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# Will a ball slide down a lumpy hill over the same path it rolls down the hill?

Suppose I have a lumpy hill. In a first experiment, the hill is frictionless and I let a ball slide down, starting from rest. I watch the path it takes (the time-independent trail it follows).

In the next experiment, the hill stays the same shape, but the ball now rolls without slipping down the hill. Assume there is no deformation of the ball or of the hill, no micro-sliding of the contact surface, and no other forms of rolling resistance. Energy is conserved. I release the ball from rest at same point on the hill. I track the path the rolling ball takes.

Does the rolling ball follow the same path as the sliding ball, but more slowly, or does it sometimes follow a different path?

Note: There may be some "trick" answers. For example, if the hill curves significantly on scales similar to the radius of the ball, the ball could get "wedged in" somewhere, or two parts of the ball could contact the hill at the same time. Let's assume that geometrically, the shape of the hill is such that it is only possible to have the ball contact the hill at one point, and that the ball always contacts the hill at a single point (i.e. it never flies off).

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Good, you finally limited the conditions. (It was easy to imagine a scenario where the ball picks up so much speed it flies off the hill and lands further down.) I think at least one of answers could have taken this into account, but it seems several were assuming it couldn't. –  Mark C Nov 16 '10 at 14:51
Actually, that condition was included within minutes of when I first posted the problem (check revision history). –  Mark Eichenlaub Nov 16 '10 at 17:25

Lagrangian solution: Use Cartesian coordinates (x,y) and let the height of the hill define the potential. Each point (x,y) gives a unique position of the ball which therefore uniquely determines its potential (which is not in general equal to the height since the ball can lean over at various angles but this is a detail that does not alter the math here). Thus the potential part of the Lagrangian can be written as V(x,y).

Following the useful and obvious comment by Mark Eichenlaub that $\omega = Rv$, and noting that the kinetic energy depends only on the squares of velocity and angular momentum, and calculating that $\vec{v} = (\dot{x},\dot{y},h_x\dot{x}+h_y\dot{y})$ where $h_x,h_y$ are the slopes of the hill in the x and y directions, the kinetic energy can be written in the form
$T(x,y,\dot{x},\dot{y}) = \alpha(\dot{x}^2A+\dot{y}^2B+\dot{x}\dot{y}C)$
where $A,B,C$ depend only on the slope at a given position, and $\alpha$ depends on the moment of inertia and is different between the rolling and sliding ball. (I.e. the sliding ball has the same energy as a rolling ball with zero moment of inertia.) Note, per Qmechainc that $A,B,C$ do depend on position through the slope. Of course the potential energy depends only on the position.

The equations of motion are:
$\frac{d}{dt}\frac{\partial T}{\partial \dot{x}} = \frac{\partial (T+V)}{\partial x}$
and similarly for y.

Let $t\to t/\sqrt{\alpha}$. This makes $dt\to dt/\sqrt{\alpha}$ and $\partial \dot{x} \to \sqrt{\alpha} \partial \dot{x}$. So the operator acting on T is unaffected by this transformation:

$\frac{d}{dt}\frac{\partial }{\partial\dot{x}} \to \frac{d}{dt}\frac{\partial }{ \partial\dot{x}}$,

and since T is only quadratic in the velocities it transforms as the square of the velocities, i.e. $\dot{x}^2 \to \dot{x}^2/\alpha$:
$T\to T/\alpha$
This eliminates $\alpha$ from the equations of motion.

By the way, a similar argument shows that the answer to the question:

"Does the path a ball slides down a hill depend on the magnitude of the force of gravity?"

is also no, which is a more intuitive way of understanding the answer.

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@Carl The calculation looks right, but isn't it obvious that $\omega^2$ is proportional to $v^2$ because $v = \omega R$? I don't understand why this is a Langrangian solution, or how this dovetails into an energy argument. Energy considerations alone should only tell us about the magnitudes of $v$ and $\omega$, not their directions. –  Mark Eichenlaub Feb 6 '11 at 2:27
I'll add some more... –  Carl Brannen Feb 6 '11 at 21:11
@Carl Well done. I like this more theoretical approach. –  Mark Eichenlaub Feb 7 '11 at 7:02
Dear @Carl Brannen: Even when, e.g., $A$ only depends indirectly on position $(x,y)$ via height and slope, one must still use the chain rule, so I don't see why $\frac{\partial T}{\partial x}$ and $\frac{\partial T}{\partial y}$ should be zero, as is assumed in the answer (v5). –  Qmechanic Apr 23 '11 at 16:06
@Qmech; Yes, you're right. Fortunately it doesn't effect the argument! I'll edit it. –  Carl Brannen Apr 25 '11 at 0:07

I think the answer is that yes, they do follow the same path. I'll try to show this by doing some kinematics, then simply calculating the accelerations. I could use someone to check it over carefully. I have had to use some heuristics in a couple of places because my vocabulary for discussing this problem is limited. I don't see fundamental flaws in the physics, but please point them out (as well as typos) if you see them.

• $m$ = mass of ball
• $I$ = moment of inertia of ball
• $R$ = radius of ball
• $\vec{R}$ = vector pointing from the center of the ball to the point of contact with the hill
• $\vec{g}$ = acceleration due to gravity
• $\vec{g}_h$ = acceleration due to gravity projected into the plane of the hill
• $\vec{\omega}$ = angular velocity of the ball
• $\vec{v}$ = velocity of the ball
• $\vec{\tau}$ = torque on the ball
• $\vec{f}$ = force of friction exerted on the ball by the hill
• $\vec{a}$ = acceleration of the ball
• $\vec{a}_h$ = acceleration of the ball projected in the plane of the hill
• subscript "r" on the above refers to rolling ball
• subscript "s" refers to sliding ball
• Energy is conserved in both cases, so the two balls must have the same kinetic energy as they pass through any given point. The kinetic energy of the sliding ball is $1/2 mv_s^2$, and the kinetic energy of the rolling ball is $1/2(mv_r^2 + I\omega^2)$. Using $|\omega| = |v|/R$, some algebra yields $(v_s/v_r)^2 = 1+I/mR^2$. This tells us that if the rolling ball follows the same path as the sliding ball, it does so by doing everything slower by this set amount. Let's let $v_r/v_s = c$.

We want to know whether the balls move on the same path. They will do this if they always point the same direction when they get to the same place. If we allow their velocities to be functions of position, they follow the same path if $\vec{v_r}(\vec{x}) = c\vec{v_s}(\vec{x})$. We know this will hold for the velocities' magnitudes, so this equation is expressing the condition that the velocities' directions are the same.

Let's introduce a parameters $\lambda_r$, $\lambda_s$ along the balls' paths. If they will follow the same path, the $\lambda$'s can be identical. Make $\lambda$ equal to the distance the ball travels. Differentiating both sides of the previous condition with respect to $\lambda$ and applying the chain rule gives $d\vec{v_r}/dt*dt/d\lambda = c d\vec{v_s}/dt*dt/d\lambda$. These expressions are accelerations and velocities, and so this can be rewritten $\vec{a_r}/|v_r| = c\vec{a_s}/|v_s|$. The definition of $c$ allows us to write this as $\vec{a_r} = c^2\vec{a_s}$ or $\vec{a_s} = (1 + I/mR^2)\vec{a_r}$. As long as this condition holds, the assumption that the $\lambda$'s are the same is justified, the velocities of the balls will change the same way as they move through space, and the paths will be the same.

First we'll look at the component of acceleration normal to the hill. Imagine that at some point on the hill both balls are moving in the same direction. Then we take a circle that is tangent to the path the balls are following right at that point, like a loop coming up out of the hill. The acceleration of the balls as they follow their paths at this point is just their centripetal acceleration going around this circle, and that's $v^2/r$. Since the ratio of the velocities is $c$, the ratio of the centripetal accelerations is $c^2$, which is exactly the ratio specified in the condition that the accelerations must satisfy. This means the projections of the accelerations normal to the hill "work out" in terms of the balls following the same path.

Next, we need the acceleration of the sliding ball projected into the hill, $\vec{a_{sh}}$. There is only one force that has any component in the hill, and that's gravity. So we have $\vec{a_{sh}} = \vec{g_h}$.

The rolling ball has two forces that contribute in the plane of the hill - gravity and friction. This gives $\vec{a_{rh}} = \vec{g_h} + \vec{f}/m$.

To find $\vec{f}$, we'll make two requirements of it. One is what's above - that $\vec{f}$ is what's needed to add to gravity to make the acceleration. Another is that $\vec{f}$ contributes a torque about the center of mass of the ball. This torque can be determined kinematically though, using only knowledge of $\vec{a}$. There will be only one friction force capable of satisfying both conditions.

The torque due to friction is $\vec{\tau} = \vec{R}\times\vec{f} = I d\vec{\omega}/dt$. To find another expression involving $d\vec{\omega}/dt$, note that the bottom of the ball is moving backwards relative to the center of mass, so $\vec{\omega} \times \vec{R} = -\vec{v_r}$. Differentiate with respect to time. $d\vec{R}/dt\times\vec{\omega} + \vec{R}\times d\vec{\omega}/dt = \vec{a_r}$. Plugging in for $d\vec{\omega}/dt$ gives $d\vec{R}/dt\times\vec{\omega} + \vec{R} \times (1/I* \vec{R} \times \vec{f}) = \vec{a_r}$. We are only interested in the component of $\vec{a_r}$ in the plane of the hill. The term $d\vec{R}/dt\times\vec{\omega}$ is normal to the hill because both vectors being crossed are in the plane of the hill. The term $\vec{R} \times (1/I *\vec{R} \times \vec{f})$ is completely in the plane of the hill because it's perpendicular to $\vec{R}$. A vector identity lets us simplify this to $1/I(\vec{R}(\vec{R}\cdot\vec{f}) - \vec{f}(\vec{R}\cdot\vec{R})) = \vec{a_{rh}}$. Use that $\vec{R} \cdot \vec{f} = 0$ because they are perpendicular, so $\vec{f} R^2/I = \vec{a_{rh}}$.

We have two expressions for $\vec{a_{rh}}$, so equate them. $\vec{f}R^2/I = \vec{g_h} + \vec{f}/m$. This gives $\vec{f} = -\vec{g_h}+mI/(R^2m+I)$. Plugging this back into the equation for acceleration gives $\vec{a_{rh}} = \vec{g_h}(1 - I/(R^2m+I))$. Comparing to the sliding ball, $\vec{a_{sh}} = (1 + I/mR^2)\vec{a_{rh}}$. This is the condition on the accelerations we sought to demonstrate.

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I haven't read all of this because you have a fundamental flaw right at the beginning. You say energy is conserved in both cases. Well, it isn't in the case with friction (unless you plan to account for the energy of all the universe). I think this should be obvious to anyone who has seen e.g. a football rolling on a grass plane. It certainly stops at some moment: the kinetic energy having been transformed to heat through the thermal interaction of football and ground atoms. (to be continued) –  Marek Nov 15 '10 at 10:47
More concisely: the case with friction is a real-world example. In the real world, energy goes away in the form of heat and systems attain stable equilibria and stop moving at some moment. So this is certainly different from the frictionless case: imagine a valley and roll the ball down the hill: it goes down the valley, through the bottom, climbs up the second hill and then goes back, moving in this manner indefinitely as a kind of oscillator. This is not what we observe in the real world. –  Marek Nov 15 '10 at 10:47
I know that in the real world things slow down due to friction, but this is just a theoretical example. The friction here does no work because the ball rolls without slipping, i.e. there is no relative motion between the bottom of the ball and the hill. This may be unrealistic, but we can still analyze this ideal case. –  Mark Eichenlaub Nov 15 '10 at 11:32
Nice proof! I still suspect that there should be a more conceptual way of seeing all this, and I need to go over the argument a few more times, but this was definitely not what I was expecting... –  j.c. Nov 15 '10 at 23:54
@Tim I like the direction you're going. Sliding is simply a result of a mismatch between the translational and angular velocities of the ball, so it is not obvious to me that friction must point against other forces on the ball in general. However, since gravity produces zero torque about the center of mass it is not too hard to intuit that a force opposing gravity can be a solution (and therefore would be the solution) in this special case. Would you be interested in writing your thoughts up as an answer? –  Mark Eichenlaub Nov 18 '10 at 7:16

A ball rolling down the hill will go slower than a ball sliding down the hill, because of its angular momentum. So if it rolls down the hill and into a banked curve, it will take the curve lower on the bank, and thus have a different path.

EDIT: as the other answers show, this is in fact incorrect, for the reason that I suggested in my comment.

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Because how far up the bank you go is a balance between centrifugal force (dependent on speed) and gravity (independent of speed). Unless maybe there's also a force coming from changing the moment of inertia of the ball. If this third force occurs, then when you send a hollow ball and a solid ball at the same rpm around the inside of a bowl, the hollow ball would be closer to the rim. That would be a neat demonstration. –  Peter Shor Nov 14 '10 at 12:46
I finally understand your point, but I think it actually works out that they behave the same when banking. Although the rolling ball goes more slowly, the friction force (which provides a torque about the ball's center and induces the precession you mentioned) is just enough to hold the ball up. Specifically, I put the balls in a cone that opens upward at a fixed angle $\theta$. I imagined them orbiting at a fixed distance $l$ up the side of the cone. Then I worked out the velocity needed to maintain this orbit. I found a ratio of velocities $(v_s/v_r)^2 = 1 + I/mR^2$. –  Mark Eichenlaub Nov 15 '10 at 1:31
$v_s$ and $v_r$ and the slipping and rolling velocities. $I$ is the moment of inertia, $m$ the mass, and $R$ the radius of the ball. But this ratio is exactly what we expect! The rolling ball has kinetic energy $1/2[mv_r^2 + I(v_r/r)^2]$ and the slipping ball has kinetic energy $1/2mv_s^2$. When we set this kinetic energies equal and equate the masses, we again get $(v_s/v_r)^2=1+I/mR^2$. Hence, the rolling ball needs to go more slowly to bank at the same height, but it does in fact go more slowly by just the right amount. –  Mark Eichenlaub Nov 15 '10 at 1:35
@Marek, if it will clarify the question, please just imagine I never used the word "friction", and simply asked about the difference between perfect sliding and rolling motions. –  Mark Eichenlaub Nov 15 '10 at 12:01
@Tim Calculate it. –  Mark Eichenlaub Nov 17 '10 at 7:43

In one dimension, the answer is that the path must be the same, since the only possible thing that could go wrong is that the rolling ball might not make it over a hill that the sliding ball does (or vice versa). However, this will never happen due to conservation of energy.

I tried for a bit to put the general case into Lagrangian dynamics (there the question is formulated as whether the projections of $\vec{q}(t)$ will be the same, given two different Lagrangians) but then I remembered that a ball rolling without slipping is a non-holonomic system which makes the problem non-integrable, and I gave up. Maybe I stopped too soon.

I suspect there is a clever solution to the problem; my intuition is that the potential energy term depends only on position and this implies that the trajectories will also be independent of the speed of the ball too. I'll think a bit more about this.

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Thanks for the link. I encountered the same basic problem. At least now I know what it's called! –  Mark Eichenlaub Nov 13 '10 at 19:47
"Maybe I stopped too soon.": I think you did, this case altough at first being non-holonomic can be put in a holonomic form quite easily. –  Cedric H. Nov 14 '10 at 13:18
@Cedric H. I encourage you to write your own answer! Aside: at the moment I think I'm convinced by Peter Shor's argument in the comments to his answer. –  j.c. Nov 14 '10 at 19:00
Mark's detailed answer I think solves this question now. –  j.c. Nov 15 '10 at 23:54
You may not need to completely integrate the problem to show that the paths are the same. For example, suppose you use (x,y) to describe the points of possible contact for the ball. The rotational energy will depend on the derivatives (in a rather complicated manner unless you assume that R << the worst curvature). What you need is to show that dy/dx is the same for the two paths. And I think that dy/dx may arise when you simplify a differential equation with no t dependence. –  Carl Brannen Feb 5 '11 at 23:16

1) Problem. Assume that the ball at all times touches the hill in precisely one point of contact $P$. Let the ball have radius $R$, have mass $m$, be spherically symmetric (but not necessarily with uniform density), and with moment of inertia $$I~=~(\alpha-1) mR^2,$$ where $\alpha\geq 1$ is a dimensionless constant. The value $I=\frac{2}{5}mR^2$ ($\alpha=\frac{7}{5}$) corresponds to a ball with uniform density. The point is now, as Carl Brannen observes, that a rolling ball with no sliding and zero moment of inertia $I=0$ ($\alpha=1$) corresponds to a ball sliding without friction and without spinning. From now on, we therefore only have to consider rolling without sliding for various value of $I$. The question can hence be reformulated as follows.

Question. Is the (time-independent) path of the rolling ball independent of the moment of inertia $I$ if the ball starts from rest with no initial spinning normal to the surface?

We assert the following two theorems.

Theorem 1. The answer is in general No.

This is basically because the ball can on a generic surface spontaneously acquire a spin component $\omega_{\! N}$ perpendicular to the surface.

Theorem 2. The answer is Yes, if furthermore the Hessian ${\bf H}$ in each point is proportional to the $2\times 2$ unit matrix ${\bf 1}_{2\times 2}$, $${\bf H} ~\propto~ {\bf 1}_{2\times 2}. \qquad (1)$$

The Hessian ${\bf H}$ is defined in eq. (2) below. We will show that condition (1) implies that that the ball cannot spontaneously acquire a spin component $\omega_{\! N}$ perpendicular to the surface.

2) Center-of-mass surface. Let the surface of the hill be described by a smooth function $x^3=h(x^1,x^2)$. Since we assume that the ball with radius $R$ continuous to touch the hill in precisely one point $P$, the center-of-mass (CM) of the ball will lie on a curve $x^3=f(x^1,x^2)$, where the distance, between the two graphs

$$\mathrm{graph}(h)~=~\{(x^1,x^2,x^3)\in \mathbb{R}^3 \mid x^3=h(x^1,x^2)\},$$

and

$$\mathrm{graph}(f)~=~\{(x^1,x^2,x^3)\in \mathbb{R}^3 \mid x^3=f(x^1,x^2)\},$$

is $R$ everywhere. (Formulated more precisely: The distance from one graph to an arbitrary point $P$ on the other graph is always $R$, independent of the point $P$.) We will next choose the horizontal center-of-mass coordinates $(x^1,x^2)$ as generalized coordinates for the ball. The CM $x^3$-coordinate is then just $f(x^1,x^2)$, which determines the potential energy

$$V(x^1,x^2)~=~mgf(x^1,x^2)$$

of the ball. We are not able to describe the orientation of the ball only with the help of the generalized coordinates $(x^1,x^2)$. Since we don't describe the orientation of the ball, we will avoid non-holonomic constraints. Next note that friction and normal force do no work on the ball, and that the total mechanical energy $E=T+V$ of the ball is conserved.

3) Newtonian analysis. A spin normal to the surface is not captured by the configuration space variables $(x^1,x^2,\dot{x}^1,\dot{x}^2)$ alone, so the Lagrangian $(x^1,x^2,\dot{x}^1,\dot{x}^2)$ formulation is incomplete if there is such spin. Here we will first analyze the general system using Newton's laws without assuming any particular surface or initial conditions.

Center-of-mass point CM: $${\bf r}_{cm}~:=~(x^1,x^2, f(x^1,x^2)) ~=~{\bf r}_p + R{\bf n}, \qquad x^3_{cm}~=~f(x^1,x^2).$$

Contact point $P$:

$${\bf r}_p~:=~{\bf r}_{cm}- R{\bf n}, \qquad \dot{\bf r}_{cm}~=~\dot{\bf r}_p + R\dot{\bf n}, \qquad \dot{\bf r}_{cm} \perp {\bf n} \perp \dot{\bf r}_p.$$

Tangent vectors to the surface in the CM point:

$${\bf r}_{cm,1}~:=~ (1,0,f_1), \qquad {\bf r}_{cm,2}~:=~(0,1,f_2),$$ $$\dot{\bf r}_{cm}~=~\sum_{i=1}^2{\bf r}_{cm,i}\dot{x}^i, \qquad f_i~:=~\frac{\partial f}{\partial x^i}, \qquad i=1,2.$$

Normal vector to the surface in the CM point:

$${\bf N}~:=~{\bf r}_{cm,1} \times {\bf r}_{cm,2}~=~(-f_1,-f_2,1).$$

Length of normal vector:

$$N~:=~|{\bf N}|~=~\sqrt{1+\sum_{i=1}^2 f_i f_i}~>~0,$$

Hessian:

$${\bf H}~:=~\left[\begin{array}{cc} f_{11} &f_{12} \cr f_{21} &f_{22} \end{array}\right], \qquad f_{ij}~:=~\frac{\partial^2 f}{\partial x^i\partial x^j}, \qquad i,j=1,2. \qquad (2)$$

Time derivative of Normal vector:

$$\dot{N}_i~=~-\sum_{j=1}^2 f_{ij} \dot{x}^j, \qquad i=1,2, \qquad \dot{N}_3~=~0, \qquad\dot{N}~=~ \sum_{i,j=1}^2\frac{f_i f_{ij} \dot{x}^j}{N}.$$

Unit normal vector:

$${\bf n}~:=~ \frac{{\bf N}}{N}, \qquad |{\bf n}|~=~1, \qquad n_i ~=~- \frac{f_i}{N}, \qquad i=1,2,\qquad n_3 ~=~ \frac{1}{N},$$

$$\dot{\bf n}~=~ \frac{\dot{\bf N}}{N} -\frac{\dot{N}}{N^2} {\bf N}, \qquad \dot{\bf n} \perp {\bf n} .$$

In components:

$$\dot{n}_k~=~ -\sum_{j=1}^2\frac{f_{kj} \dot{x}^j}{N} +\sum_{i,j=1}^2\frac{f_i f_{ij} \dot{x}^j}{N^3} f_k ~\stackrel{(4)}{=}~ -\sum_{j=1}^2\frac{f_{kj} \dot{x}^j}{N}-\dot{n}_3 f_k, \qquad k=1,2,\qquad (3)$$ $$\dot{n}_3~=~ -\sum_{i,j=1}^2\frac{f_i f_{ij} \dot{x}^j}{N^3} ~=~\sum_{k=1}^2\dot{n}_k f_k. \qquad (4)$$

Rolling/No slip condition: $$\dot{\bf r}_{cm}~=~{\bf \omega} \times R{\bf n} ~=~{\bf \omega}_{\parallel} \times R{\bf n}, \qquad \dot{\bf r}_{cm} \perp {\bf n}, \qquad {\bf \omega}_{\parallel}~=~ \frac{\bf n}{R}\times\dot{\bf r}_{cm}, \qquad (5)$$ $$|\dot{\bf r}_{cm}|~=~R|{\bf \omega}_{\parallel}| , \qquad {\bf \omega}~=~{\bf \omega}_{\parallel}+{\bf \omega}_{\perp}, \qquad |{\bf \omega}|^2~=~|{\bf \omega}_{\parallel}|^2+\omega_{\! N}^2.$$

Normal component of the angular velocity ${\bf \omega}$: $${\bf \omega}_{\perp} ~=~\omega_{\! N} {\bf n}, \qquad \omega_{\! N}~:=~{\bf \omega}\cdot {\bf n} ~=~ \frac{\omega_3-\sum_{k=1}^2\omega_k f_k}{N}. \qquad (6)$$

Rolling/No slip condition (5) in components: $$\dot{x}^i~=~\frac{R}{N} \sum_{j=1}^2\epsilon^{ij} ( \omega_j +\omega_3 f_j), \qquad i=1,2, \qquad \epsilon^{12}~:=~+1. \qquad (7)$$

Total angular momentum around origin $0$: $${\bf J}_0~=~{\bf L}_0+{\bf S}, \qquad {\bf L}_0 ~=~ {\bf r}_{cm}\times m \dot{\bf r}_{cm}, \qquad {\bf S}~=~I {\bf \omega}.$$

Total angular momentum around contact point $P$: $${\bf J}_p~=~{\bf L}_p+{\bf S}, \qquad {\bf L}_p ~=~ R{\bf n}\times m \dot{\bf r}_{cm}, \qquad {\bf J}_0-{\bf J}_p~=~{\bf L}_0-{\bf L}_p~=~{\bf r}_p \times m\dot{\bf r}_{cm} .$$

Newton's 2nd law: $$m\ddot{\bf r}_{cm}~=~{\bf F}_n + m{\bf g}, \qquad {\bf g}~=~(0,0,-g). \qquad (8)$$

Angular Newton's 2nd law around origin $0$: $$\dot{\bf J}_0~=~\tau_0~=~{\bf r}_p\times{\bf F}_n +{\bf r}_{cm}\times m{\bf g} .$$

Angular Newton's 2nd law around contact point $P$: $$\dot{\bf J}_p ~=~\tau_p~=~R{\bf n}\times m{\bf g} .$$

Angular Newton's 2nd law around CM: $$I \dot{\bf \omega}~=~\dot{\bf S}~=~\tau_{cm}~=~-R{\bf n}\times {\bf F}_n ~\stackrel{(8)}{=}~R{\bf n}\times m({\bf g}-\ddot{\bf r}_{cm}), \qquad \dot{\bf \omega} \perp {\bf n}. \qquad (9)$$

Equations (5) and (9) yield the equations of motion for the angular velocity ${\bf \omega}$:

$$(mR^2+I) \dot{\bf \omega} ~=~R{\bf n}\times m({\bf g} - {\bf \omega} \times R\dot{\bf n}) ~=~R{\bf n}\times m {\bf g} + mR^2\omega_{\! N} \dot{\bf n} . \qquad (10)$$

The equations (10) of motion for the angular velocity ${\bf \omega}$ in components:

$$\alpha \dot{\omega}_i ~=~\frac{g}{RN}\sum_{j=1}^2\epsilon^{ij}f_j +\omega_{\! N} \dot{n}_i, \qquad i=1,2, \qquad \alpha \dot{\omega}_3~=~\omega_{\! N} \dot{n}_3. \qquad (11)$$

The equations of motion for the normal component $\omega_{\! N}$ of the angular velocity:

$$\dot{\omega}_{\! N}~\stackrel{(9)}{=}~{\bf \omega}\cdot \dot{\bf n} ~=~ \sum_{i=1}^2\omega_i\dot{n}_i +\omega_3\dot{n}_3 ~\stackrel{(3)+(4)}{=}~-\sum_{i,j=1}^2\frac{\omega_i f_{ij} \dot{x}^j}{N} +N\omega_{\! N} \dot{n}_3$$ $$~\stackrel{(7)}{=}~-\frac{R}{N^2}\sum_{i,j,k=1}^2\omega_i f_{ij}\epsilon^{jk}(\omega_k+\omega_3 f_k) + N\omega_{\! N} \dot{n}_3 .\qquad (12)$$

Note that the first term in eq. (12) vanish iff condition (1) is satisfied:

$${\bf H} ~\propto~ {\bf 1}_{2\times 2} \qquad \Leftrightarrow \qquad \forall{\bf \omega}\in\mathbb{R}^2 : \sum_{i,j,k=1}^2 \omega_i f_{ij}\epsilon^{jk}\omega_k~=~0.$$

Warning: Condition (1) is not equivalent to saying that every point is umbilical. We assume from now on that condition (1) is satisfied. The main punch line of above Newtonian analysis is the following.

The condition (1) and the equations (11) and (12) of motion show that both rates of change of the vertical component $\dot{\omega}_3$ and the normal component $\dot{\omega}_{\! N}$ are linear combinations of $\omega_3$ and the normal component $\omega_{\! N}$. Thus since we impose that $\omega_3$ and $\omega_{\! N}$ should vanish at an initial time, they will also vanish at all later times.

4) Kinetic Energy. Since there is no slipping and no spinning normal to the surface at any time, cf. condition (1), we have enough information to fully describe the total kinetic energy $T$ in terms of the configuration space variables $(x^1,x^2,\dot{x}^1,\dot{x}^2)$. The total kinetic energy is

$$T ~=~ T_{cm}+T_{rot} ~=~ \alpha T_{cm}, \qquad \alpha~:=~1+\frac{I}{mR^2},$$ $$T_{cm}~=~\frac{m}{2}|{\bf v}|^{2}, \qquad {\bf v} ~=~ \left(\dot{x}^1,\dot{x}^2,\sum_{i=1}^2\frac{\partial f}{\partial x^i}\dot{x}^i\right).$$

5) Maupertuis principle. At this point, let us recall Jacobi's formulation of Maupertuis principle. Ultimately, this approach does not solve the problem, but it does offer a conceptional understanding. The principle states that the ball chooses (among virtual same-energy-paths) a path that extremizes

$$\int_{P_i}^{P_f} \sqrt{T} \ \mathrm{d}s,$$

for fixed initial and final position, $P_i$ and $P_f$, respectively. Here the metric

$$\mathrm{d}s^2~=~\sum_{i,j=1}^2 g_{ij}\mathrm{d}x^i \mathrm{d}x^j$$

arises from the total kinetic energy

$$T~=~\frac{1}{2}\sum_{i,j=1}^2 g_{ij}\dot{x}^i\dot{x}^j ~=~ \frac{1}{2}\left(\frac{ds}{dt}\right)^2, \qquad g_{ij}~=~ \alpha m \left(\delta_{ij}+ \frac{\partial f}{\partial x^i}\frac{\partial f}{\partial x^j} \right).$$

Now, an overall constant multiplicative factor $\alpha$ will clearly not affect the extremization process. Nevertheless, this does not solve the original problem, where we don't want to assume a specific (i.e., fixed) final position $P_f$. So Maupertuis principle does only cover the case, where the end points $P_f$ are assumed to be the same a priori. Then it states that the intermediate paths must also be the same.

6) Lagrange equations. To solve the original question, it turns out to be better to just examine Lagrange equations directly,

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}^i}\right) ~=~ \frac{\partial L}{\partial x^i}, \qquad L~=~T-V.$$

The total kinetic energy $T$ becomes independent of $\alpha$ under a scaling of time $t\to \sqrt{\alpha} t$. Also note that the Lagrange equations are invariant under the time scaling $t\to \sqrt{\alpha}t$. Thus the solution will be the same up to a rescaled time parameter and choice of initial conditions. This is also Carl Brannen's point. Interestingly, it is important that the initial velocity is assumed to be zero to achieve the same paths in the two situations, as the following simple example illustrates.

Finally, if we compare with the Newtonian analysis, the equation (7) is indeed invariant under a combined scaling $t\to \sqrt{\alpha}t$ and $\omega_i\to \omega_i/\sqrt{\alpha}$. However $\alpha$ is only scaled away from equation (11) if $\omega_{\! N}=0$. This is the main reason behind Theorem 1 and 2.

7) Example. Let the hill be an inclined plane with inclination $45^{\circ}$, and let the $x^1$-axis be in the "uphill direction". We take the function $f(x^1,x^2)=x^1$. Let the initial position be $(x^1_i,x^2_i)=(0,0)$, and the initial velocity be $(\dot{x}^1_i,\dot{x}^2_i)=(0,v_i)$ along the $x^2$-axis. Then the metric becomes

$$g_{ij} ~=~ \alpha m \left[\begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array} \right].$$

The equations of motion are

$$2\alpha m \ddot{x}^1 ~=~ -mg, \qquad \alpha m \ddot{x}^2 ~=~0.$$

The solution is a parabola

$$x^1(t) ~=~ -\frac{g}{4\alpha} t^2, \qquad x^2(t)~=~v_i t.$$

The paths are only the same (i.e., independent of $\alpha$), if the initial velocity $v_i$ is zero.

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Edit: the example below doesn't work (as a counterexample). I realized that thanks to the conservation of the angular momentum in the direction perpendicular to the slope the motion decouples also in the second case. Now I am starting to believe that the result holds in general and am going to try to find a simple conceptual proof of this. I think Mark's argument is correct but it looks like an accidental mathematical result that is not at all obvious physically. In any case: thanks for the interesting food for thought!

Ok, here's a particularly easy example where trajectories (probably) differ. Just consider an inclined plane and suppose that the initial velocity of the ball is perpendicular to the slope. In the case where the ball is just sliding, the momentum in the perpendicular direction will be conserved. In particular, these two directions (perpendicular and along the slope) can be treated independently and you obtain a parabola as the trajectory. But when the ball is rotating the equations of motion are quite complicated (I will try to solve them later to complete the argument) and it would surprise me hugely if the result was again parabola (not to mention the same one).

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I really do hope there's a simple proof. This problem has been on my mind for three months. –  Mark Eichenlaub Nov 16 '10 at 23:29

I posed the problem to my physics club, and someone pointed out that the calculation is easier if we do it around the point of contact with the hill.

Let the ball have moment of inertia $I$ about a point on its edge, and make $\vec{\omega}$ the angular velocity around the point where the ball contacts the hill. We'll try to show that the path doesn't change when you change the moment of inertia with fixed radius.

A ball sliding down the hill is equivalent to a ball rolling down the hill with all its mass located at the center of the ball, so if we show that all rolling balls (with different moments of inertia but the same radius) follow the same path, the sliding ball follows that path, too.

Introduce a parameter $\lambda$ measuring the length along the spatial path of the ball, so that when we take a derivative with respect to $\lambda$, we are looking at how values change in space rather than in time.

The direction of the path is described by a unit tangent vector $\vec{v}/v$. The path curves in space, and this is described by the derivative of the tangent vector with respect to lambda. We want to show that this is independent of $I$. The quantity we're examining is

$$\frac{\textrm{d}}{\textrm{d}\lambda} \frac{\vec{v}}{v} = \frac{\textrm{d}\vec{v}/\textrm{d}\lambda}{v} - \frac{\vec{v}}{v^2} \frac{\textrm{d}v}{\textrm{d}\lambda}$$

We are going to describe the physics in terms of torque and angles, so let's define $\vec{R}$ as a vector pointing from the point of contact with the hill to the center of the ball, and write

$$\vec{v} = \vec{\omega}\times\vec{R}$$

This lets us do some calculus and algebra to re-express the change in the tangent vector, which we previously wrote in terms of $\vec{v}$, in terms of $\vec{\omega}$. I get

$$\frac{1}{\omega R}\left( \left( \frac{\textrm{d}\vec{\omega}}{\textrm{d}\lambda} - \frac{\vec{\omega}}{\omega} \frac{\textrm{d}\omega}{\textrm{d}\lambda}\right) \times \vec{R} + \vec{\omega} \times \frac{\textrm{d}\vec{R}}{\textrm{d}\lambda}\right)$$

The final term on the right is clearly independent of $I$ because it depends the radius, which is assumed to be the same. We'll drop that term in future discussion because we're only interested in whether this change in the tangent is independent of $I$. For the same reason, we don't care that the stuff in parentheses is being crossed with $\vec{R}$, or divided by $R$, so we'll simplify to

$$\frac{1}{\omega}\left( \frac{\textrm{d}\vec{\omega}}{\textrm{d}\lambda} - \frac{\vec{\omega}}{\omega} \frac{\textrm{d}\omega}{\textrm{d}\lambda} \right)$$

Now we need to look at the physics. The goal is to find expressions for $\textrm{d}\vec{\omega}/\textrm{d}t$ and $\textrm{d}\omega/\textrm{d}t$.

The torque $\vec{\tau}$ is the same for all symmetric balls of equal mass and radius, and we have

$$I \frac{\textrm{d}\vec{\omega}}{\textrm{d}t} = \vec{\tau}$$

Using the chain rule,

$$\frac{\textrm{d}\vec{\omega}}{\textrm{d}\lambda} = \frac{\textrm{d}\vec{\omega}}{\textrm{d}t}\frac{\textrm{d}t}{\textrm{d}\lambda} = \frac{\textrm{d}\vec{\omega}}{\textrm{d}t}\frac{1}{v} = \frac{\textrm{d}\vec{\omega}}{\textrm{d}t}\frac{1}{\omega R}$$

$$\frac{\textrm{d}\omega}{\textrm{d}\lambda} = \frac{\textrm{d}\omega}{\textrm{d}t}\frac{1}{\omega R}$$

Using

$$\omega = (\vec{\omega}\cdot\vec{\omega})^{1/2}$$

gives

$$\frac{\textrm{d}\omega}{\textrm{d}t} = \frac{\textrm{d}\vec{\omega}/\textrm{d}t\cdot\vec{\omega}}{\omega}$$

This lets us plug things back in to the expression we were examining. Simplifying the algebra,

$$\frac{1}{\omega}\left( \frac{\textrm{d}\vec{\omega}}{\textrm{d}\lambda} - \frac{\vec{\omega}}{\omega} \frac{\textrm{d}\omega}{\textrm{d}\lambda} \right) = \frac{1}{I\omega^2 R}\left(\vec{\tau} - \frac{\vec{\omega}(\vec{\tau}\cdot\vec{\omega})}{\omega^2} \right)$$

$I \omega^2$ is the twice the kinetic energy, and therefore a function only of the position. It doesn't change as we change $I$. $\vec{\tau}$ is also independent of $I$. Finally, $\frac{\vec{\omega}(\vec{\tau}\cdot\vec{\omega})}{\omega^2}$ is also independent of $I$, because although changing $I$ changes the size of $\omega$, in this expression all the units of $\omega$ cancel out.

We conclude that we can take two balls with different spherically-symmetric mass distributions and the same radius and roll them down the same path over the hill. The sliding ball is identical to the limit where $I \to m R^2$.

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@Mark: The right side of the last equation simplifies to $\frac{\vec{n} \times\vec{g}}{v^2}$ where $\vec{n}$ is a unit vector of $\vec{R}$ and $v$ is speed of the center of the ball. Now, at a same height, balls with different inertial moments should have different speeds and different values in the right side of the last equation at the same height. Or i misunderstood something? –  Martin Gales Feb 7 '11 at 12:03
@Martin Where do you get the claim that the right hand side simplifies to that expression? They don't even have the same units. –  Mark Eichenlaub Feb 7 '11 at 12:47
@Mark: Note that $\frac{d\lambda}{dt}=v=\omega R$ and a correct equation should be $\frac{1}{\omega}\frac{d\vec{\omega}}{d\lambda}=\frac{\vec{R}\times m \vec{g}}{RI\omega^2}$ –  Martin Gales Feb 8 '11 at 6:06
@Martin $\lambda$ is just a parameter that is smooth and monotonic increasing as you go down the path, hence I wrote a proportionality. We could set it up so your expression in your second comment is correct, but that expression is different from the one in your first comment. The expression in your second comment is still independent of $I$. –  Mark Eichenlaub Feb 8 '11 at 6:44
@Mark: If $\lambda$ is a dimensionless parameter then the expression simplifies to $\frac{1}{\omega}\frac{d\vec{\omega}}{d\lambda}\propto\frac{\vec{R}\times\vec{g}}{v^2}$ The right side is still different say for a solid sphere and a hollow sphere. –  Martin Gales Feb 8 '11 at 7:01

This certainly isn't a rigorous answer, but it should be 'almost certainly not':

Pointwise, gravity will pull with a force in the direction of the gradient of the height function projected onto hill's surface, while friction will push in a direction opposing the motion of the ball.

If the hill is lumpy, the direction of this projection of the hill's geometric gradient, and thus, the direction of the gravitational force, will vary with position in a way different from the variation in the direction of the ball's path. Thus the direction of the net force on the ball will depend on whether or not you have the no-slip condition 'turned on'.

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Why does friction necessarily point in a direction opposing the motion of the ball? Remember there is no relative velocity between the part of the ball touching the hill and the hill itself. –  Mark Eichenlaub Nov 13 '10 at 21:14
But there {\bf is} a relative motion between the center of mass of the ball and the hill itself--if you look at the thing from the perspective of the contact point, this induces a torque on the ball that causes the rolling. –  Jerry Schirmer Nov 13 '10 at 21:20
But the torque is related to the angular acceleration, not the angular velocity. For example, when the ball is rolling over a flat plane at constant speed the friction is zero. –  Mark Eichenlaub Nov 13 '10 at 21:24
Ick. You're right. And the torque is induced by the gravity, so it should be collinear with the gravitational force. Comment rescinded. –  Jerry Schirmer Nov 13 '10 at 21:27

Although my intuition was wrong (like Peter Shor's) and while I admire both Mark Eichenlaub's (self) and Carl Brannen's detailed and exact answers, what does it to me is the intuitive response (with correction based on comments) that Jerry Schirmer gave, although I would take the opposite conclusion.

The simple fact that the reaction force needed to create the lossless friction must be exactly opposed to the angular acceleration (which can only be caused by the effects of gravity) means that the accelerations are colinear. That's it really, no need to say more, so they follow the same path.

By the way, kudos for a very well defined problem. As the minute the ball flies or touches in two points, the friction thing will go mad with infinite acceleration, or break the rule, and the ball's spin will dramatically alter its course (not counting on atmospheric effects either ;-)).

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• It the original trajectory, as seen from above, is straight, then friction will not curve it because it's always parallel and opposite to velocity, and velocity is always parallel to the trajectory. The only difference is that the ball will have less energy and it will not go the same distance. Sooner or later it will stop due to friction.

• If the original trajectory, as seen from above, is curved, then the trajectory will be changed because in a curve velocity and trajectory are not parallel. If you need an example think about a ball going around a bowl. Without friction, the ball will keep on going on a stable "orbit" around the bowl. With friction it will spiral down until it stops.

Edited due to a possible different interpretation of the question.

If the only effect of friction is to make the ball roll. The ball will acquire moment of inertia that will oppose any change in direction. If the trajectory was initially straight, nothing should change. If the trajectory was curved, it would follow a different one.

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the friction is just there so the balls rolls without slipping. no energy dissipated. –  Mark Eichenlaub Nov 13 '10 at 21:09
actually the question says that the ball slows down. –  Sklivvz Nov 13 '10 at 21:10
also, it the ball slides in the first case and rolls in the second, surely some energy goes into making the ball roll, which changes the trajectory. –  Sklivvz Nov 13 '10 at 21:14
Mark, first of all the definition of trajectory is here and it's exactly the path taken by the ball. google.co.uk/… –  Sklivvz Nov 13 '10 at 21:32
@Sklivvz Okay, "trajectory" and "path" mean the same. It is not obvious to me that the rolling ball will follow a different trajectory simply because it has some moment of inertia. If I interpret this argument correctly, it simply states that the acceleration of the rolling ball is less. However, the rolling ball is moving more slowly, so it needs less acceleration to follow the same path. –  Mark Eichenlaub Nov 13 '10 at 21:50

The principle of Maupertuis solves this problem with no work. The path in both cases is the extremum of

$$\int ds \sqrt{A - \phi(x(s))}$$

After dividing by a suitable quantity with units mass, with the only difference between the two cases the constant you divide by, which determines the velocity. So the paths are the same if the sphere starts from rest at some point.

The nonholonomic nature of the rolling sphere constraint is not important at all, because it just means that the sphere's orientation depends on the path to get to a certain point. But nothing in the problem depends on the orientation of the sphere, because the mass distribution inside is assumed to be spherically symmetric. The state space has the orientation and velocity of the sphere, but any two orientations are identical, and the extremal paths extremize the Maupertuis integral without reference to the orientation, which just goes along for the ride. The kinetic and potential energies are determined entirely by the position and velocity of the sphere at any position, along with the no-slip condition.

This problem is very non-obvious, and is a good illustration of the power of the Maupertuis formulation.

Maupertuis' principle is formulated between fixed endpoints, and fixed energy, not a fixed time. This is like all Lagrangian principles, where the endpoints are the variables you keep fixed as you vary the path.

Does this mean that Maupertuis principle solves a different problem, perhaps one with fixed endpoints? Absolutely not (this ridiculous idea was suggested below). Consider starting a ball sliding at the top of a frictionless hill with some velocity. It travels down the hill to the final endpoint, whatever that may be.

The resulting trajectory is an extremal point of the Maupertuis integral between the endpoints. Therefore it is also an extremal point for the rolling problem between the two endpoints. Now you rescale the velocity between the two problems, and you find the solution to the rolling problem with a rescaled initial velocity is the same as the solution to the sliding problem.

A good reference for is

• Captain Obvious, Introduction to Lagrange principles, with additional commentary by Dr. Duh.
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Maupertuis' principle (as explained in H. Goldstein "Classical Mechanics" and my answer) keeps the endpoints fixed, and therefore does not address the problem at hand, where we do not want to impose conditions on where the ball ends up. –  Qmechanic Sep 25 '11 at 13:51
@Qmechanic: This is incorrect--- this is the usual Lagrangian business of fixing the endpoints. The other endpoint is just a surrogate for the initial velocity. –  Ron Maimon Sep 25 '11 at 15:33
Qmechanic: So what? the path is the same given fixed initial and final endpoints, this means that for any path for the sliding ball, there is a corresponding path for the rolling ball, and all it does is rescale the velocity, including the initial velocity. This is what the problem asked. For the special case of vanishing initial velocity, it vanishes in both the rolling and sliding case, because the thing inside the square root is zero at the beginning. –  Ron Maimon Sep 25 '11 at 16:21
@RonMaimon I've done a search on Amazon.com for that reference in your answer, but I can't find it; is it an old rare book? –  Physiks lover Nov 10 '14 at 21:48
The difficult part is showing that the constraint for no slipping and its angular momentum doesn't affect the path taken by the center of mass when it's allowed to slip. –  Physiks lover Nov 12 '14 at 0:11

In this answer I would like to give an explicit example where the paths do depend on the moment of inertia $I$, cf. Theorem 1 in my other answer. I will use the same notation here. Let the ball have radius $R=1$, and with initial conditions

$$(x_i^1, x_i^2)~=~(0,1),\qquad (\dot{x}_i^1, \dot{x}_i^2)~=~(0,0), \qquad (\omega_{1i},\omega_{2i},\omega_{3i})=(0,0,0) ,$$

at initial time $t_i=0$. Let the hill be such that the center-of-mass of the ball continues to lie on the surface

$$x^3~=~f(x)~=~ x^1 + \frac{\epsilon}{2} (x^2)^2, \qquad |\epsilon| \ll 1.$$

In this example I will only calculate the leading $\epsilon$ behavior in the beginning of the path. Obviously, if $\epsilon=0$, the ball will continue downhill along the path $x^2-1=0$, and the angular velocities $\omega_1$, $\omega_3$, and $\omega_{\! N}$ would continue to be zero for all times, so these quantities must at least be of order ${\cal O} (\epsilon)$. The length $N$ of the normal vector ${\bf N}$ is

$$N~=~\sqrt{2+(\epsilon x^2)^2} ~=~N_{\epsilon}+\frac{\epsilon^2(x^2-1)}{N}+\frac{\epsilon^2(x^2-1)^2}{2N} + {\cal O} (\epsilon^5)~=~N_{\epsilon}+ {\cal O} (\epsilon^3),$$

where I have defined the constant

$$N_{\epsilon}~:=~\sqrt{2+\epsilon^2},$$

so that

$$\dot{N}~=~\frac{\epsilon^2}{N}x^2\dot{x}^2+ {\cal O} (\epsilon^5) ~=~{\cal O} (\epsilon^3).$$

One may also show that the normal component $\omega_{\! N}$ of the angular velocity is of order ${\cal O} (\epsilon^2)$:

$$\omega_{\! N}~=~-\frac{\omega_k f_k}{N}+{\cal O} (\epsilon^5)~=~{\cal O} (\epsilon^2).$$

Equation (11) from my other answer then shows that

$$\alpha \dot{\omega}_3~=~\omega_{\! N} \dot{n}_3 ~=~-\frac{\omega_{\! N} \dot{N}}{N^2}~=~{\cal O} (\epsilon^5) \qquad \Rightarrow \qquad \omega_3={\cal O} (\epsilon^5).$$

Equations (7) and (11) from my other answer yield

$$N\dot{x}^i ~=~\epsilon^{ij}\omega_j+{\cal O} (\epsilon^5), \qquad \alpha \dot{\omega}_j ~=~\frac{g }{N}\epsilon^{jk}f_k -\frac{\omega_{\! N}}{N}\dot{f}_j +{\cal O} (\epsilon^5),$$

so

$$\underbrace{\frac{\epsilon^2}{N}x^2\dot{x}^2\dot{x}^i}_{{\cal O} (\epsilon^3)} +N\ddot{x}^i+{\cal O} (\epsilon^5) ~=~\epsilon^{ij}\dot{\omega}_j ~=~- \frac{g}{\alpha N}f_i -\underbrace{\frac{\omega_{\! N}}{\alpha N}\epsilon^{ij}f_{jk}\dot{x}_k}_{{\cal O} (\epsilon^4)}+{\cal O} (\epsilon^5).$$

Hence to fourth order,

$$\ddot{x}^1 ~=~-\frac{g}{\alpha N^2} -\underbrace{\frac{\epsilon^2}{N^2}x^2\dot{x}^2\dot{x}^1}_{{\cal O} (\epsilon^3)} -\underbrace{\frac{\epsilon\omega_{\! N}}{\alpha N^2}\dot{x}^2}_{{\cal O} (\epsilon^4)} +{\cal O} (\epsilon^5),$$

$$\ddot{x}^2 ~=~-\frac{\epsilon g}{\alpha N^2} x^2 -\underbrace{\frac{\epsilon^2}{N^2}(\dot{x}^2)^2}_{{\cal O} (\epsilon^4)} +{\cal O} (\epsilon^5),$$

with solution to second order,

$$x^1(t)~=~-\frac{gt^2}{2\alpha N^2}+{\cal O} (\epsilon^3),$$

$$x^2(t)~=~\cos(\frac{t}{N}\sqrt{\frac{\epsilon g}{\alpha}})+{\cal O} (\epsilon^3) ~=~1-\frac{\epsilon gt^2}{2\alpha N^2} +\frac{1}{6}\left(\frac{\epsilon gt^2}{2\alpha N^2}\right)^2+{\cal O} (\epsilon^3).$$

To second order, the path is still independent of $\alpha$,

$$x^2~=~\cos(\sqrt{-2 \epsilon x^1})+{\cal O} (\epsilon^3) ~=~1+\epsilon x^1 +\frac{1}{6}(\epsilon x^1)^2+{\cal O} (\epsilon^3),$$

i.e., a parabola. So we are not done yet. (It turns out that the effect we are after shows up at fourth order.) The angular velocities are

$$\omega_2~=~N\dot{x}^1 +{\cal O} (\epsilon^5)~=~-\frac{gt}{\alpha N}+{\cal O} (\epsilon^3),\qquad$$

$$\omega_1~=~-N\dot{x}^2 +{\cal O} (\epsilon^5) ~=~\sqrt{\frac{\epsilon g}{\alpha}} \sin(\frac{t}{N}\sqrt{\frac{\epsilon g}{\alpha}})+{\cal O} (\epsilon^3)$$ $$~=~\frac{\epsilon gt}{\alpha N} - \frac{t^3}{6N^3}\left(\frac{\epsilon g}{\alpha }\right)^2+{\cal O} (\epsilon^3),$$

$$\omega_{\! N}~=~-\frac{\omega_k f_k}{N}+{\cal O} (\epsilon^5) ~=~ -\frac{2\epsilon^2x^1 \dot{x}^1}{3}+{\cal O} (\epsilon^3) ~=~ -\frac{t^3}{3N^4}\left(\frac{\epsilon g}{\alpha }\right)^2+{\cal O} (\epsilon^3).$$

We conclude that $\omega_{\! N}$ is not identically zero. This observation is very important. It shows that the ball can spontaneously acquire a non-zero spin component $\omega_{\! N}$ normal to the surface even when it started from rest with no spinning to begin with.

Substituting back the second order solution into

$$\frac{1}{N^2}~=~\frac{1}{N^2_{\epsilon}}-\frac{\epsilon^2(x^2-1)}{2} -\frac{\epsilon^2(x^2-1)^2}{4}+ {\cal O}(\epsilon^5)$$ $$~=~\frac{1}{N^2_{\epsilon}}+\frac{\epsilon^3x^1}{4}-\frac{\epsilon^4(x^1)^2}{24}+ {\cal O}(\epsilon^5).$$

and the equations of motions yield

$$\ddot{x}^1 ~=~-\frac{g}{\alpha}\left(\frac{1}{N^2_{\epsilon}}+\frac{\epsilon^3x^1}{4}-\frac{\epsilon^4(x^1)^2}{24}\right) -\frac{\epsilon^3}{2}(1+\epsilon x^1) (\dot{x}^1)^2 +\frac{\epsilon^4}{3\alpha}x^1(\dot{x}^1)^2+{\cal O} (\epsilon^5),$$

$$\ddot{x}^2 ~=~-\frac{\epsilon g}{\alpha} \left(\frac{1}{N^2_{\epsilon}}+\frac{\epsilon^3x^1}{4}-\frac{\epsilon^4(x^1)^2}{24}\right) \left( 1+\epsilon x^1 +\frac{1}{6}(\epsilon x^1)^2 \right) -\frac{\epsilon^4}{2}(\dot{x}^1)^2+{\cal O} (\epsilon^5),$$

with solution to fourth order

$$x^1(t)~=~-\frac{gt^2}{2\alpha N^2_{\epsilon}} -\frac{\epsilon^3}{12} \left(\frac{gt^2}{2\alpha N^2_{\epsilon}}\right)^2+\frac{\epsilon^4}{45}\left(\frac{17}{4}-\frac{2}{\alpha} \right)\left(\frac{gt^2}{2\alpha N^2_{\epsilon}}\right)^3 +{\cal O} (\epsilon^5),$$

$$x^2(t)~=~1-\frac{\epsilon gt^2}{2\alpha N^2_{\epsilon}} +\frac{1}{6 }\left(\frac{\epsilon gt^2}{2\alpha N^2_{\epsilon}}\right)^2 -\frac{1}{90 }\left(\frac{\epsilon gt^2}{2\alpha N^2_{\epsilon}}\right)^3+{\cal O} (\epsilon^5).$$

Notice how most of the $\alpha$ dependences appear in the combination $t^2/\alpha$, so that a scaling $t\to \sqrt{\alpha}t$ would remove these $\alpha$ dependences. However, the unbalanced $\frac{1}{\alpha}$ in $x^1(t)$ at order ${\cal O}(\epsilon^4)$, which can be traced to a non-zero $\omega_{\! N}$ at order ${\cal O}(\epsilon^2)$, breaks this pattern.

Conclusion: The path depends on $\alpha=1+ I/mR^2$.

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text:
These are my notes from week 1 of MIT course 7.88j: Protein Folding and Human Disease, held by Dr. Jonathan King on February 5, 2015.

### Introduction

This class focuses on a central unsolved question in biology: how does amino acid sequence direct the folding of proteins? How does cellular machinery assist? We will also discuss human pathologies associated with failures of protein folding. Because the central question here is unanswered, we still do not know what areas of knowledge will prove most relevant to solving it. Dr. King predicts that the problem will be solved - that protein structure will be predictable from sequence alone - within 5-10 years.

The fact that we cannot predict structure from sequence has several important consequences:

• difficulty in achieving native fold when expressing cloned proteins - for instance, recovering natively folded recombinant insulin from E. coli is extremely laborious
• limited progress in de novo design of proteins
• inability to rationally design drugs for protein-folding diseases

In addition to canonical protein deposition diseases such as Alzheimer’s and Parkinson’s, we’ll also consider cataracts, which arise from pathological aggregation of lens proteins. Cataracts are usually not listed as cause of death, despite being responsible for considerable mortality due to accidents associated with reduced visual acuity. One should not think of protein-folding disorders as being diseases of age, even though their incidence is age-dependent. Many environmental insults - say, inhalation of coal dust (black lung) or cigarette smoking - require decades of exposure to cause disease, meaning their incidence too is age-dependent, but age does not cause the disease.

### Survey of proteins

One of the smaller, simpler prokaryotes, Methanococcus, has only 1,735 protein-coding genes, with average length 287 amino acids - compare to E. coli, which has 4,285 protein coding genes with average length 317.

How do long protein-coding sequences form over evolution? There are a handful of examples where it has been demonstrated through phylogenetic approaches that two shorter proteins got fused into one. In some cases, two enzymes in the same pathway fused together and ended up doing both jobs.

W and C are the rarest, and L, A and G the most common amino acids in proteins. See this table for frequencies of all amino acids in vertebrates.

A few di- or tri-peptide sequences are rare but very famous - for instance, polyQ, polyFG (in nuclear pore complex proteins), polyP, and G-P-hydroxyproline which is the collagen repeat. But on average, amino acid sequence in proteins appears largely random (or we might say, high information entropy), which has been recognized even from some of the smallest genomes that have been sequenced [Fraser 1995].

Even a 300 amino acid protein has 20300 possible sequences, which is larger than the number of atoms in the universe. This means that the protein folding problem cannot be solved by a lookup table. The set of all sequences that exist in any organism anywhere on earth is but a vanishing subset of possible sequences.

### A brief history of empirical protein structure determination

The first growth of protein crystals was achieved by J.D. Bernal and Dorothy Hodgkin (1936-8; see 1964 Nobel Prize in Chemistry). The first 3-D protein structure to be solved was myoglobin, solved via X-ray crystallography by Max Perutz and John Kendrew (1956-8; see Nobel Prize in Chemistry 1962). As of today, Protein Data Bank (PDB) contains 106,293 empirically determined structures (though the number of unique proteins is much smaller).

Until the structure was solved, people expected that myoglobin would possess, at the monomer level, the elegant symmetry of the myoglobin crystal. In fact, its structure is utterly asymmetric. The original structure is available as [PDB# 1MBN]. I show it here in yellow, with the heme in red and sticks shown for the side chains of two histidines coordinating it:

fetch 1mbn
bg_color white
hide everything
show cartoon
color yellow
show sticks, organic
color red, organic
show sticks, resi 64 or resi 97

Above: The first protein structure ever solved.

How did proteins acquire such prosaic folds? The structure of DNA was solved >10 years before that of myoglobin was. DNA was helical, and some proteins were already known to be helical, so some people hypothesized a direct templating of DNA conformation onto protein conformation. This theory fell out of favor as experimental data failed to support it - for instance, you couldn’t add DNA to protein and change its conformation. When mRNA was discovered, the theory was revised to posit that mRNA created a conformational template for protein. This too proved experimentally unsupportable. The next theory was that the ribosome physically organizes the nascent polypeptide. This was ruled out quickly, however, as it became clear that ribosomes are all pretty much the same, lacking the diversity of the proteins they produce. Finally, we were left with the “no template” model, which proposed that protein folding proceeded spontaneously. Yet this model remained unpopular for many years, apparently because it didn’t seem very conceptually elegant or exciting.

Eventually, experimental support for the “no template” model arrived in the form of the Anfinsen experiment, the first results of which were published in [Anfinsen 1961]. Anfinsen and colleagues at NIH purified RNAse from the bovine pancreas, an enzyme responsible simply for breaking down the RNA found in dietary grass. Bovine pancreas RNAse comprises 124 amino acids or ~13.7 kDa, and its structure is now known to contain 37% β-sheet, 23% α-helix, as well as coils, turns, and 8 cysteines forming 4 disuflide bonds. Consider the relationship between the number of cysteines and the number of possible permutations of disuflide bonds you could form:

# of cysteines # of disulfide bonds # of combinations
4 2 3
6 3 15
8 4 105
10 5 945

So in theory, RNAse could form 105 different disulfide-bonded isomers. But in fact, for RNAse as well as for all disulfide bond-containing proteins now known, only one isomer is actually produced.

Bovine pancrease RNAse was appealing as an experimental substrate for several reasons. It is highly expressed, and available cheaply in large amounts from slaughterhouses. It is extremely stable to protease digestion, a wide range of pH, and a wide range of temperature. It evolved to be super stable, because it operates in the cow gut at pH ~2 and in the presence of high concentrations of trypsin and chymotrypsin. These properties could be leveraged to purify RNAse. Until then, the only proteins that had been appreciably purified were very abundant blood proteins - usually hemoglobin. In fact, Perutz got his myoglobin from whale blood. RNAse had yet another desirable property: it is an enzyme with readily assayable activity. By measuring cleavage of RNA in vitro you could measure how much enzyme function you had recovered.

By this time, chemists already knew that thiol (-SH) groups were the most reactive groups on amino acids. Anfinsen and colleagues therefore expected that the disulfide bonds would prove responsible for the catalytic activity of RNAse. They hypothesized that if they reduced the disulfide bonds with mercaptoethanol, they could abolish RNAse’s activity. You could count the number of reduced (-SH) cysteines using a colorimetric assay based on a small molecule that selectively binds -SH. They were unable to completely reduce all eight cysteines with mercaptoethanol, so they turned to 8M urea. In the process, they discovered that high molarity urea denatures proteins. After they had finally denatured and reduced all four disulfide bonds of RNAse, the catalytic activity was gone. They then dissolved the protein at 1 mg/mL into a phosphate buffer, and then very gradually bubbled in an oxidizing reagent, or alternately dialyzed out the urea and mercaptoethanol. Once the solution turned sufficiently oxidizing, about 15% of the original catalytic activity was recovered. This was taken as evidence that the information required for protein folding was encoded in the polypeptide chain itself, because the protein was able to “remember” and reassume its native conformation.

While this represented an enormous breakthrough, it also led to several misconceptions which took decades to break down. One conclusion that many people drew from the Anfinsen experiment is that the native state of a protein is the lowest free energy state. This attitude may have retarded progress in the field, as it implied that a protein’s search for its native state is simply combinatoric rather than following a specific and dissectable pathway. Cyrus Levinthal eventually calculated the number of possible folds for a protein, and found that the number was astronomical, and there was no way a protein could sample every state. Even after this calculation was published, though, it was years before people accepted the empirical evidence that, indeed, a great many proteins simply do not adopt a native fold spontaneously. Many researchers tried to get actin and tubulin to spontaneously refold in solution, each thinking they were the first to try, then failing, and keeping their unpublishable failure a secret, thus allowing the next fool to do the same. It took many decades of failure before it was discovered that actin and tubulin can fold only with the help of specific chaperones, which will be introduced later this semester. Another, later, realization has been that the native state is not necessarily the lowest energy. In attempts to purify proteins, it was often the case that a large amount of aggregated protein would be discarded in the pellet, and only the properly folded soluble protein would be studied and reported. Calculations regarding free energy and so on were all done on only the soluble fraction, with the rest swept under the rug. It has been a long road to the realization that the amyloid state of many proteins is exceptionally stable.

Indeed, even for RNAse, one of Anfinsen’s students later discovered that re-folding RNAse resulted in the production of plenty of species containing incorrect combinations of cysteines being bonded to each other - these species would pellet out and had been ignored in the original experiment. That student went on to discover disulfide isomerase, an enzyme whose physiological function is to reorganize disulfide bonds into the correct configuration.

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text:
on July 22, 2011 by in Public Health, Statistical Methods, Comments (0)

# Basic Statistics for Epidemiology: Prevalence vs Incidence

Author Sara Muller, University of Keele (based on material from Health Knowledge)

Epidemiology vs Clinical Medicine

There are lots of definitions of epidemiology, but the one below is fairly comprehensive:

“…the study of the distribution and determinants of health-related states or events in specified populations, and the application of this study to control of health problems”

In short, epidemiology is sort of the who what, when, why, how and where of disease.  It doesn’t involve any experimentation or clinical trials, so it is really of interest to public health, because it is just about what happens.

A concept most people are probably familiar with is ‘clinical medicine’, where interest is in the individual patient and you want to see large and clinical relevant changes in that patient. In the case of a clinical trial, you might be interested in these relatively large changes in a select group of people. For example: Will this painkiller reduce the level of pain in this patient so that they are comfortable?

Epidemiology differs from clinical medicine, because the unit of interest is the population, not the individual. In epidemiology, we are interested in small changes that have an effect on the population level. For example, the slight raising of your cholesterol leads to only a very slight increased risk of you dying from a coronary heart disease, but if everyone in a population increased their cholesterol level very slightly, there would be a big jump in the total number of CHD deaths in the population.

Factors Affecting Health

There are many factors that can affect health.

At the biological level, our genetic heritage may make us susceptible to specific conditions such as hypertension, sickle cell anaemia, Cancer and haemophilia.

Then there is how we live – how much we exercise, what we eat, what we drink and what we might smoke. That some societies seem to live longer than others probably reflects differences in the interaction of these factors.

Then there is the environment. Hippocrates, ‘the father of medicine’, was one of the first to identify the environment as determinant. He noted that the seasons had an impact and the influence of where people lived.  It was a prophetic suggestion, later borne out in modern society. Along with the innovations of Industrial Revolution came urbanisation and in particular problems of squalid housing, poor sanitation and polluted water supplies. Concern at the combined effect of these factors gave rise to the first Public Health Act of 1848.

More latterly, the shrinking of the earth through advances in transport has now made it easier than ever for individuals – and diseases – to move freely across the globe. Whilst economic progress has many advantages there can be positive and negative impacts on health.

We often have to think about these more in epidemiology and public health than in other areas of medicine because we do not control for these things, e.g. in a RCT, we randomise out many of the effects

Quantifications of disease in population

Absolute counts of the incidence of disease is important, but just relying on counts makes it difficult for comparisons to be made. The bigger a population, the larger the number of cases is likely to be. One way to get around this is to consider the number of cases in relation to the population as a whole. Several possible approaches are possible and we shall review some of the more frequently encountered methods.

Ratio (also known as odds)

Calculating the ratio is the foundation for the calculation of many other measures.

a number/another number

where the two numbers are separate e.g. number of males, number of females and the same person is not counted in each group and the size of both or either group can change.

It allows the comparison between the number of people with disease in one population with the number of people with the disease in another population.  It doesn’t have any unit.

Let’s take an example of the number of deaths in males and females. We have the total number of deaths in each gender.

2005 all cause mortality

Females                      269,368

Males                          243,324

Divide number of deaths in females, by number of deaths in males to get the female to male death ratio.

$269,368/243,324 = 1.107$

Proportion

Proportion is the fraction of a population who have a characteristic of interest.  Those who are included in the numerator are also included in the denominator.  As the numerator is a subgroup of the denominator the value is always between 0 and 1.  Values can be expressed as a percentage by multiplying by 100.  Again this is a dimensionless quantity, so it has no units.

Examples of proportion include:

• Proportion of men in a population
• Proportion of people in a sample with diabetes

Using the same example as before, let’s consider the proportion of deaths that are in females, rather than the ratio of female to male deaths.

The proportion of deaths that were in females:

$269368/512692 = 0.525$

We can alternatively present this as 52.5%.

Rate

Rates provide a common time frame and unit of population. They allow a direct comparison of frequency of disease. Two requirements for the rate calculation are: (1) time frame and; (2)  a unit of population.

$Frequency of observed event in a given time period/total number in whom event might occur in that time period.$

The denominator includes all those who are eligible to appear in the numerator i.e., those at risk of the observed event.

Time frame is important, especially if your event of interest is death: everyone will die at some point, so the time period has to be defined to get a sensible answer

An example is based around a group of 742 people with knee pain. They were followed up for 3 years after completing a survey. During the follow-up, the group as a whole consulted their GPs 202 times for knee surgery.

Rate of consultation for knee surgery was:

$202/742*3 = 202/2226 = 0.091 consultation per person per year.$

Prevalence Rate

The prevalence rate is a measure of the proportion of a population affected by a specific condition in a specified time period.  Prevalence itself is the number of people in a population who have the disease of interest in a particular time period.  This isn’t very useful, as the larger the number of people in the population, the more people might have the disease, so a prevalence rate (often referred to simply as prevalence) is what we really want to know.

$Number of cases of disease in given time period/total number in population in that time period$

Be careful to specify your total population properly in terms of socio-demographic and environmental factors. In looking at the prevalence of cervical cancer, you wouldn’t want to include men in the denominator as the prevalence estimate would be far too low.

Prevalences are usually given in three ‘types’:

1. Point prevalence: relates to prevalence with respect to a specific point in time – Did you have an asthma attack on Monday?
2. Period prevalence: related to prevalence over a defined period of time – Did you have an asthma attack in January?

An example is based on the number of responders to a survey reporting knee pain in the past month.

All responders to a survey aged 50+                                                   787

Responders reporting knee pain in the past month                           287

The one prevalence rate of knee pain in the responders of the survey:

$2874/7878 = 0.365$

Alternatively, this can be presented as 36.5% or 365 per 1,000 responders

Incidence Rate

Incidence is the number of new cases of a disease in a population.  As when considering prevalence, because of different population sizes, it is usual to consider a rate.

$Number of new cases of disease in given time period/total number in population at risk at that time period$

For the calculation:

• The time period should be specified, as the number of incident cases can be made arbitrarily large or small depending on the length of the time period being considered.
• ‘At Risk’ population.  People who already have the disease at the start of the time period are not included in the denominator – if they already have the disease, they are not at risk of developing it.

Depending on your question, you might want to consider first ever instances of disease, or just new cases during your time period.  For example, if you consider back pain, you might want any flu where that episode of flu started during your 1 month time period of interest, or you might want any flu that started during your 1 month time period and where that was the first time that person had had flu.  This decision will change who is a member of our ‘at risk’ population.  Others ways of defining your ‘at risk’ group might be age, immunisation status, gender (don’t look at cervical cancer in men!)

Responders to BHPS with new CVD in 2007                                       264

Responders to BHPS with no new CVD in 2007                               1309

The one year incidence in CVD in responders to this survey is:

$264 /13097+264 = 0.020 per year$

Alternatively, this can be expressed as 2.0% per year or 20 per 1000 responders per year

Prevalence vs Incidence

So hopefully the difference between prevalence and incidence rates are clear, but they are related – along with the average duration of disease.  The following relationships hold:

Low incidence, long duration – chronic diseases, e.g. asthma, diabetes

High incidence, short duration – acute, common diseases, e.g. cold, chicken pox in children

Preventative measures might lower incidence, e.g. vaccination, public health campaigns, whereas clinical interventions may reduce decrease duration, or decrease mortality, resulting in an increase in disease duration.

This whole process can be seen in the following diagram.

Prevalence rates are generally used to describe the extent of a disease in a particular population whereas incidence rates look at the rate at which new cases of disease develop.  Whilst prevalence can be affected by how long people live with a condition, incidence does not take this into account

Prevalence is descriptive, often demonstrating public health ‘need’. On the other hand, incidence is useful for studying the causes of disease (the aetiology) or to look at the order in which events occur.  In disease with long durations and very low levels of incidence, there may be little difference between prevalence and incidence.


text:
GR 8677927796770177 | # Login | Register

GR8677 #22
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Electromagnetism$\Rightarrow$}Lorentz Transformation

When an electric field is Lorentz-transformed, afterwards, there might be both a magnetic and electric field (in transverse components). Or, more rigorously, one has,

$\begin{eqnarray} E_x'&=& E_x\\ E_y'&=& \gamma(E_y-vB_z)\\ E_z'&=& \gamma(E_z+vB_y)\\ B_x' &=& B_x\\ B_y' &=& \gamma(B_y + E_z v/c^2)\\ B_z' &=& \gamma(B_z-vE_y/c^2),
\end{eqnarray}$

for motion in the $x$ direction.

(A) Obviously not. Suppose initially, one has just $E_x$, afterwards, there's still $E_x'$.

(B) True, as can be seen from the equations above.

(C) Not true in general. Suppose $\vec{E}=E_x\hat{x}+E_y\hat{y}+E_z\hat{z}$. Afterwards, the transverse components are off by a $\gamma$ even if there's no B field to start with.

(D) Nope.

(E) Mmmm... no need for gauge transformations.

Alternate Solutions
 There are no Alternate Solutions for this problem. Be the first to post one!
djh101
2014-08-27 16:38:09
You can eliminate almost all of the answers without really knowing anything about relativity.

A. Assume observer 1 = observer 2. According to this, an identity transformation will transform an electric field into a magnetic field.
C. From basic E&M, you should know that there's a little bit of symmetry going on between electricity and magnetism. With only that in mind, you can eliminate both C and D. They can't both be true, so by symmetry, neither is true.
D. See above.
wittensdog
2009-09-25 17:53:58
In general, what happens is that the field tensor transforms with two factors of whatever matrix corresponds to the given Lorentz transformation. The conditions on the matrices that are in the Lorentz group are not terribly strict, and certainly we know from the simple case of a boost along the x-axis that the components can mix into each other (think about the usual space and time four-vector, and how it transforms).

In general, for a Lorentz transformation, whatever four-vector or tensor you have can have its components mix together pretty generally.

Also, at least in the classical realm, fields NEVER depend on gauge in any way. The issue of gauge fixing only ever comes up when you start working with potentials, which are not measurable. So immediately throw out anything involving gauge dependence when asked about fields.
FortranMan
2008-10-17 23:15:56
For more on this topic, look around for Relativistic Electrodynamics, specifically in Griffiths section 12.3. Imagine the case of a large parallel-plate capacitor moving perpendicular to their surface areas, and imagine what the Lorentz contraction does to the charge distributions on the plates. This ends up increasing the electric field that's perpendicular to the direction of motion. Note for a parallel-plate capacitor situated such their motion is parallel to their surfaces, this contraction does not occur and the field remains unchanged.
grae313
2007-10-07 15:50:57
On choice (E), does the Lorentz transformation already assume the Lorentz guage?
 rorytheherb2008-10-06 10:22:16 yeah I'm looking at griffiths EM and it's not really illuminating this particular question. i knew of course that E and B mix in S.T.R., but when I saw choice E, I recalled the Lorentz gauge and got duped. Can anyone clarify why how we can say E and B mix _without_ first specifying a gauge?
 dean2008-10-09 21:00:55 I'm not sure there's any relevance at all. Gauge transformations talk about potentials, not fields. A (the vector potential) may change depending on what gauge you're using, but B does not, and can not, else we'd have a physically different situation. Lorentz transformations talk about fields directly in different inertial frames.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$


text:
12:05 AM
@Kurt It's not his day. ;)

@Qrrbrbirlbel Well, now is a new day ...

12:45 AM
@Kurt bad habit to get in to, you could end up being like @egreg

@DavidCarlisle LOL, I think not. @egreg and you knew mutch more about TeX/LaTeX than I. I think I was lucky ...

9 hours later…
9:42 AM

10:01 AM
@AndrewStacey :-)

@percusse Fortunately, when I originally posted that "bug" (to pgf-users) then I also posted another oddity which did turn out to be a genuine bug. So my initial contact with Till Tantau is a score draw, I deem.

@AndrewStacey He is quite hyped about fixing bugs/closing tickets lately, so I'm sure he is happy anyway.

2 hours later…
11:59 AM
@percusse I can't think of a way to do it directly in TeX, but the tables could be stored in SQL and use datatooltk to pull the sorted data into a datatool file.
@Dan Try texdoc glossaries-user or texdoc glossariesbegin or look at the glossaries samples directory or try the article on the LaTeX community site. You don't need Perl to uses glossaries you can use makeindex directly.

12:26 PM
@DavidCarlisle I call upon your expertise to answer the latest comment on my answer tex.stackexchange.com/a/129521/86

1 hour later…
1:40 PM

@AndrewStacey That's not funny
@AndrewStacey we should get some l3 regex submitted to the syntax highlighter for this site (which uses some google api if I recall correctly)

2:08 PM
@PauloCereda When you have finished answering @AndrewStacey's question above you can submit some syntax rules there, now you've found it:-)

@DavidCarlisle A vim style? No thanks. :)

3:05 PM
@NicolaTalbot thanks! that was very helpful!

1 hour later…
4:05 PM
@DavidCarlisle Damn, even on Lua questions you are faster than me!

@JosephWright oh I know lots about lua:-)
2

@DavidCarlisle: can I suggest an improvement to your answer? :)
I was gonna answer it, but you were faster. :)

@JosephWright Actually I'm on the lookout for tikz questions (that @Qrrbrbirlbel doesn't get to first) as I only need half a dozen more to get a silver badge will probably do for my new year's resolution
@PauloCereda edit away:-)

@DavidCarlisle Nope. :) You can also go with Lua's tonumber method enclosed on every TeX argument. :) Say, tonumber(#1) ensures the argument will get a numeric representation. On a second thought, perhaps the parenthesis approach is cleaner and more efficient.

@PauloCereda I think I'd say you shouldn't coerce here: non-numerical argument = error

4:13 PM
@JosephWright Good point. :)

@JosephWright Is it ok to make a feature request for one of your package using this chat room?

4:34 PM
Just for fun:
2

I am teaching an elementary student. He has a homework as follows. There are 16 students who use either bicycles or tricycles. The total number of wheels is 38. Find the number of students using bicycles. I have 3 solutions as follows. Using a single variable. Let $x$ be the number of s...

4:56 PM
@JosephWright: assert(type(#1) == "number", "we need numbers!") :)

@PauloCereda OK, you could of course do that, but how far do you go?
@Philipp Yes

@JosephWright Not that far. :)
2

Are there any technical reasons that have killed the DVI format as an output for TeX engines? I know that the microtype package works better with pdflatex, but is there any technical reason for microtype not supporting DVI better? Perhaps the main practical reason for not using DVI is that pict...

I expect some witty answer from @DavidCarlisle.
@JosephWright: what's your Mac wallpaper?

@JosephWright In this answer (tex.stackexchange.com/a/130300/6870) to my question about "related entries" in BibLaTeX the issue of mutually relating references came up which leads to a LaTeX overflow when used with biblatex-chem but is ok with standard styles since built-in biblatex styles have the ability to detect related entry loops. Could you add this functionality to biblatex-chem as well, please?

@Philipp I can certainly look at it: probably the weekend

@JosephWright Mac wallpaper, please. :) Tell me it's an epic HD wallpaper. :P

5:10 PM
@JosephWright Thanks, that's great.
@PauloCereda Maybe it is even this one (concertposter.org/-2013Mar/Epic/Epic-Hero-Over-Drop.jpg) :)

@Philipp in 3D! :P
I have a bunch of wallpapers, transitioning every 15 minutes.
@JosephWright: ^^ :P

@PauloCereda Even more epic then ;)

@DavidCarlisle's wallpaper:

@PauloCereda And this one for bad days when you simply want to get in an angry mood :)

:10982601 \$ vim
bash: vim: command not found

5:22 PM
@DavidCarlisle <3
@Philipp C-x wallpaper :)

@PauloCereda I voted for "Leave Open", by the way. (There is one close vote for "primarily opinion-based".)

@Qrrbrbirlbel Is there a "leave open" option?! Oh my. :)
I need more rep. :)

@PauloCereda In the Review section, yes.

@Qrrbrbirlbel Ah review. :)

@PauloCereda ... and one vote closer to the next badge. ;)

5:26 PM
Maybe @David can teach me how to harvest rep by adding % to the end of some line. :)
@Qrrbrbirlbel ooh! :)

@PauloCereda That's egreg's speciality 2000 bonus hrmph.

@DavidCarlisle :)

6:00 PM
@PauloCereda not very witty I fear
@PauloCereda ?

@DavidCarlisle Nice answer. :) I'll upvote it ASAP. :)
@DavidCarlisle A font which incorporates Braille in the typeface. :) Not my cup of tea, but a nice idea. :)

@PauloCereda yes I just wondered how you found the video or if you helped set it up or...

@DavidCarlisle Ah I saw it in a Gizmodo post on my RSS reader. :)

6:29 PM
your point about separating the pictures is one of the best supporting dvi; if you want to check beforehand that a figure will actually reproduce on a press, having them separate is much simpler and more reliable. as one of my colleagues has observed, once a figure is embedded in a pdf file, all you have is hamburger. — barbara beeton 53 mins ago
Gotta love barbara's comment. :)

2 hours later…
8:15 PM
Hello

@leo Hi
What are you studying?

Right now, Galois theory
:-)
Well, I'm not studying right now. I'm taking that course
I need to do my first script
Where can I learn?
I think it must be in Perl or Python

@leo I'm touring with my bike. ;-) Today a climb to a 1460m high mountain in the Apennines, with a magnificent view over the “pianura padana” (the Po river plain).

@egreg Here I am. :)
@leo Tell me about your script. :)

@egreg Cool!!! You are a fortunate man! Could you share a photo? :-)

8:27 PM
@leo Here's the view (but taken in the winter)

@Paulo Hi! I' ve never done a script. Now I must translate some question in a tex file to GIFT format in order to use them in Moodle

@leo In a clear day like that you can see a large part of the Alps

@leo Hm I see. :) Do you know if the standalone class can help you?

So I wanted to automate the process and learn scripting :-)

@leo: which OS do you use? :)

8:32 PM
@egreg It looks very nice indeed. It makes me wish to be there
@PauloCereda Windows, but I'm have been trying Ubuntu for about 3 months

@leo There are too many pylons, unfortunately, because of the mountain's very good position.

@PauloCereda Did you interpreted GIFT as GIF? I ask because the class you suggest is useful to produce images

@leo Yep, I thought it was GIF. :) Isn't it? :)
aaaaaaaaah it is a Moodle format.

@PauloCereda Indeed ! :) I was looking for a link to the explanation
a question in that format will look like
Enunciado de la Pregunta.
{
= Opción 1. Respuesta Correcta.
~ Opción 2. Respuesta Incorrecta.
.
.
.
~ Opción n. Respuesta Incorrecta.
}
there are some guys who made a little program tex2Gift, but it seems is not available
they are from the Universidad Nacional de Colombia

@leo: in that case, I think Python might be very useful for you. :) The question now is, we might need to have some specs in order to read the .tex file and say, "hey, this line is the question, the next one is the answer..." and so on. :)

8:40 PM
@egreg We are plenty of those these days

@leo: see Code 4 of the paper you linked. :) We might need something like that: the specs on which our script will parse the .tex file. :)
Thankfully my Spanish is not so rusty. :)

@PauloCereda exactly. I think the question can be enclosed by a question environment. The choices by choices environment (inside the question) which behaves just like an enumerate. Instead of \item a \choice, and a \correct command to mark the right choice. What do you mean by specs?

@leo The semantics of your document. :)
This thingy you just said. :)

@PauloCereda I've hear that it's easier to someone who speak portuguese to understand spanish, than the other way around
@PauloCereda Oh I see. :)
@PauloCereda well the exam document class have exactly those commands

@leo Once you define how your .tex file will be, the GIFT generation is not that complicated. :)

9:09 PM
@PauloCereda It seems the way to is by using the exam document class. That way we can do things like:
\begin{questions}
\question One of these things is not like the others; one of these
things is not the same. Which one is different?
\begin{oneparchoices}
\choice John
\choice Paul
\choice George
\choice Ringo
\CorrectChoice Socrates
\end{oneparchoices}
\end{questions}
And the algorithm in the paper I linked can be simplified. It will no longer need to know how many questions are there. The loop will stop when it reach \end{questions}

10:01 PM
why print "Mary had a little lamb," gives Syntax error in Python IDLE?

10:14 PM
@leo But not in regular python. (I have no idea; I've never used IDLE.)

Look at:
G:\Documents and Settings\Leo\Escritorio>python mary.py
File "mary.py", line 2
print "Mary had a little lamb,"
^
SyntaxError: invalid syntax

G:\Documents and Settings\Leo\Escritorio>
that was the output of in the terminal
:-(

@leo Does print("stuff") work?

@Qrrbrbirlbel yes
I'm just coping and pasting from a tutorial

@leo That's what I did. ;)
searched Wikipedia for IDLE and Python, found a link to an guide
and scrolled until I found print

@leo Could it be another case of the wrong quotation marks?
(I.e. in the guide it has 'smart' quotes.)

10:25 PM
@Qrrbrbirlbel I'm following this one

@AlanMunn nope. It has "

@leo Yes, I just looked at your guide.

11:20 PM
@leo: print "xxx" raises an error in the new 3.x series of Python (it's valid for 2.x though), maybe that's why.
Being print a function, they decided to require parentheses. :)

Can somebody recommend a guide to writing packages? (No, I don't want to deal with .dtx files.)
Or should I?

@Qrrbrbirlbel dtx files are just packages where the comments are latex markup rather than plain text, so they make for nicer documentation

@DavidCarlisle I am currently reading the chapter "Dokumentieren eigener LaTeX-Pakete" (the last one before the appendix) from the Companion ...

@Qrrbrbirlbel reading German is a lot harder than writing a package

@DavidCarlisle really? :-)
@PauloCereda that explains everything. It must be an old tutorial

11:35 PM
@DavidCarlisle Haha ... it shouldn't be too hard to transfer one's .sty to a proper documented package? Saves me the hassle if it won't be accepted (or it doesn't work at all).

@Qrrbrbirlbel I always start undocumented, then once it's more or less working just use an editor to add the % \begin{macrocode} lines at all the main gaps then get round to adding some words.

@DavidCarlisle That's what I thought.

@Qrrbrbirlbel texdoc clsguide dtxtut dtxgallery docstrip

11:53 PM
@texenthusiast I liked David's response more ...

@Qrrbrbirlbel from the horse mouth :)

@Qrrbrbirlbel Write code first, document it later. :)
And remember: there are no bugs, only random undocumented features. :)

@PauloCereda All of my code's features are undocumented ...

@Qrrbrbirlbel I know that feeling. :)


text:
# Numbers

## Numbers

The computer can also do regular math that you see in school. Use Console.WriteLine to print out the result from the math expressions. No quotation marks are needed for numbers!

Console.WriteLine(5 + 4);
Console.WriteLine(6 * (9 - 7) / 3);

Here is the full list of math symbols that you can use:

OperatorDescriptionOperatorDescription
+Add\ | Divide |
-Subtract%Modulo (remainder)
*Multiply(,)Parenthesis