Please help in part of derivation in retarted potential formula.

[yungman]

This is part of the derivation that I cannot do, this is to show:

$$\nabla'\cdot\vec J_{(\vec r\;',t_r)} =-\frac {\partial \rho_{(\vec r\;',t_r)}}{\partial t} -\frac 1 c \frac {\partial \vec J_{(\vec r\;',t_r)}}{\partial t_r}\cdot(\nabla'\eta) \;\hbox { where } \;\eta = |\vec r - \vec r\;'| \;\hbox { ,}\; t_r= t-\frac {\eta}{c} \;\hbox { and }\; c=\frac 1 {\sqrt{\mu_0\epsilon_0}}$$

The book claimed the first term $\rho$ is from continunity equation. I just don't know how to get this.

$\nabla$ is respect to $\vec r$ and $\nabla'$ is respect to $\vec r\;'$

I have no problem solving:

$$\nabla\cdot\vec J_{(\vec r\;',t_r)} = \frac {\partial J_x}{\partial t_r} \frac { \partial t_r}{\partial x} + \frac {\partial J_y}{\partial t_r} \frac { \partial t_r}{\partial y} +\frac {\partial J_z}{\partial t_r} \frac { \partial t_r}{\partial z} \;=\;-\frac 1 c \left ( \frac {\partial J_x}{\partial t_r} \frac { \partial \eta}{\partial x} + \frac {\partial J_y}{\partial t_r} \frac { \partial \eta}{\partial y} +\frac {\partial J_z}{\partial t_r} \frac { \partial \eta}{\partial z}\right ) \;=\; -\frac 1 c \frac {\partial \vec J}{\partial t_r} \cdot (\nabla \eta)$$

I thought it is just the same:

$$\nabla'\cdot\vec J_{(\vec r\;',t_r)} = \frac {\partial J_x}{\partial t_r} \frac { \partial t_r}{\partial x'} + \frac {\partial J_y}{\partial t_r} \frac { \partial t_r}{\partial y'} +\frac {\partial J_z}{\partial t_r} \frac { \partial t_r}{\partial z'} \;=\;\frac 1 c \left ( \frac {\partial J_x}{\partial t_r} \frac { \partial \eta}{\partial x'} + \frac {\partial J_y}{\partial t_r} \frac { \partial \eta}{\partial y'} +\frac {\partial J_z}{\partial t_r} \frac { \partial \eta}{\partial z'}\right ) \;=\; -\frac 1 c \frac {\partial \vec J}{\partial t_r} \cdot (\nabla \eta)$$


Please help.

Thanks

 

[yungman]

Also $$\vec A =\frac {\mu_0}{4\pi} \int \frac {\vec J_{(\vec r\;',t_r)}}{\eta} d\tau'$$

Means the potential A at the field point (x,y,z) at time t is the result of the current density at the source point (x',y',z') at the retarded time $t_r$. Then:

$$\nabla\cdot\vec J_{(\vec r\;',t_r)} = \frac {\partial J_ {x'} }{\partial x} + \frac {\partial J_{y'}}{\partial y} +\frac {\partial J_{z'}}{\partial z}$$

Instead of

$$\nabla\cdot\vec J_{(\vec r\;',t_r)} = \frac {\partial J_ {x} }{\partial x} + \frac {\partial J_{y}}{\partial y} +\frac {\partial J_{z}}{\partial z}$$

Shown in the book. Because J is located at the source point (x',y',z') instead of at the field point (x,y,z). Am I correct?

Thanks

 

[yungman]

Anyone please? I am stuck! At least one of the two post!

Thanks

Alan

 

[Born2bwire]

You understand the charge continuity equation, right? It is one of the constitutive equations of electrodynamics. If you have a net current over a closed volume, then this means that there is a net flow of charges into or out of the volume and thus the charge density in the volume must be changing with time. You can easily use integral forms with the divergence theorem to recast this more explicitly but reflection upon the physical meaning of the divergence operator should show the same phenomenon.

That being said, this is just an example of the chain rule. You need to remember that the retarded time is also dependent upon your source point. So if you keep your observation point fixed (r and t) and you vary your source point (r') then you will vary both r' and t'. Thus, when you take the derivative of your retarded potentials you need to make use of the chain rule.

Thus,

$$\frac{\partial f(y(x'),z(x'))}{\partial x'} = \frac{\partial f(y(x'),z(x'))}{\partial y(x')} \frac{\partial y(x')}{\partial x'} + \frac{\partial f(y(x'),z(x'))}{\partial z(x')} \frac{\partial z(x')}{\partial x'}$$

Using this with the charge continuity equation:

$$\left[ \nabla' \cdot \mathbf{J} \right]_{ret} \nabla' \cdot \mathbf{x'} = \left[ \nabla' \cdot \mathbf{J} \right]_{ret} = \nabla' \cdot \left[ \mathbf{J} \right]_{ret} - \left[ \frac{ \partial \mathbf{J} }{\partial t'} \right]_{ret} \nabla' \left( t' \right) = -\frac{\partial \rho_{ret}}{\partial t} - \left[ \frac{ \partial \mathbf{J} }{\partial t'} \right]_{ret} \nabla' \left( t' \right)$$

 

[yungman]

You understand the charge continuity equation, right? It is one of the constitutive equations of electrodynamics. If you have a net current over a closed volume, then this means that there is a net flow of charges into or out of the volume and thus the charge density in the volume must be changing with time. You can easily use integral forms with the divergence theorem to recast this more explicitly but reflection upon the physical meaning of the divergence operator should show the same phenomenon.

That being said, this is just an example of the chain rule. You need to remember that the retarded time is also dependent upon your source point. So if you keep your observation point fixed (r and t) and you vary your source point (r') then you will vary both r' and t'. Thus, when you take the derivative of your retarded potentials you need to make use of the chain rule.

Thus,

$$\frac{\partial f(y(x'),z(x'))}{\partial x'} = \frac{\partial f(y(x'),z(x'))}{\partial y(x')} \frac{\partial y(x')}{\partial x'} + \frac{\partial f(y(x'),z(x'))}{\partial z(x')} \frac{\partial z(x')}{\partial x'}$$

Using this with the charge continuity equation:

$$\left[ \nabla' \cdot \mathbf{J} \right]_{ret} \nabla' \cdot \mathbf{x'} = \left[ \nabla' \cdot \mathbf{J} \right]_{ret} = \nabla' \cdot \left[ \mathbf{J} \right]_{ret} - \left[ \frac{ \partial \mathbf{J} }{\partial t'} \right]_{ret} \nabla' \left( t' \right) = -\frac{\partial \rho_{ret}}{\partial t} - \left[ \frac{ \partial \mathbf{J} }{\partial t'} \right]_{ret} \nabla' \left( t' \right)$$

Thanks so much for taking the time to help. I was going to give up on this already!

I still need to think more about this, but one question, why don't I need to do this if the divergence is at the field point r? This is given from the book:

$$\nabla\cdot\vec J_{(\vec r\;',t_r)} = \frac {\partial J_x}{\partial t_r} \frac { \partial t_r}{\partial x} + \frac {\partial J_y}{\partial t_r} \frac { \partial t_r}{\partial y} +\frac {\partial J_z}{\partial t_r} \frac { \partial t_r}{\partial z} \;=\;-\frac 1 c \left ( \frac {\partial J_x}{\partial t_r} \frac { \partial \eta}{\partial x} + \frac {\partial J_y}{\partial t_r} \frac { \partial \eta}{\partial y} +\frac {\partial J_z}{\partial t_r} \frac { \partial \eta}{\partial z}\right ) \;=\; -\frac 1 c \frac {\partial \vec J}{\partial t_r} \cdot (\nabla \eta)$$

I don't understand this equation:

$$\left[ \nabla' \cdot \mathbf{J} \right]_{ret} \nabla' \cdot \mathbf{x'} = \left[ \nabla' \cdot \mathbf{J} \right]_{ret}$$



Also can you help on post #2? I disagree with the book because even you take the divergence respect to the field point ($\nabla \cdot \vec J$), the J is always locates at the source point (x',y',z') so it should be Jx', jy', jz' instead. Am I correct?

Thanks so much.

Alan

 

[yungman]

can anyone help, I don't quite get the post #4.

Thanks

 

[Born2bwire]

Eh, I don't have time to work it out exactly but I fudged the LHS obviously since the divergence of the position vector is not always unity. Maybe it needs to be scaled by the norm of the vector, I don't know. The point is that there is a factor there but it always comes out to be one. Maybe the proper way of showing it was from a tensor product.

The point is though is that it is just an application of the chain rule. The problem is that you need to think about how you are differentiating the functions.

The retarded current is a vector function that goes something like,

$$\left[ \mathbf{J} \right]_{ret} = \left[ \mathbf{J} (\mathbf{r}', t') \right]_{ret} = \mathbf{J} (\mathbf{r}', t-R/c)$$

The retarded time is dependent upon your observation point just as the position vector. So when you take the divergence, are you taking the divergence with respect to changing r' and constant t', changing r' and changing t'(r'), etc. These divergences are different.

So what are we interested in when we work these Maxwell equations?

In Maxwell's equations we relate the fields in response to the retarded sources (assuming the Lorenz Gauge). For example,

$$\mathbf{E}(\mathbf{r},t) = k_{something} \int d\mathbf{r}' \frac{1}{R} \left[ -\nabla' \rho - \frac{1}{c^2} \frac{ \partial \mathbf{J}} {\partial t'} \right]_{ret}$$

This comes from the fact that the effective source for the electric field is

$$\mathbf{s}(\mathbf{r},t) =-\nabla \rho - \frac{1}{c^2} \frac{ \partial \mathbf{J}} {\partial t}$$

So we find the retarded Green's function and use that in an integral with the source and we find that the retarded Green's function forces the source in the resulting integral to be acting with the retarded time (which we of course expect).

So essentially, we are interested in things like

$$\left[ \nabla' \rho \right]_{ret}$$

where we are taking the gradient with respect to changing r' but constant t'. But we do not want to work in that case. We want the gradient to be respect to changing r' and changing t'(r'). That is, we want to express the resulting fields using

$$\nabla' \left[ \rho \right]_{ret}$$

So, we use the chain rule with:

$$\nabla' \left[\rho\right]_{ret} = \left[ \nabla' \rho \right]_{ret} + \left[ \frac{\partial \rho}{\partial t'} \right]_{ret} \nabla' (t-R/c)$$

where taking the derivative outside the brackets means that both x' and t' can vary while inside the brackets x' varies and t' is fixed.

That is, we take the derivative with respect to x' on a function that is dependent upon y(x') and z(x'). The first term we take the derivative of the function with respect to y(x') assuming z(x') is constant. The second term we take the derivative of the function with respect to z(x') assuming y(x') is constant.

So, y(x') = x' and z(x') = t'. Now, the derivative with respect to y(x') is the same as taking the derivative with respect to x' and we just assume that the retarded time is fixed (and since y(x') = x' there is no additional factor from the \partial y(x') / \partial (x') ). The derivative with respect to z(x') is the same as taking the derivative with respect to t' with the gradient of t' as the additional factor.

EDIT: As for your second posts, I do not understand what the subscripts x, x', etc. mean.