0–forms (functions)#
This article is part of a series of systematic calculations of combinations of the Hodge star operator and the exterior derivative.
Here, we perform these calculations for a function. A generic scalar field is denoted \(f\).
\(⋆f\)#
Applying the Hodge star results directly in:
\[⋆ f = - f dt ∧ dx ∧ dy ∧ dz\]
\(df\)#
Applying the exterior derivative results directly in:
\[d f = ∂_t f dt + ∂_x f dx + ∂_y f dy + ∂_z f dz\]
\(d⋆f\)#
Applying the exterior derivative results directly in:
\[d ⋆ f = 0\]
\(⋆df\)#
Applying the Hodge star results directly in:
\[\begin{split}⋆ d f = \left[ \begin{aligned}
∂_t f dx ∧ dy ∧ dz \\
∂_x f dt ∧ dy ∧ dz \\
∂_y f dt ∧ dz ∧ dx \\
∂_z f dt ∧ dx ∧ dy \\
\end{aligned} \right]\end{split}\]
\(⋆d⋆f\)#
Applying the Hodge star results directly in:
\[⋆ d ⋆ f = 0\]
\(d⋆df\)#
\[d ⋆ d f = (∂_t^2 - ∂_x^2 - ∂_y^2 - ∂_z^2) f \; dz ∧ dt ∧ dx ∧ dy\]
Calculations
Apply the exterior derivative
\[\begin{split}d ⋆ d f = d \left[ \begin{aligned}
∂_t f dx ∧ dy ∧ dz \\
∂_x f dt ∧ dy ∧ dz \\
∂_y f dt ∧ dz ∧ dx \\
∂_z f dt ∧ dx ∧ dy \\
\end{aligned} \right]\end{split}\]
Expand the exterior derivative
\[\begin{split}d ⋆ d f = \left[ \begin{aligned}
∂_t ∂_t f dt ∧ dx ∧ dy ∧ dz \\
∂_x ∂_x f dx ∧ dt ∧ dy ∧ dz \\
∂_y ∂_y f dy ∧ dt ∧ dz ∧ dx \\
∂_z ∂_z f dz ∧ dt ∧ dx ∧ dy \\
\end{aligned} \right]\end{split}\]
Simplify and conclude
\[d ⋆ d f = (∂_t^2 - ∂_x^2 - ∂_y^2 - ∂_z^2) f \; dz ∧ dt ∧ dx ∧ dy\]