Commuting the Lorentz transformation with derivative

In summary, the commutativity of terms in the transformed d'Alembert operator, obtained by applying a Lorentz transformation, is key in proving its invariance. This holds true as the Lorentz group is a global symmetry group, meaning the elements do not depend on the coordinates. However, if the transformation involves a rotation, the components of lambda will depend on the coordinates, possibly leading to a failure in commutativity.
  • #1
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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  • #2
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.
 
  • #3
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.
 

1. What is the Lorentz transformation?

The Lorentz transformation is a mathematical equation used in special relativity to describe the relationship between space and time for two observers in different inertial frames of reference.

2. What does it mean to commute the Lorentz transformation?

Commuting the Lorentz transformation means rearranging the mathematical equation so that the operations are performed in a different order. This allows for easier calculations and a better understanding of the relationship between space and time.

3. Why is it important to commute the Lorentz transformation with derivative?

Commuting the Lorentz transformation with derivative allows for a more accurate and precise calculation of the relationship between space and time in special relativity. It also helps to simplify the equation and make it easier to solve.

4. What is the process for commuting the Lorentz transformation with derivative?

The process for commuting the Lorentz transformation with derivative involves using the chain rule to differentiate the equation and then rearranging the terms to get the derivative on one side of the equation. This allows for a more simplified and easier to solve equation.

5. Can the Lorentz transformation be commuted with other mathematical operations?

Yes, the Lorentz transformation can be commuted with other mathematical operations, such as integration, to further simplify the equation and make it easier to solve. However, it is important to follow the correct mathematical rules and use the appropriate equations for the specific situation.

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