# General relativity: Why don't these two differentials commute?

I encountered some metric today defined by $$ds^2 = - \left( 1 - \frac{2GM}{r} \right) dv^2 + (dv dr + dr dv) + r^2 d \Omega^2$$

In all education I've done until now $$dv dr = dr dv.$$ Why is this no longer the case? I suspect this has something to do with tensors, but I am not sure why.

• As an side, usually (e.g. as in multivariable calculus), the implicit product of differential forms is meant to be the wedge product, and they anti-commute: $\mathrm{d}x \mathrm{d}y = -\mathrm{d}y \mathrm{d}x$. You didn't notice this multivariable calculus, because whenever you swapped the order of integration, the sign flip due to reversing the order of the two differentials gets cancelled out by the sign flip due to reversing the orientation of the region you integrate over. – user5174 Apr 6 '17 at 14:40
• Usually $dxdy$ stands for the symmetrization of $dx\otimes dy$, then of course $dxdy=dydx$. Sometimes $dxdy$ stands for the tensor product, then you need to write it as in the book. – MBN Apr 6 '17 at 21:10

1. More generally, a metric tensor $$\mathbb{g} ~\in~ \Gamma\left( {\rm Sym}^2(T^{\ast}M)\right)$$ is a section in the symmetric tensor product $${\rm Sym}^2(T^{\ast}M)~=~T^{\ast}M\odot T^{\ast}M$$ over the cotangent bundle $$T^{\ast}M$$. In other words, $$\mathbb{g}$$ is a symmetric $$(0,2)$$ covariant tensor field.

2. In a coordinate chart $$U\subseteq M$$, it takes the form $$\mathbb{g}|_U ~=~ g_{\mu\nu}\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu},$$ with the manifest rule $$\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}~=~\mathrm{d}x^{\nu}\odot \mathrm{d}x^{\mu},$$ cf. OP's question.

• Out of curiosity why not use $\otimes$ for the tensor product? – JamalS Apr 5 '17 at 20:18
• The notation $\odot$ (or $\vee$) denotes the symmetric $\otimes$ tensor product in the same way that $\wedge$ denotes the antisymmetric $\otimes$ tensor product. – Qmechanic Apr 5 '17 at 20:21
• It might be worth emphasizing the fact that the symmetry of $g$ implies that $$g_{\mu\nu}\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu} = g_{\mu\nu}\mathrm{d}x^{\mu}\otimes \mathrm{d}x^{\nu}$$ which allows one to get away with ambiguity about the product. (so long as one does not try to use the wedge product of the forms) – user5174 Apr 6 '17 at 14:43
• @Hurkyl Does that mean that Qmechanic is saying dv dr = dr dv? Thanks – Mikkel Rev Apr 6 '17 at 16:02
• @MariusJonsson: I take Qmechanic's intent as being more about describing the object being calculated and adding more precision to the notation for clarity, rather than about making any sort of assertion about the right way to interpret an implicit product of differentials. My comment is because I think the answer would be even better if it explained the phenomenon that so many people are writing the formula in a way that usually means to invoke the tensor product. – user5174 Apr 6 '17 at 16:11

That is the Eddington-Finkelstein metric (one of the forms) and it can indeed be written as:

$$ds^2 = - \left( 1 - \frac{2GM}{r} \right) dv^2 + 2dv dr + r^2 d \Omega^2$$

Why your book doesn't write it that way I don't know - you'd have to ask the author.

• I'm confused, there seem to be conflicting statements about whether $dvdr = drdv$ here, are you saying that $dvdr$ here isn't a tensor product? Thanks – user95137 Apr 5 '17 at 17:31
• There is a formal sense in which $\mathrm{d} v \, \mathrm{d} r$ is a tensor product (of two 1-forms $\mathrm{d} v$ and $\mathrm{d} r$), but in the context of GR this can safely be forgotten. It is fine to treat these objects merely as coordinate increments, and $\mathrm{d} v \,\mathrm{d} r$ as their (ordinary) product. The only reason the author might have written it as they have is to emphasise that both $g_{rv}$ and $g_{vr}$ are non-zero (and indeed both equal to 1). – gj255 Apr 5 '17 at 19:11

This is the Eddington-Finkelstein metric for a Schwarzschild geometry, and the differentials do commute. I believe they put the line element in this form in order to highlight the that the components of the metric are

$$g_{vv}=\left(1-\frac{2GM}{r}\right),\hspace{0.5cm}\,g_{vr}=g_{rv}=-1$$ $$g_{\theta\theta}=-r^2,\hspace{0.5cm}g_{\phi\phi}=-\sin^2{\theta}$$

(Note that I'm using a $(+,-,-,-)$ signature.) If this is an introductory text, then the author may have done this to show that off-diagonal elements of the metric have to be halved when being read off from the line element.

I hope this helps!

• Differentials do not commute. Tensor products are not commutative. – AccidentalFourierTransform Apr 5 '17 at 16:52
• @AccidentalFourierTransform I agree that tensor products do not commute but $dv$ $dr$ doesn't stand for the tensor product here. To ensure you that it doesn't stand for the tensor product, I would like to point out that the left-hand side of the expression is a scalar and the summation of two tensor products would yield a tensor, not a scalar. – Dvij Mankad Apr 5 '17 at 17:04
• @Electrodynamist what? $\mathrm ds^2$ is a tensor, and so are $\mathrm d v$, $\mathrm dr$, etc. The proper notation is $\mathrm ds^2=g_{\mu\nu}\mathrm dx^\mu\otimes\mathrm dx^\nu$. Both the r.h.s. and the l.h.s. are tensors. – AccidentalFourierTransform Apr 5 '17 at 17:08
• @AccidentalFourierTransform Ok. Then you are correct. I never knew this way of thinking about the metric. I always thought of $ds^2$ as a scalar invariant interval. – Dvij Mankad Apr 5 '17 at 17:25
• In my GR lecture we were told that $ds^2$ is to be thought of as a quadratic form. One could write $ds^2 (V)=g_{\mu \nu}V^{\mu}V^{\nu}$ with only one argument, so in this notation Electrodynamist would be right. I think common practice is not to commit to one view by not precisely defining the objects mathematically. – Adomas Baliuka Apr 5 '17 at 22:31

In general, tensors with "down" indices are maps from the space of vectors${}^{1}$ to the space of functions. So, a tensor $\bf T$ is a function that takes two vectors and spits out a real number, and you can think of it in the "index-free" manner as:

$${\bf T}\left({\vec v_{1}}, {\vec v_{2}}\right) = f(x^{a})$$

There is also a further restriction on $T$: it is multilinear in both of its coordinates. So, if you multiply either vector by a scalar, you multiply $f$ by the same scalar; and if you replace either vector by a sum of two other vectors, the result is the sum of the tensor acting on the individual vectors. This forces $T$ to be representable as a matrix, which leads us to the familiar notation:

$$f(x^{c}) = v_{1}^{a}T_{ab}v_{2}^{b}$$

Now, it should be obvious that ${\bf T}$ can, in principle, act differently on $v_{1}$ and $v_{2}$, and this happens if $T_{ab}$ is not symmetric.

Now, what does this have to do with differentials? Well, remember that we can express the basis of the vector space with partial derivatives, so that:

$${\vec v} = v^{a}\frac{\partial}{\partial x^{a}}$$

which makes $\vec v$ acting on a function $f$ the directional derivative of $f$ along $\vec v$. So, partial derivatives are the basis of our "up" space of vectors. What is the basis of the "down" space of vectors? Well, we need something that has an index, let's call it $e^{a}$, we also need:

$$\frac{\partial}{\partial x^{a}}e^{b} = \delta^{b}{}_{a}$$

Well, if we choose $e^{a} = dx^{a}$, this should be obviously true. So, just like we can express:

$${\vec v} = v^{a}\frac{\partial}{\partial x^{a}}$$

we can also express:

$${\bf T} = T_{ab}dx^{a}dx^{b}$$

And the only way we can also have a $\bf T$ that acts differently on its first and second argument is if we also have $dx^{a}dx^{b} \neq dx^{b}dx^{a}$ when $a \neq b$

Finally, note that a lot of this is moot, because metric tensors are defined to be symmetric: the basal requirement for a dot product is that ${\vec v} \cdot {\vec w} = {\vec w} \cdot {\vec v} = g_{ab}v^{a}w^{b} = g_{ab}w^{a}v^{b}$

${}^{1}$ Yes, I know we're talking about vector fields and tensor fields, rather than vectors and tensors, but let's not complicate this by making that distinction right now.

Written out explicitly (see this Wikipedia article):

$$v=t+r^{*}$$

$$dr^{*}=(1-{\frac {2GM} r})^{-1}dr$$

so

$$dv=dt+dr^{*}=dt+(1-{\frac {2GM} r})^{-1}dr$$

Then

$$dvdr=(dt+(1-{\frac{2GM} r})^{-1}dr)dr=dtdr+(1-{\frac{2GM}{r}})^{-1}d^2r$$

and$$drdv=dr(dt+(1-{\frac{2GM} r})^{-1}dr)=drdt+(1-{\frac{2GM} r})^{-1}dr^2$$

wich are equal because the infinitesimal increments $dr$ and $dt$ are infinitesimal increments in numbers and $r$ is a number too.

So $$dvdr=drdv$$

which is why the metric in the article has a $2dvdr$ term.

In multilinear algebra, there are several ways to construct new vector spaces from old ones: for a given (real, finite dimensional) vector space $V$, we can construct its dual $V^\ast$, given vector spaces $V$ and $W$ we can construct $V\otimes W$, and we can construct $V\odot V$, $V\wedge V$ (and many more of course). Also these constructions can be repeated.

Typically the latter, $V\odot V$, $V\wedge V$, are constructed by mathematicians as quotients of $V\otimes V$, which works much more generally than for finite dimensional vector spaces over a field. In our special case however (over a field), they can also be constructed as subspaces of $V\otimes V$, in the symmetric case the subspace generated by elements of the form $v\otimes v' + v'\otimes v$, and the alternating case by elements of the form $v\otimes v' - v'\otimes v$.

Physicists, by historical accident, by the lack of need for more generality or by the emphasis on calculation, focus on the second approach. If you make use of several natural isomorphisms like $V^{\ast\ast} \cong V$, $\text{Hom}(V,V) \equiv V^\ast\otimes V$, you get quite far in describing all kinds of derived vector spaces purely in terms of tensor products and duals. When $V = T_xM$, the tangent space of a manifold, the elements of the resulting objects are called tensors. One ugly artefact of this approach that I am aware of, is that the one-dimensional top-exterior power $\Lambda^nV$ of an $n$-dimensional vector space $V$ is viewed as something that is almost like a scalar, but not quite; its elements are called tensor densities.

Another approach popular under physicists is to consider the new vector spaces as multilinear maps $V\times\cdots V\times V^\ast\times\cdots\times V^\ast\to\Bbb R$. If you allow restrictions on these maps, like symmetry or antisymmetry in some of the arguments, and you make use of $V\otimes W \cong W\otimes V$, you can describe most tensors of interests to physicists.

Finally, a metric tensor is a section $g$ of $T^\ast M\odot T^\ast M$, like Qmechanic said. Here $T^\ast M$ can be seen as the union of all $T_x^\ast M$ (along with a given differentiable structure), and a section assigns an element $g(x)\in T_x^\ast M\odot T_x^\ast M$ to each $x\in M$. Depending on your explicit construction, you view elements of $T_x^\ast M\odot T_x^\ast M$ as being generated by $\alpha\otimes\beta + \beta\otimes\alpha$, for $\alpha,\beta\in T_x^\ast M$, i.e. $drdv + dvdr$ in your case, or as products $\alpha\odot\beta$ or $\alpha\beta$ with the understanding that $\alpha\odot\beta = \beta\odot\alpha$ and $\alpha\beta = \beta\alpha$.