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**The question:**

Show that the Lorentz condition ∂µAµ =0 is expressed as d∗ A =0.

Where A is the four-potential and * is the Hodge star, d is the exterior differentiation.

Show that the Lorentz condition ∂µAµ =0 is expressed as d∗ A =0.

Where A is the four-potential and * is the Hodge star, d is the exterior differentiation.

**In four-dimensional space, we know that the Hodge star of one-forms are the followings.**

**3. My attempt**

Since the four potential one-form is

**
**

Therefore we have

**
**

Then d*A = 0 is equivalent of saying

However, the actual Lorentz gauge potential is

I don't know why there is a sign difference?