Motion of a Particle in Three Dimensions

We first start with the concept of Torque.  Torque can be viewed as a twisting force.  If you have a lever or length "r" and you pull on the lever and turn it about its pivot, the force that is acting upon the lever is the force that is perpendicular to the lever arm.  So if the angle between the lever and your arm is theta, then the component perpendicular to the lever arm is..

This can be expressed as the vector product between the position vector of the lever arm and the Force vector.

Imagine we have a plane with a central axis segment of AB.  If we apply a force to the plane to spin it around the axis of rotation, only the force perpendicular to the applied force on the surface contributes to the motion.  In other words, only the perpendicular component of the force contributes.  If we define the angle theta as the angle between the x projection of the force vector on the plane, (the only force moving the object as previously stated) and the position vector of the lever arm, then geometrically...

The sign of torque is defined by the right hand rule. Perpendicular to the plane pointing up is defined as positive and pointing down is negative.  From this definition we can expect that the force component parallel to the rotation axis will contribute no torque, so let us dot our definition with the normal vector of the plane.

From this definition, F no longer needs to be decomposed into its components and its sign is automatically determined.

From here I will now deviate to describe work in three dimensions.  In one dimension, work is defined as dW=Fdx.  This is the same for three dimensions except in every dimension.  So for y it would be dW=Fdy and so on.  This can be expressed as..

Therefore work is defined as the line integral over the path or travel of a force F.

Polar coordinates are very useful in analyzing physical problems.  In many cases they tend to be a lot less cumbersome than standard Cartesian coordinates.  Polar coordinates has a unit vector r and a unit vector theta, instead of the standard x and y.  The unit vector r and theta are given by the following relations. (I will also differentiate them to get a useful property.)

Therefore the position vector is simply..

The velocity vector is..

We now differentiate again to find acceleration.

An interesting thing to note is when the acceleration in the radial direction is equal to zero, corresponding to centripetal acceleration, you obtain..

This should be a familiar formula!

We now move into three dimensions.  A useful coordinate system here is cylindrical coordinates.  These are exactly the same as polar coordinates except we keep z as the third component.  Therefore the position, velocity and acceleration vectors in cylindrical coordinates are..

As you can see, we pretty much just tacked on the z component.

If we have the following force in cylindrical coordinates, and we differentiate..

You must remember that the unit vector r and theta are both functions of theta!  The z unit vector is not!

We now focus our attention to spherical coordinates.  As you may already know, the x, y and z components of these coordinates (As well as other useful formulas) are...

So, we know that we can represent any vector rho with..

Substituting the spherical values in we get..

Now if we call this vector R then the unit vector rho is just the partial derivative of R with respect to rho.  And same for theta and phi.  Of course, all these vectors must be divided by their magnitude so they will have unit length.  Remember, partial derivatives give a vector in the direction of change! Think about this concept and you will understand why the following formulas make sense.

A similar approach can be done using the cylindrical coordinates, all one has to do is change the position vector.

Very elegant indeed!!

We now want to find the velocity and acceleration of a position vector in spherical coordinates.  Before we do that however, let us differentiate the spherical unit vectors to get some useful relationships we will need to use.

Now we can differentiate the position vector in spherical coordinates...

If you have a force expressed in spherical coordinates..

Then the derivative with respect to time is...

Now that we have all these formulas for spherical and cylindrical coordinates, lets apply them in a physical sense.

Starting with Newtons second law, we can express it in three dimensions using vectors.

Deriving the expression for the impulse momentum theorem is pretty much exactly the same as in one dimension.

We now derive the expression for the work kinetic energy theorem.  We could proceed as we did earlier, multiplying both sides of Newtons Second Law by velocity and doing some manipulation.  However, it is much more elegant to take the dot product with 2v to both sides of the second law..

We now move our focus to angular momentum.  Imagine we have a particle in a circular motion.  Its momentum is mv, where v is a vector.  The angular momentum about the point O (Which will be the rotation axis) is the component of momentum acting in the direction of the unit vector theta.  The particle has mass "m", its angular momentum is then..

Let us now differentiate this with respect to t...

Angular momentum is like torque, its value is dependent on where you define the point O to be. The point O in our case will always be the axis of rotation.  Recalling the radial and angular vector components for polar coordinates, the equation of motion becomes (with the force represented in polar coordinates as well)...

We now have the angular momentum theorem for motion in a plane. For one more additional topic on angular momentum, we want to find it around the point O.

So the time rate of change of the vector angular momentum is equal to the torque acting upon it.  This is exactly the result we just derived except this could be applied to three dimensions.

If we wanted to model a harmonic oscillator in three dimensions such as an atom in a lattice, it would be exactly the same in the two dimensional case.  This is because the force "F" is only dependent on the position in the components direction.  For example, the force acting  in the x direction is only dependent on x, and same for y and z.  This means we could solve each component separately from each other, or do them all at once in vector form..

Using vectors, we can express the motion of projectiles in three dimensions. When a projectile is in motion, the only force acting upon it is the force of gravity if we neglect, for the time being, air resistance.  Hence, the equation of motion is..

These equations should look familiar! A more realistic model is one where the particle experiences air resistance. (We will assume that the air drag coefficient b is the same in all directions however the problem is no more difficult if we assume they are not the same.)

These differential equations can be solved by general methods, it shouldn't be too bad.

Our next objective is to define a definition for potential energy in three dimensions.  This definition should be analogous to the one dimensional case and only depend on the coordinates.  However, it is very different than  the one dimensional case.  In one dimension, to get from point A to point B, there is only one possible path.  In two and three dimensions there are infinite ways to get from point A to point B.

Where "r sub s" is a standard point chosen for the starting point, since we are only interested in the change in potential energy.  Since there are an infinite number of ways to get from point A to B, the integral on the right hand side MUST be independent of path for this definition to be correct!  In other words, F must be a conservative force.

Let us assume that F is indeed a conservative force, then the infinitesimal potential energy change is...

We now seek a condition to test F for to see if it is indeed a conservative force.  We know that the cross product of any vector is equal to the zero vector.  If we apply the cross product to the del operator we know that it too will be equal to the zero vector. We will therefore cross both sides of the above equation with the del operator.

Since we have assumed the existence of V, this represents a necessary condition that must be fulfilled in order to deduce that a potential function exists and F is conservative.  If F is indeed conservative, then the work in moving a particle from point A to B and then back from point B to A is equal to zero.  This is because it takes the same amount of work to move the same displacement (remember, displacement is different than distance!). Therefore the work from going from B to A is the negative of going from A to B.  Hence their sum is zero. This fact can be confirmed by stokes theorem.

When the force is conservative, we can express the work done as the difference between the two potential energy values at the two points.

Conservation of energy!

We now can use this to create an energy integral for motion in three dimensions..

This is the reason why F is called a conservative force, because it leads to the conservation of energy, meaning there are no dispersion forces that leads to loss of energy.

If the force F is a function of position and time, then we can show that..

All this can be proves but F is no longer a conservative force because it now not only depends on position, but time as well.  This means that the equation for conservation of energy does not hold for this case and we cannot call F a conservative force.  Remember, it is called a conservative force not because the force equals the gradient of the potential, but because it leads to the conservation of energy principle!

We now move onto the concept of a central force.  A central force is a force which only acts either toward or away from a fixed center.  An example of this type of force is the gravitational force.  A central force in spherical coordinates is..

Breaking this down into the x, y and z components...

Is a central force conservative?  Let's find out by computing the curl.

This pattern follows for each component and they all equal to zero, thereby proving that if a force is a central force, then it is conservative.  By this, we now also proved that gravity is a conservative force.  A central force represents an attractive force if F(r)<0, and represents a repulsive force if F(r)>0.  Usually in nature, a central force is inversely proportional to the square of the radius, such as the gravitational force.

So what is the angular momentum of a body under a central force?  We know the field has no twisting or curling since the curl of F is equal to zero, so an object in a central force only experiences radially inward or outward forces.  Therefore there is no torque acting upon it.  From the torque-angular momentum theorem from earlier..

So the angular momentum of a body in a central force remains constant.  Many central forces arise in nature and it is because of this fact of constant angular momentum, that angular momentum is an important concept.

What we will now show, is that a particle acted on by a central force will have a path of motion that lies in a single plane containing the center of the force.  To show this, let the position r_0 and the velocity v_0 be the initial conditions at t_0, and choose the x axis through the intial position r_0 of the particle, and the z axis perpendicular to the initial velocity v_0.  Then we have...

The equations of motion in rectangular coordinates are...

Since the initial position and velocity in the z direction is zero, a suitable solution is z(t)=0, meaning the motion takes place entirely in the x and y plane.  Using the equations of motion in polar coordinates, we can once again arrive to this importance fact of constant angular momentum.

Integrating the first equation of motion with respect to t, we end up getting the conservation of energy.  This fact is evident because since F(r) is always conservative (proved earlier), the integral of it is the negative of a potential function, more over the integral of the left hand side is the kinetic energy in polar coordinates. 

Solving this expression for r dot, we obtain...

This equation would then be evaluated and solved for r(t).  To obtain the motion in the theta direction, we use the equation for the time derivative of theta..

Going back to the first equation of motion in cylindrical coordinates, we can use the definition of angular momentum to get the following relationship..

So what exactly happened here.  Why is the the potential energy function in quotes.  This is because in line 4 of the above work, this represents the force F(r) plus a 'fictitious' centrifugal force.  Why is centrifugal force fictitious? This is because the term added to both sides is not a force at all but part of mass times acceleration.  There is no actual force acting on the object, but rather an effect based purely on the objects acceleration.  To understand this, consider the following. 

Imagine you are in a car and going fairly fast.  Now imagine you turn a corner fast.  You will feel yourself pushed into the door.  One would argue that this is because of centrifugal force, but is there really a force pushing on you into the door?  No there is not, the reason for this pushing is because of your inertia, your body resists to changes of motion.  When your car is going straight and then turns, your body still wants to go straight and continues to do so as your car turns.  In order to make you turn, you must travel straight as far as you can and be stopped by the car.  (Either by the seat belt or the car door, or friction.)  At this point, you can no longer travel straight and the resulting force pulls you to the side.  Think about it, it will make sense!

This is why centrifugal force is considered a fictitious force, it is actually a product of acceleration and mass.  In the above mathematics, we are treating it as a fake force in order to get an expression for a potential function.  That is why V(r) is in quotes, because it isn't entirely a potential function because it contains a term that is the integral of a term that is not a force.

Notice that this term of 'V'(r) appears in our boxed in solution to find r(t) above.  It is at this point why we realize the importance of constant angular momentum in a central force.  It is because of this constancy, that we are able to completely eliminate theta from the equation of motion and reduce it to one dimensional motion.

Often times, the big integral in the above boxed equation to find r(t) is very complicated or impossible to solve, and then solving for r(t) presents another challenge.  It is because of this reason that we attempt to describe the motion with respect to the angle theta.  Using the following expression from earlier, and the following substitution we can obtain..

Finding the motion of a particle based on angle theta is very common in astrophysics. Say we have a object in orbit and we want to find out the area the object sweets out in a given time.  If we approximate the section of a curved surface to be the following.  (This argument comes from finding the area of a function in polar coordinates..)

Then using the definition of angular momentum...

Where "T" is the period and is just the integral evaluated at the limits 0 and t.  So an object in orbit influenced by a central force, will sweep out equal areas in equal times.  This is evident by the fact that the derivative of the area with respect to time is constant.  This is one of Keplers Laws.

The importance of radial forces comes when the force is inversely proportional to the square of the distance.  These types of forces come up very frequently in nature.  A force of this form and its potential function look like..

For the gravitational force, K=-Gm1m2, where the negative indicates an attractive force. For the electrostatic force between two charges, K=q1q2, where the value can either be positive or negative depending on the charges in question.

If we combine this expression for the potential energy, with the the fictitious centrifugal force term, we obtain the "potential energy" for an object in orbit.  Depending on the values of K and whether or not L=0, we can obtain different curves for the plot of this function.  For the gravitational force K<0 and L is not equal to zero. In this case, the potential energy actually has a minimum value.  To find this, we simply differentiate the potential function and set equal to zero.

(I apologize to anyone who may be reading this, as I will not be able to fabricate the graph for this potential function, the image for this makes it a lot easier to understand.  As I mainly use this blog for notes, sometimes going and making images is impractical since people wont really be reading.)

For this value, the graphical location of the point is a minimum, corresponding to one radius, meaning a circular orbit.  If V was a little higher, the straight line would cut through the graph twice, corresponding to two values of r, meaning an elliptical orbit.  

If we take the gravitational force being ku^2, (Remember what u equals! "1/r"!) Plugging this into the boxed differential equation above, we can easily obtain the general solution..

These are the minimum and maximum radii of the elliptical orbit.  Looking at the graph of the potential function, it is easily seen that above the point of a circular orbit, the potential will oscillate between two point (radii).  To find these turning points, we must remember that the kinetic energy (Which we will not denote as T since are using a K already.) is equal to zero at these points. (Just like throwing a ball in the air and having K=0 at the top of its travel.)  Therefore...

An ellipse is defined as the curve traced by a particle moving so that the sum of its distances from two fixed points F,F' is constant.  Where F and F' are the foci of the ellipse.  If r represents the distance from F to any point on the edge of the ellipse and r' represents the distance from F' to the same point, then by the definition of an ellipse..

Where "a" is half the distance of the major axis. (Largest Diameter)  From the geometry described above, using the cosine law we get... (Again, sorry for no picture!)

Where epsilon is the eccentricity of the ellipse.  If epsilon is equal to zero, the two foci have no distances between them and the ellipse is now a circle.  So eccentricity is simply a measure of how elliptical an orbit is.  As epsilon approaches 1, the ellipse turns into a parabola or straight line.

Substituting r' from the first equation into the cosine law..

This is the equation of an ellipse in polar coordinates with the origin at one focus.

A hyperbola is defined as the curve traced by a particle moving so that the difference of its distances from two fixed foci F, F' is constant.  Where F and F' are the focus of the "side ways" parabola  on each side.  The equation of an ellipse is also the equation of a hyperbola, except the eccentricity is now greater than 1.  Also the hyperbola has two branches, each branch is "taken care of" in the above equation by the +1 in the denominator, when it is the left branch it is +1, and when it is the right branch is it -1.

The asymptotes of the hyperbola make an angle "alpha" through the center of the two branches and has a value of theta where the radii equal infinity.

A parabola is defined as the curve traced out by a particle moving so that its distance from a fixed line, the directrix, equals its distance from a fixed focus F.  (Sorry again for no picture!!)

The equations of all three conic sections can now be written as one equation with four different cases.

B<-A cannot occur because r would then not be positive for any value of theta.  Note that in all cases, the eccentricity is equal to A/|B|  We could decide to shift the axis by introducing a phase constant in the cosine term..

This is the EXACT same form as the solution to our previous differential equation for motion in an orbit!  Comparing this equation with the differential equation..

Starting from the equation for 'V'(r), if you solve for r, you obtain..

Using the equation for an ellipse, we can find the turning points and then compare that equation with the one we just obtained.

If we wish to find the length of half the semi-major axis,b (This will come in handy soon!) then from the geometry of an ellipse. (Sorry again for no picture! But these are my notes!)

This is Kepler's Third Law.