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.
The binomial series
is
- 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