Key concepts from Physics 201

Physics 202 draws signiﬁcantly from the material of Physics 201, and the Engineering statics and dynamics courses are not an adequate substitute for 201. If you have not taken 201, please read through this review of the minimal set of topics needed for 202. I have not included rotational rigid-body dynamics in the review because most of the dynamical applications found in 202 involve point-particles with electro-magnetic forces acting on them. ©2004 Jeffrey R. Schmidt

Vectors are denoted with bold-faced characters wherever possible.

Vectors

Most of the quantities that you will work with in this course have direction as well as a magnitude associated with them. Such quantities are called vectors. A quantity with no direction associated with it, that is unaltered by any changes in coordinate referential, is called a scalar. The most important scalars in non-relativistic physics are mass and temperature. Typical vectors are forces, positions, velocities and momenta. The common representation of a vector is as a point in space such as in the ﬁgure

The point is speciﬁed by giving its x and y coordinates, either as an ordered pair (triple) or as multiples of unit vectors. We could specify the vector in the ﬁgure by giving Cartesian coordinate of its tip

v = (vx, vy) = vxi + vyj

v = (v , v , v ) = v i+v j+v k x y z x y z

from the Pythagorean theorem we obtain its length (magnitude)

∘ ---------- ∣v ∣ = v2 + v2 x y

∘ --2-----2------2- ∣v ∣ = v x + vy + vz

and its direction relative to the x-axis

vy θ = tan - 1 --- vx

(thats inverse tangent, not one over tangent).

In terms of the magnitude and direction we can express the components as

vx = ∣v ∣ cos θ, vy = ∣v ∣ sin θ

Vectors as algebraic objects add component by component

a = (a , a ), b = (b , b ) x y x y

then

a + b = (ax + bx, ay + by)

and similarly in three dimensions

a = (ax, ay, az), b = (bx, by, bz)

a+b = (ax+bx, ay+by, az+bz)

It will prove to be very useful to be able to compute the angle between two vectors, from the ﬁgure below we see that

- 1 ay-- - 1 by θab = θa - θb = tan - tan ax bx

take the cosine of both sides of this equation and use

cos(x - y) = cos(x) cos(y) + sin(x) sin(y)

-AxBx----- -AyBy----- cos( θab) = cos( θa- θb) = cos( θa) cos( θb)+sin( θa) sin( θb) = + ∣A ∣∣B ∣ ∣A ∣∣B ∣

Deﬁne the dot product of two vectors, which produces a scalar

a ⋅ b = axbx + ayby

a ⋅ b = a b + a b + a b x x y y z z

and we see that the cosine of the angle between two vectors is

a ⋅ b cos( θab) = -------- ∣a∣∣b ∣

which is very useful since it involves only the components of the two vectors.

Remember at all times that there are only two allowed vector arithmetic operations that are legal (at this point);

1. Scalar multiplication

s r = s (rx, ry, rz) = (srx, sry, srz)

for

r a vector, s a scalar

2. Vector addition

a + b = (a , a , a ) + (b , b , b ) = (a + b , a + b , a + b ) x y z x y z x x y y z z

The length of a vector is computed via the Pythagorean theorem;

∘ ----------------- ∣a ∣ = a2x + a2y + a2z

Find a vector ⊥ to another vector

Given a vector in two dimensions such as

a = (a , a ) x y

any vector b perpendicular to it (θ_ab =

) is a vector such that

-a-⋅-b-- cos θab = 0 = = 0, a ⋅ b = 0 ∣a ∣∣b ∣

Write this out in components

a b + a b = 0 x x y y

This is one equation in two unknowns b_x,b_y, and so cannot be uniquely solved. As long as both components a_x and a_y are non-zero, we simply pick a value for b_x and proceed;

ax Let bx = 1, ax + ayby = 0, by = - ---- ay

and therefore

a b = (1, - -x-) a y

However if b ⊥ a, then so is sb for s any scalar, and we could even pick s = a_y, making

sb = (a , - a ) ⊥ a = (a , a ) y x x y

In three dimensions there is a whole plane of vectors ⊥ to

a = (a , a , a ), a , a , a ⁄= 0 x y z x y z

and so we proceed in steps. Let

b = (b , b , b ), a ⋅ b = a b + a b + a b = 0, if b ⊥ a x y z x x y y z z

There will be many such vectors, so for example let b_x = 0, b_y = 1;

a 0 + a + a b = 0, b = - --y y z z z a z

and then any vector

say sb = (0, s, - -----) ⊥ a az

This vector lies in the yz-plane. We could also set b_y = 0, b_x = 1; to get

sax sb = (s, 0, - -----) ⊥ a az

another vector perpendicular to a but lying in the xz-plane.

Write a vector as a sum of vectors

Let

a = (a , a ), b = (b , b ) x y x y

be two non-collinear vectors. They are non-collinear if

a ⁄= s b

Pick another vector

c = (cx, cy)

and determine the two scalars α,β such that

c = α a + β b

We call this resolving vector c into a and b. Write it out in components

(cx, cy) = α(ax, ay) + β(bx, by)

use the scalar multiplication rule

(cx, cy) = ( αax, αay) + (βbx, βby)

use the vector addition rule

(cx, cy) = ( αax + βbx, αay + βby)

and look at the two resulting equations

c = αa + βb , c = αa + βb x x x y y y

which is two equations in two unknowns, which we solve; in the ﬁrst solve for β

c - αa β = --x--------x b x

and put it into the second

cx - αax cy = αay + ( ------------)by bx

and get α by itself;

c b - c b c b - c b = α(a b - a b ), α = --y-x------x--y- y x x y y x x y a b - a b y x x y

and go back and get β;

c a - c a β = -y--x------x--y- a b - a b y x x y

You will make extensive use of your abilities to solve two equations in two unknowns in this course.

Dot products

The dot product (or scalar product) can be neatly summarized by the rules

i ⋅ i = j ⋅ j = k ⋅ k = 1

i ⋅ j = j ⋅ i = 0, i ⋅ k = k ⋅ i = 0, k ⋅ j = j ⋅ k = 0

and the fact that it is distributive;

a ⋅ (b + c) = a ⋅ b + a ⋅ c

Demonstration of these facts from the deﬁnition

a ⋅ b = axbx + ayby + azbz

I will leave to you.

Perpendicularity

Two vectors a and b are ⊥ to each other if there is a θ_ab = 90 degree angle between them, making

cos θ = 0, a ⋅ b = 0 ab

Collinearity

Two vectors a and b are collinear if there is a scalar s such that

a = sb, (a , a , a ) = (sb , sb , sb ) x y z x y z

Solve these three equations

a a a s = --x-= -y- = --z b b b x y z

To test for collinearity, we compute all three ratios, if they are not all the same, the vectors are not collinear. Remember that you can divide numbers, such as individual components of vectors, but you cannot divide by the vectors themselves.

Newton’s laws

Newton’s three dynamical laws, established empirically (via observation) are as follows;
1. A body in uniform motion will remain so unless acted upon by forces.
2. An agent applying a force on a body experiences an equal and opposite reaction force exerted upon it by the body.
3. The vector sum of the forces on a body equal it’s mass times acceleration

∑ F = ma i i

The third law has the ﬁrst as a consequence, if no forces act on a body then

dv 0 = ma = m ---- dt

which can be integrated

∫ v(t) dv = 0 dt, v dv = 0, v(t) = v0 0

which is precisely what is meant by uniform motion; the velocity has no change over time, the direction and magnitude remain constant.

There are two main categories of application-problems; given a collection of forces acting on a mass m, determine its acceleration vector, and given a partial set of forces and a bodies acceleration, ﬁnd all forces acting on it. The strategy for solving both types is the same;
Step 1 set up a coordinate system of your choice into which we resolve the components of all vectors;
Step 2 Write Newton’s third law for every mass involved in the problem in component form using the laws of vector mathematics;

m a = m (ax, ay, az) = (max, may, maz)

∑ ∑ ∑ ∑ ∑ = Fi = (Fi,x, Fi,y, Fi,z) = ( Fi,x, Fi,y, Fi,z) i i i i i

ﬁnally obtaining three equations for each mass

∑ ∑ ∑ ma = F , ma = F , ma = F x i,x y i,y z i,z i i i

Step 3 we simply solve these equations for any unknowns present in them.

Example

A box slides freely with no friction on a tabletop. A force F, insufficient to lift the box, is applied as shown. Find the acceleration and the force exerted on the table.
Using up at positive y-direction and right as positive x, draw all forces acting on the box. The table exerts a force N upward, which by the second law equals the force exerted on the table downwards.

Decompose F into components

F = (∣F ∣ cos θ, ∣F ∣ sin θ) = (F , F ) x y

then by noting that if the block accelerates in contact with the table

a = (a , a ) = (a, 0) x y

then

M a = N + Fg + F

M (a, 0) = (N, 0) + (0, - M g) + (F ∣ cos θ, ∣F ∣ sin θ)

and so

∣F ∣ cos θ ∣F ∣ cos θ = M a, a = ----------- and N = M g - ∣F ∣ sin θ M

Circular motion

Consider a mass m tied to a string, forced to travel in a circle horizontally at constant speed v, radius R. If in time t the mass travels arc length s along the circle, it sweeps out an angle θ. Then since

s = R θ, θ in radians

differentiate

ds d θ v = ----= R ---- = R ω dt dt

where ω is the angular speed or frequency and has units of radians per second. Remember that

360o = 2π rad

Impose a coordinate system on the ﬁgure and write down the position vector after integrating the above equation to

θ = ωt

then

r(t) = (R cos ωt, R sin ωt)

and the velocity is

d v(t) = ---r(t) = ( - R ω sin ωt, R ω cos ωt) dt

Notice that

r ⋅ v = 0

This is a consequence of the fact that

r(t) = R (cos ωt, sin ωt) = Rer

and e_r is a unit vector;

2 2 er ⋅ er = cos ωt + sin ωt = 1

Calculate derivatives;

d-- -d- d-- (er ⋅ er) = 2er ⋅ ( er) = 1 = 0 dt dt dt

The moral of the story; the derivative of a unit vector is perpendicular to the vector itself.
Therefore the position vector is radial and the velocity vector is tangential to the circle. Now compute the acceleration

d a(t) = ---v(t) dt

2 2 (- R ω cos ωt, - R ω sin ωt)

- ω2r(t)

This acceleration, called centripetal acceleration, points towards the center of the circle. In any problem involving circular motion at constant speed, the acceleration is given by this expression, which has magnitude

2 2 v-- ac = ω R = R

Example

In the problem below, M₂ slides on a table-top, presumed frictionless; ﬁnd the speed that M₂ must move in a circle at in order to supply enough string tension to support M₁ against gravity.
Write Newton’s laws, we want M₁ to have no acceleration so

F = (0, - M g) = - M gj g 1 1

T = (0, T ) = T j

and so

M10 = Fg + T, (0, 0) = (0, - M1g) + (0, T ), T = M1g

and for M₂ all relevant vectors point in the ±e_r direction;

2 v T = - T er, a = - acer = - ---er R

and

F = - M gj, N = N j g 2

and therefore

2 v-- M2a = Fg + N + T, - M2 er = N j - M2gj - T er R

so that

┌│ --------- v2 ││ M gR M2 --- = T , v = ∘ -------- R m

Work and Energy

The dynamical equations relating cause (forces) to effect (acceleration) in particle dynamics are technically second order differential equations

ma = F

2 d-r(t)-- m 2 = F(r) dt

and in general second order differential equations are much harder to solve than ﬁrst order equations. The process of transforming this second order equation into a ﬁrst order one involves a single integration, which gives rise to a constant of integration called the energy.
Consider

ma = F

Now dot the velocity vector into both sides

ma ⋅ v = F ⋅ v

and use the identity

d-- d-- 1- a ⋅ v = ( v) ⋅ v = ( v ⋅ v) dt dt 2

We have manipulated Newton’s law into

d 1 dr ---(--mv ⋅ v) = mv ⋅ a = F ⋅ v = F ⋅ ---- dt 2 dt

This can be integrated

∫ tf d 1 ∫ tf dr ---( --mv ⋅ v) dt = F ⋅ ----dt t0 dt 2 t0 dt

let

dr-- dr-- r0 = r(t0), rf = r(tf ), v0 = ( )(t0), vf = ( )(tf ) dt dt

be the initial and ﬁnal positions (and velocities) of the body as it is displaced along some path, while being acted upon by the force F, then

∫ tf dr-- ∫ rf t F ⋅ dt = r F ⋅ dr 0 dt 0

and we arrive at the Work-Energy theorem;

∫ ∫ tf -d- 1- 1- 1- rf t0 ( mv ⋅ v) dt = mv(tf ) ⋅ v(tf ) - mv(t0) ⋅ v(t0) = r0 F ⋅ dr dt 2 2 2

The left hand side is called the change in kinetic energy, the right is called the work done by the force F.

Many types of dynamical problems can be solved by either the use of Newton’s laws or by the use of this formula. The advantage of the work-energy theorem is that it does not require much vector arithmetic, and certain types of forces are incapable of doing work. For example consider a body sliding along a surface. The normal force is always perpendicular to the displacement dr, and so

N ⋅ dr = 0

If the force is conservative, we can apply the chain rule of calculus to the integral on the right-side of the work-energy theorem. A force is conservative if there exists a function called the potential V (r) such that application of the chain rule results in

d ∂V dx ∂V dy ∂V dz - ---V (x, y, z) = - ---------- ---------- --------- dt ∂x dt ∂y dt ∂z dt

dx dy dz = F ---- + F ----+ F ----= F ⋅ v x dt y dt z dt

This will mean that the components of F can be gotten directly from V ;

∂V--- ∂V--- ∂V--- F = (- , - , - ) = - ∇V ∂x ∂y ∂z

This may not be possible to do for a given force. Friction is a good example of a non-conservative force; there is no such function V (r) because technically the force of friction depends on both position r and velocity v of an object.

For a conservative force,

d 1 d ---( -mv ⋅ v) = F ⋅ v = - ---V (x, y, z) dt 2 dt

-d- 1- ( mv ⋅ v + V (r)) = 0 dt 2

which can be integrated. Let the position of the moving body at time t₁ and

r = r(t ), and r = r(t ) 1 1 2 2

respectively. Integrate both sides of the above equation accordingly

∫ d 1 ∫ d t2---( --mv ⋅ v) dt = - t2 ---V (r(t)) dt t1 dt 2 t1 dt

∫ r2 = - dV (r) r1

= - V (r(t2)) + V (r(t1)) = - V (r2) + V (r1)

The left side is

∫ t2 d-- 1- m-- m-- t ( mv ⋅ v)dt = v(t2) ⋅ v(t2) - v(t1) ⋅ v(t1) 1 dt 2 2 2

putting it together we ﬁnd

m m ---v(t2) ⋅ v(t2) + V (r2) = ---v(t1) ⋅ v(t1) + V (r1) 2 2

Since the quantity

1- 2 E = mv + V (x, y, z) 2

has the same value at t₂ as it does at time t₁, and these times were arbitrary, it must in fact be a constant. This is an integration constant, since we integrated F = ma once to get it. The constant is called the energy E, and consists of two parts; kinetic

1- 2 2 2 K E = mv , v = ∣v ∣ = v ⋅ v 2

energy associated with motion. The potential energy V (r) is energy associated with spatial position, and can be written several different ways. First of all it’s derivatives represent the components of conservative forces

∂V ∂V ∂V Fx = - ----, Fy = - -----, Fz = - ----- ∂x ∂y ∂z

Multiply each of these by v_x,v_y,v_z respectively and add them

dx dy dz dr Fx ----+ Fy ----+ Fz ---- = F ⋅ ---- dt dt dt dt

and integrate

∫ ∫ t2 dr-- r2 dV--(r)-- t F ⋅ dt = - r dt 1 dt 1 dt

= - (V (r ) - V (r )) = - W 2 1

where we deﬁne the work done against the force F to be

∫ ∫ t2 dr-- r2 W = - t1 F ⋅ dt = - r1 F ⋅ dr dt

which for conservative forces becomes

= V (r ) - V (r ) 2 1

The total energy is conserved, it merely changes form, from kinetic to potential or kinetic to work, as the particle undergoes its prescribed motion.

Example

Compute the work done against the force

F = kx i

moving a particle from (0, 0) to (1,c) along the path

y = cx2

against this force.
The ﬁrst step in performing the integral is to parameterize the path. This step turns what appears to be a many-variable integral into a single variable integral. We will use the parameterization

x = t, y = ct2

If we eliminate t we get y = cx². Now we write the r vector for the points on the curve, and compute the velocity vector

d r = (x, y) = (t, ct2), v = ---r = (1, 2ct) dt

Next we calculate the force vector at every point of the path

F = kx i = (kx, 0) = (kt, 0)

and ﬁnally the dot-product

F ⋅ v = kt ⋅ 1 + 2ct ⋅ 0 = kt

Now we integrate.

∫ (1,c) ∫ 1 dr ∫ 1 1 1 1 F ⋅ dr = F ⋅ ----dt = kt dt = --k(1 - 0 ) (0,0) 0 dt 0 2

The parameterization method is virtually fool-proof. You will get the same result no matter what parameterization you use, so you cannot make the wrong choice. Try that out, let

x = t2, y = ct4, r = (t2, ct4), v = (2t, 4ct3)

and

F = (kx, 0) = (kt2, 0)

put this into the integral and churn out an answer; it will again be

When does a force not have a potential?

If one cannot ﬁnd a function V such that

∂V ∂V Fx = - -----, Fy = - ----- ∂x ∂y

then the force is not conservative and has no potential. A good example is

F = - yi + xj = (- y, x)

We attempt to solve

∫ ∂V--- Fx = - y = - , - V = - y dx = - yx + C (y) ∂x

(since we integrate with respect to x, our “constant” of integration could depend on anything but x, including y). This potential does not give the correct F_y;

∂V ∂ ′ Fy = x ⁄= - -----= - ----(yx - C (y)) = - x + C ∂y ∂x

On the other hand the force

F = yi + xj = ( - y, x)

is conservative and does have a potential function;

∂V ∫ Fx = y = - -----, - V = y dx = yx + C (y) ∂x

checking that this gives the correct F_y;

∂V--- ∂--- ′ Fy = x = - = - ( - yx - C (y)) = x + C ∂y ∂x

which works ﬁne of C(y) = C so C′ = 0. The potential that correctly gives both force components is

V (x, y) = - xy + C

Example

Find the work done by

F = yi + xj = (y, x)

when we move a mass from (0, 0) to (a,b) along the curve

x y = b(--)2 a

Solution

W = - (V (a, b) - V (0, 0)) = - ((- ab + C ) - (- 0 + C )) = ab

(The work done against the force by the mover of the object is the opposite of this).

Example

Find the work done by

F = - yi + xj = (- y, x)

when we move a mass from (0, 0) to (a,b) along the curve

x y = b(--)2 a

Solution We have no choice but to integrate. Parameterize the curve

2 x(t) = a t, y(t) = b t , 0 ≤ t ≤ 1

Find the velocity

dx vx(t) = ---- = a, vy(t) = 2bt dt

ﬁnd the force at any point on the curve

Fx = - y = - bt2, Fy = x = at

and put it together; t₀ = 0, t_f = 1;

∫ ∫ W = 1 ( - bt2, at) ⋅ (a, 2bt) dt = 1abt2 dt = ab-- 0 0 3

Conservation of Momentum

Consider a system of particles, three for example, in which some forces (F_1,ext,F_2,ext,F_3,ext) are externally applied, and all other forces are internal (inter-particle). The force exerted in M₁ by M₂ is F₁₂. By Newton’s second law this is equal and opposite to the force F₂₁ exerted on M₂ by M₁.

Write the equation of motion for each mass

M1a1 = F1,ext + F12 + F13

M a = F + F + F 2 2 2,ext 21 23

M3a3 = F3,ext + F32 + F31

Add all three of these, and note that all of the inter-particle forces cancel pairwise since

Fij + Fji = 0

we ﬁnd that

∑ M a + M a + M a = F + F + F = F 1 1 2 2 3 3 1,ext 2,ext 3,ext i,ext i

and since

2 d a1 = ---2-r1 dt

we can rewrite this as

2 d---- ∑ 2 (M1r1 + M2r2 + M3r3) = F1,ext + F2,ext + F3,ext = Fi,ext dt i

2 d----M1r1----+--M2r2----+--M3r3--- = (M1 + M2 + M3) 2( ) dt M1 + M2 + M3

d2 ∑ = Mtotal ----rCOM = Fi,ext dt2 i

In other words, the acceleration of the center of mass of a collection of particles is due only to the sum of external forces acting on the collection. The center of mass is the ”balancing point” of the collection, it can be found by applying a gravitational force to the collection, regarded as a rigid body, and locating the single point of support where a force can be applied to hold the body stationary against gravity, with no rotation or acceleration of the collection. The formula for the center of mass is

∑ ∫ i Miri rdm rCOM = -∑-------- = -∫------ i Mi dm

What if there are no external forces acting on the collection? All forces are inter-particle and cancel in the sum, and we ﬁnd

d2 ----(M1r1 + M2r2 + M3r3) = 0 dt2

which can be integrated once

2 ∫ t2 d ---2-(M1r1 + M2r2 + M3r3) dt = 0 t1 dt

to give

M v (t ) + M v (t ) + M v (t ) = M v (t ) + M v (t ) + M v (t ) 1 1 1 2 2 1 3 3 1 1 1 2 2 2 2 3 3 2

where

-d- vi = ri dt

This is in the form of a conservation law, a quantity evaluated at one time equals its value at all times. This is probably the most important conservation law in physics, and is called conservation of momentum. Deﬁne the momentum of a mass m with velocity v to be

p = mv

then in the absence of externally applied forces, a collection of particles evolves subject to the constraint

∑ p (t ) = ∑ p (t ) i 1 i 2 i i

and Newton’s law for the center of mass motion if external forces are present can be written as

-d- ∑ ∑ ( pi) = Fi,ext dt i i

Oscillations

Oscillatory motion is periodic motion; the pattern of displacements of a body under the inﬂuence of forces is periodic of period T if

r(t + T ) = r(t)

Newton’s laws in one dimension can easily be reduced from a second order equation for the case of an external force depending only on position

2 d--x- m 2 = Fext(x) dt

to a ﬁrst order equation by a simple transformation equivalent to constructing a conserved energy; multiply both sides by dx dt

2 d--x-dx-- dx-- 1- -d- dx-- 2 m 2 = F (x) = m ( ) dt dt dt 2 dt dt

and if there exists a function V (x) such that

dV--- F (x) = - dx

then the chain rule allows us to write the differential equation as

1 d dx d --m ---(----)2 + ---V (x) = 0 2 dt dt dt

since

d dV dx ---V (x) = --------- dt dx dt

and we can integrate directly by introducing a constant of integration C = E. This is a conserved quantity, also called a ﬁrst integral since it was obtained by an integration;

m-- dx-- 2 ( ) + V (x) = E 2 dt

and invert this to get

∫ ′ x(t) ∘-------dx----------- t - 0 = x(0) 2- ′ m (E - V (x ))

which we perform and again invert to obtain x(t).

For the case of a force exerted by a spring;

1 2 Fx = F (x) = - kx, V (x) = --kx 2

we can perform this integral, and the result is a periodic function. Starting with

∫ ′ x(t) --------dx----------- t = x(0) ∘ 2 1 ′2 m-(E - 2-kx )

make the variable substitution

┌ ----- ┌ ----- ││ ││ ′ │∘ 2E--- ′ │∘ 2E--- x = sin θ, dx = cos θ d θ k k

and the integral becomes

∘ 2E-- ┌ ---- ┌ ---- ∫ θ(t) -k- cos θ dθ ││ m ∫ θ(t) ││ m t = ∘----------------------- = ∘ --- dθ = ∘ --- (θ(t) - θ0) θ0 -2 (E - E sin2 θ) k θ0 k m

We solve this equation;

┌│ ---- ││ k θ(t) = θ0 + ∘ --- t m

and insert it into the equation for x;

┌│ ----- ┌│ ----- ┌│ ---- ││ 2E ││ 2E ││ k x(t) = ∘ -----sin θ = ∘ ----- sin (θ + ∘ ---t) k k 0 m

The result is periodic motion, since the sine function is periodic. Assuming that the period is T, then

x(t + T ) = x(t)

implies that

┌│ ----- ┌│ ---- ┌│ ----- ┌│ ---- ││ 2E ││ k ││ 2E ││ k ∘ ----- sin (θ0 + ∘ --- t) = ∘ ----- sin (θ0 + ∘ --- (t + T )) k m k m

and we can use the trig identity

┌│ ---- ┌│ ---- ┌│ ---- ┌│ ---- ┌│ ---- ││ k ││ k ││ k ││ k ││ k sin (θ + ∘ ---(t+T )) = sin (θ + ∘ ---t) cos (∘ ---T )+cos (θ + ∘ --- t) sin (∘ ---T ) 0 m 0 m m 0 m m

Periodicity will be met if

┌│ ---- ┌│ ---- ││ k ││ k cos (∘ ---T ) = 1, sin (∘ ---T ) = 0 m m

and this is true if

┌│ ---- ┌ ---- ││ k ││ m ∘ ---T = 2π, T = 2 π ∘ --- m k

We often say that the motion repeats with frequency

┌ ---- 1 ω ││ k f = ---= ---, ω = │∘ --- T 2π m

Lets create a master formula; suppose that we have two conditions at t = 0;

x(0) = x0, v(0) = x˙(0) = v0

then x(t) = C sin(θ₀ + ω t) reduces to

x0 = C sin θ0, v0 = C ω cos θ0

dividing

x0 ω - 1 x0 ω ------ = tan θ0, θ0 = tan ------ v0 v0

instead, square and add

┌│ ----------- v2 ││ v2 x2 + --0-= C 2 sin2 θ + C 2 cos2 θ = C 2, C = ∘ x2 + --0- 0 ω2 0 0 0 ω2

and the master formula is

┌│ ----------- ││ v20 x0 ω x(t) = ∘ x20 + ---- sin( ωt + tan - 1 -----) ω2 v0

How are these factors (amplitude and phase) related to the energy of the oscillating mass? The total mechanical energy is the sum of kinetic and potential, and is conserved;

1 2 1 2 1 2 1 2 Emech = -mv (t) + -kx (t) = --mv (0) + -kx (0) 2 2 2 2

Multiply by two, divide by k

2 2 2 2E--- v-(t)-- 2 v--(t)- 2 v--(0)- 2 2 = k- + x (t) = 2 + x (t) = 2 + x (0) = C k m ω ω

and we discover that the oscillator energy depends only on its amplitude

┌│ ----------- ┌│ ----- ││ v20 x0 ω ││ 2E x0 ω x(t) = ∘ x20 + ---- sin(ωt + tan - 1 -----) = ∘ -----sin( ωt + tan - 1 ------) ω2 v0 k v0

Example

Very small amplitude deviations from equilibrium (zero-force condition) for any potential will result in simple harmonic motion. For example a commonly used inter-molecular potential (Morse) is

V (x) = 2ae - 2ax - ae - ax

Show that a mass subjected to this potential has no force acting on it if it is located at x₀ = 1 a ln 4, and that if it is allowed to deviate slightly from this position by amount x, it undergoes oscillations of frequency

┌│ --------------------------- ┌│ ---- ││ 8a3e - 2ax0 - a3e - ax0 ││ k d2 ω = ∘ ---------------------------= ∘ ---, k = -----V (x) ∣x=x m m dx2 0

This is done by ﬁrst locating the equilibrium position (where F_x = 0) by

d F (x ) = 0 = - ----V (x) ∣ = - V ′(x ) x eq dx xeq eq

and by expanding V (x) in a power series about that point

′ 1 2 ′′ V (x) = V (xeq) + (x - xeq) V (xeq) + ---(x - xeq) V (xeq) + ⋅ ⋅ ⋅ 2!

1 = V (x ) + ---(x - x )2 V ′′(x ) + ⋅ ⋅ ⋅ eq 2! eq eq

Using this formula for the potential the equation of motion becomes

2 d--x- -d-- ′′ max = m 2 = - V (x) = - (x - xeq) V (xeq) + ⋅ ⋅ ⋅ dt dx

Now let X = x - x_eq, and we obtain a harmonic oscillator equation

2 ┌│ ----------- d X ′′ ││ V ′′(xeq) m ------ = - V (xeq) X, X (t) = A sin (∘ -----------t - φ) dt2 m

┌│ ----------- ││ V ′′(xeq) x(t) = xeq + A sin (∘ -----------t - φ) m

and so any potential with an equilibrium point and a non-zero second derivative will possess simple harmonic motions of small amplitude.

Newton’s universal law of gravitation

Isaac Newton was the ﬁrst person that we know of to have made the direct cause and effect link between the motion of planets around the sun, and the force of gravity. By physical and purely mental experimentation Newton determined that the force of gravity exerted on mass M₁ by mass M₂ satisﬁes

(r1 - r2) F12 = - M1M2G ------------- ∣r1 - r2 ∣3

N m2 G = 6.67 × 10 - 11------- kg2

This was an amazing deduction that can easily be taken for granted today, but at the time at which it was made, it ranked as the most important physical discovery of the century. The magnitude of the force between two masses varies with the inverse square of the distance between them, and is always attractive, pulling one mass towards the other. This force acts vectorially just like any other vector forces.

Example

Four identical masses m are held in place at the corners (0, 0), (0,d), (d, 0), and (d,d) of a square. Compute the force on the mass placed at the origin.

(0, 0) - (0, d) (0, 0) - (d, 0) (0, 0) - (d, d) F = - m2G (------------------ + ------------------ + ------√----------- ) d3 d3 ( 2d)3

which can be simpliﬁed a bit to

m2G 1 1 F = ------((1 + √----), (1 + √---)) d2 8 8

We can construct a potential energy function for the gravitational ﬁeld by equating the difference in potential energy between points at r_i and r_f with the work done against the gravitational ﬁeld in moving from the ﬁrst to the second point.

∫ rf mM----Gr--- WFg = - (V (rf ) - V (ri)) = ri - 3 ⋅ dr r

where the mass M at the origin exerts gravitational forces on the test mass m that we are moving through the force ﬁeld of stationary mass M. We can rewrite this by introducing the vector derivative or gradient

∂ ∂ ∂ ∂ ∂ ∂ ∇ = (----, ----,----) = i----+ j----+ k ---- ∂x ∂y ∂z ∂x ∂y ∂z

and by noting that

x dx + y dy + z dz r ⋅ dr 1 1 --√---------------------- = --------= - (∇ √-------------------) ⋅ dr = - (∇ --) ⋅ dr ( x2 + y2 + z2)3 r3 x2 + y2 + z2 r

and the integral becomes an integral of an exact differential

∫ rf M--mG---- - (V (rf ) - V (ri)) = ri ∇( - ) ⋅ dr r

M--mG---- rf mM----G-- M--mG---- = - ∣ri = - + r ∣rf ∣ ∣ri ∣

we discover that the point mass M placed at the origin produces a gravitational potential energy ﬁeld whose value depends only on how far one is from the origin

M--mG---- V (r) = V (r) = - ∣r ∣

Example

Find the gravitational force exerted by a ring of mass m radius R on a point mass M held a distance z from the center.
Divide the ring into point masses of length Rdθ and mass ρRdθ,

--m--- ρ = 2 πR

The fragment at location

r = (R cos θi + R sin θj) 1

is a distance √ -2----2-
z + R

from our mass, and so exerts

-√-dmM-----G----- dF = 2 2 3 (R cos θi + R sin θj - zk) ( z + R )

on it. Two of three integrals are zero;

∫ 2π --ρ√Rd---θM---G--- -ρ2√-πRM-----G--z- F = 0 2 2 3(R cos θi + R sin θj - zk) = - 2 2 3 k ( z + R ) ( z + R )

-√mM-----G--z---- = - 2 2 3 k ( z + R )

Example

Compute the gravitational potential energy function for a ring of mass m radius R, at a distance z from its center.
The potential is much easier to compute than the force by integration. In the setup of the previous example, the fragment d at r₁ = R cos θi + R sin θj produces

√dm---M---G--- dV (z) = - 2 2 z + R

at the location of the mass M, and so this is the potential energy of the pair of charges M and dm. Add up the potential energies of M with all of the ring fragments;

∫ dm M G mM G V (z) = - √-------------= - √------------- z2 + R2 z2 + R2

for the total. Furthermore we can compute F_z from the fact that the gravitational force is conservative;

∂V (z) mM G z F = - ---------= - -√--------------- z ∂z ( z2 + R2)3

in agreement with the previous example.

Equilibrium thermodynamics

A quasi-static process is a change of state that occurs so slowly that the system remains in equilibrium during the process. A process is reversible if it can be carried out in reverse order; the initial state gotten from the ﬁnal by reversing all steps. This requires time reversal insensitivity of the system in the succession of states along the process path. We will see that the entropy concept is related to time-direction sensitivity of the system, and that a reversible process is one that takes the system quasi-statically from one equilibrium state to another isentropically.
Is the propagation of a sound wave through a gas a quasi-static process? Absolutely not, since the pressure of the gas is not a global variable, it has spatial variations P = P(x,t). This brings us to an important juncture, where we ask the question “what kinds of processes and systems will thermodynamics deal with”? We will study only quasi-static processes, such that the entire system possesses global pressures and temperatures, in other words we will exclude (for now) any such processes for which T = T(x,t) and or P = P(x,t). This will allow us to map out our processes of study on a PV diagram (or BI diagram for magnetic systems).

For a gas, the state being completely speciﬁed by volume V and pressure P, a ﬁnite volume expansion accomplished in ﬁnite time is irreversible; during the expansion large local variations in the pressure and temperature will be created, and these will not disappear until a ﬁnal equilibrium state is reached. More importantly, they cannot be perfectly replicated if we try to undo the expansion. The only ﬁnite-volume change processes that can be regarded as reversible are those that take an inﬁnite time to accomplish, so that during the process we can regard the gas at any time as being in a static or quasi static state with a constant P and T throughout.
An equation of state for a gas is a constraint between the states of two gases that are at thermal equilibrium. If gas 1 with state (V ₁,P₁) is in thermal equilibrium with gas 2 in state (V ₂,P₂) then there is some function Θ_i such that

Θ1(V1, P1) = Θ2(V2, P2)

We call Θ_i the empirical temperature and declare two gases in thermal contact to be in a state of thermal equilibrium (not exchanging heat energy) if they have the same empirical temperatures. Empirical temperature Θ is a function of absolute temperature T as determined by kinetic theory. We will use absolute temperature as the empirical temperature.
Inversion of the equilibrium condition leads to the equation of state

P = f (V, Θ(T )) = P (V, T )

The collection of states (P,V ) for which T is a constant is called an isotherm. In an adiabatic or isentropic process a change of state occurs during which the system is thermally isolated, with no contact with another system. Adiabatic processes are those for which no heat is exchanged.

The work done by a gas during a change of state is the integral

∫ ΔW = Γ P dV

This is easy to understand from the deﬁnition of pressure; the force exerted per unit area by a piston-head conﬁning the gas to a piston of cross sectional area A and normal direction n;

F--⋅-n- P = A

The work done in moving the piston-head by amount dx in the direction n is

F ⋅ n dW = F ⋅ (dx n) = (-------)(A dx) = P dV A

where dV is the accompanying volume change of the gas.
Other types of work include electrical, magnetic and chemical. We study these in 201 but only mention them here.

Work is one of only two means of getting energy into or out of a system. In all forms of dynamics, energy is conserved if all forms are included in the balance. We now state the principle of conservation of energy in the context of thermodynamics, and divide energy into two categories; mechanical (work) and non-mechanical (heat).

First Law of thermodynamics. In thermodynamics we recognize heat as a form of energy, and so there are only two ways to change the energy of a system; we do work on it or give it heat;
1. A system A has a unique function U_A(P,V ) the internal energy, such that during an adiabatic process from (P₁,V ₁) to (P₂,V ₂)

ΔW = UA(P1, V1) - UA(P2, V2) = - ΔUA

is the work done by the system ( a convention).
2. The heat absorbed by A during a state change is

QA = ΔUA - ΔW

and for an inﬁnitismal quasi-static change

δQ = dU + P dV

and

∫ U (P2, V2) - U (P1, V1) = Γ (δQ - P dV )

This is all merely energy conservation, identifying heat as an energy form. We will actually use this as our deﬁnition of heat energy.
The Joule experiment established heat as a form of energy. In this experiment a sealed, insulated container of mercury was vigorously stirred, and its temperature monitored. It’s temperature rose in direct proportion to the amount of work done on it, indicating a transformation of work energy into heat energy.

The fundamental processes studied in equilibrium thermodynamics of gases are cyclic processes; a conﬁned gas is allowed to expand, contract, absorb heat and shed heat in a process that begins with the gas in some state (V ₁,P₁) and ends with the gas in the same state. The fundamental laws of thermodynamics were established by studying cyclic processes.
The most important is the Carnot cycle.
A Carnot cycle is a reversible cyclic process consisting of two isotherms and two adiabatic processes undergone by the auxiliary system.
From (P₁,V ₁) to (P₃,V ₃) the auxiliary system (presumably a gas) absorbs heat Q₁ at empirical temperature Θ₁(T) from a heat bath (primary system) at this temperature.
From (P₂,V ₂) to (P₄,V ₄) the system dumps heat Q₂ into a heat bath (secondary) at Θ₂(T). Let all subprocesses be quasi-static. Compute the heat in and out,

∫ V Q = 3P (V, T )dV + U (V , Θ (T )) - U (V , Θ (T )) 1 V1 1 3 1 1 1

∫ V2 Q2 = P (V, T2)dV + U (V2, Θ2(T )) - U (V4, Θ2(T )) V4

and on the two adiabatic lines the heat in is zero

∫ V 0 = 2 P dV + U (V , Θ (T )) - U (V , Θ (T )) V3 2 2 3 1

∫ V 0 = 4 P dV + U (V , Θ (T )) - U (V , Θ (T )) V1 4 2 1 1

Subtract the ﬁrst two equations and add in the last two

∫ ∫ ∫ ∫ ∮ V3 V2 V4 V1 Q1 - Q2 = V P dV + V P dV + V P dV + V P dV = c P dV 1 3 2 4

or in other words

Q - Q = ΔW 1 2

This implies that over the entire cycle

ΔU = 0

but this is just a consequence of the ﬁrst law, U depends on P and V and the gas returns to its original values after a cycle.
Let the efficiency of the cycle be the ratio of work extracted to input heat

W Q2 ε = ---- = 1 - ---- Q1 Q1

Kelvin’s theorem; it is impossible for the efficiency of a cyclic process to be equal to one. For the Carnot cycle, application of Kelvin’s law allows us to show that Q2-
Q1 depends only on the empirical (therefore absolute) temperatures of the isotherms, in fact we show that

Q2-- T2-- = Q1 T1

in Physics 201, and therefore for a Carnot cycle

∮ δQ--- Q1-- Q2-- C = + - = 0 T T1 T2

and therefore the quantity δQT--

must be a state variable, since it is unchanged over the course of a cyclic process. This is called the entropy

dQ dS = ----- T

and we have

Carnot’s Theorem. Let C_R be any closed reversible cycle, then

∮ δQ--- CR = 0 T

This is trivially proven by covering C_R with a mesh of isotherms and adiabatic processes and building up the cycle as a chain of inﬁnitismal adiabatics and isotherms since we have proven it for a Carnot process. If we compute this line integral around each little Carnot-cell that covers an arbitrary cycle, the integrations over the isothermal and adiabatic segments shared by two cells cancel, leaving only the integrations over the segments forming the arbitrary cycle.

Deﬁne a new function of state (a function that depends only on the state of the material, not how it got into that state), called the entropy

δQ S = S(V, T ), dS = ----- T

Then the entropy change during a process that begins at point 1 in the P -V plane and ends at point 2 is process independent and can be expressed as

∫ 2 δQ S(V2, T2) - S(V1, T1) = ----- 1 T

with

∮ δQ--- = 0 T

We now deﬁne several state functions; all are expressions of the energy of the system, and each one is an appropriate form of the energy useful for computation when a speciﬁc set of conditions are met.
1. Helmholtz Free Energy

F = U - T S

2. Gibbs Free Energy

G = U - T S + P V

The Gibbs free energy per particle is the chemical potential, which is the work done (or needed) to add particles to the system through chemical processes.
3. Enthalpy

H = G + T S

Chemist call the enthalpy the heat of reaction, and it is the heat liberated or absorbed in chemical processes.
F-Theorem; In a mechanically isolated system at constant temperature, F never increases and thermal equilibrium is a state of extremal F.
G-Theorem; For a system at constant T and P, G never increases and equilibrium is a state of extremal G.

Ideal gas thermodynamics

We will now consider the thermodynamics of the ideal gas of molecules whose distribution of molecular velocities is given by the Maxwell-Boltzmann distribution function. All of the physics is governed by a pair of equations of state, these are

3- P V = nRT , U = n CV T = nRT 2

in which C_V =

R =

N_Ak is the molar heat capacity for a monatomic ideal gas. The constant N_A = 6.023 × 10²³ is Avagadro’s constant, R = 8.314 J
mole-K-

is the ideal gas constant, and k is Boltzmann’s constant.
From these two equations, all of the thermodynamic functions that one could possibly need can be determined. Start with the ﬁrst law

dU = δQ - dW = δQ - P dV

and consider a reversible process

dU = T dS - P dV

Into this insert both equations of state

nRT---- dU = n CV dT , P = V

dV--- dT--- dV--- n CV dT = T dS - nRT , dS = n CV + nR V T V

This is the master entropy equation (the ﬁrst form) for most molar thermodynamics problems studied in Physics 201.

Example

Compute the increase in entropy of an ideal gas that is heated from T₁ to T₂ at constant volume.
Use the equation above with dV = 0;

∫ ∫ dT--- S2 T2 dT--- T2-- dS = nCV , S1 dS = S2 - S1 = T1 nCV = nCV ln T T T1

Example

Compute the increase in entropy of an ideal gas that is allowed to expand from V ₁ to V ₂ at constant temperature.
Use the equation above with dT = 0;

∫ ∫ dV--- S2 V2 dV--- V2-- dS = nR , S1 dS = S2 - S1 = V1 nR = nR ln V V V1

Example

Explore the connection between entropy and exhaustion of spontaneity; consider a sample of one mole of ideal gas atoms trapped in a perfectly insulated cylinder of volume V ₀ = 5.0 ℓ.

There are six partitions, between each is an additional 1 ℓ of empty space. If partition p₁ is pulled out, compute the entropy change of the gas. Now each p_i is pulled out in turn and in order. Compute ΔS in each case. Is the succession of entropy changes increasing or decreasing? Which free expansion triggered by removing a partition would you judge as being “most” spontaneous? Removal of the partitions can do no work on the gas. Solve this problem by using the previous two examples.

Example

Compute the increase in internal energy of an ideal gas that is allowed to expand from V ₁ to V ₂ at constant temperature.
There is no change in internal energy, since U depends only on T. If T does not change, neither does U.

Example

Compute the increase in internal energy of an ideal gas that is heated from T₁ to T₂ at constant volume.
Use

∫ U ∫ T ndU = nC dT , 2dU = U - U = 2 nC dT = C (T - T ) V U1 2 1 T1 V V 2 1

Example

Compute the increase in internal energy of an ideal gas that is allowed to expand reversibly and adiabatically from V ₁ to V ₂.
Start with

dU = T dS - P dV = - P dV (dS = 0)

and recognize that we simply need to compute the temperature change, since ΔU = nC_V ΔT. Insert the ideal gas law

nRT---- dT--- -nR----dV--- 2-dV--- dU = nCV dT = - dV, = - = - V T nCV V 3 V

and integrate

∫ ∫ T2 dT--- T2-- 2- V2 dV--- 2- V2-- T1 = ln = - V1 = - ln T T1 3 V 3 V1

rearrange;

T 3 V 2 2 2 ln --2- = ln --1-, T V 3 = T V 3 T 3 V 2 2 2 1 1 1 2

and we get

V 2 ΔU = nC (T - T ) = nC (T (--1) 3 - T ) V 2 1 V 1 V 1 2

If you use the ideal gas law to eliminate the temperature in the so-called adiabatic relation, you get a second adiabatic relation

2 2 5 5 3 3 3 3 T2V 2 = T1V 1 , P2V 2 = P1V 1

for a monatomic ideal gas.

Example

Compute the change in Helmholtz free energy if an ideal gas is allowed to isothermally expand from V ₁ to V ₂.
Start with

dU = δQ - P dV = T dS - P dV

and use

F = U - T S, dF = dU - T dS - S dT = - S dT - P dV

For isothermal expansion dT = 0, and insert the ideal gas law

∫ dV--- V2 dV--- V2-- dF = - P dV = nRT , F2 - F1 = - nRT V = - nRT ln V 1 V V1

Example

Compute the change in Gibbs free energy if an ideal gas is isothermally re-pressurized from P₁ to P₂.
Start with

dU = δQ - P dV = T dS - P dV

and use

G = U - T S+P V, dG = dU - T dS - S dT +P dV +V dP = - S dT +V dP

For isothermal expansion dT = 0, and insert the ideal gas law

∫ P2 nRT---- P2-- G2 - G1 = P dP = nRT ln 1 P P1

Example

Compute the heat that ﬂows into a gas during a constant pressure heating process in which the temperature rises from T₁ to T₂.
Start with

dU = T dS - P dV = δQ - P dV, δQ = dU + P dV

in any constant pressure ideal gas process

nRT nR dT nRT dV dT dV dP = 0 = d( ------) = --------- - -----------, -----= ----- V V V 2 T V

therefore inserting this and the ideal gas law;

V-- 3- δQ = dU + P dV = dU + P dT = dU + nR dT = nR dT + nR dT T 2

This leads us to

∫ 5- ΔQ = δQ = nR(T2 - T1) = CP (T2 - T1) 2

nRT---- ----J------ P = , R = 8.31 V mole K

This was experimentally established along with the internal energy equation of state

dU = n CV dT

The ideal gas is particularly simple since its internal energy depends only on its absolute temperature. For monatomic ideal gases we ﬁnd experimentally that C_V = 3
2

R =

N_Ak per mole.