# Orthogonality relation four-vectors

If $$A,B,C,D$$ are four-vectors and $$A,B,C$$ form an orthogonal hypersurface to $$D$$, and $$det(g_{\mu\nu})=g$$ where $$g_{\mu\nu}$$ are the metric components, then it is true that $$\begin{equation} \sqrt{-g}A^{\alpha}B^{\beta}C^{\gamma}\epsilon_{\alpha\beta\gamma\delta}=cte \cdot D_{\delta} \end{equation}$$

Where $$cte$$ is a constant. I dont understand how this equation is derived.

Making an analogy with 3D euclidean linear algebra, if $$A,B$$ are vectors that create a plane orthogonal to $$C$$ then $$A \times B=C$$, or, in components $$\begin{equation} A_{i}B_{j}\epsilon^{ijk}=C^{k} \end{equation}$$

Which looks very similar to the equation above. However, I don't understant where the $$\sqrt{-g}$$ comes from (I suppose it comes from the fact that this is in general a curved space, but I don't see it clearly enough). And I don't understand why a constant is added in the right side

• The constant is presumably there because you haven’t imposed any normalization conditions on the vectors. And I don’t see the point of the $\sqrt{-g}$. The four vectors live at some point of the manifold, and the value of $\sqrt{-g}$ at that point is just a number that can be absorbed into the constant. If you stated where you found this equation, I suspect we might discover that there is more to this than you have described. Aug 5, 2019 at 5:01

Orthogonality is ensured even without the metric determinant on the left. However, it becomes necessary if you wish the length of the resultant vector $$D$$ to be preserved under coordinate changes.

To see this, contract the expression with $$D^\delta$$ and identify the lhs with the Riemannian volume form

$$\sqrt{-g}\;\varepsilon_{\alpha\beta\gamma\delta}A^\alpha B^\beta C^\gamma D^\delta = \sqrt{-g}\,(dx^1\wedge dx^2\wedge dx^3\wedge dx^4)(A,B,C,D) = \mathrm{const}\,D^\delta D_\delta$$ where $$\wedge$$ denotes the exterior product of the coordinate one-forms.

The appearance of the metric determinant ensures the expression to be invariant under coordinate changes, i.e. the length of $$D$$ doesn't depend on the choice of basis.

The minus sign under the square root comes about because you are (presumably) dealing with a Lorentzian metric. More generally one would write $$\sqrt{|g|}$$.

The constant on the rhs is in my opinion meant to be a constant independent of the point on the manifold. You may choose it however to scale $$D$$.

• Thank you... do you know why the minus sign inside the square root? And I didnt understand the $^$ notation, is it the outer product between the 1-forms? Aug 5, 2019 at 11:31
• @JuanPabloArcila See the edit. Aug 5, 2019 at 12:48