Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; g.
We’ve seen that for a smooth metric, we can define a locally flat coordinate system at a given event in which the metric is the flat spacetime metric . Although we specify the value of the metric tensor at that point, we haven’t imposed any conditions on its derivatives with respect to the various coordinates. If we impose the addition constraint that all the derivatives of the metric are zero, then we have a locally inertial frame or LIF.
It’s not immediately obvious that we can impose this restriction, so let’s see how we can demonstrate this. We start with some arbitrary metric which doesn’t satisfy the LIF conditions and consider the transformation to another metric on which we try to impose these conditions. As usual, the transformation is
Each of the partial derivatives is a function of the primed coordinates so, for a region close to the event point , we can expand these derivatives in Taylor series:
where the coefficients are defined in terms of and its derivatives, all evaluated at :
We can also think of the unprimed metric as a function of the primed coordinates, and expand it in a Taylor series as well:
We can now substitute these two series into 1 and collect terms. Due to the large number of indices floating about, it’s easier to use a condensed notation for this step. We can temporarily drop the indices to get
The transformation now becomes, up to second order terms:
The last equation is the Taylor expansion of where every factor is now a function of . That is
We therefore have, taking each term separately and restoring the indices:
We won’t actually try to solve these equations, but what we need to do is see if we have enough freedom to impose the conditions above, that is, that and . To do this, we need to see how many independent equations each of these equations provides.
Consider 13 first. Each index and has 4 possible values, but because we have to exclude permutations of and . We can get these using binomial coefficients:
For 15, the symmetry of means we have (10 choices for and , times the 4 different values of ) independent equations. Finally, for 17, since the order of partial derivatives doesn’t matter and the metric is symmetric, we have independent equations.
The original metric is assumed to be given, so we can’t change that, so how many variables can we solve for in these equations? The transformation is effectively determined (up to second order) by the coefficients , and , so we need to figure out how many of these are independent of each other.
For every choice of the two indices gives an independent quantity, so there are 16 different variables. Thus we have 6 more degrees of freedom than we need to solve 13 and set .
For , the order of the partial derivatives doesn’t matter so there are 10 possible choices of and for each value of , giving . Having chosen the in solving 13, we can plug these into the 40 equations specified by 15 and solve for the 40 coefficients. Therefore, in principle, we can set .
When it comes to , again the order of the 3 partial derivatives doesn’t matter, so for each value of , we can have , or two indices equal with the third different, or all three indices different. The total number of choices for each value of is then
To get the middle term, note that is the number of ways of choosing 2 numbers out of 4, but it assumes there are only 2 slots into which these numbers can be placed. In our case, one of the numbers is used twice, so this term would imply that, for example, 1-1-2 and 2-2-1 are just permutations of the same combination, whereas we want them to be distinct, so we have to multiply by 2. The total number of independent is therefore . The fact that there are 100 independent equations specified by 17 means that we can’t impose values on all of the so in general even if we require 80 of them to be zero, the other 20 may have non-zero values. It turns out that this is a result of the inherent curvature of the spacetime, so it’s not surprising that we can’t get rid of that.
The LIF imposes stricter conditions on the local metric than the locally flat frame we mentioned at the start. The latter just specifies the metric at a given point, while the LIF specifies both the metric and its derivatives.