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#+title: inverse square
#+author: Preston Pan
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* Derivation
Say you're doing some physics modeling, and you want to describe a force that has the following properties (and one more that I will introduce later,
and many of which inherit from the definition of a [[id:c1e836e3-a80c-459d-8b68-396fa1687177][central force]]):
1. The force happens at a distance between two particles.
2. The force gives both the particles an equal amount of force but in opposite directions (Newton's Third Law).
3. The force magnitude only depends on the distance between the two particles in question, and maybe some other property intrinsic to both the particles.
4. If you draw a straight line between the two points, the force vector has to be parallel to that line.
5. The force works in the same way no matter where in the universe you are.

We will call our mysterious force field \(\vec{f(\vec{r})}\). We also want to consider two particles, and for simplicity we can say these
particles have no volume, so they are just points. These point particles we will call \(P_{1}\) and \(P_{2}\). For simplicity, we will also
consider two dimensions instead of three, realizing that generalizing to arbitrary dimensions in euclidean space is trivial in this case.
Now, the direction of force must be solely dependent on the orientation of the other particle, which follows from property 4.
We call this direction \(\hat{r}\) and the distance between them \(r\); the vector that represents the direction and the distance together
we are going to define to be \(\vec{r}\).

If we imagine drawing \(P_{1}\) and \(P_{2}\) randomly on a piece of paper, we can imagine using a compass to draw a circle by centering at \(P_{1}\)
and putting the end of the compass at \(P_{2}\). If we imagine putting \(P_{2}\) at any other point on the perimeter of this circle, the force magnitude remains the
same by property 3. This is what is called radial symmetry. Note that this is true in three dimensions as well, only in three dimensions
we will have to use a sphere. Now, let's say there's another particle further away than \(P_{2}\) called \(P_{3}\). That particle will also have a circle, but this
circle will have a total circumference larger than that of \(P_{2}\). Here we introduce the last property: let's say that these two circles stop becoming hypothetical
and start becoming an infinite amount of particles. If we ignore property 2 for a second, the total force felt by the ring at \(P_{2}\) should be the same as the
total force felt by the ring at \(P_{3}\). In other words, the field gives each ring around it an equal amount of force, or in more technical terms, the flux though
all the possible concentric rings is the same, or the flux through a given concentric ring is independent of \(\vec{r}\).

Think about the implications of this last rule. The circumference of a circle is \(2\pi r\), which means if both rings are getting the same amount of force, in the
case of the \(P_{3}\) ring, we are distributing the same amount of force to a greater amount of particles (it is farther away and the amount of particles is going
to be proportional to the circumference). In other words, we are dividing some amount of force \(k\) by the circumference of a circle \(2\pi r\), or in other words,
the /field intensity/ drops off proportionally to \(\frac{1}{r}\).
Therefore, our force has to look something like:
\begin{align*}
\frac{k}{2\pi r}\hat{r}
\end{align*}
but in three dimensions, that isn't quite right.

Spheres have a surface area of \(4\pi r^{2}\). In three dimensions, instead of dividing among the circumference of a circle, we are dividing among the surface area
of a sphere. Therefore, in three dimensions we get:
\begin{align*}
\frac{k}{4\pi r^{2}}\hat{r}
\end{align*}

by property 3, there might be some property intrinsic to each particle in question that makes the force vector have a higher magnitude (like mass for gravity), so we have to scale
this force by our measurement of that property. By property 2, the force must be the same magnitude but opposite direction on the two particles. In order to
preserve that property, this force must scale by both particles' properties:
\begin{align*}
\vec{f}(\vec{r}) = \frac{kP_{1}P_{2}}{4\pi r^{2}}\hat{r}
\end{align*}
in this article I am going to consider this \(4\pi\) constant as a part of \(k\), so the final form will be:
\begin{align*}
\frac{kP_{1}P_{2}}{r^{2}}\hat{r}
\end{align*}
And from 6 properties that seem like reasonable constraints, we derive the inverse square law.

To formalize the sixth property by taking the flux through a sphere and showing it doesn't depend on \(\vec{r}\):
\begin{align*}
kP_{1}P_{2}\oint\frac{\hat{r}}{r^{2}} \cdot d\vec{a} = kP_{1}P_{2}\frac{1}{r^{2}}\int da = k\frac{P_{1}P_{2} 4\pi r^{2}}{r^{2}} = constant
\end{align*}
Which is a result we will use later.

Also, from here on out, we will be dealing with a field that is not dependent on \(P_{2}\); we will be dealing with systems with many particles and therefore
simply not considering \(P_{2}\) and multiplying it in at the end will be more convenient. Additionally, usually we simply don't know the value of \(P_{2}\);
we usually in the real world are given a system that can interact with arbitrary values of \(P_{2}\), or in other words, in the real world systems often
interact with a lot of different objects, so we want to consider the field of the system independent of the unknown other object that might interact with it,
which leads us to:

* Definition
Inverse square fields for point particles are fields which have the form:
\begin{align*}
\frac{k\sigma}{r^{2}}\hat{r}.
\end{align*}
where \(\vec{r}\) is the distance between the source particle (the particle that gives the force) and test particle (the particle that receives the force),
and sigma defines the strength of the source point. This would be mass in gravity or charge in electrodynamics.

Inverse square laws follow the superposition principle, and therefore:
\begin{align*}
\vec{f}_{total} = \sum_{i = 0}^{n}\vec{f}_{i}
\end{align*}
Which means statements made about fields of single point particles is most likely true of systems of more than one particle.
Inverse square fields also have a continuous distribution form:
\begin{align*}
\vec{f}(\vec{r''}) = \int_{space}\frac{k\sigma(\vec{r'})}{r^{2}}\hat{r}d\tau
\end{align*}
Where sigma becomes a smooth distribution. The integral is for a continuous distribution of points that follow the inverse square law, where
the field generated by each individual point contributes in a weighted sum.

\(\vec{r'}\) is displacement vector of the field source, whereas \(\vec{r''}\) is the displacement of the test particle.  \(\vec{r} := \vec{r''} - \vec{r'}\).
* Divergence of Inverse Square Fields
Using the [[id:4bfd6585-1305-4cf2-afc0-c0ba7de71896][del operator]]:
\begin{align*}
\vec{\nabla} \cdot \vec{f(\vec{r''})} = \int_{space}k\sigma(\vec{r'}) * \vec{\nabla} \cdot \frac{\hat{r}}{r^{2}}d\tau
\end{align*}
In this example, I will only be using the Cartesian del operator because although the calculation is harder
the math needed in order to understand Cartesian coordinates is lesser.
If we are using the Cartesian del operator, we must also use Cartesian coordinates. Therefore, we do unit
conversions to replace \(\vec{r}\):
\begin{align*}
\vec{\nabla} \cdot \vec{f} = k * \int_{space} \sigma(\vec{r'}) * \vec{\nabla} \cdot \frac{1}{x^{2} + y^{2} + z^{2}}\frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^{2} + y^{2} + z^{2})^{\frac{1}{2}}}d\tau
\end{align*}
If we can solve for one of these partial derivative terms, we can solve for the other two by symmetry:
\begin{align*}
= k\int_{space} \sigma(\vec{r'}) * \vec{\nabla} \cdot (\frac{x\hat{i}}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}} + \frac{y\hat{j}}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}} + \frac{z\hat{k}}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}})d\tau \\
= k\int_{space}\sigma(\vec{r'})(\frac{\partial}{\partial x}\frac{x}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}} + \frac{\partial}{\partial y}\frac{y}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}} + \frac{\partial}{\partial z}\frac{z}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}})d\tau
\end{align*}
We use the chain rule to solve for this:
\begin{align*}
\frac{\partial}{\partial x}x(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} = (x^{2} + y^{2} + z^{2})^{-3/2} - x * 2x * \frac{3}{2} (x^{2} + y^{2} + z^{2})^{-\frac{5}{2}} \\
= (x^{2} + y^{2} + z^{2})^{-3/2} - 3x^{2}(x^{2} + y^{2} + z^{2})^{-\frac{5}{2}}
\end{align*}
Then it will look like this inside the integral:
\begin{align*}
k\int_{space}\sigma(\vec{r'})((x^{2} + y^{2} + z^{2})^{-3/2} - 3x^{2}(x^{2} + y^{2} + z^{2})^{-\frac{5}{2}} + (x^{2} + y^{2} + z^{2})^{-3/2} - 3y^{2}(x^{2} + y^{2} + z^{2})^{-\frac{5}{2}} + (x^{2} + y^{2} + z^{2})^{-3/2} - 3z^{2}(x^{2} + y^{2} + z^{2})^{-\frac{5}{2}})d\tau
\end{align*}
If we factor out the \((x^{2} + y^{2} + z^{2})^{-\frac{5}{2}}\) term and collecting the like terms:
\begin{align*}
k\int_{space}\sigma(\vec{r'})(3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} - 3(x^{2} + y^{2} + z^{2})(x^{2} + y^{2} + z^{2})^{-\frac{5}{2}})d\tau \\
= k\int_{space}(3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}} - 3(x^{2} + y^{2} + z^{2})^{-\frac{3}{2}})d\tau \\
= 0.
\end{align*}
So is the divergence of this field zero? Well, not exactly. In order to understand why, we must ask: What happens to the divergence at \(\vec{0}\)?
On first glance, it seems clearly undefined. After all, we're dividing by zero. However, after some amount of inspection, the assumption that our field's divergence
is zero everywhere is incoherent. As we know from section one, he surface integral over a sphere of a point charge (flux) is a constant,
but if the [[id:44e65b69-e5d5-464a-b1f3-8a914e1b7e9e][divergence theorem]] is true, then the surface integral should also be zero, not constant. Because this flux is not \(\vec{r}\) dependent and therefore
is the same no matter how small the concentric ring is, the flux must come from the origin.

We can model this behavior with the [[id:90574fea-88f4-4b80-9cda-32cff0bcb76d][dirac delta]] distribution. Our actual divergence is:
\begin{align*}
k\sigma(\vec{r'})4\pi\delta^{3}(\vec{r})
\end{align*}
So now we can take the volume integral of this quantity:
\begin{align*}
\vec{\nabla} \cdot \vec{f}(\vec{r''}) = 4\pi k\int_{space}\delta^{3}(\vec{r''} - \vec{r'})\sigma(\vec{r'})d\tau \\
= 4\pi k\sigma(\vec{r''})
\end{align*}
As an analogy, if we have a point mass (an object with finite mass in a single point), the density of any volume containing the point will be zero, but
the gravitational field generated by that point mass will not be zero. Obviously just taking the density blindly will not accurately account for the
actual distribution of mass within space. You need to use something like the [[id:90574fea-88f4-4b80-9cda-32cff0bcb76d][dirac delta]] distribution to model this behavior. Also, since the \(k\) constant
includes the term \(\frac{1}{4\pi}\), we can just say:
\begin{align*}
\vec{\nabla} \cdot \vec{f}(\vec{r''}) = k\sigma(\vec{r''})
\end{align*}

Where \(k\) in this case is a constant without the \(4\pi\) term. By the [[id:44e65b69-e5d5-464a-b1f3-8a914e1b7e9e][divergence theorem]]:
\begin{align*}
\int_{V}\vec{\nabla} \cdot \vec{f}d\tau = \int_{S}\vec{f} \cdot d\vec{a} = k\int_{V}\sigma(\vec{r''})d\tau = k\sigma_{enc.}
\end{align*}

* Curl of Inverse Square Fields
Because an inverse square field is a special case of a [[id:c1e836e3-a80c-459d-8b68-396fa1687177][central force]], this result is also the same as for all general central fields. That is:
\begin{align*}
\vec{\nabla} \times \vec{f} = \vec{0}
\end{align*}
Which implies the field in question is a [[id:6f2aba40-5c9f-406b-a1fa-13018de55648][conservative force]]:
\begin{align*}
\oint \vec{f} \cdot d\vec{l} = 0
\end{align*}
It also implies:
\begin{align*}
\vec{f} = -\vec{\nabla}V
\end{align*}
Where \(V\) is known as the potential of this field.

* Potential of Inverse Square fields
Now we want to find the specific potential scalar field for inverse square fields. In an inverse square field,
given the identity:
\begin{align*}
\int_{\vec{a}}^{\vec{b}}\vec{f} \cdot d\vec{l} = V(\vec{a}) - V(\vec{b})
\end{align*}
We can set a reference point as \(\vec{a}\) -- something analogous to setting sea level or setting your coordinate system. Here, we set it infinitely far away and integrate
in a straight line such that the path is parallel to \(\vec{r}\), with respect to a point particle at the origin:
\begin{align*}
k\sigma\int_{\vec{\infty}}^{\vec{b}}\frac{1}{r^{2}}\hat{r} \cdot d\vec{l} = V(\vec{\infty}) - V(\vec{b}) \\
k\sigma\int_{\infty}^{b}\frac{1}{r^{2}}dr = V(\vec{\infty}) - V(\vec{b}) \\
-\frac{\sigma}{4\pi \epsilon_{0}r}\Big|^{b}_{\infty} = - \frac{k\sigma}{\infty} + \frac{k\sigma}{b} = \frac{k\sigma}{b} \\
V(\vec{r}) := \frac{k\sigma}{r}
\end{align*}
It is easy to prove that \(V\) follows the superposition principle. Thus, we can define a continuous distribution for the potential field as well:
\begin{align*}
V(\vec{r}) := k\int_{space}\frac{\sigma(r')}{r}d\tau
\end{align*}
Note that because this field does not require keeping track of vector orientation, it is significantly easier to solve for \(V\) then convert to \(\vec{f}\). Additionally,
setting a reference point to something that is not infinity would be valid as well -- we just choose infinity because it cancels off the constant term. However,
the /difference/ in potentials is absolute and does not require any constant adjustment. Then, the divergence of inverse
square fields can be reformulated with the [[id:65004429-a6b7-41f2-8489-07605841da3d][Laplacian]] operator:
\begin{align}
\label{}
\nabla^{2}V(\vec{r''}) = k\sigma(\vec{r''})
\end{align}