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How do the scalar potential $\phi(x,t)$ and vector potential $A(x,t)$ behave under the parity and time-reversal transformations?

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Recall the relations $\textbf{E}=-\boldsymbol{\nabla} V(\textbf{r},t)-\frac{\partial}{\partial t}\textbf{A}(\textbf{r},t)$ and $\textbf{B}=\boldsymbol{\nabla}\times \textbf{A}(\textbf{r},t)$. Now, $\textbf{E}$ is an polar vector i.e., under parity $\textbf{E}\rightarrow -\textbf{E}$, and $\textbf{B}$ is an axial vector i.e., under parity $\textbf{B}\rightarrow\textbf{B}$. Since, $\boldsymbol{\nabla}$ is odd under parity, $\textbf{A}$ must also be odd under parity. Similarly, $V(\textbf{r},t)$ is also odd under parity for $\textbf{E}$ to be a polar vector.

To understand that the behaviour of $\textbf{E}$ under parity one can use the formula $$\textbf{E}(\textbf{r})=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\textbf{r}^\prime)d^3\textbf{r}^\prime}{|\textbf{r}-\textbf{r}^\prime|}$$ from Electrostatics. Note that, under parity $d^3\textbf{r}^\prime\rightarrow -d^3\textbf{r}^\prime$ and the integrand is invariant under parity. Therefore, under parity $\textbf{E}\rightarrow -\textbf{E}$. To derive the behaviour of $\textbf{B}$ under parity use the fact that the current density $\textbf{J}\rightarrow -\textbf{J}$ under parity.

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The scalar potential V does not change sign under parity . I think it must have even parity. It can be shown from the expression of electric field. $\textbf{E}=-\boldsymbol{\nabla} V -\frac{\partial}{\partial t}\textbf{A}$

I think this equation should remain invariant under parity. For that V should not change sign under parity.

Under parity,

$\textbf{E}\rightarrow - \textbf{E}$

$\textbf{B}\rightarrow \textbf{B}$

$\boldsymbol{\nabla}\rightarrow - \boldsymbol{\nabla} $

$\textbf{A}\rightarrow -\textbf{A}$

$\textbf{V}\rightarrow \textbf{V}$

Under time reversal operation,

$\textbf{E}\rightarrow \textbf{E}$

$\textbf{B}\rightarrow -\textbf{B}$

$\boldsymbol{\nabla}\rightarrow \boldsymbol{\nabla} $

$\textbf{A}\rightarrow -\textbf{A}$

$\textbf{V}\rightarrow \textbf{V}$

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    $\begingroup$ Welcome to Physics SE! Please add some detail to your answer. For example, you could give an expression for the electric field and demonstrate that $V$ does not change under parity. $\endgroup$ – Chris Oct 3 '18 at 14:30

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