Confusion over Einstien summation convention and metric tensors.

1. Oct 22, 2011

Lyalpha

My understanding of the Einstein Summation convention is that you sum over the repeated indices. But when I look at the metric tensor for a flat space I know that

g$^{λ}_{λ}$ = 1

But the summation convention makes me think that it should equal the trace of the matrix g_{μσ}. So it should be the number of dimensions of the space?

2. Oct 22, 2011

Pengwuino

$g_{\alpha}^{\; \alpha} = N$ where N is the dimension of the space. Why do you think it's 1?

3. Oct 22, 2011

Lyalpha

A$^{\nu}$ = g$^{μ \nu}$A$_{μ}$

A$_{μ}$ = g$_{μ\nu}$A$^{\nu}$

A$_{μ}$ = g$_{μ\nu}$g$^{\nu\rho}$A$_{\rho}$

g$_{μ\nu}$g$^{\nu\rho}$ = g$^{\rho}_{\mu}$

g$^{\rho}_{\mu}$ = 1 for μ = ρ

g$^{\rho}_{\mu}$ = 0 for μ $\neq$ ρ

taken from Dirac's book

4. Oct 22, 2011

Pengwuino

Yes but $g^{\rho}_{\mu}$ is not a summation. It simply indicates what the components are. So it's saying when $\rho = \mu$, the component is 1, else it is 0. However, $g^{\rho}_{\rho} = \sum^{N}_{\alpha = 1} g_{\alpha}^{\alpha}$.

It's like how sometimes people get confused with the delta function $\delta_{ij}$. In a 3-dimensional space, you can have $\delta_{ij} = 1$ if $i=j$, else it is 0. However, this is different from $\delta_{ii}$ which is the actual summation which is equal to 3.

5. Oct 22, 2011

Lyalpha

ok thanks

6. Oct 22, 2011

Lyalpha

I asked the question because I'm getting g$^{λ}_{λ}$ after some contractions of a certain curvature tensor. For Example I have a term g$^{μ}_{σ}$ and I'm contracting the tensor by setting μ and σ to λ. The term will now equal the dimension D?

7. Oct 22, 2011

Pengwuino

Yes it will equal D if you're contracting the metric in the way you asked. This is true only for the metric, however.

Last edited: Oct 22, 2011
8. Oct 22, 2011

WannabeNewton

Not in general. If $g_{\mu \nu } = \delta _{\mu \nu }$ on the n - manifold in question then $g^{\lambda }_{\lambda } = \delta ^{\lambda }_{\lambda } = n$. But this doesn't have to be true for arbitrary metrics on their respective manifolds.

9. Oct 22, 2011

Matterwave

No, in general (at least in GR, I don't know about some really weird cases or w/e),

$$g^\mu_\nu=\delta^\mu_\nu$$

This is because the contravariant metric must be the inverse of the covariant metric. The two must be inverses of each other or else you won't get the same vector back if you lower it's indices and then raise them.

The trace of the Kronecker delta is obviously the dimension, and therefore so is the trace of the (1,1) form of the metric.

10. Oct 22, 2011

Pengwuino

No, in general it is true if $g_{ab}$ and $g^{ab}$ are inverses of each other. Remember, $g_{a}^{a} = \sum_{i,j=1}^N g_{ij}g^{ij} =N$

11. Oct 22, 2011

WannabeNewton

Well it still isn't true in general because what if g isn't diagonal. But yeah I get your point if g is diagonal.

12. Oct 22, 2011

Dickfore

ORLY? Where did you see that?

13. Oct 22, 2011

Pengwuino

This is in general, the metric does not need to be diagonal since $g_{ab}$ and $g^{ab}$ by definition are inverses of each other. If $g^{ab}$ is the inverse of $g_{ab}$, then by definition, $g_a^b = \sum_{i=1}^n g_{ai}g^{ib} = \delta_{ab}$. Remember, this $g_a^b$ component is a n-term summation and the inverse metric must be defined so this is true.

14. Oct 22, 2011

Lyalpha

But what if it's the fundamental metric tensor? Where g$_{00}$ is -1 and the rest are 1? Would I get D-2 ?

15. Oct 22, 2011

Matterwave

Here:

$$A_\mu=g_{\mu\nu}A^\nu$$

$$A^\mu=g^{\mu\nu}A_\nu$$

Which implies:

$$A^\mu=g^{\mu\nu}g_{\nu\tau}A^\tau$$

If it is true that $$A^\mu=\delta^\mu_\tau A^\tau$$

Then it must be that:

$$g^{\mu\nu}g_{\nu\tau}=g^\mu_\tau=\delta^\mu_\tau$$

16. Oct 22, 2011

Matterwave

No, you can only trace the (1,1) form of the metric. You can't trace the purely covariant metric. the (1,1) form of the metric, almost by definition, MUST be diag(1,1,1,1...). See my proof above.

17. Oct 22, 2011

WannabeNewton

So you're telling me that for the metric $g_{\mu \nu } = \begin{pmatrix} 1 + 4c^{2}p^{2} & 2cp\\ 2cp & 1 \end{pmatrix}$ that $g_{\mu \nu }g^{\mu \nu } = 2$ (c = const.).

18. Oct 22, 2011

Pengwuino

Yes. And this is because the inverse metric must be defined such that it is true. The inverse for that metric won't just look like 1 over the components of the metric because that does not satisfy $g_{am}g^{mb} = \delta_{ab}$

19. Oct 22, 2011

Matterwave

Yes, WannabeNewton, see my previous proof...why does this bother you?

20. Oct 22, 2011

WannabeNewton

yeah my bad forgot to include $2g_{12}g^{12}$ lol

21. Oct 22, 2011

Lyalpha

I'm starting with a conformally flat space with the interval ds$^{2}$ = e$^{2φ}$ η$_{ab}$ dx$^{a}$ dx$^{b}$ , where $\eta$$_{ab}$ is a fundamental metric tensor of D-dimensional flat space. I'm trying to find the usual stuff (Christoffel symbols, curvature tensor, ricci tensor, curvature scalar). When it comes to contracting the curvature tensor I get a few $\eta$$^{c}_{c}$ terms. What would they be in the case I described?

22. Oct 22, 2011

Pengwuino

It's D. You know what the components of the metric and inverse metric are for such a simple case. Do the actual summation for even D = 2 and you'll see $\eta_n^{\; n} = D$

23. Oct 23, 2011

Lyalpha

But isn't $\eta$$^{n}_{n}$ equal to $\eta$$^{0}_{0}$ + $\eta$$^{1}_{1}$ + $\eta$$^{2}_{2}$ + ... + $\eta$$^{D}_{D}$ where $\eta$$^{0}_{0}$ = -1 and $\eta$$^{1}_{1}$, $\eta$$^{2}_{2}$, $\eta$$^{D}_{D}$ are 1?

24. Oct 23, 2011

Pengwuino

No $\eta_{00} = -1$ but $\eta_0^0 = 1$. $\eta_0^0 = \sum_{i=0}^D \eta_{0i}\eta^{i0} = 1$

25. Oct 23, 2011

Phrak

Careful with the negative signs. -12+12+12+12=4=D