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Deriving equations of motion in spherical coordinates

  1. Sep 15, 2013 #1
    1. The problem statement, all variables and given/known data

    OK, we've been asked to derive the equations of motion in spherical coordinates. According to the assignment, we should end up with this:

    $$
    \bf \vec{v} \rm = \frac{d \bf \vec{r} \rm}{dt} = \dot{r} \bf \hat{r} \rm + r \dot{\theta}\hat{\boldsymbol \theta} \rm + r \dot{\phi}\sin \theta \bf \hat{\boldsymbol\phi}\rm
    $$

    In this case θ is the angle from the z axis and phi is the angle in the x-y plane.

    Now, if I take it that position $$\bf \vec{r} \rm = r \bf \hat{r}$$ and say $$ \bf \hat{r} \rm = \bf \hat{x} \rm \sin\theta \cos\phi + \bf \hat{y} \rm \sin \theta \sin \phi + \bf \hat{z}\rm \cos\theta \\
    \hat{\boldsymbol\theta} = \bf \hat{x} \rm \cos\theta \cos\phi + \bf \hat{y} \rm \cos \theta \sin \phi - \bf \hat{z}\rm\sin\theta\\
    \hat{\boldsymbol\phi} = \bf \hat{x} \rm (-\sin\phi) + \bf \hat{y} \rm \cos \phi\\
    $$

    now maybe I am making this more complex than it is. And maybe it's just a notation problem (I really hate the dot notation sometimes because I feel it obscures things, but I need to know it, I know).

    If we assume that when r changes, [itex]\phi[/itex] and [itex]\theta[/itex] and their unit vectors stay the same, then we can safely say that [itex]\frac{d \hat{\boldsymbol\phi}}{dr} = 0[/itex] and [itex]\frac{d \hat{\boldsymbol\theta}}{d r} = 0.[/itex] (someone please tell me if i am wrong).

    If we do the same thing with changing θ and [itex]\varphi[/itex] though, the result is different. hen we change θ, r has to change because it changes direction, and when we change [itex]\varphi[/itex] r has to change because it changes direction in that case also.

    When I take the derivative of [itex]\hat{r}[/itex] with respect to [itex]\vartheta[/itex], I get the following:

    $$\frac{d \bf \hat{r}}{d\theta} = \bf \hat{x} \rm \cos\theta \cos\phi + \bf \hat{y} \rm \cos \theta \sin \phi - \bf \hat{z}\rm\sin\theta $$

    which as it happens also is equal to [itex] \hat{\boldsymbol{\theta}}[/itex]

    Now, if I look at [itex]\bf \vec{r} \rm = r \bf \hat{r}[/itex] and take the derivative w/r/t time, I should get [itex]\frac{d \bf \vec{r} \rm}{dt} = r \frac{d\bf \hat{r}}{dt} + \frac{dr}{dt}\bf \hat{r} \rm[/itex]

    I notice that this happens (and some of this is just seeing the notation):
    $$
    \frac{d \bf \vec{r} \rm}{dt} = r \frac{d\bf \hat{r}}{dt} + \frac{dr}{dt}\bf \hat{r} \rm = \dot{r} \bf \hat{r} \rm + r \hat{\boldsymbol \theta} \rm $$

    ANd I feel like I am almost there. But I am having trouble making that last step. I am getting a bit frustrated because I can't seem to make the differentiation work the way it does in the text and I haven't found a derivation online that matches up with anything I have seen in class. Again, maybe it's just the notation used.

    Any help is most appreciated. Thanks.

    EDIT: unit vector phi expression corrected.
     
    Last edited: Sep 15, 2013
  2. jcsd
  3. Sep 15, 2013 #2

    vela

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    Just look at your expressions for the unit vectors. ##r## doesn't appear at all, so changing it leaves the unit vectors unchanged.

    You mean ##\hat{r}##, not ##r##, right?

    Note that you calculated ##d\hat{r}/d\theta## in one case and ##d\hat{r}/dt## in the other.
     
  4. Sep 15, 2013 #3

    CAF123

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    The unit vector for ##\hat{\phi}## is incorrect, check this.

    Yes, that's correct.
    You have that ##\dot{\hat{r}} = \hat{\theta}## which is incorrect. You forgot the chain rule.
     
  5. Sep 15, 2013 #4

    pasmith

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    In which case [itex]\mathbf{r} = r(\sin\theta \cos \phi \hat x + \sin\theta \sin\phi \hat y + \cos\theta \hat z)[/itex], and
    [tex]
    \hat r = \sin\theta \cos\phi \hat x + \sin\theta \sin\phi \hat y + \cos \theta \hat z \\
    \hat \theta = \cos\theta \cos\phi \hat x + \cos\theta \sin\phi \hat y - \sin\theta \hat z \\
    \hat \phi = -\sin\phi \hat x + \cos\phi \hat y
    [/tex]

    Those derivatives should be partial.

    No; if you fix [itex]r[/itex] then you are confined to a sphere of radius [itex]r[/itex]. [itex]\hat r[/itex] is always normal to the sphere and [itex]\hat \theta[/itex] and [itex]\hat \phi[/itex] are tangent to the sphere. All three unit vectors are functions of [itex]\theta[/itex] and [itex]\phi[/itex] and independent of [itex]r[/itex]. (Actually, as you see from above, [itex]\hat \phi[/itex] depends only on [itex]\phi[/itex]).

    Start with
    [tex]
    \mathbf{r} = r\hat r
    [/tex]
    and differentiate with respect to time:
    [tex]
    \dot{\mathbf{r}} = \dot r \hat r + r \dot {\hat r}
    [/tex]
    Now calculate [itex]\dot{\hat r}[/itex] from the definition using the product rule and the chain rule. Then collect terms involving [itex]\dot \theta[/itex] and [itex]\dot \phi[/itex]. You should then get the given answer.
     
  6. Sep 15, 2013 #5
    So when I look at the change in the unit vector with respect to time, I have to consider it as [itex]\vec{r}=r(\theta + d\theta, \phi+ d\phi)\hat{r}[/itex]. Is that the case? So I could rewrite this as [itex]d\hat{r} = d\theta \hat{\theta} + d \phi hat{\phi}[/itex]
     
  7. Sep 15, 2013 #6
    So when I look at the change in the unit vector with respect to time, I have to consider it as [itex]\vec{r}=r(\theta + d\theta, \phi+ d\phi)\hat{r}[/itex]. Is that the case? So I could rewrite this as [itex]d\hat{r} = d\theta \hat{\theta} + d \phi \hat{\phi}[/itex]

    I was thinking that in two dimensions I did this:

    [itex]d\hat{r} = d\theta \hat{\theta}[/itex]

    and multiplied through by [itex]\frac{d\theta}{dt}[/itex] and got [itex]\frac{d\hat{r}}{d\theta} = \hat{\theta}\frac{d \theta}{dt}[/itex]
     
  8. Sep 15, 2013 #7

    vela

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    That doesn't work units-wise, does it?
     
  9. Sep 15, 2013 #8

    CAF123

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    I do not quite follow. What you want to find is ##\dot{\hat{r}}##. Since ##\hat{r} = \hat{r}(\theta, \phi)##, the differential is as follows:$$\frac{d \hat{r}}{dt} = \frac{\partial \hat{r}}{\partial \theta} \frac{d \theta}{dt} + \frac{\partial \hat{r}}{\partial \phi} \frac{d \phi}{dt},$$
    Sub this into ##\dot{\vec{r}}## and this should give you your answer.
     
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