This is the final section of the mathematical section part
of this report. Having some concept of the basics of the curvilinear system, we
are now in position to proceed with the concept of the Riemann Tensor and the
Ricci Tensor.

We first start off with the Riemann Tensor. Let us consider a following figure,

In the figure above we can parallel
transport the vector $\vec{V}$ two ways around the loop from $A$ to $D$. Now
the question we could ask ourselves is that if the vector being transported
match up. In the general we would say that the vectors do match up and we
wouldn’t doubt our conscience. But this in general wouldn’t be correct, since
we are so much familiar to using the Cartesian system of the coordinate. In
general there could be various different geometries beside the usual Euclid’s
geometry. In the various other geometries, the measure of the difference in the
vector is indeed the measure of the deviation from the usual Cartesian coordinate
or similarly known as the Riemann curvature tensor of the system.
If we look at the above diagram, in
general the mathematical statement equivalent is as follows,

After the partial differential and substituting $V_{n}$ to $V_{m}$
i.e. substituting the dummy index $n$ to $m$, we have the following,

Here, $R_{\alpha \beta \gamma }^{m}$ is called the curvature tensor or the Riemann Curvature Tensor and is equal to,

**RICCI TENSOR**

Now that we have established the
Riemann tensor as the notion of the curvature, we may want to relate it to the
stress-energy tensor. Unfortunately a $4^{th}$ rank tensor cannot equal a $2^{nd}$
rank one, so we are motivated to consider $2^{nd}$ rank tensor built from
Riemann curvature tensor.

If we look expand the curvature
tensor, it has a forbidding $256$ components. Luckily several symmetries reduce
these substantially. These symmetries reduce the number of independent
component to $20$. And these symmetries also mean that there is only one
independent contraction to reduce to the second rank tensor or the Ricci
tensor.

Notice here the indices being contracted; contracting the
other indices gives us zero. We can thus write the expression for the Ricci
tensor as follows,

If we further contract the Ricci tensor we have the quantity
the Ricci scalar.

We can ask ourselves the question
what if the Ricci tensor is zero?

The definite answer would be that it
implies the space is flat. That is to say that there exists a transformation
property between the curvilinear system concerned and the Cartesian system.

Extra note here..I hope this is helpful..

Extra note here..I hope this is helpful..

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