Some important ideas to review for Theoretical Mechanics

Plane polar coordinates (r, theta).

• be able to sketch a point (or a vector from the origin to the point) with coordinates r,theta
• from the sketch, write down x and y coordinates in terms of r and theta

Cylindrical coordinates (3-dimensional version of plane polar coords, with z axis added)

Spherical coordinates (r, theta, phi)

• Theta is the 'polar' angle from the pole (z-axis) to the point
• The 'azimuthal' angle phi is located in the x-y plane
• Point z-coordinate is a projection onto the z-axis: z = r cos theta
• The point projected into the x-y axis has magnitude r sin theta
• This must then be projected onto x and y axes, using the angle phi, as in plane polar coordinates
• You should be able to draw a sketch of a vector from the origint to (r,theta,phi)
• From this sketch you should be able to quickly obtain the x,y,z coordinates in terms of r, theta, & phi.

Useful Series Expansions

• If at one point P we know the value of a function f, and the values of its derivatives at that point, we can find the value of a function f at any point near point P. The taylor series allows us to do this.
  
• In one dimension, the taylor series of f(x) with respect to point x=a is
  
     f(x) = c0+ c1 (x-a) + c2 (x-a)^2 + ...
  
     where c0 = f(@x=a), c1 = df/dx(@ x=a), c2 = d^2f/dx^2(@x=a), etc.
  
• When x is extremely close to a, then x-a is extremely small. In this case, terms beyond c1 (x-a) are super-tiny, which lets us write
  
     f(x) = f(@x=a) + df/dx(@x=a) (x-a) to very high precision.
  
  or if we think of (x-a) as dx,
  
     f(x) = f(a) + dx f '(a)+ terms too tiny to consider
  
  
• In 3 dimensions, the taylor series of  f(x,y,z) with respect to point (a,b,c) is
  
    f(x,y,z) = f(@(a,b,c)) + (x-a)*df/dx(@(a,b,c)) + (y-b)*df/dy(@(a,b,c))+(z-c)*df/dz(@(a,b,c))
                    + higher powers of (x-a), (y-b), and z-c).
  
  If we think of x-a as dx, y-b as dy, and z-c as dz, this can be written as
  
    f(x,y,z) = f(a,b,c) + dx*df/dx(a,b,c) + dy*df/dy(a,b,c)+dz*df/dz(a,b,c)
                    + higher powers of (x-a), (y-b), and z-c).
  
  And if we regard f(x,y,z)-f(a,b,c) as df (a tiny change in the function f) we can write this as
  
  [1]   df = df/dx(a,b,c) dx + df/dy(a,b,c) dy + df/dz(a,b,c) dz + terms which vanish as dx, dy, dz get small
  
  In general, if f is a function of variables u,v, and w (which could be r, theta, and phi in spherical coordinates, for example), then [1] can be written in the limit of vanishingly small du, dv, and dw as
  
     df = df/du du + df/dv dv + df/dw dw.
  
  Note: derivatives above should all be partials, and not ordinary derivatives.
  
  If you think of the taylor series as a fundamental of calculus, you can derive many useful results from it, as we have done here.

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The binomial series is

(a+b)^n = a^n +a^(n-1)b (n/1!) + a^(n-2) *b^2 (n*(n-1)/2!) + ...

• The binomial series is finite if n is a positive integer. For example (a+b)^2 + a^2 + 2ab + b^2.
• This series is infinite if n is not a positive integer

If a=1, and b=x where x<<1, the series converges rapidly:
   (1+x)^n = 1 + nx + n(n-1) x^2/2! + ...

Dot product of two vectors V1 and V2 .

• The dot product of vector V1 and vector V2 is the projection of V1 on V2 times the length of V2.
• (The dot product is the projection of one vector on another times the length of the vector projected upon.)
• The dot product is a scalar, and can be positive or negative.
• You should be able to say which of the pairs in the sketch has the largest dot product, and which has the smallest.

• For unit vectors i, j, and k, it should be clear that i.j = i.k = j.k = 0.
• You should be able to write V1 in terms of its components (V1x, V1y, V1z) and i, j, and k.
• You should then be able to show that V1.V2 = V1x V2x + V1y V2y + V1z V2z.
• The dot product is also written in terms of magnitudes and the angle theta between the vectors.
  Find a coordinate system where it is easy to show the dot product of V1 and V2 in terms of the
  vector magnitudes and the angle theta between the two vectors. [Find a place for the x-axis in this coordinate system so the calculation becomes simple.]

• The 'square' of a vector is the dot product of the vector with itself.
• The square any vector is the dot product of the vector with itselff
• The square of the difference of two vectors V1 and V2 is then written as
        (V1-V2)^2 = (V1-V2).(V1-V2) = V1.V1 + V2.V2 - 2 V1.V2
• One way to write the magnitude of a vector is as the square root of the vector squared, or as the square root of the dot product of the vector with itself
• With all this in mind, you should be able to write down the magnitude of the difference of two vectors V1 and V2,
  |V2-V1| in terms of V1, V2, and their dot product. (This is used a lot in potential theory, and in electromagnetic theory.)

Gradient of a scalar (combines derivative, and dot product)

• I like to think of the relation df = grad f . dr as an identity, defining the gradient operator.
• df is a tiny change in some function f, and dr is a tiny change in position.
• (What kind of change in f do we get when dr is perpendicular to grad f?)
• (What kind of change in f do we get when dr is parallel to grad f?)
• in any coordinate system df gets expressed in terms of partial derivatives (the derivatives below are supposed to be partials, though they look like ordinary derivatives) with respect to the coordinates
• Just above, we used the taylor series to show in xyz, df = df/dx dx + df/dy dy + df/dz dz
• And we used the taylor series to show in cylindrical coordinates, df = df/dz dz + df/dr dr + df/dphi dphi
• in xyz, dr = i dx + j dy +k dz. From the equation at the top we see that in xyz,
  
       grad f = i df/dx + j df/dy + k df/dz
  
  
• You should be able to write down df in cylindrical coordinates in terms of r, z, and phi
• Then you should be able to write down dr in cylindrical coordinates using unit vectors
  k, r^ and phi^ .
• From the last two and the identity for the gradient, you should be able to work out the
  gradient operator in cylindrical coordinates.

Vector derivatives

• Background. The product rule for derivatives is (fg)' = f g' + f ' g .
  
  Applying this enables you to show that
  
  
• (A.B)' = A'.B + A.B'
  [derivative of dot product with respect to a scalar variable, like time]
  
• (AxB)' = A'xB + AxB'
  
• and many other derivatives involving vectors.

• Read M&T sects 1.13 & 1.14 on the derivative of a vector with respect to a scalar (like time)
• You know that the unit vectors i, j, and k are constant (their time derivatives are zero)
• What about plane polar coordinates?
• Are the time derivatives of r^, and theta^ equal to zero?
• (When r changes is there a change in r^ or theta^?)
• (When theta changes is there a change in r^, or theta^?)
• What about in cylindrical coordinates?
• Are the time derivatives of r^, phi^ and k equal to zero?
• (When r changes is there a change in r^, phi^ or k ?)
• (When z changes is there a change in r^, phi^ or k ?)
• (When phi changes is there a change in r^, phi^ or k ?)
• Exercise: work out the time derivative of the dot product of two vectors V1 and V2: c = V1.V2
  The dot product in xyz can be expressed as the sum of three products. Each of the three terms