# What does "invariant" mean in Kretschmann invariant?

1. Sep 15, 2015

### dyn

Hi. I have a couple more questions in my quest to self-study GR.
1 - I have some notes where the Kretschmann Invariant is defined as Rk = RabuvRabuv and is given in Schwarzschild coordinates as Rk = 48u2/r6 . My notes say this is an invariant field so that its value at any point as evaluated in one frame is the same when evaluated in any other frame. But its value depends on r so when switching from one frame to another the value of r will change which means Rk will change. What is wrong with my thinking ?

2 - I came across the following in a "derivation" of the Einstein field equations : gvv = 4. I don't understand how or why this is true. As far as I know the components of g are normally not constants. On a similar issue what is ηvv where η is the Minkowski metric. For a general tensor A does Avv = A ? and what is A ? Is it a scalar ? an invariant ?

Thanks in anticipation for any help

2. Sep 15, 2015

### Staff: Mentor

No, it won't. The $r$ here is not the radial coordinate; it's the "areal radius", i.e., $\sqrt{A / 4 \pi}$, where $A$ is the physical area of a 2-sphere containing the given event. So $r$ itself is an invariant, as it appears in the formula.

It is true that, if you change to a chart in which $r$ is not represented directly as a coordinate (for example, the Kruskal chart), the Kretschmann invariant will look different as a formula if you insist on writing it solely in terms of coordinates. But $r$, the areal radius invariant, can still be expressed in terms of the coordinates in any chart, so you can always recover the form of $R_k$ that you wrote down.

The formula $g^v{}_v$ is shorthand for $g^0{}_0 + g^1{}_1 + g^2{}_2 + g^3{}_3$. Each of these terms, in its turn, can be expanded as, for example, $g^0{}_0 = g^{0u} g_{u0} = g^{00} g_{00} + g^{01} g_{10} + g^{02} g_{20} + g^{03} g_{30}$. If you substitute the formula for the inverse metric $g^{uv}$ (as the matrix inverse of the metric $g_{uv}$) and work through the algebra, you will see that each of these sums comes out to $1$, i.e., $g^0{}_0 = 1$, and similarly for the other three terms in the sum. So $g^v{}_v$ is the sum of four terms each of which equals $1$, i.e., it's $4$.

This works the same as above, except that the sums are much simpler because the Minkowski metric is diagonal and all of the diagonal elements are $\pm 1$. So it's much easier to show that the sum works out to $4$.

If you want to define it that way, yes. For a general tensor there isn't much you can say about the trace (which is the general term for $A^v{}_v$ for a general tensor), except the items below.

Yes to both. In fact the trace is the simplest scalar invariant that you can obtain from a 2nd-rank tensor.

3. Sep 16, 2015

### bcrowell

Staff Emeritus
Here are a couple of ways of evaluating $g^\nu_\nu$ that may be easier.

Method 1: It's notated as a scalar (all indices are contracted), so its value is the same when evaluated in any coordinate system. Therefore you're free to evaluate it in Minkowski coordinates, which is easy.

Method 2: Or, note that $g^\mu_\nu=g^{\mu\kappa}g_{\kappa\nu}$ (using g to raise an index on itself). But the right-hand side is matrix multiplication, and the matrix representations of $g^{\mu\kappa}$ and $g_{\kappa\nu}$ are matrix inverses of each other, which tells us that as a matrix, $g^\mu_\nu$ is the identity matrix, whose trace is 4.

4. Sep 16, 2015

### dyn

Thanks for your replies. So gvv = ηvv = 4 . Does this value only apply to these 2 metric tensors? Does the inverse argument given above apply to other tensors eg Rv v . If the argument only applies to the two metric tensors what is special about them compared to other rank 2 tensors ?

5. Sep 16, 2015

### Staff: Mentor

Yes.

No. See below.

The metric tensor (the Minkowski metric $\eta_{uv}$ is really just a special case of the general metric $g_{uv}$) is the tensor that is used to raise and lower the indexes of all tensors, including itself. So for a general tensor $R_{uv}$, raising one index gives $R^u{}_v = g^{uw} R_{wv}$, which is not the product of a matrix with its own inverse, so taking the trace won't give 4 . But if the tensor is $g$ itself, then raising one index does give the product of a matrix with its own inverse, as bcrowell showed, so taking the trace will give 4.