# Why does (current density) dJ/dt=0?

1. Sep 1, 2014

### iScience

could someone explain to me why the the time derivative of current density J would be zero?

more specifically..

$$\mu_0\frac{∂\vec{J}}{∂t}=0$$

trying to derive maxwell's wave equation for E-field and i guess one of the assumptions i have to make is the above. but if i'm making a wave, i thought i had to have accelerating charges in which case why would the above be true?

2. Sep 1, 2014

### rude man

The Maxwell equation you are referring to is ∇ x H = j + ∂D/∂t

where

H = B0
j = current density vector
D0E.

Fact: e-m waves are created by virtue of the ∂D/∂t term, not the j term.

BTW I am not aware that ∂j/∂t has to be zero.

3. Sep 1, 2014

### iScience

yea that's the equation i'm referring to (the corrected ampere's law).

I agree that a constant J would not produce a wave, however ∂J/∂t should. and here's where i get ∂J/∂t:

if we set this equation equal to faraday's law in differential form:

$$\frac{∂}{∂t}(\vec{\nabla} \times \vec{B})= \frac{∂}{∂t}(\mu_0(\vec{J}+\epsilon_0\frac{∂\vec{E}}{∂t}))$$

$$\nabla \times \frac{∂\vec{B}}{∂t}=-\vec{\nabla} \times (\vec{\nabla} \times E)$$

$$\frac{∂}{∂t}(\mu_0(\vec{J}+\epsilon_0\frac{∂\vec{E}}{∂t}))=-\vec{\nabla} \times (\vec{\nabla} \times E)$$

this is the place i'm referring to. why does the $\mu_0 \frac{∂\vec{J}}{∂t}$ term disappear?

4. Sep 1, 2014

### ehild

$$\vec{\nabla} \times \vec{B}= \mu_0\left(\vec{J}+\epsilon_0\frac{∂\vec{E}}{∂t}\right)$$

and

$$\vec{\nabla} \times \vec{E}= -\frac{∂\vec{B}}{∂t}$$

Take the curl of the second one,

$$\vec{\nabla}\times\vec{\nabla} \times \vec{E}= -\frac{∂}{∂t}\vec{\nabla}\times\vec{B}$$

and substitute the first one
--->

$$\vec{\nabla}\times\vec{\nabla} \times \vec{E}= -\frac{∂}{∂t}\mu_0\left(\vec{J}+\epsilon_0\frac{∂\vec{E}}{∂t}\right)$$

ehild

5. Sep 1, 2014

### iScience

exactly what i did..

still don't see how that term disappears

6. Sep 1, 2014

### ehild

What do you mean that J disappears? It is still there.

ehild