# Vector Calculus with Maxwell's Equations

#### CAF123

Gold Member
1. The problem statement, all variables and given/known data
Consider the following representation of Maxwell's eqns: $$\nabla \cdot \underline{E} =0,\,\,\, \nabla \cdot \underline{B} = 0,\,\,\, \nabla \times \underline{E} = -\frac{\partial \underline{B}}{\partial t}, \,\,\,\frac{1}{\mu_o}\nabla \times \underline{B} = \epsilon_o \frac{\partial \underline{E}}{\partial t}.$$

By considering $\nabla \times (\nabla \times \underline{E})$, use the above and an appropriate vector identity to deduce $$\left(\frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2\right) \underline{E} = 0.$$

Therby deduce an expression for $c$. Find a similar eqn for B.

(Note: This question is from a Calculus course)

3. The attempt at a solution

So, considering $\nabla \times (\nabla \times \underline{E})$ I rewrote this as (which I believe is the 'appropriate vector identity' : $\nabla(\nabla \cdot \underline{E}) - (\nabla^2) \underline{E}.$ However, I believe both these terms vanish because of the equations given. The other thing I tried more directly was subbing in what we have for $\nabla \times \underline{E}$, to get $\nabla \times \left(-\frac{\partial \underline{B}}{\partial t}\right).$ It looks like I could then use the eqn above involving del cross B, but I am not sure whether I can just move the ∂/∂t around. (Probably not)

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#### pasmith

Homework Helper
3. The attempt at a solution

So, considering $\nabla \times (\nabla \times \underline{E})$ I rewrote this as (which I believe is the 'appropriate vector identity' : $\nabla(\nabla \cdot \underline{E}) - (\nabla^2) \underline{E}.$ However, I believe both these terms vanish because of the equations given.
Certainly $\nabla (\nabla \cdot \underline{E})$ vanishes, but you can't conclude that $\nabla^2 \underline{E}$ vanishes.

The other thing I tried more directly was subbing in what we have for $\nabla \times \underline{E}$, to get $\nabla \times \left(-\frac{\partial \underline{B}}{\partial t}\right).$ It looks like I could then use the eqn above involving del cross B, but I am not sure whether I can just move the ∂/∂t around. (Probably not)
I think you are entitled to assume that $\underline{B}$ is sufficiently smooth that
$$\frac{\partial}{\partial t} (\nabla \times \underline{B}) = \nabla \times \frac{\partial \underline{B}}{\partial t}.$$

Now you have two expressions for $\nabla \times (\nabla \times \underline{E})$; what happens if you compare them?

#### CAF123

Gold Member
Certainly $\nabla (\nabla \cdot \underline{E})$ vanishes, but you can't conclude that $\nabla^2 \underline{E}$ vanishes.
$$(\nabla \cdot \nabla)\underline{E} = \nabla \cdot (\nabla E) .$$ I see this is not zero now, I must have confused the part in brackets as a dot product before.

I think you are entitled to assume that $\underline{B}$ is sufficiently smooth that
$$\frac{\partial}{\partial t} (\nabla \times \underline{B}) = \nabla \times \frac{\partial \underline{B}}{\partial t}.$$
How can you be sure this is correct? What identity are you using?

#### Dick

Homework Helper
How can you be sure this is correct? What identity are you using?
pasmith is just using that the time derivative commutes with the space derivatives. For example, $\frac{\partial}{\partial t}\frac{\partial}{\partial x}=\frac{\partial}{\partial x}\frac{\partial}{\partial t}$. You can always do that in physics. It can fail for some functions that don't have enough continuous derivatives, but I wouldn't worry about that.

#### CAF123

Gold Member
Hi Dick,
Is it sensible to write:$$\frac{\partial}{\partial t} \left (\nabla \times \underline{B} \right) = \frac{\partial \nabla}{\partial t} \times \underline{B} + \nabla \times \frac{\partial B}{\partial t}.$$
And since $\nabla$ does not depend on time, that term disappears?

Once I do the rerrangment I get that $$\frac{1}{c^2} = \mu_o \epsilon_o \Rightarrow c = \pm \frac{1}{\sqrt{\mu_o \epsilon_o}}$$. I know c represents the speed of light, but in the question it is given as a constant. Should I therefore include the $\pm$?

I'll try the next part later today. But as far as I can tell by doing some of it in my head, it looks wholly symmetric, so the eqn derived for E will have any E replaced with B.

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#### Dick

Homework Helper
Hi Dick,
Is it sensible to write:$$\frac{\partial}{\partial t} \left (\nabla \times \underline{B} \right) = \frac{\partial \nabla}{\partial t} \times \underline{B} + \nabla \times \frac{\partial B}{\partial t}.$$
And since $\nabla$ does not depend on time, that term disappears?

Once I do the rerrangment I get that $$\frac{1}{c^2} = \mu_o \epsilon_o \Rightarrow c = \pm \frac{1}{\sqrt{\mu_o \epsilon_o}}$$. I know c represents the speed of light, but in the question it is given as a constant. Should I therefore include the $\pm$?

I'll try the next part later today. But as far as I can tell by doing some of it in my head, it looks wholly symmetric, so the eqn derived for E will have any E replaced with B.
No, that's not very accurate. You aren't differentiating a product, you are interchanging derivatives. And when you talk about a 'speed' you are pretty definitely talking about something positive.

#### CAF123

Gold Member
No, that's not very accurate. You aren't differentiating a product, you are interchanging derivatives.
If I was to replace the del operator with some other vector, then the statement would be correct. So basically I cannot do what I did above because del is an operator, right?

And when you talk about a 'speed' you are pretty definitely talking about something positive.
I realise this, but given that the question says c is a constant, we are not meant to know that the quantity is indeed the speed of light. The only reason I know it to be the speed of light is because I have seen it in another course. So, I take the $\pm$ since it is only given that c is a constant. (or I can just say in my answer that I have seen this before and know it to be the speed of light and a speed is always +ve)

#### Dick

Homework Helper
If I was to replace the del operator with some other vector, then the statement would be correct. So basically I cannot do what I did above because del is an operator, right?

I realise this, but given that the question says c is a constant, we are not meant to know that the quantity is indeed the speed of light. The only reason I know it to be the speed of light is because I have seen it in another course. So, I take the $\pm$ since it is only given that c is a constant. (or I can just say in my answer that I have seen this before and know it to be the speed of light and a speed is always +ve)
You can do what you did, I just think it's sloppy thinking. You should be a little more careful with operators than with numbers. And if you don't know that c is a speed the negative answer is just as good as positive. Since it only occurs in the equation as c^2, it hardly matters.

#### CAF123

Gold Member
You can do what you did, I just think it's sloppy thinking. You should be a little more careful with operators than with numbers.
I am not sure what you mean by 'sloppy thinking'? I know that you have to be careful using the BAC-CAB identity for del. Could you explain more why it is sloppy? Is it just because I am taking the time derivative of an operator?

Many thanks.

#### Dick

Homework Helper
I am not sure what you mean by 'sloppy thinking'? I know that you have to be careful using the BAC-CAB identity for del. Could you explain more why it is sloppy? Is it just because I am taking the time derivative of an operator?

Many thanks.
Yes, taking the derivative of an operator should require you to say what the definition of an operator derivative is. And you aren't really differentiating a 'product' either. So what 'product rule' are you using? The answer is correct but I think the lack of any sort of formalism is what I would call 'sloppy'. Seems much simpler to just regard it as interchanging the order of partial derivatives. That is a proper theorem, even has a name and limits to its validity. It's called Clairaut's theorem.

#### CAF123

Gold Member
Yes, taking the derivative of an operator should require you to say what the definition of an operator derivative is. And you aren't really differentiating a 'product' either. So what 'product rule' are you using? The answer is correct but I think the lack of any sort of formalism is what I would call 'sloppy'. Seems much simpler to just regard it as interchanging the order of partial derivatives. That is a proper theorem, even has a name and limits to its validity. It's called Clairaut's theorem.
Yes, i've heard of Clairaut's theorem. So is it the case that $\frac{\partial}{\partial t}$ commutes with $\frac{\partial}{\partial x}$ in the del operator, provided B is continuous
on it's domain.

#### Dick

Homework Helper
Yes, i've heard of Clairaut's theorem. So is it the case that $\frac{\partial}{\partial t}$ commutes with $\frac{\partial}{\partial x}$ in the del operator, provided B is continuous
on it's domain.
Actually you need B to have continuous second partial derivatives, but in this kind of problem you just assume everything is as differentiable as it needs to be.

#### CAF123

Gold Member
I have two similar questions related to this material:
1)Let E be a vector field such that $∇ \times E = 0$. Show $∇(a \cdot E) = (a \cdot \nabla)E$ given that a is a constant vector.

I have got this down to $∇(a \cdot E) = (a \cdot \nabla)E + (E \cdot \nabla)a$ so I must argue that the latter term is necessarily 0. I thought I could write $(E \cdot \nabla)a = E \cdot (\nabla a)$, but a is a vector quantity so I am not sure. Any ideas?

I'll ask my other question later to avoid clutter.

#### Dick

Homework Helper
I have two similar questions related to this material:
1)Let E be a vector field such that $∇ \times E = 0$. Show $∇(a \cdot E) = (a \cdot \nabla)E$ given that a is a constant vector.

I have got this down to $∇(a \cdot E) = (a \cdot \nabla)E + (E \cdot \nabla)a$ so I must argue that the latter term is necessarily 0. I thought I could write $(E \cdot \nabla)a = E \cdot (\nabla a)$, but a is a vector quantity so I am not sure. Any ideas?

I'll ask my other question later to avoid clutter.
If you're not sure write out what these things are in components.

#### CAF123

Gold Member
If you're not sure write out what these things are in components.
Ok, so $$= \left[\frac{\partial a_1}{\partial x_1}e_1 + \frac{\partial a_2}{\partial x_2}e_2 + \frac{\partial a_3}{\partial x_3}e_3 \right].$$ I see that this makes sense. Strange.. I remember recalling reading some other thread on PF that the gradient of a vector is defined as something else...

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#### Dick

Homework Helper
Ok, so $$= \left[\frac{\partial a_1}{\partial x_1}e_1 + \frac{\partial a_2}{\partial x_2}e_2 + \frac{\partial a_3}{\partial x_3}e_3 \right].$$ I see that this makes sense. Strange.. I remember recalling reading some other thread on PF that the gradient of a vector is defined as something else...

#### CAF123

Gold Member
Hi Dick,
I have another question relating to this material:
Consider some vector field $\underline{u}(\underline{r}(t),t)$. Find the total derivative $\frac{du}{dt}$. What I have is $$\frac{du}{dt} = \frac{\partial u}{\partial r} \frac{dr}{dt} + \frac{\partial u}{\partial t}\frac{dt}{dt}$$

This is a 'show that' question so what I need to show is that $\frac{\partial u}{\partial r} \frac{dr}{dt} = (u \cdot \nabla) u$

Using your previous hint, I wrote this in components: $$\left[\frac{\partial u}{\partial x_1} \frac{dx_1}{dt}e_1 + \frac{\partial u}{\partial x_2}\frac{dx_2}{dt}e_2 + \frac{\partial u}{\partial x_3}\frac{dx_3}{dt}e_3 \right]$$

From this it seems that I have $(\nabla \cdot u)u$ instead? Thank you.

#### Dick

Homework Helper
Hi Dick,
I have another question relating to this material:
Consider some vector field $\underline{u}(\underline{r}(t),t)$. Find the total derivative $\frac{du}{dt}$. What I have is $$\frac{du}{dt} = \frac{\partial u}{\partial r} \frac{dr}{dt} + \frac{\partial u}{\partial t}\frac{dt}{dt}$$

This is a 'show that' question so what I need to show is that $\frac{\partial u}{\partial r} \frac{dr}{dt} = (u \cdot \nabla) u$

Using your previous hint, I wrote this in components: $$\left[\frac{\partial u}{\partial x_1} \frac{dx_1}{dt}e_1 + \frac{\partial u}{\partial x_2}\frac{dx_2}{dt}e_2 + \frac{\partial u}{\partial x_3}\frac{dx_3}{dt}e_3 \right]$$

From this it seems that I have $(\nabla \cdot u)u$ instead? Thank you.
That is really confusing. What is $\frac{\partial u}{\partial r}$ supposed to mean if r is a vector? Is that actually how they stated the problem? If not, what is the actual statement of the problem?

#### CAF123

Gold Member
That is really confusing. What is $\frac{\partial u}{\partial r}$ supposed to mean if r is a vector? Is that actually how they stated the problem? If not, what is the actual statement of the problem?
Question: The velocity of an element of fluid at position r(t) at time t is described by a vector field $\underline{u}(\underline{r}(t),t)$. Use the chain rule to show that the total derivative of this velocity field with respect to time is$$\frac{du}{dt} = \frac{\partial u}{\partial t} + (u \cdot \nabla)u$$

#### Dick

Homework Helper
Question: The velocity of an element of fluid at position r(t) at time t is described by a vector field $\underline{u}(\underline{r}(t),t)$. Use the chain rule to show that the total derivative of this velocity field with respect to time is$$\frac{du}{dt} = \frac{\partial u}{\partial t} + (u \cdot \nabla)u$$
That doesn't look right at all. How can u appear twice in the result? This looks something like http://en.wikipedia.org/wiki/Material_derivative I think the result should be something more like $$\frac{du}{dt} = \frac{\partial u}{\partial t} + (\frac{dr}{dt} \cdot \nabla)u$$.

#### CAF123

Gold Member
But isn't $$\frac{d\underline{r}}{dt} = \underline{u}?$$
(How I wrote it initially was how it appeared in the question.)

#### Dick

Homework Helper
But isn't $$\frac{d\underline{r}}{dt} = \underline{u}?$$
(How I wrote it initially was how it appeared in the question.)
I guess I don't see why dr/dt would be u. You can't really have u appearing twice in that last term. If you multiply u by 2 then the first two terms get multiplied by u, but the last one gets multiplied by 4. All the terms need to scale the same way.

#### CAF123

Gold Member
If you multiply u by 2 then the first two terms get multiplied by u, but the last one gets multiplied by 4.
Why?

So assuming that we have $dr/dt$, I don't think that resolves my problem in post #17.

#### Dick

Homework Helper
Why?

So assuming that we have $dr/dt$, I don't think that resolves my problem in post #17.
Because it has 2 u's in it. They both double. What you posted in 17 looks a lot like what I wrote later if you drop the e1, e2 and e3. What exactly is the problem again once you correct one of the u's to dr/dt?

#### CAF123

Gold Member
The reason I had the e1,e2,e3 was that $$\frac{dr}{dt} = \frac{dx_1}{dt}e_1 + \frac{dx_2}{dt}e_2 + \frac{dx_3}{dt}e_3$$ I think I must have the basis vectors here to recover the del term: $\nabla = \sum_1^3 \frac{\partial}{\partial x_i} e_i.$

So with this, from post 17, looking at $$\left[ \frac{\partial u}{\partial x_1} \frac{dx_1}{dt} e_1 + \frac{\partial u}{\partial x_2} \frac{dx_2}{dt} e_2 + \frac{\partial u}{\partial x_3} \frac{dx_3}{dt}e_3 \right],$$ it seems to make more sense if the supposed relation is $(\nabla \cdot u) dr/dt$ rather than $(u \cdot \nabla) dr/dt?$

"Vector Calculus with Maxwell's Equations"

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