Proof of a property of the derivative of 4-velocity

In General Relativity, An Introduction for Physicists, in between equations 8.7 and 8.8, the authors make use of the property that the contraction of the 4 velocity gives the speed of light squared ($$u_{\nu}u^{\nu} = c^2$$), so the partial derivative will be zero. They also claim $$\left(\partial_{\mu} u^{\nu}\right) u_{\nu}+u^{\nu}\left(\partial_{\mu} u_{\nu}\right)=2\left(\partial_{\mu} u^{\nu}\right) u_{\nu}=0$$ where I assume the second part of the sum must be equal to the first (hence the 2). Is this true by raising or lowering the indices with the metric? How would I prove this statement?

Edit:

I can't just contract the $$u^{\nu}\left(\partial_{\mu} u_{\nu}\right)$$ with the metric without it being part of an equation. Also, it's not necessarily valid that I can contract a metric inside a derivative because it is not true that the derivative of that metric are zero. (I eventually want to generalise this statement to covariant instead of partial derivatives. )

• This is SR, not GR, and it doesn't have anything to do with differential geometry. Is this true by raising or lowering the indices with the metric? Yes. How would I prove this statement? Have you tried doing what you said? – user4552 Nov 12 '19 at 23:15
• I'm going to use it to prove something about the einstein field equations, so yes it's quite GR related, please try to be less condescending. – Andrew Hardy Nov 12 '19 at 23:27
• This discussion in the book is clearly written for flat spacetimes... $g = \eta$, so the metric can pass freely through partial derivatives. – OkThen Nov 12 '19 at 23:41
• I can't just contract the uν(∂μuν) with the metric without it being part of an equation. Not sure why you think this. This is a manipulation like changing $\sqrt{ x^2}$ to $|x|$. Also, it's not necessarily valid that I can contract a metric inside a derivative because it is not true that the derivative of that metric are zero. (I eventually want to generalise this statement to covariant instead of partial derivatives. ) The generalization of this property to GR would use a covariant derivative, and there it would again be true that the derivative of the metric is zero. – user4552 Nov 12 '19 at 23:54
• Just to clarify, it would be true that the covariant derivative of the metric is zero (assuming metric compatibility) – lux Nov 13 '19 at 14:47

1 Answer

In contractions you can lower one index if you bring another one up. This is a really useful property: \begin{align}A^\mu B_\mu&=A^\mu g_{\mu\nu}B^\nu\\&=A_\nu B^\nu\\&=A_\mu B^\mu\end{align} Partial derivatives are bit less obvious though. Looking at your second term: \begin{align}u^\nu(\partial_\mu u_\nu)&=u^\nu(\partial_\mu g_{\nu\lambda}u^\lambda)\\ &=u^\nu g_{\nu\lambda}\partial_\mu u^\lambda+u^\nu u^\lambda\partial_\mu g_{\nu\lambda}\\ &=u_\nu \partial_\mu u^\nu+u^\nu u^\lambda\partial_\mu g_{\nu\lambda} \end{align} In the Minkowski metric $$\partial_\mu g_{\nu\lambda}=0$$ so partials commute with the lowering of indices. In GR this is generally not the case. This last term is often referred to as the non-tensorial part, because it is what prevents the partial derivative from transforming like a tensor. So in GR we generally have $$(\partial_\mu u^\nu)u_\nu \neq u^\nu(\partial_\mu u_\nu)$$ If you replace the partials by covariant derivatives (like Ben mentioned) you can commute the metric $$\nabla_\lambda g_{\mu\nu}=g_{\mu\nu}\nabla_\lambda$$ so you would get a similar identity.