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Merzbacher's problem of a particle in a magnetic field

  1. Feb 18, 2010 #1
    1. The problem statement, all variables and given/known data

    This is Problem 4.3 from Quantum Mechanics, Merzbacher 3rd ed. pg. 78.

    2. Relevant equations

    [tex]\mathbf{B}=\mathbf{\nabla}\times\mathbf{A}[/tex]

    3. The attempt at a solution

    I am massively confused by this problem. The first step would be to obviously start taking the cross products, but I'm not even sure how to do that. For example, in the equation given in the problem, one of the cross products is:

    [tex]\mathbf{\hat{k}}\times\mathbf{r}[/tex]

    I do vector products by constructing a determinant and simplifying. But in this case we have [itex]\mathbf{\hat{k}}[/itex], so we only have the [itex]\hat{k}[/itex] component (i.e. the i and j components are zero). Then [itex]\mathbf{r}[/itex] is obviously shorthand for x, y and z, so it looks like we have a 2x3 matrix, so I can't take the determinant!

    I really don't even know where to get started on this, any help, advice, suggestions, anything at all would be deeply appreciated.

    Thanks yall

    IHateMayonnaise
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 18, 2010 #2

    gabbagabbahey

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    Is it? You are ultimately looking to find the curl of [itex]\textbf{A}[/itex]; do you really need to simplify the cross products first to do that?:wink:

    I'm not sure what you mean by this. The position vector is always given in Cartesian Coordinates by [tex]\textbf{r}=x\mathbf{\hat{i}}+y\mathbf{\hat{j}}+z\mathbf{\hat{k}}[/itex].

    Which means,

    [tex]\mathbf{\hat{k}}\times\textbf{r}=\begin{vmatrix}\mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ 0 & 0 & 1 \\ x & y & z \end{vmatrix}=-y\mathbf{\hat{i}}+x\mathbf{\hat{j}}[/tex]

    However, there is a much less intensive method for finding [itex]\textbf{B}[/itex]. When I look at your expression for [itex]\textbf{A}[/itex], I see the product of a vector function, [tex]\mathbf{\hat{k}}\times\textbf{r}[/itex] and a scalar function, [tex]\frac{\Phi}{2\pi}(\mathbf{\hat{k}}\times\textbf{r})^{-2}[/tex]. Isn't there a product rule you've come across for taking the curl of a product between a vector function and a scalar function?...In fact, there are several product rules that will help you greatly here.
     
  4. Feb 18, 2010 #3
    That was an extremely helpful post, thank you so much!

    As for that product rule.. my trusty reference is Griffiths. You say that I am looking for a product rule that involves taking the curl of a product of a vector and a scalar. That would be product rule #7 (From G):

    [tex]\mathbf{\nabla} \times (f \mathbf{A}) = f(\mathbf{\nabla} \times \mathbf{A}) - \mathbf{A} \times (\mathbf{\nabla}f)[/tex]

    But I'm not taking the curl of anything, to my understanding the curl of A, for example, is

    [tex]\mathbf{\nabla} \times \mathbf{A}[/tex]

    But I don't have a del operator. To me it just looks like I need to take a vector product between a scalar and a vector. What am I missing?


    EDIT: Also, since when is:

    [tex]
    \frac{\Phi}{2\pi}(\mathbf{\hat{k}}\times\textbf{r} )^{-2}
    [/tex]

    a scalar? I may be sleep deprived, or perhaps insane, but a scalar multiplied by a vector is still a vector


    EDIT2: Whoops, first question was kind of stupid, the magnetic field (B) is defined as the curl of the magnetic vector potential (A). i.e.,

    [tex]
    \mathbf{B}=\mathbf{\nabla}\times\mathbf{A}
    [/tex]

    so

    [tex]
    \mathbf{B}=\mathbf{\nabla}\times\mathbf{A}=\mathbf{\nabla}\times\Phi\mathbf{\hat{k}}\times\mathbf{r}/[2\pi(\mathbf{\hat{k}}\times\mathbf{r})^2]
    [/tex]
     
    Last edited: Feb 18, 2010
  5. Feb 18, 2010 #4

    gabbagabbahey

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    You might want to use a different letter than [itex]\textbf{A}[/itex] so that you don't confuse it with the vector potential. But yes, that's the rule I was referring to.

    Ermmm... [itex]\textbf{B}=\mathbf{\nabla}\times\textbf{A}[/itex] perhaps?


    Zzzzz....get some sleep!:zzz::wink:

    [tex](\mathbf{\hat{k}}\times\textbf{r} )^2\equiv(\mathbf{\hat{k}}\times\textbf{r} )\cdot(\mathbf{\hat{k}}\times\textbf{r} )[/tex]

    The dot product of two vectors is a scalar.
     
  6. Feb 18, 2010 #5
    Heh, nope sorry not tonight. Thanks for your help though, now that I know how to find B I have to figure out how to do the rest of the problem...

    Thanks!!!!
     
  7. Feb 18, 2010 #6
    I'm still not getting it, apparently (when to use the product rule). SO, I have:
    [tex]
    \mathbf{B}=\mathbf{\nabla} \times \mathbf{A}=\mathbf {\nabla}\times \left [ \Phi\mathbf{\hat{k}}\times\frac{\mathbf{r}}{[2\pi(\mathbf{\hat{k}}\times\mathbf{r})^2]}\right ]
    [/tex]

    Focusing on this part first:
    [tex]\left [ \Phi\mathbf{\hat{k}}\times\frac{\mathbf{r}}{[2\pi(\mathbf{\hat{k}}\times\mathbf{r})^2]}\right ]
    [/tex]

    the bottom turns out to be [itex]2(x-y)[/itex]. So we have

    [tex]
    \left [ \Phi\mathbf{\hat{k}}\times\frac{\mathbf{r}}{[4\pi(x-y)]}\right ]
    [/tex]

    Which is equal to

    [tex]
    \left [ \Phi\mathbf{\hat{k}}\times\frac{x\mathbf{\hat{i}}+y\mathbf{\hat{j}}+z\mathbf{\hat{k}}}{[4\pi(x-y)]}\right ]
    [/tex]

    Which is still taking the cross product of a vector with a vector. Griffith's product rule no 8 would work but it breaks it down into four terms; I did it by brute force (no tricks) and got:

    [tex]

    \left[-\frac{\Phi}{4\pi}\left(\frac{x+y}{(x-y)^2}\right)\right]\mathbf{\hat{k}}

    [/tex]

    Which, to me, doesn't look much like the magnetic field of a long thin solenoid with flux along z-axis. Halp :(
     
  8. Feb 18, 2010 #7

    gabbagabbahey

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    No it doesn't,

    [tex](\mathbf{\hat{k}}\times\textbf{r} )^2\equiv(\mathbf{\hat{k}}\times\textbf{r} )\cdot(\mathbf{\hat{k}}\times\textbf{r} )=(-y\mathbf{\hat{i}}+x\mathbf{\hat{j}})\cdot(-y\mathbf{\hat{i}}+x\mathbf{\hat{j}})=y^2+x^2[/tex]


    I'd start by defining the functions,

    [tex]f\equiv\frac{\Phi}{2\pi(\mathbf{\hat{k}}\times\textbf{r} )^2}\;\;, \;\;\;\textbf{U}=\mathbf{\hat{k}}\times\textbf{r}[/tex]

    You then have [itex]\textbf{B}=\mathbf{\nabla}\times\textbf{A}=\mathbf{\nabla}\times(f\textbf{U})[/itex] and you can apply the product rule.

    P.S. If you are pulling an all-nighter, I've found that taking a 15-20 min catnap increases your cognitive abilities much more than any readily available stimulant.
     
  9. Feb 18, 2010 #8
    Okay thanks again. But whatever answer I get, how will that resemble the field of a long thing solenoid? Isn't the equation for that just [itex]B=\mu_o n I[/itex] ? Since the field is uniform inside and zero outside
     
  10. Feb 18, 2010 #9

    gabbagabbahey

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    Right, and what is the flux (through a single loop) of that solenoid if it has radius [itex]\rho[/itex]?

    Edit: After taking couple of minutes to work out the answer, it hit me that this problem will be much easier in cylindrical coordinates. Using the notation of Griffiths (except with [itex]\{\mathbf{\hat{i}},\mathbf{\hat{j}},\mathbf{\hat{k}}\}[/itex] instead of [itex]\{\mathbf{\hat{x}},\mathbf{\hat{y}},\mathbf{\hat{z}}\}[/itex]),

    [tex]\textbf{r}=s\mathbf{\hat{s}}+z\mathbf{\hat{k}}\implies \mathbf{\hat{k}}\times\textbf{r}=s( \mathbf{\hat{k}}\times\mathbf{\hat{s}})=s \mathbf{\hat{\phi}}[/itex]

    And so your vector potential is just [tex]\textbf{A}=\frac{\Phi}{2\pi s}\mathbf{\hat{\phi}}[/itex]
     
    Last edited: Feb 18, 2010
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