General Relativity Tutorial III: Coordinate Transformation and Covariant Derivative

In the previous tutorial, we explored the metric tensor and basis. In this tutorial, we will delve into the concept of tensors and their properties related to coordinate transformations. 

A tensor is a multidimensional array that obeys certain transformation rules under coordinate transformations. A vector is an example of a tensor, specifically one with rank 1. One of the key properties of a vector is that it possesses both magnitude and direction. The significance of direction lies in the fact that a vector can be transformed under a coordinate transformation. 

Let's consider the basis vectors denoted by $g_i$, which can be expressed as $g_i = \frac{\partial \vec{r}}{\partial x^i}$. We also have another basis, denoted by $g'_i=\frac{\partial \vec{r}}{\partial x'^i}$. 

By applying the chain rule, we can establish the following relationship between the two bases: 

$\frac{\partial \vec{r}}{\partial x^j} = \frac{\partial \vec{r}}{\partial x'^k} \frac{\partial x'k}{\partial x^j}  = g'_k \frac{\partial x'^k}{\partial x^j}$, 

Now, let's apply this coordinate transformation to two types of vectors: covariant and contravariant vectors. 

To do this, let's consider a vector $\vec{u}$. As we learned in the previous tutorial, we can express this vector in two different bases: one as $u^i g_i$ and the other as $u_i g^i$. Here, $u_i$ is called the contravariant component, and $u^i$ is the covariant component. 

Contravariant component: 

For the contravariant component, the coordinate transformation should satisfy the following relationship: 

$\vec{u} = u^i g_i = u'^k g'_k$ 

Hence, the contravariant component can be obtained as: 

$u^i g_i = u^i \frac{\partial x'^k}{\partial x^i} g'_k = u'^k g'_k$, which yields $u'^k = \frac{\partial x'^k}{\partial x^i} u^i$.

Covariant component:

$$\stackrel{}{}$$

On the other hand, for the covariant component, the coordinate transformation leads to:

$\vec{u} = u_j g^j = u'_k g'^k$,

which can be written as:

$ u_j g^j g'_i =  u'_k g'^k g'_i $  $\Rightarrow u_j \frac{\partial x^j}{\partial x'^i}  = u'_k \delta'^k_i = u'_i$.

Therefore, we find: 

$ u'_i = u_j \frac{\partial x^j}{\partial x'^i}$. 

Consequently, the two components transform differently under coordinate transformations. Now, what about a rank-2 tensor? Let's consider the relation between $T^{kl}$ and $T'^{ij}$. The contravariant components $T^{kl}$ of the rank-2 tensor should transform as follows:

$T'^{ij} = \frac{\partial x'^i}{\partial x^k} \frac{\partial x'^j}{\partial x^l} T^{kl}$.

This equation describes the transformation of rank-2 tensor components. Tensors must satisfy this transformation rule. The reason why this relation is important is that the laws of physics should remain unchanged under coordinate transformations. For instance, let's consider Newton's Second Law, $\vec{F} = m\vec{a}$. 

If we change the coordinate system, the equation can be written as $\vec{F'} = m \vec{a}'$. Both of these vector equations should satisfy the coordinate transformation, ensuring that the laws of physics can be described from different observers' perspectives while still maintaining their validity. 

Einstein also considered this mathematical structure of coordinate transformations. His field equations in general relativity are described using rank-2 tensors. However, before explaining those equations, it is important to consider another significant property of tensors.

Derivative of tensor is NOT a tensor

Let's consider the derivative of a tensor and its coordinate transformation:

$\partial_\nu A^\mu \longrightarrow \partial'_\nu A'^\mu$.

We would expect the coordinate transformation to be given by:

$\partial_\nu A^\mu = \frac{\partial x^\lambda}{\partial x'^\nu} \frac{\partial x'^\mu}{\partial x^\rho} \partial_\lambda A^\rho $.

Unfortunately, when we calculate the derivative term, we encounter an issue:  

$ \partial'_\nu A'^\mu = \frac{\partial}{\partial x'^\nu} \left( \frac{\partial x'^\mu}{\partial x^\rho} A^\rho \right) = \frac{\partial x^\lambda}{\partial x'^\nu} \frac{\partial x'^\mu }{\partial x^\rho} \partial_\lambda A^\rho + \frac{\partial^2 x'^\mu}{\partial x'^\nu \partial x^\rho} A^\rho$.

The additional term that appears breaks the coordinate transformation of the given vector. Therefore, the derivative of a tensor is not a tensor because it fails to satisfy the coordinate transformation.

Covariant Derivative

To overcome this problem, we introduce the concept of the covariant derivative:

$D_\mu A^\nu = \partial_\mu A^\nu - \Gamma^\nu_{\mu \lambda} A^\lambda $, 

here $\Gamma^\nu_{\mu \lambda}$ is defined as the `Christoffel symbol' and is given by:

$\Gamma^\nu_{\mu \lambda} = \frac{1}{2} g^{\nu \rho} [ \partial_\lambda g_{\mu \rho} + \partial_\mu g_{\lambda \rho} - \partial_\rho g_{\mu \lambda}$ ].

The inclusion of the Christoffel symbol plays a crucial role in eliminating the extraneous term that arises during the coordinate transformation of the derivative vector. Consequently, the covariant derivative ensures the preservation of coordinate transformations, thus maintaining its consistency with tensor analysis. It is worth noting that the covariant derivative corresponds to the first derivative in tensor analysis, while the concept of the second derivative will be further elucidated in our upcoming tutorial.

Then, let us meet at the next tutorial as we delve deeper into the intricacies of tensors and their fascinating properties for Einstein Field equations.