Law conservation of charge relativistic

[Petar Mali]

\frac{\partial \rho}{\partial t}+div\vec{j}=0

In Deckart coordinate system

\frac{\partial j_x}{\partial x}+\frac{\partial j_y}{\partial _y}+\frac{\partial j_z}{\partial _z}+\frac{\partial (c\rho)}{\partial (ct)}=0

definition

divA^{\mu}=\frac{\partial A^{\mu}}{\partial x^{\mu}}
scalar (invariant)

Why I define divergence like that? Is there some certain rules for that?

j^{\mu}=(j_x,j_y,j_z,c\rho)=(\vec{j},c\rho)

\frac{\partial j^{\mu}}{\partial x^{\mu}}=0

divj^{\mu}=0

Now is satisfied

j_{\mu}=(-\vec{j},c\rho)

Can I interprate this like time inversion. Changing od indeces, think of that?

What is with

divj_{\mu}=?

 

[Mentz114]

I'm not sure what you're asking, but if charge is conserved then \frac{\partial \rho}{\partial t}=0 and so \nabla j_\mu=0. Physically this means that the amount of charge entering a volume of space must be equal to the amount leaving the volume. Or, expressed as a surface integral, the current crossing the boundary of the volume is zero. Hence

<br /> \nabla j_\mu=0 \rightarrow \nabla( -j_\mu)=0<br />

just means that reversing the current leaves the divergence unchanged. If \frac{\partial \rho}{\partial t}\ne 0 then reversing the current changes the sign of \frac{\partial \rho}{\partial t}.

See Stoke's theorem.

 

[Phrak]

You cannot obtain the conservation of charge simply by proposing that

<br /> j^{\mu}=(j_x,j_y,j_z,c\rho)=(\vec{j},c\rho)<br />

is a perfectly good contravariant vector in four dimensions. And it is.

You're on the right track. It just doesn't lead to conservation of charge. You need to combine Ampere's law and Gauss's law to get charge conservation. Normally it is called the "charge continuity equation" that you might find in wikipedia.

(Mentz114, I'm still thinking over your posts in another thread, thinking of something intelligent to add.)

 

[Petar Mali]

My question is bold in the text and also after them is very interesting sign ?

 

[Petar Mali]

@ Phrak

<br /> divj^{\mu}=0<br />

This is law conservation of charge. I don't see why it wasn't finished?

 

[Phrak]

You should drop the dot product convention and adhere to valid tensor operations in the four dimensions of spacetime.

With this, in simplest form, the charge continuity equation is

D_\mu J^\mu = 0

where
D_\mu = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}, -\frac{\partial}{c\partial t})[/itex]<br /> and<br /> J^\mu = (j_x, j_y, j_z, c\rho)[/itex]&lt;br /&gt; &lt;br /&gt; The result is a scalar. It could be any scalar, but it is zero because both charge and current are derivatives of the electric and magnetic fields. It happens to be zero because partial derivatives commute. &lt;br /&gt; &lt;br /&gt; \frac{\partial}{\partial x^i} \frac{\partial}{\partial x^j} -&amp;lt;br /&amp;gt; \frac{\partial}{\partial x^j} \frac{\partial}{\partial x^i}=0[/itex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; You can find all of this is wikipedia somewhere where you would find that J is a derivative of the Faraday tensor.

 

[Phrak]

I don't understand. Ask again.

 

[Petar Mali]

I have problems with definition of divergence and curl (rot)

<br /> divA^{\mu}=\frac{\partial A^{\mu}}{\partial x^{\mu}}<br />

rotA_{\mu}=\frac{\partial A_{\nu}}{\partial x^{\mu}}-\frac{\partial A_{\mu}}{\partial x^{\nu}}

Why I define this like that?

In one mathematical physics book I find

(rotA_i)_j=\frac{\partial A_i}{\partial x^j}-\frac{\partial A_j}{\partial x^i}

 

[Phrak]

Yes, divergence and curl are very confusing operators at first. In more advanced mathematics they finally make sense.

They make more sense understood as a dot product and a cross product.

The partials,

\nabla = \left( \frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} \right)
is a vector in three dimensions.

A vector such as

\left( A\frac{\partial}{\partial x},B\frac{\partial}{\partial y},C\frac{\partial}{\partial z} \right)
obeys all the rules of a vector space over the field of real numbers.

 

[Phrak]

@ Phrak

<br /> divj^{\mu}=0<br />

This is law conservation of charge. I don't see why it wasn't finished?

Now I understand. Yes, this is conservation of charge.

 

[Dale]

I'm not sure what you're asking

My question is bold in the text and also after them is very interesting sign ?

Petar Mali, when someone points out that your question is confusing it will be more productive to clarify your intention rather than to point out some trivial formatting. For example the following question:

Changing od indeces, think of that?

Is indeed in bold and does have a question mark afterwards, but it is gramattically incorrect and meaningless despite the presence of the correct formatting. It is not that your questions could not be found, but that they could not be understood.