# Commuting the Lorentz transformation with derivative

In the process of proving that the d'Alembert operator

View attachment 31306

is invariant under Lorentz transformations, it was required to commute two terms in the following expression for the transformed operator, which was obtained by switching the index on the first del to the contravariant position, applying the Lorentz transformation to each del, then switching the index on the second del to the covariant position:

View attachment 31307

Now, so long as the last lambda and the second last del terms can commute, then all the Lorentz and metric terms collapse into the Kronecker delta, and the proof is complete (after switching the index positions to obtain the original form). But what justifies the ability to commute them, i.e. why is the following true (or is it at all?):

View attachment 31308

If the Lorentz transformation is just a boost in some arbitrary direction, then I see why it holds - the components of lambda then do not depend explicitly on the time or space coordinates, thus the derivative can commute with that term. However if the transformation involves a rotation, the components of lambda will depend explicitly on the coordinates, so it seems that commutativity would fail.

What am I missing?

EDIT:
It seems these attachments failed to upload...here are the intended equations, in order:
http://www.use.com/ccd724f4259a55704e5f?tc=no#photo=1

Last edited: