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Deriving equations of motion in spherical coordinates

  • Thread starter Emspak
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  • #1
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Homework Statement



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:

Answers and Replies

  • #2
vela
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Homework Statement



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 $$ \bf \hat{r} \rm = \bf \hat{x} \rm \sin\theta \cos\phi + \bf \hat{y} \rm \sin \theta \sin \phi + \bf \hat{z}\rm \cos\theta \\
\hat{\boldsymbol\theta} = \bf \hat{x} \rm \cos\theta \cos\phi + \bf \hat{y} \rm \cos \theta \sin \phi - \bf \hat{z}\rm\sin\theta\\
\hat{\boldsymbol\phi} = \bf \hat{x} \rm (-\sin\phi) + \bf \hat{y} \rm \cos \phi \sin \phi \\
$$

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

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

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.
Note that you calculated ##d\hat{r}/d\theta## in one case and ##d\hat{r}/dt## in the other.
 
  • #3
CAF123
Gold Member
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88
Now, if I take it that position $$\bf \vec{r} \rm = r \bf \hat{r}$$ and say $$ \bf \hat{r} \rm = \bf \hat{x} \rm \sin\theta \cos\phi + \bf \hat{y} \rm \sin \theta \sin \phi + \bf \hat{z}\rm \cos\theta \\
\hat{\boldsymbol\theta} = \bf \hat{x} \rm \cos\theta \cos\phi + \bf \hat{y} \rm \cos \theta \sin \phi - \bf \hat{z}\rm\sin\theta\\
\hat{\boldsymbol\phi} = \bf \hat{x} \rm (-\sin\phi) + \bf \hat{y} \rm \cos \phi \sin \phi \\
$$
The unit vector for ##\hat{\phi}## is incorrect, check this.

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]
Yes, that's correct.
I notice that this happens (and some of this is just seeing the notation):
$$
\frac{d \bf \vec{r} \rm}{dt} = r \frac{d\bf \hat{r}}{dt} + \frac{dr}{dt}\bf \hat{r} \rm = \dot{r} \bf \hat{r} \rm + r \hat{\boldsymbol \theta} \rm $$
You have that ##\dot{\hat{r}} = \hat{\theta}## which is incorrect. You forgot the chain rule.
 
  • #4
pasmith
Homework Helper
1,738
410

Homework Statement



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.
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]

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).
Those derivatives should be partial.

If we do the same thing with changing θ and [itex]\varphi[/itex] though, the result is different. When 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.
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.
 
  • #5
243
1
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]
 
  • #6
243
1
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]
 
  • #7
vela
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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]
That doesn't work units-wise, does it?
 
  • #8
CAF123
Gold Member
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88
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]
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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