# Homework Help: Derivation of amperes law

1. May 28, 2012

### aaaa202

Hi, I have some trouble understanding my book's derivation of Ampere's law. I have attached the derivation, and would like to ask the following. In equation 5.47 they throw away the first term because, ∇xJ=0. I don't understand why that is. Their argument is that J does not depend on the coordinates (x,y,z), but I don't see why that is. What if you wanted the magnetic field inside the mass distribution?

2. May 28, 2012

### robertjford80

you forgot to attach the derivation

3. May 28, 2012

### aaaa202

yes oops :) Here it comes

#### Attached Files:

• ###### derivationampere.png
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4. May 28, 2012

### Ray Vickson

J depends on (x',y',z') because the primed coordinates refer to points of the charge/current distribution, not to the "observation" point (x,y,z). Imagine chopping up the current region into a large number (N) of little blocks $$B_i = (x_i', x_i' + \Delta x) \times (y_i', y_i' + \Delta y) \times (z_i', z_i' + \Delta z),\; i = 1, 2, \ldots, N.$$ The total field at the point (x,y,z) is the field at (x,y,z) from block B1, plus the field from block B2 , plus ... plus the field from block BN. When you make the blocks smaller and smaller an increasingly numerous, your field formula converges to an expression given by an integral over (x',y',z'). The integrand also contains the values of x, y and z in it, but the integration is over (x',y',z'). When you differentiate with respect to x, that derivative just comes under the integration sign because it is really no different from saying familiar things like
$$\frac{d}{dx} \int_0^1 x e^y \, dy = \int_0^1 \frac{d}{dx} (x e^y) \, dy = \int_0^1 e^y \, dy,$$ for example.

RGV

5. May 29, 2012

### aaaa202

I still don't see it to be honest. J(x',y',z') denotes the volume current density at the point (x',y',z'), where the origin is arbitrarily chosen - right?
(x,y,z) denotes the position relative to the point r, at which we are calculating the field.

Well then how can it not matter J, when we change (x,y,z). That is just another coordinate frame, but it still contains the volume current just shifted in space!!

In electrostatics I certainly never saw this happens - what's the difference?

6. May 29, 2012

### Ray Vickson

Of course it happens in electrostatics---there IS no difference! Have you never before solved a problem by changing from rectangular to polar coodinates, for example? If you did, you would have expressed both the "observation" point (x,y,z) in polars and the "source" point (x',y',z') in polars---in the _same_ coordinate system. If you rotate the coordinate system you need to rotate both (x,y,z) and (x',y',z') in the same way. These are just two different points in the same coordinate frame. If you shift the origin to a new location, so that (x,y,z) becomes (z-a,y-b,z-c), then you need to shift the source by the same amount, so that (x',y',z') becomes (x'-a, y'-b, z'-c).

RGV

7. May 29, 2012

### aaaa202

Well exactly. So there is a bond between the coordinates, right? Well how can you then say that the curl doesn't depend on one of the coordinates, when the others are bonded to them by a mathematical relation? - I hope my question is clear now - maybe I'm just confusing myself completely...

8. May 29, 2012

### Ray Vickson

Let me give you a very simple example (not of the curl, but in electrostatics instead---the principles are exactly the same). Suppose we have a charge of magnitude 1 which is distributed uniformly along the interval [0,1] along the x-axis. Suppose we want the electrical potential V(x,y,z) due to this charge. We could express it as an integral along the line segment (which would involve an integration dummy variable x', for example), but just to make sure there is no confusion, say we _approximate_ the charge distribution by 10 equally-spaced charges of 1/10, at the mid-points 0.05, .15, .25, ..., .95 of the 10 equal subintervals; basically, we approximate the integral by a finite sum of 10 terms. The potential of the 10 charges is given by $$V(x,y,z) = \frac{1}{4 \pi \epsilon_0} U(x,y,z),$$ where
$$U(x,y,z) = \frac{0.1}{\sqrt{(x-0.05)^2 + y^2 + z^2}} + \frac{0.1}{\sqrt{(x-.15)^2 + y^2 + z^2}} + \cdots + \frac{.1}{\sqrt{(x-.95)^2 + y^2 + z^2}}.$$ In this approximation to the potential, the x-component of the electrical field
$E_x = - \partial/\partial x V(x,y,z)$ has the form
$$E_x = \frac{1}{4 \pi \epsilon_0} \left[ \frac{.1 (x - .05)}{[(x-.05)^2 + y^2 + z^2]^{3/2}} + \cdots \right].$$ We do not differentiate the numbers 0.1, 0.05, 0.15, etc.; we just differentiate x. That is what is happening in the treatment you cite.

Note that we could have written
$$U(x,y,z) = \sum_{x'} \frac{q_{x'}}{\sqrt{(x - x')^2 + y^2 + z^2}},$$ where the summation is over $x' \in \{ 0.05, 0.15, 0.25, \ldots, 0.95 \},$ but x' just plays the role of a dummy variable (and could be given any other name instead). When we take the derivative, x' is not differentiated.

Note: Of course, if I change the coordinate system I need to change the x' as well. For example, if I shift the origin one unit up along the y-axis, the charges would now lie along the line {(x', -1), 0 ≤ x' ≤ 1} and I would need to change the terms in U(x,y,z) accordingly.

RGV

Last edited: May 29, 2012
9. May 30, 2012

### aaaa202

Okay, I think I understand what you mean now - I have been a bit slow.
But my confusion also arose from the fact that they sometimes switch coordinates for the divergence at the cost of a minus sign. Like on the attached picture - is that really consistent with what you say?

#### Attached Files:

• ###### divergence.png
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10. May 30, 2012

### Ray Vickson

$$\frac{d}{dx} \frac{1}{\sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}} = - \frac{d}{dx'} \frac{1}{\sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}} .$$

RGV

11. May 30, 2012

### aaaa202

Well I'm back to scratch. In that example above differentiating with respect to the primed coordinates clearly isn't zero, even though the potential U(x,y,z) is a function of the regular coordinates.
If you agree, what is then the difference between this and the example in my book, where the curl (which is just a bunch of derivatives) is equated to zero because, "it doesn't depend on the unprimed (x,y,z)".

12. May 30, 2012

### Ray Vickson

I cannot figure out what your problem is. You have a function of the form
$$F = g(x',y',z')f(x-x',y-y',z-z'),$$
and you want to compute (for example) $\partial F/\partial x.$ This is given by
$$\frac{\partial}{\partial x} F = g(x',y',z') \frac{\partial}{\partial x} f(x-x',y-y',z-z'),$$
just by using the rules of elementary calculus.

RGV

13. May 31, 2012

### aaaa202

I cannot see why you can't see my problem.

You have a function of the form:

V(x,y,z), which also contains the variables (x',y',z'). In the derivation in my book ∂V/∂x' = 0, because V doesn't depend on x' according to that derivation. But indeed they do in your example... Can you really NOT see a conflict in what is being said?

14. May 31, 2012

### vela

Staff Emeritus
I think I see what's confusing you. You have something like
$$U(x,y,z) = \int \frac{\rho\,d\tau'}{\sqrt{(x - x')^2 + (y-y')^2 + (z-z')^2}},$$ (We'll pretend the charge density ρ is constant to avoid having to differentiate it though that assumption leads to other problems.) On the one hand, you think
$$\frac{\partial U(x,y,z)}{\partial x'} = 0.$$ On the other hand, you have
\begin{align*}
\frac{\partial}{\partial x'}\int \frac{\rho}{\sqrt{(x - x')^2 + (y-y')^2 + (z-z')^2}}\,d\tau'
&= \int \frac{\partial}{\partial x'}\frac{\rho}{\sqrt{(x - x')^2 + (y-y')^2 + (z-z')^2}}\,d\tau' \\
&= \int \big(-\frac{\partial}{\partial x}\frac{\rho}{\sqrt{(x - x')^2 + (y-y')^2 + (z-z')^2}}\big)\,d\tau' \\
&= -\frac{\partial}{\partial x}\int \frac{\rho}{\sqrt{(x - x')^2 + (y-y')^2 + (z-z')^2}}\,d\tau' \\
&= -\frac{\partial}{\partial x} U(x,y,z)
\end{align*} which generally doesn't equal 0. So what's going on?

The resolution to the paradox lies in the fact that the primed variables are dummy variables, so they only have meaning inside the integral. In particular, it doesn't make sense to ask what $\partial U/\partial x'$ is because x' has no meaning outside of the integral. Another way to put it is an x' outside of the integral is not the same variable as x' inside of the integral.

15. May 31, 2012

### aaaa202

So is it all in all because even though (x,y,z) refers to a point where we could have J≠0 just as well as J=0, then the way we describe the volume current density it is already described by another coordinate system (x',y',z'), which completely has its own life and completely describes the volume current density. But wouldn't it be possible to just let (x,y,z)=(x',y',z') and then get easier results? Obviously not, since noone does it, but can you give a counterexample?

16. May 31, 2012

### Ray Vickson

Let me try one more time. Your first question, initially, was why an x-derivative did not apply to J; the answer was that J does not depend on x, so an x-derivative of J is zero. Then you seem to have introduced a new question arising from the fact that (for some reason) your cited source now wants to take an x'-derivative. That is a _separate_ issue.

Let's go through my original electrostatic example again. I gave a 10-point approximation to the actual potential; there, there is NO x' anywhwere to be differentiated. We could have gotten a better approximation by using 100 points, even better by using a million points, etc., but in every case there would still be no issue about differentiation. To get an exact answer we need to take the limit of infinitely many infinitesimal charges --> we need an integral, and here it is:
$$U(x,y,z) = \int_0^1 \rho(w) f(x-w,y,z)\, dw,$$
where
$$f(X,Y,Z) = (X^2 + Y^2 + Z^2)^{-1/2}.$$
In the original example the charge density was constant (ρ(w) = 1 for 0 < w < 1) but we can work more generally. As before, we have
$$U_x(x,y,z) \equiv \frac{\partial}{\partial x} U(x,y,z) = \int_0^1 \rho(w) \frac{\partial}{\partial x} f(x-w,y,z) \, dw.$$ If, for some reason, we would like to convert this to another form, we might try to manipulate the w-integration, using integration by parts, for example. So, let's do that: WARNING: new issue!

We have $\partial f(x-w,y,z)/ \partial x = -\partial f(x-w,y,z)/\partial w,$ so
$$U_x(x,y,z) = -\int_0^1 \rho(w) \partial f(x-w,y,z)/\partial w \, dw = -\int_0^1 \partial [\rho(w) f(x-w,y,z)]/\partial w \, dw + \int_0^1 f(x-w,y,z) \partial \rho(w)/\partial w \, dw,$$ which is just integration by parts. The first term just looks at the integrand $\rho(w) f(x-w,y,z)$ at the boundary points w=0 and w=1, while the second term may (we hope) be a simpler thing to work with than the original integral. In the 3-dimensional case (where we have $\rho (w_1,w_2,w_3)$ the boundary terms in the integration by parts will become a surface integral with respect to the other two variables w2 and w3, and so we might be able to link it to some physical law or characteristics involving surface integrals.

So, the outline is:
(1) the original problem did not involve derivatives with respect to x', y', z', because these were DUMMY variables, and the quantity U or E or B or whatever, does not contain any x', y', or z' in the final formula. This is no different from saying that I have $$\frac{d}{dx} \int_0^1 x^2 \, dx = 0,$$ because the quantity $\int_0^1 x^2 \, dx = 1/3$ just does not contain x---that is, 1/3 is not a function of x. The fact that before getting the 1/3 we had a formula containing the symbol x is irrelevant: we could equally have asked for $$\frac{d}{dx} \int_0^1 w^2 \, dw$$ because what name we choose for the dummy integration variable is irrelevant.
(2) For some type of integration expressions we can choose to manipulate the formula by moving the derivative wrt x into a derivative wrt the dummy variable, then doing integration by parts, for example. This issue is separate from (1).

There: that's my last words on this topic.

RGV

Last edited: May 31, 2012
17. May 31, 2012

### vela

Staff Emeritus
As Ray said earlier, the primed and unprimed variables don't correspond to separate coordinate systems. They're the coordinates of two different points within the same coordinate system — same set of axes, two different points. The point $\vec{r} = (x, y, z)$ is the location where you're calculating the field or potential. The point $\vec{r}' = (x', y', z')$ is the location of charge or current which contributes to the field or potential at $\vec{r}$. To get the field or potential at $\vec{r}$, you have to sum the contributions of every bit of charge or current, which is why you integrate over $\vec{r}'$.

You might want to go back and review how the variables are defined. Griffiths, if i I recall correctly, explicitly went over his convention to avoid this sort of confusion.