Projections, Constraints, and Gradients

I’m a big fan of thinking about wrong ideas. One wrong idea I’ve been spending a lot of time thinking about for over a year is one that I came across while working through a piece of curriculum intended for physics majors to learn more about this relationship

F = –Grad(U)

The materials had students think about a rock on a mountain, and how the force on the rock would be related to the slope of the mountain at that point. The  materials went on to have students think about a topographic map, where the height of the mountain could be described everywhere with some function z = f(x,y) and the potential energy by U = mgz. They had students eventually build up to considering this expression

– mg * [(∂f/∂x) x-hat + (∂f/∂y) y-hat ] ,

which would appear to be the a “projected gradient” of some sort.

In this post, I want to explain why this expression is (1) not a gradient and (2) not a force.

So, let’s imagine an object moving on some randomly shaped “frictionless” surface.

And let’s say that we can describe this surface as before, with z = f(x,y). This surface could be a ramp, or a bowl, or a complicated ice sculpture. It doesn’t matter, but I want to consider the case that the object can’t fall off or fly off. Let’s also imagine that gravity is involved and that the gravitational potential energy of the object is

U= mgz, as was assumed in the rock on the hill case.

So by substitution we can write

U = mg f(x,y), as was done in the curriculum.

Now if there was no surface involved (i.e., the object was just free-falling in gravity), then we could certainly relate  the force due to gravity  to the potential energy in the following way

 Fg = -∂U/∂z z-hat= -mg z-hat.

However, that’s not going to be true with our surface constraining the motion. Given this, however, I do agree it’s tempting to consider the expression

– mg * [(∂f/∂x) x-hat + (∂f/∂y) y-hat ],

which involves taking some the partial derivatives of U with respect to x and y (instead of z). The question is, is this a force or perhaps a projection of a force? It’s certainly tempting to think that this is, because it looks like the gradient of the potential, and it looks to be happening in the x-y plane.

It turns out this isn’t a force, because taking the partial derivatives in this way does not constitute a gradient. The reason it’s not a proper gradient is interesting (to me) and has to due with the fact that taking partial derivatives like this only makes a gradient in the special case of a euclidean geometry. To do so, I will have to convince you the transformation of U(z) to U(x,y) is very different than what it at first seems.  So, we’ll have to back track a bit.

See, in a 3-dimensional euclidean space, we measure distance by the following metric:

ds² = dx² + dy² + dz²

But what we have is different. We have a 2D surface embedded in a 3d space. Given that  z = f(x,y), we can write the differential dz = (∂f/∂x)  dx + (∂f/∂y) dy and substitute that into our 3D metric to get the following metric for the 2D surface:

ds² = dx² + dy² + [(∂f/∂x)  dx + (∂f/∂y) dy]²

This can be simplified and rewritten as

ds² = (1+α²) dx² + (1+β²) dy² + αβ dx dy

where I’ve just named ∂f/∂x  as α and ∂f/∂y as β.

OK, so now have this 2D non-euclidean surface described by the above way of measuring distances. It is weird, because you have these correction factors such as α² and β² that augment the distance traveled depending on the slope of the surface where you are measuring. We also have this weird cross term that involves dx dy. These terms are what make space non-euclidean. It’s like your ruler grows and shrinks and even twists as you move around the surface. Or, if you will, those corrections exist because we are only looking at x and y and the correction factors add back in the distances in the z-direction.

So the following questions arise:

1. How do free particles move on such surfaces?
2. Can we calculate forces by taking the gradient of the function U = f(x,y)

OK. OK. Let’s step back again and clarify why these are the two important questions:

In a Euclidean geometry, free particles can be describe by the follow equation

m d²r/ dt² = 0 → This is just Newton’s first law.

and many types of forces can be calculated  using by F = – Grad (U)

This allows us to write Newton’s 2nd law as,  m d²r/ dt² =  –Grad (U)

All of this works for Euclidean geometries and for forces described by conservative vector fields–but not all conservative forces, just forces described by conservative vector fields. There’s a subtle difference, because Normal forces are conservative (i.e. they do no work), but they cannot be described by a conservative vector field.

OK. So, what we’d like to do is write a similar equation for particles on our surface. The problems we’ll end up running into are these:

1. Free particles on our 2D surface can’t be described by d²r/ dt² = 0, and
2. We’re going to have to be smart about how we take our gradient.

I want to take up these two obstacles in two other posts. So, next I’ll talk about free particle motion in non-euclidean spaces. And then I’ll talk about gradients and external forces in non-euclidean spaces. Hopefully, I’ll get to a fourth post where I talk about why classical mechanics and general relativity are the same thing. And last, I’d like to talk about the mistakes students made in this curriculum, especially the ones that got me thinking about these issues so deeply. Wish me luck.