How to show something transforms as a covector?

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Discussion Overview

The discussion revolves around how to demonstrate that the partial derivatives transform as covectors under a boost in the x direction. Participants explore the mathematical framework and rules involved in this transformation, particularly focusing on the use of the chain rule and the properties of covariant vectors.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests using the calculus rule relating the transformed partial derivative \(\partial_{x'}\) to the original \(\partial_x\).
  • Another participant expresses skepticism about the covariant nature of the derivatives, questioning the use of total versus partial derivatives and suggesting that a gradient would be more appropriate.
  • A later reply clarifies that the derivatives in question are indeed partial derivatives and seeks to understand the application of the chain rule in this context.
  • Further elaboration includes a mathematical expression showing how the transformation of coordinates leads to a sum that aligns with the law of covariant transformation.

Areas of Agreement / Disagreement

Participants do not reach a consensus; there are competing views on whether the derivatives can be considered covariant vectors, and the discussion remains unresolved regarding the application of the chain rule and the nature of the derivatives involved.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the nature of the derivatives and the specific conditions under which the transformation is being analyzed. The dependence on definitions of covariant vectors and the mathematical steps involved in the transformation are also not fully resolved.

Baggio
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Considering a boost in the x direction, how do you show that

(d/dx,d/dy,d/dz,d/dt)

transforms as a covector?


thanks
 
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I assume you mean partial derivs.
Use the calculus rule relating \partial_x' to \partial_x.
 
Baggio said:
Considering a boost in the x direction, how do you show that

(d/dx,d/dy,d/dz,d/dt)

transforms as a covector?


thanks
I don't see that this is a covariant vector. If it was the -gradient then I'd see that (you used total derivatives rather than partial derivative. I've never seen such an object). If it was then I'd use the chain rule for partial differentiation.

Pete
 
yes sorry they are partials... why is the chain rule used.. I don't see how that leads to the solution

thanks
 
Baggio said:
yes sorry they are partials... why is the chain rule used.. I don't see how that leads to the solution

thanks

[tex]\frac{\partial}{\partial x'} = \frac{\partial x}{\partial x'}\frac{\partial}{\partial x} + \frac{\partial y}{\partial y'}\frac{\partial}{\partial y} + \frac{\partial z}{\partial z'}\frac{\partial}{\partial z} + \frac{\partial t}{\partial t'}\frac{\partial}{\partial t}[/tex]

Now change coordinates (x, y, z, t) - > (x1, x2, x3, x4)

Then the above sum can be written as a sum which is identical to the law of covariant transformation.

Pete
 
Last edited:

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