- #1
DavideGenoa
- 155
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Hello, friends! My textbook, Gettys's Physics, says that the Lorenz gauge choice uses the magnetic vector potential $$\mathbf{A}(\mathbf{x},t):=\frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{y},t-c^{-1}\|\mathbf{x}-\mathbf{y}\|)}{\|\mathbf{x}-\mathbf{y}\|}d^3y $$and the electric potential $$V(\mathbf{x},t):=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{y},t-c^{-1}\|\mathbf{x}-\mathbf{y}\|)}{\|\mathbf{x}-\mathbf{y}\|}d^3y $$which are such that $$\nabla^2\mathbf{A}(\mathbf{x},t)-\varepsilon_0\mu_0\frac{\partial^2 \mathbf{A}(\mathbf{x},t)}{\partial t^2}=-\mu_0\mathbf{J}(\mathbf{x},t)$$but the book does not prove how ##\mathbf{A}## satisfies this equality.
How can we prove that it satisfies ##\nabla^2\mathbf{A}-\varepsilon_0\mu_0\frac{\partial^2 \mathbf{A}}{\partial t^2}=-\mu_0\mathbf{J}##, as well as Maxwell's equations such as $$\nabla\cdot(\nabla\times\mathbf{A})=0$$ $$\oint_{\partial^+\Sigma} \nabla V\cdot d\mathbf{x}=-\frac{d}{dt}\int_{\Sigma} (\nabla\times\mathbf{A})\cdot d\mathbf{S}$$ $$\int_{\partial^+ \Sigma}(\nabla\times\mathbf{A})\cdot d\mathbf{x}=\mu_0\int_{\Sigma} \mathbf{J}\cdot d\mathbf{S}+\mu_0\varepsilon_0\frac{d}{dt}\int_{\Sigma}\nabla V\cdot d\mathbf{S}?$$
I heartily thank you for any answer!
How can we prove that it satisfies ##\nabla^2\mathbf{A}-\varepsilon_0\mu_0\frac{\partial^2 \mathbf{A}}{\partial t^2}=-\mu_0\mathbf{J}##, as well as Maxwell's equations such as $$\nabla\cdot(\nabla\times\mathbf{A})=0$$ $$\oint_{\partial^+\Sigma} \nabla V\cdot d\mathbf{x}=-\frac{d}{dt}\int_{\Sigma} (\nabla\times\mathbf{A})\cdot d\mathbf{S}$$ $$\int_{\partial^+ \Sigma}(\nabla\times\mathbf{A})\cdot d\mathbf{x}=\mu_0\int_{\Sigma} \mathbf{J}\cdot d\mathbf{S}+\mu_0\varepsilon_0\frac{d}{dt}\int_{\Sigma}\nabla V\cdot d\mathbf{S}?$$
I heartily thank you for any answer!
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