# 8 Differential Equations

**(Relevant Section of the Textbook: Only a portion of 13.4: Motion in Space, Velocity and Acceleration. Not the whole section)**

One of the most striking uses for the theory of paramteric curves is in describing the world around us. Indeed the original inventions of calculus, parametric curves, and modern physics all can be traced to the same source: Issac Newton and his desire to predict the future. Newton’s big insight was that this lofty goal could be converted into something concrete: the study of differential equations of parametric curves.

## 8.1 Projectile Motion

If I drop an apple from my hand above the ground, what will it do next? Can I *predict* the future using some mathematical regularities of nature? Of course - I can predict it *qualitatively*- the apple will fall. But this isn’t what we (or Newton) are after. We seek an *exact prediction* - of exactly where the apple will be at every point in time after we drop it.

Finding such a prediction sounds daunting indeed: but the first step towards a solution is to realize that what we are after - the exact location \((x,y,z)\) of the apple at each time - is a parametric curve!

\[\mathrm{apple}(t)=(x(t),y(t),z(t))\]

[The three words that I’ve *never* managed to spell correctly are *business*, *necessary*, and *Gall] Thus, we seek a method of determining \(x,y\) and \(z\) as functions of \(t\). Of course this is the hard part! We need something to grab on to: and luckily, the key insight was already discovered by Galileo nearly 500 years ago.

**Every object accelerates downwards by the same amount, regardless of what its made of or how big it is**

The story usually goes that this was discovered by dropping items from the Leaning Tower of Pisa - but Galileo actually did something much more precise, rolling objects down ramps of various slopes and measuring their positions over time consistently by counting heartbeats. In the end, we learned that the *acceleration vector* of our curve can be written

\[\vec{a}=\langle 0,0,-g\rangle\]

Where \(g\) is a constant (the amount that gravity accelerates things: here on earth its 9.8 in metric units). It’s in the \(z\) component as gravity acts vertically, and its negative because gravity accelerates *downwards*.

And it is in this humble equation that all of kinematics lies. We can fully derive the position \(p(t)\) of our apple at any future time from this, and calculus. Remember, acceleration is the second derivative of position, so really

\[\vec{a}=\langle x^{\prime\prime}(t),y^{\prime\prime}(t),z^{\prime\prime}(t)\rangle\]

Equating this to the vector \(\langle 0,0,-g\rangle\) gives us a system of *three differential equations* one for each of our mystery functions \(x,y,z\). And we can solve these! Starting with \(x\), we start wtih \(x^{\prime\prime(t)}\), so we integrate once to get rid of one derivative:

\[x^\prime(t)=\int x^{\prime\prime}(t)dt=\int 0\, dt= C\]

Where \(C\) is a constant of integration. But since \(x^\prime\) measures *velocity* in the \(x\) direction, this is saying that the \(x\)-velocity is *constant in time* and is equal to \(C\). Thus we should probably rename this constant easier to interpret: let’s call it \(v_x\). Thus we know \(x^\prime(t)=v_x\), and to get \(x\) we integrate once more:

\[x(t)=\int x^\prime(t)\,dt=\int v_x dt= v_x t+C\] Where again \(C\) is a constant of integration. Here we see that at time \(t=0\) we have \(x(t)=C\), so this constant actually tells us the *initial position* so we should rename it something more memorable too, let’s go with \(x_0\). This is the entire equation for \(x\):

\[x(t)=v_x t + x_0\]

And since \(y\) has the same differential equation \(y^{\prime\prime}(t)=0\), we know it will have a similar solution \[y(t)=v_yt+y_0\]

This leaves us only to sovle the \(z\)-equation \(z^{\prime\prime} (t)=-g\). Again, we integrate twice

\[z^\prime(t)=\int z^{\prime\prime}(t)\,dt=\int (-g)\, dt=-gt+v_z\] Where here we’ve already named the constant something useful since we know that it represents the *initial velocity* in the z-direction. Integrating again, \[z(t)=\int z^\prime(t)dt=\int -gt+v_z\, dt = \frac{-1}{2}gt^2+v_zt+z_0\]

But now we know all of \(x,y,z\) in terms of six constants: three velocities \(v_x,v_y,v_z\) specifying how fast the apple is moving when we let go at \(t=0\), and three positions \(x_0,y_0,z_0\) specifying where the apple is at \(t=0\). From here, everything is completely determined

\[\mathrm{apple}(t)=\pmat{v_x t+x_0\\ v_y t + y_0\\ \frac{-1}{2}g t^2+v_z t+ z_0}\]

All of these formulas from early physics courses are really just the simple vector equation \(\vec{a}=\langle 0,0,-g\rangle\) unpacked.

## 8.2 Universal Gravitation

Newton built a grand new theory of gravitation that realizes the above story as a mere approximation occuring near the Earth’s surface. Newton describes all of mechanics in terms of *forces*, where we can use a force to figure out acceleration via Newton’s law setting them proportional: \[\vec{F}=m\vec{a}\]

### 8.2.1 Deriving Newton’s Gravitational Law

Newton conjectured a behavior for the force law of gravity between a two masses, say of size \(m\) and \(M\) to work as follows: the force on the mass \(m\) has

- Its
*direction*points from the location of mass \(m\) to the location of mass \(M\). - Its
*magnitude*is propotional to the product of the masses \(Mm\) - Its
*magnitude*is inversely proportional to the distance between the masses.

Can we turn these ideas into a concrete formula using our knowledge of vectors and operations? First, let’s set up our coordinates \(x,y,z\) in space. Let’s put one of our masses - say \(M\) at the point \((0,0,0)\) because that’s easy to work with. Then our goal is to compute the force at the point \(m\): we don’t know where \(m\) is relative to \(M\), so we can just call its location \(r=(x,y,z)\).

The *direction* of a vector is usually specified by giving a unit vector - so we need to describe the *unit vector pointing from \((x,y,z)\) to the origin*. We know how to find a vector from one point to another: its ending point minus starting point:

\[\mathrm{end}-\mathrm{start}=(0,0,0)-(x,y,z)=(-x,-y,-z)=-\vec{r}\]

Now to make this a unit vector, we divide by its length: \[\vec{\|r\|}=\sqrt{x^2+y^2+z^2}\] So finally we get the unit vector \(\hat{u}\) pointing from \(m\) to \(M\) is \[\hat{u}=\frac{-r}{\|r\|}\]

Now we need to deal with the *magnitude*. Here Newton says that we know two things: the magnitude is proportional to the product \(Mm\), and its *inversely proportional* to the square of the distance between the masses. How do we write down the square of the distance? Well, \(r\) is a vector reaching from \((0,0,0)\) to \((x,y,z)\) so its length is the distance between \(M\) and \(m\): thus the distance squared is

\[\|r\|^2=x^2+y^2+z^2\]

Thus, whatever the force vector \(F\) is, we know its magnitude is proportional to \(Mm/\|r\|^2\): this means there’s some constant \(G\) for which \[\|\vec{F}\|=\frac{GMm}{\|r^2\|}\]

But now we are done! We know both the magnitude and direction of \(F\), so we can directly write down a vector \(F\) satisfying Newton’s requirements:

\[F=\left(\frac{GMm}{\|r\|^2}\right)\hat{u}=\left(\frac{GMm}{\|r\|^2}\right)\frac{-\vec{r}}{\|r\|}\]

We can simplify this a bit by combining the three factors of \(\|r\|\) we see in the denominator, and expand it out by plugging in \(r=(x,y,z)\) if we like:

\[\begin{align*} \vec{F}&=\frac{-GMm}{\|r\|^3}\vec{r}\\ &=\frac{-GMm}{(x^2+y^2+z^2)^{3/2}}\langle x,y,z\rangle \end{align*}\]

Combining this with \(F=ma\) we see that the two small \(m\)’s cancel: this is exactly Galileo’s insight that the mass of the falling body doesn’t affect its motion!

\[a = \frac{-G M}{\|r\|^{3/2}}\vec{r}\]

Which, when expanded out in \(x,y,z\) gives a more complicated set of equations than we’ve dealt with so far:

\[\pmat{x^{\prime\prime}\\ y^{\prime\prime}\\ z^{\prime\prime}}=\frac{-GM}{(x^2+y^2+z^2)^{3/2}}\pmat{x\\ y\\ z}\]

This set of equations is *much* harder to solve than our earlier case, where we approximated that the earth was so big and we were so close that we could treat the acceleration as only occuring in one direction. Here, the \(x,y,\) and \(z\) all show up in all of the equations! It was an incredible triumph of Newton to actually *solve* these, and show that planets orbit the sun on Ellipses - a fact that had been known since Kepler from observation, but had no theroetical understanding. Look in our textbook, in the last section of Chapter 13, for a rather in-depth treatment of this!

Here I do not want to try and actually sovle this system, but rather showcase another technique - sometimes its possible to partially *guess a solution*, and then use an equation to fix things up until it works!

### 8.2.2 The Length of a Year

Imagining a physical system like the earth orbiting the sun - we know that our particlar orbit isn’t perfectly circular, but it sure is close! This might make us wonder “are there any perfectly circular orbits”? Let’s put our circle in the \(xy\) plane, so \(z=0\). If the circle is radius \(R\), we would be proposing an orbit of the form

\[r(t)=(R\cos t, R\sin t, 0)\]

Does this solve our equation? To check, we need to plug these in for \(x,y,\) and$ z$ to see if it comes out true. Taking the second derivative gives us the proposed left side:

\[\langle x^{\prime\prime}, y^{\prime\prime},z^{\prime\prime}\rangle = \langle -R \cos t, -R\sin t, 0\rangle\]

Now computing the right hand side, we first note that \(x^2+y^2+z^2=R^2\) (since the path is a circle of radius \(R\)) This makes it easier to simplify things:

\[\frac{-GM}{R^{3}}\langle R\cos t,R\sin t, 0\rangle\]

If this circle *was* a solution: these two sides should be equal! But are they? No, unfortunately not. They have the right functions, and the right sign, but one side has a jumble of constants out from that the other does not!

How can we fix this? It turns out we overlooked one thing: the *speed* of our planets orbit! We just used sine and cosine directly - which traverse the unit circle at *unit speed*. This made our orbit traced out at speed \(R\): meaning that the farther our planet was from the sun the faster it was going! This isn’t anything like what we see in the world around us (Earth takes 1 year, whereas pluto goes much slower and takes 120). So, we shoud try to “adjust” our solution by allowing the speed to vary. As we saw when working with helices in class, we can change the speed by multipying the time paramter by a constant. So our next guess is

\[r(t)=(R\cos(kt),R \sin(kt),0)\]

Re-running the same math, everythign is the same except the second derivative now has a factor of \(k^2\) out from (one \(k\) pops out from each derivative, by the chain rule). Thus we have

\[k^2\pmat{- R\cos(kt)\\ - R\sin(kt)\\ 0}=\frac{GM}{R^{3}}\pmat{ -R\cos(kt)\\ -R\sin(kt)\\0 }\]

These are equal precisely when the constants are equal: thus we have found \[k^2=\frac{GM}{R^{3}}\hspace{0.5cm}\implies\hspace{0.5cm}k=\sqrt{GM/R^{3}}\]

After this adjustment, we managed to describe a precise circular orbit for a planet around a star! What can we learn from this? Well, the functions \(\cos t\) and \(\sin t\) repeat every \(2\pi\), so the functions \(\cos(kt)\) and \(\sin(kt)\) repeat every \(2\pi/k\) time units. This means the length of a year for our planet at radius \(R\) is

\[\mathrm{Year}=\frac{2\pi}{k}=\frac{2\pi}{\sqrt{\frac{GM}{R^{3}}}}=\frac{2\pi}{\sqrt{GM}}R^{3/2}\]

That is, for a circular orbit there’s a relationship between the year length, the radius of the orbit, and the mass of the planet. And since realistic orbits are near-ish to circles, this should approximately hold for other orbits (in fact, doing more math we figure out it *exatly* holds, if you replace the radius with the longer axis of an ellipse).

Why is this important? Well - year length and radius are two things that we can measure from earth with telescopes: but masses of objects that are light years away are much harder to pin down. And this relationship gives us a way to find the mass without ever leaving our home!

\[M=\frac{4\pi^2}{G}\frac{R^3}{\mathrm{Year}}\]

Astrophysicts can then take this information to learn quite alot: once you get the star’s mass, you can use this plus one other measurement (the radial velocity, or slow movement of the star) to find the planet’s mass itself!

### 8.2.3 Living in 3-dimensions

(You are not responsible for this, so I will not write much detail here: but I may discuss it a bit in class, if I have time).

### 8.2.4 The 3-body Problem

(You are not responsible for this, so I will not write much detail here: but I may discuss it a bit in class, if I have time).

## 8.3 Electromagnetism

The force of gravity isn’t the only thing that can be understood by Newton’s paradigm of forces and accelerations. In the 1800s a mathematical understanding of electricity and magnetism began to come together. We learned that in addition to mass, certain particles have a *charge*, which lets them interact with the Electirc Field \(\vec{E}\), and the Magnetic Field \(\vec{B}\).

Each of these fields is described by a vector at each point in space, describing how strong (and in which direction) electiricty and magnetism point there.

PIC MAGNETIC FIELD

But how *does* a charged particle interact with these fields? The right answer turns out to be the Lorentz Force Law:

\[\vec{F}=q(\vec{E}+\vec{v}\times \vec{B})\] Where \(q\) is the particles charge and \(v\) is the particle’s veloicty!

In general particles are exposed to both electric and magnetic fields, but to give us some practice working with parametric curves, we will consider the two cases separately.

### 8.3.1 Charges in an Electric Field

First, consider the case that we have a particle in a constant electric field \[\vec{E}=\langle E_x,E_y,E_z\rangle\] How does the particle behave? Recalling that \(F=ma\) and \(a\) is the second derivative of position, this gives us a system of equations:

\[m\pmat{x^{\prime\prime}\\ y^{\prime\prime}\\ z^{\prime\prime}}=q\pmat{E_x\\ E_y\\ E_z}\]

Dividng by \(m\), we see that this system of equations just says that the acceleration of \(x,y,\) and \(z\) are all constant! But we know how to solve this type of equation: we saw it with our earliest gravity model. For instance, in the \(x\)-direction we have

$\(x^{\prime\prime} = -\frac{q E_x}{2 m}t^2+v_xt+x_0\) And similarly for the \(y\) and \(z\) components. Thus we can predict the path of particles in an eletric field: they get accelerated (speed up), and follow parabolas!

### 8.3.2 Charges in a Magnetic Field

What about a particle in a constant magnetic field? Here the equation becomes \(m\vec{a}=q\vec{v}\times B\). Setting \(B=\langle B_1,B_2,B_3\rangle\), we can write this equation out in full

\[m\pmat{x^{\prime\prime}\\ y^{\prime\prime}\\ z^{\prime\prime}}=q\pmat{x^\prime \\ y^\prime \\ z^\prime }\times\pmat{B_x\\ B_y\\ B_z}\]

If we multiply out that cross product, things will get messy quick and it will definitely look difficult to solve this equation! But - as we are imagining a constant magnetic field anyway - why don’t we arrange it so that its pointing just along the \(z\)-axis? Then our magnetic vector becomes \(\langle 0,0,B_z\rangle\), and the cross product simplifes:

\[v\times B = \left|\begin{matrix}\ihat & \jhat &\khat\\ x^\prime & y^\prime & z^\prime\\ 0 & 0& B_z\end{matrix}\right|=\langle y^\prime B_z, - x^\prime B_z,0\rangle\]

Thus, in this special case we get the equations

\[m\pmat{x^{\prime\prime}\\ y^{\prime\prime}\\ z^{\prime\prime}} = qB_z\pmat{y^\prime \\ -x^\prime \\ 0}\]

The last of these equations is easy to solve: \(z^{\prime\prime}=0\) means that \(z(t)=v_zt+z_0\) as for our simple model of gravity. So, \(z\) travels linearly at a constant speed. But what do \(x\) and \(y\) do? This is a bit tricky, as all we know is how the second derivatives relate to the first!

To reduce some of the primes, lets write \(v_x\) and \(v_y\) for \(x^\prime\) and \(y^\prime\). Then, dividing by \(m\) and setting the big constant \(qB_z/m=k\) to help simplify notation further, we have the system of equations

\[\begin{matrix}v_x^\prime = kv_y\\ v_y^\prime = -kv_x\end{matrix}\]

How can we solve these? Look what happens if we differentiate the second equation to get \(v_y^{\prime\prime}=-kv_x^\prime\): now the right hand side has \(v_x^\prime\) in it, so we can substitute the first equation in!

\[v_y^{\prime\prime}=-kv_x^\prime = -k(kv)y=-k^2v_y\]

So, whatever \(v_y\) is, its a function whose second derivative is a negative multiple of itself! We know functions like this - sine and cosine! More specifically, \(\sin(kt)\) and \(\cos(kt)\) since we need that \(k^2\) out front.

With this insight, we can quickly find a solution to our equations \[v_x = \sin(kt)\hspace{1cm}v_y=\cos(kt)\]

But these aren’t the solutions we are after - remember these are \(x^\prime\) and \(y^\prime\) - we want \(x\) and \(y\)! So, we must integrate once more:

\[x=\int x^\prime dt = \int \sin(kt)\,dt = \frac{-1}{k}\cos(kt)+x_0\] \[y=\int y^\prime dt = \int \cos(kt)\,dt = \frac{1}{k}\sin(kt)+y_0\]

Putting it all together, what do we have?

\[\pmat{x\\ y\\ z }=\pmat{\frac{-\cos (kt)}{k}\\ \frac{\sin(kt)}{k}\\ v_z t}+\pmat{x_0\\ y_0\\ z_0}\]

What is this? This is a helix!! So **charged particles move on helices through magnetic fields**. What are the radius of the circles that this helix traces out? They’re \(1/k\) where remember \(k=qB_z/m\) is our constant. But this means that \(k\) is the *curvature* of these circles! (Recall last lecture - we derived that the curvature of a circle was the reciprocal of its radius)

This gives us another piece of information to understand particles: **the curvature of the circle a charged particle traces out in a magnetic field is \(qB/m\)** If we know the magnetic field and either the mass or the charge - we can solve for the other so long as we can measure the radius of the helix! This is a way to actually meausre the mass of elementary particles!

But there’s still one secret this equation has yet to give up: our constant \(k\) here is a multiple of the charge, and it also shows up *inside* the sine and cosine. What happens to a parameteric curve like \((\cos t,\sin t)\) when you multiply \(t\) by a negative number? It reverses in direction! Thus, the direction a particle traces out a helix depends on its charge: **positively and negatively charged particles spiral in different directions in the same magnetic field**

### 8.3.3 The Discovery of Antimatter

This is how we discovered antimatter!

## 8.4 Weather

(You are not responsible for this topic either)

\[\pmat{x^\prime \\ y^\prime \\ z^\prime}=\pmat{\sigma y - x\\ rx-y-xz\\ xy-bz}\]

Here are some solutions to the Lorenz equations of \(x,y,z\).

Of course this is *much* easier to interpret what is going on if we instead plot these solutions as a vector curve in three dimensions!

Here’s a program that solves the Lorenz equations in real time, so you can see an initially rather collected set of initial conditions quickly spread out all over the place and get mixed up: this is why it is hard to predict the weather!