Vector components in polar coordinates

[PFuser1232]

The magnitude of the parallel component of the time derivative of a vector $\vec{A}$ is given by:
$$|\frac{d\vec{A}_{\parallel}}{dt}| = |\frac{dA}{dt}|$$
Where $A$ is the magnitude of the vector.
Can we write the actual derivative in vector form as $\frac{dA}{dt} \hat{A}$? Notice how I dropped the absolute value sign.
The reason I'm asking this is when I tried deriving acceleration in polar coordinates, I split both $\Delta \vec{v}_r$ and $\Delta \vec{v}_{θ}$ into components parallel and perpendicular to each of the vectors, then I divided each of the terms by $\Delta t$ and took the limit as $\Delta t →0$. I ran into a problem, though. The scalars $v_r$ and $v_{θ}$ do not represent magnitudes, but scalar components that can be positive, negative, or zero. So, for the term $\frac{d\vec{v}_{r\parallel}}{dt}$, I wrote $\frac{d|v_r|}{dt} \hat{v}_r$ where $\hat{v}_r$ is a unit vector always pointing in the direction of $\vec{v}_r$. I then realized that if $v_r$ is positive, $\hat{v}_r$ and $\hat{r}$ point in the same direction. If $v_r$ is negative, $\hat{v}_r$ and $\hat{r}$ point in opposite directions, giving the same result in both cases, namely $\frac{dv_r}{dt} \hat{r}$. Is this correct?
My book does not place emphasis on the fact that $v_r$ may be negative and arrives at the desired result much quicker.

 

[BiGyElLoWhAt]

Maybe I'm missing something, what's the problem with the radial component being negative? dv_r/dt rhat seems fine to me.
I'm also not sure I know what you mean by " If vr is negative, vr and r point in opposite directions giving the same result in both cases" rhat doesn't point in the direction of vr, it points in the direction of r. So it shouldn't give the same result in both cases.

 

[PFuser1232]

Maybe I'm missing something, what's the problem with the radial component being negative? dv_r/dt rhat seems fine to me.
I'm also not sure I know what you mean by " If vr is negative, vr and r point in opposite directions giving the same result in both cases" rhat doesn't point in the direction of vr, it points in the direction of r. So it shouldn't give the same result in both cases.

$\hat{v}_r$ is defined to be in the direction of $\vec{v}_r$, which itself can point either in the $+\hat{r}$ or $-\hat{r}$ direction.
The reason I did this is because the original equation for a generic time-varying vector involves $A$, the magnitude of $\vec{A}$, which is always positive or zero.
$v_r$ is not the magnitude of $\vec{v}_r$. Its absolute value is.
Is there any other way to arrive at this result?

 

[Chestermiller]

What does the "parallel component of the vector A" mean? Is there somehow a velocity involved, and does A represent a vector somehow associated with a moving particle?

Chet

 

[PFuser1232]

What does the "parallel component of the vector A" mean? Is there somehow a velocity involved, and does A represent a vector somehow associated with a moving particle?

Chet

We can always split an increment in $\vec{A}$, $\Delta \vec{A}$, into two vector components. One parallel to $\vec{A}$ and the other perpendicular to $\vec{A}$.
I guess the problem with my reasoning is that it breaks down when $v_r = 0$ [which may very well be the case], since the function $|v_r|$ is not differentiable at $v_r = 0$.

 

[Chestermiller]

We can always split an increment in $\vec{A}$, $\Delta \vec{A}$, into two vector components. One parallel to $\vec{A}$ and the other perpendicular to $\vec{A}$.

Ah, of course. Thanks. So the parallel component is:

$$\left(\frac{d\vec{A}}{dt}\centerdot \frac{\vec{A}}{|\vec{A}|}\right)\frac{\vec{A}}{|\vec{A}|}=\frac{d|\vec{A}|}{dt}\frac{\vec{A}}{|\vec{A}|}$$

The magnitude of this vector is $\left|\frac{d|\vec{A}|}{dt}\right|$

So the vector itself is given by:

$\left|\frac{d|\vec{A}|}{dt}\right|\left(\frac{\frac{d|\vec{A}|}{dt}}{\left|\frac{d|\vec{A}|}{dt}\right|}\frac{\vec{A}}{|\vec{A}|}\right)$

The term in parenthesis is the direction of the unit vector.

Chet

 

[PFuser1232]

Ah, of course. Thanks. So the parallel component is:

$$\left(\frac{d\vec{A}}{dt}\centerdot \frac{\vec{A}}{|\vec{A}|}\right)\frac{\vec{A}}{|\vec{A}|}=\frac{d|\vec{A}|}{dt}\frac{\vec{A}}{|\vec{A}|}$$

The magnitude of this vector is $\left|\frac{d|\vec{A}|}{dt}\right|$

So the vector itself is given by:

$\left|\frac{d|\vec{A}|}{dt}\right|\left(\frac{\frac{d|\vec{A}|}{dt}}{\left|\frac{d|\vec{A}|}{dt}\right|}\frac{\vec{A}}{|\vec{A}|}\right)$

The term in parenthesis is the direction of the unit vector.

Chet

Which simplifies to $\frac{d|\vec{A}|}{dt} \hat{A}$. Now, how exactly do we apply this to the case of radial velocity? If we apply this formula to the case of radial velocity (replacing $\vec{A}$ by $\vec{v}_r$, we get $\frac{d|v_r|}{dt} \hat{v}_r$. We can express this in terms of the polar base vector $\hat{r}$ by considering what happens when $v_r > 0$ and when $v_r < 0$, giving us $\frac{dv_r}{dt} \hat{r}$. This, however, only holds for nonzero $v_r$ since the time derivative of $|v_r|$ doesn't exist when $v_r = 0$. How do we resolve this?

 

[Chestermiller]

I don't quite understand. Are you saying that you have a purely radial velocity vector, or you saying that the velocity vector has components in the radial and circumferential directions, and that these components of the particle velocity are changing as the particle moves, and that you want to find the radial component of the particle acceleration?

Chet

 

[PFuser1232]

I don't quite understand. Are you saying that you have a purely radial velocity vector, or you saying that the velocity vector has components in the radial and circumferential directions, and that these components of the particle velocity are changing as the particle moves, and that you want to find the radial component of the particle acceleration?

Chet

The change in velocity, $\Delta \vec{v}$ has two components, $\Delta \vec{v}_r$ (radial) and $\Delta \vec{v}_{θ}$ (tangential). Each component is split into two components (parallel and perpendicular to $\vec{v}_r$ and $\vec{v}_{θ}$).

 

[Chestermiller]

The change in velocity, $\Delta \vec{v}$ has two components, $\Delta \vec{v}_r$ (radial) and $\Delta \vec{v}_{θ}$ (tangential). Each component is split into two components (parallel and perpendicular to $\vec{v}_r$ and $\vec{v}_{θ}$).

OK. If this is what you want (I hope I understand), then
$$\vec{v}=v_r\vec{i}_r+v_{θ}\vec{i}_θ$$
where $\vec{i}_r$ and $\vec{i}_θ$ are functions of θ(t).
So, $$\frac{d\vec{v}}{dt}=\frac{dv_r}{dt}\vec{i}_r+v_r\frac{d\vec{i}_r}{dθ}\frac{dθ}{dt}+\frac{dv_θ}{dt}\vec{i}_θ+v_θ\frac{d\vec{i}_θ}{dθ}\frac{dθ}{dt}$$
So, $$\frac{d\vec{v}}{dt}=\frac{dv_r}{dt}\vec{i}_r+v_r\frac{dθ}{dt}\vec{i}_θ+\frac{dv_θ}{dt}\vec{i}_θ-v_θ\frac{dθ}{dt}\vec{i}_r$$
So, $$\frac{d\vec{v}}{dt}=\left(\frac{dv_r}{dt}-v_θ\frac{dθ}{dt}\right)\vec{i}_r+\left(v_r\frac{dθ}{dt}+\frac{dv_θ}{dt}\right)\vec{i}_θ$$
Is this what you were asking for?

Chet

 

[PFuser1232]

OK. If this is what you want (I hope I understand), then
$$\vec{v}=v_r\vec{i}_r+v_{θ}\vec{i}_θ$$
where $\vec{i}_r$ and $\vec{i}_θ$ are functions of θ(t).
So, $$\frac{d\vec{v}}{dt}=\frac{dv_r}{dt}\vec{i}_r+v_r\frac{d\vec{i}_r}{dθ}\frac{dθ}{dt}+\frac{dv_θ}{dt}\vec{i}_θ+v_θ\frac{d\vec{i}_θ}{dθ}\frac{dθ}{dt}$$
So, $$\frac{d\vec{v}}{dt}=\frac{dv_r}{dt}\vec{i}_r+v_r\frac{dθ}{dt}\vec{i}_θ+\frac{dv_θ}{dt}\vec{i}_θ-v_θ\frac{dθ}{dt}\vec{i}_r$$
So, $$\frac{d\vec{v}}{dt}=\left(\frac{dv_r}{dt}-v_θ\frac{dθ}{dt}\right)\vec{i}_r+\left(v_r\frac{dθ}{dt}+\frac{dv_θ}{dt}\right)\vec{i}_θ$$
Is this what you were asking for?

Chet

Yes, that's the result I'm looking for. I'm trying to understand the derivation geometrically, though, in the fashion I have described above. That is, splitting velocity into four components and taking the limit as $\Delta t$ approaches zero.

 

[Chestermiller]

OK. The key to this is taking into account that the unit vectors are functions of θ. And, if there's a finite change in θ, say before you take the limit, you need to express the unit vectors at the new point in terms of the unit vectors at the starting point and the angle change.

Chet

 

[PFuser1232]

OK. The key to this is taking into account that the unit vectors are functions of θ. And, if there's a finite change in θ, say before you take the limit, you need to express the unit vectors at the new point in terms of the unit vectors at the starting point and the angle change.

Chet

Which is what I did. The thing is, there seems to be a discontinuity. My goal is not to express the derivative in terms of the unit vector $\hat{v}_r$, but to express it in terms of $\hat{r}$. To go from $\frac{d|v_r|}{dt} \hat{v}_r$ to $\frac{dv_r}{dt} \hat{r}$ I need to consider the change in direction of $\hat{v}_r$ that results from the sign change of $v_r$. If $v_r$ is positive, both unit vectors point in the same direction and we get the desired result, namely $\ddot{r} \hat{r}$. If $v_r$ is negative, the unit vectors point in opposite directions but we still get the same result. If $v_r$ is zero, then $\frac{d|v_r|}{dt}$ does not even exist.

 

[Chestermiller]

Which is what I did. The thing is, there seems to be a discontinuity. My goal is not to express the derivative in terms of the unit vector $\hat{v}_r$, but to express it in terms of $\hat{r}$. To go from $\frac{d|v_r|}{dt} \hat{v}_r$ to $\frac{dv_r}{dt} \hat{r}$ I need to consider the change in direction of $\hat{v}_r$ that results from the sign change of $v_r$. If $v_r$ is positive, both unit vectors point in the same direction and we get the desired result, namely $\ddot{r} \hat{r}$. If $v_r$ is negative, the unit vectors point in opposite directions but we still get the same result. If $v_r$ is zero, then $\frac{d|v_r|}{dt}$ does not even exist.

I wish I could help (did I help at all?), but I'm just not able to understand. Sorry.

Chet

 

[BiGyElLoWhAt]

Are you getting caught up on adding a d theta component (perpendicular) and how it doesn't produce a radius of equal length? If that's the case, you need to realize that d theta is actually curved, but we approximate it as being straight when taking the limit. Not sure if that helps or not.