Can the Time Derivative of a Vector Change if its Magnitude or Angle Decreases?

[PFuser1232]

In chapter 1 of the book "Introduction to Mechanics" by Kleppner and Kolenkow, the derivative of a generic vector $\vec{A}$ is discussed in terms of decomposing an increment in $\vec{A}$, $Δ\vec{A}$, into two perpendicular vector vectors; one parallel to $\vec{A}$ and the other perpendicular to $\vec{A}$
Two equations are then derived:
$$|\frac{d\vec{A}_{perp}}{dt}| = A\frac{dθ}{dt}$$
$$|\frac{d\vec{A}_{par}}{dt}| = \frac{dA}{dt}$$
Do these equations fail if $θ$ or $A$ decrease with time (since the magnitude of a vector is always positive or zero)?

 

[Svein]

Do these equations fail if θθ or AA decrease with time (since the magnitude of a vector is always positive or zero)?

No. A is allowed to decrease with time. When you use polar coordinates, you introduce a sort of singularity when A = 0. When A tries to go through 0, θ changes value abruptly (by π). A will keep on being positive.

 

[PFuser1232]

No. A is allowed to decrease with time. When you use polar coordinates, you introduce a sort of singularity when A = 0. When A tries to go through 0, θ changes value abruptly (by π). A will keep on being positive.

If A decreases with time, then $\frac{dA}{dt}$ is negative, right? But it also happens to be the magnitude of the vector I described earlier.

 

[Chestermiller]

Just because dA/dt is negative doesn't mean that A is negative.

Chet

 

[PFuser1232]

Just because dA/dt is negative doesn't mean that A is negative.

Chet

Of course. But the vector I was referring to wasn't $\vec{A}$, I was talking about the derivative of the vector component of $\vec{A}$ parallel to $\vec{A}$ (left hand side).

 

[Chestermiller]

Of course. But the vector I was referring to wasn't $\vec{A}$, I was talking about the derivative of the vector component of $\vec{A}$ parallel to $\vec{A}$ (left hand side).

The component of the derivative of $\vec{A}$ parallel to $\vec{A}$ also has a direction. That direction is either $+\frac{\vec{A}}{|\vec{A}|}$ or $-\frac{\vec{A}}{|\vec{A}|}$, depending on the sign of dA/dt. The vector component of the derivative of the vector $\vec{A}$ with respect to time in the direction parallel to $\vec{A}$ is:

$$\frac{dA}{dt}\frac{\vec{A}}{|\vec{A}|}$$

Even if dA/dt is negative, the magnitude of this vector component is |dA/dt| (positive). The unit vector describing the direction of this component is equal to this vector divided by |dA/dt|. So the negative sign is associated with the unit vector and not the magnitude.

Chet

 

[PFuser1232]

The component of the derivative of $\vec{A}$ parallel to $\vec{A}$ also has a direction. That direction is either $+\frac{\vec{A}}{|\vec{A}|}$ or $-\frac{\vec{A}}{|\vec{A}|}$, depending on the sign of dA/dt. The vector component of the derivative of the vector $\vec{A}$ with respect to time in the direction parallel to $\vec{A}$ is:

$$\frac{dA}{dt}\frac{\vec{A}}{|\vec{A}|}$$

Even if dA/dt is negative, the magnitude of this vector component is |dA/dt| (positive). The unit vector describing the direction of this component is equal to this vector divided by |dA/dt|. So the negative sign is associated with the unit vector and not the magnitude.

Chet

So if we want to be more technical, we should put absolute value bars on the right hand side of the equation, right?
What about $A\frac{d\theta}{dt}$? Does the same argument hold?

 

[Chestermiller]

So if we want to be more technical, we should put absolute value bars on the right hand side of the equation, right?
What about $A\frac{d\theta}{dt}$? Does the same argument hold?

Yes.

 

[PFuser1232]

Thanks! I am very curious as to where (and why) the author left out the absolute value bars in the derivation. I have attached two screenshots to this post.

 

[Chestermiller]

Thanks! I am very curious as to where (and why) the author left out the absolute value bars in the derivation. I have attached two screenshots to this post.

He probably didn't think of it in the way that you have thought about it.

Chet

 

[vanhees71]

Ok, you forgot to tell us the all-clarifying point given in the book. The author considers a function $\vec{A}:\mathbb{R} \rightarrow V$ with $|\vec{A}|=\text{const}$. This implies that
$$|\vec{A}|^2=\vec{A} \cdot \vec{A}=\text{const} \; \Rightarrow \; \dot{\vec{A}} \cdot \vec{A}=0,$$
i.e., that $\vec{A} \perp \dot{\vec{A}}$.

In more physical terms: If you take $\vec{A}$ as the position vector of a particle, $|\vec{A}|=\text{const}$ means that the particle is moving along a circle (or spherical shell). Then it's velocity is always perpendicular to the position, i.e., the radius vector of the circle/shell. Or more geometrically: The tangent vector to a circle or spherical shell is always perpendicular to the radius vector of this circle/shell. This is also well known from elementary Euclidean geometry!

 

[PFuser1232]

Ok, you forgot to tell us the all-clarifying point given in the book. The author considers a function $\vec{A}:\mathbb{R} \rightarrow V$ with $|\vec{A}|=\text{const}$. This implies that
$$|\vec{A}|^2=\vec{A} \cdot \vec{A}=\text{const} \; \Rightarrow \; \dot{\vec{A}} \cdot \vec{A}=0,$$
i.e., that $\vec{A} \perp \dot{\vec{A}}$.

In more physical terms: If you take $\vec{A}$ as the position vector of a particle, $|\vec{A}|=\text{const}$ means that the particle is moving along a circle (or spherical shell). Then it's velocity is always perpendicular to the position, i.e., the radius vector of the circle/shell. Or more geometrically: The tangent vector to a circle or spherical shell is always perpendicular to the radius vector of this circle/shell. This is also well known from elementary Euclidean geometry!

I agree. But how does this change anything?

 

[vanhees71]

Change what? It's just the short and elegant analytical-geometry proof of the fact that the radius vector of a circle or a spherical shell is always perpendicular to all tangent vectors on this curve/surface.

 

[PFuser1232]

Change what? It's just the short and elegant analytical-geometry proof of the fact that the radius vector of a circle or a spherical shell is always perpendicular to all tangent vectors on this curve/surface.

My bad, I misunderstood your post; I thought you were hinting at some relationship between what you were saying and what I was discussing earlier regarding absolute values.