# Interpretation of Derivative of Metric = 0 in GR - Learning from Wald

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• LarryL
In summary: But that doesn't mean the metric tensor is constant. In summary, the covariant derivative of the metric tensor being equal to zero does not necessarily mean that the metric tensor is constant throughout spacetime. The covariant derivative is chosen as the unique connection that preserves the metric, but this does not imply constant curvature. Differentiating the metric is measuring the rate of change of the tensor field relative to itself, and this is something that one might intuitively expect to be zero.
LarryL
I am trying to learn GR, primarily from Wald. I understand that, given a metric, a unique covariant
derivative is picked out which preserves inner products of vectors which are parallel transported.
What I don't understand is the interpretation of the fact that, using this definition of the derivative,
the covariant derivative of the metric tensor = 0.

In my mind, the covariant derivative applied to any tensor is telling us something about
how the tensor changes from point to point.
Isn't saying that the covariant derivative of the metric tensor = 0 (everywhere, correct?) the same thing
as saying the metric tensor is "constant" (in some sense) throughout spacetime?
But, of course, it does not have to be constant. And wouldn't that imply constant curvature
(or zero curvature) everywhere? This, of course, must be a wrong interpretation (and Riemann and Ricci
tensors are clearly nonzero, and they more directly relate to curvature).

So what is wrong with my interpretation of a derivative as a measure of change?
thanks.

This stack exchange answer casts some light on it. The derivative is chosen from amongst the many available connections as the (provably unique) connection that gives zero when applied to the metric. Preserving inner products of parallel-transported vectors - the officially stated criterion for selection of the connection - is mathematically equivalent to preserving the metric.

As you observe, the constancy of the derivative at zero does not entail constant curvature, because the metric tensor is not itself a curvature tensor.

In the approach described on stack exchange the derivative is selected based on the metric. But it can also be done the other way around. IIRC, Lee does that in his book 'Riemannian Manifolds'. He defines various types of connection and then says that, given an affine connection on a manifold, a Riemannian metric is the unique symmetric 2-tensor that has a certain property in relation to the connection. I don't have the book here right now, so don't remember the property, but it's probably equivalent to giving a zero derivative of the tensor.

Here's another hokey, folksy way of looking at it, that may or may not mean anything or have any validity: using a connection to differentiate a tensor is - in a very loose sense - measuring the rate of change of the tensor field relative to the connection. But since there is a unique relation between affine connections and their corresponding metric tensors, we could say that differentiating the metric is measuring the rate of change of the tensor field relative to itself. And that is something that one might intuitively expect to be zero.

LarryL said:
Isn't saying that the covariant derivative of the metric tensor = 0 (everywhere, correct?) the same thing as saying the metric tensor is "constant" (in some sense) throughout spacetime?

No. One way to look at it, heuristically, is that the point of the covariant derivative is to tell how a vector or tensor changes "relative to the metric"; i.e., it is intended to filter out changes that are due to the way the geometry of spacetime changes, and focus on changes that are due to the way the vector or tensor itself changes. But of course all changes in the metric tensor are changes in the way the geometry of spacetime changes, so if you filter out those changes, you're left with zero change.

vanhees71

## 1. What is the significance of the derivative of metric being equal to 0 in general relativity (GR)?

The derivative of metric being equal to 0 in GR indicates that the spacetime is flat and does not have any curvature. This is a key concept in GR, as it allows for the mathematical description of gravity as the curvature of spacetime.

## 2. How does understanding the derivative of metric = 0 help in understanding GR?

Understanding the derivative of metric = 0 helps in understanding GR by providing a mathematical basis for describing the effects of gravity on spacetime. It allows for the prediction of how objects will move and interact in a gravitational field.

## 3. What does the derivative of metric = 0 tell us about the structure of spacetime?

The derivative of metric = 0 tells us that spacetime is homogeneous and isotropic, meaning that it has the same properties at every point and in every direction. This is a fundamental assumption in GR and is necessary for solving Einstein's field equations.

## 4. How does the concept of the derivative of metric = 0 relate to the Einstein field equations?

The concept of the derivative of metric = 0 is directly related to the Einstein field equations, as it is one of the conditions that must be satisfied for these equations to hold. It is a manifestation of the principle of equivalence, which states that gravitational effects are equivalent to the effects of acceleration in an inertial frame.

## 5. Are there any real-world applications of understanding the derivative of metric = 0 in GR?

Yes, there are many real-world applications of understanding the derivative of metric = 0 in GR. For example, it is used in the development of GPS technology, as the accurate tracking of time and position relies on precise understanding of the effects of gravitational fields on spacetime. It also has implications in astrophysics, where the curvature of spacetime can affect the behavior of objects such as black holes and galaxies.

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