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Here is a question from a problem sheet I found which I'm going to use to illustrate a point:

The $\color{red}{\text{metric}}$ on a unit sphere is $${ds}^2={d \theta}^2+{\sin}^2\theta\, {d\phi}^2\tag{1}$$ where $\theta$ and $\phi$ are spherical polar coordinates. $u$ and $v$ are vector fields on the sphere with components $$u^\theta=0,\,\,\,u^\phi=1-\cos\theta,\qquad v^\theta =1,\,\,\,v^\phi = 0$$ Evaluate $u\cdot u$, $v\cdot v$ and $u\cdot v$


I am not interested in the answer to this question as I know I can calculate these dot products. I am simply questioning the terminology of the word I marked in red. It was my understanding that the ${ds}^2$ is known as the "invariant interval", and may not even be a distance, for example, it could measure the time between two events. I also know that the ${ds}^2$ may be written in terms of a 'metric (Minkowski) tensor', in other words, $${ds}^2=\eta_{\mu\nu}{dx^\mu}{dx^\nu}$$ where $\eta_{\mu\nu}=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}.\,$ Eqn $(1)$ can also be re-written in terms of a metric tensor.

This is also confirmed by Wikipedia where the metric is denoted by $g_{\mu\nu}$. So if the metric is not ${ds}^2$, then why is ${ds}^2$ being called the metric in the question?


I have already looked at this question and this question asked on this site, an answer given to the latter seems to suggest that there is no difference between ${ds}^2$ and $g_{\mu\nu}$, but I don't understand the reasoning.

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  • $\begingroup$ $ds^2$ is technically the invariant interval (here, invariant under Lorentz transformation), not the metric. The metric is $g_{\mu\nu}$ and is required to construct the invariant interval $\endgroup$ Jun 13 at 21:22

2 Answers 2

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If you want to be super systematic about language and not overloading the terminology, you can say the following. Fix a smooth $n$-dimensional manifold $M$.

  • A pseudo-Riemannian metric tensor field on $M$ is by definition a smooth $(0,2)$-tensor field on $M$, which is pointwise symmetric, non-degenerate, and has the same signature (number of plus and minus in the Sylvester matrix form) at every point. Such a tensor field is typically denoted by the symbol $g$ (so for each point $p\in M$, $g_p:T_pM\times T_pM\to\Bbb{R}$ is a bilinear, symmetric non-degenerate function which eats two tangent vectors and outputs a real number).

Now, in physics, it is common to abbreviate terminology in the following ways. First, the adjective pseudo-Riemannian is often omitted because in GR we care exclusively with Lorentzian signature (1 plus, and $n-1$ minus, or the other way around), and since everyone knows it's about Lorentzian signature, we'd rather not beat the already dead horse. Next, the phrase "tensor field" is often shortened (by abuse of language) to just "tensor" because... well that's just the way things are. So, you may hear $g$ being referred to as "the metric tensor $g$". This is of course incorrect (but standard) terminology since the word field tells us there is one at every point in the manifold. The next abbreviation is to omit the word 'tensor' in this description, and simply speak of "the metric $g$". In Physics, people won't have any trouble understanding what you mean, but in math, a very common source of confusion for students is with the use of 'metric' in 'metric tensor field' in the context of Riemannian manifolds and 'metric space'. Now, one can always introduce a coordinate chart $(U,x=(x^1,\dots, x^n))$, and in this chart, we can write \begin{align} g|_U&=g_{ab}\,dx^a\otimes dx^b, \end{align} where $g_{ab}:U\to\Bbb{R}$ are smooth functions, namely $g_{ab}(p):=g\left(\frac{\partial}{\partial x^a}(p),\frac{\partial}{\partial x^b}(p)\right)$.

Ok, now given the object $g$ as above, we can define the following object, called the quadratic form associated to $g$. This I shall denote as $Q_g$, and it is a function $Q_g:TM\to\Bbb{R}$ defined as $Q_g(v)= g(v,v)$, for all $v\in TM$. So, you take any tangent vector $v$, and plug it into $g$ twice. The following is a basic linear algebra fact: we can recover $g$ from $Q_g$, in the following sense. For any $p\in M$ and $v,w\in T_pM$, we have \begin{align} g(v,w)&=\frac{Q_g(v+w)-Q_g(v-w)}{4}\tag{$*$}. \end{align} Think of the analogous statement for multiplication of real numbers. If I have two real numbers $x,y$, then from the sum/difference of squares formula, $xy=\frac{(x+y)^2-(x-y)^2}{4}$. The general form $(*)$ above is called the polarization identity. Thus, given any symmetric $(0,2)$ tensor field, we can define a corresponding quadratic form, and conversely given any quadratic form, we can define a symmetric $(0,2)$ tensor field which has that as the quadratic form.

Because of this equivalence (going back and forth between $(0,2)$ tensor (fields) and quadratic form (fields)) some would consider this a reasonable overload of terminology, and start referring to $Q_g$ as "the metric".

The above terminology of "quadratic form associated to $g$" is how a mathematician would phrase it. In terms of a coordinate chart $(U,x)$, this would be written as \begin{align} Q_g|_U&=g_{ab}\,dx^a\,dx^b. \end{align} The meaning of the product $dx^a\,dx^b$ on the right is as follows. The object $Q_g$ takes a tangent vector $v\in T_pM$ and outputs the number $Q_g(v)=g_{ab}(p)\,dx^a(v)\cdot dx^b(v)$ (recall that $dx^a$ is a 1-form so it takes a vector $v\in T_pM$ as input and outputs $dx^a(v)\in\Bbb{R}$ as output; this number is often denoted as $v^a$, and called the "$a^{th}$ component of $v$ with respect to the coordinate-induced basis $\left\{\frac{\partial}{\partial x^i}(p)\right\}_{i=1}^n$ of $T_pM$").

Motivated by the coordinate expression on the right and classical terminology, in Physics, we refer to $Q_g$ as the infinitesimal squared distance, or also as the line element (associated to $g$), and in SR/GR also the (infinitesimal) spacetime interval (the adjective 'infinitesimal' referring to the fact that it is at the level of tangent spaces), and instead of the notation $Q_g$, it is much more common to use the notation $ds^2$, (even though it is not the exterior derivative $d(s^2)$ of the square of a function $s^2$, nor is it the product of a 1-form $ds$ with itself in any manner). This is just symbolic and suggestive notation. In coordinates, we thus write \begin{align} Q_g|_U\equiv ds^2|_U=g_{ab}\,dx^a\,dx^b, \end{align} where $\equiv$ means 'same thing in different notation'. See this MSE answer of mine for more details about the notation involving tensor products and symmetrized tensor products and quadratic forms.


In your specific example, you're dealing with Riemannian signature (all plus). So, being slightly more systematic with the language, I would say the metric tensor field $g$ on the unit sphere $S^2$ is such that if you restrict it to the domain of the spherical coordinate mapping, then \begin{align} g&=d\theta\otimes d\theta+\sin^2\theta\,d\phi\otimes d\phi. \end{align} (but take a look at the above MSE answer of mine; if you use the symmetrized tensor product notation, you can write this also as $g=d\theta^2+\sin^2\theta\,d\phi^2$). Equivalently, you could say the line element on the sphere is such that when restricted to the sphere, it equals \begin{align} ds^2=d\theta^2+\sin^2\theta\,d\phi^2 \end{align} (the RHS now being interpreted as a quadratic form).

At the end of the day, they're giving you the same information.


If it were me, I prefer to keep a terminological distinction between $g$ (pseudo-Riemannian metric tensor field, or if I'm working in RIemannian geoemtry, I'd abbreviate this to "Riemannian metric", or if I'm doing GR, I'd say "Lorentzian metric") and $Q_g\equiv ds^2$ (which I'd prefer to call the Quadratic form associated to $g$, or just the line element). Having said this, because of the linear algebra fact mentioned above, it isn't super necessary (once you have learnt the definitions) to be so strict with maintaining the distinction (and for people who know this fact, it is so obvious that they may even blur the distinction between a symmetric $(0,2)$ tensor field and its associated quadratic form, so they may start writing stuff like $g=ds^2=g_{ab}\,dx^a\,dx^b$).

One final thing I'll mention is that sometimes you'll see statements like "the metric $g_{ab}$". This can be interpreted in two ways. The first is you have $g$ as above, and you're fixing a coordinate chart $(U,x)$ as above, and considering the component functions $\{g_{ab}\}_{a,b=1}^n$; in this sense identifying a tensor field with its component functions with respect to some coordinates is an abuse/overload of language, and I would strongly caution against it unless you know precisely what you're talking about. Alternatively, it is also common to use the abstract index notation in which the symbol $g_{ab}$ denotes the actual $(0,2)$ tensor field $g$ (I have my reasons for using this notation only occasionally, but it's logically fine).

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  • $\begingroup$ What beautiful answer! Almost poētic! $\endgroup$
    – Felicia
    Jun 14 at 1:24
  • $\begingroup$ This answer is way too complicated for a beginner, which is what OP is. So I'm not sure how much this will benefit them. $\endgroup$
    – Ishan Deo
    Jun 14 at 20:06
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    $\begingroup$ @IshanDeo The OP clearly asked "I am not interested in the answer to this question as I know I can calculate these dot products. I am simply questioning the terminology of the word I marked in red", which to me suggests they're not satisfied with the typical usage of terminology in Physics, and the only way I see to address this is to be precise with terminology, and explain where the 'shortcuts' in Language are being made, and why. $\endgroup$
    – peek-a-boo
    Jun 14 at 20:09
  • $\begingroup$ The only real math idea in my answer is the polarization identity which is just a glorified way of writing $xy=\frac{(x+y)^2-(x-y)^2}{4}$. The rest is pretty much an explanation of the language used in physics, math and the bridge between them (where I'm of course assuming OP knows what tensor fields are). But by all means, if you have a way of simplifying this, I'd be interested to hear it. $\endgroup$
    – peek-a-boo
    Jun 14 at 20:11
  • $\begingroup$ @peek-a-boo But do they know what a tensor field is, or even what a manifold is? If they do, then your answer elucidates the answer in an excellent fashion. However, I'm not sure they do, and that is my entire point. But I see that your answer was accepted, and so my concerns are moot. $\endgroup$
    – Ishan Deo
    Jun 15 at 12:11
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The interval is just a convenient way to give a metric, since from

$$ds^2 = g_{\mu\nu} dx^\mu dx^\nu$$

you can just read off the metric components. It's especially useful when the metric has a lot of zeros, since you don't have to write those terms.

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    $\begingroup$ Thank you for your answer, so you are saying both ${ds}^2$ and $\eta_{\mu\nu}$ are "metrics"? $\endgroup$
    – FutureCop
    Jun 13 at 18:10
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    $\begingroup$ If you're careful about vocabulary, no. Just see this as a shortcut for "an indirect way to define the metrics". People can be lazy. ;) $\endgroup$
    – Miyase
    Jun 13 at 18:12
  • $\begingroup$ @Miyase Well, okay, but laziness leads to confusion, and general relativity is confusing enough as it is. I wish people would just call ${ds}^2$ the "invariant interval" which is what it was called when I learned special relativity. Then the word "metric" can be reserved for $g_{\alpha\beta}$ $\endgroup$
    – FutureCop
    Jun 13 at 18:18
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    $\begingroup$ @FutureCop While we're being careful, $g_{\alpha\beta}$ is the $(\alpha\beta)$-component of the metric in an as-yet-unspecified basis. The metric $g$ itself is a basis-independent object. $\endgroup$
    – J. Murray
    Jun 13 at 18:55
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    $\begingroup$ @FutureCop Sure, I'm not arguing that $\mathrm ds^2$ should be called the metric. Just pointing out that there's plenty of confusion among students who see expressions like $\mathrm{Tr}(AB)=A^\mu_{\ \ \nu} B^\nu_{\ \ \mu} = B^\nu_{\ \ \mu} A^\mu_{\ \ \nu} = \mathrm{Tr}(BA)$ and wonder why you can freely exchange $A^\mu_{\ \ \nu}$ and $B^\mu_{\ \ \nu}$, since matrix multiplication doesn't commute - not understanding that $A$ and $B$ are matrices, but $A^\mu_{\ \ \nu}$ and $B^\mu_{\ \ \nu}$ are numbers. $\endgroup$
    – J. Murray
    Jun 13 at 19:26

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