How can one understand intuitively that without magnetic field, the potential can still exist? Also $\nabla^2 \Phi=−\rho/\epsilon_0$. If charge density is $0$,$\Phi$ is non zero. How can potential still exist without charge?
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If $\mathbf{A}=\nabla\phi$ where $\phi$ is an arbitrary scalar field, then we have $$\nabla\times\mathbf{B}=\nabla\times\nabla\phi=0.$$ Since $\phi$ is arbitrary, it is easy to choose a $\phi$ that makes $\mathbf{A}\neq0$. $\mathbf{B}$ is physical. $\mathbf{A}$ is simply a mathematical tool to make calculation about magnetic field easier. It does not have any physical meaning in itself.
Regarding electric potential, same thing goes. It is merely a mathematical tool to simplify equations in certain cases.
Using these auxiliary fields is similar to the following use of $a$:
Suppose we have an equation $$x+f(x)+g(x)=\left(\frac{c_1+c_2}{c_3}+c_4\right)^{c_5}+c_6c_7$$ where $c_i,(i=1,2,...,7)$ are constants. We can simplify this equation by defining the auxiliary parameter $$a\equiv\left(\frac{c_1+c_2}{c_3}+c_4\right)^{c_5}+c_6c_7$$ The equation then becomes $$x+f(x)+g(x)=a$$ In the consequent calculations, you need not to deal with the large number of $c_i$ that cause inconvenience.
Yufei
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