# Explicit calculation using metric tensor

I'm trying to calculate the following quantity:

$$S_{\bar{\alpha} \bar{\beta}} \equiv \left( \Lambda^{-1} \right)^\alpha_{ \ \bar{\alpha}} \left( \Lambda^{-1} \right)^\beta_{ \ \bar{\beta}} \eta_{\alpha \beta}$$

I want to do it explicitly, writing out the sums over $\alpha, \beta$ and substituting values for known coefficients of the metric tensor and the inverse Lorentz transform tensor. So, I write:

$$S_{\bar{\alpha} \bar{\beta}} = \sum_\alpha \left( \Lambda^{-1} \right)^\alpha_{ \ \bar{\alpha}} \{ \eta_{\alpha 0 } \left( \Lambda^{-1} \right)^\beta_{ \ \bar{\beta}} + ... \}$$

Now I want to substitute for the stuff in the braces. I know the tensor elements, or rather I know their matrix form. What's throwing me off are the upper and lower indices in the Lorentz matrix. I want to write:

$$\eta_{\alpha 0} = -\delta_{\alpha 0}$$ $$\eta_{\alpha i} = \delta_{\alpha i}$$

And for the Lorentz tensor:

$$\left( \Lambda^{-1} \right)^0_{ \ \bar{\beta}} = \gamma \delta^0_{\ \bar{\beta}} + v\gamma \delta^1_{\ \bar{\beta}}$$ $$\left( \Lambda^{-1} \right)^1_{ \ \bar{\beta}} = v\gamma \delta^0_{\ \bar{\beta}} + \gamma \delta^1_{\ \bar{\beta}}$$ $$\left( \Lambda^{-1} \right)^{2,3}_{ \ \bar{\beta}} = v\gamma \delta^{2,3}_{\ \bar{\beta}}$$

Is this right? If I proceed with substituting to the sum, I'm getting terms like this:

$$\delta_{\alpha 0} \delta^0_{\ \bar{\beta}} \quad \mbox{and} \quad \delta_{\alpha 1} \delta^0_{\ \bar{\beta}}$$

This I don't know what to do with. If these were normal Kronecker deltas then it wouldn't be a problem but the upper and lower indices confuse me, I'm fairly new to this. Can these above mentioned 'cross terms' be simplified somehow? Or are the expresions for Lorentz tensor elements that I've written wrong to begin with?

• I carried on with the calculation, now getting terms like these: $\delta^1_{ \ \bar{\alpha}}\delta^1_{ \ \bar{\beta}}.$ Is this notation correct? I want it to stand for the matrix with a $1$ in position $(1,1)$ and zeroes everywhere else. Then the answer I got is correct (I get the metric tensor again). – Spine Feast Oct 28 '15 at 19:14
• What definition of $\Lambda$ have you been given? Usually the lorentz group is defined as the set of linear operators that leave $\eta$ invariant, in which case it's trivial. You could also rewrite your equation as a matrix equation, which is MUCH easier to work with. (You end up with $S=A\eta A$ where $A=\Lambda^{-1}$ and these are just matrices) – user12029 Oct 28 '15 at 19:54