# Covariant derivative of metric tensor

• Damidami
In summary, the covariant derivative of the metric tensor is not D(g) = 0. It is D(g) = 2a^2sin(\theta)cos(\theta) - (0+0) - (0+0)
Damidami
Hi, I'm trying to verify that the covariant derivative of the metric tensor is D(g) = 0.
But I have a few questions:
1) This is a scalar 0 or a tensorial 0? Because it is suposed that the covariant derivative of a (m,n) tensor is a (m,n+1) tensor, and g is a (0,2) tensor so I think this 0 should be a (0,3) tensor. Am I right?

2) Following the MTW book in the example of page 341 we have:
$$\Gamma^{\theta}_{\phi\phi} = -sin(\theta)cos(\theta)$$
$$\Gamma^{\phi}_{\phi\theta} = cos(\theta)/sin(\theta)$$
and all the other $$\Gamma = 0$$

But when I try to verify the covariant derivative of the metric tensor for the component $$g_{\phi\phi;\theta}$$ it doesn't give me 0, but instead:

$$g_{\phi\phi;\theta} = g_{\phi\phi,\theta} - \Gamma^{k}_{\theta\phi} g_{k\phi} - \Gamma^{k}_{\theta\phi} g_{\phi k} = 2a^2sin(\theta)cos(\theta) - (0+0) - (0+0) = 2a^2sin(\theta)cos(\theta) \neq 0$$

I checked it a lot of times and am not sure if this is a conceptual error or a procedure error.
Can someone clarify this to me?

Damidami said:
Hi, I'm trying to verify that the covariant derivative of the metric tensor is D(g) = 0.
But I have a few questions:
1) This is a scalar 0 or a tensorial 0? Because it is suposed that the covariant derivative of a (m,n) tensor is a (m,n+1) tensor, and g is a (0,2) tensor so I think this 0 should be a (0,3) tensor. Am I right?
Yup, your reasoning is correct: it would be the type (0,3) tensor with every component zero.

2) Following the MTW book in the example of page 341 we have:
$$\Gamma^{\theta}_{\phi\phi} = -sin(\theta)cos(\theta)$$
$$\Gamma^{\phi}_{\phi\theta} = cos(\theta)/sin(\theta)$$
and all the other $$\Gamma = 0$$

But when I try to verify the covariant derivative of the metric tensor for the component $$g_{\phi\phi;\theta}$$ it doesn't give me 0, but instead:

$$g_{\phi\phi;\theta} = g_{\phi\phi,\theta} - \Gamma^{k}_{\theta\phi} g_{k\phi} - \Gamma^{k}_{\theta\phi} g_{\phi k} = 2a^2sin(\theta)cos(\theta) - (0+0) - (0+0) = 2a^2sin(\theta)cos(\theta) \neq 0$$

I checked it a lot of times and am not sure if this is a conceptual error or a procedure error.
Can someone clarify this to me?
I don't know what the actual question is (since I don't have the book to hand) but I can't quite follow your last line. I get $$g_{\phi\phi;\theta}=g_{\phi\phi,\theta}-g_{k\phi}\Gamma^{k}_{\phi\theta} - g_{\phi k}\Gamma^{k}_{\phi\theta}= g_{\phi\phi,\theta}-2g_{\phi\phi}\Gamma^\phi_{\phi\theta}$$

Is this what you get before plugging in the values?

Damidami said:
Hi, I'm trying to verify that the covariant derivative of the metric tensor is D(g) = 0.
But I have a few questions:
1) This is a scalar 0 or a tensorial 0? Because it is suposed that the covariant derivative of a (m,n) tensor is a (m,n+1) tensor, and g is a (0,2) tensor so I think this 0 should be a (0,3) tensor. Am I right?

2) Following the MTW book in the example of page 341 we have:
$$\Gamma^{\theta}_{\phi\phi} = -sin(\theta)cos(\theta)$$
$$\Gamma^{\phi}_{\phi\theta} = cos(\theta)/sin(\theta)$$
and all the other $$\Gamma = 0$$

But when I try to verify the covariant derivative of the metric tensor for the component $$g_{\phi\phi;\theta}$$ it doesn't give me 0, but instead:

$$g_{\phi\phi;\theta} = g_{\phi\phi,\theta} - \Gamma^{k}_{\theta\phi} g_{k\phi} - \Gamma^{k}_{\theta\phi} g_{\phi k} = 2a^2sin(\theta)cos(\theta) - (0+0) - (0+0) = 2a^2sin(\theta)cos(\theta) \neq 0$$

I checked it a lot of times and am not sure if this is a conceptual error or a procedure error.
Can someone clarify this to me?

I do get zero. Note that $$\Gamma^\phi_{\theta \phi}$$ is not zero, it is equal to
$$\Gamma^\phi_{\phi \theta}$$ !

Indeed that was the mistake, I forgot the symetry of the chistoffel symbols, it gives 0 now!
:)

## What is the covariant derivative of a metric tensor?

The covariant derivative of a metric tensor is a mathematical operation that is used to differentiate a tensor field, such as the metric tensor, with respect to another tensor field, such as a vector field. It takes into account the curvature of the underlying space and ensures that the resulting derivative is invariant under coordinate transformations.

## Why is the covariant derivative of a metric tensor important?

The covariant derivative of a metric tensor is important because it allows us to define a notion of parallel transport for vectors and tensors on a curved space. This is essential for understanding the behavior of physical quantities in general relativity and other theories of gravity.

## How is the covariant derivative of a metric tensor calculated?

The covariant derivative of a metric tensor is calculated using a formula that involves the Christoffel symbols, which are coefficients that describe the curvature of the underlying space. The specific formula depends on the dimensionality of the space and the metric tensor being used.

## What is the relationship between the covariant derivative and the metric tensor?

The covariant derivative of a metric tensor is closely related to the metric tensor itself. In fact, the covariant derivative of the metric tensor is equal to zero if and only if the metric tensor is constant. This property is known as metric compatibility and is a key feature of the covariant derivative.

## Can the covariant derivative of a metric tensor be extended to higher dimensions?

Yes, the covariant derivative of a metric tensor can be extended to higher dimensions. In fact, the formula for the covariant derivative becomes more complicated as the dimensionality of the space increases. However, the underlying principles and properties of the covariant derivative remain the same in higher dimensions.

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