Matter & Interactions I, Week 11

As (almost) usual, I’m writing this on the Monday after the week in question.

This week we hit chapter 5, which is packed full of interesting physics and mathematics! We encounter the infamous time derivative of a unit vector (aka a direction), which I have found quite mysterious because of the rather hand waving ways it’s treated in standard texts. M&I used to introduce a unit vector’s derivative as an angular velocity operating on the unit vector with a cross product, but that approach is now absent. It’s been replaced with a derivation relying on similar triangles, but the good news is that it’s relativistically valid so you get that with the package.

Let’s talk about a vector’s derivative. The traditional textbooks present a coordinate-based expression for a vector’s time derivative, namely the sum of the time derivatives of the vector’s components in some basis. However, this is essentially useless for geometric purposes, and I’m trying very hard to incorporate goemetric reasoning into the course.

Let’s write a momentum vector “factored” into a magnitude and direction:

\vec{p} = \left\lVert\vec{p}\right\rVert \hat{p}

Now apply the usually product rule from first semester calculus:


The first term represents the change in \vec{p} due to a change in its magnitude, and note that it is in the same direction as the original vector. It’s parallel to \vec{p}. Recall from the momentum principle that another name for  \dfrac{\mathrm{d}\vec{p}}{\mathrm{d}t} is \vec{F}_{\mathrm{net}} so the first term is the component of \vec{F}_{\mathrm{net}} that is parallel to \vec{p}. Geometrically, a force applied parallel to a momentum changes the momentum’s magnitude. That’s the first term.

The second term represents the change in \vec{p} due to a change in its direction, and I argue that it must be perpendicular to the original vector. (A simple proof by contradiction assuming uniform circular motion can be used to justify this. If there were a component of change parallel to \vec{p} then the momentum’s magnitude, which contradicts the assumption of uniform circular motion. Therefore, there can be no component of change parallel to the original vector, and the only other option, for uniform circular motion, is that any change must be perpendicular to the original vector.) Next, I argue that a vector’s changing direction can be thought of as a rotation around an axis. In class, we stood up, held an arm outstretched, and slowly turned in place to our left. Our arms swept out a plane around our bodies, and the bodies played the role of the axis around which our arms rotated. Turning to the left defines the “positive” direction of rotation, merely a convention. Rate at which the vector rotates defines the magnitude of a new quantity, angular velocity. We define the direction of the angular velocity to be given by the thumb of the right hand when its fingers wrap around the axis in the newly defined “positive” direction. This is weird! We’re defining a new vector quantity, angular velocity, in terms of its magnitude and direction separately. Now, how does the direction of the angular velocity and the direction of the original momentum vector relate to the direction of \dfrac{\mathrm{d}\hat{p}}{\mathrm{d}t}? We define it to be the direction of the right hand’s thumb when its fingers point in the direction of the angular velocity and the palm points in the direction of the original vector. This is all encoded in this expression for the second term:

\left\lVert\vec{p}\right\rVert\dfrac{\mathrm{d}\hat{p}}{\mathrm{d}t}=  \left\lVert\vec{p}\right\rVert\ \vec{\omega}\times\hat{p}

Factor \vec{\omega} into its magnitude and direction.

\left\lVert\vec{p}\right\rVert\dfrac{\mathrm{d}\hat{p}}{\mathrm{d}t}=  \left\lVert\vec{p}\right\rVert \left\lVert\vec{\omega}\right\rVert\hat{\omega}\times\hat{p}

where we’re taking the symbol \hat{\omega}\times\hat{p} to represent the hand machinations.

Now, rewrite \left\lVert\vec{\omega}\right\rVert as \dfrac{\left\lVert\vec{v}\right\rVert}{R} , an easily reasoned relation, and rewrite \vec{p} as Newton would have written it, m\left\lVert\vec{v}\right\rVert. The final result becomes

\left\lVert\vec{p}\right\rVert\dfrac{\mathrm{d}\hat{p}}{\mathrm{d}t}=m \dfrac{\left\lVert\vec{v}\right\rVert^2}{R}\hat{\omega}\times\hat{p}

which is one of the most important results in introductory physics. This is a simplified version of a more complicated derivation I have that relies on a brute force way of calculating the time derivative of a unit vector and employing a vector identity (BAC-CAB) to write the result as a triple cross product (really a double cross product because there are two operations, not three), and recognizing one of those cross products as the definition of angular velocity; it just quite literally falls right out from the basics and it’s so beautiful that I want to write it up for The Physics Teacher and intend to do so. Geometrically, a force applied perpendicularly to a momentum changes the momentum’s direction. That’s the second term.

About three years ago, it occured to me that we can think of \vec{\omega}\times as an operator with one slot that, when filled with a unit vector, returns its time derivative as the product of an angular velocity’s magnitude and a direction given by a cross product.

\dfrac{\mathrm{d}(\_)}{\mathrm{d}t} = \vec{\omega}\times (\_) = \left\lVert\vec{\omega}\right\rVert\hat{\omega}\times (\_)

\dfrac{\mathrm{d}\hat{a}}{\mathrm{d}t} =\left\lVert\vec{\omega}\right\rVert\hat{\omega}\times\hat{a}

(In preparation for things to come, I’m changing the way I think about vectors and treating them as “slotted machines” a la Misner, Thorne, and Wheeler. I have much more to say about this in future posts.)

Also in this chapter, students encounter the dot product for the first “official” time. However, I have already exposed them to it early in the course in a discussion of the concept of projecting vectors onto bases. The choice of basis is arbitrary, and in this chapter students see that one uses a momentum vector as a basis. In other words, students use the dot product to find components of force parallel and perpendicular to a given momentum. The dot product itself picks out the parallel component of one vector relative to another (scaled by a magnitude, etc.) and one way of getting the perpendicular component is by merely subtracting the parallel component from the original vector. However, there is a way to get the perpendicular component using, again, a triple cross product. It’s a straightforward derivation based on a common vector identity (BAC-CAB used in reverse) but I’ve not yet had the guts to present it. Maybe next year.

Comments and feedback welcomed!


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