Help with Tensors: Using Einstein Summation Convention

[user1139]

Homework Statement

Show that the d'Alembertian of the scalar $$1/R^2=0$$

Relevant Equations

I am suppose to use this expression $$\partial_{\mu}R=\frac{\eta_{\mu\nu} x^{\nu}}{R}$$ to help show

Assuming Einstein summation convention, suppose $$R^2=\eta_{\mu\nu}x^{\mu}x^{\nu}$$

I was able to show that $$\partial_{\mu}R=\frac{\eta_{\mu\nu} x^{\nu}}{R}$$ by explicitly doing the covariant component of the four-gradient and using the kronecker tensor.

However, how do I use the equation expressed in the second paragraph to show that $$\eta^{\alpha \beta}\partial_{\alpha}\partial_{\beta}\frac{1}{R^2}=0$$? I tried $$R\rightarrow \frac{1}{R^2}$$ but I got expressions containing $$x^{\nu}x_{\alpha}$$ which will not give me 0 unless I assume $$x^{\nu}$$ and $$x_{\alpha}$$ are orthogonal which I think is wrong to do so.

 

[PeroK]

One idea is to differentiate $1 \equiv \frac 1 {R^2} R^2$.

 

[user1139]

I am not sure if that helps though.

 

[PeroK]

I am not sure if that helps though.

Have you made progress by other means?

 

[user1139]

Unfortunately, no.

 

[PeroK]

Unfortunately, no.

Try this: $$0 = \partial_{\beta}(\frac{1}{R^2}{R^2}) = \partial_{\beta}(\frac 1 {R^2})R^2 + \frac 1 {R^2} \partial_{\beta}(R^2)$$ $$\partial_{\beta}(\frac 1 {R^2}) = -\frac{1}{R^4}\partial_{\beta}(R^2)$$ Then differeniate the first equation again by $\partial_{\alpha}$: $$0 = \partial_{\alpha}\partial_{\beta}(\frac{1}{R^2}{R^2}) = \partial_{\alpha} \big [ \partial_{\beta}(\frac 1 {R^2})R^2 + \frac 1 {R^2} \partial_{\beta}(R^2) \big ]$$ And see whether you can make progress.

 

[PeroK]

PS try also to show that $$\partial_{\alpha} R^2 = 2x_{\alpha}$$