# Motion in the Projected Parabola

OK, so building off my last post, the question I want to work through is, “How do free particles move in spaces that are non-euclidean?”

Let’s use Andy’s suggestion of an object sliding around on a parabola, and we’ll project that parabola onto the x-axis, to see what motion is like on the 1D projected parabolic space. But let’s take away gravity, because we want to consider free particles that are subject to no ‘external’ forces.

We know of course that in the 2D world, the particle would move with constant speed from one side of the parabola to the other. The reason it would move with constant speed is because, without gravity, there is no way for the object to change its kinetic energy, or direction for that matter.

What I want you to imagine is you are looking down into the parabola from above, and you can only see what’s happening in the x-direction. It’s like you have no depth perception. To you, the ball would seem to speed up as the ball goes down the parabola and slow down as it goes back up the parabola. That’s because some of the speed is “hidden” in the y-direction, which you can’t see. Keep this in mind, because our mathematics has to jive with this intuition.

**Getting the Projected Parabolic Metric**

So let’s take the parabola y = 1/2 x²

As I said in my previous post we typically measure distances like this

ds² = dx² + dy²

But we can transform this 2D euclidean space to our 1D non-euclidean space by using the fact that

dy = x dx

Plugging that into ds² = dx² + dy² gives

ds² = (1+ x²) dx²

Recognize that the term “x” is just the value of the slope of parabola at the point x, so it’s completely consistent with what I’ve talked about in the first post.

**Newton’s 1st Law in the Projected Parabola**

Now, I’ve already established that the object moves with constant speed in the 2D euclidean world. So let’s set a value for that velocity and just call it “v”.

With a velocity v, the distance in the 2D geometry would be s = v t or ds = v dt, where s is measure along the arc length of the parabola.

This allows us to write

v² dt² = (1+ x²) dx²

which I’ll rearrange to be

(dx/dt)² = v² / (1+ x²)

or

dx/dt = v / Sqrt (1+x²)

Here is where we should check out intuition, because the value dx/dt is the velocity in the x-direction. I said that the ball appear faster near x =0, because there none of the speed is hidden in the y-direction. Plugging in x =0, we get dx/dt = v. That’s good, because that’s when none of the velocity is hidden. If we take the limit as x gets really large, the velocity goes to zero. And that also makes sense because the parabola gets so steep that you can’t see any of the velocity–it’s all hidden in the y-direction.

OK, what we might want to do is write Newton’s 1st laws for a 1D parabolic geometry. So what we want is to rewrite this expression so that it looks like

d²x/dt² + Correction[x, dx/dt] = 0

And we can do this by taking a time derivative of our equation for dx/dt and rearranging.

d²x/dt² – v * d/dt [ 1/ Sqrt (1+x(t)²)] = 0

d²x/dt² + v * x / (1+x²)^{3/2} *dx/dt = 0

Now this is pretty ugly, but it puts it in a form like Newton’s 1st law. Our corrections term is the messy part, because it is non-linear function of x and also depends on dx/dt. It might be tempting to think of this whole term as velocity and position dependent “force”, but we don’t have to. I want to stress that this is just how particles move in our 1D projected parabolic space without forces. I’ll show you later how we can add forces.

The truth is, it would have been prettier to not take that derivative, and cast this in term of energy. One way of doing this is to multiply our expression for (dx/dt)² by 1/2 m, to get

1/2 m (dx/dt)² = KE / (1+ x²)

where KE is just a constant.

**Adding Back in Gravity (the easy way)**

To add back in gravity, we just have to realize that KE is no longer a constant in the 2D Euclidean space. But since the total energy in 2D euclidean space is still constant, we can write

KE = Total Energy – PE = TE – mgy = TE – mg (1/2 x²)

1/2 m (dx/dt)² = (TE – mgy) / (1+ x²)

1/2 m (dx/dt)² = (TE – 1/2 mg x²) / (1+ x²)

where TE is just a constant expressing the total energy in the 2D Euclidean space.

I’m pretty sure this works. I suppose we’ll see.

**Adding Back in Gravity (the hard way)**

This will be the subject of another post, but basically we’ll start with our Modified Newton’s 1st Law expression, and we’ll have to figure out how to take a gradient in a non-euclidean space.

As an aside, I’d like to point out that SO far, we haven’t had to use calculus of variations or Langrangians.

Just a quick comment about your last point. I think you are actually doing calculus of variations because it looks so similar. I guess I’m not 100% sure about that (chock that up to pain meds and crappy hospital internet connections, I guess), You calculations also look a lot like the calculations involved in virtual work approaches. Again, not quite sure.

This is really cool stuff, Brian. Thanks for the mental distraction, I needed it!

Yeah, all approaches have to similar on some grounds, because if they are correct they lead to the same place. I *think*, in this case, I avoid doing any calculus of variations by importing ds = v dt from the euclidean space. That is, I steal something I know about the kinematics in the euclidean world. With calculus of variations approach I’d be trying to minimize s(x, dx/dt) without that assumption… they must be equivalent, but I think only in certain cases. For example, this approach won’t quite work when going from 3D to 2D because the condition of constant v (or KE) is not enough constraints to get two equations of motion. I’ll have to think whether I can get around the calculus of variations for that scenario.