PHY5200 F07

Chapter 4: Energy

Reading

Taylor 4.3-4.5 (today) and 4.5-4.7 (Monday).

Recall

Determining U(r)

We can use the definition of work to determine the form of U(r) for a given force. Choose a reference point to be point 1, where we define the potential to be zero.

U(r) - U(1) = U(r) = -W(1-->r) = -∫1rF(r')⋅dr'.

Example: Some Example Potential Energy Functions

Near the earth's surface the force of gravity is nearly constant, F=-mg khat. The potential energy is U(r) = ∫1rmgkhat⋅dr' = mg(z - z1) = mgh, where z and z1 are the vertical coordinates of the limits, and, if we pick the reference point to be the surface of the earth, h is the height above the surface.

At greater distances from the earth (or any other celestial object), the force of gravity is given by Newton's law of universal gravitation, F = GmM/r² rhat, where m and M are the masses of the test object and the gravitating body, r is the distance between their centers, and G is the constant of gravitation. The potential energy is U(r) = -GmM ∫1r(1/r'²)r'hat⋅dr' = GmM(1/r - 1/r1).

The spring force (in the x direction) is given by F=-kx. The potential energy is U(x) = k ∫1x x'dx' = ½k(x² - x1²).

If we choose r=r1, then the potential energy must be 0, independent of the path taken to go from and return to the starting point:

loop F(r')⋅dr' = 0

This integral is known as a loop integral, and Stokes Theorem relates it to an integral over any surface bounded by the loop

loop F(r')⋅dr' = ∫surface (∇×F)⋅dnhat = 0

Since the loop is arbitrary, so is the surface, resulting in the conclusion that ∇×F = 0 everywhere for the force to be conservative (to have a potential energy function).

U(r) - U(1) = U(r) = -W(1-->r) = -∫1rF(r')⋅dr'.
∇×F = 0 everywhere for a force to be conservative

The Curl

The curl of F is written ∇×F because it behaves like the cross product of the del operator with F, where we use

∇ ≡ ihat ∂/∂x + jhat ∂/∂y + khat ∂/∂z
When written this way, the partial derivatives aren't to be evaluated. They represent the operation that will be performed once the operator acts on a vector function. Note also that these are partial derivatives not full derivatives.

The curl of F is a vector. This assumes that F is a function of the the spatial coordinate, r, F = F(r). So the curl can also be a function of the spatial coordinate. This is a compact notation for a complex object: first F is a function of r, and it is a vector (so it is like three separate functions of r, F(r) = ( F1(r), F2(r), F3(r) ) = ( F1(r1, r2, r3), F2(r1, r2, r3), F3(r1, r2, r3) ); second, the curl is a vector function of partial derivatives of F, ∇×F. To be more explicit, we must choose a coordinate system.

In cartesian coordinates, the curl of F=(Fx,Fy,Fz) is

∇×F = (∂Fz/∂y - ∂Fy/∂z , ∂Fx/∂z - ∂Fz/∂x , ∂Fy/∂x - ∂Fx/∂y)

If F is expressed in different coordinates (cylindrical or spherical) then the curl has a different form. The expressions for the 3 components are printed on the inside-back-cover of the text.

Force as the Gradient of Potential Energy

Let's go back to the definition of potential energy and see if we can come up with the inverse relation between potential energy and force.

U(r2) - U(r1) = ∫r1r2 F(r')⋅dr'

Now what if r2 is infinitisemally displaced from r1, r2 = r1 + dr = r + dr so that I can drop the subscript. Then the left hand side is the differential change in potential energy

dU = U(r+dr) - U(r) = U(x+dx, y+dy, z+dz) - U(x, y, z) = F(r)⋅dr = Fx(r)dx + Fy(r)dy + Fz(r)dz

If we have a function of one dimension, g(x), then we write the differential of g in terms of its derivative dg = g(x+dx) - g(x) = (dg/dx)dx. We'd like a generalization of this for two, three, or more dimensions, and that looks like

dU = ∂U/∂x dx + ∂U/∂y dy + ∂U/∂z dz

Time Dependent Potential Energy

It is possible to write a potential energy that depends on time, that is, U = U(r,t). The force is still given by F = -∇U, and will still satisf ∇×F = 0, but now the force can also depend on time, F = F(r,t). If the potential energy depends on time, then energy is not conserved. To see this, note that the differential change in U is

ΔU = (∂U/∂x)Δx + (∂U/∂y)Δy + (∂U/∂z)Δz + (∂U/∂t)Δt

and the differential change in kinetic energy is still

ΔT = dT/dt Δt = (mvvdot)Δt = F⋅Δr.

Since F = -∇U, we can write ΔU = -F⋅Δr + (∂U/∂t)Δt. Adding ΔU and ΔT, we see that the force term cancels, leaving

ΔU + ΔT = Δ(T+U) = ΔE = (∂U/∂t)Δt

including the definition of mechanical energy E = T+U. Taking the limit as Δt goes to zero yields

∂E/∂t = ∂(T+U)/∂t = ∂U/∂t

The rate of change of mechanical energy equals the rate of change of the potential energy.


© 2007 Robert Harr