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Vector Calculus with Maxwell's Equations

  1. Feb 22, 2013 #1

    CAF123

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    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)

    Many thanks for any advice.
     
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  3. Feb 22, 2013 #2

    pasmith

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    Certainly [itex]\nabla (\nabla \cdot \underline{E})[/itex] vanishes, but you can't conclude that [itex]\nabla^2 \underline{E}[/itex] vanishes.


    I think you are entitled to assume that [itex]\underline{B}[/itex] is sufficiently smooth that
    [tex]\frac{\partial}{\partial t} (\nabla \times \underline{B}) = \nabla \times \frac{\partial \underline{B}}{\partial t}.[/tex]

    Now you have two expressions for [itex]\nabla \times (\nabla \times \underline{E})[/itex]; what happens if you compare them?
     
  4. Feb 22, 2013 #3

    CAF123

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    $$
    (\nabla \cdot \nabla)\underline{E} = \nabla \cdot (\nabla E) .$$ I see this is not zero now, I must have confused the part in brackets as a dot product before.

    How can you be sure this is correct? What identity are you using?
     
  5. Feb 22, 2013 #4

    Dick

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    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.
     
  6. Feb 23, 2013 #5

    CAF123

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    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.
     
    Last edited: Feb 23, 2013
  7. Feb 23, 2013 #6

    Dick

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    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.
     
  8. Feb 23, 2013 #7

    CAF123

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    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)
     
  9. Feb 23, 2013 #8

    Dick

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    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.
     
  10. Feb 23, 2013 #9

    CAF123

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    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.
     
  11. Feb 23, 2013 #10

    Dick

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    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.
     
  12. Feb 23, 2013 #11

    CAF123

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    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.
     
  13. Feb 23, 2013 #12

    Dick

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    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.
     
  14. Feb 25, 2013 #13

    CAF123

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    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.
     
  15. Feb 25, 2013 #14

    Dick

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    If you're not sure write out what these things are in components.
     
  16. Feb 25, 2013 #15

    CAF123

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    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...
     
    Last edited: Feb 25, 2013
  17. Feb 25, 2013 #16

    Dick

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    That would be grad(a) alright.
     
  18. Feb 26, 2013 #17

    CAF123

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    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.
     
  19. Feb 26, 2013 #18

    Dick

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    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?
     
  20. Feb 26, 2013 #19

    CAF123

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    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$$\frac{du}{dt} = \frac{\partial u}{\partial t} + (u \cdot \nabla)u$$
     
  21. Feb 26, 2013 #20

    Dick

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