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

• happyparticle
In summary: So, in summary, the conversation discusses using the Ampere-Maxwell equation to solve a problem, but the person is unsure of how to begin and how to isolate certain parts of the equation. They are advised to take the curl on both sides of the equation and to make sure to account for spherical coordinates. The conversation also mentions the need to look at the denominator in the equation and correctly identify all vector components. Eventually, the person realizes their mistake and the conversation ends with a summary of the equation and a clarification of the vector components.
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?

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.

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.

EpselonZero said:
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.

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

EpselonZero said:
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?"

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}##

EpselonZero said:
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?

##\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##

EpselonZero said:
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?

Ah, no I didn't know.

EpselonZero said:
##\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?

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

EpselonZero said:
\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,
$$\vec A=A_r~\hat r+A_{\theta}~\hat{\theta}+A_{\phi}~\hat {\phi}.$$
EpselonZero said:
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?

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

Last edited by a moderator:
kuruman said:
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 ##

EpselonZero said:
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.

EpselonZero said:
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.)

vela said:
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.

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.

vela said:
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.

kuruman said:
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.

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.

vela said:
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##.

PhDeezNutz and vanhees71

## What is the Ampere-Maxwell law?

The Ampere-Maxwell law is a fundamental law in electromagnetism that relates the magnetic field to the electric current and electric displacement. It states that the circulation of the magnetic field around a closed loop is equal to the sum of the current passing through the loop and the time rate of change of the electric displacement through the loop.

## How does the Ampere-Maxwell law relate to the concept of "showing that B = ∇ × A = 0"?

The Ampere-Maxwell law can be used to show that the curl of the magnetic field, ∇ × B, is equal to the current density, J, plus the time derivative of the electric displacement, ∂D/∂t. By setting the electric displacement to zero, we can simplify the equation to ∇ × B = J, which shows that the curl of the magnetic field is solely dependent on the current density.

## Why is it important to show that B = ∇ × A = 0 using the Ampere-Maxwell law?

Showing that B = ∇ × A = 0 using the Ampere-Maxwell law is important because it helps us understand the relationship between the magnetic field and the electric current. It also demonstrates the fundamental principle of electromagnetism that the magnetic field is generated by the flow of electric current.

## What are the steps to show that B = ∇ × A = 0 using the Ampere-Maxwell law?

The steps to show that B = ∇ × A = 0 using the Ampere-Maxwell law are as follows:

1. Write out the Ampere-Maxwell law equation: ∇ × B = J + ∂D/∂t
2. Set the electric displacement, D, to zero since we are trying to show that B = ∇ × A = 0
3. Use the vector identity ∇ × (∇ × A) = ∇(∇ · A) - ∇2A to simplify the equation to ∇ × B = J
4. Rearrange the equation to solve for B: B = ∇ × A = 0

## What are some real-world applications of the Ampere-Maxwell law and showing that B = ∇ × A = 0?

The Ampere-Maxwell law and the concept of showing that B = ∇ × A = 0 have many real-world applications in the field of electromagnetism. Some examples include the design of electric motors and generators, understanding the behavior of electromagnetic waves, and the development of technologies such as MRI machines and particle accelerators. Additionally, the Ampere-Maxwell law is essential in the study of plasma physics and the behavior of charged particles in magnetic fields.

Replies
5
Views
2K
Replies
32
Views
872
Replies
5
Views
1K
Replies
8
Views
2K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
2
Views
812
Replies
5
Views
2K
Replies
9
Views
1K