# What is Dirac's reasoning when showing the curvature vanishing implies we can choose rectilinear coordinates?

In section 12 of Dirac's book "General Theory of Relativity" he is justifying the name of the curvature tensor, which he has just defined as the difference between taking the covariant derivative twice in two different orders. He first shows that if space is flat then we can choose rectilinear coordinates, and then the metric tensor is constant, hence the curvature tensor $$R_{\mu\nu\rho\sigma}$$ vanishes. I can see that.

Then for the other direction he assumes the curvature tensor vanishes, and shows (over several steps) that this means we can specify a scalar field $$S$$ and hence a covector field $$S_{,\mu}$$ such that $$S_{,\mu\nu}=\Gamma^\sigma_{\mu\nu}S_{\sigma}$$ Then he says we can take "four independent scalars" that satisfy this, and we take them to be the coordinates $$x^{\alpha'}$$ "of a new system of coordinates". Okay, I am still following, I think.

Then he looks at the transformation of the metric tensor $$g_{\mu\lambda}=g_{\alpha'\beta'}x^{\alpha'}_{,\mu}x^{\beta'}_{,\lambda }$$ and then takes several steps (which I can follow) to deduce that $$g_{\alpha'\beta'}$$ is constant, which, again, I can follow, but now I have a problem:

He then says, "Thus we have a flat space referred to rectilinear coordinates." But where in the reasoning has he used/specified/assumed rectilinear coordinates? When he picks the "new system of coordinates" surely he needs to say, "... but not spherical polar coordinates" for example?

• Even in flat space the spherical coordinates metric is not constant, I think a constant metric implies rectilinear coordinates. Or maybe another way to say it is an orthonormal coordinate frame is necessarily rectilinear in flat space? Commented May 31, 2023 at 7:52
• @LewisKirby I think in general what Dirac gets are what he refers to earlier as "Oblique" coordinates meaning in general with cross terms of the form $dx^{\mu}dx^{\nu}$ with $\mu\neq\nu$. Dirac's GR book is very condensed and he leaves a lot as an implicit exercise for the reader :) so I agree, it needs to be shown that the choice of the four independet scalars he mentions may well be rectilinear coordinates (w/o the cross terms)
– Amit
Commented May 31, 2023 at 9:26
• @Amit it does have one distinct advantage for newbies; his "operational" definition of tensors is the simplest summary that I know of! Commented May 31, 2023 at 12:31
• @m4r35n357 In general I think it's amazing how much Dirac does manage to convey in such a short text, and whatever is left for the reader is worthwhile working out anyway. My only criticism would be that the first chapters which comprise the mathematical introduction, aren't using very modern math (DG), but then perhaps that is just because of when it was written
– Amit
Commented May 31, 2023 at 13:10
• "surely he needs to say" - somehow, I'm convinced Dirac didn't feel the same in the slightest :) You know that story about a person in one of his lectures saying "I don't understand the equation on the top right corner"? There was a moment of awkward silence, and when the moderator interrupted it by asking Dirac if he wanted to address the question, Dirac said something along the lines of "Oh, I didn't realize that was a question. I thought it was a comment." Commented May 31, 2023 at 23:21

Suppose that we have $$n$$ independent variables $$x^\mu$$ and $$m$$ "dependent variables" $$y^i$$ and an overdetermined system$$\frac{\partial\phi^i}{\partial x^\mu}(x)=F^i_{\mu}(x,\phi(x)) \qquad(\ast)$$of partial differential equations (PDEs), where $$y^i=\phi^i(x)$$ are $$m$$ unknown functions of the $$n$$ variables.

Equations like this are often called "total differential equations", "Pfaffian equations" or "Frobenius-type equations" or something like those.

A solution of ($$\ast$$) is an $$m$$-component function $$y^i=\phi^i(x)$$ such that when substituted into ($$\ast$$), it becomes an identity. Note that the $$\phi^i(x)$$ do not need to be defined on the entire domain of the differential equation, just on some open subset of $$\mathbb R^n$$.

Furthermore, a solution $$y^i=\phi^i(x)$$ satisfies the initial value problem (IVP) $$(x_0,y_0)$$ if $$y_0^i=\phi^i(x_0)$$.

The differential equation ($$\ast$$) is said to be integrable if for every pair of points $$(x_0,y_0)$$, there is a solution satisfying this IVP.

Equations of the above type are closely related to ordinary differential equations (ODEs), for example you can prove that if an IVP is solvable, then the solution is reasonably unique. You can prove this by projecting the PDE ($$\ast$$) onto a curve connecting the initial point to any other point and solving the projected ODE. Then we get a "potential problem", i.e. if the given IVP is not solvable, it is because the solutions of the projected ODE depends on the shape of the curve connecting the two points.

We can define the functions$$Q^i_{\mu\nu}(x,y)=\frac{\partial F^i_\nu}{\partial x^\mu}-\frac{\partial F^i_\mu}{\partial x^\nu}+F^j_\mu\frac{\partial F^i_\nu}{\partial y^j}-F^j_\nu\frac{\partial F^i_\mu}{\partial y^j},$$then we have a

Theorem: The PDE system ($$\ast$$) is integrable if and only if $$Q^i_{\mu\nu}\equiv 0$$.

This theorem is called the Frobenius integrability theorem. In modern mathematics, this theorem is usually considered in the context of differential geometry, where it is formulated in terms of certain systems of vector fields or differential forms, but those modern formulations are equivalent to this one.

The direction that if ($$\ast$$) is integrable, then $$Q^i_{\mu\nu}=0$$ is easy to prove, because$$Q^i_{\mu\nu}(x,\phi(x))=\frac{\mathrm d}{\mathrm d x^\mu}\left(F^i_\nu(x,\phi(x))\right)-\frac{\mathrm d}{\mathrm dx^\nu}\left(F^i_\mu(x,\phi(x))\right)=\frac{\partial^2\phi^i}{\partial x^\mu\partial x^\nu}-\frac{\partial^2\phi^i}{\partial x^\nu\partial x^\mu}=0,$$ i.e. the symmetry of second partial derivatives forces $$Q^i_{\mu\nu}$$ to vanish along any solution, and then if the equation is integrable, we have a solution passing through every pair of points $$(x,y)$$, hence $$Q^i_{\mu\nu}$$ vanishes altogether. The converse direction is proven in eg. Lovelock, Rund: Tensors, Differential Forms and Variational Principles .

Now let's walk through Dirac's proof, or at least a closely related variant of it step by step.

First, let's fix a couple of terms. The proof works in any dimension and in any metric signature, so let's say that the index of $$g_{\mu\nu}$$ is $$s$$ if the matrix $$(g_{\mu\nu})$$ has $$s$$ negative (and thus $$n-s$$ positive) eigenvalues, and let $$(\eta_{\mu\nu})=\mathrm{diag}(-1,...,-1,1,...,1)$$ be the "generalized Minkowski symbol", where the number of $$-1$$'s is $$s$$ and the number of $$+1$$'s is $$n-s$$.

Then, a coordinate system $$x^\mu$$ is rectilinear if $$g_{\mu\nu}(x)=c_{\mu\nu}$$, i.e. the components of the metric tensor are constants.

A coordinate system is Cartesian if $$g_{\mu\nu}=\eta_{\mu\nu}$$, so Cartesian coordinate systems are special rectilinear systems.

The Riemannian space is flat if each point has a finite neighborhood on which a Cartesian coordinate system can be defined (Dirac ignores this, but coordinates on a manifold are usually only locally definable, and a flat Riemannian space is not necessary the same "globally" as Euclidean space).

Observation: A Riemannian space is flat if and only if each point lies in the domain of a rectilinear coordinate system.

Why? Because in a rectilinear coordinate system $$x^\mu$$, the metric is constant, $$g_{\mu\nu}=c_{\mu\nu}$$, and we know from linear algebra that a constant symmetric matrix is always transformable into diagonal form, so that there is a constant matrix $$A^{\mu}_{\ \nu}$$ such that $$c_{\mu\nu}A^\mu_{\ \rho}A^\nu_{\ \sigma}=\eta_{\rho\sigma}$$. But then we can define new coordinates, by$$x^\mu=A^\mu_{\ \nu}y^\nu,$$ with $$\partial x^\mu/\partial y^\nu=A^\mu_{\ \nu}$$, so that$$\mathrm ds^2=c_{\mu\nu}\mathrm dx^\mu\mathrm dx^\nu=c_{\mu\nu}A^\mu_{\ \rho}A^\nu_{\ \sigma}\mathrm dy^\rho\mathrm dy^\sigma=\eta_{\rho\sigma}\mathrm dy^\rho\mathrm dy^\sigma.$$

It is thus sufficient to prove that if $$R^\rho_{\ \sigma\mu\nu}=0$$, then we can find rectilinear coordinates near each point.

Because $$\Gamma^\rho_{\ \mu\nu}$$ can be expressed homogeneously in $$\partial_\rho g_{\mu\nu}$$ and also the latter can be expressed homogeneously in terms of $$\Gamma_{\rho\mu\nu}$$, it follows that a coordinate system is rectilinear if and only if the Christoffel symbols vanish in that system.

Thus, to prove the existence of rectilinear coordinates, we must be able to transform the Christoffel symbols into zero.

The transformation rule is$$\Gamma^\rho_{\ \mu\nu}=\frac{\partial x^\rho}{\partial y^\sigma}\frac{\partial y^\kappa}{\partial x^\mu}\frac{\partial y^\lambda}{\partial x^\nu}\Gamma^\sigma_{\ \kappa\lambda}\{y\}+\frac{\partial^2y^\sigma}{\partial x^\mu\partial x^\nu}\frac{\partial x^\rho}{\partial y^\sigma}$$

Since in the $$y^\mu$$ system we want the Christoffels to vanish, we insert $$\Gamma^\sigma_{\ \kappa\lambda}\{y\}=0$$ and rearrange things a bit to get$$\frac{\partial^2 y^\sigma}{\partial x^\mu\partial x^\nu}=\Gamma^\rho_{\ \mu\nu}\frac{\partial y^\sigma}{\partial x^\rho}.$$ We want to find $$n$$ independent functions $$y^\sigma$$ that solves this PDE. This is almost like $$(\ast)$$, but the equation is for second, rather than first derivatives. We overcome this obstacle by introducing new variables.

We can replace this equation with a pair $$\frac{\partial y^\sigma}{\partial x^\mu}=p^\sigma_\mu \\ \frac{\partial p^\sigma_\nu}{\partial x^\mu}=\Gamma^{\rho}_{\ \mu\nu}p^\sigma_\rho, \qquad (\ast\ast)$$ which is now of the same type as ($$\ast$$) in the set of variables $$y^\sigma,p^\sigma_\nu$$. The integrability conditions are given by forming the antisymmetric part of the second partial derivatives of the variables (which must vanish), while substituting from the equations themselves to eliminate the first derivatives. We get:$$\frac{\partial^2 y^\sigma}{\partial x^\mu\partial x^\nu}-\frac{\partial y^\sigma}{\partial x^\nu\partial x^\mu}=\left( \Gamma^\rho_{\ \mu\nu}-\Gamma^\rho_{\ \nu\mu}\right)p^\sigma_\rho=0 \\ \frac{\partial^2 p^\sigma_\rho}{\partial x^\mu\partial x^\nu}-\frac{\partial ^2 p^\sigma_\rho}{\partial x^\nu\partial x^\mu}=R^\tau_{\ \rho\mu\nu}p^\sigma_\tau=0.$$

The first equation is identically satisfied because the Christoffel symbols are symmetric, and the second equation is satisfied if $$R^\rho_{\ \sigma\mu\nu}=0$$.

Hence, if the curvature tensor is zero, we can solve ($$\ast\ast$$) with any initial conditions$$y^\mu(x_0)=y_0^\mu,\quad p^\sigma_\mu(x_0)=A^\sigma_\mu$$ we'd like. The first initial condition is just the value of the point $$x_0$$ in the new $$y^\mu$$ coordinates, but the second initial value fixes the initial coordinate transformation matrix $$p^\sigma_\mu=\partial y^\sigma/\partial x^\mu$$. We may choose $$p^\sigma_\mu(x_0)$$ such that $$\det(p(x_0))\neq 0$$, then by continuity, the determinant will be nonzero in some neighborhood of $$x_0$$, and thus the functions $$y^\mu$$ are independent and therefore specify a new coordinate system near $$x_0$$.

In this coordinate system, the equations $$\Gamma^\sigma_{\ \mu\nu}\{y\}=0$$ are satisfied, since the coordinates $$y^\mu$$ have been constructed precisely so that the Christoffels vanish in them. But this means that $$y^\mu$$ is a rectilinear coordinate system, hence the space is flat.

Actually, we can go a bit further and choose $$p^\sigma_\mu(x_0)=A^\sigma_\mu$$ such that this matrix diagonalizes the metric $$g_{\mu\nu}(x_0)$$ at $$x_0$$. Then, because the metric coefficients in the new system are constants, it follows that the metric has been brought into the canonical form $$\eta_{\mu\nu}$$ everywhere in the domain of the new coordinate system.

To reiterate, the coordinate system $$y^\mu$$ is rectilinear because the Christoffel symbols vanish in them, and this is true by construction, i.e. we constructed the new coordinates so that this is true.

• Thank you so much for this -- you've given an answer in language that I can understand (mostly!) and pretty much answers my question. So I've marked this as the answer Commented Jun 1, 2023 at 13:49

If you are looking for a deeply mathematical explanation, please refer to @Bence Racskó's answer, I think it really is very good and insightful. I want to do something different here, I want to show how this "machinery" works for a specific example, thereby hoping to encourage a bit of an intuitive understanding of why it works as well.

I am also currently working through Dirac's GR book, and while I didn't have much trouble just accepting that a constant metric enables us to pick rectilinear coordinates, I did want to work out a concrete example to convince myself that at least in principle his method for constructing such coordinates works.

What I did was start with a flat metric that describes the familiar polar coordinates in $$2$$ dimensions:

$$g_{\mu\nu}dx^{\mu}dx^{\nu} = dr^2+r^2d\varphi^2$$

Calculated the Christoffel symbols, the only non-zero ones are:

$$\Gamma^{r}{} _{\varphi\varphi} = -r$$ $$\Gamma^{\varphi}{} _{r\varphi} = \Gamma^{\varphi}{} _{\varphi r} = \frac{1}{r}$$

And, following the recipe, used the derived relation:

$$x^{\alpha '}_{,\mu\nu} = \Gamma^{\sigma}_{\mu\nu}x^{\alpha'}_{,\sigma}$$

To obtain differential equations for the two (since we're in $$2d$$) independent scalars describing the coordinate system. Plugging in the calculated Christoffel symbols, what I got was:

\begin{align*} \frac{\partial^2{x^{\alpha'}}}{\partial r \partial \varphi} = \frac{1}{r}\frac{\partial{x^{\alpha'}}}{\partial{\varphi}}\tag{1} \\ \frac{\partial^2{x^{\alpha'}}}{\partial \varphi^2} = -r\frac{\partial{x^{\alpha'}}}{{\partial{r}}}\tag{2} \end{align*}

It took me a bit too long probably to realize, the simplest solution to the above equations is given by:

$$x^1 = r\cos\varphi$$ $$x^2 = r\sin\varphi$$

Which is nothing but the coordinate transformation from polar to cartesian coordinates! Now, after all this work this becomes a bit more relevant to your question, note that we could might as well pick:

$$x^1 = r\cos\varphi - r\sin\varphi$$ $$x^2 = r\sin\varphi$$

Which is not a choice of orthogonal$$^{*}$$ coordinates, however this choice still satisfies the same differential equations $$(1)$$ & $$(2)$$. Such coordinates are what Dirac refers to in Chapter $$2$$ as "Oblique" coordinates. Indeed, using the more familiar symbols $$u:=x^1$$, $$v:=x^2$$ , $$x:=r\cos\varphi$$ , $$y:=r\sin\varphi$$ the above transformation takes the form of one from cartesian to oblique coordinates:

$$u = x-y$$ $$v = y$$

Which you may verify have associated with them the non diagonal, constant metric:

$$g_{\mu\nu}dx^{\mu}dx^{\nu} = du^2+2dv^2+2dvdu$$

So, while this is by no means a proof that rectilinear coordinates can always be chosen, I hope it helps showing how this machinery works. In particular, note that in our second choice of the oblique coordinates, we have used a linear combination of the more elementary solutions for the PDE. Once more, please refer to @Bence Racskó to complete the mathematical background for your question, as he rightly points out, this part can be simply understood as the linear algebra fact that

a constant symmetric matrix is always transformable into diagonal form

$$^{*}$$ I've edited this from "rectilinear" to "orthogonal" after realizing that, at least by some conventions, rectilinear coordinates aren't necessarily orthogonal.

• Thanks for this, it all adds to the understanding. I was trying something similar myself but couldn't get past a couple of places where you could. Commented Jun 1, 2023 at 13:50

Dirac is not just the specialist in General Relativity. Of course, one cannot deduce a cartesian system from $$\mathcal R =0$$.

The algebraic picture came to life (Cartan, Eisenhart, Wheeler) when Dirac was very old already.

The commutator of the covariant derivatives

$$dx^i \wedge dx^k\ \left(\ ( \nabla_i \nabla_k -\nabla_k \nabla_i )\left( a^l(x) e(x)_l \right)\ \right) = dx^i \wedge dx^k \ e_s (\mathcal R^s_t)_{i,k} v^t$$

is nothing else but a rotation matrix $$\mathcal R^s_t$$ applied to the base frame $$e(x)_i$$ when transported around a small rectangle, whose sides are given by the the coordinate projectors $$dx^i$$ on tangent vectors $$v^k e_k$$.

Observe, that $$\mathcal R^s_t$$ does not differentiate the vector at all. It evaluates only the movement of the frames axis, that is expressed by the Christoffel maps on neighbouring frames while transported infintesimally around in a given 2-plane.

Conclusion: If $$\mathcal R^s_t$$=0 in any 2-plane, one immediately can spray out a flat n-spherical system of coordinates, metric linear in radius by integrating all radial geodesics and connect all points of the same radius with an euclidean $$n-1$$ sphere.

In mathematical terms: The commutator =0 is the integrability condition for the existence of a global coordinate system in flat $$R^n$$. The condition is of the same weak form as eg $$\partial_t B +\nabla \times E=0$$ in electrodynamics, which is a prerequisite for the existence of the vector potential, but does not prefere any gauge or coordinate system.

• I think this is a good answer but it may be using concepts which are too advanced for the OP and for the context of the question. In particular I know just from familiarity that Dirac doesn't explain in his book how the last two indices of the Riemann tensor are equivalent to a rotation matrix.
– Amit
Commented May 31, 2023 at 14:17
• @Amit, you are correct, I'm afraid this answer is beyond me, but thank you. Commented May 31, 2023 at 16:12