Why Does (\partial_{\mu}\phi)^2 Equal (\partial_{\mu}\phi)(\partial^{\mu}\phi)?

[noahcharris]

I just came across this in a textbook: $(\partial_{\mu}\phi)^2 = (\partial_{\mu}\phi)(\partial^{\mu}\phi)$

Can someone explain why this makes sense? Thanks.

 

[fzero]

I just came across this in a textbook: $(\partial_{\mu}\phi)^2 = (\partial_{\mu}\phi)(\partial^{\mu}\phi)$

Can someone explain why this makes sense? Thanks.

The right-hand side is a short-hand way of writing
$$\sum_{\mu=0}^d (\partial_{\mu}\phi)(\partial^{\mu}\phi).$$
More generally, this shorthand is called the Einstein summation convention, commonly used whenever vectors and tensors appear, where a sum on a repeated index is assumed from the context of the expression. The left-hand side is a further shorthand, where if the quantity is supposed to be a scalar from the context, then it is assumed that the index is summed over with an appropriate metric tensor to raise one of the indices in the square. This is a shorthand that is more likely to be confusing and is used much less frequently than the Einstein summation convention itself.

 

[ShayanJ]

As you should know, $\partial_\mu \phi$ is a covariant first rank tensor, so you may name it $S_\mu$. Now considering $S^2\equiv S_\mu S^\mu$, will give you what you want.

 

[HallsofIvy]

Specifically, that is the Einstein summation convention for tensors, a notational convention: The same index, both as a subscript and a superscript is interpreted as a summation index.