I've been recently been going back over the basics of GR, differential geometry in particular. I was watching one of Susskind's lectures and did not understand the argument made here (26:33 - 35:40).

In short, the argument goes as follows (I think): we have some generic metric ${ g }_{ m n }^{ ' }\left( y \right)$. Suppose we have a coordinate transformation that takes ${ g }_{ m n }^{ ' }\left( y\right) \rightarrow { g }_{ m n }\left( x\right)$ such that ${ g }_{ m n }\left( X\right) ={ \delta }_{ m n }$ for a particular point $x=X$.

Susskind wants to show that, in general, the first derivatives, ${ \partial }_{ r }{ g }_{ m n }\left( x\right)$, can be chosen to be zero, but the second derivatives, ${ \partial }_{ r }{ \partial }_{ s }{ g }_{ m n }\left( x \right) $, can not (that is, at $x=X$).

He does this by looking at the expansion of $x$ in terms of $y$ about the point $x=X$. For simplicity, he chooses ${ X }^{ m }=0$ and for the $x$ and $y$ coordinate systems to have the same origin: $${ x }^{ m }={ a }_{ r }^{ m }{ y }^{ r }+{ b }_{ rs }^{ m }{ y }^{ r }{ y }^{ s }+{ c }_{ rst }^{ m }{ y }^{ r }{ y }^{ s }{ y }^{ t }+\dots$$

The argument is (again, I think) that because (for the case of a four-dimensional space) ${ \partial }_{ r }{ g }_{ m n }\left( x\right)=0$ is 40 equations and ${ b }_{ rs }^{ m }$ consists of 40 variables, we can always choose values of ${ b }_{ rs }^{ m }$ that satisfy the equations. Meanwhile, ${ \partial }_{ r }{ \partial }_{ s }{ g }_{ \mu \nu }\left( x \right) =0$ is 160 equations, but ${ c }_{ rst }^{ m }$ consists of only 80 variables, so we do not have enough free parameters to force the second derivatives to all vanish.

The problem is that I simply don't see why the existence of 40 variables in that expansion means that we can satisfy the 40 equations. Is the connection a simple one or do I just have to do something like grind out the values of the derivatives at $x=X$ using the series expansion?

  • $\begingroup$ It's not really a rigorous argument, because you would need to show that the equations are all independent and stuff like that. The real argument is just the construction of Riemann normal coordinates, which you can find in books. $\endgroup$
    – Javier
    Jul 6, 2020 at 2:00
  • $\begingroup$ Yep, wrote them out and saw how the equations could be satisfied. It was just that the way Susskind presented the argument made me think that is was clear through some sort of inspection. $\endgroup$ Jul 6, 2020 at 16:08

1 Answer 1


Hint: A simplifying trick is to realize that we may also assume without loss of generality that $a^m_r=\delta^m_r$. Then each of the 40 conditions becomes a linear equation in precisely one $b^m_{rs}$ variable. The new coordinates are known as Riemann normal coordinates.

  • $\begingroup$ I suppose we can rotate coordinates so that ${x}^{m}={y}^{m}$ for small $y$? Then do I just write out what ${ \partial }_{r}{g}_{ m n }\left( x\right)$ is in terms of $y$ and ${g}_{m n}^{ ' }\left( y \right)$ to see this? I end up with a series of terms (some proportional to ${g}_{m n}^{'}\left( y \right)$ and the others ${ \partial }_{r}^{'}{g}_{m n}^{'}\left( y \right)$), some constant in $y$ proportional to products of $a$ and $b$, some linear in $y$ proportional to products of $a$, $b$ and $c$, and then the rest are higher order contributions involving $a$, $b$, $c$, $d$ and so on. [1/2] $\endgroup$ Jul 6, 2020 at 3:44
  • $\begingroup$ Then setting ${ \partial }_{r}{g}_{ m n }\left( x\right)=0$ at $x=y=0$ removes all the higher-order terms and just leaves those constant terms of products of $a$ and $b$, but as $a$ is trivial we actually just end up with forty equations involving the components of $b$ (and they're linear in $b$ too)? I dropped the variable indices when writing it all out as it got rather long, but I suppose if I end up with forty equations involving terms constant or linear in all of the components of $b$, then assuming they're all independent, I can (in principle) solve for the components of $b$. [2/2] $\endgroup$ Jul 6, 2020 at 3:57
  • 1
    $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Jul 6, 2020 at 5:54

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