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In general relativity, we introduce the line element as $$ds^2=g_{\mu \nu}dx^{\mu}dx^{\nu}\tag{1}$$ which is used to get the length of a path and $dx$ is an infinitesimal displacement But for a manifold $M$, we define the metric tensor as $$g=g_{\mu \nu}dx^{\mu}dx^{\nu}\tag{2}$$ where $dx$ is the basis of $T^*M$. My question is that what are difference between $ds^2$ and $dx$ in both cases? Do they have the same meaning? How we can identify if we consider the spacetime as manifold?

Qmechanic
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Mahtab
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2 Answers2

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That notation is unfortunate, seeing as those $dx$'s do not refer to covectors. Physicists use the formula $$ds^2=g_{mn}dx^mdx^n \tag{1} \label{1}$$ to define the metric tensor because they mean to convey the notion that the metric tensor gives a sense of arc length locally. So, they introduce the notation $dx^m$ to signify this supposed infinitesimal displacement. This is fine as far as intuition is concerned; that is the real physical content of the metric. However, formally, it is not quite correct.

The metric $g$ loosely speaking is a function on the tangent space (a bilinear one), that gives you a notion of inner product on your manifold. It takes two vectors in the tangent space, $T_pM$, and outputs a real number that is thir inner product. That means the better notation is $$g(V,W)=g_{mn}V^mW^n$$ where we denote the components of the vectors $V,W \in T_pM$ by $V^m$. Thus you see that your $dx$'s are meant to be components of vectors, not of covectors.

Now, $ds^2$ is also an unfortunate notation, although not quite as much. You see, if you have a curve on the manifold, $\gamma: I \subset \mathbb{R} \rightarrow M$, and a metric at every tangent space, you can define the arclength of that path as $$s=\int_I \sqrt{g(\gamma',\gamma')} dt \tag{2} \label{2}$$ This is completely analogous to the usual formula for arclength for curves in $\mathbb{R}^n$ $$s=\int_I \sqrt{\gamma' \cdot \gamma'}dt$$ but the dot product is promptly generalized by the metric. Thus it is tempting to note $$ds=\sqrt{g(\gamma',\gamma')} dt$$ and to call $$ds^2=g(\gamma',\gamma') dt^2$$ But this of course is merely simbolic. Expressions like $ds^2$ don't make much sense. The real meaning of $ds$ is given by equation $\ref{2}$. So when people write $ds^2$ they mean that, if you had a curve of which velocity had components $dx^m$, then the "square" of the arclength of that curve would be given by equation $\ref{1}$ (or rather, really, equation $\ref{2}$).

In short, the $dx^m$ are not basis covectors, rather, they're components of vectors that physicist write that way because they want to convey they are in some way local objects, and $ds^2$ is a symbolic representation of the arclength of a certain path on the manifold

Lourenco Entrudo
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    $dx^i$ are definitely covectors (covector fields on an open set to be precise). – peek-a-boo Aug 23 '23 at 06:22
  • @peek-a-boo No, in this context they aren't. The metric is a function on vectors, not covectors. If they were, the indices would be down, and the inverse metric would be used, seeing as it is the inner product on the cotagent space – Lourenco Entrudo Aug 23 '23 at 08:41
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    you’re confusing a tensor for its components. $dx^1$ is a covector field, so is $dx^2$, so is $dx^3$, and so on up to $dx^{\dim M}$. And your final sentence about $dx^m$ being components of a vector is definitely wrong. Given a tangent vector $v$, the number $dx^m(v)$ is the $m^{th}$ component of the vector $v$ relative to the cooridnate-induced basis $\left{\frac{\partial}{\partial x^i}\right}_{i=1}^{\dim M}$, but $dx^m$ itself is a covector field. – peek-a-boo Aug 23 '23 at 08:59
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    If you want to go the index route, then $dx^1$ is the (locally defined) covector field with indices $(dx^1){\mu}=\delta^1{,\mu}$, next, $dx^2$ is the covector field with components $(dx^2){\mu}=\delta^2{,\mu}$ and so on. Also, you’re misunderstanding my objection. I never said the metric is not a (bilinear) function on vectors. – peek-a-boo Aug 23 '23 at 09:05
  • @peek-a-boo I am aware of the definition of the dual basis, and of the difference between tensors and their components, but the $dx^m$ in this post are most definitely not covectors. The one place where you could introduce them is in writing $g=g_{mn} dx^m \otimes dx^n$. But in the arclength formula they really are supposed to stand for vector components, because the metric acts on vectors to produce scalars. – Lourenco Entrudo Aug 23 '23 at 09:29
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    They still are covectors, and $dx^m,dx^n$ is also shorthand notation for the symmetrized tensor product. At this point, I’ll stop discussing basic differential geometry. – peek-a-boo Aug 23 '23 at 09:58
  • @peek-a-boo I don't know what else to tell you. They're not covectors. If they were, the notation $ds^2$ would just be moronic, because $g_{mn}dx^mdx^n$ would just be the metric – Lourenco Entrudo Aug 23 '23 at 10:26
  • @peek-a-boo anyway this is getting lengthy. You're welcome to move this to chat if you want to – Lourenco Entrudo Aug 23 '23 at 10:26
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It is the same principle from polar coordinates in the plane, except that in the general case the manifold has non-zero curvature. In a given point $(r,\theta)$, the basis vectors are $\mathbf e_r$ and $\mathbf e_{\theta}$. A infinitesimal displacement can be obtained by Pythagoras: $ds^2 = (dr\mathbf e_r)\cdot(dr\mathbf e_r) + (d\theta \mathbf e_{\theta})\cdot(d\theta \mathbf e_{\theta}) = g_{11}(dX^1)(dX^1) + g_{22}(dX^2)(dX^2)$

Where $\mathbf e_r\cdot \mathbf e_r = g_{11}$, $\mathbf e_{\theta}\cdot \mathbf e_{\theta} = g_{22}$, $dr = dX^1$ and $d\theta = dX^2$

The metric tensor is the function that takes 2 times a contravariant infinitesimal vector and delivers a scalar. The $dX^i$ are the contravariant components of this vector, and $ds^2$ is the scalar generated by the function.

Claudio Saspinski
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