0
$\begingroup$

I am following these notes (A Practical Introduction to Differential Forms, by Alexia E. Schulz and William C. Schulz 2013) and on page 67 (pdf page number 71) there is an expression for the codifferential of the flattened 4-potential. They have:

$$\delta A=\star d\star A$$

With $A=-\phi cdt+A_1dx+A_2dy+A_3dz$, which gives:

$$\delta A=\star d\left(-\phi dx\wedge dy\wedge dz+A_1 cdt\wedge dy\wedge dz\ldots\right)$$

So:

$$\delta A=\star \left(-\frac{1}{c}\frac{\partial\phi}{\partial t} cdt\wedge dx\wedge dy\wedge dz+\frac{\partial A_1}{\partial x} dx\wedge cdt\wedge dy\wedge dz\ldots\right)$$

Which they claim is:

$$\delta A=\star \left(\left(-\frac{1}{c}\frac{\partial\phi}{\partial t} +\frac{\partial A_1}{\partial x} \right)cdt\wedge dx\wedge dy\wedge dz\right)$$

In the second term how does $dx\wedge cdt\wedge dy\wedge dz\rightarrow cdt\wedge dx\wedge dy\wedge dz$? Is it because we are commuting a time and space 1-form in the wedge product?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic ♦
    Commented Oct 20, 2023 at 19:24
  • $\begingroup$ The wedge product shouldn't care about causality issues. Are you sure the signs are correct after acting with this $\star$ operator? $\endgroup$
    – P-A
    Commented Oct 23, 2023 at 8:00

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.