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Section 4.2 Applications of Multiple Integration

Subsection 4.2.1 Definitions for Mass and Moments

I've done a bunch of volume calculations so far. Similar but more practical is the calculation of the mass of a solid object. If the object has a fixed desnity, then its mass is just this constant density multiplied by volumes. However, sometimes density is variables. The first application I want to introduce for multiples integrals is calculating the mass of an object with variable density.

Definition 4.2.1.

If \(\rho(x,y,z)\) is a density function, integrable on a region \(S \in \RR^3\text{,}\) the mass of the object \(S\) is found from by integrating the density function.

\begin{equation*} m = \int_S \rho dV \end{equation*}

This can also be done in \(\RR^2\) for flat objects with variable density depending on only \(x\) and \(y\text{.}\) Such objects are normally refered at laminae (the singular is lamina).

A similar interesting physical problem is the problem of centre of mass. The centre of mass of an rigid object (or system) is the point in space where linear forces on the object can be accuratly modeled as forces on a point-mass. In particular, forces on the centre of mass do not cause rotational acceleration. The calculation of centre of mass involves the calculation of so-called first moments. These definition vary by dimension. First I will defined moments for a lamina in \(\RR^2\text{.}\)

Definition 4.2.2.

Let \(S\) be a region in \(\RR^2\) with density \(\rho(x,y)\) and mass \(m\text{.}\) Its first moments are defined b the following integrals.

\begin{align*} M_x \amp = \int_S y \rho dA\\ M_y \amp = \int_S x \rho dA \end{align*}

Then the coordinates of the centre of mass are written \((\bar{x}, \bar{y})\) and calculated from the moments.

\begin{align*} \bar{x} \amp = \frac{M_x}{m}\\ \bar{y} \amp = \frac{M_y}{m} \end{align*}

If \(S\) is a solid region in \(\RR^3\) with density \(\rho(x,y,z)\) and mass \(m\text{,}\) then its first moments are defined by the following integrals.

\begin{align*} M_{yz} \amp = \int_S x \rho dV\\ M_{xz} \amp = \int_S y \rho dV\\ M_{xy} \amp = \int_S z \rho dV \end{align*}

The coordinates of the centre of mass are written \((\bar{x}, \bar{y}, \bar{z})\) and calculated from the moments.

\begin{align*} \bar{x} \amp = \frac{M_{yz}}{m}\\ \bar{y} \amp = \frac{M_{xz}}{m}\\ \bar{z} \amp = \frac{M_{xy}}{m} \end{align*}

Centre of mass and first moments are important physical properties that deal with linear acceleration; for the purpose of linear acceleration, the object acts like a point-mass at the centre of mass. However, there are also moments involved in rotational movement and acceleration. These are called second moments or moments of intertia.

Definition 4.2.3.

Let \(S \subset \RR^2\) be a laminae with density function \(\rho\text{.}\) Its second moments are calcaulted by the following integrals.

\begin{align*} I_x \amp = \int_S y^2 \rho dA\\ I_y \amp = \int_S x^2 \rho dA\\ I_0 \amp = \int_S (x^2+y^2) \rho dA \end{align*}

These moment measure the resistance to rotation: \(I_x\) is the resistance to rotation about the \(x\) axis; \(I_y\) is the resistance to rotation about the \(y\) axis; and \(I_0\) is resistance to rotation about the origin. The three radii of gyration are calcluated from the second moments.

\begin{align*} \bar{\bar{x}}^2 \amp = \frac{I_x}{m}\\ \bar{\bar{y}}^2 \amp = \frac{I_y}{m}\\ \bar{\bar{R}}^2 \amp = \frac{I_0}{m} \end{align*}

The radii of gyration are similar to centre of mass for movement. The object acts as a point mass at radius \(\bar{\bar{R}}\) for rotation about the origin. Similar, it acts like a point mass at radius \(\bar{\bar{x}}\) for rotation about the \(x\) axis and radius \(\bar{\bar{y}}\) for rotation about the \(y\) axis.

Let \(S\) be a solid object in \(\RR^3\) with density function \(\rho\text{.}\) Its second moments are calculated by the following integrals.

\begin{align*} I_x \amp = \int_S (y^2 + z^2) \rho dV\\ I_y \amp = \int_S (x^2 + z^2) \rho dV\\ I_z \amp = \int_S (x^2 + y^2) \rho dV \end{align*}

Its radii of gyration are calculated from the second moments.

\begin{align*} \bar{\bar{x}}^2 \amp = \frac{I_x}{m}\\ \bar{\bar{y}}^2 \amp = \frac{I_y}{m}\\ \bar{\bar{z}}^2 \amp = \frac{I_z}{m} \end{align*}

The object acts like a point mass at radius \((\bar{\bar{x}}\) for purposes of rotation around the \(x\) axis, and similarly for the other two axes.

Subsection 4.2.2 Mass and Moments Examples

Consider a quarter circle of radius \(a\) in the first quadrant with density function \(\rho = k\sqrt{x^2 + y^2}\text{.}\) What is its centre of mass? I need to calculate the mass, the first moments, and then the coordinates of the centre of mass. Since there is circular geometry here, I do all the integrals in polar coordinates; in thses coordinates the density function is \(\rho = kr\text{.}\) In each integral, I have to remember the Jacobian of polar coordinates.

\begin{align*} m \amp = \int_0^{\frac{\pi}{2}} \int_0^a krrdr d\theta\\ \amp = \frac{\pi}{2} \frac{kr^3}{3} \Bigg|_0^a = \frac{\pi ka^3}{6}\\ \bar{x} \amp = \frac{1}{m} \int_0^{\frac{\pi}{2}} \int_0^a y \rho dA = \frac{1}{m} \int_0^{\frac{\pi}{2}} \int_0^a kr^3 \sin \theta dr d\theta\\ \amp = \frac{k}{m} \int_0^{\frac{\pi}{2}} \sin \theta d\theta \int_0^a r^3 dt\\ \amp = \frac{k}{m} \left( -\cos \theta \right) \Bigg|_0^{\frac{\pi}{2}} \frac{r^4}{4} \Bigg|_0^a = \frac{6}{\pi a^3} 1 \frac{a^4}{4} = \frac{3a}{2\pi}\\ \bar{y} \amp = \frac{3a}{2\pi} \end{align*}

I don't actually have to calculate the second moment; \(\bar{x} = \bar{y}\) due to the symmetry of the situation. Whenever possible, I like to look for symmetries that can save me some calculation. The centre of mass is found at \(\left( \frac{3a}{2\pi}, \frac{3a}{2\pi} \right)\text{.}\)

Now consider a lamina with \(y \in [-1,1]\) and \(x\) bounded between \(\pm y^4\) with \(\rho = 1\text{.}\) What is the area and moment of intertia about the \(x\)-axis? I need to calculate the mass and then the second moments. I do these integrals in cartesian coordinates.

\begin{align*} A \amp = \int_{-1}^1 \int_{-y^4}^{y^4}1 dx dy\\ \amp = \int_{-1}^1 2y^4 dy = \frac{2y^5}{5} \Bigg|_{-1}^1 = \frac{4}{5}\\ I_x \amp = \int_{-1}^1 \int_{-y^4}^{y^4} y^2 dx dy\\ \amp = \int_{-1}^1 xy^2 \Bigg|_{-y^4}^{y^4} = \int_{-1}^1 2y^6 dy = \frac{2y^7}{7} \Bigg|_{-1}^1 = \frac{4}{7} \end{align*}

Now I want to compare this lamina to a rectangle of height \(2\) and width \(\frac{2}{5}\text{,}\) which has the same area. I can quickly calculate the moment of inertia for the rectangle.

\begin{equation*} I_x = \int_{-1}^1 \int_{-\frac{1}{5}}^\frac{1}{5} y^2 dx dy = \frac{4}{15} \end{equation*}

The previous shape has twice the resitance to rotation compare with the rectangle, even though it has the same cross-section area. The reasons for the higher moment of intertia is that the area is distributed futher away from the origin; things that are farther away are harder to cause to rotate. This physical fact demonstrated in this example partially explains the use of \(I\) beams in construction: they have more resistance to twisting and shear forces than a rectangular cross-section beam of the same size or weight, since the material is distributed further away from the central axis of a twist or shear force.

Now consider a hemisphere above the \(xy\) plane with radius \(a\) and density \(\rho = kz\text{.}\) I will calculate all the properties here: mass, centre of mass, and moments of inertia. Since I am calculating over a hemisphere, I'll be using spherical coordinates. Wherever possible, I'll use symmetry to simplify things. In particular, for first moments, if the mass is equally distributed on either side of an axis, that moment must be zero due to symmetry.

\begin{align*} m \amp = \int_0^{2\pi} \int_0^{\pi/2} \int_0^a k (r\cos \phi) r^2 \sin \phi dr d\phi d\theta\\ \amp = 2\pi k \int_0^{\pi/2} \frac{r^4}{4} \Bigg|_0^a \cos \phi \sin \phi d\phi\\ \amp = \frac{2\pi ka^4}{4} \int_0^{\pi/2} \frac{\sin 2\phi}{2} d\phi\\ \amp = \frac{\pi ka^4}{2} \left( \frac{-\cos2\phi}{4} \right) \Bigg|_0^{\pi/2} = \frac{\pi ka^4}{8} ( -\cos \pi + \cos 0) = \frac{\pi ka^4}{4}\\ M_{yz} \amp = 0 \text{ due to symmetry. }\\ M_{xz} \amp = 0 \text{ due to symmetry. }\\ M_{xy} \amp = \int_D z \rho dV\\ \amp = \int_0^{2\pi} \int_0^{\pi/2} \int_0^a kr^2 \cos^2\phi r^2 \sin \phi dr d\phi d\theta.\\ \amp = 2\pi k \int_0^a r^4 dr \int_0^{\pi/2} \cos^2 \phi \sin \phi d \phi\\ \amp = \frac{2\pi k a^5}{5} \left( \frac{-\cos^3 \phi}{3} \right) \Bigg|_0^{\pi/2} = \frac{2\pi k a^5}{15}\\ \bar{x} \amp = 0\\ \bar{y} \amp = 0\\ \bar{z} \amp = \frac{\frac{2\pi ka^5}{15}}{\frac{\pi ka^4}{4}} = \frac{8a}{15} \end{align*}

The centre of mass is at \((0,0, \frac{8a}{15})\text{.}\) Now I will calculate the moments of inertia.

\begin{align*} I_x \amp = \int_D (y^2 + z^2) \rho dV = \int_D (y^2 + z^2) z k dV\\ \amp = k \int_0^{2\pi} \int_0^{\pi/2} \int_0^a r^2 (\sin^2 \phi \sin^2 \theta + \cos^2 \phi) r \cos \phi r^2 \sin\phi drd\phi d\theta\\ \amp = k \int_0^a r^5 dr \int_0^{2\pi} \int_0^{\pi/2} \left( \sin^3 \phi \cos \phi \sin^2 \theta + \cos^3 \phi \sin \phi \right) d\phi d\theta\\ \amp = \frac{ka^6}{6} \left[ \left( \frac{\sin^4 \phi}{4} \right) \Bigg|_0^{\pi/2} \left( \frac{\theta}{2} - \frac{\sin 2\theta}{4} \right) \Bigg|_0^{2\pi} + \left( \frac{-\cos^4 \phi}{4} \right) \Bigg|_0^{\pi/2} 2\pi \right]\\ \amp = \frac{ka^6}{6} \left[ \left( \frac{1}{4} - 0 \right) \pi _ 2\pi \left( \frac{1}{4} - 0 \right) \right] = \frac{\pi ka^6}{8}\\ I_y \amp = \frac{\pi ka^6}{8} \text{ by symmetry }\\ I_z \amp = \int_D (x^2 + y^2) \rho dV\\ \amp = k \int_0^{2\pi} \int_0^{\pi/2} \int_0^a r^2 \sin^2 \phi r \cos \phi r^2 \sin \phi dr d\phi d\theta\\ \amp = \frac{2\pi ka^6}{6} \int_0^{\pi/2} \sin^3 \phi \cos \phi d\phi\\ \amp = \frac{\pi ka^6}{3} \frac{\sin^4 \phi}{4} \Bigg|_0^{\pi/2} = \frac{\pi ka^6}{12}\\ \bar{\bar{x}}^2 \amp = \frac{\frac{\pi ka^6}{8}}{\frac{\pi ka^4}{4}} = \frac{a^2}{2}\\ \bar{\bar{x}} \amp = \frac{a}{\sqrt{2}}\\ \bar{\bar{y}}^2 \amp = \frac{a^2}{2}\\ \bar{\bar{y}} \amp = \frac{a}{\sqrt{2}}\\ \bar{\bar{z}}^2 \amp = \frac{\frac{\pi ka^6}{12}}{\frac{\pi ka^4}{4}} = \frac{a^3}{3}\\ \bar{\bar{z}} \amp = \frac{a}{\sqrt{3}} \end{align*}

The centre of rotation is at \(\left( \frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}, \frac{a}{\sqrt{3}} \right)\text{.}\)

Now consider a parabaloid bounded by \(z = b(x^2 + y^2)\) with height \(h\) and density \(\rho = 1\text{.}\) What are its moments of inertia? I will use cylindrical coordinates, which are well suited to describe the parabaloid. However, I need a bound on the radius term at the height \(h\text{.}\) The equation of the parabaloid in cylindrical coordinates is \(z = br^2\) and when \(z=h\text{,}\) I see that \(r = \sqrt{\frac{h}{b}}\text{.}\) This is the outer bound on radius. With contstant bounds on angle and radius, I can let \(z\) range from the parabaloid graph \(z=br^2\) to the constant height \(h\text{.}\)

\begin{align*} m \amp = \int_S 1 dV = \int_0^{2\pi} \int_0^{\sqrt{\frac{h}{b}}} \int_{br^2}^h r dz dr d\theta\\ \amp = \int_0^{2\pi} d\theta \int_0^{\sqrt{\frac{h}{b}}} z \Bigg|_{br^2}^h rdr\\ \amp = 2\pi \int_0^{\sqrt{\frac{h}{b}}} (h-br^2)rdr \\ \amp = 2\pi \left( \frac{hr^2}{2} - \frac{br^4}{4} \right) \Bigg|_0^{\sqrt{\frac{h}{b}}} = 2\pi \left( \frac{h^2}{2b} - \frac{bh^2}{4b^2} \right) \\ \amp = 2\pi \frac{h^2}{b} \left( \frac{1}{2} - \frac{1}{4} \right) = \frac{\pi h^2}{2b}\\ I_z \amp = \int_0^{2\pi} \int_0^{\sqrt{\frac{h}{b}}} \int_{br^2}^h (x^2 + y^2) r dz dr d\theta\\ \amp = \int_0^{2\pi} \int_0^{\sqrt{\frac{h}{b}}} \int_{br^2}^h r^3 dz dr d\theta\\ \amp = 2\pi \int_0^{\sqrt{\frac{h}{b}}} \left( r^3h - r^5 b \right) dr\\ \amp = 2\pi \left( \frac{hr^4}{4} - \frac{br^6}{6} \right) \Bigg|_0^{\sqrt{\frac{h}{b}}}\\ \amp = 2\pi \left( \frac{h^3}{4b^2} - \frac{bh^3}{6b^3} \right) = \frac{2\pi h^3}{b^2} \left( \frac{1}{4} - \frac{1}{6} \right) = \frac{\pi h^3}{6b^2}\\ \bar{\bar{z}}^2 \amp = \frac{I_z}{m} = \frac{\frac{\pi h^3}{6b^2}}{\frac{\pi h^2}{2b}} = \frac{h}{3b}\\ \bar{\bar{z}} \amp = \sqrt{\frac{h}{3b}}\\ I_x \amp = \int_0^{2\pi} \int_0^{\sqrt{\frac{h}{b}}} \int_{br^2}^h (y^2 + z^2) r dz dr d\theta\\ \amp = \int_0^{2\pi} \int_0^{\sqrt{\frac{h}{b}}} \int_{br^2}^h (r^2 \sin^2 \theta + z^2) r dz dr d\theta\\ \amp = \int_0^{2\pi} \int_0^{\sqrt{\frac{h}{b}}} r^3 \sin^2 \theta \left( h- br^2 \right) + r \left( \frac{h^3}{3} - \frac{b^3r^6}{3}\right) dr d\theta\\ \amp = \int_0^{2\pi} \frac{r^4}{4} h \sin^2 \theta - \frac{r^6}{6} b \sin^2 \theta + \frac{r^2}{2} \frac{h^3}{3} - \frac{r^8}{8} \frac{b^3}{3} \Bigg|_0^{\sqrt{\frac{h}{b}}} d \theta\\ \amp = \int_0^{2\pi} \frac{h^2}{4b^2} h \sin^2 \theta - \frac{h^3}{6b^3} b \sin^2 \theta + \frac{h^4}{6b} - \frac{h^4 b^3}{24 b^4} d\theta\\ \amp = \frac{h^3}{b^2} \int_0^{2\pi} \sin^2 \theta d \theta \left( \frac{1}{4} - \frac{1}{6} \right) + \frac{2\pi h^4}{b} \left( \frac{1}{6} - \frac{1}{24} \right)\\ \amp = \frac{h^3}{12b^2} \left( \frac{\theta}{2} - \frac{\sin 2\theta}{4} \right) \Bigg|_0^{2\pi} + \frac{3\pi h^4}{12b}\\ \amp = \frac{h^3 \pi}{12b^2} + \frac{3\pi h^4}{12b} = \frac{h^3 \pi}{12b} \left( \frac{1}{b} + 3h \right)\\ \bar{\bar{x}}^2 \amp = \frac{ \frac{h^3\pi}{12b} \left( \frac{1}{b} + 3h \right)}{\frac{\pi h^2}{2b}} = \frac{h}{6} \left( \frac{1}{b} + 3h \right) \\ \bar{\bar{x}} \amp = \sqrt{\frac{h}{6} \left( \frac{1}{b} + 3h \right)}\\ \bar{\bar{y}} \amp = \bar{\bar{x}} \text{ by symmetry } \end{align*}

As a final example for this section, I can prove a a nice theorem from physics. This is a fact that is used all through high school and first year physics: large objects (like stars, planets, etc) are treated as point masses for the purpose of the force of gravity. This fact seemed intuitive, but now I actually want to formally prove it. I can't, unfortunately, construct a general proof at this time, but I will prove it for the special case of a sphere.

I'll just prove the theorem for the special case of a sphere. The force of gravity between two masses \(m_1\) and \(m_2\) is

\begin{equation*} F = \frac{Gm_1m_2}{r^2} \end{equation*}

I'll assume the source of the gravitational attraction sits at \((0,0,c)\) where \(c \gt a\) is larger than \(a\) the radius of the sphere. I'll also assume that \(\rho\) is the constant density of the sphere. Then \(\rho dV\) is an infinitesimal piece of mass in the sphere, and I can write \(\omega\) for the distance from the infinitesimal mass to the gravitational source.

The force on the infinitesimal mass is

\begin{equation*} F dV = \frac{Gm\rho dV}{\omega^2}\text{.} \end{equation*}

I'll write this in polar coordinates.

\begin{equation*} F dV = \frac{Gm\rho r^2 \sin \phi dr d\phi d\theta}{\omega^2}\text{.} \end{equation*}

By symmetry of the sphere, all lateral forces will cancel. I only care about the \(z\) component of the force. Let \(\alpha\) be the angle from the \(z\) axis of the line from \((0,0,c)\) to the infinitesimal mass. Then I can rewrite the force again.

\begin{equation*} F_z dV = \frac{Gm\rho r^2 \sin \phi \cos \alpha dr d\phi d\theta}{\omega^2} \end{equation*}

Now I'll do some trigonometry. Consider the triange with vertices \((0,0,c)\text{,}\) \((0,0,0)\) and the location of our infinitesimal mass. The angle at \((0,0,c)\) is \(\alpha\) by definition, and likewise the angle at \((0,0,0)\) is \(\phi\text{.}\) The side lengths are \(c\text{,}\) \(\omega\) and \(r\text{,}\) also by definition. I will also use the cosine law.

\begin{align*} \omega^2 \amp = r^2 + c^2 - 2rc \cos \phi\\ \omega \amp = \sqrt{r^2 + c^2 - 2rc \cos \phi} \end{align*}

The length \(c\) along the \(x\) axis can be divided into two pieces, so that \(c = \omega \cos \alpha + r \cos \phi\text{.}\) I can solve for \(\cos \alpha\text{.}\)

\begin{equation*} \cos \alpha = \frac{ c - r \cos \phi}{\omega} \end{equation*}

I can replace \(\omega\) with the square root expression.

\begin{equation*} \cos \alpha = \frac{ c - r \cos \phi}{\sqrt{r^2 + c^2 - 2rc \cos \phi}} \end{equation*}

Now I can replace both \(\cos \alpha\) and \(\omega\)in the force expression.

\begin{equation*} F_z dV = \frac{Gm\rho r^2 \sin \phi}{r^2 + c^2 - 2rc\cos \phi} \left( \frac{c - r\cos \phi}{\sqrt{r^2 + c^2 - 2rc \cos \phi}} \right) dr d\phi d\theta \end{equation*}

The total force is this integral of this infinitesimal force over the sphere.

\begin{equation*} F_z = \int_0^{2\pi} \int_0^{\pi} \int_0^a \frac{Gm\rho r^2 \sin \phi}{r^2 + c^2 - 2rc\cos \phi} \left( \frac{c - r\cos \phi}{\sqrt{r^2 + c^2 - 2rc \cos \phi}} \right) dr d\phi d\theta \end{equation*}

I'll do some substitution here, more or less reversing the trigonometry. I leave \(r\) and \(\theta\) alone, but write \(\omega^2 = r^2 + c^2 - 2rc\cos\phi\) with the cosine law as before.

\begin{equation*} 2\omega d\omega = 2rc \sin \phi d\phi \implies r \sin \phi d\phi = \frac{\omega}{c} d\omega \end{equation*}

The bounds are \(\omega(0) = c-r\) and \(\omega(\pi) = c+r\text{.}\) Finally, I can solve to get \(r\cos \phi = \frac{r^2 + c^2 - \omega^2}{2c}\text{.}\) I use this as a substitution in the \(\phi\) integral.

\begin{align*} F_z \amp = \int_0^{2\pi} \int_0^{\pi} \int_0^a \frac{Gm\rho r}{(r^2 + c^2 - 2rc\cos \phi)} \frac{(c-r\cos \phi)}{\sqrt{r^2+c^2-2rc\cos\phi}} (r\sin \phi d\phi) dr d\theta\\ \amp = \int_0^{2\pi} \int_0^a \int_{c-r}^{c+r} \frac{Gm\rho r}{\omega^2} \frac{c - \frac{r^2 + c^2 - \omega^2}{2c}}{\omega} \frac{\omega}{c} d\omega dr d\theta\\ \amp = \int_0^{2\pi} \int_0^a \int_{c-r}^{c+r} \frac{Gm\rho r}{\omega^2} \frac{2c^2 - r^2 - c^2 + \omega^2}{2c^2} \omega dr d\theta\\ \amp = \frac{2\pi Gm \rho}{2c^2} \int_0^a \int_{c-r}^{c+r} r \left( \frac{c^2 -r^2 +\omega^2}{\omega^2} \right) d \omega dr\\ \amp = \frac{\pi Gm \rho}{c^2} \int_0^a \int_{c-r}^{c+r} r \left( (c^2 -r^2) \frac{1}{\omega^2} + 1 \right) d \omega dr\\ \amp = \frac{\pi Gm \rho}{c^2} \int_0^a r \left( (c^2-r^2) \frac{-1}{\omega} + \omega \right) \Bigg|_{c-r}^{c+r} dr\\ \amp = \frac{\pi Gm\rho}{c^2} \int_0^a (rc^2 - r^3) \left( \frac{1}{c-r} - \frac{1}{c+r} \right) + r (c+r-(c-r)) dr\\ \amp = \frac{\pi G m \rho}{c^2} \int_0^a \frac{-r(c^2-r^2)(-2r)}{c^2-r^2} + 2r^2 dr\\ \amp = \frac{\pi G m \rho}{c^2} 2r^2 + 2r^2 dr\\ \amp = \frac{\pi G m \rho}{c^2} 4r^2 dr = \frac{4\pi Gm \rho a^3}{3c^2} = \frac{Gm \left( \frac{\rho 4 \pi a^3}{3} \right)}{c^2} \end{align*}

The expression in brackets is the mass of the sphere and \(c\) is the distance from the centre of mass to the gravitational sources. This is exactly the expression we wanted: it is the force due to a point mass at the origin with mass equal to the total mass of the sphere.

In the proof, the two masses were seperated from each other. I could instead consider hollow sphere, with outside radius \(a\) and inside radius \(b\text{,}\) and a point mass at \((0,0,c)\) with \(c\lt b\text{,}\) so that the point mass is inside the sphere. What is the force of gravity on that point mass? The set-up is almost the same; the only difference is that the bounds on \(\omega\) are reversed in sign. Let me do the final calculation with this adjustment.

\begin{align*} F_z \amp = \int_0^{2\pi} \int_a^b \int_{r-c}^{r+c} \frac{Gm\rho r (c^2 -r^2 + \omega^2)}{2\omega^2 c^2} d\omega dr d\theta\\ \amp = \frac{\pi G m \rho}{c^2} \int_a^b (r^3-rc^2) \frac{1}{\omega} \Bigg|_{r-c}^{r+c} + r\omega \bigg|_{r-c}^{r+c} dr\\ \amp = \frac{\pi G m \rho}{c^2} \int_a^b r(r^2 -c^2) \left( \frac{1}{r+c} - \frac{1}{r-c} \right) + 2rc dr\\ \amp = \frac{\pi G m \rho}{c^2} \int_a^b r(r^2 -c^2) \frac{-2c}{r^2-c^2} + 2rc dr\\ \amp = \frac{\pi G m \rho}{c^2} \int_a^b -2rc + 2rc dr = \frac{\pi G m \rho}{c^2} \int_a^b 0 dr = 0 \end{align*}

Everything cancels out, and I reach a fairly strange conclusion: no matter the location of a point mass is inside a hollow sphere (of uniform density), it experiences no force of gravity.

Subsection 4.2.3 Moments and Probability Distributions

Let me recall some definitions about continuous probability from previous courses.

Definition 4.2.9.

A probability distribution on an integrable set \(D \subset \RR^n\) is a function \(\rho: D \rightarrow [0, \infty)\) such that

\begin{equation*} \int_D \rho dV = 1\text{.} \end{equation*}

The set \(D\) is called the set of states; each point represents the state of the system. For any integrable subset \(A \subset D\text{,}\) the probability that the state of the system is in the subset \(A\) is the integral over that subset.

\begin{equation*} P(A) = \int_A \rho dV \end{equation*}

Definition 4.2.10.

An integrable function \(f: D \rightarrow \RR\) is called an observable or measurable; for each state, an observable measure some property of that state. The expectation value of the observable \(f\) is the continuous version of the average value of \(f\) over the state. It is written \(\langle f\rangle\) and calculated by integration.

\begin{equation*} \langle f\rangle = \int_D f \rho dV \end{equation*}

If \(D \subset \RR^3\text{,}\) then \(D\) may be a domain of positions. These points in \(D\) are the states of a probabilistic system: are simply the possible places where a particle may be found. Then I cancalculate the expectation of each coordinate, using the definition.

\begin{align*} \langle x\rangle \amp = \int_D x \rho dV\\ \langle y\rangle \amp = \int_D y \rho dV\\ \langle z\rangle \amp = \int_D z \rho dV\\ \langle r \rangle = \langle \sqrt{x^2+y^2+z^2}\rangle \amp = \int_D \sqrt{x^2+y^2+z+^2} \rho dV \end{align*}

The first three of these are the expectation values of each coordinate of position and the last is the expectation value of the distance to the origin.

Alternatively, with \(D \subset \RR^3\text{,}\) \(\rho\) could be an ordinay mass-density. I can think of this as the probability of finding mass at each point in the set. In that context, \(\langle x\rangle\text{,}\) \(\langle y\rangle\) and \(\langle z\rangle\) are just the first moment and the coordinates of the centre of mass. That also makes sense: the centre of mass is the ‘average’ location of the mass of the object.

The second moments are also expectation values. For example, \(\langle x^2 + y^2\rangle\) is the average (square) distance from the \(z\) axis. Objects further away from the \(z\) axis are more difficult to rotate around that axis, so this naturally measure the resistance to rotation around the \(z\) axis. Likewise for \(\langle x^2 + z^2\rangle\) about the \(y\) axis and \(\langle y^2 + z^2\rangle\) about the \(x\) axis.

The use of the term ‘moment’ is historical. The entire subject of continuous probability can be developed using the terminology of moments. For the rest of notes, I'll use the term ‘expectation value’ instead of ‘moment’.

Definition 4.2.11.

The standard deviation of a observable is a measure of the width of the distribution. It can be expressed as the expectation value of the distance from the average. In practice, I use the square of distance and calculate the square of standard deviation (much like the pythagorean theorem). Let \(f\) be a observable. Its standard deviation \(\sigma_f\) is calculate by the following integral.

\begin{equation*} \sigma^2_f = \int_D (f-\langle f\rangle)^2 \rho dV \end{equation*}

With some algebra, I can uncover some interesting properties of standard deviation.

\begin{align*} \sigma_f^2 \amp = \int_D (f - \langle f\rangle)^2 \rho dV\\ \amp = \int_D (f^2 - 2 f\langle f\rangle + \langle f\rangle^2)\rho dV\\ \amp = \int_D f^2 \rho dV - 2 \int_D f \langle f\rangle \rho dV + \int_D \langle f\rangle^2 \rho dV\\ \amp = \langle f^2\rangle - 2 \langle f\rangle \int_D f \rho dV + \langle f\rangle^2 \int_D \rho dV\\ \amp = \langle f^2\rangle - 2 \langle f\rangle \langle f\rangle + \langle f\rangle^2 = \langle f^2\rangle - \langle f\rangle^2\\ \sigma_f^2 \amp = \langle f^2\rangle - \langle f\rangle^2 \end{align*}

The last line above is an important identity for standard deviation and expectation values.

Subsection 4.2.4 Quantum Mechanics

One of the most well-known applications of continuous probability is quantum mechanics. The whole field is built on the assumption that states of physical systems are probabilistically determined and that we should study the physics of the system by studying the probability distribution.

The state of a physical system in quantum mechnics is measured by a function \(\Psi\text{,}\) called a wave function. \(\Psi\) itself is not exactly the probability density; quantum mechanics works with a \(\CC\)-valued function, which adds another layer of complexity. With \(\CC\)-valued functions, there is an operation called complex conjugation, which changes the sign of the imaginary piece of the function. It is written with a bar, so \(\bar{\Psi}\) is the conjugate. Then, expectation values in quantum mechanics are given by the following integral, where \(D\) is the domain of states and \(f\) is a observable on that domain of states.

\begin{equation*} \langle f\rangle = \int_D \bar{\Psi} f \Psi dV \end{equation*}

All observables in quantum mechancs are calcualted from the wave function. Determining the behaviour of the wave function over time is the goal of the discpline; that behaviour is given by the famour Schrodinger equation. In this equation \(V\) is a potential energy function on the space of states, \(\hbar\) is a constant, \(m\) is the mass and \(\imath\) is the imaginary number, with \(\imath^2 = -1\text{.}\)

\begin{equation*} \imath \hbar \frac{\del \Psi}{\del t} = \frac{-\hbar}{2m} \nabla^2 \Psi + V \Psi \end{equation*}

I'll assume that \(\Psi(x,t)\) has only one variable of position. (The following derivation works in three variables of position with roughly the same steps, but the notation becomes much more challenging). The expectation value for position is \(\langle x\rangle\text{.}\) What is momentum? It should be (up to a mass term) the rate of change of position. But the only available sense of position is the expectation value. Therefore, I should try to calculate \(\frac{\del \langle x\rangle}{\del t}\text{.}\) I use the compatibility of integration and differentiation to exchange the operations. (There are theorems, which I've omitted in this course, which allow this exchange).

\begin{equation*} \frac{\del \langle x\rangle}{\del t} = \frac{\del}{\del t} \int_{\RR} \bar{\Psi} x \Psi dx = \int_{\RR} \frac{\del}{\del t} x \bar{\Psi}\Psi dx = \int_{\RR} x \frac{\del}{\del t} \left( \bar{\Psi}\Psi \right) dx \end{equation*}

I need to calculate this derivative term. I make use of the Schrodinger equation to change the time derivatives into space derivatives. (This is really what the Schrodinger equation is all about!)

\begin{align*} \frac{\del}{\del t} \left( \bar{\Psi}\Psi \right) \amp = \bar{\Psi} \frac{\del}{\del t} \Psi + \Psi \frac{\del}{\del t} \bar{\Psi}\\ \amp = \bar{\Psi} \frac{1}{\imath \hbar} \left[ \frac{-\hbar^2}{2m} \frac{\del^2}{\del x^2} \Psi + V \Psi \right] + \Psi \frac{-1}{\imath \hbar} \left[ \frac{-\hbar^2}{2m} \frac{\del^2}{\del x^2} \bar{\Psi} + V \bar{\Psi} \right]\\ \amp = \frac{\imath \hbar}{2m} \bar{\Psi} \frac{\del^2 \Phi}{\del^2 x} - \frac{\imath V \Phi \bar{\Phi}}{\hbar} - \frac{\imath \hbar}{2m} \Psi \frac{\del^2 \bar{\Phi}}{\del x^2} + \frac{\imath V \Psi \bar{\Psi}}{\hbar}\\ \amp = \frac{\del}{\del x} \left[ \frac{\imath \hbar}{2m} \left( \bar{\Psi} \frac{\del \Psi}{\del x} - \frac{\del \bar{\Psi}}{\del x} \Psi \right) \right]\\ \frac{d\langle x\rangle}{dt} \amp = \frac{\imath \hbar}{2m} \int x \frac{\del}{\del x} \left( \bar{\Psi} \frac{\del \Psi}{\del x} - \frac{\del \bar{\Psi}}{\del x} \Psi \right) dx \end{align*}

Here I integrate by parts.

\begin{equation*} = \frac{\imath \hbar}{2m} x \left( \bar{\Psi} \frac{\del \Psi}{\del x} - \frac{\del \bar{\Psi}}{\del x} \Psi \right) \Bigg|_{-\infty}^{\infty} - \frac{\imath \hbar}{2m} \int \left( \bar{\Psi} \frac{\del \Psi}{\del x} - \frac{\del \bar{\Psi}}{\del x} \Psi \right) dx \end{equation*}

The first term, when evaluated on the bounds, decays to zero due to the normalization limits. Then I integrate by parts again. In the second integration by parts, half the terms will cancell.

\begin{align*} \amp = \frac{-\imath \hbar}{m} \int \bar{\Psi} \frac{\del \Psi}{\del x} dx = \frac{\hbar}{\imath m} \int \bar{\Psi} \frac{\del \Psi}{\del x} dx\\ \langle p\rangle \amp = \frac{d \langle x \rangle}{dt} = \int \bar{\Psi} \frac{\hbar}{\imath} \frac{\del}{\del x} \Psi dx = \left\langle \frac{\hbar}{\imath} \frac{\del}{\del x} \right\rangle \end{align*}

This is something new: the expectation value of an operator instead of a function. This is well defined because the operator acts on the wave function. This leads to a general correspondence.

Expectation Values \(\rightarrow\) Operators on Wave Functions
Position \(\rightarrow\) Multiplication by \(x\)
Momentum \(\rightarrow\) Operator \(\frac{\hbar}{\imath} \frac{\del}{\del x}\)

Any observable in quantum mechanics can be reduced to an operator \(F\) on the space of wave functions. Its expectation value is its integral.

\begin{equation*} \langle F\rangle = \int_D \bar{\Psi} (F\Psi) dV \end{equation*}

Moreover, all operators in quantum mechanics can be derived from combination of the position and momentum opreators.

Subsection 4.2.5 The Uncertainty Principle

The study of quantum mechanics thus becomes the study of operators on wave functions. This is a mathematically intense study, leading to whole new branches of mathematics focused on operator algebra. One of the first and most important question concerning operator is this: given two operators, do they commute? That is, if \(F\) and \(G\) are operators, is \(F(G\Psi) = G(F\Psi)\text{?}\)

Definition 4.2.12.

For any objects with multiplication \(a\) and \(b\text{,}\) the commutator bracket is defined as

\begin{equation*} [a,b]= ab - ba\text{.} \end{equation*}

The two items commute if and only is \([a,b] = 0\text{.}\)

I use the commutator bracket to study the commutativity of operators. In studying operator algebra and continuous probability, I can derive the following inequality. (I don't have the time or machinery for the proof, unfortunately).

The commutator bracket is still an operator, so it has an expectation value. Let's look at position and momentum, to see if they commute. Let \(f\) be a test function (something to act upon, for an operator).

\begin{align*} \left[x, \frac{\hbar}{\imath} \frac{\del}{\del x} \right] f \amp = x \frac{\hbar}{\imath} \frac{\del f}{\del x} - \frac{\hbar}{\imath} \frac{\del}{\del x} (xf)\\ \amp = x \frac{\hbar}{\imath} \frac{\del f}{\del x} - x \frac{\hbar}{\imath} \frac{\del f}{\del x} - \frac{\hbar}{\imath} f\\ \amp = \frac{-\hbar}{\imath} f\\ [x, \frac{\hbar}{\imath} \frac{\del}{\del x} ] \amp = \frac{-\hbar}{\imath} = \imath \hbar \end{align*}

So I know the commutator of position and momentum. What is its expectation value?

\begin{equation*} \left\langle \imath \hbar \right\rangle = \int \bar{\Psi} \imath \hbar \Psi dx = \imath \hbar \int \bar{\Psi} \Psi dx = \imath \hbar \end{equation*}

This makes some sense: this operator is simply multiplication by a constant; it doesn't change the wave functions at all. It's expectation is simply itself — it is a constant.

Then I apply the Theorem 4.2.13 to get the next result, where \(x\) stands for position and \(p\) for momentum.

\begin{equation*} \sigma_x \sigma_p \geq \frac{\hbar}{2} \end{equation*}

What does this mean? The \(\sigma\) is the standard deviation: it measures how wide the probability is for each of the measurements. This can be thought of as error. If a \(\sigma\) is very small, we have a very precise observable. If \(\sigma\) is large, the possible values in a reasonable probabity are much larger. The product of these two \(\sigma\) is the product of the uncertain in the measurement of position and moment. This is the famous uncertainty princple. Because the operators do not commute, I can't measure them both precisely at the same time. That fact comes directly the the operator mathematics; this makes it intrinsic to this quantum mechanical model. The uncertainly is not a problem of measurement, but a mathematical fact of the systems.

This can be thought of in terms of particple/wave duality. A particle, as a probability distribution, is just a single peak. A wave, as a probability distribution, has a sinusoidal graph. Momentum is essentially wavelength in this interpretation. A particple has a definite position: the single peak is located somewhere. However, with no repetition, it has no wavelength. A wave has a wavelength, but since the graph extends outward, it has no fixed position. The uncertainly principle reflects how elementary objects have both wave and particple like behaviour, one to the exclusion of the other.

This commutator analysis works for all operators. Two operators are compatible observables if they commute. If they do not, a version of the uncertainly princples holds for them.

This leads to some of the standard philosophical problems of measurement in Quantum Mechanics. What does a measurement do? Why does it collapse a wave function? Is it human observation? Machine observation? Consciousness?

Historically, there were three main camps. The Realist camp claimed that things existed in reality and that probability is an illusion and weakness of the model. The Orthodox camp said that nothing exists before a measurement. The system is a probability–nothing more or less. The Agnostic camp said that before measurement, any such question is meaningless, since measurement is all we have to interact with the universe. Many feel that none of the three answers are entirely satisfactory and the mystery of quantum mechanics remains with us.