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# Einstein Tensor

This is the fifth in a series of articles about tensors, leading up to the theory of general relativity. The last article about parallel transport and geodesics finished with an epic section about the Riemann tensor. This article builds up from the metric tensor through the Riemann tensor to the Einstein tensor, the thing on the left side of the equation for the general theory of relativity.

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## Metric Tensor

I introduced metric tensors in the article about covariant derivatives. They are two dimensional tensors related to a coordinate system. They have many useful properties. Perhaps the most useful is the ability to define lengths between points. For example, in flat Euclidean three dimensional space, the metric tensor is the 3x3 identity matrix:

$$g_{ab} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

To find the length of a vector in 3 dimensional space, you can multiply twice by this matrix. For example, imagine you are setting at a table (which may very well be true) in a room. Space-time is probably quite flat across that room (unless you are in a massive cave very deep underground and close to the Earth's core). You could draw a vector from a corner of the room to the corner of your desk. In meters you might measure the (x,y,z) distances as:

$$V_a = \begin{bmatrix}2 & 1 & 0.8 \end{bmatrix}$$

To measure the length of this vector (squared) multiply twice by g:

length of $$V$$ squared = $$V^a g_{ab} V^b = \begin{bmatrix}2 \\ 1 \\ 0.8 \end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}2 \\ 1 \\ 0.8 \end{bmatrix} = 2^2 + 1^2 + 0.8^2 = 5.64$$

In equations in books, the length of the vector is often given as ds, and this means the length is:

$$ds^2 = dx^2 + dy^2 + dz^2$$

Which is roughly the same forumla as Pythagoras calculated 2500 years ago.

## Metric Tensor in Minkowski Space Time

However, we are not living in a flat Euclidean space, even in a nice small room far from a center of gravity. We are living in Minkowski spacetime. This was first mentioned in the article on special relativity. Minkowski space time has 4 dimensions, time and the three spatial dimensions. Its metric is:

$$g_{ab} = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Which means that vector lengths are calculated by:

$$ds^2 = - (c * dt)^2 + dx^2 + dy^2 + dz^2$$

Note that the time dt is multiplied by the speed of light c. This is so that all the units are in terms of meters. When dt=1, c * dt is 1 light second, a distance of 299,792,458 meters. In 1922 Einstein first published the book The Meaning of Relativity based on his lectures, and he uses dl for this instead. Though other books I've seen just always explicitly put the c next to the dt, and others just put dt and assume the c is there. Einstein's book also puts the time last instead of first. And other sources (like the text book I'm reading) make the time positive and the spatial dimensions negative:

$$ds^2 = (c * dt)^2 - dx^2 - dy^2 - dz^2$$

Einstein's book also puts the square root of negative one in there sometimes so that everything becomes an addition:

$$ds^2 = dx^2 + dy^2 + dz^2 + (i * c * dt)^2$$

I'll stick to the way it's done on the Wikipedia page for the Minkowski metric with time first and negative as above. The maths ends up roughly the same whichever version you use.

This now represents the distance between two events rather than two points. This distance will be negative for two events occurring at the same place but at a different times (a timelike vector), and positive for events occurring at the same time but in different places (a spacelike vector).

## Lorentz Tranformations

This is a bit of a tangent, but it shows how intimate Minkowski spacetime and relatively are. In the special relativity article, I also talked about the Lorentz transformations. These transformations describe how space and time seem to warp and distort when an observer watches an object going by really really really fast. When an object moves in the x-direction, the observed coordinate changes are:

$$t' = \gamma (t - \frac {vx}{c^2})$$

$$x' = \gamma (x - vt)$$

$$y' = y$$

$$z' = z$$

Where $$\gamma$$ is the Lorentz factor which is zero for every day speeds, but gets infinitely large as an object approaces light speed:

$$\gamma = \frac1{\sqrt{1-\frac{v^2}{c^2}}}$$

It is easy to show that the length calculated above is invariant under a Lorentz transformation. I say "easy" but it does involve a lot of algebra:

$$s'^2 = - (ct')^2 + x'^2 + y'^2 + z'^2 =$$ $$- \gamma^2 (ct - \frac {cvx}{c^2}) ^2 + \gamma^2 (x - vt)^2 + y^2 + z^2 =$$ $$\gamma^2 (-c^2 t^2 + 2 \frac {c^2tvx}{c^2} - \frac {v^2x^2}{c^2} + x^2 - 2xvt + v^2t^2) + y^2 + z^2 =$$ $$\gamma^2 (- c^2 t^2 + v^2t^2 - \frac {v^2x^2}{c^2} + x^2) + y^2 + z^2 =$$ $$\gamma^2 (- c^2 t^2 (1 - \frac {v^2}{c^2}) + x^2 (1 - \frac {v^2}{c^2}) + y^2 + z^2 =$$ $$- (ct)^2 + x^2 + y^2 + z^2 = s^2$$

We can now turn these Lorentz transformations into a matrix, which (because the t and x are linear) will be both a coordinate transformation matrix and a Jacobian. To do that though, we'll need to use relativistic units where the velocity is replaced by a fraction of the speed of light $$\beta = \frac{v}{c}$$. For example, at one quarter the speed of light $$\beta = 0.25$$. At the same time, we'll multiply both sides of the equation for time by the speed of light c. This now corresponds with what we did above, so that all coordinates will represent distances. It means that the left side of the time equation is now $$ct'$$:

$$ct' = \gamma (ct - c \frac{v}{c} \frac{x}{c}) = \gamma (ct - \beta x)$$

$$x' = \gamma (x - \frac{v}{c} c t) = \gamma (x - \beta c t)$$

Here is the Lorentz transformation matrix:

$$\begin{bmatrix} ct' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} ct \\ x \\ y \\ z \end{bmatrix}$$

The inverse equations have positives instead of negatives:

$$ct = \gamma (ct' + \beta x')$$

$$x = \gamma (x' + \beta c t')$$

And the inverse Lorentz transformation matrix is therefore:

$$\begin{bmatrix} ct \\ x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \gamma & \gamma \beta & 0 & 0 \\ \gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} ct' \\ x' \\ y' \\ z' \end{bmatrix}$$

Now we can also show that the metric is invariant under a Lorentz transformation. The metric tensor has two covariant dimensions and so is transformed by multiplying by S twice.

$$g'_{ab} = g_{cd} S^c_a S^d_b$$

For this it will be useful to put $$\gamma^2$$ in terms of $$\beta$$:

$$\gamma^2 = \frac1{1-\frac{v^2}{c^2}} = \frac1{1-\beta^2}$$

S is the matrix that converts from primed coordinates back to the original coordinate system, the second matrix above. So now we will multiply g by S twice. In these multiplications, we multiply the columns of the first S times the columns of the second S. Here's some tensor algebra:

$$g'_{ab} = g_{cd} S^c_a S^d_b = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \gamma & \gamma \beta & 0 & 0 \\ \gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \gamma & \gamma \beta & 0 & 0 \\ \gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} =$$ $$\begin{bmatrix} -\gamma & -\gamma \beta & 0 & 0 \\ \gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \gamma & \gamma \beta & 0 & 0 \\ \gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} =$$ $$\begin{bmatrix} - \gamma^2 + \gamma^2 \beta^2 & - \gamma^2 \beta^2 + \gamma^2 \beta^2 & 0 & 0 \\ - \gamma^2 \beta^2 + \gamma^2 \beta^2 & - \gamma^2 \beta^2 + \gamma^2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} =$$ $$\begin{bmatrix} \gamma^2 (\beta^2 - 1) & 0 & 0 & 0 \\ 0 & \gamma^2 (1 - \beta^2) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} =$$ $$\begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} = g_{ab}$$

## Raising and Lowering Indices

This is even more of a tangent, but will be needed later on in this article. Another very useful property of the metric tensor is that it kind raise and lower the indices on a vector/tensor. It can turn a contravariant vector like $$V^a$$ into a covaraint vector $$V_a$$. This goes back to the first article about tensor types, but I now understand it much better.

Firstly, the terms "contravariant" and "covariant" refer to how vectors change when you move from one coordinate system to another. Therefore they are meaningless (I think) if you always stay in the same coordinate system. For example, consider the vector [2,2]. It could represent anything - the score in a football match, or the number of human arms and legs currently in my living room. Or it could describe the width and length of a desk in meters. We can plot this vector:

Let's say we wanted to convert that into feet. We can transform this vector into a feet-based coordinate system by mutliplying it by a simple matrix (called T), which contains the number 3.281, the number of feet in a meter:

$$V^{\text{feet}} = T V^{\text{meters}} = \begin{bmatrix}3.281 & 0\\0 & 3.281\end{bmatrix} \begin{bmatrix}2 \\ 2 \end{bmatrix} = \begin{bmatrix}6.562 \\ 6.562 \end{bmatrix}$$

That matrix is the Jacobian, also known as the inverse transformation matrix. It is a table of partial derivatives. For example the top left entry specifies that the feet width changes by 3.281 for every 1 meter width:

$$J = T = \frac{\partial \text {feet}} {\partial \text {meters}} = \begin{bmatrix} \frac{\partial \text {width in feet}} {\partial \text {width in meters}} & \frac{\partial \text {width in feet}} {\partial \text {length in meters}} \\ \frac{\partial \text {length in feet}} {\partial \text {width in meters}} & \frac{\partial \text {length in feet}} {\partial \text {length in meters}} \end{bmatrix} = \begin{bmatrix}3.281 & 0\\0 & 3.281\end{bmatrix}$$

We can also plot this, but note that I have kept the vector the same, but just changed the spacing of the coordinate grid. This reflects what happens to real desks. They don't change size just because you measure them in feet rather than meters. The desk won't accommodate any more clutter:

Because the vector transforms in this particular way, with the feet on top and meters on bottom in the Jacobian, it is called contravariant. The "contra" is because the numbers get bigger while the grid (aka measurements aka basis vectors) gets smaller. The vectors vary contrary to the coordinate basis. Contravariant vectors are written with a raised index like $$V^a$$.

Symbollically the matrix T is also written with indices like a and b. These respresent summation indices. And usually the coordinate system we are going to is indicated with an apostrophe or a tilde, and mathematicians have a great love of the letter x. So the symbols actually look like this where $$\equiv$$ means "equivalent notation":

$$J \equiv T^a_b \equiv \frac {\partial x'^a} {\partial x^b} \equiv \frac {\partial \tilde x^a} {\partial x^b}$$

But the same set of numbers [2,2] can mean many other things. Continuing with the furniture theme, it could be a way of computing the perimeter of the desk, by multiplying the width/length of the desk by this vector. Imagine a desk 1.5 meters wide and 3 meters long and calculate its perimeter:

perimeter = $$\begin{bmatrix}2 & 2 \end{bmatrix} \begin{bmatrix}1.5 \\ 3 \end{bmatrix} = 2*1.5 + 2*3 = 9$$

To transform this vector into feet, we need to multiply by the inverse of the Jacobian matrix above, also known as the transformation matrix S. In this case, for every 1 foot width, the meters changed by about 0.3:

$$V_{\text{perimeter in feet}} = V_{\text{perimeter in meters}} S = \begin{bmatrix}2 & 2 \end{bmatrix} \begin{bmatrix}0.305 & 0\\0 & 0.3.05\end{bmatrix} = \begin{bmatrix}0.610 \\ 0.610 \end{bmatrix}$$

We can compute perimeter again and we get the same answer (in meters). The table is 1.5m=4.922ft wide and 3m=9.843ft long:

perimeter = $$\begin{bmatrix}0.61 & 0.61 \end{bmatrix} \begin{bmatrix}4.9 \\ 9.8 \end{bmatrix} = 0.61*4.922 + 0.61*9.843 = 9$$

This vector is an example of a linear function. Because it transforms in this way, with the meters on top and feet on bottom, it is called covariant. The "co" is because the numbers get smaller while the grid gets smaller. The vectors co-vary in the same direction as the coordinate basis. Covariant vectors are written with a lowered index like $$V_a$$.

Symbolically, this matrix is the inverse of the J and T above and can be written as:

$$J^{-1} = S^a_b = \frac {\partial x^a} {\partial x'^b} = \frac {\partial x^a} {\partial \tilde x^b}$$

So contravariant vectors are coordinates and covariant vectors are functions. The main thing to note is that the vector [2,2] by itself is neither a contravariant nor covariant. It only becomes contravariant or covariant when we transform it. If I described to you a wonderful new furniture coordinate system and told you about my new vector [2,2], even that wouldn't be enough information. You might naturally ask if I was referring to an actual piece of furniture or some sort of linear function. (If I said "an actual desk", then philosophically, I suppose it could be said that [2,2] became contravariant at that very instant, but that's only because it implied that we could then convert it into other measurements.)

Philosophy aside, we have now reached the crucial part of this section. What does it mean to raise or lower an index? In other words, how do we get from $$V^a$$ to $$V_a$$?

If we are dealing with just one lonely coordinate system where nobody cares if you are contravariant or covariant, then $$V^a = V_a$$. It is only when the vectors are transformed that the distinction matters. So a better question is how do we get from $$\tilde V^a$$ to $$\tilde V_a$$?

We can do that in a few steps. Firstly, to get from our feet width/length $$V^a$$ to meters width/length $$\tilde {V^a}$$, we multiplied by T:

$$\tilde V^a = T^a_b V^b$$

To go in the other direction from meters to feet, we mulitply by the inverse of T which is S:

$$V^a = S^a_b \tilde V^b$$

And now we are back to $$V^a$$ in the original meters coordinate system. But in this coordinate system contravariant and covariant are the same so $$V^a$$ is the same as $$V_a$$. Eg those numbers [2,2] could equally represent the dimensions of a desk or a perimeter function. Now if we multiply V by S again, we go back into the feet coordinate system, but this time as a covariant vector:

$$\tilde V_a = V_b S^b_a$$

So the way to convert from a contravariant to covariant vector is to multiply by S twice. I'm not sure about the lettering on this one:

$$\tilde V_a = \tilde V^c S S$$

And since this equation now only has elements from the feet coordinate system, we can drop the ~:

$$V_a = V^c S S$$

This also helps to show that $$V_a = V^a$$ in a single coordinate system. Because we could imagine transforming from the meters system to itself, in which case S is the identity matrix and doesn't change V at all.

Now back to the metric tensor. It is actually quite similar to the perimeter vector because it also performs a function, but it has two covariant dimensions, so it perform its function twice and is written as $$g_{ab}$$. In a single flat untransformed coordinate system (like the meters one) it is the identity matrix:

$$g_{ab} = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}$$

Since it has two covariant dimensions the metric tensor transforms by multiplying by S twice:

$$\tilde g_{ab} = g_{cd} S^c_a S^d_b$$

And this is exactly what we had to do to turn a contravariant vector into a covariant! So in fact, we can use the metric tensor for that (again I'm not sure about the lettering):

$$\tilde V_a = \tilde V^c S S = \tilde V^b g_{cd} S^c_a S^d_b = \tilde V^c \tilde g_{ca}$$

The substitution works because $$g_{cd}$$ is the identity matrix, the matrix equivalent of the number 1. This means that multiplying by the metric tensor converts a contravariant coordinate into a covariant linear function. That is a rough mathematical and philosophical justification for how the metric tensor can raise and lower indices. I also used some of this example to answer a Maths Stackexchange question.

## Gamma

So far I've only talked about the metric, but now we must progress. The article on covariant derivatives introduced Gamma, the Greek letter $$\Gamma$$ used to label the Christoffel symbols in equations. Gamma is the thing in the covariant derivative equation which fixes ordinary tensor differentiation and makes it work across complex coordinate transformations. To remind you, the covariant derivative is:

$$\nabla V = \partial V + \Gamma V$$.

There are actually two kinds of $$\Gamma$$. They are creatively named the Christoffel symbols of the first kind and Christoffel symbols of the second kind. Both of them are based on the metric tensor that I've been going on about so much $$g_{ab}$$. The first kind has three covariant (lowered) indices and looks like:

$$\Gamma_{cab} = \frac12 (\partial_a g_{bc} + \partial_b g_{ac} - \partial_c g_{ab})$$

Each of those indices c, a and b can take on four different values. Gamma is a three-dimensional tensor, and since we are now in a four-dimensional space, Gamma has 4x4x4=64 entries. However, since the a and b indices are swappable (it is symmetric with respect to the lower covariant indices), it has at most 10*4 = 40 independent entries. In this case "independent" means that we only have to calculate 40 of the entries; all the rest are equal to entries we've already calculated.

Up till now, we've only dealt with the second kind. This is the same as the first kind but with the c index raised by contracting with the inverse of the metric tensor:

$${\Gamma^c}_{ab} = \frac12 g^{ck} (\partial_a g_{bk} + \partial_b g_{ak} - \partial_k g_{ab} )$$

The index k in this is the free variable or the index of summation. We can expand the formula above by assigning each of the four coordinate values to t=0, x=1, y=2, z=3 to k. The resulting equation has 12 terms, each involving one element of g and its inverse:

$$\Gamma^{c}_{ab} = \frac12 g^{0c} \partial_{a} g_{0b} + \frac12 g^{1c} \partial_{a} g_{1b} +$$ $$\frac12 g^{2c} \partial_{a} g_{2b} + \frac12 g^{3c} \partial_{a} g_{3b} + \frac12 g^{0c}$$ $$\partial_{b} g_{0a} + \frac12 g^{1c} \partial_{b} g_{1a} + \frac12 g^{2c} \partial_{b} g_{2a} +$$ $$\frac12 g^{3c} \partial_{b} g_{3a} - \frac12 g^{0c} \partial_{0} g_{ab} - \frac12 g^{1c}$$ $$\partial_{1} g_{ab} - \frac12 g^{2c} \partial_{2} g_{ab} - \frac12 g^{3c} \partial_{3} g_{ab}$$

You can see all 40 independent $$\Gamma_{cab}$$ entries here.

## The Riemann Tensor

The Riemann tensor is a four dimensional tensor which measures the curvature of spacetime. It can be expressed with in terms of $$\Gamma$$ of the second kind as:

$$R^{a}_{bcd} = \partial_{c} \Gamma^{a}_{bd} - \partial_{d} \Gamma^{a}_{bc} + \Gamma^{e}_{bd} \Gamma^{a}_{ec} - \Gamma^{e}_{bc} \Gamma^{a}_{ed}$$

In the four dimensions of spacetime it has 4x4x4x4=256 entries. Similar to above, in this equation e is a free variable and we can expand the forumla by assigning t=0, x=1, y=2, z=3 to e and summing. Each of the $$\Gamma$$ multiplications on the right turns into 4 terms:

$$R^{a}_{bcd} = \partial_{c} \Gamma^{a}_{bd} - \partial_{d} \Gamma^{a}_{bc} + \Gamma^{0}_{bd} \Gamma^{a}_{0c} + \Gamma^{1}_{bd} \Gamma^{a}_{1c} + \Gamma^{2}_{bd} \Gamma^{a}_{2c} + \Gamma^{3}_{bd} \Gamma^{a}_{3c} - \Gamma^{0}_{bc} \Gamma^{a}_{0d} - \Gamma^{1}_{bc} \Gamma^{a}_{1d} - \Gamma^{2}_{bc} \Gamma^{a}_{2d} - \Gamma^{3}_{bc} \Gamma^{a}_{3d}$$

Remember that each of those $$\Gamma$$s turns into 12 terms. The first two Riemann terms are partial derivatives of $$\Gamma$$. Since each $$\Gamma$$ term multiplies two terms (one with $$g^{ab}$$ and the other a derivative of $$g_{ab}$$), these will expand into 24 separate terms each by the chain rule for derivatives. The remainaing eight Riemann terms multiply a 12 part sum times a 12 part sum, and so lead to 144 terms each. So expanding the above will lead to 2*24 + 8*12*12 = 1200 terms. A few of these cancel out and we are left with 1000 separate terms.

Here is one of the non-zero entries fully expanded. Some of the 1000 terms combine with each other and so it leaves 280 terms, each involving g's inverse times a derivative or second derivative of g.

R0001 = ¼ g00 g01 ∂0g00 ∂0g11 - ½ g00 g01 ∂0g00 ∂1g01 + ¼ g00 g01 ∂1g00 ∂1g00 + ¼ g00 g02 ∂0g00 ∂0g12 - ¼ g00 g02 ∂0g00 ∂1g02 - ¼ g00 g02 ∂0g00 ∂2g01 + ¼ g00 g02 ∂1g00 ∂2g00 + ¼ g00 g03 ∂0g00 ∂0g13 - ¼ g00 g03 ∂0g00 ∂1g03 - ¼ g00 g03 ∂0g00 ∂3g01 + ¼ g00 g03 ∂1g00 ∂3g00 + ½ g00 g11 ∂0g01 ∂0g11 - g00 g11 ∂0g01 ∂1g01 + ½ g00 g11 ∂1g00 ∂1g01 + ½ g00 g12 ∂0g01 ∂0g12 - ½ g00 g12 ∂0g01 ∂1g02 - ½ g00 g12 ∂0g01 ∂2g01 + ½ g00 g12 ∂0g02 ∂0g11 - g00 g12 ∂0g02 ∂1g01 + ½ g00 g12 ∂1g00 ∂1g02 + ½ g00 g12 ∂1g01 ∂2g00 + ½ g00 g13 ∂0g01 ∂0g13 - ½ g00 g13 ∂0g01 ∂1g03 - ½ g00 g13 ∂0g01 ∂3g01 + ½ g00 g13 ∂0g03 ∂0g11 - g00 g13 ∂0g03 ∂1g01 + ½ g00 g13 ∂1g00 ∂1g03 + ½ g00 g13 ∂1g01 ∂3g00 + ½ g00 g22 ∂0g02 ∂0g12 - ½ g00 g22 ∂0g02 ∂1g02 - ½ g00 g22 ∂0g02 ∂2g01 + ½ g00 g22 ∂1g02 ∂2g00 + ½ g00 g23 ∂0g02 ∂0g13 - ½ g00 g23 ∂0g02 ∂1g03 - ½ g00 g23 ∂0g02 ∂3g01 + ½ g00 g23 ∂0g03 ∂0g12 - ½ g00 g23 ∂0g03 ∂1g02 - ½ g00 g23 ∂0g03 ∂2g01 + ½ g00 g23 ∂1g02 ∂3g00 + ½ g00 g23 ∂1g03 ∂2g00 + ½ g00 g33 ∂0g03 ∂0g13 - ½ g00 g33 ∂0g03 ∂1g03 - ½ g00 g33 ∂0g03 ∂3g01 + ½ g00 g33 ∂1g03 ∂3g00 - ¼ g01 g01 ∂0g00 ∂1g11 + ¼ g01 g01 ∂0g11 ∂1g00 - ½ g01 g02 ∂0g00 ∂1g12 + ½ g01 g02 ∂0g12 ∂1g00 - ½ g01 g03 ∂0g00 ∂1g13 + ½ g01 g03 ∂0g13 ∂1g00 - ½ g01 g11 ∂0g01 ∂1g11 + ¼ g01 g11 ∂0g11 ∂0g11 + ¼ g01 g11 ∂1g00 ∂1g11 - ½ g01 g12 ∂0g01 ∂2g11 - ½ g01 g12 ∂0g02 ∂1g11 + ½ g01 g12 ∂0g11 ∂0g12 + ¼ g01 g12 ∂1g00 ∂2g11 + ¼ g01 g12 ∂1g11 ∂2g00 - ½ g01 g13 ∂0g01 ∂3g11 - ½ g01 g13 ∂0g03 ∂1g11 + ½ g01 g13 ∂0g11 ∂0g13 + ¼ g01 g13 ∂1g00 ∂3g11 + ¼ g01 g13 ∂1g11 ∂3g00 - ½ g01 g22 ∂0g02 ∂2g11 + ¼ g01 g22 ∂0g12 ∂0g12 - ¼ g01 g22 ∂1g02 ∂1g02 + ½ g01 g22 ∂1g02 ∂2g01 + ¼ g01 g22 ∂2g00 ∂2g11 - ¼ g01 g22 ∂2g01 ∂2g01 - ½ g01 g23 ∂0g02 ∂3g11 - ½ g01 g23 ∂0g03 ∂2g11 + ½ g01 g23 ∂0g12 ∂0g13 - ½ g01 g23 ∂1g02 ∂1g03 + ½ g01 g23 ∂1g02 ∂3g01 + ½ g01 g23 ∂1g03 ∂2g01 + ¼ g01 g23 ∂2g00 ∂3g11 - ½ g01 g23 ∂2g01 ∂3g01 + ¼ g01 g23 ∂2g11 ∂3g00 - ½ g01 g33 ∂0g03 ∂3g11 + ¼ g01 g33 ∂0g13 ∂0g13 - ¼ g01 g33 ∂1g03 ∂1g03 + ½ g01 g33 ∂1g03 ∂3g01 + ¼ g01 g33 ∂3g00 ∂3g11 - ¼ g01 g33 ∂3g01 ∂3g01 + ½ g01 ∂00g11 - g01 ∂01g01 + ½ g01 ∂11g00 - ¼ g02 g02 ∂0g00 ∂1g22 + ¼ g02 g02 ∂0g22 ∂1g00 - ½ g02 g03 ∂0g00 ∂1g23 + ½ g02 g03 ∂0g23 ∂1g00 - g02 g11 ∂0g01 ∂1g12 + ½ g02 g11 ∂0g01 ∂2g11 + ¼ g02 g11 ∂0g11 ∂0g12 + ¼ g02 g11 ∂0g11 ∂1g02 - ¼ g02 g11 ∂0g11 ∂2g01 + ½ g02 g11 ∂1g00 ∂1g12 - ¼ g02 g11 ∂1g00 ∂2g11 - ½ g02 g12 ∂0g01 ∂1g22 - g02 g12 ∂0g02 ∂1g12 + ½ g02 g12 ∂0g02 ∂2g11 + ¼ g02 g12 ∂0g11 ∂0g22 + ¼ g02 g12 ∂0g12 ∂0g12 + ½ g02 g12 ∂0g12 ∂1g02 - ½ g02 g12 ∂0g12 ∂2g01 + ¼ g02 g12 ∂1g00 ∂1g22 + ¼ g02 g12 ∂1g02 ∂1g02 - ½ g02 g12 ∂1g02 ∂2g01 + ½ g02 g12 ∂1g12 ∂2g00 - ¼ g02 g12 ∂2g00 ∂2g11 + ¼ g02 g12 ∂2g01 ∂2g01 - ½ g02 g13 ∂0g01 ∂1g23 + ½ g02 g13 ∂0g01 ∂2g13 - ½ g02 g13 ∂0g01 ∂3g12 - g02 g13 ∂0g03 ∂1g12 + ½ g02 g13 ∂0g03 ∂2g11 + ¼ g02 g13 ∂0g11 ∂0g23 - ¼ g02 g13 ∂0g11 ∂2g03 + ¼ g02 g13 ∂0g11 ∂3g02 + ¼ g02 g13 ∂0g12 ∂0g13 + ¼ g02 g13 ∂0g12 ∂1g03 - ¼ g02 g13 ∂0g12 ∂3g01 + ¼ g02 g13 ∂0g13 ∂1g02 - ¼ g02 g13 ∂0g13 ∂2g01 + ¼ g02 g13 ∂1g00 ∂1g23 - ¼ g02 g13 ∂1g00 ∂2g13 + ¼ g02 g13 ∂1g00 ∂3g12 + ¼ g02 g13 ∂1g02 ∂1g03 - ¼ g02 g13 ∂1g02 ∂3g01 - ¼ g02 g13 ∂1g03 ∂2g01 + ½ g02 g13 ∂1g12 ∂3g00 + ¼ g02 g13 ∂2g01 ∂3g01 - ¼ g02 g13 ∂2g11 ∂3g00 - ½ g02 g22 ∂0g02 ∂1g22 + ¼ g02 g22 ∂0g12 ∂0g22 + ¼ g02 g22 ∂0g22 ∂1g02 - ¼ g02 g22 ∂0g22 ∂2g01 + ¼ g02 g22 ∂1g22 ∂2g00 - ½ g02 g23 ∂0g02 ∂1g23 + ½ g02 g23 ∂0g02 ∂2g13 - ½ g02 g23 ∂0g02 ∂3g12 - ½ g02 g23 ∂0g03 ∂1g22 + ¼ g02 g23 ∂0g12 ∂0g23 - ¼ g02 g23 ∂0g12 ∂2g03 + ¼ g02 g23 ∂0g12 ∂3g02 + ¼ g02 g23 ∂0g13 ∂0g22 + ¼ g02 g23 ∂0g22 ∂1g03 - ¼ g02 g23 ∂0g22 ∂3g01 + ¼ g02 g23 ∂0g23 ∂1g02 - ¼ g02 g23 ∂0g23 ∂2g01 - ¼ g02 g23 ∂1g02 ∂2g03 + ¼ g02 g23 ∂1g02 ∂3g02 + ¼ g02 g23 ∂1g22 ∂3g00 + ¼ g02 g23 ∂1g23 ∂2g00 - ¼ g02 g23 ∂2g00 ∂2g13 + ¼ g02 g23 ∂2g00 ∂3g12 + ¼ g02 g23 ∂2g01 ∂2g03 - ¼ g02 g23 ∂2g01 ∂3g02 - ½ g02 g33 ∂0g03 ∂1g23 + ½ g02 g33 ∂0g03 ∂2g13 - ½ g02 g33 ∂0g03 ∂3g12 + ¼ g02 g33 ∂0g13 ∂0g23 - ¼ g02 g33 ∂0g13 ∂2g03 + ¼ g02 g33 ∂0g13 ∂3g02 + ¼ g02 g33 ∂0g23 ∂1g03 - ¼ g02 g33 ∂0g23 ∂3g01 - ¼ g02 g33 ∂1g03 ∂2g03 + ¼ g02 g33 ∂1g03 ∂3g02 + ¼ g02 g33 ∂1g23 ∂3g00 + ¼ g02 g33 ∂2g03 ∂3g01 - ¼ g02 g33 ∂2g13 ∂3g00 + ¼ g02 g33 ∂3g00 ∂3g12 - ¼ g02 g33 ∂3g01 ∂3g02 + ½ g02 ∂00g12 - ½ g02 ∂01g02 - ½ g02 ∂02g01 + ½ g02 ∂12g00 - ¼ g03 g03 ∂0g00 ∂1g33 + ¼ g03 g03 ∂0g33 ∂1g00 - g03 g11 ∂0g01 ∂1g13 + ½ g03 g11 ∂0g01 ∂3g11 + ¼ g03 g11 ∂0g11 ∂0g13 + ¼ g03 g11 ∂0g11 ∂1g03 - ¼ g03 g11 ∂0g11 ∂3g01 + ½ g03 g11 ∂1g00 ∂1g13 - ¼ g03 g11 ∂1g00 ∂3g11 - ½ g03 g12 ∂0g01 ∂1g23 - ½ g03 g12 ∂0g01 ∂2g13 + ½ g03 g12 ∂0g01 ∂3g12 - g03 g12 ∂0g02 ∂1g13 + ½ g03 g12 ∂0g02 ∂3g11 + ¼ g03 g12 ∂0g11 ∂0g23 + ¼ g03 g12 ∂0g11 ∂2g03 - ¼ g03 g12 ∂0g11 ∂3g02 + ¼ g03 g12 ∂0g12 ∂0g13 + ¼ g03 g12 ∂0g12 ∂1g03 - ¼ g03 g12 ∂0g12 ∂3g01 + ¼ g03 g12 ∂0g13 ∂1g02 - ¼ g03 g12 ∂0g13 ∂2g01 + ¼ g03 g12 ∂1g00 ∂1g23 + ¼ g03 g12 ∂1g00 ∂2g13 - ¼ g03 g12 ∂1g00 ∂3g12 + ¼ g03 g12 ∂1g02 ∂1g03 - ¼ g03 g12 ∂1g02 ∂3g01 - ¼ g03 g12 ∂1g03 ∂2g01 + ½ g03 g12 ∂1g13 ∂2g00 - ¼ g03 g12 ∂2g00 ∂3g11 + ¼ g03 g12 ∂2g01 ∂3g01 - ½ g03 g13 ∂0g01 ∂1g33 - g03 g13 ∂0g03 ∂1g13 + ½ g03 g13 ∂0g03 ∂3g11 + ¼ g03 g13 ∂0g11 ∂0g33 + ¼ g03 g13 ∂0g13 ∂0g13 + ½ g03 g13 ∂0g13 ∂1g03 - ½ g03 g13 ∂0g13 ∂3g01 + ¼ g03 g13 ∂1g00 ∂1g33 + ¼ g03 g13 ∂1g03 ∂1g03 - ½ g03 g13 ∂1g03 ∂3g01 + ½ g03 g13 ∂1g13 ∂3g00 - ¼ g03 g13 ∂3g00 ∂3g11 + ¼ g03 g13 ∂3g01 ∂3g01 - ½ g03 g22 ∂0g02 ∂1g23 - ½ g03 g22 ∂0g02 ∂2g13 + ½ g03 g22 ∂0g02 ∂3g12 + ¼ g03 g22 ∂0g12 ∂0g23 + ¼ g03 g22 ∂0g12 ∂2g03 - ¼ g03 g22 ∂0g12 ∂3g02 + ¼ g03 g22 ∂0g23 ∂1g02 - ¼ g03 g22 ∂0g23 ∂2g01 + ¼ g03 g22 ∂1g02 ∂2g03 - ¼ g03 g22 ∂1g02 ∂3g02 + ¼ g03 g22 ∂1g23 ∂2g00 + ¼ g03 g22 ∂2g00 ∂2g13 - ¼ g03 g22 ∂2g00 ∂3g12 - ¼ g03 g22 ∂2g01 ∂2g03 + ¼ g03 g22 ∂2g01 ∂3g02 - ½ g03 g23 ∂0g02 ∂1g33 - ½ g03 g23 ∂0g03 ∂1g23 - ½ g03 g23 ∂0g03 ∂2g13 + ½ g03 g23 ∂0g03 ∂3g12 + ¼ g03 g23 ∂0g12 ∂0g33 + ¼ g03 g23 ∂0g13 ∂0g23 + ¼ g03 g23 ∂0g13 ∂2g03 - ¼ g03 g23 ∂0g13 ∂3g02 + ¼ g03 g23 ∂0g23 ∂1g03 - ¼ g03 g23 ∂0g23 ∂3g01 + ¼ g03 g23 ∂0g33 ∂1g02 - ¼ g03 g23 ∂0g33 ∂2g01 + ¼ g03 g23 ∂1g03 ∂2g03 - ¼ g03 g23 ∂1g03 ∂3g02 + ¼ g03 g23 ∂1g23 ∂3g00 + ¼ g03 g23 ∂1g33 ∂2g00 - ¼ g03 g23 ∂2g03 ∂3g01 + ¼ g03 g23 ∂2g13 ∂3g00 - ¼ g03 g23 ∂3g00 ∂3g12 + ¼ g03 g23 ∂3g01 ∂3g02 - ½ g03 g33 ∂0g03 ∂1g33 + ¼ g03 g33 ∂0g13 ∂0g33 + ¼ g03 g33 ∂0g33 ∂1g03 - ¼ g03 g33 ∂0g33 ∂3g01 + ¼ g03 g33 ∂1g33 ∂3g00 + ½ g03 ∂00g13 - ½ g03 ∂01g03 - ½ g03 ∂03g01 + ½ g03 ∂13g00 - ½ ∂0g00 ∂1g00 - ∂0g01 ∂1g01 - ∂0g02 ∂1g02 - ∂0g03 ∂1g03 + ½ ∂0g11 ∂0g01 + ½ ∂0g12 ∂0g02 + ½ ∂0g13 ∂0g03 + ½ ∂0g00 ∂1g00 + ½ ∂0g02 ∂1g02 - ½ ∂0g02 ∂2g01 + ½ ∂0g03 ∂1g03 - ½ ∂0g03 ∂3g01 + ½ ∂1g00 ∂1g01 + ½ ∂1g02 ∂2g00 + ½ ∂1g03 ∂3g00

## Riemann Tensor Symmetries

Although the Riemann Tensor has 256 entries, there are a few symmetries, and it actually has only 20 independent entries.

For example, the indices c and d are skew symmetric. Which means that:

$$R^{a}_{bcd} = -R^{a}_{bdc}$$

This skew symmetry boils down each box of 4x4=16 entries into just 6, leaving a total of 4x4x6=96 independent entries. The skew symmetry can be easily seen by expanding the two parts above:

$$R^{a}_{bcd} = \partial_{c} \Gamma^{a}_{bd} - \partial_{d} \Gamma^{a}_{bc} + \Gamma^{e}_{bd} \Gamma^{a}_{ec} - \Gamma^{e}_{bc} \Gamma^{a}_{ed}$$

$$R^{a}_{bdc} = \partial_{d} \Gamma^{a}_{bc} - \partial_{c} \Gamma^{a}_{bd} + \Gamma^{e}_{bc} \Gamma^{a}_{ed} - \Gamma^{e}_{bd} \Gamma^{a}_{ec}$$

The other main symmetry is only easily visible in a locally inertial frame (LIF). Remember that special relativity deals with inertial frames, in other words non-accelerating. And the whole original point of g, $$\Gamma$$ and the Riemann tensor is to convert between coordinate systems. So imagine that you are sitting in one coordinate system (a spaceship) travelling past another one (your mate Henry's spaceship).

If you are flying alongside your friend at exactly the same speed, direction and acceleration, then you are locally at rest with him. You are in the same inertial frame, and g will be the Minkowski metric as above. $$\Gamma$$ which is based on the first derivatives of g will be 0.

If you are accelerating at the same rate as Henry, but your speed or direction is different, then the first derivatives of g and therefore $$\Gamma$$ may no longer be zero, but the second derivatives of g will be zero. In this case some of the terms in the Riemann tensor like $$\partial_0 \partial_2 g_{01}$$ will be zero.

On the other hand if you are moving at a different speed and accelerating at different rates (you have a Spaceship 5000 after all, much faster than Henry's Spaceship 3000), then the second derivatives of g will not be zero either.

There's one more possibility. You could be accelerating at different rates, but for an instant, you could have the same speed and direction. This could happen if you have your foot on the gas and are overtaking Henry. For a split second you are at rest with Henry, and the first derivatives of your g will be zero, but not your second derivatives. This is known as the locally inertial frame (I think).

It means that any term containing a first derivative of g will be zero, which includes $$\Gamma$$. Then the last two parts of the Riemann tensor will be zero:

$$R^{a}_{bcd} = \partial_{c} \Gamma^{a}_{bd} - \partial_{d} \Gamma^{a}_{bc} + 0 + 0$$

For the first two terms $$\Gamma$$ is expanded by the chain rule and only the second derivaties are kept, so it becomes:

$$R^{a}_{bcd} = \frac12 g^{ak} \partial_c (\partial_b g_{dk} + \partial_d g_{bk} - \partial_k g_{bd} ) - \frac12 g^{ak} \partial_d (\partial_b g_{ck} + \partial_c g_{bk} - \partial_k g_{cd} )$$

To fully realise the symmetry, we then lower the index a, and $$g_{ak}$$ and $$g^{ak}$$ cancel to leave:

$$R_{abcd} = g_{ak} R^{k}_{bcd} = \frac12 \partial_c (\partial_b g_{da} + \partial_d g_{ba} - \partial_a g_{bd} ) - \frac12 \partial_d (\partial_b g_{ca} + \partial_c g_{ba} - \partial_a g_{cd} ) =$$ $$\frac12 (\partial_c \partial_b g_{da} + \partial_c \partial_d g_{ba} - \partial_c \partial_a g_{bd} - \partial_d \partial_b g_{ca} - \partial_d \partial_c g_{ba} + \partial_d \partial_a g_{cd} ) =$$ $$\frac12 (\partial_c \partial_b g_{ad} - \partial_c \partial_a g_{bd} - \partial_d \partial_b g_{ac} + \partial_d \partial_a g_{cd} )$$

Notice in this equation that the a and b index are interchangeable, which means they are skew symmetric. This means only 6x6=36 independent terms. And there are further more subtle symmetries that reduce this to 20. Those further symmetries involve adding multiple terms together, such as:

$$R_{adbc} = - R_{abcd} - R_{acdb}$$

By "independent" terms, it means that we only have to calculate 20 of them, and then we can figure out the rest by adding and subtracting other terms. Although we have only shown this for a LIF with a lowered a index, it still applies to the full R. It's just that in the full R the indices a and b and not skew-symmetric. Instead there are 236 terms which are much more complicated combinations of the other 20 indepedent terms.

You can see all 20 independent Riemann entries here in the LIF and with a lowered.

## Ricci Tensor

The Ricci curvature tensor is formed by contracting the a and c index in the Riemann tensor. It represents the amount by which the volume of a ball differs in the new coordinate system. I don't know how to visualise this or what is really means. The formula is:

$$R_{bd} = R^{a}_{bad}$$

This tensor is symmetric and so has 10 independent entries:

$$R_{ab} = R_{ba}$$

Because the Riemann tensor's entries each have 1000 terms, each Ricci entry could have 4000, though various terms combine so there are at most 1014 separate terms and 52 in a LIF.

Here is the first entry for the Ricci tensor in a LIF:

R00 = - ½ g11 ∂00g11 + g11 ∂01g01 - ½ g11 ∂11g00 - g12 ∂00g12 + g12 ∂01g02 + g12 ∂02g01 - g12 ∂12g00 - g13 ∂00g13 + g13 ∂01g03 + g13 ∂03g01 - g13 ∂13g00 - ½ g22 ∂00g22 + g22 ∂02g02 - ½ g22 ∂22g00 - g23 ∂00g23 + g23 ∂02g03 + g23 ∂03g02 - g23 ∂23g00 - ½ g33 ∂00g33 + g33 ∂03g03 - ½ g33 ∂33g00

You can see all 10 independent Ricci entries here in a LIF. Also shown is the first entry outside of a LIF, eg including the first derivatives of the metric tensor g. The Ricci tensor still looks symmetrical even outside of a LIF.

## Ricci Scalar

The Ricci scalar is formed by raising one of the Ricci tensor's indexes and then contracting it. It represents a single number which compares the volume of a ball in this coordinate system versus a flat Euclidean space.

$$R = g^{ak} R_{ak}$$

Since each entry in the Ricci tensor has up to 1014 terms, and there are two contractions, the Ricci scalar could have up to 16224 terms. After combinations it has 2286 terms based on g (assuming my computer program worked). In a LIF there are 150 terms.

Here is the Ricci scalar in a LIF:

R = - g00 g11 ∂00g11 + 2 g00 g11 ∂01g01 - g00 g11 ∂11g00 - 2 g00 g12 ∂00g12 + 2 g00 g12 ∂01g02 + 2 g00 g12 ∂02g01 - 2 g00 g12 ∂12g00 - 2 g00 g13 ∂00g13 + 2 g00 g13 ∂01g03 + 2 g00 g13 ∂03g01 - 2 g00 g13 ∂13g00 - g00 g22 ∂00g22 + 2 g00 g22 ∂02g02 - g00 g22 ∂22g00 - 2 g00 g23 ∂00g23 + 2 g00 g23 ∂02g03 + 2 g00 g23 ∂03g02 - 2 g00 g23 ∂23g00 - g00 g33 ∂00g33 + 2 g00 g33 ∂03g03 - g00 g33 ∂33g00 + g01 g01 ∂00g11 - 2 g01 g01 ∂01g01 + g01 g01 ∂11g00 + 2 g01 g02 ∂00g12 - 2 g01 g02 ∂01g02 - 2 g01 g02 ∂02g01 + 2 g01 g02 ∂12g00 + 2 g01 g03 ∂00g13 - 2 g01 g03 ∂01g03 - 2 g01 g03 ∂03g01 + 2 g01 g03 ∂13g00 - 2 g01 g12 ∂01g12 + 2 g01 g12 ∂02g11 + 2 g01 g12 ∂11g02 - 2 g01 g12 ∂12g01 - 2 g01 g13 ∂01g13 + 2 g01 g13 ∂03g11 + 2 g01 g13 ∂11g03 - 2 g01 g13 ∂13g01 - 2 g01 g22 ∂01g22 + 2 g01 g22 ∂02g12 + 2 g01 g22 ∂12g02 - 2 g01 g22 ∂22g01 - 4 g01 g23 ∂01g23 + 2 g01 g23 ∂02g13 + 2 g01 g23 ∂03g12 + 2 g01 g23 ∂12g03 + 2 g01 g23 ∂13g02 - 4 g01 g23 ∂23g01 - 2 g01 g33 ∂01g33 + 2 g01 g33 ∂03g13 + 2 g01 g33 ∂13g03 - 2 g01 g33 ∂33g01 + g02 g02 ∂00g22 - 2 g02 g02 ∂02g02 + g02 g02 ∂22g00 + 2 g02 g03 ∂00g23 - 2 g02 g03 ∂02g03 - 2 g02 g03 ∂03g02 + 2 g02 g03 ∂23g00 + 2 g02 g11 ∂01g12 - 2 g02 g11 ∂02g11 - 2 g02 g11 ∂11g02 + 2 g02 g11 ∂12g01 + 2 g02 g12 ∂01g22 - 2 g02 g12 ∂02g12 - 2 g02 g12 ∂12g02 + 2 g02 g12 ∂22g01 + 2 g02 g13 ∂01g23 - 4 g02 g13 ∂02g13 + 2 g02 g13 ∂03g12 + 2 g02 g13 ∂12g03 - 4 g02 g13 ∂13g02 + 2 g02 g13 ∂23g01 - 2 g02 g23 ∂02g23 + 2 g02 g23 ∂03g22 + 2 g02 g23 ∂22g03 - 2 g02 g23 ∂23g02 - 2 g02 g33 ∂02g33 + 2 g02 g33 ∂03g23 + 2 g02 g33 ∂23g03 - 2 g02 g33 ∂33g02 + g03 g03 ∂00g33 - 2 g03 g03 ∂03g03 + g03 g03 ∂33g00 + 2 g03 g11 ∂01g13 - 2 g03 g11 ∂03g11 - 2 g03 g11 ∂11g03 + 2 g03 g11 ∂13g01 + 2 g03 g12 ∂01g23 + 2 g03 g12 ∂02g13 - 4 g03 g12 ∂03g12 - 4 g03 g12 ∂12g03 + 2 g03 g12 ∂13g02 + 2 g03 g12 ∂23g01 + 2 g03 g13 ∂01g33 - 2 g03 g13 ∂03g13 - 2 g03 g13 ∂13g03 + 2 g03 g13 ∂33g01 + 2 g03 g22 ∂02g23 - 2 g03 g22 ∂03g22 - 2 g03 g22 ∂22g03 + 2 g03 g22 ∂23g02 + 2 g03 g23 ∂02g33 - 2 g03 g23 ∂03g23 - 2 g03 g23 ∂23g03 + 2 g03 g23 ∂33g02 - g11 g22 ∂11g22 + 2 g11 g22 ∂12g12 - g11 g22 ∂22g11 - 2 g11 g23 ∂11g23 + 2 g11 g23 ∂12g13 + 2 g11 g23 ∂13g12 - 2 g11 g23 ∂23g11 - g11 g33 ∂11g33 + 2 g11 g33 ∂13g13 - g11 g33 ∂33g11 + g12 g12 ∂11g22 - 2 g12 g12 ∂12g12 + g12 g12 ∂22g11 + 2 g12 g13 ∂11g23 - 2 g12 g13 ∂12g13 - 2 g12 g13 ∂13g12 + 2 g12 g13 ∂23g11 - 2 g12 g23 ∂12g23 + 2 g12 g23 ∂13g22 + 2 g12 g23 ∂22g13 - 2 g12 g23 ∂23g12 - 2 g12 g33 ∂12g33 + 2 g12 g33 ∂13g23 + 2 g12 g33 ∂23g13 - 2 g12 g33 ∂33g12 + g13 g13 ∂11g33 - 2 g13 g13 ∂13g13 + g13 g13 ∂33g11 + 2 g13 g22 ∂12g23 - 2 g13 g22 ∂13g22 - 2 g13 g22 ∂22g13 + 2 g13 g22 ∂23g12 + 2 g13 g23 ∂12g33 - 2 g13 g23 ∂13g23 - 2 g13 g23 ∂23g13 + 2 g13 g23 ∂33g12 - g22 g33 ∂22g33 + 2 g22 g33 ∂23g23 - g22 g33 ∂33g22 + g23 g23 ∂22g33 - 2 g23 g23 ∂23g23 + g23 g23 ∂33g22

## Einstein Tensor

Nearly there! The Einstein tensor is a combination of the Ricci tensor and the Ricci scalar. It is written with an upper case G, like this:

$$G_{ab} = R_{ab} - \frac12 g_{ab} R$$

The Einstein tensor also expresses the curvature of space-time, but I really can't say anything more specific than that. After this whole adventure, it still seems very mysterious to me. Maybe it will become clearer later.

Like g and R, G is also symmetric. You can see all 10 independent Einstein tensor entries here in the LIF.

## Conclusion

This article has built up from Gamma and the Riemann tensor so the Einstein tensor, which is the thing on the left side of the general equation of relativity. When I first started learning about general relativity, I had a vague notion that if I could just see the equation written out in full with all its dependencies, then I could get an idea how to solve it. This article and this companion tensor equations page has attempted to do that. It shows how the Einstein tensor can be written only in terms of the metric tensor g and its inverse. Unfortunately, it is a lot more complicated than I ever imagined.