of distance up to a sign$^\dagger$, so the distance function should still be symmetric. We use this kind of mathematical object because it is a good model that makes predictions that agree with experiments. We expect differential geometry with a pseudo-Riemannian manifold to be a good model because the spacetime geometry we experience appears to be smooth and behaves like a manifold. However, you might notice nothing I'm saying is a proof it has to work like this -- if you want to explore the implications of the tensor field in general relativity having an antisymmetric part, you could join a whole field of researchers studying modifications of general relativity, but almost all modifications do not end up working for one reason or another and randomly changing an assumption in the math with no underlying physical motivation is usually a starting point of failure.

$^\dagger$ Depending on your sign convention, the spacetime interval between two points in the tangent space reduces to plus or minus the normal Euclidean distance on surfaces of constant time (and you can always do a (local) Lorentz boost to make two spacelike separated points lie on a surface of constant time (in the tangent space), and the interval is invariant under Lorentz boosts). Similarly, the spacetime interval reduces to minus or plus a 1-dimensional distance (time interval) on lines of constant spatial coordinates (and you can always do a (local) Lorentz boost to make two timelike separated points have the same spatial coordinates (in the tangent space)). And, of course, the spacetime interval is zero for all null separated distances.

Answer 5

The argument seems to be circular:

Firstly, one should be precise on what kind of multiplication between $dx^\alpha$ and $dx^\beta$ is considered in

$$ds^2~=~\sum_{\alpha,\beta}\eta_{\alpha\beta} ~dx^{\alpha}~dx^{\beta}$$

• Although $dx^\alpha$ and $dx^\beta$ are differential forms, the multiplication is clearly not meant to be the wedge product $dx^\alpha \wedge dx^\beta$, since this would imply that $ds^2 \in \Gamma(T^*\mathcal M^{\otimes 2})$ would be totally antisymmetric.

• Considering the tensor product $dx^\alpha \otimes dx^\beta$ makes sense since $(dx^\alpha \otimes dx^\beta)_{\alpha\beta}$ form a $C^\infty(\mathcal M)$ basis of the space of $(0,2)$-tensor fields $\Gamma(T^*\mathcal M^{\otimes 2})$. However, the statement

only the sum $\eta_{\alpha\beta}+\eta_{\beta\alpha}$ ever appears

would be wrong in this case because in general $dx^\alpha \otimes dx^\beta \neq dx^\beta \otimes dx^\alpha$.

• Last, the multiplication could also mean the symmetric tensor product $dx^\alpha \odot dx^\beta$. Since, $dx^\alpha \odot dx^\beta = dx^\beta \odot dx^\alpha$, the statement about $\eta_{\alpha\beta}+\eta_{\beta\alpha}$ would then be correct. However, the reasoning is somehow circular since the $C^\infty(\mathcal M)$-span of $(dx^\alpha \odot dx^\beta)_{\alpha \beta}$ are precisely the symmetric $(0,2)$-tensor fields. So, we are essentially saying that we can assume that the tensor is symmetric if the tensor is symmetric.

Statement about coordinate expressions

A correct statement would be the following: All tensor fields of the form $$ T = \sum_{\alpha \beta} \eta_{\alpha \beta} dx^{\alpha} \odot dx^{\beta}$$ can also be expressed by $$ T = \sum_{\alpha \beta} \frac{1}{2}(\eta_{\alpha \beta}+ \eta_{\beta \alpha}) dx^{\alpha} \odot dx^{\beta}$$ The reason is of course that the spanning set $(dx^\alpha \odot dx^\beta)_{\alpha \beta}$ is not linearly independent.

This fact however is purely mathematical and does not provide a physical reason, why the metric in physics must be symmetric (Such arguments can be formulated).

What is the meaning of $ds^2$:

The Notation $ds^2$ looks as if it means either

• $s$ is a differential (1)-form and $ds^2$ means $(ds)\otimes (ds) = (ds) \odot (ds)$,
• $s$ is a differential form and $ds^2 := d(s \wedge s)$.

However, neither of the interpretations is correct. $ds^2$ is just a different name for coordinate free version of metric tensor field $g \in \Gamma(T^*\mathcal M^{\odot 2})$.