Volume integral of current density

[daudaudaudau]

Hi.

Can anyone tell me what the volume integral of the current density is? I find it strange, but G.D. Mahan uses it in his book on page 30. He claims that this is in fact the current. I have attached the particular page.

 

[naima]

I think that the volume integral of a "something"-density is the "something". (same as for a mass density)

 

[daudaudaudau]

I think that the volume integral of a "something"-density is the "something". (same as for a mass density)

Can you refer my to any other author, who takes the volume integral of the current density and claims that is a current? The current density has units A/m^2, so shouldn't you integrate it over a surface to get the current?

 

[Calrik]

I think that the volume integral of a "something"-density is the "something". (same as for a mass density)

That's not very revealing.

Density is related by the volume and mass of any set of "atoms" in a sphere or any other shape for that matter, dynamic is just a vector or another axis.

Simply it is an equation of volume/m^2 and volume as per :

4/3 pi r^3

Integrate both with limits surely? You can stick in an I or two but relatively speaking there is no point current wise.

You could just use Maxwell/Faraday equation though.

\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}

Just engineer it for your use.

\mathbf{J} (\mathbf{r}, t) = \int_{-\infty}^t \mathrm{d}t' \int \mathrm{d}^3\mathbf{r}' \; \sigma(\mathbf{r}-\mathbf{r}', t-t') \; \mathbf{E}(\mathbf{r}',\ t') \, ,

http://en.wikipedia.org/wiki/Current_density

Main article: Continuity equation

Because charge is conserved, the net flow out of a chosen volume must equal the net change in charge held inside the volume:

\int_S{ \mathbf{J} \cdot \mathrm{d}\mathbf{A}} = -\frac{\mathrm{d}}{\mathrm{d}t} \int_V{\rho \; \mathrm{d}V} = - \int_V{\left( \frac{\partial \rho}{\partial t} \right) \mathrm{d}V}\ ,

where ρ is the charge density per unit volume, and dA is a surface element of the surface S enclosing the volume V. The surface integral on the left expresses the current outflow from the volume, and the negatively signed volume integral on the right expresses the decrease in the total charge inside the volume. From the divergence theorem,

\int_S{ \mathbf{J} \cdot \mathrm{d}\mathbf{A}} = \int_V{(\mathbf{\nabla} \cdot \mathbf{J}) \mathrm{d}V}\ .

Hence:

\int_V{(\mathbf{\nabla} \cdot \mathbf{J}) \mathrm{d}V}\ = - \int_V{\left( \frac{\partial \rho}{\partial t} \right) \mathrm{d}V}\ .

Because this relation is valid for any volume, no matter how small, no matter where located:

\nabla \cdot \mathbf{J} = - \frac{\partial \rho}{\partial t}\ ,

which is called the continuity equation.[5][6]

Differentials/integrals are called that because they are based on axioms that say 0=a and a therefore can = x. Hence PDE. If we assume there is a point where nothing changes we can assume there are points where something changes if not then it is static.

 

[naima]

When you integrate over a surface you get a quantity (the flow). Here it is not the integral of the dot product but
r*dot product. r is a vector.
So you do not find the same thing : you get a vector.
The aim is to see the current as a vectorial operator
which is done with i[H,P]
It is also the first time I see that.
I have still to convince myself that there is no pb with the
integration by parts.

 

[Calrik]

When you integrate over a surface you get a quantity (the flow). Here it is not the integral of the dot product but
r*dot product. r is a vector.
So you do not find the same thing : you get a vector.
The aim is to see the current as a vectorial operator
which is done with i[H,P]
It is also the first time I see that.
I have still to convince myself that there is no pb with the
integration by parts.

In this case there is, but it's because to integrate by parts you have to assume something ie the laws of calculus are true for all cases of x. Otherwise you are absolutely right.

 

[blenx]

In classical electrodynamics, the volume integral of the current density is the time derivative of the electric dipole moment of the system:

\int {{\boldsymbol{J}}{{\text{d}}^3}V} = \sum\limits_i^{} {\int {{{\boldsymbol{J}}_i}{{\text{d}}^3}V} } = \sum\limits_i^{} {{q_i}{{\boldsymbol{v}}_i}} = \sum\limits_i^{} {{q_i}{{{\boldsymbol{\dot x}}}_i}} = {\boldsymbol{\dot d}}

 

[daudaudaudau]

In classical electrodynamics, the volume integral of the current density is the time derivative of the electric dipole moment of the system:

\int {{\boldsymbol{J}}{{\text{d}}^3}V} = \sum\limits_i^{} {\int {{{\boldsymbol{J}}_i}{{\text{d}}^3}V} } = \sum\limits_i^{} {{q_i}{{\boldsymbol{v}}_i}} = \sum\limits_i^{} {{q_i}{{{\boldsymbol{\dot x}}}_i}} = {\boldsymbol{\dot d}}

Yes. What is the physical interpretation of such a quantity? Is it some sort of "total current"?

 

[blenx]

Yes. What is the physical interpretation of such a quantity? Is it some sort of "total current"?

It is approximately the product of the current and the displacement between the centers of the positive and negetive charges, viz. {\boldsymbol{\dot d}} \sim I\Delta {\boldsymbol{l}}.