# Question about basic formalism of GR and the metric tensor

I really don't know much about GR, but I've come across a few rough sketches of its formalism in my DG books. I'm trying to piece it together to get a very basic intuition of what spacetime is in GR. Spacetime will be a $4$ dimensional, smooth manifold $M$, equipped with a symmetric, nondegenerate, $(0,2)$-tensor field $g$. Let $p\in M$ and $U$ be a coordinate neighborhood containing $p$. Let $x^0,...,x^3$ be local coordinates for $p$. Then we can define a basis {$\partial_{0},...,\partial_{3}$} at $T_pM$. Then the dual basis for $T_p^*M$ is given by {$dx^0,...,dx^3$}. Define $g_{ij}=\langle\partial_{i},\partial_{j}\rangle$, a $C^{\infty}$ function of $p$, where $\langle .,.\rangle_p$ is some "inner-like" product defined on $T_pM\cong\mathbb{R}^4$. (I use inner-like because $g$ is only nondegenerate, not positive definite, and I don't know what to call it.) Then setting $g_p(X,Y)=\langle X,Y\rangle_p$, $g_p$ defines an inner-like product on $T_pM$. If $X,Y\in T_pM$, with $X=\sum_{i=0}^3a^i\partial_{i}$ and $Y=\sum_{i=0}^3b^i\partial_{i}$, then using linearity of $g_p$, $g_p(X,Y)=\sum_{i,j=0}^3a^ib^j\langle\partial_{i},\partial_{j}\rangle=\sum_{i,j=0}^3g_{i,j}dx^i\otimes dx^j(X,Y)=g_{ij}dx^i\otimes dx^j(X,Y)$. Hence $g=g_{ij}dx^i\otimes dx^j$. $(M,g)$ can be seen as a "metric-like" space (again, not sure what to call it) where we can define "generalized distance" of a point $q\in M$ from $p$ by $ds^2=g_{ij}dx^idx^j$. If we choose units so $c=1$, let $M=\mathbb{R}^4$ with cartesian coordinates $x^0=t,...,x^3$, and $g_{ij}=\eta_{ij}:=\left\{ \begin{array}{lcc} -1 ,& i=j=0 \\ \\ 0, & i\neq j \\ \\ 1, & i=j=1,2,3 \end{array} \right.$

then we get special relativity with $T_pM=\mathbb{R}^4$ with origin $p$. But if spacetime is curved, and $q$ is very close to $p$, then measuring $ds^2$ can be very closely approximated by using $\eta$ at $T_pM$. In this sense, special relativity is a local approximation of GR.

I know that was a bit long-winded, but is this basically the setup for GR? If there's any misunderstandings, I'd appreciate it if someone could mention them.

Thanks

• That's more or less the idea behind GR's mathematical formalism, but the heart of GR is the Einstein equation for $g_{ab}$, which relates curvature to the matter distribution. – Jerry Schirmer Apr 2 '15 at 17:49
• And I wouldn't say that the setup is complete until you start at least talking about the kinematics you get from the geodesic equation. – Jerry Schirmer Apr 2 '15 at 17:56
• Thanks! I know there's much more to GR than this, but I was just trying to piece together a very naive idea of what spacetime is and how the metric tensor acts on it, based on my limited reading of DG textsm, where GR is mentioned only in passing. – user153582 Apr 2 '15 at 18:24

When we talk about the geometry of GR, it is understood that the manifold of spacetime is not a Riemannian one, but rather a Lorentzian manifold. This means that the metric is not positive definite. With this understanding, we call $g(.,.):=\langle.,.\rangle$ an inner product as usual. This lack of positive definiteness has many consequences. It is the reason behind the rich study of causal structure. Here are two striking consequences: there exist nontrivial curves of $0$ proper time (Lorentz $\mathrm{d}s^2$). We identify these with the paths of light rays. On a Riemannian manifold, a geodesic is ideally a minimum of length. However, in GR a geodesic is (barring such things as conjugate points) a maximum of proper time.

You are correct that GR and SR are locally related. In particular, the Equivalence Principle says (amongst other things) that the laws of SR hold in GR when one is on a geodesic. To get GR equations from SR ones, one simply transforms into non-freefall coordinates. This usually amounts to $\eta_{ab}\rightarrow g_{ab}$ and $\partial_a\rightarrow\nabla_a$. Note that this is not foolproof: there are certain equations in GR that involve various curvature terms. Since the curvature is identically zero in SR, we cannot recover these equations directly$^1$. (Mathematically, the Equivalence Principle states that there is a local homeomorphism between the GR spacetime and Minkowski spacetime, which just says that spacetime is a manifold.) (Note further that there are many different formulations of this principle. I am working off of the formulation given in N. Straumann, General Relativity (2013), section 2.3 "Spacetime as a Lorentzian Manifold".)

You are, however, missing a critical piece of the puzzle. Here we quote R.M. Wald, General Relativity (1984):

The entire content of general relativity may be summarized as follows: Spacetime is a manifold $M$ on which there is defined a Lorentz metric $g_{ab}$. The curvature of $g_{ab}$ is related to the matter distribution in spacetime by Einstein's equation.

The entire structure and dynamics of GR is contained in this sentence. We conclude by showing that the equation of motion of a free particle can be derived from the field equations. (I have seen some sources say that the geodesic hypothesis is necessary for the formulation of GR. I will show that it is not.) In SR we have the energy-momentum tensor of a pressureless perfect fluid as $T_{ab}=\rho u_a u_b$. This is trivially extended to GR: $T_{ab}=\rho u_a u_b$. Note that $u_a u^a(:=\langle u,u\rangle)=-1$. The Einstein equations tell us that the covariant divergence of this tensor vanishes, $\nabla^a T_{ab}=0$. Thus $u_b\nabla^a(\rho u_a)+\rho u_a\nabla^a u_b=0$. Taking the covariant divergence of $u_au^a=-1$ gives $u^a\nabla^b u_a=0$. Now contract $\nabla^a T_{ab}=0$ with $u^b$. We find $\nabla^a(\rho u_a)=0$, whence $u^b\nabla_b u_a=0$. We have thus shown that a piece of noninteracting dust (i.e. a free particle) obeys the geodesic equation. This is the kinematics that Jerry Schirmer spoke of in the comments above.

$^1$ Here is an example. Let $A_a$ denote the electromagnetic 4-potential and $j^a$ the electromagnetic 4-current. $R_{ab}$ is the Ricci tensor as usual. Then $$\nabla_b\nabla^b A^a-R^a{}_b A^b=-4\pi j^a$$ holds in the Lorentz gauge. In SR, the curvature is trivially zero, and the covariant derivatives reduce to ordinary partial derivatives. Thus, in SR, $$\Box A^a=-4\pi j^a$$ where $\Box$ is the d'Alembertian. Indeed, one may derive this equation within the framework of SR electromagnetism. Simply replacing $\partial\rightarrow\nabla$ does not work. In this case, this happens because second order covariant derivatives do not commute, unlike their flat space counterparts.

Your post has many details, but overall you were a little vague about the signature (++++, +++-, ++--, +---, ----) of the metric. In fact it only came up when you said the metric is close to $\eta$ when two points are close. And that could be interpreted ina way that isn't true. For any point there exists a coordinate system where the metric is very close to $\eta,$ but not every metric will be close to $\eta,$ that isn't even true in SR if you allow arbitrary coordianates.

The next problem is that you omit any connection to stress energy, simply put, spacetime has to curve a certain way that is consistent with the stress energy present. This puts constraints on what kinds of spacelike surfaces are allowed and tells the system how to evolve. Just like with electromagnetism, the equation $\vec \nabla \cdot \vec B=0$ constrains the initial data, and the equations $\vec \nabla\times \vec B=\mu_0\left(\vec J + \epsilon_0\frac{\partial \vec E}{\partial t}\right)$ and $\vec \nabla\times \vec E=-\frac{\partial \vec B}{\partial t}$ tell it how to evolve. If you don't do this then you can have warp drives and such because you can write down pretty much any manifold. Not any manifold, that would defeat the point of your question, but highly unphysical ones. And not mentioning that stress-energy constrains initial conditions and tells it how to evolve would be like forgetting to mention that lack of magnetic charge constrains initial data (i.e. $\vec \nabla \cdot \vec B=0$) and that currents change hows fields evolve e.g. $\frac{1}{\epsilon_0}\left(\frac{1}{\mu_0}\vec \nabla\times \vec B-\vec J\right)=\frac{\partial \vec E}{\partial t}.$

Finally, I do object to you assuming your manifold is smooth. This poses additional constraints for unphysical (and purely mathematical) reasons. And such constraints are not always benign. For instance analyticity is stronger than smoothness, but sometimes forcing a solution to be analytic requires that regions of time travel (Closed Timelike Curves) arise within chronology horizons even if solutions to Einstein's Field Equations exist with the same initial conditions that do not have time travel. Just assuming something for mathematical reasons that force a time machine into existence when otherwise it was possible for no time machine is about as big an ask as you can make. So analyticity is too much. I don't know exactly the price to be paid for mere smoothness, but any physical cost is too high when there is no physical reason, so just make it differentiable enough and don't insist on more just because some mathematician or student didn't want to keep track of how differentiable something was.

And if you did instead want to go the route of smoothness because you just like the math, why not just use synthetic differential geometry and make an axiomatic system out of the whole thing.