What are Good Books on Tensors for Understanding Einstein's Field Equation?

[mathwonk]

Ok, so I think of it this way now: in physics there are various operations on (fields of) vectors taking them linearly or multi-linearly to other (fields of) vectors, and it is useful for calculations to represent these operations as tensor fields, hence entirely in terms of tensor products of dx^j and ∂/∂x^I in local coordinates.

Moreover, as long as the value at a point of the output of the operation, depends only on the values of the inputs at this point, (which is not usually true for derivative operations), this can always be done, using these rules: (finite dimensions only!)

1) L(X,Y) ≈ X*(tens)Y, for all X,Y,
2) and (XtensY)* ≈ X*(tens)Y*.
3) Also, Bil(XxY, Z) ≈ L(X(tens)Y, Z).

Hence, Bil(XxY,Z) ≈ L(X(tens)Y ,Z) ≈
(XtensY)*(tens)Z ≈ X*(tens)Y*(tens)Z.

alternately,
Bil(XxY, Z) ≈ L(X, L(Y,Z)) ≈ X*(tens)L(Y,Z) ≈ X*(tens)Y*(tens)Z.

Similarly, Trilin(XxYxZ,W) ≈ X*(tens)Y*(tens)Z*(tens)W.
(the Riemann curvature tensor lives here.)

Remark: Although the derivative of a vector field at a point wrt another vector field, depends also on input values at nearby points, miraculously, certain commutator relations involving derivatives, e.g. the Riemann curvature, do depend only on the input values at the point, hence are actually tensors.

[to paraphrase another member, please take all this with a grain of salt.]

remark: rule 3) actually follows from rules 1),2) as follows:
Bil(XxY, Z) ≈ L(X,L(Y,Z))≈ X*(tens)L(Y,Z) ≈ X*(tens)Y*(tens)Z ≈ (XtensY)*(tens)Z ≈ L(X(tens)Y, Z).

 

[mathwonk]

To someone like me, i.e. me, who knows nothing of this, and never heard of Einstein's field equations before, wikipedia looks actually useful.

It is an equation between two tensors of type (0,2). (I have been calling them type (2,0)), i.e. 2-covariant tensors, linear combinations of dx^j(tens)dx^k.

Since the dot product is such a tensor, I think of such a tensor as a generalization of a dot product, i.e. a generalization of a Riemannian metric. Since it defines a bilinear pairing on pairs of tangent vectors, it tells us something about the shape of space.

The left side is a sum of two terms, each of which is derived from the given Riemannian metric on space, more precisely it is entirely determined (up to some constant multipliers) by that metric and its first and second derivatives. Thus the left side is a certain measure of the geometric curvature of space.
This is set equal to a tensor on the right side that is called the "stress-energy" tensor, something apparently determined by matter, radiation, and force fields, in space(time). Thus the equation says a certain geometric measure of the shape of space(time) is determined by the physical attributes or content of space(time).

This equation involves three tensors, so to understand it, one should address each one of them individually. One of the tensors is just (a scalar multiple of) the Riemannian metric tensor, so the first place to start is understanding that.

Then the second tensor on the same side, is the "Einstein" tensor, which is determined by the metric, and its first and second derivatives. It involves a certain "averaging" of the Riemann curvature tensor. I.e. it is a sum of the "Ricci" curvature, and the "scalar" curvature multiplied by the metric. Both of these curvatures are computed as a sort of "trace" starting from the Riemann curvature tensor.

I.e. thinking of that Riemann curvature tensor as a sort of matrix whose entries are bilinear forms, the Ricci curvature is the trace of that matrix, hence itself a bilinear form. Then thinking of that bilinear form as the composition of the metric with a linear map in the first variable, one further takes the trace of that map, a scalar.

In coordinates, one gets the (0,2) Ricci curvature as a contraction of the (1,3) Riemann curvature; and then one gets a scalar by further contracting this (0,2) tensor against the "inverse" of the metric (0,2) tensor, which apparently makes it a (2,0) tensor.(?) [as a mathematician, it looks to me as if matrix transposes are involved, rather than matrix inverses.]

Apparently, the thing to focus on, after the metric tensor, is the Riemann curvature tensor, and then its descendants, the Ricci and scalar curvatures. Finally one would study the stress -energy tensor.

I would suggest studying closely the covariant derivative, and how it leads to the Riemann curvature operator, and finally how that operator is expressed using tensor notation. I.e. in a sense, tensors are just a notation to express this concept, and understanding the concept seems to be the more crucial matter.

To a simple person like myself, second order mixed partial derivatives are equal in flat Euclidean space, but not in curved space. Thus the difference in the values obtained by taking second derivatives in different orders in two given directions, measures curvature of space in the direction of the plane they span.
A careful, intrinsic, elaboration of this idea, using "covariant derivatives", defines the Riemann curvature operator, which can be expressed as a tensor of "type (1,3)", hence contracts to a tensor of type (0,2), and further, using the metric, to a scalar.

As a quick check on the equations, what happens if the space is geometrically flat? Then it seems the Einstein tensor is zero, since it measures the curvature. Thus the left side seems to reduce to a constant multiple of the metric. What then is the right side, the stress energy tensor? Why is it also a multiple of the metric?

Apologies for posting on something I do not know anything at all about. This is obviously just speculation. Thanks to you all for introducing me to what tensors are used for in real life.

 

[Orodruin]

This equation involves three tensors, so to understand it, one should address each one of them individually. One of the tensors is just (a scalar multiple of) the Riemannian metric tensor, so the first place to start is understanding that.

That would be the cosmological constant term. You can either keep it on the LHS or move it to the RHS and consider it part of the stress-energy tensor (corresponding to a cosmological constant dark energy component).

Then the second tensor on the same side, is the "Einstein" tensor, which is determined by the metric, and its first and second derivatives. It involves a certain "averaging" of the Riemann curvature tensor. I.e. it is a sum of the "Ricci" curvature, and the "scalar" curvature multiplied by the metric. Both of these curvatures are computed as a sort of "trace" starting from the Riemann curvature tensor.

I.e. thinking of that Riemann curvature tensor as a sort of matrix whose entries are bilinear forms, the Ricci curvature is the trace of that matrix, hence itself a bilinear form. Then thinking of that bilinear form as the composition of the metric with a linear map in the first variable, one further takes the trace of that map, a scalar.

In coordinates, one gets the (0,2) Ricci curvature as a contraction of the (1,3) Riemann curvature; and then one gets a scalar by further contracting this (0,2) tensor against the "inverse" of the metric (0,2) tensor, which apparently makes it a (2,0) tensor.(?) [as a mathematician, it looks to me as if matrix transposes are involved, rather than matrix inverses.]

When we talk about the metric inverse, we are generally referring to the natural map between tangent vectors and dual vectors implied by the metric. For example, given a tangent vector $V \in T_p M$, we have a corresponding dual vector $\omega_V \in T^*_p M$ given by
$$
\omega_V(X) = g(X,V)
$$
for all $X$. In other words, we are considering a map $g: T_p M \to T^*_p M$ implied by the metric. When there is a metric we will usually forego the distinction between tangent and dual vectors and talk about $V$ and $\omega_V$ interchangeably, just with contravariant/covariant components. We would refer to the components of $\omega_V$ as $V_a = g_{ab} V^b$.

In this sense, a (0,2) tensor can also be considered a map $g: T_pM \to T^*_pM$ and the inverse $g^{-1}$ would then be a (2,0) tensor, being a map $g^{-1}:T^*_pM \to T_pM$ such that
$$
g^{-1}(\xi, \omega_V) = \xi(V)
$$
for all dual vectors $\xi$.

Apparently, the thing to focus on, after the metric tensor, is the Riemann curvature tensor, and then its descendants, the Ricci and scalar curvatures. Finally one would study the stress -energy tensor.

The metric tensor is the focus all of the time. The Riemann curvature tensor, and therefore also the Ricci tensor and Ricci scalar, are generally written in terms of the metric components and their derivatives. This makes the Einstein field equation a non-linear coupled system of partial differential equations for the metric components. The stress-energy tensor on the right hand side depends on the matter content that you put into the physical system.

The Einstein field equations generally follow from extremising the Einstein-Hilbert action with additional terms describing the cosmological constant and the matter content. The EFEs drop out when you vary the action with respect to the metric components (or, more naturally, the inverse metric components). In addition to these equations, you will also need the equations of motion for the matter fields entering through the stress energy tensor. These are acquired by varying the action with respect to the matter fields.

I would suggest studying closely the covariant derivative, and how it leads to the Riemann curvature operator, and finally how that operator is expressed using tensor notation. I.e. in a sense, tensors are just a notation to express this concept, and understanding the concept seems to be the more crucial matter.

Yes, this is a crucial part of understanding general relativity. Not only because of the fact that the covariant derivative is necessary to understand the EFEs, but also because it is absolutely required to understand it in order to interpret the results once you have a solution.

In addition, you will ideally need the concepts of the Lie derivative, Killing fields, and derived symmetries to help you along in finding solutions to the EFEs as well as to study the dynamics of test particles in the resulting spacetime.

As a quick check on the equations, what happens if the space is geometrically flat, and there is no input from the stress-energy?

Then you get the Minkowski space of special relativity - at least locally (up to non-trivial topologies globally).

 

[mathwonk]

Orodruin to the rescue!
As Orodruin makes clear, a map from T to T*, (where T is the tangent space), corresponds to a tensor of type (0,2), say by rule 1), post #31, since L(T,T*) ≈ T*(tens)T*. Thus its inverse is a map from T* to T, and by the isomorphism L(T*,T) ≈ T**(tens)T ≈ T(tens)T, the inverse (not the transpose) corresponds to a tensor of type (2,0).

In fact, if as I said, the matrix of the (0,2) form is represented by the composition of (the matrix of) a map and (the matrix of) the metric, then to recover the matrix of that map, we would compose the (0,2) form with the inverse of the metric.

Finally, in this setting, all the matrices involved are symmetric, so transposes would not change anything. More to the point, taking the transpose of a map does not change the type of the corresponding tensor, (again by rule 1)).