Vector Calculus with Maxwell's Equations

[CAF123]

Homework Statement

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)

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)

Many thanks for any advice.

 

[pasmith]

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]

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]

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]

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.

 

[Dick]

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]

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 realize 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]

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 realize 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]

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]

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]

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]

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]

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]

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]

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...

 

[Dick]

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...

That would be grad(a) alright.

 

[CAF123]

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]

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]

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]

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]

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

 

[Dick]

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]

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]

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]

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 $(u \cdot \nabla) dr/dt?$

 

[Dick]

The things you are writing down are products of two vectors, since u is already a vector. The result should just be a vector. How does $\frac{dr}{dt} \cdot (\nabla u)$ sound? The $(\nabla u)$ part will be a product of two vectors or as the wikipedia article said, a 'tensor derivative', but the dot product with dr/dt will cut that down to a single vector.

 

[pasmith]

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 $$\frac{du}{dt} = \frac{\partial u}{\partial t} + (\frac{dr}{dt} \cdot \nabla)u$$.

Yes, and then one substitutes \dot r(t) = u(r(t),t) to obtain the result. The material derivative is what you get when you differentiate with respect to time under the assumption that position is a function of time whose derivative is the local fluid velocity.

I guess I don't see why dr/dt would be u.

Because it's stated to be so in the question: "The velocity of an element of fluid at position r(t) at time t is described by a vector field u(r(t),t)".

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.

There is no basis, either in physics or mathematics, for such a requirement here. So long as the expression is dimensionally consistent (as it is) there is no problem in it being non-linear.

 

[Dick]

Perhaps I got sucked into this without completely understanding the context. I meant to answer just the original posted question. Apologies for my confusion. Please take over, I've really been hoping somebody else would step in.

 

[pasmith]

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 $\frac{\partial u}{\partial r} \frac{dr}{dt} = (u \cdot \nabla) u$

There is an abuse of notation in which one can write \frac{\partial}{\partial \mathbf{x}} to mean \nabla, but this is best avoided.

The chain rule for a real-valued function f(t, x_1(t), \dots, x_n(t)) is
<br /> \frac{\mathrm{d}f}{\mathrm{d}t} = \frac{\partial f}{\partial t} + \sum_{k=1}^n<br /> \frac{\partial f}{\partial x_k} \frac{\mathrm{d}x_k}{\mathrm{d}t}<br />
The sum on the right can be interpreted as \frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} \cdot<br /> \nabla f if the x_i are treated as cartesian coordinates.

For a vector-valued function, it's best to look component-by-component in cartesian coordinates, and if you do you will discover that
<br /> \frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} = \frac{\partial \mathbf{u}}{\partial t} + \sum_{k=1}^n<br /> \frac{\partial \mathbf{u}}{\partial x_k} \frac{\mathrm{d}x_k}{\mathrm{d}t}
where the sum can be interpreted as (\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} \cdot \nabla)\mathbf{u} in cartesian coordinates.

 

[CAF123]

I have the comment on my work that the gradient of a vector is 'ill defined'.
So when I wrote $\nabla(\underline{a})$ and gave the expression $$\frac{\partial a_1}{\partial x_1}e_1 + \frac{\partial a_2}{\partial x_2}e_2 + \frac{\partial a_3}{\partial x_3}e_3$$that was wrong