Vector Divergence: Are the Expressions True?

In summary, the conversation is about calculating the divergence of a given current density in the context of electromagnetic theory. The expressions for the two divergence operators, ##\nabla' \cdot \vec{J}## and ##\nabla \cdot \vec{J}##, are discussed and the correct expressions are given as sums of partial derivatives of the current density with respect to the position coordinates. The question of whether these expressions are true is also raised.
  • #1
Arman777
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Homework Statement
##\nabla \cdot \vec{J}## and ##\nabla' \cdot \vec{J}$## ?
Relevant Equations
I am doing EMT and I am trying to calculate the divergence of this current density given as,

$$\vec{J}(\vec{r}', t_r) = \vec{J}(\vec{r}', t - \frac{|\vec{r}-\vec{r}'|}{c})$$

for ##\vec{r} = (x,y,z)## and ##\vec{r'} = (x',y',z')##
Now We have two divergence operator,

$$\nabla' = \frac{\partial}{\partial x'}\vec{i} +\frac{\partial}{\partial y'}\vec{j} +\frac{\partial}{\partial z'}\vec{k}$$ and

$$\nabla = \frac{\partial}{\partial x}\vec{i} +\frac{\partial}{\partial y}\vec{j} +\frac{\partial}{\partial z}\vec{k}$$
Do I have to write something like,

$$\nabla' \cdot \vec{J} = \frac{\partial J^m(r')}{\partial x'^m} + \frac{\partial J^m(t_r)}{\partial x'^m}$$

$$\nabla \cdot \vec{J} = \frac{\partial J^m(r')}{\partial x^m} + \frac{\partial J^m(t_r)}{\partial x^m} = \frac{\partial J^m(t_r)}{\partial x^m}$$

Are these expressions true ?
 
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  • #2
Everything seems fine to me
 
  • #3
Arman777 said:
Homework Statement:: ##\nabla \cdot \vec{J}## and ##\nabla' \cdot \vec{J}$## ?
Relevant Equations:: I am doing EMT and I am trying to calculate the divergence of this current density given as,

$$\vec{J}(\vec{r}', t_r) = \vec{J}(\vec{r}', t - \frac{|\vec{r}-\vec{r}'|}{c})$$

for ##\vec{r} = (x,y,z)## and ##\vec{r'} = (x',y',z')##
Now We have two divergence operator,

$$\nabla' = \frac{\partial}{\partial x'}\vec{i} +\frac{\partial}{\partial y'}\vec{j} +\frac{\partial}{\partial z'}\vec{k}$$ and

$$\nabla = \frac{\partial}{\partial x}\vec{i} +\frac{\partial}{\partial y}\vec{j} +\frac{\partial}{\partial z}\vec{k}$$

Do I have to write something like,

$$\nabla' \cdot \vec{J} = \frac{\partial J^m(r')}{\partial x'^m} + \frac{\partial J^m(t_r)}{\partial x'^m}$$

$$\nabla \cdot \vec{J} = \frac{\partial J^m(r')}{\partial x^m} + \frac{\partial J^m(t_r)}{\partial x^m} = \frac{\partial J^m(t_r)}{\partial x^m}$$

Are these expressions true ?
I don't understand why you have a sum of derivatives of ##J^m##. There is only one set of ##J^m##, and each is a function of both position and time.
 
  • #4
Well yes that's the kind of the problem I am not sure how to express those things
 
  • #5
Arman777 said:
Homework Statement:: ##\nabla \cdot \vec{J}## and ##\nabla' \cdot \vec{J}$## ?
Relevant Equations:: I am doing EMT and I am trying to calculate the divergence of this current density given as,

$$\vec{J}(\vec{r}', t_r) = \vec{J}(\vec{r}', t - \frac{|\vec{r}-\vec{r}'|}{c})$$

for ##\vec{r} = (x,y,z)## and ##\vec{r'} = (x',y',z')##
Now We have two divergence operator,

$$\nabla' = \frac{\partial}{\partial x'}\vec{i} +\frac{\partial}{\partial y'}\vec{j} +\frac{\partial}{\partial z'}\vec{k}$$ and

$$\nabla = \frac{\partial}{\partial x}\vec{i} +\frac{\partial}{\partial y}\vec{j} +\frac{\partial}{\partial z}\vec{k}$$

Do I have to write something like,

$$\nabla' \cdot \vec{J} = \frac{\partial J^m(r')}{\partial x'^m} + \frac{\partial J^m(t_r)}{\partial x'^m}$$

$$\nabla \cdot \vec{J} = \frac{\partial J^m(r')}{\partial x^m} + \frac{\partial J^m(t_r)}{\partial x^m} = \frac{\partial J^m(t_r)}{\partial x^m}$$

Are these expressions true ?
Ok, the correct expressions are

EDIT: This is a better way to write it
$$\nabla' \cdot \vec{J} = \sum_{m=1}^3 \frac{\partial J^m(\vec{r},\vec{r'})}{\partial x'^m}$$
and
$$\nabla \cdot \vec{J} = \sum_{m=1}^3 \frac{\partial J^m(\vec{r},\vec{r}')}{\partial x^m} $$
 
Last edited:

1. What is vector divergence?

Vector divergence is a mathematical operation that measures the rate of flow of a vector field from a given point. It is represented by the symbol ∇ · V, where ∇ is the nabla operator and V is the vector field.

2. How is vector divergence calculated?

Vector divergence is calculated by taking the dot product of the nabla operator (∇) and the vector field (V). This results in a scalar value that represents the rate of flow of the vector field from a given point.

3. What are the applications of vector divergence?

Vector divergence has many applications in physics and engineering, particularly in the fields of fluid dynamics and electromagnetism. It is used to describe the behavior of fluids and electromagnetic fields, and is an important tool in understanding the motion of particles in these systems.

4. Are the expressions for vector divergence always true?

Yes, the expressions for vector divergence are always true. However, they may not always be applicable in certain situations, such as when dealing with discontinuous or singular vector fields.

5. Can vector divergence be negative?

Yes, vector divergence can be negative. A negative divergence value indicates that the vector field is converging towards the given point, while a positive value indicates that the vector field is diverging away from the given point.

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