Commuting the Lorentz transformation with derivative

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SUMMARY

The discussion centers on the commutation of terms in the context of the d'Alembert operator's invariance under Lorentz transformations. It is established that both Lorentz boosts and spatial rotations belong to the Lorentz group SO(3,1), which is a global symmetry group. This means that the components of the transformation, represented by lambda, are constants and do not depend on the coordinates, thus allowing the derivatives to commute with these terms. The confusion regarding the dependence of lambda on coordinates during rotations is clarified, affirming that commutativity holds true in both cases.

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  • Understanding of the d'Alembert operator in the context of special relativity
  • Familiarity with Lorentz transformations and the Lorentz group SO(3,1)
  • Knowledge of derivative operations in tensor calculus
  • Basic concepts of global symmetry groups in physics
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elemental09
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In the process of proving that the d'Alembert operator

https://www.physicsforums.com/attachments/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:

https://www.physicsforums.com/attachments/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?):

https://www.physicsforums.com/attachments/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:
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I don't understand your argument why a rotation from SO(3) would depend on the coordinates, but a boost from SO(3,1) not. You should pay attention to that, because it will solve your problem.

The Lorentz group SO(3,1) or Poincare group ISO(3,1) is a global symmetry group. That means that the elements don't involve the coordinates. So, performing a Lorentz rotation (whether a boost or spatial rotation) and taking a derivative commute.
 
A silly mistake of mine - that's true, the components Lambda are just constants that depend on the speed of light, rotation angle, and relative velocity of the two frames. Thanks. Problem solved.
 

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