Divergence in Spherical Coordinate System by Metric Tensor

AI Thread Summary
The discussion centers on the divergence of vector fields in spherical coordinates, highlighting discrepancies between the user's results and standard electrodynamics forms. The user struggles with the transformation of components and the distinction between covariant and contravariant components, leading to confusion about their dimensions. It is clarified that the user's calculations utilize holonomic basis vectors, while textbooks employ normalized basis vectors, which affects the divergence expression. The correct transformation between these components is essential to align with textbook results, ultimately leading to a proper divergence formula. Understanding these distinctions is crucial for accurate application in electrodynamics.
Astrocyte
Messages
15
Reaction score
4
Homework Statement
I want to get a familiar form of spherical coordinate system's divergence formula from the metric tensor.
Relevant Equations
metric tensor, covariant derivative
1620289371095.png

The result equation doesn't fit with the familiar divergence form that are usually used in electrodynamics.
I want to know the reason why I was wrong.
My professor says about transformation of components.
1620289527170.png

But I cannot close to answer by using this hint, because I don't have any idea about "x".
Even I'm confused whether covariant component is correct or contravariant one does.
 
Physics news on Phys.org
In your setting all the three components of A have same dimension on L or not? In familiar case you showed their components have same dimension on L and distinction of covariant and contravariant components are not clear to me there.

For an instance a special case of ##\mathbf{A}=d\mathbf{r}## you may take
A_r=dr,A_\theta =d\theta,A_\phi=d\phi
A^r=dr,A^\theta =r^2 d\theta,A^\phi=r^2 \sin^2\theta\ d\phi
In familiar vector analysis vector components are somewhat between, i.e.
A_r=dr,A_\theta =r\ d\theta,A_\phi=r\ sin\theta \ d\phi
All the components have dimension of L. You see the coefficient ##r## differ between them.
 
Last edited:
The difference between your results and the ones usually found in textbooks on electrodynamics simply is that in the above calculuation you use the holonomic (in this case orthogonal) basis vectors, while in the textbooks they use the corresponding normalized (i.e., orthonormal) basis vectors (3-beins).

It's easier to see calculating the vectors explicitly in terms Carstesian components:
$$\vec{x}=\begin{pmatrix} r \sin \vartheta \cos \varphi \\ r \sin \vartheta \sin \varphi \\ r \sin \vartheta \end{pmatrix}.$$
The holnomous basis is simply given as the partial derivatives wrt. the spherical coordinates,
$$\vec{b}_r = \partial_r \vec{x} = \begin{pmatrix} \sin \vartheta \cos \varphi \\ \sin \vartheta \sin \varphi \\ \sin \vartheta \end{pmatrix}.$$
$$\vec{b}_{\vartheta}=\partial_{\vartheta} \vec{x} = \begin{pmatrix} r \cos \vartheta \cos \varphi \\ r \sin \vartheta \sin \varphi \\ -r \cos \vartheta \end{pmatrix}.$$
$$\vec{b}_{\varphi} = \partial_{\varphi} \vec{x}= \begin{pmatrix} -r \sin \vartheta \sin \varphi \\ r \sin \vartheta \cos \varphi \\ 0 \end{pmatrix}.$$
The vectors are orthogonal but not normalized. In your standard Ricci calculus these are the basis vectors you use.

In the E&M textbooks you usually use the corresponding normalized vectors. The norms are
$$|\vec{b}_r|=1=\sqrt{g_{rr}}, \quad |\vec{b}_{\vartheta}|=r=\sqrt{g_{\vartheta \vartheta}}, \quad |\vec{b}_{\varphi}|=r \sin \vartheta=\sqrt{g_{\varphi \varphi}}.$$

So your vector fields are decomposed as
$$\vec{A}=A^r \vec{b}_r + A^{\vartheta} \vec{b}_{\vartheta} + A^{\varphi} \vec{b}_{\varphi}.$$
In the textbooks it's
$$\vec{A}=\tilde{A}^r \vec{e}_r + \tilde{A}^{\vartheta} \vec{e}_{\vartheta} + \tilde{A}^{\varphi} \vec{e}_{\varphi}.$$
To get the "textbook expression" for the divergence you simply have to express the ##A_j## by the ##\tilde{A}_j##. That's easy: Just express the ##\vec{e}_j## with the ##\vec{b}_j##:
$$\vec{A}=\tilde{A}^r \vec{b}_r + \frac{\tilde{A}^{\vartheta}}{r} \vec{b}_{\vartheta} + \frac{\tilde{A}^{\varphi}}{r \sin \vartheta} \vec{b}_{\varphi}.$$
So you have
$$A^r=\tilde{A}^r, \quad A_{\vartheta}=\frac{\tilde{A}^{\vartheta}}{r}, \quad A^{\varphi}=\frac{\tilde{A}_{\varphi}}{r \sin \vartheta}.$$
In your expression you rather use the covariant components, i.e., using your covariant metric components,
$$A^r=A_r=\tilde{A}^r, \quad A_{\vartheta}=r^2 A^{\vartheta}=r \tilde{A}^{\vartheta}, \quad A_{\varphi}=r^2 \sin^2 \vartheta A^{\varphi}=r \sin \vartheta \tilde{A}^{\varphi}.$$

Plug this in your equation for the divergence, you get
$$\vec{\nabla} \cdot \vec{A}=\frac{1}{r^2}(r^2 \tilde{A}_r) + \frac{1}{r \sin \vartheta} \partial_{\vartheta}(\sin \vartheta \tilde{A}^{\vartheta}) + \frac{1}{r \sin \vartheta} \partial_{\varphi} \tilde{A}^{\varphi},$$
as it should be.

Also note that in the "textbook formalism" ##\tilde{A}^{j}=\tilde{A}_{j}##, because here your basis is orthonormal ("3-bein formalism") and thus ##\tilde{g}_{jk}=\delta_{jk}##.
 
  • Like
Likes anuttarasammyak and etotheipi
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top