The Ampere's Law is [tex] \nabla \times B = \mu J [/tex] and Gauss's Law is [tex] \nabla \cdot E = \frac{1}{\epsilon} \rho [/tex](adsbygoogle = window.adsbygoogle || []).push({});

Since J is current density, is it right to say that, [tex] J = \frac{d}{dt} \rho [/tex] in general?

I am abit confused, since I know that a current four-vector, [tex] (\rho , J) [/tex] is similar to a spacetime four-vector [tex] (t, x) [/tex]. But, x is not [tex] \frac{d}{dt} t [/tex]

Also, does a non-zero J automatically implies a non-zero [tex] \rho [/tex] ?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Ampere's Law and Gauss's Law

**Physics Forums | Science Articles, Homework Help, Discussion**