Show that ##B = \nabla \times A = 0## using Ampere-Maxwell law

[happyparticle]

Homework Statement

I have a charge Q on a sphere at the origin. At t=0 the sphere explode and create a current density $\vec{J}(\vec{r}) = J(r)\hat{r}$. I have to prove that $B = \nabla \times A = 0$ using ampere Maxwell law

Relevant Equations

$\nabla \times B = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{d \vec{E}}{dt}$
$\vec{A}(\vec{r},t) = \frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}', t_r)}{|r - r'|} d \tau '$
$t_r = t - \frac{|r - r'|}{c}$

Since I have to use Ampere-Maxwell equation I don't see how to begin the problem. I don't know how to right hand side can be 0 and I don't see how to isolate $\nabla \times A$ on the left hand side. Is it purely mathematic?

 

[kuruman]

You have the equation for the vector potential as an integral of the current density. Just take the curl on both sides of the equation. Needless to say that all derivatives in the curl are with respect to unprimed coordinates which means that you can move them past the integral sign. Also don't forget to look up the curl in spherical coordinates.

 

[happyparticle]

I have $\vec{nabla} \times \vec{A} = \vec{\nabla} \times (\frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}', t_r)}{|r - r'|} d \tau ')$ Since There's no unprimed vector is it why the curl is 0? I feel is too simple.

 

[kuruman]

I have $\vec{nabla} \times \vec{A} = \vec{\nabla} \times (\frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}', t_r)}{|r - r'|} d \tau ')$ Since There's no unprimed vector is it why the curl is 0? I feel is too simple.

Look again in the denominator. Actually it is simple but not for the reason you think.

 

[happyparticle]

There's no vector in the dominator or I made a mistake and r' should be a vector.

 

[kuruman]

There's no vector in the dominator or I made a mistake and r' should be a vector.

The correct expression is $$\vec{\nabla} \times \vec{A} = \vec{\nabla} \times \left (\frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}', t_r)}{|\vec r - \vec {r}'|} d \tau '\right).$$ Just because you didn't put vector signs over $r$ and $r'$ in the denominator doesn't mean that they shouldn't be there. The point is that the unprimed derivatives do not vanish identically when you take the curl.

Consider the vector in the integrand $\dfrac{\vec{J}(\vec{r}', t_r)}{|\vec r - \vec {r}'|}$. That is what you want to take the curl of with respect to unprimed coordinates. Don't forget that you are given that $\vec{J}(\vec{r}', t)={J}(r', t)~\hat r$.

Ask yourself the question, "What is the vector that depends on unprimed coordinates the curl of which I must take?"

 

[happyparticle]

I really don't see anything else, but since there's not $\phi$ or $\theta$ the curl in spherical coordinates should be 0. I don't see what you mean when you say to look the denominator.

If I take the curl of $\dfrac{\vec{J}(\vec{r}', t_r)}{|\vec r - \vec {r}'|}$ I have $0 \hat{r} + 0 \hat{\theta} + 0 \hat{\phi}$

 

[kuruman]

I really don't see anything else, but since there's not $\phi$ or $\theta$ the curl in spherical coordinates should be 0. I don't see what you mean when you say to look the denominator.

If I take the curl of $\dfrac{\vec{J}(\vec{r}', t_r)}{|\vec r - \vec {r}'|}$ I have $0 \hat{r} + 0 \hat{\theta} + 0 \hat{\phi}$

Can you show the details?

 

[happyparticle]

$\nabla \times \frac{\vec{J}(r')\hat{r}}{|\vec{r} - \vec{r}'|} = \frac{1}{r sin \theta}(\frac{\partial (A_{\phi} sin \theta )}{\partial \theta} - \frac{\partial A_{\theta}}{\partial \phi}) + \frac{1}{r} (\frac{1}{sin \theta} \frac{\partial A_r}{\partial \phi - \frac{\partial (rA_{\phi})}{\partial r}}) + \frac{1}{r} \frac{\partial (rA_{\theta})}{\partial r} - \frac{\partial A_r}{ \partial \theta} = 0$

 

[vela]

I really don't see anything else, but since there's not $\phi$ or $\theta$ the curl in spherical coordinates should be 0. I don't see what you mean when you say to look the denominator.

If I take the curl of $\dfrac{\vec{J}(\vec{r}', t_r)}{|\vec r - \vec {r}'|}$ I have $0 \hat{r} + 0 \hat{\theta} + 0 \hat{\phi}$

You do realize that $\vec r = (r \sin\theta \cos\phi, r \sin\theta\sin \phi, r \cos\theta)$, right?

 

[happyparticle]

Ah, no I didn't know.

 

[kuruman]

$\nabla \times \frac{\vec{J}(r')\hat{r}}{|\vec{r} - \vec{r}'|} = \frac{1}{r sin \theta}(\frac{\partial (A_{\phi} sin \theta )}{\partial \theta} - \frac{\partial A_{\theta}}{\partial \phi}) + \frac{1}{r} (\frac{1}{sin \theta} \frac{\partial A_r}{\partial \phi - \frac{\partial (rA_{\phi})}{\partial r}}) + \frac{1}{r} \frac{\partial (rA_{\theta})}{\partial r} - \frac{\partial A_r}{ \partial \theta} = 0$

This doesn't say much to me. What are $A_r~$, $A_{\theta}$ and $A_{\phi}$ in this specific case?

 

[happyparticle]

This doesn't say much to me. What are $A_r~$, $A_{\theta}$ and $A_{\phi}$ in this specific case?

My bad $A = \dfrac{\vec{J}(\vec{r}', t_r)}{|\vec r - \vec {r}'|}$

However, I didn't know about $\vec r = (r \sin\theta \cos\phi, r \sin\theta\sin \phi, r \cos\theta)$

 

[kuruman]

My bad $A = \dfrac{\vec{J}(\vec{r}', t_r)}{|\vec r - \vec {r}'|}$

I asked you for three components $A_r$, $A_{\theta}$ and $A_{\phi}$ which is what you used in your expression for the curl. The subscripts indicate that these are in spherical, not Cartesian, coordinates. That is,
$$\vec A=A_r~\hat r+A_{\theta}~\hat{\theta}+A_{\phi}~\hat {\phi}.$$

However, I didn't know about $\vec r = (r \sin\theta \cos\phi, r \sin\theta\sin \phi, r \cos\theta)$

That's the radial vector $r$ in the Cartesian representation. In terms of unit vectors,
$$\vec r=r\sin\!\theta \cos\!\phi~\hat x+r\sin\!\theta \sin\!\phi~\hat y+r\cos\!\theta~\hat z.$$In the spherical representation in terms of unit vectors, one would write$$\vec r=r~\hat r+0*\hat{\theta}+0*\hat {\phi}.$$Do you see where this is going?

 

[happyparticle]

I'm a bit confuse. I have $\nabla \times (\frac{\hat{r}}{|\vec{r'} - \vec{r}|})$

 

[happyparticle]

I asked you for three components Ar, Aθ and Aϕ

Since I don't have any $\phi, \theta$, $A_{\theta} = A_{\phi} = 0$ and $A_r = \frac{\hat{r}}{|\vec{r'} - \vec{r}|}$
And $\frac{1}{r sin \theta}(\frac{\partial (A_{\phi} sin \theta )}{\partial \theta} - \frac{\partial A_{\theta}}{\partial \phi})\hat{r} =0$

 

[kuruman]

Since I don't have any $\phi, \theta$, $A_{\theta} = A_{\phi} = 0$ and $A_r = \frac{\hat{r}}{|\vec{r'} - \vec{r}|}$
And $\frac{1}{r sin \theta}(\frac{\partial (A_{\phi} sin \theta )}{\partial \theta} - \frac{\partial A_{\theta}}{\partial \phi})\hat{r} =0$

That is correct for the radial component of the curl. What about the $\hat {\theta}$ and $\hat {\phi}$ components of the curl? Why are they zero? Complete the proof.

 

[vela]

I'm a bit confuse. I have $\nabla \times (\frac{\hat{r}}{|\vec{r'} - \vec{r}|})$

Shouldn't that be $\nabla \times (\frac{\hat{r}'}{|\vec{r'} - \vec{r}|})$? (Note the prime in the numerator.)

 

[kuruman]

Shouldn't that be $\nabla \times (\frac{\hat{r}'}{|\vec{r'} - \vec{r}|})$? (Note the prime in the numerator.)

I agree with the prime in the numerator but I disagree with the prime in the denominator which should be $|\vec r-\vec {r'}|$ because it is the magnitude of the vector from the source to the point of interest.

 

[vela]

I noticed the OP switched the order for some reason, but it shouldn't make a difference mathematically as $\lvert \vec r - \vec r' \rvert = \lvert \vec r' - \vec r \rvert$.

The change from $\vec r$ to $\vec r'$ in the numerator, however, does suggest you can't assume $A_\phi = A_\theta = 0$ anymore.

 

[kuruman]

I noticed the OP switched the order for some reason, but it shouldn't make a difference mathematically as $\lvert \vec r - \vec r' \rvert = \lvert \vec r' - \vec r \rvert$.

Yes, but $\vec {\nabla}\lvert \vec r - \vec r' \rvert = -\vec {\nabla}\lvert \vec r' - \vec r \rvert$. It would be confusing not to stick to convention.

 

[vela]

Yes, but $\vec {\nabla}\lvert \vec r - \vec r' \rvert = -\vec {\nabla}\lvert \vec r' - \vec r \rvert$. It would be confusing not to stick to convention.

That can't be true. You can't get two different answers by applying the same operation to equal quantities. If you work out the two gradients, they come out to be equal.

 

[happyparticle]

Even if there's no $\phi$ and $\theta$ I need to show that they are 0?
I couldn't say $\vec r=r~\hat r+0*\hat{\theta}+0*\hat {\phi}$?
Then show that $r\hat{r} = 0$ as I did.

 

[kuruman]

That can't be true. You can't get two different answers by applying the same operation to equal quantities. If you work out the two gradients, they come out to be equal.

Of course it isn't true. I was thinking of $\nabla '=-\nabla$.