Physics 16 – Howard Georgi

Good news and bad news

Other staff

TFs – Adilet Imambekov, Yevgeny Kats,Josh Lapan

Textbook author - David Morin

What is classical mechanics?

Degrees of freedom

The Art of Theoretical Physics

Motion, trajectories and F = ma

F = ma implies two initial conditions per degree offreedom

Two initial conditions per degree of freedom impliesF = ma

So what?

Finding trajectories numerically

Forces of the form F (t)

What is classical mechanics?

Classical mechanical system = Anything that doesn’tinvolve quantum mechanics

Coordinates = the numbers (possibly dimensional)that describe the system at a given fixed time.

Configuration = q(t) is a number or vector consistingof values of a set of independent coordinatessufficient to uniquely describe the system at a giventime - Mathematical snapshot

The problem of classical mechanics is to put thesesnapshots together into a Movie!

Degrees of freedom

# of degrees of freedom = # of independentcoordinates required to specify the configuration

system coordinate # of DOFs

point mass on a track ` (distance along the track) 1

point mass on a flat surface (x, y) 2

point mass in 3-d ~r = (x, y, z) 3

rigid body in 3-d ~r of center + 3 angles 6

2 masses + massless spring ~r1 and ~r2 6

2 masses + massive spring ~r1, ~r2 and spring ∞

# of degrees of freedom may depend on the questionwe ask and the accuracy we require!

Degrees of freedom

# of degrees of freedom = # of independentcoordinates required to specify the configuration

system coordinate # of DOFs

point mass on a track ` (distance along the track) 1

point mass on a flat surface (x, y) 2

point mass in 3-d ~r = (x, y, z) 3

rigid body in 3-d ~r of center + 3 angles 6

2 masses + massless spring ~r1 and ~r2 6

2 masses + massive spring ~r1, ~r2 and spring ∞

# of degrees of freedom may depend on the questionwe ask and the accuracy we require!

Degrees of freedom

# of degrees of freedom = # of independentcoordinates required to specify the configuration

system coordinate # of DOFs

point mass on a track ` (distance along the track) 1

point mass on a flat surface (x, y) 2

point mass in 3-d ~r = (x, y, z) 3

rigid body in 3-d ~r of center + 3 angles 6

2 masses + massless spring ~r1 and ~r2 6

2 masses + massive spring ~r1, ~r2 and spring ∞

# of degrees of freedom may depend on the questionwe ask and the accuracy we require!

hockey puck - sliding on the ice

q = (x, y)

hockey puck - sliding on the ice

q = (x, y)

comes off the ice on a slap shot

q = (x, y, z)

hockey puck - sliding on the ice

q = (x, y)

comes off the ice on a slap shot

q = (x, y, z)

rotatesq = (x, y, z, θ)

hockey puck - sliding on the ice

q = (x, y)

comes off the ice on a slap shot

q = (x, y, z)

rotatesq = (x, y, z, θ)

deformsq = (x, y, z, θ, · · ·)

The Art of Theoretical Physics

the ability to identify and focus on the parameters ofa system that are important in a given situation andto a given accuracy is much more important thanfancy mathematical gymnastics — and useful infields very different from physics

one of the most important things you can get out ofthis course

Motion, trajectories and F = ma

Trajectory = a possible motion of the system - aparticular movie that makes physical sense

Described by giving q(t) - the coordinates as afunction of time, t

We know in our bones that a snapshot is not enoughto tell us what a system is doing – to specify q(t), itis not enough to give q(t0) - an ∞ # of trajectoriesgo through any particular configuration.

Must also give the velocity at t0 - the derivatives ofthe coordinates - but then that is enough!

Motion, trajectories and F = ma

Trajectory = a possible motion of the system - aparticular movie that makes physical sense

Described by giving q(t) - the coordinates as afunction of time, t

We know in our bones that a snapshot is not enoughto tell us what a system is doing – to specify q(t), itis not enough to give q(t0) - an ∞ # of trajectoriesgo through any particular configuration.

Must also give the velocity at t0 - the derivatives ofthe coordinates - but then that is enough!

Motion, trajectories and F = ma

Trajectory = a possible motion of the system - aparticular movie that makes physical sense

Described by giving q(t) - the coordinates as afunction of time, t

We know in our bones that a snapshot is not enoughto tell us what a system is doing – to specify q(t), itis not enough to give q(t0) - an ∞ # of trajectoriesgo through any particular configuration.

Must also give the velocity - the derivatives of thecoordinates - at t0 - but then that is enough!

Notation for derivatives

q̇(t) ≡ dq(t)dt

remember that q(t) may have multiple components

for the hockey puck on the ice

q̇(t) = (ẋ(t), ẏ(t)) =

(dx(t)

dt,dy(t)

dt

)

for 3-dimensions

q̇(t) = ~̇r(t) ≡ d~r(t)dt

=

(dx

dt,dy

dt,dz

dt

)

Notation for derivatives

q̇(t) ≡ dq(t)dt

remember that q(t) may have multiple components

for the hockey puck on the ice

q̇(t) = (ẋ(t), ẏ(t)) =

(dx(t)

dt,dy(t)

dt

)

for 3-dimensions

q̇(t) = ~̇r(t) ≡ d~r(t)dt

=

(dx

dt,dy

dt,dz

dt

)

Notation for derivatives

q̇(t) ≡ dq(t)dt

remember that q(t) may have multiple components

for the hockey puck on the ice

q̇(t) = (ẋ(t), ẏ(t)) =

(dx(t)

dt,dy(t)

dt

)

for 3-dimensions

q̇(t) = ~̇r(t) ≡ d~r(t)dt

=

(dx

dt,dy

dt,dz

dt

)

Notation for derivatives

q̇(t) ≡ dq(t)dt

remember that q(t) may have multiple components

for the hockey puck on the ice

q̇(t) = (ẋ(t), ẏ(t)) =

(dx(t)

dt,dy(t)

dt

)

for 3-dimensions

q̇(t) = ~̇r(t) ≡ d~r(t)dt

=

(dx

dt,dy

dt,dz

dt

)

two initial conditions

per degree of freedom

The trajectory, or subsequent motion,

of a mechanical system is completely

determined if we know the

configuration, q and the velocity, q̇ at

some given time.

How do we actually find the trajecotries? Newton’ssecond law tells us that a single point particle withcoordinate q = x satisfies

F = ma = mẍ a ≡ d2x

dt2≡ ẍ

m - the mass - is a fixed property

F - the force - depends on what the system is doing -

F (x(t), ẋ(t), t)

if the trajectory is to be determined if we know xand ẋ, F can’t depend on ẍ or higher derivative

How do we actually find the trajecotries? Newton’ssecond law tells us that a single point particle withcoordinate q = x satisfies

F = ma = mẍ a ≡ d2x

dt2≡ ẍ

m - the mass - is a fixed property

F - the force - depends on what the system is doing -

F (x(t), ẋ(t), t)

if the trajectory is to be determined if we know xand ẋ, F can’t depend on ẍ or higher derivatives

Newton’s second law is a formula for acceleration

a(t) = ẍ(t) =1

mF (x(t), ẋ(t), t)

thus a second order differential equation

This is not a course on differential equations!

But we will learn to solve a few simple ones. Also, Fdepends on several variables, and we will useconcepts from multivariable calculus. But nothingdeep or complicated.

Deep question - why a formula for accleration?

Newton’s second law is a formula for acceleration

a(t) = ẍ(t) =1

mF (x(t), ẋ(t), t)

thus a second order differential equation

This is not a course on differential equations!

But we will learn to solve a few simple ones. Also, Fdepends on several variables, and we will useconcepts from multivariable calculus. But nothingdeep or complicated.

Deep question - why a formula for accleration?

F = ma implies two initial conditions perdegree of freedom

This follows from mathematics - a second orderdifferential equation for x has a family of solutions.The solutions can be uniquely picked out by giving xand ẋ at a given time.

Example:F (x, ẋ, t) = F0 = m a

x(t) =F02m

t2 + ẋ(0) t + x(0)

Easy to check that F = ma is satisfied

F = ma implies two initial conditions perdegree of freedom

This follows from mathematics - a second orderdifferential equation for x has a family of solutions.The solutions can be uniquely picked out by giving xand ẋ at a given time.

Example:F (x, ẋ, t) = F0 = m a

x(t) =F02m

t2 + ẋ(0) t + x(0)

Easy to check that F = ma is satisfied

F = ma implies two initial conditions perdegree of freedom

This follows from mathematics - a second orderdifferential equation for x has a family of solutions.The solutions can be uniquely picked out by giving xand ẋ at a given time.

Example:F (x, ẋ, t) = F0 = m a

x(t) =F02m

t2 + ẋ(0) t + x(0)

Easy to check that F = ma is satisfied

x(t) =F02m

t2 + ẋ(0) t + x(0)

x(t) =F02m

t2 + ẋ(0) t + x(0)

ẋ(t) =F0m

t + ẋ(0)

x(t) =F02m

t2 + ẋ(0) t + x(0)

ẋ(t) =F0m

t + ẋ(0)

ẍ(t) =F0m

x(t) =F02m

t2 + ẋ(0) t + x(0)

ẋ(t) =F0m

t + ẋ(0)

ẍ(t) =F0m

Nothing special about t = 0

x(t) =F02m

(t− t0)2 + ẋ(t0) (t− t0) + x(t0)

just a reorganization of the constants

Two initial conditions per degree of freedomimplies F = ma

That is, suppose that you have some physical systemand you don’t know that it satisfies Newton’s secondlaw, F = ma, but you do know, for any given x(t0)and ẋ(t0), how to find the trajectory x(t) thatsatisfies these conditions. How?

The trajectory x(t) depends on four things - t, t0,and the initial values x0 = x(t0) and v0 = ẋ(t0).Some function of four variables describes x(t) - giveit a name

x(t) = G(x0, v0, t0, t

)= G

(x(t0), ẋ(t0), t0, t

)

You can get F = ma, assuming only that x(t) is asmooth function (otherwise you have no hope).

x(t) = G(x0, v0, t0, t

)= G

(x(t0), ẋ(t0), t0, t

)

Differentiating twice in t and then setting t0 = t givesa second order differential equation

ẍ(t) =∂2

∂t2G

(x(t0), ẋ(t0), t0, t

)

=∂2

∂t2G

(x(t0), ẋ(t0), t0, t

)∣∣∣t0=t

=1

mF

(x(t), ẋ(t), t

)

x(t) = G(x(t0), ẋ(t0), t0, t

)= G

(x(t), ẋ(t), t, t

)

ẋ(t) =∂

∂tG

(x(t0), ẋ(t0), t0, t

)

=∂

∂tG

(x(t0), ẋ(t0), t0, t

)∣∣∣t0=t

ẍ(t) =∂2

∂t2G

(x(t0), ẋ(t0), t0, t

)

=∂2

∂t2G

(x(t0), ẋ(t0), t0, t

)∣∣∣t0=t

=1

mF

(x(t), ẋ(t), t

)

So what?

Have we learned WHY F = ma? No - because wedon’t know why our mantra - two initials conditionsper degree of freedom - is true. In fact, I don’t reallythink it is, at least not exactly, and we will comeback to this at the end of the course and talk moreabout why F = ma.

So what?

Have we learned WHY F = ma? No - because wedon’t know why our mantra - two initials conditionsper degree of freedom - is true. In fact, I don’t reallythink it is, at least not exactly, and we will comeback to this at the end of the course and talk moreabout why F = ma.

So what?

Have we learned WHY F = ma? No - because wedon’t know why our mantra - two initials conditionsper degree of freedom - is true. In fact, I don’t reallythink it is, at least not exactly, and we will comeback to this at the end of the course and talk moreabout why F = ma.

Should you take Physics 16?

Hard but humame

Help is available and the staff will work as hard asyou do

Fun, cooperative work

You can probably all handle it

But you need to think about the whole schedule

Should you take Physics 16?

Hard but humame

Help is available and the staff will work as hard asyou do

Fun, cooperative work

You can probably all handle it

But you need to think about the whole schedule

Should you take Physics 16?

Hard but humame

Help is available and the staff will work as hard asyou do

Fun, cooperative work

You can probably all handle it

But you need to think about the whole schedule

Should you take Physics 16?

Hard but humame

Help is available and the staff will work as hard asyou do

Fun, cooperative work

You can probably all handle it

But you need to think about the whole schedule

Should you take Physics 16?

Hard but humame

Help is available and the staff will work as hard asyou do

Fun, cooperative work

You can probably all handle it

But you need to think about the whole schedule

Should you take Physics 16?

Hard but humame

Help is available and the staff will work as hard asyou do

Fun, cooperative work

You can probably all handle it

But you need to think about the whole schedule

Finding trajectories numerically

Taylor expansion (my personal favorite)

x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·2nd term is just the definition of derivative

ẋ(t0) =x(t0 + ∆t)− x(t0)

∆t+ · · ·

x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·We would like to iterate to find x(t0 + 2∆t)

x(t0 + 2∆t) = x(t0 + ∆t) + ∆t ẋ(t0 + ∆t) + · · ·- but do we know ẋ(t0 + ∆t)?

Finding trajectories numerically

Taylor expansion (my personal favorite)

x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·2nd term is just the definition of derivative

ẋ(t0) =x(t0 + ∆t)− x(t0)

∆t+ · · ·

x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·We would like to iterate to find x(t0 + 2∆t)

x(t0 + 2∆t) = x(t0 + ∆t) + ∆t ẋ(t0 + ∆t) + · · ·- but do we know ẋ(t0 + ∆t)?

Finding trajectories numerically

Taylor expansion (my personal favorite)

x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·2nd term is just the definition of derivative

ẋ(t0) =x(t0 + ∆t)− x(t0)

∆t+ · · ·

x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·We would like to iterate to find x(t0 + 2∆t)

x(t0 + 2∆t) = x(t0 + ∆t) + ∆t ẋ(t0 + ∆t) + · · ·- but do we know ẋ(t0 + ∆t)?

Finding trajectories numerically

Taylor expansion (my personal favorite)

x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·

We would like to iterate to find x(t0 + 2∆t)

x(t0 + 2∆t) = x(t0 + ∆t) + ∆t ẋ(t0 + ∆t) + · · ·

- but do we know ẋ(t0 + ∆t)

Finding trajectories numerically

Taylor expansion (my personal favorite)

x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·

We would like to iterate to find x(t0 + 2∆t)

x(t0 + 2∆t) = x(t0 + ∆t) + ∆t ẋ(t0 + ∆t) + · · ·

- but do we know ẋ(t0 + ∆t)?

Finding trajectories numerically

Taylor expansion (my personal favorite)

x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·

We would like to iterate to find x(t0 + 2∆t)

x(t0 + 2∆t) = x(t0 + ∆t) + ∆t ẋ(t0 + ∆t) + · · ·

- but do we know ẋ(t0 + ∆t) ??????

The point is we can also keep track of ẋ

x(t1) = x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·

ẋ(t1) = ẋ(t0 + ∆t) = ẋ(t0) + ∆t ẍ(t0) + · · ·

ẍ(t0) = a(t0) =1

mF (x(t0), ẋ(t0), t0)

The point is we can also keep track of ẋ

x(t1) = x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·

ẋ(t1) = ẋ(t0 + ∆t) = ẋ(t0) + ∆t ẍ(t0) + · · ·

= ẋ(t0) +∆t

mF (x(t0), ẋ(t0), t0)

→ x, ẋ at t1 = t0 + ∆t in terms of x and ẋ at t0 Nowwe can iterate. Physicists proof for 2 initialconditions - we have to keep track of both x and ẋ tomake use of F = ma

The point is we can also keep track of ẋ

x(t1) = x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·

ẋ(t1) = ẋ(t0 + ∆t) = ẋ(t0) + ∆t ẍ(t0) + · · ·

= ẋ(t0) +∆t

mF (x(t0), ẋ(t0), t0)

→ x, ẋ at t1 = t0 + ∆t in terms of x and ẋ at t0 Nowwe can iterate. Physicists proof for 2 initialconditions - we have to keep track of both x and ẋ tomake use of F = ma

The point is we can also keep track of ẋ

x(t1) = x(t0 + ∆t) = x(t0) + ∆t ẋ(t0) + · · ·

ẋ(t1) = ẋ(t0 + ∆t) = ẋ(t0) + ∆t ẍ(t0) + · · ·

= ẋ(t0) +∆t

mF (x(t0), ẋ(t0), t0)

→ x, ẋ at t1 = t0 + ∆t in terms of x and ẋ at t0 Nowwe can iterate. Physicists proof for 2 initialconditions - we have to keep track of both x and ẋ tomake use of F = ma

The point is we can also keep track of ẋ

x(t2) = x(t1 + ∆t) = x(t1) + ∆t ẋ(t1) + · · ·

ẋ(t2) = ẋ(t1 + ∆t) = ẋ(t1) + ∆t ẍ(t1) + · · ·

= ẋ(t1) +∆t

mF (x(t1), ẋ(t1), t1)

→ x, ẋ at t2 = t1 + ∆t in terms of x and ẋ at t1 Nowwe can iterate. Physicists proof for 2 initialconditions - we have to keep track of both x and ẋ tomake use of F = ma

The point is we can also keep track of ẋ

x(t3) = x(t2 + ∆t) = x(t2) + ∆t ẋ(t2) + · · ·

ẋ(t3) = ẋ(t2 + ∆t) = ẋ(t2) + ∆t ẍ(t2) + · · ·

= ẋ(t2) +∆t

mF (x(t2), ẋ(t2), t2)

→ x, ẋ at t3 = t2 + ∆t in terms of x and ẋ at t2 Nowwe can iterate. Physicists proof for 2 initialconditions - we have to keep track of both x and ẋ tomake use of F = ma

Forces of the form F (t)

a(t) =d

dtv(t) =

1

mF (t)

v(t) =∫ t

t0dt′

1

mF (t′) + constant

The dt′ means that t′ is a completely independent“dummy” variable

t′ is a symbol for a new independent variable — not“the derivative of t”

f ′(x) means the derivative with respect to x of thefunction f(x)

f(x′) means the function f of the variable x′

f ′(x′) means the derivative with respect to thevariable x′ of the function f(x′)

Forces of the form F (t)

a(t) =d

dtv(t) =

1

mF (t)

v(t) =∫ t

t0dt′

1

mF (t′) + constant

The dt′ means that t′ is a completely independent“dummy” variable

You should never ever write an equation like

A(t) =∫ t

t0dtB(t)

No consistent meaning can be assigned to thiscollections of symbols. See that it doesn’t appear onyour problem sets or tests!

∫ tt0

dt′ B(t) 6=∫ t

t0dt′ B(t′)

Both meaningful, just different

∫ tt0

dt′ B(t) = (t− t0) B(t)

a(t) =d

dtv(t) =

1

mF (t)

v(t) =∫ t

t0dt′

1

mF (t′) + constant

set t = t0 and the integral term vanishes

a(t) =d

dtv(t) =

1

mF (t)

v(t) =∫ t

t0dt′

1

mF (t′) + constant

v(t0) = constant

This is why the constants that appear in initialconditions are sometimes called “integrationconstants”

v(t) =∫ t

t0dt′

1

mF (t′) + v(t0)

Equivalent - going backwards:

∫ tt0

dt′1

mF (t′) =

∫ tt0

dt′ a(t′)

=∫ t

t0dt′

d

dt′v(t′) = v(t)− v(t0)

find x(t) by repeating the procedure

x(t) =∫ t

t0dt′ v(t′) + x(t0)

insert what we just found for v(t)

find x(t) by repeating the procedure

x(t) =∫ t

t0dt′ v(t′) + x(t0)

x(t) =∫ t

t0dt′

[∫ t′

t0dt′′

1

mF (t′′) + v(t0)

]+ x(t0)

Note that now we need TWO dummy variables

find x(t) by repeating the procedure

x(t) =∫ t

t0dt′ v(t′) + x(t0)

x(t) =∫ t

t0dt′

[∫ t′

t0dt′′

1

mF (t′′) + v(t0)

]+ x(t0)

x(t) =∫ t

t0dt′

∫ t′

t0dt′′

1

mF (t′′) + v(t0)(t− t0) + x(t0)

find x(t) by repeating the procedure

x(t) =∫ t

t0dt′ v(t′) + x(t0)

x(t) =∫ t

t0dt′

[∫ t′

t0dt′′

1

mF (t′′) + v(t0)

]+ x(t0)

x(t) =∫ t

t0dt′

∫ t′

t0dt′′

1

mF (t′′) + v(t0)(t− t0) + x(t0)

Where it starts

find x(t) by repeating the procedure

x(t) =∫ t

t0dt′ v(t′) + x(t0)

x(t) =∫ t

t0dt′

[∫ t′

t0dt′′

1

mF (t′′) + v(t0)

]+ x(t0)

x(t) =∫ t

t0dt′

∫ t′

t0dt′′

1

mF (t′′) + v(t0)(t− t0) + x(t0)

motion with initial velocity

find x(t) by repeating the procedure

x(t) =∫ t

t0dt′ v(t′) + x(t0)

x(t) =∫ t

t0dt′

[∫ t′

t0dt′′

1

mF (t′′) + v(t0)

]+ x(t0)

x(t) =∫ t

t0dt′

∫ t′

t0dt′′

1

mF (t′′) + v(t0)(t− t0) + x(t0)

The effect of the force - this is the only term thatcontributes to the equation of motion

For F (t) = F0

x(t) =∫ t

t0dt′

∫ t′

t0dt′′

1

mF (t′′) + v(t0)(t− t0) + x(t0)

=∫ t

t0dt′

∫ t′

t0dt′′

F0m

+ v(t0)(t− t0) + x(t0)

=F0m

∫ tt0

dt′∫ t′

t0dt′′ + v(t0)(t− t0) + x(t0)

=F0m

∫ tt0

dt′ (t′ − t0) + v(t0)(t− t0) + x(t0)

=F02m

(t− t0)2 + v(t0)(t− t0) + x(t0)

For F (t) = F0

x(t) =∫ t

t0dt′

∫ t′

t0dt′′

1

mF (t′′) + v(t0)(t− t0) + x(t0)

=∫ t

t0dt′

∫ t′

t0dt′′

F0m

+ v(t0)(t− t0) + x(t0)

=F0m

∫ tt0

dt′∫ t′

t0dt′′ + v(t0)(t− t0) + x(t0)

=F0m

∫ tt0

dt′ (t′ − t0) + v(t0)(t− t0) + x(t0)

=F02m

(t− t0)2 + v(t0)(t− t0) + x(t0)

For F (t) = F0

x(t) =∫ t

t0dt′

∫ t′

t0dt′′

1

mF (t′′) + v(t0)(t− t0) + x(t0)

=∫ t

t0dt′

∫ t′

t0dt′′

F0m

+ v(t0)(t− t0) + x(t0)

=F0m

∫ tt0

dt′∫ t′

t0dt′′ + v(t0)(t− t0) + x(t0)

=F0m

∫ tt0

dt′ (t′ − t0) + v(t0)(t− t0) + x(t0)

=F02m

(t− t0)2 + v(t0)(t− t0) + x(t0)

For F (t) = F0

x(t) =∫ t

t0dt′

∫ t′

t0dt′′

1

mF (t′′) + v(t0)(t− t0) + x(t0)

=∫ t

t0dt′

∫ t′

t0dt′′

F0m

+ v(t0)(t− t0) + x(t0)

=F0m

∫ tt0

dt′∫ t′

t0dt′′ + v(t0)(t− t0) + x(t0)

=F0m

∫ tt0

dt′ (t′ − t0) + v(t0)(t− t0) + x(t0)

=F02m

(t− t0)2 + v(t0)(t− t0) + x(t0)

More next time - remember the mantra -

two initial conditions

per degree of freedom