What was Maxwell Thinking?

Conservation is one of the basic assumptions of physics.  The basic assumption is that for any kind of “stuff,” the amount of stuff going into any region is equal to the increase in stuff in that region. Using the formalism of vector calculus there is a clear way of stating this conservation law for the specific case of charge.

Let \rho be a charge density function, dependant on position \vec{r} and time t. \partial_t \rho represents the rate of change of charge density with respect to time. Using the standard basis vectors x_1,x_2,x_3 the spatial partial derivatives \partial_{x_1} \rho, \partial_{x_2} \rho, \partial_{x_3} \rho represent the rate of change in the respective directions.

The basic statement of the Continuity Equation makes a statement about any region.  It says that the sum of all of the partial derivatives is zero, or:

\partial_{x_1} \rho + \partial_{x_2} \rho + \partial_{x_3} \rho + \partial_t \rho = 0.

If we call time x_0, then we can write this a simpler way:

\sum_{\mu = 0}^3 \partial_{x^\mu} \rho = 0

If we assume that any greek index is being summed from 0 to 3, and any latin index is being summed from 1 to 3, we can get a more compact continuity equation:

\partial_{\mu} \rho = 0

Now, let’s look at the laws of nature that Maxwell learned in his physics classes.

The first one was that the partial derivatives of the electric field coming out of a region is proportional to the charge density.

\partial_1 E_1  + \partial_2 E_2 + \partial_3 E_3 \propto \rho

Using our new, compact notation where latin indices represent summing over all the spatial variables (1 to 3), we can express this as:

\partial_i E_i \propto \rho

Other authors often use the notation div \vec{E} \propto \rho, but I don’t.

 

 

 

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s