Vector Divergence: Are the Expressions True?

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The discussion centers on the correctness of expressions related to the divergence of a current density vector field in electromagnetic theory. The original expressions presented for the divergences, involving sums of derivatives of the current density, were questioned for their accuracy. It was clarified that the correct form should express the divergence as a sum over the three spatial dimensions, incorporating both position and time dependencies of the current density. The final consensus is that the divergence should be written as sums of partial derivatives with respect to the respective coordinates. This ensures clarity and correctness in representing the divergence of the current density.
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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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Everything seems fine to me
 
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.
 
Well yes that's the kind of the problem I am not sure how to express those things
 
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:

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