- #1

yucheng

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- Homework Statement
- Vanderlinde, Classical Electromagnetic Theory

- Relevant Equations
- N/A

Divergence formula

$$\vec{\nabla} \cdot \vec{A}= \frac{1}{\sqrt{G}} \frac{\partial}{\partial q^{j}} (A^{j} \sqrt{G})$$

If we express it in terms of the components of ##\vec{A}## in unit basis using

$$A^{*j} = \sqrt{g^{jj}} A^{j}$$

, we get $$\vec{\nabla} \cdot \vec{A}= \frac{1}{\sqrt{G}} \frac{\partial}{\partial q^{j}} (A^{*j} \sqrt{g^{jj} G})$$

However, did the author intend

$$A^{j} = \sqrt{g^{jj}} A^{*j}$$

(quick check: with this one can directly substitute to get the correct divergence formula)? Because

$$\vec{A} = A^{j} \vec{e}_j = A^{*j} \hat{e}_j = A^{*j} \frac{1}{\sqrt{g_{jj}}} \sqrt{g_{jj}} \hat{e}_j = A^{*j} \sqrt{g^{jj}} \, \vec{e}_j$$

, since $$\vec{e}^{i} \cdot \vec{e}_{i} = \hat{e}^{i} \cdot \hat{e}_{i} \sqrt{g^{ii}g_{ii}} = \sqrt{g^{ii}g_{ii}} = 1$$

$$\vec{\nabla} \cdot \vec{A}= \frac{1}{\sqrt{G}} \frac{\partial}{\partial q^{j}} (A^{j} \sqrt{G})$$

If we express it in terms of the components of ##\vec{A}## in unit basis using

$$A^{*j} = \sqrt{g^{jj}} A^{j}$$

, we get $$\vec{\nabla} \cdot \vec{A}= \frac{1}{\sqrt{G}} \frac{\partial}{\partial q^{j}} (A^{*j} \sqrt{g^{jj} G})$$

However, did the author intend

$$A^{j} = \sqrt{g^{jj}} A^{*j}$$

(quick check: with this one can directly substitute to get the correct divergence formula)? Because

$$\vec{A} = A^{j} \vec{e}_j = A^{*j} \hat{e}_j = A^{*j} \frac{1}{\sqrt{g_{jj}}} \sqrt{g_{jj}} \hat{e}_j = A^{*j} \sqrt{g^{jj}} \, \vec{e}_j$$

, since $$\vec{e}^{i} \cdot \vec{e}_{i} = \hat{e}^{i} \cdot \hat{e}_{i} \sqrt{g^{ii}g_{ii}} = \sqrt{g^{ii}g_{ii}} = 1$$

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