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naima said:I think that the volume integral of a "something"-density is the "something". (same as for a mass density)
naima said:I think that the volume integral of a "something"-density is the "something". (same as for a mass 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:
[itex]\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}\ ,[/itex]
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,
[itex]\int_S{ \mathbf{J} \cdot \mathrm{d}\mathbf{A}} = \int_V{(\mathbf{\nabla} \cdot \mathbf{J}) \mathrm{d}V}\ .[/itex]
Hence:
[itex]\int_V{(\mathbf{\nabla} \cdot \mathbf{J}) \mathrm{d}V}\ = - \int_V{\left( \frac{\partial \rho}{\partial t} \right) \mathrm{d}V}\ .[/itex]
Because this relation is valid for any volume, no matter how small, no matter where located:
[itex]\nabla \cdot \mathbf{J} = - \frac{\partial \rho}{\partial t}\ ,[/itex]
which is called the continuity equation.[5][6]
naima said: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.
blenx said:In classical electrodynamics, the volume integral of the current density is the time derivative of the electric dipole moment of the system:
[tex]\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}}[/tex]
daudaudaudau said:Yes. What is the physical interpretation of such a quantity? Is it some sort of "total current"?
The volume integral of current density is a mathematical concept used in electromagnetism to calculate the total current passing through a specific volume of space. It takes into account the magnitude and direction of the current at every point within the volume.
The volume integral of current density is calculated by integrating the current density vector over the volume of interest. This involves taking the dot product of the current density vector and the differential volume element, and then summing the values over the entire volume.
The volume integral of current density is significant because it allows us to determine the total current passing through a given volume, which is useful for understanding and analyzing the behavior of electromagnetic fields. It is also a fundamental concept in the study of electromagnetism and is used in many practical applications, such as designing electrical circuits and devices.
The units of the volume integral of current density depend on the units of the current density vector and the differential volume element. In the SI system, the units are typically ampere (A) per meter squared (m^2).
Yes, the volume integral of current density can be negative. This can occur if the current density vector and the differential volume element have opposite directions, resulting in a negative dot product. It is important to consider the direction of the current when interpreting the sign of the volume integral of current density.