About the derivation of Lorentz gauge condition

[QuantumRose]

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.

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

(Where

)

However, the actual Lorentz gauge potential is

(Where

)

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

 

[George Jones]

However, the actual Lorentz gauge potential is
View attachment 219844 (Where View attachment 219845 )

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

Aren't the indices "upstairs", here?

 

[QuantumRose]

Aren't the indices "upstairs", here?

Opps, such a silly mistake... thank you!

 

[dextercioby]

The correct name is Lorenz, not Lorentz. Please, tell that to your instructor, too.