Calculate the Electric and Magnetic field for this Multipole Radiation

[Lambda96]

Homework Statement

Calculate the electric and magnetic field

Relevant Equations

none

Hi

I have problems with the following task

I now wanted to try to calculate the vector potential, which according to my professor's script is defined as follows:

$$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int d^3\mathbf{x}' \frac{\mathbf{j}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}$$

I have now visualized the problem as follows in 2D with regard to the x and z axes

I'm not so sure, but does the electric as well as the magnetic field consist of a superposition when the particle is at the position $z_1=a$ and $z_2=-a$ and would look like $E=E_1+E_2$ and $B=B_1+B_2$?

Then the following $\mathbf{x'}_1=\left(\begin{array}{c} 0 \\\ 0 \\ a \end{array}\right)$ and $\mathbf{x'}_2=\left(\begin{array}{c} 0 \\\ 0 \\ -a \end{array}\right)$ would hold and thus $| \mathbf{x} -\mathbf{x'}_1 |=\sqrt{x^2+y^2+(z-a)^2}$

The electrical current density is defined as follows $\mathbf{j}=\rho \mathbf{v}$ and for this task:

$$\mathbf{j} = Q \cos(\omega t) \left[\delta(z - a) - \delta(z + a)\right] \delta(x) \delta(y) \cdot v \mathbf{e}_z$$

The vector potential can then be calculated as follows:

$$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int_{-\infty}^{\infty} d^3\mathbf{x}' \frac{Q \cos(\omega t) \left[\delta(z - a) - \delta(z + a)\right] \delta(x) \delta(y)}{\sqrt{x^2+y^2+(z-a)^2}}\cdot v \mathbf{e}_z$$

Would now simplify the potential because $\mathbf{x'}_1=\left(\begin{array}{c} 0 \\\ 0 \\ a \end{array}\right)$ as follows

$$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int_{-\infty}^{\infty} d\mathbf{z}' \frac{Q \cos(\omega t) \left[\delta(z - a) - \delta(z + a)\right] \delta(x) \delta(y)}{\sqrt{x^2+y^2+(z-a)^2}}\cdot v \mathbf{e}_z$$

 

[Lambda96]

I've just noticed, shouldn't it be?

$$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int_{-\infty}^{\infty} d\mathbf{z}' \frac{Q \cos(\omega t) \left[\delta(z - a) - \delta(z + a)\right]}{z-a}\cdot v \mathbf{e}_z$$

 

[TSny]

I'm not so sure, but does the electric as well as the magnetic field consist of a superposition when the particle is at the position $z_1=a$ and $z_2=-a$ and would look like $E=E_1+E_2$ and $B=B_1+B_2$?

The charge accumulations $q(t)$ at $z = a$ and $z = -a$ are not due to a single particle moving back and forth. Instead, there is a time-dependent current $I(t)$ between the points $z = -a$ and $z = a$.

The current $I(t)$ for $|z|<a$ does not depend on $z$ since otherwise there would be charge accumulations at points between $z = -a$ and $z = a$.

So, the current is $I = I(t)$ for $|z| < a$ and $I = 0$ for $|z| > a$. Use the Heaviside step function $H$ to express the current for all $z$ as $$I(z, t) = I(t) \left[ H(z-a) - H(z+a) \right].$$ Edit: The right side should read $I(t) \left[ H(z+a) - H(z-a) \right]$. See post #5.

Note that $I(t) = \dot q(t) = \dfrac {d }{dt} Q \cos(\omega t) = -\omega Q \sin \omega t$.

The current density vector is then $$\mathbf{j}(z,t) = I(z, t) \delta(x) \delta(y) \hat z = -\omega Q \sin \omega t\left[ H(z-a) - H(z+a) \right] \delta(x) \delta(y) \hat z.$$Edit: The right side should have $\left[ H(z+a) - H(z-a) \right]$ instead of $\left[ H(z-a) - H(z+a) \right]$.

Verify that this expression for $\mathbf j$ satisfies the charge continuity equation $$\nabla \cdot \mathbf j = - \frac{\partial \rho(t, \mathbf x)}{\partial t}.$$ Recall that the derivative of the step function is a delta function: $\dfrac{dH(z)}{dz} = \delta(z)$.

 

[Lambda96]

Thank you TSny for your help and explanation

I have tried to calculate $\nabla \cdot \mathbf{j}=-\frac{\partial \rho(t, \mathbf{x})}{\partial t}$ and get the same result with a sign error. Have I done something wrong in the calculation?


$$\nabla \cdot \mathbf{j}=-\frac{\partial \rho(t, \mathbf{x})}{\partial t}$$
$$- \omega Q \sin(\omega) \left[ \delta(z-a)-\delta(z+a) \right] \delta(x)\delta(y)=\omega Q \sin(\omega) \left[ \delta(z-a)-\delta(z+a) \right] \delta(x)\delta(y)$$

I have now tried to calculate the potential with $\mathbf{j}(z,t) = -\omega Q \sin \omega t\left[ H(z-a) - H(z+a) \right] \delta(x) \delta(y) \hat z$ with the formula from post 2

$$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int_{-\infty}^{\infty} d\mathbf{z}' \frac{-\omega Q \sin \omega t\left[ H(z-a) - H(z+a) \right]}{z-a}$$

Unfortunately, the integral cannot be solved, which is why I assume that I have done something wrong

 

[TSny]

Thank you TSny for your help and explanation

I have tried to calculate $\nabla \cdot \mathbf{j}=-\frac{\partial \rho(t, \mathbf{x})}{\partial t}$ and get the same result with a sign error. Have I done something wrong in the calculation?

The error is mine.

Instead of $I(z, t) = I(t) \left[ H(z-a) - H(z+a) \right]$, it should be $I(z, t) = I(t) \left[ H(z+a) - H(z-a) \right]$.

Check that $$H(z+a) - H(z-a) = \begin{cases}
1 &\text{if } |z| < a \\
0 &\text{if } |z|>a\\
\end{cases}$$ Thus, $$\mathbf{j}(z,t) =-\omega Q \sin \omega t\left[ H(z+a) - H(z-a) \right] \delta(x) \delta(y) \, \hat {\mathbf z}.$$ Or,
$$\mathbf{j}(t)= \begin{cases}
-\omega Q \sin \omega t \delta(x) \delta(y) \hat {\mathbf z} &\text{if } |z| < a \\
\,\,\,\,0 &\text{if } |z|>a\\
\end{cases}$$

I have now tried to calculate the potential with $\mathbf{j}(z,t) = -\omega Q \sin \omega t\left[ H(z-a) - H(z+a) \right] \delta(x) \delta(y) \hat z$ with the formula from post 2

$$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int_{-\infty}^{\infty} d\mathbf{z}' \frac{-\omega Q \sin \omega t\left[ H(z-a) - H(z+a) \right]}{z-a}$$

Unfortunately, the integral cannot be solved, which is why I assume that I have done something wrong

The formula for $\mathbf{A}$ is $$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int d^3\mathbf{x}' \frac{\mathbf{j}(\mathbf{x}', t -|\mathbf{x} - \mathbf{x}'|/c)}{|\mathbf{x} - \mathbf{x}'|}$$ Show that this reduces to $$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{-a}^a dz' \frac{I(t -|\mathbf{x} - z' \hat{\mathbf z}|/c)}{|\mathbf{x} - z' \hat{\mathbf z}|} \, \hat {\mathbf{z}}$$ You are only interested in the far zone where $|\mathbf{x}| >> a$. So you can make appropriate approximations for $|\mathbf{x} - z' \hat{\mathbf z}|$ in the integrand. Hopefully, you've seen this type of approximation before.

 

[Lambda96]

Thank you for your help and the explanation TSny

I have now started to show the following:
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int d^3\mathbf{x}' \frac{\mathbf{j}(\mathbf{x}', t -|\mathbf{x} - \mathbf{x}'|/c)}{|\mathbf{x} - \mathbf{x}'|} \quad \rightarrow \quad \mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{-a}^a dz' \frac{I(t -|\mathbf{x} - z' \hat{\mathbf z}|/c)}{|\mathbf{x} - z' \hat{\mathbf z}|} \, \hat {\mathbf{z}}$$

Unfortunately, I don't quite understand why $t \quad \rightarrow \quad \frac{t -|\mathbf{x} - z' \hat{\mathbf z}|}{c}$ holds, so I have now left the t as it is

$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int d^3\mathbf{x}' \frac{\mathbf{j}(\mathbf{x}', t )}{|\mathbf{x} - \mathbf{x}'|}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int d^3\mathbf{x}' \frac{-w Q \sin(\omega t) \left[H(z+a)-H(z-a)\right]\delta(x')\delta(y') \mathbf{\hat z}}{|\mathbf{x} - \mathbf{x}'|}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int d^3\mathbf{x}' \frac{-w Q \sin(\omega t) \left[H(z+a)-H(z-a)\right]\delta(x')\delta(y') \mathbf{\hat z}}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} dx' dy' dz'\frac{-w Q \sin(\omega t) \left[H(z+a)-H(z-a)\right]\delta(x')\delta(y') \mathbf{\hat z}}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{- \infty}^{\infty} dz'\frac{-w Q \sin(\omega t) \left[H(z+a)-H(z-a)\right] \mathbf{\hat z}}{\sqrt{x^2+y^2+(z-z')^2}}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{-a}^{a} dz'\frac{-w Q \sin(\omega t) \mathbf{\hat z}}{\sqrt{x^2+y^2+(z-z')^2}}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{-a}^{a} dz'\frac{-w Q \sin(\omega t) \mathbf{\hat z}}{|\mathbf{x} - z' \hat{\mathbf z}|}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{-a}^{a} dz'\frac{I(t) \mathbf{\hat z}}{|\mathbf{x} - z' \hat{\mathbf z}|}$$

Would that be correct?

 

[TSny]

I have now started to show the following:
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int d^3\mathbf{x}' \frac{\mathbf{j}(\mathbf{x}', t -|\mathbf{x} - \mathbf{x}'|/c)}{|\mathbf{x} - \mathbf{x}'|} \quad \rightarrow \quad \mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{-a}^a dz' \frac{I(t -|\mathbf{x} - z' \hat{\mathbf z}|/c)}{|\mathbf{x} - z' \hat{\mathbf z}|} \, \hat {\mathbf{z}}$$

Unfortunately, I don't quite understand why $t \quad \rightarrow \quad \frac{t -|\mathbf{x} - z' \hat{\mathbf z}|}{c}$ holds, so I have now left the t as it is

The time argument for $\mathbf{j}$ in the integrand needs to be the "retarded time" $\dfrac{t -|\mathbf{x} - z' \hat{\mathbf z}|}{c}$. A change in current density at time $t'$ at point $z'$ in the antenna is not "felt" at the distant field point $\mathbf x$ until the later time $t = t' + \dfrac{|\mathbf{x} - z' \hat{\mathbf z}|}{c}$. The time delay $\dfrac{|\mathbf{x} - z' \hat{\mathbf z}|}{c}$ is the "propagation time" for a signal traveling at light speed to get from $z' \hat{\mathbf z}$ to $\mathbf{x}$.

In other words, $\mathbf{A}(\mathbf x, t)$ depends on what the source at $z'$ was doing at the earlier ("retarded") time $t '= t - \dfrac{|\mathbf{x} - z' \hat{\mathbf z}|}{c}$.

Since different points of the antenna have different values of $z'$, the retarded time for a fixed field-point-time $t$ is a function of $z'$. This is the main complication in doing the integration.

$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int d^3\mathbf{x}' \frac{\mathbf{j}(\mathbf{x}', t )}{|\mathbf{x} - \mathbf{x}'|}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int d^3\mathbf{x}' \frac{-w Q \sin(\omega t) \left[H(z+a)-H(z-a)\right]\delta(x')\delta(y') \mathbf{\hat z}}{|\mathbf{x} - \mathbf{x}'|}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int d^3\mathbf{x}' \frac{-w Q \sin(\omega t) \left[H(z+a)-H(z-a)\right]\delta(x')\delta(y') \mathbf{\hat z}}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} dx' dy' dz'\frac{-w Q \sin(\omega t) \left[H(z+a)-H(z-a)\right]\delta(x')\delta(y') \mathbf{\hat z}}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{- \infty}^{\infty} dz'\frac{-w Q \sin(\omega t) \left[H(z+a)-H(z-a)\right] \mathbf{\hat z}}{\sqrt{x^2+y^2+(z-z')^2}}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{-a}^{a} dz'\frac{-w Q \sin(\omega t) \mathbf{\hat z}}{\sqrt{x^2+y^2+(z-z')^2}}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{-a}^{a} dz'\frac{-w Q \sin(\omega t) \mathbf{\hat z}}{|\mathbf{x} - z' \hat{\mathbf z}|}$$
$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{-a}^{a} dz'\frac{I(t) \mathbf{\hat z}}{|\mathbf{x} - z' \hat{\mathbf z}|}$$

Would that be correct?

Yes, that all looks good except for the need to use the retarded time in the integrand.

 

[Lambda96]

Thank you TSny for your help and the explanation regarding the retarded time , I have now inserted this into my expression

I've been thinking about the approximation and unfortunately I can't get any further. The only thing I can think of would be a binomial expansion, as I often had this in my calculus course, so

$$|\mathbf{x}-z' \mathbf{\hat z}| \approx |\mathbf{x}| \Bigl( 1- \frac{z z'}{|\mathbf{x}|^2} \Bigr)$$

Did you mean this approximation?

 

[TSny]

I've been thinking about the approximation and unfortunately I can't get any further. The only thing I can think of would be a binomial expansion, as I often had this in my calculus course, so

$$|\mathbf{x}-z' \mathbf{\hat z}| \approx |\mathbf{x}| \Bigl( 1- \frac{z z'}{|\mathbf{x}|^2} \Bigr)$$

Did you mean this approximation?

Yes, that's it.

If we let $r = |\mathbf x$|, then your result can be expressed as $$|\mathbf{x}-z' \mathbf{\hat z}| \approx r ( 1- \frac{z z'}{r^2} )$$ A nice way to express this is to note that $\dfrac {z}{r} = \cos\theta$, where $\theta$ is the angle between the z-axis and $\mathbf x$. Thus, $$|\mathbf{x}-z' \mathbf{\hat z}| \approx r ( 1- \frac{z' \cos \theta}{r} ) = r - z' \cos \theta$$
Note that $|\mathbf{x}-z' \mathbf{\hat z}|$ occurs in two places in the integrand for $\mathbf A(\mathbf x, t)$: (1) in the denominator and (2) in the retarded time for the argument of $\mathbf j$. In the far zone, we only need to keep terms where $|\mathbf A|$ falls off with distance $r$ as $1/r$ and we can ignore terms for which $|\mathbf A|$ falls off as $1/r^2$ or faster. This helps with making the approximation for the denominator. You'll also need to work on getting the correct approximation for the argument of $\mathbf j$.

 

[Lambda96]

Thanks again for your help and explanations TSny

I have now started with the denominator. I now have the expression $|\mathbf{x}-z' \mathbf{\hat z}| \approx r ( 1- \frac{z' \cos \theta}{r} ) = r - z' \cos \theta$ since we are interested in the far zone, the following applies $r \gg z'$ which means that I can simply ignore the term $z' \cos \theta$ and the integral looks like this:

$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{-a}^{a} dz'\frac{- \omega Q \sin(\omega(t-\frac{|\mathbf{x}-z' \mathbf{\hat z}|}{c}))}{r}$$

Unfortunately I still have some problems with the numerator, I have now used the approximation in the sine, which then looks like this:

$$\omega \left( t - \frac{\lvert \mathbf{x} - z' \hat{\mathbf{z}} \rvert}{c} \right) \approx \omega \left( t - \frac{r }{c} + \frac{z' \cos\theta}{c} \right)=\omega \left( t - \frac{r}{c} \right) + \frac{\omega z' \cos\theta}{c}$$

If the $\omega$ didn't appear in $\frac{\omega z' \cos\theta}{c}$, I would have simply argued that this term can simply be omitted, as it is extremely small, because $\omega$ could be a large number, which is why it no longer works. Then I wanted to use $\sin(A+B)=\cos(B)\sin(A)+\cos(A)\sin(B)$, but that doesn't get rid of the term $\sin(\cos(\theta))$, it actually makes it worse

 

[TSny]

I have now started with the denominator. I now have the expression $|\mathbf{x}-z' \mathbf{\hat z}| \approx r ( 1- \frac{z' \cos \theta}{r} ) = r - z' \cos \theta$ since we are interested in the far zone, the following applies $r \gg z'$ which means that I can simply ignore the term $z' \cos \theta$ and the integral looks like this:

$$\mathbf{A}(\mathbf{x}, t) = \frac{1}{c} \int_{-a}^{a} dz'\frac{- \omega Q \sin(\omega(t-\frac{|\mathbf{x}-z' \mathbf{\hat z}|}{c}))}{r}$$

Looks good.

Unfortunately I still have some problems with the numerator, I have now used the approximation in the sine, which then looks like this:

$$\omega \left( t - \frac{\lvert \mathbf{x} - z' \hat{\mathbf{z}} \rvert}{c} \right) \approx \omega \left( t - \frac{r }{c} + \frac{z' \cos\theta}{c} \right)=\omega \left( t - \frac{r}{c} \right) + \frac{\omega z' \cos\theta}{c}$$

Good.

If the $\omega$ didn't appear in $\frac{\omega z' \cos\theta}{c}$, I would have simply argued that this term can simply be omitted, as it is extremely small, because $\omega$ could be a large number, which is why it no longer works.

Right. You need to keep $\frac{\omega z' \cos\theta}{c}$. Very good.

Then I wanted to use $\sin(A+B)=\cos(B)\sin(A)+\cos(A)\sin(B)$,

Good.

but that doesn't get rid of the term $\sin(\cos(\theta))$, it actually makes it worse

When integrating over $z'$, $\theta$ is fixed.
Since you are integrating from $-a$ to $a$, consider if $\sin(\frac{\omega z' \cos\theta}{c})$ and $\cos(\frac{\omega z' \cos\theta}{c})$ are even or odd functions of $z'$.

 

[Lambda96]

Thanks again for your help TSny and the tip that $\cos$ and $\sin$ are even and odd functions.

If I have not miscalculated, I have got the following result:

$$\mathbf{A}(r,t)=\frac{-2Q \omega \sin\Bigl(\frac{w z' \cos(\theta)}{c} \Bigr)\sin\Bigl(\omega \Bigl( \frac{t-r}{c}\Bigr)\Bigr)}{r \cos(\theta)}$$

 

[TSny]

If I have not miscalculated, I have got the following result:

$$\mathbf{A}(r,t)=\frac{-2Q \omega \sin\Bigl(\frac{w z' \cos(\theta)}{c} \Bigr)\sin\Bigl(\omega \Bigl( \frac{t-r}{c}\Bigr)\Bigr)}{r \cos(\theta)}$$

Getting close. But, I see some typos and other errors:

(1) $\mathbf{A}(r,t)$ should be $\mathbf{A}(r, \theta, t)$.

(2) You need a unit vector to specify the direction of $\mathbf {A}$.

(3) Your expression does not have the correct dimensions for $\mathbf{A}$.

(4) The argument for the second sine function has a typo.

(5) Your result should not depend on $z'$. You integrated over $z'$ (hopefully).

 

[Lambda96]

Thanks again for your help TSny

I have now corrected all the points you listed, except for point (3) where I am not sure if the expression now has the correct dimension:

$$\mathbf{A}(r,t, \theta)=\frac{-2Q \omega \sin\Bigl(\frac{w a \cos(\theta)}{c} \Bigr)\sin\Bigl(\omega \Bigl( \frac{ct-r}{c}\Bigr)\Bigr)}{r \cos(\theta)} \mathbf{\hat z}$$

 

[TSny]

I have now corrected all the points you listed, except for point (3) where I am not sure if the expression now has the correct dimension:

$$\mathbf{A}(r,t, \theta)=\frac{-2Q \omega \sin\Bigl(\frac{w a \cos(\theta)}{c} \Bigr)\sin\Bigl(\omega \Bigl( \frac{ct-r}{c}\Bigr)\Bigr)}{r \cos(\theta)} \mathbf{\hat z}$$

Very close.

Regarding the dimensions, pick a formula that makes it easy to see what the dimensions should be. For example $$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int d^3\mathbf{x}' \frac{\mathbf{j}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}.$$

 

[Lambda96]

Tanke for your help TSny

The individual dimensions (SI) of the terms are as follows:

$$c=\frac{m}{s}, \quad |\mathbf{x} - \mathbf{x}'|=m, \quad \mathbf{j}=\frac{A}{m^2}$$

If I now insert this into the equation from your post 15, I get $\frac{s}{m} \frac{\frac{A}{m^2}}{m}=\frac{A s}{m^4}$, but if I'm not mistaken I should get $\frac{V s}{m}$.

 

[TSny]

We are working in Gaussian units where $$\mathbf{A}(\mathbf{x}) = \frac{1}{c} \int d^3\mathbf{x}' \frac{\mathbf{j}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}.$$ Consider dimensions rather than units. Using square brackets for indicating the dimension of a quantity, $$[\mathbf A] = \frac{1}{[c]} [d^3x] [\mathbf j] \frac 1 {[x]} = \frac{1}{[L]/[T]} [L]^3 [\mathbf j] \frac 1 {[L]}$$ where $[L]$ is dimension of length and $[T]$ is dimension of time. From the definition of current density $\mathbf j$, work out $[\mathbf j]$ in terms of $[Q], [L],$ and $[T]$. Here, $[Q]$ is the dimension of charge. Thus, reduce $[\mathbf A$] to some combination of $[Q], [L]$, and $[T]$.

Compare with the dimensions of your result for $\mathbf A$ in post $14$.

 

[renormalize]

The individual dimensions (SI) of the terms are as follows:
$$c=\frac{m}{s}, \quad |\mathbf{x} - \mathbf{x}'|=m, \quad \mathbf{j}=\frac{A}{m^2}$$

Don't forget to include the $d^3\mathbf{x}'=m^3$ term!

 

[Lambda96]

Thanks again for all your help and explanations TSny also thanks renormalize for your help

Now I see where I made the mistake

 

[WWGD]

Thanks again for all your help and explanations TSny also thanks renormalize for your help

Now I see where I made the mistake

Only in PF you get (re)normalized help!