QuantumRose
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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.
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?
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?
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