Reference: Moore, Thomas A., *A General Relativity Workbook*, University Science Books (2013) – Chapter 6; Problem 6.7.

In special relativity, a general four-vector transforms like

If we take the derivative of with respect to proper time, and assume that is a function of position, then

How does this transform? We get

In order for to be a four-vector, the first term in the last line would need to be zero. Since and are arbitrary this means that

That is, the coordinate partial derivatives must be independent of position. In general this isn’t true, but in flat space, coordinate transformations are given by the Lorentz transformations which are independent of position, so in flat space, is a four-vector.

If we have a scalar function of position , then

Doing the transform, we get

However, since scalars are invariant so

So in this case, the derivative of a scalar field is a valid scalar under transformation.

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