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Advanced Classical Mechanics/Rigid Bodies

Definition

 * In a rigid body, no part of the body moves relative to another part of the body regardless of the force applied.
 * A rigid body is equivalent to a system of particles restricted to have fixed distances apart.

There are two methods for solving for the motion of rigid bodies.

 * 1) Write out the kinetic energy $$T$$ and use the Lagrangian to derive the equations of motion
 * 2) Use Euler's equations to get the motion of the center of mass and the rotation.

The Motion of a Rigid Body
The restrictions of rigid body motion require that distance between any two particles within the body remain constant. For example

$$ l_{ij}^2 = \left ( \vec r_i - \vec r_j \right )^2 = \left (x_i - x_j \right)^2 + \left (y_i - y_j \right)^2 + \left (z_i - z_j \right)^2 $$

will be constant during the motion. Let's take the time derivative of this quantity,

$$ \frac{d l_{ij}^2}{d t} = 2 \left ( \vec r_i - \vec r_j \right ) \cdot \left ( \vec v_i - \vec v_j \right ) $$

For this to vanish either $$\vec v_i - \vec v_j$$ must vanish or the dot product must vanish. The first corresponds to a translation of the object. For the dot product to vanish, the velocity difference must be perpendicular to the position difference. Specifically,

$$ \vec v_i - \vec v_j = \vec \omega_{ij} \times \left ( \vec r_i - \vec r_j \right ). $$

At first it might appear that you can have a different $$\vec \omega_{ij}$$ for each pair of particles. It turns out that this is not the case. Let's calculate

$$ \vec v_i - \vec v_k = \left ( \vec v_i - \vec v_j \right ) - \left ( \vec v_k - \vec v_j \right ) = \vec \omega_{ij} \times \left ( \vec r_i - \vec r_j \right ) - \vec \omega_{kj} \times \left ( \vec r_k - \vec r_j \right ). $$

On the other hand we could have calculated it as follows,

$$ \vec v_i - \vec v_k = \vec \omega_{ik} \times \left ( \vec r_i - \vec r_k \right ) = \vec \omega_{ik} \times \left ( \vec r_i - \vec r_j \right ) - \vec \omega_{ik} \times \left ( \vec r_k - \vec r_j \right ). $$

For these expressions to be equal for arbitrary values of $$\vec r_i$$, $$\vec r_j$$ and $$\vec r_k$$, the three values of $$\vec \omega$$ must be equal to each other so we will drop the subscripts and call this quantity the angular velocity of the rotation.

To summarize, a rigid body has two types of motion:

 * 1) Translation of the entire body, and
 * 2) Uniform rotation about a particular axis (the direction of $$\vec \omega$$ and an angular speed given by the magnitude of $$\vec \omega$$.

Angular Velocity
We can use some point within or outside the body as the origin and denote its position at any time by $$\vec r_0$$. The velocity of a particular part of the body is

$$ \vec v = \vec v_0 + \vec \omega \times \left (\vec r - \vec r_0 \right ) $$

where $$\vec r$$ is the location of the part of the body and $$\vec v_0$$ is the velocity of the origin.

Let's move the origin to $$\vec r_1$$. The velocity of the particle does not change so we have

$$ \vec v = \vec v_0 + \vec \omega \times \left (\vec r - \vec r_0 \right ) = \vec v_1 + \vec \omega \times \left (\vec r - \vec r_1 \right ) $$

so

$$ \vec v_1 = \vec v_0 + \vec \omega \times \left ( \vec r_1 - \vec r_0 \right ). $$

The vector $$\vec \omega$$ is constant with respect to changes in the origin. A useful example of this is a wheel that rolls without slipping.

Moment of Inertia
Rigid bodies are such an important part of classical mechanics that we have developed special techniques to calculate their kinetic energies and angular momenta. Let's calculate the kinetic energy of a mass in rigid body motion. Its velocity is

$$ \vec v_i = \vec v_0 + \vec \omega \times \vec r_i $$

where we have written the coordinates relative to the origin ($$\vec r_0$$).

Kinetic Energy
Its kinetic energy is

$$ T_i = \frac{1}{2} m_i \left ( \vec v_i \cdot \vec v_i \right ) = \frac{1}{2} m_i \left ( \vec v_0 + \vec \omega \times \vec r_i  \right ) \cdot \left ( \vec v_0 + \vec \omega \times \vec r_i  \right ). $$

and multiplying out the terms we get

$$ T_i = \frac{1}{2} m_i \left [ v_0^2 + 2 \vec v_0 \cdot \left ( \vec \omega \times \vec r_i \right )  + \left ( \vec \omega \times \vec r_i\right )^2 \right ]. $$

We can simplify the result using the following fact

$$ \left( \vec A \times \vec B \right )^2 = A^2 B^2 \sin^2 \theta = A^2 B^2 \left ( 1 - \cos^2 \theta \right ) = A^2 B^2 \left[ 1 - \frac{\left (\vec A \cdot \vec B\right )^2}{A^2 B^2} \right ] = A^2 B^2 - \left (\vec A \cdot \vec B\right )^2 $$

to get

$$ T_i = \frac{1}{2} m_i \left [ v_0^2 + 2 \vec v_0 \cdot \left ( \vec \omega \times \vec r_i \right )  + \vec \omega^2 \vec r_i^2 - \left ( \vec \omega \cdot \vec r_i \right )^2 \right ]. $$

Angular Momentum
Let's look at the angular momentum of the particle, we have

$$ \vec L_i = m_i \vec r_i \times \vec v_i = m_i \vec r_i \times \vec v_0 + m_i \vec r_i \times \left ( \vec \omega \times \vec r_i \right ). $$

Here we can use the vector triple product to simplify the result

$$ \vec A \times \left ( \vec B \times \vec C \right ) = \vec B \left ( \vec A \cdot \vec C \right ) - \vec C \left ( \vec A \cdot \vec B \right ) $$

which for us yields

$$ \vec L_i = m_i \vec r_i \times \vec v_i = m_i \vec r_i \times \vec v_0 + m_i \left [ \vec \omega r^2 - \vec r \left ( \vec \omega \cdot \vec r \right ) \right ]. $$

If $$\vec v_0=0$$ then

$$ T_i = \frac{1}{2} \vec \omega \cdot \vec L_i. $$

In this case, let's write out $$\vec L_i$$ component by component, we have

$$ \vec L_i = m_i \left [ \begin{matrix} \omega_x r_i^2 - x_i \left ( \omega_x x_i + \omega_y y_i + \omega_z z_i \right ) \\ \omega_y r_i^2 - y_i \left ( \omega_x x_i + \omega_y y_i + \omega_z z_i \right ) \\ \omega_z r_i^2 - z_i \left ( \omega_x x_i + \omega_y y_i + \omega_z z_i \right ) \end{matrix} \right ] = m_i \left [ \begin{matrix} y_i^2 + z_i^2 & -x_i y_i & -x_i z_i \\ -x_i y_i & x_i^2 + z_i^2 & -y_i z_i \\ -x_i z_i & -y_i z_i & x_i^2 + y_i^2 \end{matrix} \right ] \left [ \begin{matrix} \omega_x \\ \omega_y \\ \omega_z \end{matrix} \right ] \equiv \vec \vec I_i \cdot \vec \omega $$

defining the moment of inertia matrix for that particular particle.

Putting It All Together
Because the quantities $$\vec v_0$$ and $$\vec \omega$$ are the same for every particle, we can sum up the kinetic energy and angular momentum for all of the particles to get

$$ T = \sum_i T_i = \frac{1}{2} M v_0^2 + \frac{1}{2} \vec \omega \cdot \vec I \cdot \vec \omega + M \vec v_0 \cdot \left ( \vec \omega \times \vec r_{CM} \right ) $$

where $$M$$ is the total mass, $$\vec r_{CM}$$ is the location of the centre of mass and $$\vec \vec I = \sum_i \vec \vec I_i$$ is the moment of inertia of the body about the origin. The total angular momentum is

$$ \vec L = M \vec r_{CM} \times \vec v_0 + \vec \vec I \cdot \vec \omega $$

Using the result for the scalar triple product we can show that

$$ \vec L = \frac{\partial T}{\partial \omega} $$

so the angular momentum is the generalized momentum conjugate to the angular velocity; if the potential does not depend on angle, the angular momentum is a first integral of the motion.

Principal Axes
The moment of inertia is a positive definite matrix. We know this because when we multiply it by any non-zero vector $$\vec \omega$$, we get the kinetic energy of the body that is necessarily positive. Such a matrix can always be diagonalized. From a physical point of view, that means that the eigenvector of the matrix are known as the principal axes of the body. If the angular velocity points along one of the principal axes, then the angular momentum is parallel to the angular velocity. Furthermore, if one chooses the principal axes of the body to be one's coordinate axes, the diagonal element of the moment of inertia matrix vanish, so for example,

$$ \vec L = \left [ \begin{matrix} I_{xx} \omega_x \\ I_{yy} \omega_y \\ I_{zz} \omega_z \end{matrix} \right ] $$

and the kinetic energy is

$$ T = \frac{1}{2} M v_0^2 + M \vec v_0 \cdot \left ( \vec \omega \times \vec r_{CM} \right )    +  \frac{1}{2}I_{xx} \omega_x^2 +  \frac{1}{2}I_{yy} \omega_y^2 +  \frac{1}{2} I_{zz} \omega_z^2 $$

Change of Origin
Sometimes it is easiest to calculate the moment of inertia about a particular point but you are interested in the moment of inertia about another point. There is a straightforward prescription to achieve this. First, let's use the center of mass of the system as the origin, so we have

$$ T = \frac{1}{2} M v_{CM}^2 + \frac{1}{2} \vec \omega \cdot \vec \vec I_{CM} \cdot \vec \omega $$

Let's write this same equation with respect to a new origin $$\vec r_0$$ so we have

$$ T = \frac{1}{2} M \left ( \vec v_0 + \vec \omega \times \vec r_{CM}\right )^2 + \frac{1}{2} \vec \omega \cdot \vec \vec I_{CM} \cdot \vec \omega $$

$$ = \frac{1}{2} M v_0^2 + M \vec v_0 \cdot \left ( \vec \omega \times \vec r_{CM}\right ) + \frac{1}{2} M \left( \vec \omega \times \vec r_{CM} \right )^2 + \frac{1}{2} \vec \omega \cdot \vec \vec I_{CM} \cdot \vec \omega. $$

If we compare this result to the earlier result we find

$$ \vec \omega \cdot \vec \vec I \vec \omega = \vec \omega \cdot \vec \vec I_{CM} \cdot \vec \omega + M \left( \vec \omega \times \vec r_{CM} \right )^2. $$

On a component by component basis we have

$$ \vec \vec I = \vec \vec I_{CM} + M \left [ \begin{matrix} y_{CM}^2 + z_{CM}^2 &  x_{CM} y_{CM} &  x_{CM} z_{CM} \\ x_{CM} y_{CM} & x_{CM}^2 + z_{CM}^2 & M y_{CM} z_{CM} \\ x_{CM} z_{CM} & y_{CM} z_{CM} & x_{CM}^2 + y_{CM}^2 \end{matrix} \right ] $$

To change from one origin to another you first use the equation above to get the moment of inertia relative to the centre of mass and then use it a second time to get the moment of inertia about the new origin.

Calculating a Moment of Inertia
Let's calculate the moment of inertia of an ellipsoid. Let's assume that the axes of the ellipsoid are line up along the coordinate axes. Along the $$x-$$axis the ellipsoid spans from $$-a$$ to $$a$$. Along the $$y$$ and $$z-$$axes, the bounds are $$b$$ and $$c$$ respectively.

Because the coordinate axes are the principal axes of the ellipsoid, the off-diagonal components of the matrix will vanish. We will calculate the sum of $$z^2$$ over the ellipsoid first. We have

$$ \sum_i m_i z_i^2 = \int_{\rm Ellipsoid} \rho \, z^2 \, dV = \int_{\rm Ellipsoid} \rho \, z^2 \, dx dy dz. $$

The region of integration is rather complicate in the Cartesian coordinates but this gives us a place to start. We would like to simplify things a bit by defining

$$ x = a u, y = b v, z = c w. $$

In these new coordinates the ellipsoid becomes a sphere of unit radius. We also have to include the Jacobian of the coordinate transformation

$$ \frac{\partial \left ( x, y, z \right )}{\partial \left ( u, v, w \right )} = a b c. $$

Now our integral looks like

$$ \sum_i m_i z_i^2 = \int_{\rm Unit Sphere} \rho \, a b c(c w)^2 \, du dv dw. $$

Let's calculate the density of the ellipsoid in terms of its $$M$$ and volume. The volume of the ellipsoid is

$$ V = \int_{\rm Ellipsoid} \, dx dy dz = \int_{\rm Unit Sphere} a b c \, du dv dw = \frac{4}{3} \pi a b c $$

so we can rewrite the density in the moment of inertia integral

$$ \sum_i m_i z_i^2 = \int_{\rm Unit Sphere} \frac{3 M}{4 \pi a b c} a b c(c w)^2 \, du dv dw = \frac{3 M c^2}{4\pi} \int_{\rm Unit Sphere} w^2 \, du dv dw. $$

The easiest way to integrate over a unit sphere is to use spherical coordinates. Let's do it:

$$ \sum_i m_i z_i^2 = \frac{3 M c^2}{4\pi} \int_0^1 dr \int_0^\pi d\theta \int_0^{2\pi} d\phi \, r^2 \sin\theta \, r^2 \cos^2\theta = \frac{3 M c^2}{4\pi} \frac{2}{5} \pi \int_{-1}^1 d\left (\cos \theta\right ) \cos^2 \theta = \frac{3 M c^2}{4\pi} \frac{4}{15} \pi = \frac{1}{5} M c^2. $$

By symmetry we can calculate the results for the other two coordinates. We have

$$ \sum_i m_i x_i^2 = \frac{1}{5} M a^2, \sum_i m_i y_i^2 = \frac{1}{5} M b^2 $$

and the moment of inertia matrix about the centre of mass have the following non-zero components:

$$ I_{xx} = \frac{M}{5} \left ( b^2 + c^2 \right ), I_{yy} = \frac{M}{5} \left ( a^2 + c^2 \right ), I_{zz} = \frac{M}{5} \left ( a^2 + b^2 \right ). $$

This is an example of reducing a general shape to a symmetric one through a transformation of variables. One important result of this technique is |Routh's rule.

Ellipsoid of Inertia
We saw earlier that if the velocity of the origin vanished, then we could write $$T = \frac{1}{2} \vec L \cdot \vec \omega$$. Now if there are no forces on the body, the kinetic energy of the body is conserved. If there are no forces on the body, there are no torques, so the angular momentum is conserved as well. Does this mean that the angular velocity is constant?

The answer is of course no because the equation only says that the component of the angular velocity along the direction of the angular momentum is constant. The other components can change. For example, if the angular velocity lies along one of the principal axes (eigenvectors of the moment of inertia matrix), then the angular momentum points in the same direction and the angular velocity must stay constant in direction and magnitude.

A usual picture to understand the force-free motion of a rigid body is the ellipsoid of inertia. Let's use the principal axes of the body as a coordinate frame to write the kinetic energy

$$ T = \frac{1}{2} \left (I_{xx} \omega_x^2 + I_{yy} \omega_y^2 + I_{zz} \omega_z^2 \right ). $$

The ellipsoid of inertia is the locus of values of $$\vec \omega$$ that give a kinetic energy of the body.

Ellipsoid of Inertia of an Ellipsoid
Let's calculate the shape of the ellipsoid of inertia of an ellipsoid. We have

$$ T = \frac{1}{2} \frac{M}{5} \left [ \left (b^2 + c^2\right) \omega_x^2 + \left (a^2 + c^2\right) \omega_y^2 + \left (a^2 + b^2\right) \omega_z^2  \right ], $$

so along the $$x$$, $$y$$ and $$z$$ directions, the axes of the ellipsoid of inertia are

$$ \sqrt { \frac{10 T}{M (b^2 + c^2)}}, \sqrt { \frac{10 T}{M (a^2 + c^2)}}, \sqrt { \frac{10 T}{M (a^2 + b^2 )}}. $$

The size of the ellipsoid of inertia increases as $$T$$ and decreases with the mass and size of the actual ellipsoidal body. The shape of the ellipsoid of inertia reflects the shape of the physical ellipsoid. The largest axis of the ellipsoid is also the largest axis of the corresponding ellipsoid of inertia. If two axes of the ellipsoid are the same size, the corresponding axes of the ellipsoid of inertia will be equal as well.

Motion of the Ellipsoid of Inertia
To conserve energy the angular velocity must remain on the ellipsoid of inertia and the orientation of ellipsoid of inertia determines the orientation of the body. We can use the conservation of angular momentum to orient the ellipsoid. If we remember that $$\vec L = \partial T/\partial \vec \omega$$ we see that the normal to the ellipsoid of inertia is the angular momentum; therefore, as the ellipsoid of inertia rotates about the angular velocity, it must remain tangent to a plane perpendicular to the angular momentum, the invariable plane. Furthermore, the angular velocity is restricted to move along the ellipsoid of inertia along curves of constant $$\left | \vec L \right|$$. This curve is called the polhode, and it traces the path of the angular velocity through the body. The angular velocity also traces a path in the invariable plane called the herpolhode. The herpolhode is the path of the angular velocity through space. We can also write that $$2 T=\vec \omega \cdot \vec L$$ so the centre of the ellipsoid of inertia must remain a constant distance above the invariable plane. The height of the centre of the ellipsoid is $$2T/\left |\vec L\right|$$, so when the ellipsoid is down low the angular momentum is high relative to the kinetic energy.

Keeping the ellipsoid of inertia tangent to the invariable plane preserves the direction of the angular momentum. The magnitude of the angular momentum is also conserved. This defines a second ellipsoid through the equation

$$ \left | \vec L \right |^2 = I_{xx}^2 \omega_x^2 + I_{yy}^2 \omega_y^2 + I_{zz}^2 \omega_z^2 $$

so the polhodes are the intersections of the ellipsoid of inertia with the angular momentum ellipsoid. The angular momentum ellipsoid deviates more from a sphere than the ellipsoid of inertia; this gives an idea of where the polhodes lie for small and large angular momenta.

If the body has two principal axes with equal moments of inertia, the polhodes are circles centered on the axis with the unique moment of inertia, and the herpolhodes are circles in the invariable plane. If the unique moment of inertia is larger than the others, the body is called oblate. Otherwise it is prolate -- in analogy to the spheroids with the similar properties. In this case, the free rotation of the body consists of a constant precession of the angular velocity in a circle about the angular momentum in the space frame (the space cone) and about the unique principal axis in the body frame (the body cone). In the oblate case the body cone rolls within the space cone, and in the prolate case, the body cone rolls on the outside of the space cone.

If no two moments of inertia are equal to each other, the polhodes are much more complicated. The polhodes near the axes with the largest and smallest moment of inertia are closed, but those near the intermediate axis are not; consequently, rotation about the intermediate axis is not stable as we shall see when we analyze the free rotation using Euler's equations.

Euler's Equations
So far we have tried to find a graphical description of the motion of an object without any torques. Although "the polhode rolls on the herpolhode without slipping" might paint a nice Victorian picture, it doesn't exploit the mathematical language of physics to which we are now accustomed. We know that the change in the angular momentum is equal to the torques on the body

$$ \frac{d \vec L}{d t} = \vec N. $$

Let's write the angular momentum in terms of the principal axes of the body. Here we will be looking at how the principal axes move so we will use the unit vectors $$\hat 1, \hat 2$$ and $$\hat 3$$ to denote the current direction of the principal axes of the body. We can write

$$ \frac{d \vec L}{dt} = \frac{\partial L_1}{\partial t} \hat 1 + \frac{\partial L_2}{\partial t} \hat 2 +  \frac{\partial L_3}{\partial t} \hat 3 + L_1 \frac{\partial \hat 1}{\partial t} + L_2 \frac{\partial \hat 2}{\partial t} + L_3 \frac{\partial \hat 3}{\partial t} $$

$$ = \frac{\partial L_1}{\partial t} \hat 1 + \frac{\partial L_2}{\partial t} \hat 2 +  \frac{\partial L_3}{\partial t} \hat 3 + L_1 \vec \omega \times \hat 1  + L_2 \vec \omega \times \hat 2  + L_3 \vec \omega \times \hat 3 $$

Let's define the angular momentum relative to the body axes as

$$ \vec L_b = \left [ \begin{matrix} L_1 \\ L_2 \\ L_3 \end{matrix} \right ] = \left [ \begin{matrix} \omega_1 I_1 \\ \omega_2 I_2 \\ \omega_3 I_3 \end{matrix} \right ] $$

to get

$$ \frac{d \vec L}{dt} = \frac{d \vec L_b}{d t} + \vec \omega \times \vec L_b = \vec N $$

where the torque and angular velocity are written relative to the body axes. One can imagine using Euler's equations would be quite cumbersome if one had to include the torques because one would always have to transform the torques from the inertia frame to the body frame. Let's write out these equations component by component

$$ I_1 \dot \omega_1 - \omega_2 \omega_3 \left (I_2 - I_3 \right) = N_1 $$

$$ I_2 \dot \omega_2- \omega_3 \omega_1 \left (I_3 - I_1 \right) = N_2 $$

$$ I_3 \dot \omega_3 - \omega_1 \omega_2 \left (I_1 - I_2 \right) = N_3 $$

These equations are most powerful when there are no torques, so the right-hand sides are zero. In this case we can see immediately that if any two moments of inertia differ than the angular velocity must lie along one of the principal axes to remain constant.

If two of the moments of inertia are equal to each other (let's take $$I_1=I_2$$), we can solve the torque-free Euler's equations exactly. We have

$$ I_1 \dot \omega_1 = \left ( I_1 - I_3 \right ) \omega_3 \omega_2 $$

$$ I_2 \dot \omega_2 = -\left ( I_1 - I_3 \right ) \omega_3 \omega_1 $$

$$ I_3 \dot \omega_3 = 0 $$

We can solve this with

$$ \dot \omega_1 = - \Omega \omega_2, \dot \omega_2 = \Omega \omega_1 $$

with

$$ \Omega = \frac{I_3 - I_1}{I_1} \omega_3, $$

yielding the solution

$$\omega_1 = A \cos \Omega t,\, \omega_2 = A \sin \Omega t.$$

This result coincides with the graphical picture that the polhodes of a symmetric body are circles centered on the axis with the unique moment of inertia.

Chandler Wobble
We can calculate the expected rate for the Earth to precess due to its oblatness. Let's estimate the ratio of the moments of inertia in terms of the radii of the Earth at the equator and at the poles, we have

$$ \frac{I_3 - I_1}{I_1} = \frac{2 R_{\rm eq}^2 - \left ( R_{\rm eq}^2 + R_{\rm pole}^2 \right )}{R_{\rm eq}^2 + R_{\rm pole}^2} $$

$$ = \frac{\left ( R_{\rm eq} - R_{\rm pole} \right )\left ( R_{\rm eq} + R_{\rm pole} \right )}{R_{\rm eq}^2 + R_{\rm pole}^2} \approx \frac{R_{\rm eq} - R_{\rm pole}}{R_{\rm eq}}. $$

where in the last step we have assume that the difference between the polar and equatorial radius is small. We have $$R_{\rm eq} \approx 6378 {\rm km}, R_{\rm pole} \approx 6356 {\rm km}$$, so the ratio is about $$1/290$$. What is $$\omega_3$$ for the Earth? It is $$2\pi/1 {\rm day}$$, so the period of the free precession of the Earth should be about 290 days. The period is somewhat longer about 433 days Chandler_wobble. You can check out a plot of the wobble at the Paris Observatory. Why the Earth wobbles is a bit of a puzzle since because the Earth is not a solid body, without a driving force it would have damped long ago. Since the period of the wobble is similar to a year, seasonal changes are thought to be to blame.

I first learned about the wobble while spending a summer at the Pulkovo Astronomical Observatory near St. Petersburg. I saw a plot of the location of the Earth's pole on the surface of the Earth -- it moves in a irregular circular shape with a radius of 3 to 15 meters. The wobble can be measured by accurate measurements of the locations of stars relative to landmarks on the Earth. I asked my hosts why did they need to keep track the wobble so accurately. "To aim our missiles at your cities -- of course" came the reply.

Triaxial Body
Unfortunately, we can't find such a straightforward to the solution of the motion of a triaxial body. In the symmetric case, one of the components of the angular velocity was constant with time. This is not the case for a triaxial body. We have the following Euler's equations

$$ I_1 \dot \omega_1 = \left (I_2 -I_3\right ) \omega_3 \omega_2 $$

$$ I_2 \dot \omega_2 = \left (I_3 -I_1\right ) \omega_3 \omega_1 $$

$$ I_3 \dot \omega_3 = \left (I_1 -I_2\right ) \omega_1 \omega_2. $$

Let's take $$I_3 > I_2 > I_1$$ and define the following three positive ratios of moments of inertia

$$ A = \frac{I_3-I_2}{I_1}, B = \frac{I_3-I_1}{I_2}, C = \frac{I_2-I_1}{I_3} $$

giving the following equations

$$ \dot \omega_1 = - A \omega_3 \omega_2 $$

$$ \dot \omega_2 = + B \omega_3 \omega_1 $$

$$ \dot \omega_3 = - C \omega_1 \omega_2 $$

Major Axis
Let's assume that the rotation is nearly about the major axis, so $$\omega_1 \gg\ \omega_2, \omega_3$$. We will only include the latter components of the angular velocity to first order, so we have

$$ \dot \omega_1 \approx 0, \dot \omega_2 = B \omega_3 \omega_1, \dot \omega_3 = - C \omega_2 \omega_1. $$

If we define $$\Omega = \sqrt{BC} \omega_1$$ we have the following solution for $$\omega_2$$ and $$\omega_3$$,

$$ \omega_3 = \sqrt{\frac{B}{C}} K \cos \Omega t, \omega_2 = K \sin \Omega t, $$

so the angular velocity travels in an ellipse centered on the major axis.

Minor Axis
Let's assume that the rotation is nearly about the minor axis, so $$\omega_3 \gg\ \omega_1, \omega_2$$. We will only include the latter components of the angular velocity to first order, so we have

$$ \dot \omega_3 \approx 0, \dot \omega_1 = -A \omega_3 \omega_2, \dot \omega_2 = B \omega_3 \omega_1. $$

If we define $$\Omega = \sqrt{AB} \omega_3$$ we have the following solution for $$\omega_1$$ and $$\omega_2$$,

$$ \omega_1 = \sqrt{\frac{A}{B}} K \cos \Omega t, \omega_2 = K \sin \Omega t, $$

so the angular velocity travels in an ellipse centered on the minor axis.

Intermediate Axis
Let's assume that the rotation is nearly about the intermediate axis, so $$\omega_2 \gg\ \omega_1, \omega_3$$. We will only include the latter components of the angular velocity to first order, so we have

$$ \dot \omega_2 \approx 0, \dot \omega_1 = -A \omega_3 \omega_2, \dot \omega_3 = - C \omega_1 \omega_2. $$

If we define $$\alpha = \sqrt{AC} \omega_2$$ we have the following solution for $$\omega_1$$ and $$\omega_3$$,

$$ \omega_1 = K_1 e^{\alpha t}, \omega_3 = K_3 e^{\alpha t},\, $$

so the angular velocity diverges exponetially away from the intermediate axis.

Lagrangian Treatment of Rigid-Body Motion
Although it is possible to include the effects of torques on the motion of a rigid body within the framework of Euler's equations, it is rather cumbersome. Here we will develop the Lagrangian treatment of a top -- a symmetric rigid body with a torque from the Earth's gravity.

Euler Angles
In a Lagrangian treatment it is necessary to find a set of independent coordinates to describe the position of the system at any time. For rigid bodies a convenient set of such coordinates are called the Euler angles. There are several different conventions for the Euler angles. The convention used here makes the analysis of the top simpler.

To get from one set of coordinates to the other, we have to do a series of three rotations. We imagine that the primed coordinates are the body coordinates and the unprimed coordinates are the inertia coordinates; therefore, it is straightforward to write the kinetic energy in the primed coordinates and the potential energy in the unprimed coordinates. Here are the three steps in words and in terms of a rotation matrix

$$D=\left [ \begin{matrix} \cos\phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi & 0 \\ 0        & 0        & 1 \end{matrix} \right ] $$ $$C=\left [ \begin{matrix} 1 & 0 & 0 \\ 0 & \cos\theta & \sin\theta \\ 0 & -\sin\theta & \cos\theta \end{matrix} \right ] $$ $$B=\left [ \begin{matrix} \cos\psi & \sin\psi & 0 \\ -\sin\psi & \cos\psi & 0 \\ 0        & 0        & 1 \end{matrix} \right ] $$
 * 1) Rotate by an angle $$\phi$$ about the initial $$z-$$axis.  The final position of the $$x-$$axis is called the line of nodes.
 * 1) Rotate about the line of nodes by an angle $$\theta$$.   This changes the direction of the $$z-$$axis and pitches the $$y-$$axis out of the initial $$x-y-$$plane.
 * 1) Finally rotate about the new $$z-$$axis by an angle $$\psi$$.  This takes the $$x-$$axis out of the initial $$x-y-$$plane.

Euler proved that any rotation can be decomposed into these three rotations. One can see this mathematically by constructing the rotation matrix for these three successive transformations ($$A=BCD$$) and verifying that it is the most general rotation matrix.

The Kinetic Energy
To calculate the kinetic energy of the body, we have to find the angular velocity in terms of the coordinates, $$\phi, \theta$$ and $$\psi$$ and their time derivatives. In any coordinate system, the angular velocity is directed along the axis of rotation. We can construct the total angular velocity vector using the following information:

 * 1) Changes in $$\phi$$ produce rotations about the $$z-$$axis (inertial frame).
 * 2) Changes in $$\psi$$ produce rotations about the $$z'-$$axis (body frame).
 * 3) Changes in $$\theta$$ produce rotations about the line of nodes.

We can write the angular velocity relative to the inertial frame to get

$$ \vec \omega = \left [ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right ] \dot \phi + \left [ \begin{matrix} \cos \phi \\ \sin \phi \\ 0 \end{matrix} \right ] \dot \theta + \left [ \begin{matrix} \sin\theta \sin \phi \\ -\sin\theta \cos\phi \\ \cos\theta \end{matrix} \right ] \dot \psi $$

or relative to the body frame

$$ \vec \omega = \left [ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right ] \dot \psi + \left [ \begin{matrix} \cos \psi \\ -\sin \psi \\ 0 \end{matrix} \right ] \dot \theta + \left [ \begin{matrix} \sin\theta \sin \psi \\ \sin\theta \cos\psi \\ \cos\theta \end{matrix} \right ] \dot \phi. $$

The first form is useful to visualize what the body is doing, while the second form is good for calculating the kinetic energy because we have used the primed coordinate system to characterize the location of the principal axes of the body.

For simplicity let's specialize to symmetric bodies with $$I_{x'} = I_{y'}$$, so we have

$$ T = \frac{1}{2} I_{x'} \left (\omega_{x'}^2 + \omega_{y'}^2 \right ) + \frac{1}{2} I_{z'} \omega_{z'}^2. $$

Because the primed frame is fixed to the body, the moments of inertia in the kinetic energy are constant with time. Let's first calculate

$$ \omega_{x'}^2 + \omega_{y'}^2 = \left ( \cos\psi \dot \theta + \sin \psi \sin \theta \dot \phi \right )^2 + \left ( -\sin\psi \dot \theta + \cos\psi \sin\theta \dot \phi \right )^2 $$

$$ = \cos^2\psi \dot \theta^2 + \sin^2 \psi \sin^2 \theta \dot \phi^2 + 2 \sin\psi \cos\psi \sin\theta \dot \theta \dot \phi + \sin^2 \psi \dot\theta^2 + \cos^2\psi \sin^2\theta \dot \phi^2 - 2 \sin\psi \cos\psi \sin\theta \dot\theta \dot\phi $$

$$ = \dot\theta^2 + \sin^2\theta \dot \phi^2. $$

This simplication only obtains for a symmetric body. The $$z'-$$component is

$$ \omega_{z'}^2 = \left ( \dot \psi + \dot \phi \cos\theta \right )^2. $$

Combining these results yields the kinetic energy

$$ T = \frac{I_{x'}}{2} \left ( \dot \theta^2 + \dot \phi^2 \sin^2\theta \right ) + \frac{I_{z'}}{2} \left ( \dot \psi + \dot \phi \cos\theta \right )^2 $$

The Potential Energy
The potential energy is pretty simple using the Euler angles. If $$l$$ is the distance between the pivot point and the center of mass of the top, we have

$$ V = M g l \cos\theta $$

The Lagrangian
The Lagrangian is the difference between the kinetic and the potential energy, so we have

$$ L = T - V = \frac{I_{x'}}{2} \left ( \dot \theta^2 + \dot \phi^2 \sin^2\theta \right ) + \frac{I_{z'}}{2} \left ( \dot \psi + \dot \phi \cos\theta \right )^2 - M g l \cos\theta. $$

We notice immediately that $$\psi$$ and $$\phi$$ don't appear in the Lagrangian. The system looks the same as in turns around the $$z-$$ and $$z'-$$axes. This means that we have two conserved quantities, reducing the three dimensional problem to a single dimension (we will choose this to be $$\theta$$). Because the Lagrangian does not depend on time, the Hamiltonian is conserved as well so we can use the techniques outlined in Linear Motion to understand the top.

Integrals of the Motion
To calculate the conserved momenta we have to calculate the partial derivatives of the Lagrangian with respect to the generalized velocities. We have

$$ p_\psi = \frac{\partial L}{\partial \dot \psi} = I_{z'} \left ( \dot \psi + \dot \phi \cos\theta \right ) = I_{z'} \omega_{z'} \equiv I_{x'} a $$

where the last step defines the conserved quantity $$a$$ that has units of angular velocity. Similarly for the coordinate $$\phi$$, we have

$$ p_\phi = \frac{\partial L}{\partial \dot \phi} = \left ( I_{x'} \sin^2 \theta + I_{z'} \cos^2\theta \right ) \dot \phi + I_{z'} \cos\theta \dot \psi \equiv I_{x'} b. $$

Finally we have the Hamiltonian that is also conserved,

$$ H = T + V = \frac{I_{x'}}{2} \left ( \dot \theta^2 + \dot \phi^2 \sin^2\theta \right ) + \frac{I_{z'}}{2} \left ( \dot \psi + \dot \phi \cos\theta \right )^2 + M g l \cos\theta. $$

As we mentioned earlier, these three integrals are sufficient to solve for the motion. We can use the two conserved momenta to solve for the generalized velocities, $$\dot \psi$$ and $$\dot \phi$$, in terms of the conserved quantities $$a$$ and $$b$$ and the value of the third coordinate $$\theta$$. We have

$$ \dot \phi = \frac{b - a\cos\theta}{\sin^2\theta}, $$

$$ \dot \psi = \frac{I_{x'} a}{I_{z'}} - \cos\theta \frac{b - a\cos\theta}{\sin^2\theta}. $$

Conservation of Energy
Since both the Hamiltonian and the angular velocity along the $$z'-$$axis are conserved, let's define the conserved quantity

$$ E' = E - \frac{I_{z'} \omega_{z'}^2}{2} = \frac{I_{x'} \dot \theta^2}{2} + \frac{I_{x'}}{2} \frac{\left( b -a \cos\theta \right )^2}{\sin^2\theta} + M g l \cos\theta $$

which we can rewrite as

$$ E' = \frac{I_{x'} \dot \theta^2}{2} + V'\left (\theta\right ) $$

where

$$ V'(\theta) = \frac{I_{x'}}{2} \frac{\left( b -a \cos\theta \right )^2}{\sin^2\theta} + M g l \cos\theta. $$

We can make further progress by looking at $$u=\cos\theta$$ which will simplify the potential. Let's write

$$ \dot u^2 = \sin^2 \theta \dot \theta^2 = \left ( 1 - u^2 \right ) \dot \theta^2 $$

so we can write $$E'$$ in terms of $$u$$ and its time derivative

$$ E' = \frac{I_{x'}}{2} \frac{\dot u^2}{1-u^2} + \frac{I_{x'}}{2} \frac{\left( b -a u \right )^2}{1-u^2} + M g l u. $$

What remains is to solve for $$\dot u^2$$ to get the boundaries of the motion

$$ \dot u^2 = \left (E' - M g l u \right ) \frac{2}{I_{x'}} \left ( 1 - u^2 \right ) - \left (b - a u\right )^2 $$

The Motion of a Symmetric Top
The equation for $$\dot u^2$$ is cubic so it will generally have three roots (designated at $$u_{1,2,3}$$ -- it must have at least one root because for large positive values of $$u$$ $$\dot u^2$$ is positive and for large negative values of $$u$$ $$\dot u^2$$ is negative.  As the top moves from straight up to straight down, $$u$$ ranges from 1 to -1, so only the roots within this range are important.  Furthermore, the equation for $$\dot \phi$$ has a root at $$u_0=b/a$$.

The figure shows a typical curve for $$\dot u^2$$ as a function of $$\theta$$. The motion is bounded between $$u=0.5$$ and $$u=0.8$$. Depending on the value of $$b/a$$, the motion in the $$\phi$$-direction can be monotonic or not. If $$b/a=0.8$$, the motion will have a cusp at the upper end of the range.

The figures to the right show the motion of the $$z'-$$axis as a function of time for several values of $$b/a$$ with the curve for $$\dot u^2$$ above. The bounds of the motion in all three cases is from $$u=0.5$$ to $$u=0.8$$. How can we set up a top to execute these three classes of motion? If we release the axis of the top from a stationary position, the axis will first drop (nutation) because of gravity. As it drops the angular momentum about the vertical axis must be conserved so the tip of the top starts moving sideways as well (precession) -- yielding the motion in the middle figure. On the other hand if we release the tip of the top with some motion in the direction of the precession, we will get the lower figure. Finally to get the rather freaky looking motion in the top figure, we release the top with some motion opposing the direction of the precession.

Specifically let's examine the case where we release the top from a stationary position, so $$u_0=b/a=u_2$$ where $$u_2$$ is the middle root of the $$\dot u^2$$ curve. Initially we have

$$ E' = M g l \cos \theta = M g l \frac{b}{a} $$

so we can write

$$ \dot u^2 = \left (\frac{b}{a} - u \right ) M g l \frac{2}{I_{x'}} \left ( 1 - u^2 \right ) - \left ( b - a u \right)^2 $$

$$ = \left (u_0- u \right ) a^2 \left [ \frac{2 M g l }{I_{x'} a^2} \left ( 1 - u^2 \right ) -\left ( u_0 - u \right ) \right ] $$

The Fast Top
Unfortunately, the motion of the top in general cannot be solved in closed form, but we can make some progress in understanding the motion of the so-called fast top. If we look at the expression above, we have the ratio $$ 2M g l/(I_{x'} a^2). $$ The numerator is the potential energy of the top while the denominator is the related to the kinetic energy of the top about its spin axis. The condition for a fast top is

$$ I_{x'} a^2 = I_{z'} \omega_{z'}^2 \frac{I_{z'}}{I_{x'}} \gg\ 2 M g l. $$

In a fast top, the kinetic energy is much larger than the potential energy; consequently, the extent of the nutation ($$u_0-u$$) is small, so

$$ 1 - u^2 \approx 1 - u_0^2 = \sin^2 \theta_0 $$

so we can write

$$ \dot u^2 \approx \left (u_0- u \right ) a^2 \left [ \frac{2 M g l }{I_{x'} a^2} \sin^2 \theta_0 -\left ( u_0 - u \right ) \right ]. $$

The extent of the nutation is

$$ u_0 - u_1 \approx \frac{2 M g l }{I_{x'} a^2} \sin^2 \theta_0 = \frac{I_{x'}}{I_{z'}} \frac{2 M g l }{I_{z'} \omega_{z'}^2} \sin^2 \theta_0 $$

and we can rewrite the equation for $$\dot u^2$$ as

$$ \dot u^2 = \left (u_0 - u \right ) \left (u - u_1\right ) a^2. $$

Let's take the time derivative of both sides to yield

$$ 2 \dot u \ddot u = a^2 \left [ \left ( u_0 - u\right ) \dot u - \left (u - u_1\right ) \dot u \right ] $$

$$ \ddot u = a^2 \left ( \frac{u_0+u_1}{2} - u \right ). $$

Let's define $$y=u-(u_0+u_1)/2$$ so we have

$$ \ddot y = -a^2 y $$

with the solution

$$ y = \frac{ u_1 - u_0 }{2} \cos a t $$ and $$ u = \frac{u_1+u_0}{2} + \frac{ u_1 - u_0 }{2} \cos a t. $$

The frequency of the nutation is

$$ a = \frac{I_{z'}}{I_{x'}} \omega_{z'} $$

and

$$ \dot \phi = \frac{a \left (u_0 - u\right )}{\sin^2 \theta} = \frac{M g l}{I_{z'} \omega_{z'}} \left ( 1 - \cos a t \right ) $$

and

$$ \phi = \phi_0 + \frac{M g l}{I_{z'} \omega {z'}} \left ( t - \frac{1}{a}\sin a t \right ). $$

Sleeping Top
A sleeping top is a top that is set in motion with its spin axis vertical and the axis remains vertical. When the top is vertical it is impossible to distinguish changes in $$\phi$$ from changes in $$\psi$$ so the relationship between $$\dot \phi$$ and $$a$$ and $$b$$ isn't very useful. Let's look at the equation for $$\dot u^2$$. We have

$$ \dot u^2 = \left ( E' - M g l u \right ) \frac{2}{I_{x'}} \left (1 - u^2 \right ) - \left ( b - a u \right )^2. $$

When the top points upwards, $$u=1$$. We let go of the top with $$\dot u=0$$ so we have

$$ 0 = \left ( E' - M g l \right ) \frac{2}{I_{x'}} \left (1 - 1^2 \right ) - \left ( b - a  \right )^2. $$

so

$$b=a$$ and

$$ \dot u^2 = \left ( 1 - u \right ) a^2 \left [ \left ( E' - M g l u \right ) \frac{2}{I_{x'} a^2} \left (1 + u \right ) - \left ( 1 - u \right ) \right ]. $$

Using the definition of $$E'$$ we find

$$ E' = E - \frac{I_{z'} \omega_{z'}^2}{2} = \frac{I_{x'} \dot \theta^2}{2} + \frac{I_{x'}}{2} \frac{\left( b -a \cos\theta \right )^2}{\sin^2\theta} + M g l \cos\theta = M g l $$

and

$$ \dot u^2 = \left ( 1 - u \right )^2 a^2 \left [ \frac{2 M g l}{I_{x'} a^2} \left (1 + u \right ) - 1 \right ] =  \left ( 1 - u \right )^2 a^2 \frac{2 M g l}{I_{x'} a^2 } \left [ u - \left (\frac{I_{x'} a^2}{2 M g l} - 1 \right ) \right ], $$

so there is a double root at $$u=1$$ and another root at

$$ u = \frac{I_{x'} a^2}{2 M g l} - 1. $$

If this root is at $$u>1$$, then the top will sleep, so

$$ I_{x'} a^2 > 4 M g l $$

is the condition for a sleeping top.


text:
Advanced Classical Mechanics/Small Oscillations and Perturbed Motion

In Linear Motion, we argued that all sufficiently small oscillations are harmonic. In this section we will exploit this result in several ways to understand

 * 1) The motion of systems with many degrees of freedom near equilibrium,
 * 2) The motion of systems perturbed from known solutions, and
 * 3) The motion of systems with Lagrangians perturbed from systems with known solutions.

All three of these points are applications of perturbation theory, and they all start with the harmonic oscillator.

Normal Modes
The modes of oscillation of systems near equilibrium are called the normal modes of the system. Understanding the frequencies of the normal modes of the system is crucial to design a system that can move (even if it isn't meant to). Let's look at a system with many degrees of freedom; we have

$$ L = \frac{1}{2} \sum_{i,j} T_{ij} \dot q_i \dot q_j - V \left (q_1, \ldots q_n \right ). $$

Let $$q_{0,i}$$ be an equilibrium position and expand about this point $$ q_i = q_{0,i} + \eta_i $$ so $$\dot q_i = \dot \eta _i$$.

We can expand the potential energy to give

$$ V \left (q_1, \ldots q_n \right ) = V \left (q_{0,1}, \ldots q_{0,n} \right ) + \sum_{i} \left ( \frac{\partial V}{\partial q_i} \right )_{q_{0,i}} \eta_i + \frac{1}{2} \sum_{i,j} \left ( \frac{\partial^2 V}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i\eta_j + \cdots. $$

The first term is a constant with respect to $$\eta_i$$ and constant terms do not affect the motion. The second term is zero, because $$q_{0,i}$$ is a point of equilibrium. If we drop all terms higher than the third term shown, we are left with

$$ L = \frac{1}{2} \sum_{i,j} \left ( T_{ij} \dot \eta_i \dot \eta_j - V_{ij} \eta_i \eta_j \right ) $$

where

$$ T_{ij} = T_{ij} \left ( q_{0,1}, \ldots q_{0,n} \right ) $$ and $$ V_{ij} = \left ( \frac{\partial^2 V}{\partial q_i \partial q_j} \right )_{q_{0,i}}, $$

yielding the equations of motion

$$ \sum_{j} \left ( T_{ij} \ddot \eta_j - V_{ij} \eta_j \right ) = 0 $$

This is a linear differential equation with constant coefficients. We can try the solution

$$ \eta_i = C a_i e^{-i \omega t} $$

so we have

$$ \sum_{j} \left ( V_{ij} a_j - \omega^2 T_{ij} a_j \right ) = 0. $$

This is a matrix equation such that

$$ \vec \vec A \cdot \vec a = 0 $$ with

$$ \vec a = \left [ \begin{matrix} a_1 \\ a_2 \\ \vdots \\ a_j \end{matrix} \right ] $$

and

$$ \vec \vec A = \left [ \begin{matrix} V_{11} - \omega^2 T_{11} & V_{12} - \omega^2 T_{12} & \cdots \\ V_{21} - \omega^2 T_{21} & V_{22} - \omega^2 T_{22} & \cdots \\ \vdots & & \end{matrix} \right ] $$

This equation only has a solution is $$\det \vec \vec A=0$$. This gives a $$n$$th-degree polynomial to solve for $$\omega^2$$. We will get $$n$$ solutions for $$\omega^2$$ that we can substitute into the matrix equation and solve for $$a_j$$.

Is this guaranteed to work? Yes, it turns out. Look at the equation in terms of matrices we have

$$ \vec \vec V \vec a = \omega^2 \vec \vec T \vec a. $$

The matrix $$\vec \vec V$$ is symmetric and real. The matrix $$\vec \vec T$$ should be positive definite (because a negative kinetic energy doesn't make sense). Technical issue: If $$\vec \vec T$$ has a null space, the degrees of freedom corresponding to the null space are massless and cannot be excited unless they are in the null space of $$\vec \vec V$$. Either way, you can drop the null space from both sides of the equation.

Assuming that $$\vec \vec T$$ is invertable we have

$$ \left ( \vec\vec T \right ) ^{-1} \vec \vec V \vec a = \omega^2 \vec a $$

and we have a standard eigenvalue equation. In most examples, the kinetic energy matrix will be diagonal, so it is straightforward to construct the quotient matrix and diagonize it.

Perturbations about Steady Motion
Let's say I have some solution to the equations of motion and I would like to look at small deviations from the solution. Let's $$ q_{0,i}(t) $$ satisfy

$$ \frac{d}{dt} \frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i} = 0, $$

and let's look at

$$ q_i(t) = q_{0,i}(t) + \eta_i(t) $$

where $$\eta_i$$ is small. Let's expand the entire Lagrangian to find the equations of motion for the deviations $$\eta_i$$. We have

$$ L \left (q_1, \ldots q_n \right ) = L \left (q_{0,1}, \ldots q_{0,n} ; \dot q_{0,1}, \ldots \dot q_{0,n} \right ) + \sum_{i} \left [ \left ( \frac{\partial L}{\partial q_i} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial L}{\partial \dot q_i} \right )_{q_{0,i}} \dot \eta_i \right ] + $$

$$ \frac{1}{2} \sum_{i,j} \left [ \left ( \frac{\partial^2 L}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i\eta_j + \left ( \frac{\partial^2 L}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \dot\eta_i \dot \eta_j + 2 \left ( \frac{\partial^2 L}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \eta_i \dot \eta_j \right ] + \cdots. $$

Now let's apply Lagrange's equations for the deviations

$$ \frac{d}{dt} \frac{\partial L}{\partial \dot \eta_i} - \frac{\partial L}{\partial \eta_i} = 0 $$

to give

$$ \frac{d}{dt} \left \{ \left ( \frac{\partial L}{\partial \dot q_j} \right )_{q_{0,i}} + \sum_{i} \left [ \left ( \frac{\partial^2 L}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \dot\eta_i + \left ( \frac{\partial^2 L}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \eta_i \right ] \right \} $$

$$ - \left ( \frac{\partial L}{\partial q_j} \right )_{q_{0,i}} - \sum_{i} \left [ \left ( \frac{\partial^2 L}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial^2 L}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \dot \eta_i \right ] = 0 $$

The two terms without $$\eta$$ actually cancel each other out, leaving the following equations of motion.

$$ \frac{d}{dt} \left \{ \sum_{i} \left [ \left ( \frac{\partial^2 L}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \dot\eta_i + \left ( \frac{\partial^2 L}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \eta_i \right ] \right \} $$

$$ - \sum_{i} \left [ \left ( \frac{\partial^2 L}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial^2 L}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \dot \eta_i \right ] = 0. $$

In steady motion, the partial derivatives are taken to be constant in time yielding the even simpler result

$$ \sum_{i} \left [ \left ( \frac{\partial^2 L}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \ddot\eta_i - \left ( \frac{\partial^2 L}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i \right ] = 0. $$

Again we have a linear differential equation with constant coefficients, and all of the results from the previous section carry over.

Perturbed Lagrangians
What about finding solutions to Lagrangians that are almost like ones that we have already solved? Let's say we have

$$ L = L_0 + L_1 $$

where $$L_1$$ is considered to be small compared to $$L_0$$ Let's say I have some solution to the equations of motion for $$L_0$$ and I would like to look at small deviations from the solution induced by the change in the Lagrangian. Let's say $$ q_{0,i}(t) $$ satisfy

$$ \frac{d}{dt} \frac{\partial L_0}{\partial \dot q_i} - \frac{\partial L_0}{\partial q_i} = 0, $$

and let's look at

$$ q_i(t) = q_{0,i}(t) + \eta_i(t) $$

where $$\eta_i$$ is small. Let's expand the entire Lagrangian to find the equations of motion for the deviations $$\eta_i$$. We have

$$ L \left (q_1, \ldots q_n \right ) = L \left (q_{0,1}, \ldots q_{0,n} ; \dot q_{0,1}, \ldots \dot q_{0,n} \right ) + \sum_{i} \left [ \left ( \frac{\partial L}{\partial q_i} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial L}{\partial \dot q_i} \right )_{q_{0,i}} \dot \eta_i \right ] + $$

$$ \frac{1}{2} \sum_{i,j} \left [ \left ( \frac{\partial^2 L}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i\eta_j + \left ( \frac{\partial^2 L}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \dot\eta_i \dot \eta_j + 2 \left ( \frac{\partial^2 L}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \eta_i \dot \eta_j \right ] + $$

$$ L_1 \left ( q_{0,1}, \ldots q_{0,n}; \dot q_{0,1}, \ldots \dot q_{0,n} \right ) + \sum_{i} \left [ \left ( \frac{\partial L_1}{\partial q_j} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial L_1}{\partial \dot q_j} \right )_{q_{0,i}} \dot \eta_i \right ] + \cdots. $$

Now let's apply Lagrange's equations for the deviations

$$ \frac{d}{dt} \frac{\partial L}{\partial \dot \eta_i} - \frac{\partial L}{\partial \eta_i} = 0 $$

to give

$$ \frac{d}{dt} \left \{ \left ( \frac{\partial L_0}{\partial \dot q_j} \right )_{q_{0,i}} + \left ( \frac{\partial L_1}{\partial \dot q_j} \right )_{q_{0,i}} + \sum_{i} \left [ \left ( \frac{\partial^2 L_0}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \dot\eta_i + \left ( \frac{\partial^2 L_0}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \eta_i \right ] \right \} $$

$$ - \left ( \frac{\partial L_0}{\partial q_j} \right )_{q_{0,i}} + \left ( \frac{\partial L_1}{\partial q_j} \right )_{q_{0,i}} + \sum_{i} \left [ \left ( \frac{\partial^2 L_0}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial^2 L_0}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \dot \eta_i \right ] = 0 $$

The two lowest orders terms without $$\eta$$ actually cancel each other out, leaving the following equations of motion.

$$ \frac{d}{dt} \left \{ \sum_{i} \left [ \left ( \frac{\partial^2 L_0}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \dot\eta_i + \left ( \frac{\partial^2 L_0}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \eta_i \right ] \right \} $$

$$ - \sum_{i} \left [ \left ( \frac{\partial^2 L_0}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i + \left ( \frac{\partial^2 L_0}{\partial q_i \partial \dot q_j} \right )_{q_{0,i}} \dot \eta_i \right ] = \left ( \frac{\partial L_1}{\partial q_j} \right )_{q_{0,i}} - \frac{d}{dt} \left ( \frac{\partial L_1}{\partial \dot q_j} \right )_{q_{0,i}}. $$

Let's specialize and assume that the unperturbed motion is steady so the partial derivatives of the unperturbed Lagrangian are constant in time, to obtain

$$ \sum_{i} \left [ \left ( \frac{\partial^2 L_0}{\partial \dot q_i \partial \dot q_j} \right )_{q_{0,i}} \ddot\eta_i - \left ( \frac{\partial^2 L_0}{\partial q_i \partial q_j} \right )_{q_{0,i}} \eta_i \right ] = \left ( \frac{\partial L_1}{\partial q_j} \right )_{q_{0,i}} - \frac{d}{dt} \left ( \frac{\partial L_1}{\partial \dot q_j} \right )_{q_{0,i}}. $$

which is the equation of a coupled set of driven harmonic oscillators.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Advanced Classical Mechanics/Solving Problems

 * 1) Read the problem
 * 2) Draw a diagram
 * 3) Identify the degrees of freedom
 * 4) Assign a coordinate to each degree of freedom
 * 5) Calculate $$T$$ and $$V$$ in terms of the coordinates
 * 6) Sometimes $$T$$ is difficult; use Cartesian coordinates or whatever system is convenient
 * 7) Convert to the coordinate system in (4)
 * 8) Look for cyclic coordinates
 * 9) Each one gives you a "first integral".
 * 10) A "first integral" is a first-order differential equation.  It is usually easier to solve than Lagrange's equations which are second order.
 * 11) If only one coordinate remains, you can either
 * 12) Use Lagrange's equations, or
 * 13) Use the conservation of $$H$$ if $$\partial L/\partial t=0$$
 * 14) If more than one coordinate remains, it is often easier to use Lagrange's equations for each of them, because the Hamiltonian usually couples the degrees of freedom.
 * 15) You should have as many equations as degrees of freedom. Solve them!
 * 16) Apply the boundary conditions.


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Advanced Classical Mechanics/The Principle of Least Action

Introduction
If we throw a ball through the air, what does its path look like?

Let's first look at the trajectory along the x-direction. I could imagine the ball going a bit faster at a certain time and then travel more slowly later to arrive at the same time. However, because the kinetic energy is quadratic in the velocity doing this would increase the integral of $$T$$ along the trajectory. The area below the red line and above the black line in the top-left panel is larger than the area below the black line and above the red line. This is similar for the blue path.

If we perturb the path in the y-direction we could have the ball go higher or lower. If we make the ball go higher, the integral of both $$T$$ and $$V$$ increases. On the other hand, if we make the ball go lower, the integrals decrease. The changes to the x-trajectory didn't affect the potential energy at all. Looking at the motion in the horizontal direction tells us that the integral of the kinetic energy is minimized but we are free to add a term for the potential energy as well. For the vertical direction, unless we include the potential energy in opposition to the kinetic energy we find that the integral will increase or decrease without bound as we change the peak of the curve.

Minimizing the Integral
Let's try to find the integral to minimize that gives the actual motion of the ball. Let's try

$$ S = \int_1^2 \left ( \alpha T - \beta V \right ) dt = \int_1^2 \left (  T - \beta V \right ) dt = \int_1^2 K dt $$

where we have set $$\alpha=0$$ without loss of generality and defined $$K$$ to be the combination of $$T$$ and $$K$$. What we are interested in finding is the path $$q_i(t)$$ where $$q_1=x$$ and $$q_2=y$$ that results in the minimum value of the integral $$S$$.

This seems like a place to use calculus and take a derivative, but taking a derivative lets you find the minimum of a function with respect to changes in a variable. Here we are interested in the minimum of an integral of a function with respect to changes in that function. What we have is

$$ S = S \left[ q_i(t) \right ] $$ and $$ \frac{\delta S}{\delta q_i} = 0. $$

The brackets denote that the quantity $$S$$ is a functional of the path. The funny deltas denote that we would like to take the functional derivative of $$S$$ with respect to the function $$q_i$$ and find where it is zero.

The Calculus of Variations
It appears that we have introduced a whole new type of mathematical problem. In fact we have, but to solve the problem at hand, we can use a modest extension of the calculus that you already know -- The Calculus of Variations. Let's imagine that the path $$q_i(t)$$ is indeed the one that minimizes the integral -- the true path of the ball through the air -- and look at nearby paths whose starting and ending points coincide with the true path,

$$ p_i(t) = q_i(t) + \gamma r_i(t) $$

Because $$q_i(t)$$ minimizes the integral, the value of the integral $$p_i(t)$$ shouldn't be much different. Let's write it out

$$ S\left [ p_i(t) \right ] = \int_1^2 K \left (p_i(t),{\dot p}_i(t) \right ) d t = \int_1^2 K \left ( q_i(t) + \gamma r_i(t), {\dot q}_i(t) + \gamma {\dot r}_i(t) \right ) dt $$

where $$K$$ is a function of the position and velocity of the ball. One could generalize this for functions that also depended on higher derivatives of the path. We have always imagined that difference between the true path and perturbed path is small so it is natural to use a Taylor expansion of $$K$$ to evaluate the integral

$$ S\left [ p_i(t) \right ] = \int_1^2 \left \{ K \left ( q_i(t), {\dot q}_i(t) \right ) + \gamma \sum_i \left [ \frac{\partial K}{\partial q_i} r_i(t) + \frac{\partial K}{\partial {\dot q_i}} {\dot r}_i(t) \right ] \right \} dt. $$

Because $$q_i(t)$$ minimizes the integral we must have $$d S[p_i]/d\gamma=0$$ for any deviation $$r_i(t)$$ so we must solve

$$ \int_1^2 \sum_i \left [ \frac{\partial K}{\partial q_i} r_i(t) + \frac{\partial K}{\partial {\dot q_i}} {\dot r}_i(t) \right ] dt = 0 $$

If we integrate the second term by parts we get the following integral

$$ \int_1^2 \sum_i \left [ \frac{\partial K}{\partial q_i} - \frac{d}{dt} \frac{\partial K}{\partial {\dot q_i}} \right ] r_i(t) dt = 0. $$

However since the deviation $$r_i(t)$$ is completely arbitrary the quantity in the brackets must vanish, so we find that for the path that mimimizes the integral $$S$$ the following set of differential equations holds

$$ \frac{d}{dt} \frac{\partial K}{\partial {\dot q_i}} - \frac{\partial K}{\partial q_i} = 0. $$

If we take $$K=L=T-V$$, these are simply Lagrange's equations that we derived earlier from $$F=ma$$; therefore, a particle that at every instant manages to follow the second-order differential equation $$F=ma$$ somehow also manages to minimize the integral of $$T-V$$ over time. These two different ways of looking at the same process offer insights into the connection between quantum and classical mechanics and the underpinnings of physics in general. The integral of $$T-V$$ over time is known as the action and it has units of energy multiplied by time. The realm of quantum mechanics is delineated by the constant $$\hbar$$ that also has units of energy multiplied by time. When the change in the action between the classical path and neighbouring paths is smaller than $$\hbar$$, the effects of quantum mechanics become crucial to understand what happens.

Applications
We are free to apply the results derived in the previous section to a wide variety of problems where you are trying to minimize the integral of a function. We only need to find the integrand $$K$$ and apply Lagrange's equations to get a set of differential equations to solve.

Bubbles
An example will make things a bit clearer. Let's say that you have two rings of different radii $$r_1$$ and $$r_2$$ held a distance $$l$$ apart from each other. The rings are parallel to each other and their centres lie on a line perpendicular to the rings.

We dip the rings in soapy water and a bubble forms between them. The surface of the bubble is a surface of revolution and surface tension of the bubble forces it to form a surface that minimizes the area. Essentially there is an energy proportional to the area of the surface and the bubble evolves to minimize this energy.

So we wish to minimize the area of the surface. Let's denote the crossection of the surface by $$r(x)$$ where $$r(0)=r_1$$ and $$r(l)=r_2$$. The total area is

$$ A = \int_0^l 2 \pi r \sqrt{ dr^2 + dx^2 } = \int_0^l 2 \pi r(x) \sqrt{ 1 + \left [ r'(x) \right ]^2 } d x $$

so $$ K = 2 \pi r \sqrt{ 1 + r'^2} $$, $$q_1=r$$ and $$x$$ plays the role of $$t$$. These are just names for things but the basic problem is the same we would like to minimize some functional by changing the function $$r(x)$$. Here is the Lagrange equation,

$$ \frac{d}{dx} \frac{\partial K}{\partial r'} - \frac{\partial K}{\partial r} = 0. $$

which yields

$$ \frac{d}{dx} \left [ \frac{2 \pi r(x) r'(x)}{\sqrt{1+\left [r'(x) \right ]^2}} \right ] = 2\pi \sqrt{1 + \left [ r'(x)\right]^2} $$

and after some simplification we have

$$ 2\pi \frac{r(x) r''(x) - \left [ r'(x) \right ]^2 - 1}{\left \{ 1 + \left [ r'(x) \right ]^2 \right \}^{3/2}} = 0. $$

We will solve this equation by making a guess. We have a function times its second derivative is sort of equal to its first derivative squared. If we have a plus sign between the terms, we would be tempted to use sines or cosines but since there is a minus sign, that points toward hyperbolic functions. Let's try substituting

$$ r(x) = a \cosh \left [ b \left (x - c\right) \right ] $$

into the differential equation. This yields $$a^2 b^2 = 1$$ so we have the following general solution

$$ r(x) = a\cosh \left (\frac{x-c}{a} \right ) $$

that we need to fit to the boundary conditions. The minimum radius of the bubble is $$a$$ at $$x=c$$. The total area of the bubble is

$$ A = \left. \pi a \left [ x - c + \frac{a}{2} \sinh \left ( 2 \frac{x-c}{a} \right ) \right ] \right |_0^l. $$

Some interesting things that you may find. First, if the two rings are different sizes there is only a solution when the rings are sufficiently close together. Second, even for rings of equal size there comes a point where the bubble can take a much different form that has lower area. Specifically, if the bubble collapses to a line between the centres of the rings and fills the rings themselves we have

$$A=\pi \left (r_1^2 + r_2^2 \right).$$

When this area is less than the area of the locally minimizing curve, the bubble can take a much different form and dramatically reduce its energy (it usually pops at this point). The derivation that the minimizing curve satisfied Lagrange's equations assumed implicitly that the curve was smooth and looked for local extrema. Because we could also derive Lagrange's equations from $$F=ma$$ the physical relevance of the equations is not affected by this; rather we have to be a bit careful while applying these equations to minimize integrals.


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Advanced Classical Mechanics

Welcome to the Wikiversity Advanced Classical Mechanics Learning Project, part of the School of Physics.

Project description
This Wikiversity project is for development of learning resources concerned with finding and solving the equations of motion of a system with constraints. These resources help students learn electrodynamics and general relativity, and they are also provided with a powerful (but not so easy) route to quantum mechanics.

Lessons

 * /Introduction/
 * Linear Motion
 * /Energy and Angular Momentum/
 * /Many-Particle Systems/
 * /Compound ballistic pendulum with spinning ball/
 * /Constraints and Lagrange's Equations/
 * /Solving Problems/
 * /The Principle of Least Action/
 * /Conservative Systems/
 * /Dissipative Forces/
 * /Central Forces/
 * /Rigid Bodies/
 * /Small Oscillations and Perturbed Motion/
 * /Continuum Mechanics/
 * /Hamilton's Equations/
 * /The Eikonal Approximation and Classical Particle Motion/
 * /Poisson Brackets/
 * /Phase Space/
 * /Liouville's theorem/

title: Advanced Classical Mechanics


text:
Portal:General relativity

General Relativity Topic within the Astrophysics department

=Topic description= General relativity (GR) is an advanced topic within the department of Astrophysics and requires a high level of knowledge in calculus, algebra and physics in general. Tensor fields are a critical part of general relativity.

Prerequisites

 * Topic:Special relativity (SR) - the spacetimes of GR have the spacetime of SR as a limiting case at a point in spacetime, so you must first understand SR
 * Topic:Differential geometry - you need to understand tensors, tensor fields and differential geometry in general in order to understand GR; elements of these are frequently taught within a GR course as an alternative to a full differential geometry course

Courses

 * Special relativity and steps towards general relativity (shortcut: SRepsilonGR)
 * this course includes a few elements towards elementary general relativity
 * Cosmology
 * twentieth/twenty-first cosmology is one of the main applications of general relativity

Lectures

 * General relativity
 * Introduction to general relativity
 * Inveiling the breakdown of Euclidean geometry in linearized general relativity

Learning Projects
See: Learning Projects and the Learning model.

Learning materials and learning projects can be used by multiple departments. Cooperate with other departments that use the same learning resource.
 * Tensors
 * Tensors Differential Forms and Variational Principles
 * Tidal Forces

Relevant topics: Covariance, Vectors/ Fields, Tensor Calculus, Euler-Lagrange equations, Equivalence principle, Curvature, Riemann Tensor, Gravitational Waves

Problem Reviews

 * Tensors Differential Forms and Variational Principles

External learning resources

 * The website for a GR class taught at Caltech by Kip Thorne and Lee Lindblom. Contains problems, solutions and some lecture notes.
 * The first of a sequence of videos of lectures of an "Introduction to General Relativity" class taught by Kip Thorne.
 * "Lecture Notes on General Relativity" by Sean Carroll. Coincidentally the basis for his book.
 * The General Relativity Tutorial, John Baez: online tutorials and reading list.

Wikipedia

 * Exact solutions in general relativity category
 * Deriving the Schwarzschild solution

Relativité générale

title: Portal:General relativity


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General Relativity/Tensors

Introduction
Tensors are a critical part of General Relativity and since I couldn't get three pages into a General Relativity book without encountering tensors, unless it was written by Brian Greene and didn't have any equations, Jason and I figured this was the place to start.

Currently we have started with "Tensors, Differential Forms and Variational Principles" by David Lovelock an Hanno Rund, which came highly recommended by my good friend Mark Friesel. The link below will go over the problems from each chapter. This particular page will discuss the relevance of tensors to General Relativity.

Syllabus
I don't have the book yet, so I'll defer this to someone else. I know that we'll want to talk about vectors, covectors, covariant tensors, contravariant tensors, metrics, rotation, subscripts and superscpripts, conversion between covariant and contravariant tensors, einstein notation, commmon tensors such as the stress/strain tensor. There are other things such as parallel transport and such that I'm not sure when/where we'll talk about that. - Jhouse

Problem Reviews

 * Tensors_Differential_Forms_and_Variational_Principles

title: General Relativity/Tensors


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General Relativity/Tensors/Tensors Differential Forms and Variational Principles

Please note: We can not submit copyrighted information onto this site. We'll have to have offline discussions over specific problems in the book. I am pretty sure however that we can discuss problems of our own creation that are inspired by similar problems within the books we reference. Jhouse

Chapter 1
In E $$a/b$$
 * Problem 1.1


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General Relativity/Tidal forces

Radially from an object of mass m, the tidal acceleration is $${-2Gm \over r^3}*\Delta X$$ and perpendicular to the radial line, the acceleration is $${Gm \over r^3}*\Delta X$$ (Where G is the gravitational constant and deltaX is the seperation distance of two test particles.

To calculate those, you must remember that the acceleration due to gravity is $${Gm \over r^2}$$. When you consider two test particles separated by a distance of $$\Delta X$$, the results vary depending on if they are along a common radius from the center of the earth or perpendicular to it. (Any other cases can be decomposed into a combination of those two cases). Where the two test particles are radial, the two effective radii are r and (r+\Delta X). If you look at the relative acceleration between the two particles, you get $${Gm \over (r+\Delta X)^2} - {Gm \over r^2} = { {Gmr^2-Gm(r+\Delta X)^2} \over r^2(r+\Delta X)^2} = {-2Gmr \Delta X + Gm \Delta X^2 \over r^2(r+\Delta X)^2} \sim {-2Gmr \Delta X \over r^4} = {-2Gm \over r^3}\Delta X $$

More to follow...

title: General Relativity/Tidal forces


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Talk:Medicinal plants and wild foods

There are some stubby materials on this topic on wikibooks that could be developed further. See b:Category:Weed profiles for pages that deal with wild plants. --SB_Johnny | talk 10:47, 12 October 2006 (UTC)

I would like to get more involved with helping shape the Medicinal Plants and Wild Foods area, and wanted to see if anyone is planning on moving this along that would like to work with me? Feel free to contact me. --Dixiecrat | talk 1:29 PM CDT, 21 July 2008

title: Talk:Medicinal plants and wild foods


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User:Roadrunner~enwikiversity/Researcher advising

What I'm doing....

I'm going to try make a deal with my employer to give me one week each month off, and let me telecommute in the summer. If I don't get this, I'll walk.

I plan to spend that week somewhere other than Austin, Texas, probably MIT mostly. I'll also be attending conferences.

I have enough home equity to afford this. At some point I hope to develop wikiversity to the point that I can apply for grant money, do consulting, and/or tutor students.

Roadrunner 21:20, 5 August 2006 (UTC)

Notes on University of Phoenix
UoP is a **great** place to learn how to do distance education, but I don't think that insturctors are well compensated. Other options for funding are to work at a community college as adjunct faculty.

Roadrunner 21:20, 5 August 2006 (UTC)


text: Template:PD-self

title: Template:PD-self


text:
Template talk:PD-self

b:Template talk:PD-self

title: Template talk:PD-self


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Portal:Physiology

Welcome to the Department of Physiology

 This department is responsible for the research, teaching and assessment in the biomedical area of Physiology. The department is part of the Basic sciences Division of the School of Medicine. The department is involved in the teaching of the medical curriculum, but students are welcome to join in the teaching at any time if they do not wish to follow the recommended pathway.

Physiology is a branch of science that deals with the functioning of living organisms. Physiologists study cells, tissues, organs, and the whole body in an attempt to understand how organisms function. The Department of Physiology is a content development project for Wikiversity. Participants collaborate to develop learning resources for physiology.

Students are invited to take a look at the Human Physiology Wikibook, the Medical Physiology Wikibook, the Physiology articles on Wikipedia, including Human physiology and other more specific subject divisions.

Projects, Research and Assignments
 * There are currently no projects, research, or assignments.

Department Noticeboard
 * This department is under construction. Feel free to improve upon what you see in its establishment.

Learning materials and learning projects can be used by multiple departments. Cooperate with other departments that use the same learning resource.
 * Human Physiology
 * Medical Physiology
 * Fundamentals of Neuroscience

Assessments
 * No assessments have been requested

Modules Offered
 * Physiology

title: Portal:Physiology


text: Category:Physiology

title: Category:Physiology


text:
Introduction to Strategic Studies

Part of the Strategic Studies curriculum

Strategic Studies 101 - Introduction to Strategic Studies
Welcome to Strategic Studies 101. This course is a first-year core course in the Strategic Studies curriculum. Its active learning content should be the equivalent of about 30 hours of classroom time, and it is organized into classes. Each class has a discrete learning objective and a way (or ways) to reach that objective. This will be complemented by readings that are prerequisites to each course. The course is presented in a logical order, and one class may be built upon in subsequent lessons. The order in which you perform these studies, however, is not important. If you feel you'd prefer a different order, that is up to you.

Course Content
This is the foundational course in the Strategic Studies curriculum and presents a broad overview of the subject matter. The main objective of this course is to gain familiarity with war, strategy, and the terminology that will be used throughout this degree program.

By the end of this course, you should know:
 * The differences between Tactics, Grand Tactics, Strategy, and Grand Strategy
 * The basics of military organization
 * A majority of the specialized words used in the study of warfare and strategy
 * Many of the major technical and technological innovations over the course of military history

Active participants
I will be studying this course from the notes. should anyone want to step into a mentoring role, that would be good. --Jolie 19:44, 14 October 2008 (UTC)

I will also be studying this course, among others, and am wondering if there is anyone who would want to share their expertise. Thanks --Iskid2astop 19:03, 15 February 2008 I am about to begin studying this course and I'd greatly appreciate if anyone is willing to share their expertise and resource materials. -A.T. 3rd October 2018

This appears to be very different from the Ilias system of learning. I look forward to completing studies here.

Collaborative Wikibook Project
The Study of Strategy

Online Texts

 * Military_strategy
 * Carl_von_Clausewitz
 * http://www.clausewitz.com/CWZHOME/VomKriege2/ONWARTOC2.HTML

 * The art of war

Offline Texts

 * Art, Robert J. & Waltz, Kenneth N. (Eds.), The Use of Force. Military Power and International Poitics. Lanham: Rowman & Littlefield, 2004 (last edition). ISBN 0742525570
 * Baylis, John, Cohen, Eliot A., Gray, Colin S., Wirtz, James J. (eds.), Strategy in the Contemporary World, Oxford: Oxford University Press, 2002 (ISBN 0-19-878273-X) and 2006 (ISBN 978-0-19-928978-3).
 * Betts, Richard K. (Ed.), Conflict After The Cold War. New York: Pearson-Longman, 2006. ISBN 0-321-20946-X
 * Dawson, Doyne. The Origins of Western Warfare: Militarism and Morality in the Ancient World. Boulder: Westview Press, 1996. ISBN 0-8133-2940-X
 * Freedman, Lawrence. War. Oxford: Oxford University Press, 1994. ISBN 0-19-289254-1
 * Fuller, J. F. C. Armament and History: The Influence of Armament on History from the Dawn of Classical Warfare to the End of the Second World War. New York: Da Capo Press, 1998 (originally published 1945). ISBN 0-306-80859-5
 * Gray, Colin S. Modern Strategy. Oxford: Oxford University Press, 1999. ISBN 0-19-828030-0
 * Hammond, Grant T. "The Mind of War: John Boyd and American Security". Washington: Smithsonian Books, 2001. ISBN 1-56098-941-6
 * Hart, B.H. Lidell Strategy, New York: Penguin Group, 1991 (originally published 1954). ISBN 0-452-01071-3
 * Heuser, Beatrice. The Evolution of Strategy: Thinking War from Antiquity to the Present. Cambridge: Cambridge University Press, 2010. ISBN 978-0-521-15524-3
 * Keegan, John. A History of Warfare. New York: Vintage Books, 1994. ISBN 0-679-73082-6 (London: Hutchison and Knopf editions also available, but references will be taken from the Vintage Books edition)

Classes

 * What Is Strategy? Why Study Strategy?
 * "Policy by Other Means"
 * The Fundamentals of Military Organization

Classes

 * Strategic Resources
 * Geopolitics
 * The State as Rational Actor

Classes

 * Territoriality and Warfare in the Animal Kingdom
 * Monkey Wars
 * Tribal and Ritualized Warfare

Classes

 * Cities, Agriculture, and Early Military Organization
 * The Greek City States and Rome
 * China

Classes

 * From Ritual Warfare to Battle of Annihilation
 * The Greek and Roman traditions
 * Chinese Military Philosophy

Classes

 * The Crusades and Religious Conflict
 * Religion in the 30-Years' War
 * The Beginnings of Modern Warfare

Classes

 * Colonialism and the Balance of Power
 * Re-Ritualization? Law and Positional Warfare
 * Genocide

Classes

 * Napoleon and the Levée en Masse - The Rise of National Armies
 * Guerrilla
 * The State of Military Philosophy

Classes

 * Economy and the Clash of Ideologies in the Cold War
 * Nuclear Strategy
 * Non-State Actors

Classes

 * Clausewitz vs. Jomini (and their respective ilks)
 * Sun Tzu
 * John Boyd

Course Final Project
Final - Introduction to Strategic Studies

Related lessons

 * Deterrence

Online Syllabi

 * Eliot Cohen, Strategy and Policy John Hopkins School of Advanced International Studies(Spring 2001)
 * Richard Betts War, Peace and Strategy Columbia (Fall 2000)
 * Barry R. Posen, Foundations of Security Studies Massachusetts Institute of Technology (Fall 2003)
 * Barry R. Posen, Comparative Grand Strategy and Military Doctrine Massachusetts Institute of Technology (Fall 2003)

title: Introduction to Strategic Studies


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Chain mail/Jewelry

With chain mail you can do at least those jewelry:
 * Necklaces
 * Bracelets
 * Anklets
 * Earring
 * Pendants
 * Headpieces
 * Watch wrist
 * Inlay

Here is a list of many possible weave sorted by difficulty. Their existe tutorial on the web for many of them.

title: Chain mail/Jewelry


text:
Portal:Irish

Welcome to the Irish Studies Portal

title: Portal:Irish


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Talk:Fundamental Mathematics/Arithmetic

I'm trying to create something appropriate for grade 1-3. --HappyCamper 20:59, 12 October 2006 (UTC)

Nice start!
I tried to get the entire 1-12 multiplication table on there but it would only show -9 in the box - the rest dumped onto the page in a long string! Oh well. If you know how to fix that I will put it all in (up to the usual 12, anyway). Keep it coming!sutton 17:46, 9 November 2006 (UTC)
 * Oh, the trick is to add a single space in front of the text you want placed in a box in Courier New font. I also made a table too, maybe that might be better? --HappyCamper 18:10, 9 November 2006 (UTC)

title: Talk:Fundamental Mathematics/Arithmetic


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Buddhism/North American Culture

Buddhist Culture in North America
According to a recent Higher Education Research Institute study, about one percent of American college students self-identify as Buddhist. What factors and conditions resulted to the making of this recent phenomenon?

title: Buddhism/North American Culture


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Buddhism/South Asian Culture

Buddhist Culture in South Asia
A study of Buddhist culture in Sri Lanka, India, Myanmar, Thailand, Indonesia, Cambodia & Vietnam is expected for this paper. The main focus should be on the following themes:
 * Pre-Buddhistic socio-religious conditions
 * Introduction and localization of Buddhism
 * Royal patronage and persecutions on Buddhism
 * Buddhist impact on art, architecture and literature
 * Order of the Buddhist Sangha

Recommended Reading

 * Buddhism in East Asia - S. Dutt, India, 1966
 * 2500 Years of Buddhism - P.V. Bapat, Delhi, 1971
 * Hinduism and Buddhism - C. Eliot, London, 1967
 * History of Theravada Buddhism in South-East Asia - K.L./Hazra, Delhi, 1982

title: Buddhism/South Asian Culture


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Buddhism/Basic Doctrines

Early Buddhism - Basic Doctrines
The vast literature of Buddhist Sutras outlining the various doctrines are mind boggling. The focus of this course will be to examine Buddhist doctrines as presented in the discourses of the Sutta Pitaka, with greater emphasis on the following topics:
 * The Four Noble Truths
 * The Three Signs of Existence (Tilakkhana)
 * Impermanence, No-self or Anatta, and Suffering or Dukkha.

 * The Analysis of Empirical Existence (Khandhayatanadhatu)
 * The Doctrine of Dependent Origination
 * The Doctrines of Kamma (Karma)
 * Buddhist Ethics
 * Nirvana and the Path Leading to its Realization

You are encouraged to study of the life of the Buddha - emphasizing his magnanimous qualities. This is especially useful for understanding the basic doctrines propounded in the core of Buddhist teachings.

The Doctrine of Dependent Origination

Determinism has been expressed in the Buddhist doctrine of Interdependent Origination, which states that every phenomenon is conditioned by, and depends on, the phenomena that it is not. This doctrine is common to all Schools of Buddhism. A common teaching story, called the Indra's Net, illustrates this point. A vast auditorium is decorated with mirrors and/or prisms hanging on strings of different lengths from an immense number of points on the ceiling. One flash of light is sufficient to light the entire display since light bounces and bends from hanging bauble to hanging bauble. Each bauble lights each and every other bauble. So, too, each of us is "lit" by each and every other entity in the Universe. In Buddhism, this teaching is used to demonstrate that to ascribe special value to any one thing is to ignore the interdependence of all things. Volitions of all sentient creatures determine the seeming reality in which we perceive ourself as living, rather than a mechanical universe determining the volitions which humans imagine themselves to be forming.

In the story of the Indra's Net, the light that streams back and forth throughout the display is the analog of karma. We can also view that the resulting 'illusion' is a set of probabilities. A shifting flow of probabilities for futures lies at the heart of theories associated with the Yi Jing (or I Ching, the Book of Changes). Probabilities take the center of the stage away from things and people. A kind of divine volition sets the fundamental rules for the working out of probabilities in the universe, and human volitions are always a factor in the ways that humans can deal with the real world situations one encounters. If one's situation in life is surfing on a tsunami, one still has some range of choices even in that situation. One person might give up, and another person might choose to struggle and perhaps to survive. The Yi Jing mentality is much closer to the mentality of quantum physics than to that of classical physics, and also finds parallelism in voluntarist or Existentialist ideas of taking one's life as one's project.

The Doctrines of Karma

In Buddhism, only intentional actions are karmic "acts of will". Will in philosophy refers to the quality or instance that produces conscious and intended actions. It was seen as the underlying reality of all perceptions. Buddhists subscribed to the concept of intentional actions that will be subjected to the 'Law of Karma'.

The 'Law of Karma' refers to "cause and effect", but Karma literally means "action" - often indicating intent or cause. Accompanying this usually is a separate tenet called Vipaka, meaning result or effect. The re-action or effect can itself also influence an action, and in this way, the chain of causation continues ad infinitum.

When Buddhists talk about karma, they are normally referring to karma/action that is 'tainted' with ignorance. Buddhist theory holds that every karma (every intentional action) will bear karmic fruit (produce an effect somewhere down the line). Karma is the only thing that is fundamentally real. Volitional acts drive the universe. The consequences of this view often confound our ordinary expectations - much in the way quantum physics has results that are strongly counterintuitive.

There is also a completely different type of karma that is neither good nor bad, but liberating. This karma allows for the individual to break the uncontrolled cycle of rebirth which always leads to suffering, and thereby leave samsara to permanently enter Nirvana. The Buddhist sutras explain that in order to generate liberating karma, we must first develop incredibly powerful concentration, and proper insight into the (un)reality of samsara. This concentration is akin to the states of mind required to be reborn in the Deva realm, and in itself depends upon a very deep training in ethical self-discipline. This differentiation between good karma and liberating karma has been used by some scholars to argue that the development of Tantra depended upon Buddhist ideas and philosophies. Karma is related to the notion of Buddhist rebirth - sometimes understood to be the same thing as reincarnation - which has its roots in the principle of Karma.

Recommended Reading:

 * What the Buddha Taught - Ven. Walpola Rahula
 * Buddha and His Teaching - Ven. Narada
 * Life of the Buddha - E. J.Thomas
 * Essentials of Buddhism - Ven. P Gnanarama.

Coursework
You are invited to write, in not more than 1200 words an opinion of what you think on the subject below:

Explore this statement: "The reflection in the mirror is complete in itself." When a mirror fell and broke into many pieces, how do you explain the reflections in each and every one of the fragments?

title: Buddhism/Basic Doctrines


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Buddhism/Early History

Early History of Buddhism up to the Third Century B.C.E.
An exploration of the general knowledge of the history of Buddhism from the enlightenment of the Buddha up to the Buddhist Missionary activities launched in the reign of Emperor Asoka. Special attention are paid to the following topics:

 * Expansion of Buddha’s ministry during his life time;
 * Establishment of the Bhikkhu and Bhikkhuni orders, origin and development of Buddhist monasteries;
 * Causes that led to the first three Buddhist Councils and their contribution to Buddhism and its literature;
 * Emergence of Buddhist Schools and their impact on the development of Buddhist thought;
 * Relationship between Buddhism and the state up to the period of Emperor Asoka; and
 * Spread of Buddhism beyond Indian frontiers.

Recommended Reading:

 * History of Buddhism in Ceylon - Ven. Walpola Rahula
 * Buddhist India - Rhys Davids. Calcutta, 1955
 * Early History of the Spread of Buddhism and the Buddhist Schools - N. Dutt
 * History of Buddhist Thought - E. J. Thomas, London, 1933
 * Indian Buddhism - A. K. Warder, Delhi, 1980

title: Buddhism/Early History


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Buddhism/Emergence and Basic Buddhist Teachings

Emergence of Buddhism and Basic Buddhist Teachings
Religious and philosophical background and basic Buddhist teachings embodied in the Pali Canon are examined in this module. The following topics are emphasized for the religious and philosophical background of Buddhism:
 * Pre-Buddhist paths of emancipation and traditional means of knowledge;
 * Pre-Buddhist teaching of Dhyana and Samapati;
 * Pre-Buddhist beliefs in Karma;
 * Rebirth and the concept of soul;
 * Origin, evolution and objectives of the ascetic movement;
 * Metaphysical and Non-empirical concepts;
 * Six religious teachers and their life stories and philosophies;
 * Self-indulgence and self-mortification (Kamasukhallikanuyoga and Attakilamathanuyoga); skeptical and materialist tendencies.

The following topics are emphasized for the basic teachings embodied in the Pali Canon. Three characteristics of existence; four noble truths; interdependent origination; Analysis of Khandha, Dhatu and Ayatana; Kamma; Rebirth; Buddhist Analysis of Sassatavada and Ucchedavada; Attention should also be paid to the Buddhist analysis of the following topics: world and its inhabitants; sense perception; psychology; ethics; reason, revelation and experience. Four fold analysis of questions and the Buddhist attitude towards language should also be studied here.

Recommended Reading:

 * What The Buddha Taught, W. Rahula, Buddhist Missionary Society, Kuala Lumpur, Malaysia – 1978
 * Early Buddhist Theory of Knowledge, K.N. Jayatilake, Motilal Banarsidas, Delhi -- 1980
 * Causality : The Central Philosophy of Buddhism, D.J. Kalupahana, Honolulu, The University Press of Hawaii – 1975
 * History of Buddhist thought, E.J. Thomas, London-1953
 * The Buddha’s Ancient Path, Piyadasi Thera, Taiwan-1995


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Buddhism/Geographical Expansion

Geographical Expansion of Theravada Buddhism
The focus of this module is on the inception and spread of Theravada Buddhism. This is especially so in the countries of South East Asia. The main emphasis on the following aspects:
 * The Geographical Expansion of Buddhism in the Pre-Asokan period;
 * Buddhism Emergence as a World Religion During the Reign of Emperor Asoka;
 * The Spread of Theravada Buddhism in South India and in the Regions of South East Asia (e.g. Myanmar, Thailand);
 * Religious Intercourses between Theravada Buddhist Countries; and
 * Contemporary Theravada Buddhism.

Recommended Reading:

 * History of Theravada Buddhism in South East Asia - K. L. Hazra
 * Thai Buddhism - Kenneth E. Wells
 * Buddhism in Myanmar; A Short History - Roger Biscoff, Wheel Publications
 * Buddhist India - Rhys Davids

Coursework
You are invited to do the exercise below. Using Google Map, or any other mapping platform, trace the extend of the Asokan Empire. Notice where the geographic limitations and the terrain (mountains and rivers) and what role they played in the spread of military conquest.

Investigate: what caused Asoka to have a change of heart and embrace Buddhism?

title: Buddhism/Geographical Expansion


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Buddhism/Pali Language

Pali is a Middle-Indo Aryan language which serves as the language in which the canonical texts of Theravada Buddhism are maintained. It is also the language used for chanting in the Theravada tradition.

A knowledge of source languages is vitally necessary to any serious student of religion. Without a working knowledge of these languages, the student will remain reading about a religion, even if (s)he is reading a translation of a primary source, rather than reading the bases of the religion directly without an intermediary to fill the head with doctrines and expectations about a religion's nature. The primary language of Buddhism is Sanskrit, but Pali is also used by all Theravadin Buddhists, and the Pali Canon is the core textual body of all Buddhist schools. Therefore, an understanding of Pali is very important to a person seeking a graduate-level understanding of the religion.

Requirements
The student is expected to understand the following:
 * The origins of the language
 * The origins of the Pali Canon
 * Ability to recite from memory the Pali alphabet
 * A full knowledge of pronunciation
 * The relationship between Pali and other Prakrits, particularly the mutual influence between it and Sanskrit
 * A working vocabulary
 * Culminating in an ability to read and comprehend Buddhist scriptures in the original Pali without a translation as aid

title: Buddhism/Pali Language


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Buddhism/Introduction

Introduction to Buddhism
Welcome to the introductory courses. Many of us had heard of Buddha or Buddhism in one way or another. There are quite a lot of misconceptions about Buddhism. It is the intention of this introductory course to guide you through a general understanding of Buddhism.

You may want to be introduced to the subject matter by reading about the Buddha or historic Buddha  and how Buddhism started. An overview of the history will be helpful for students attending this course. Another insight is to review the timeline of Buddhism.

The Buddha referred to his teachings simply as Dhamma-vinaya or "the doctrine and discipline". Many people have tried to categorise the teachings in various ways, trying to fit them into prevailing molds of cultural, philosophical and religious thought. Buddhism, arguably, is best viewed as an ethical system or that of a systematic social education. The system provides a guide to a way of life that leads to a very specific goal. In that way, Buddhism possesses some aspects of both religion and philosophy. Buddhism is an education by Buddha. Many masters have indicated that Buddhism neither fall in religion nor philosophical field.

There are many schools in Buddhism, and further divisions and subdivisions of the schools. You may refer to them here. It is important to bear in mind that only a few schools survived today, which will be covered in the Diploma in Buddhism course.

Of interest, it's noted that Buddhism had spawned a whole genre of Buddhist art, especially evolving from aniconic phase where there is a conscious avoiding direct representations of the Buddha. It was around the 1st century AD that Buddhist Art entered an iconic phase or that with direct representations of the Buddha.

Coursework
You are invited to write, in not more than 1200 words a topic from below:
 * Buddhism as I know it
 * The historic Buddha
 * Buddha oracle - In a playful way, we learn the main principles of Buddhism.

To experience the diversity of Buddhism beliefs and philosophy, you are invited to travel to the recommended temples listed below. You are to note and write in 1200 words what are the common underlying principles amidst the diversity presented across cultural and geographic region.
 * Borobudur http://en.wikipedia.org/wiki/Borobudur
 * Huacheng Temple http://en.wikipedia.org/wiki/Huacheng_Temple
 * Mahabodhi Temple http://en.wikipedia.org/wiki/Mahabodhi_Temple
 * Shaolin Monastery http://en.wikipedia.org/wiki/Shaolin_temple
 * Temple of the Golden Pavillion http://en.wikipedia.org/wiki/Kinkaku-ji

Introdução ao Budismo

title: Buddhism/Introduction


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Sanskrit Language

Courses
2 years with each level as quarter 3 years with each level as semester

First Level

 * Introduction to Sanskrit
 * Sanskrit as an embodiment of Reality
 * Introduction to Devanagari
 * The calligraphy
 * About pronunciation
 * Vowels
 * Consonants
 * Compound consonants
 * Romanized Sanskrit
 * Capitalization vs. doubling
 * Examples from sacred texts
 * Sanskrit grammar
 * Declination of verbs
 * Case of nouns
 * Combining rules: Sandhi
 * Sanskrit Grammar: Sutras of Panini
 * Pandits and tradition
 * The teacher/pupll relationship
 * Maintaining accuracy through the centuries
 * The six vedangas
 * Readings from sacred texts
 * Bhagavad Gita
 * Rig Veda: Mandalas 1, 9, and 10
 * Upanishads

Second Level

 * Etymology: translating Sanskrit words
 * Comparative studies of Sanskrit and English
 * Devanagari 1
 * Sanskrit Grammar 2
 * Sanskrit pronunciation 2: chanting

Third Level

 * Translating Sanskrit: sentences
 * Devanagari 2
 * Comparative studies of Sanskrit and Pāli
 * Sanskrit literature 1: poems

 * Sanskrit pronunciation 3: talking

Fourth Level

 * Writing in Sanskrit: poems
 * Devanagari calligraphy
 * Translating Sanskrit: scripture
 * Sanskrit literature 2: prose
 * Introduction to Buddhism

Fifth Level

 * Devanagari calligraphy 2
 * Comparative studies of Sanskrit and AngloSaxon
 * Sanskrit literature 3: sacred scriptures
 * Sanskrit pronunciation 4: discourse
 * Buddhist scriptures

Sixth Level

 * Devanagari calligraphy 2
 * Sanskrit history
 * Sanskrit literature 4: translating scriptures
 * Writing in Sanskrit: prose
 * Comparative studies of Sanskrit and Latin

Other

 * [[File:Spoken sanskrit 1.pdf]]

title: Sanskrit Language


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Buddhism/Schools

Introduction
There are many schools of Buddhism while the surviving schools today can be roughly grouped under the categories of Theravāda, Mahāyāna and Vajrayāna schools.

It will be useful for us to look at the circumstances that lead to the divisions among the early Buddhist schools. The division came about due to doctrinal or practical differences in the views of the Buddhist Sangha following the death of the Buddha. The initial split was between Sthaviravāda and Mahāsaṃghika occurred about 100 years after Buddha's death, due to differing views concerning the rigidity of monastic rules.

It is important for us to follow the development of various school of thoughts and how they had developed to what we have today.

List of schools

 * /Theravāda/
 * /Mahāyāna/
 * /Vajrayāna/
 * /Sthaviravāda/
 * /Mahāsaṃghika/
 * /Pudgalavāda/
 * /Pure Land/
 * /Ekayāna/
 * /Nikāya/

title: Buddhism/Schools


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Buddhism/Schools/Mahāsaṃghika

 * Mahāsaṃghika ('Majority', c. 380 BCE)
 * Ekavyahārikas (under Asoka)
 * Lokottaravāda
 * Golulika (during Asoka)
 * Bahuśrutīya
 * Prajñaptivāda
 * Cetiyavāda
 * Caitika
 * Apara Śaila
 * Uttara Śaila

title: Buddhism/Schools/Mahāsaṃghika


text:
Buddhism/Schools/Mahāyāna

Mahayana

 * Madhyamaka
 * Prāsangaka
 * Yogācāra
 * Tathagatagarbha
 * Wei-Shi (Consciousness-only school) or Faxiang (Dharma-character school)
 * Beopsang
 * Hossō
 * Sanlun (Three Treatise school)
 * Sanron
 * Daśabhūmikā (absorbed in to Huayan)
 * Huayan (Avataṃsaka)
 * Hwaeom
 * Kegon
 * Chan / Zen / Seon
 * Caodong
 * Soto
 * Linji
 * Rinzai
 * Ōbaku
 * Fuke
 * Pure Land (Amidism)
 * Jodo
 * Jodo Shin
 * Tiantai (Lotus Sutra School)
 * Cheontae
 * Tendai (also contained Vajrayana elements)
 * Nichiren
 * Nichiren Shū
 * Nichiren Shōshū
 * Nipponzan Myōhōji
 * Soka Gakkai
 * Vijñānavāda

title: Buddhism/Schools/Mahāyāna


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Buddhism/Origins

Origins of Buddhism and the Basic Concepts of Culture
A systematic knowledge on the early period of Buddhist civilization is expected with special emphasis on the topics given below : - The Buddhist theory pertaining to the nature of sentient being; - the true nature if empirical exsistence according to Buddhism; - Buddhist analysis of perception, mind and matter; - the Buddhist teaching on Critical outlook, universal love , adaptability, Democratic nature.
 * Buddhist critique of its religio-philosophic environment
 * Special Characteristics of the life of the Buddha
 * Fundamental teachings of Buddhism :
 * Social, political and economic thought in Buddhism

Recommended Reading:

 * Studies in the Origin of Buddhism, G.C. Pande , Delhi ,1983
 * Buddhist Thought in India, E.Conze, London , 1982
 * A Theory of Causality in Early Buddhism, W.S. Karunaratne , Colombo , 1988
 * Buddhist Philosophy: A Historical Analysis, D.J. Kalupahana, Honolulu , 1976

title: Buddhism/Origins


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Buddhism/Philosophy of Early History Buddhist Schools

Philosophy of Early History Buddhist Schools
This module examines comprehensively the Sthaviravada and Mahasanghika schools of Buddhism which came into existence after the second council. Their philosophical teachings are explored with special attention to the following:
 * The relationship between the philosophical trends in the Buddhist thought appearing in the Suttapitaka and the Buddhist thought of the early schools of Buddhism;
 * Different views of Dharmavada and their various revisions; analysis of mind and sense perception; analysis of matter and atomism;
 * The concept of time and the theory of momentariness;
 * Analysis and synthesis of Dharma;
 * Divisions of truths as conventional and absolute and their mutual relationship;
 * Nirvana and its sectarian interpretations.

Recommended Reading:

 * Central Conception of Buddhism - K. Stcherbatsky, Calcutta, 1956
 * Early History of the Spread of Buddhism and Buddhist School - N. Dutt, New Delhi, 1980
 * Guide Through the Abhidhamma Pitaka – Nanatiloka, Colombo, 1957
 * Indian Buddhism - A. K. Warder, Delhi, 1980
 * Points of Controversy (Kathavatthu) - Trs. S. Z. Aung and Mrs. Rhys Davids. London, 1958
 * 2500 Years of Buddhism - Ed. P. V. Bapat, Delhi, 1956
 * System of Buddhist Thought - Yamakami Sogen, Calcutta, 1912


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Buddhism/Philosophy of Mahayana Buddhist Schools

Philosophy of Mahāyāna Buddhist Schools
A critical study of the origin, expansion and philosophy of Mahayana Buddhist schools are emphasised. Reference to the original sources is done with special attention drawn to the following:
 * Canonical evidence of the causes that led to the emergence of Mahayana;
 * The interpretation of the word Mahayana and the nature of its philosophy
 * Origin and development of Madhyamaka philosophy; concept of Sunyata and its interpretations;
 * Madhyamaka viewpoint of voidness; interdependent origination; Middle path;
 * Samsara and Nirvana;
 * Dialectical method and its adaptation;
 * Nature and realization of unconditioned realities;
 * Rangtong vs. Shentong.

Special attention are paid to the concepts of Trikaya, Bodhicitta, Bodhisattva pranidhana, ideal of Bodhisattva, perfections (paramita) and ten states (dasabhumi); the causes that led to the emergence of Yogacara vignanavada; analysis of Citta matrata and Vignanamatrata; various divisions, nature and function of vignanta and how vignana differs from cittamatrata.

A general knowledge of the basic tenets and nature of Tantrayana which is an offshort of Mahayana is also explored in this module.

Recommended Reading:

 * Prajna Paramita Hrdaya Sastra (with text and notes) - Ed. E. Conze, Buddhist wisdom book, 1958
 * The Central Philosophy of Buddhism - T. R. V. Murti, London, 1955
 * Buddhist Thought in India - E. Conze, London
 * The Yogacara Idealism – A. K. Chaterjee, varanasi, 1975
 * The Bodhisatva Doctrine in Buddhist Sanskrit Literature - Har Dayal, Motilal Banarsidas, Delhi
 * Mahayana Buddhism - Nalinaksha Dutt, Mortilal Banarasidas, Delhi, 1978
 * An introduction to Mahayana Buddhism – W.M. McGovern, Delhi
 * Perfection of Wisdom - R. C. Jamieson, New York, 2000, 0670889342 0711215103
 * Meditation on Emptiness - Jeffrey Hopkins, Wisdom PublicationS