[Music]
this program is brought to you by
Stanford University please visit us at
stanford.edu
I want to work out a few examples two
examples three examples not in
particular order of ascending complexity
the first one is going to be very simple
and next one is going to be more
complicated and the third one is gonna
be even simpler than the first one I
have my reasons but the first two
examples are to show you how much easier
it is to use the Lagrange's equations
and to use the formal structure of
classical mechanics that we've learned
up till now the mindless set of
mechanical rules then to actually sit
there and try to use F equals MA for all
the parts of a system and juggle all the
different parts of the system and use
the fact that that the forces are 1 or
equal and opposite of the forces on the
other that can often be just incredibly
complicated whereas working out the
lagrangian is very simple usually now
there's a simple mathematical reason
when you work out f equals ma a is the
second derivative with respect to time
or something that means you have to
start calculating second derivatives you
have to start calculating accelerations
and that's often complicated much more
complicated than calculating velocities
which are just first derivatives if you
can calculate the velocity of a system
you can calculate its kinetic energy
very easily one-half MV squared
if you've calculated its kinetic energy
and also its potential energy you know
the Lagrangian and from that like a
mechanical fool you can sit down and
calculate the equations of motion when
you know the equations of motion you can
put them on a computer and grind out
trajectories calculating accelerations
is harder typically then what I have in
mind of course when I say calculating
accel accelerations as I have in mind
where you have some complicated set of
coordinates where your coordinates are
not simply rectangular coordinates some
angles some other kinds of things and
out of this you have to calculate
components of acceleration that's
often much much more complicated than
just calculating the kinetic energy let
me take the first example I'm going to
study is the simple pendulum then we're
gonna do a more complicated kind of
pendulum the simple pendulum is simple
enough that with a little bit of thought
and a little bit of cleverness you can
figure out how the forces on the
pendulum Bob and then use F equals MA
okay
the compound pendulum we're gonna do is
far more complicated and would boggle
the mind of anybody trying to use F
equals MA but both of them are
relatively simple not too hard to work
out if you know the Lagrangian you know
you lagrangian's alright so let's start
with the simple pendulum simple pendulum
there's just a weight hanging whoops I
don't think I like the way I drew it
let's draw it slightly different
let's draw it hanging in an angle all
right so it's connected to the ceiling
it hangs at an angle I assume that it's
connected the pendulum Bob has mass m
it's connected by a rigid rod to a pivot
at the ceiling and we'll take the rigid
rod to be so light that its mass is not
important all right so all of the mass
is concentrated at the end of the
pendulum and the angle relative to the
vertical is called theta there are two
features of the theory there are two
things in the Lagrangian potential
energy and kinetic energy potential
energy is proportional to the height
above some particular point and of
course the height will depend on the
angle as the pendulum swings the height
will change and as the height changes
the potential energy changes and then
this kinetic energy and kinetic energy
is just related to the velocity of the
pendulum Bob all right let's uh let's
first we need to specify what the length
of the pendulum is this is the the
pendulum length doesn't change it's just
a constant in this problem
let's call it our our is just a number
one meter or meter and a half or six
inches or whatever it's not going to
change the mass doesn't change the only
thing that changes is in this problem is
Theta there's the only thing which
changes with time is Theta and we want
to work out its equations of motion all
right what's the horizontal distance
here that's related to our and it's
related to theta thank you our sine
theta all right since you're so smart
what's the vertical one our cosine theta
that's this distance here let's
calculate the components the x and y
components of the velocity of the mass
they're just the time derivatives of the
vertical and horizontal components okay
so what is the O in Siddhant Li if we
can oh this is fine yeah let's calculate
the X component of velocity the X
component of velocity is just a time
derivative of our sine theta r doesn't
change only sine theta changes so the X
component of velocity
all right V is equal now there's going
to be two components to it the first
component is the X component it's R
times the derivative of sine theta
that's cosine theta times the time
derivative of theta if I want to know
the time derivative of sine theta I
differentiate with respect to theta and
then I multiply by the time derivative
of time derivative of theta so that's R
cosine theta times theta dot that's the
X component of the velocity the Y
component of the velocity
I'm counting components downward I guess
I count distance downward as positive in
this in this picture here so the
vertical component of the velocity is
the time derivative of art cosine theta
and that's R - R
sine of theta times theta dot why -
because the derivative of cosine is
minus sine so these are the two
components of the velocity of the
pendulum Bob and now if we want the
kinetic energy we just have to write
one-half MV squared so the kinetic
energy T is just 1/2 the mass times the
sums of the squares of the components of
the velocity the x squared plus B y
squared ok these are components of the
velocity in two perpendicular directions
the total velocity or the total speed is
the sums of the squares and what is that
that's going to be R squared theta dot
squared times cosine theta squared plus
sine of theta squared each component
will give an R squared and the theta dot
squared and the square of the velocity
and then one of them has cosine squared
the other has sine squared but what is
cosine squared plus sine squared it's
just one so the kinetic energy is just
1/2 the mass times R squared times theta
dot squared we can forget cosine squared
plus sine squared it's just 1 that's the
kinetic energy of the pendulum Bob what
about its potential energy the potential
energy is simply the height above some
particular point we could take that
point to be the ceiling pendulum can
swing around let's take the potential
energy to be 0
well let's see yeah let's take the
potential energy to be 0 when theta is
when the pendulum is absolutely
horizontal the point is of course it
doesn't matter at what point you choose
the potential energy to be 0 only
differences of potential energy are
important and so
the the height of the pendulum is - -
because it's hanging down - R cosine of
theta when theta is zero then the height
is minus R when theta is PI it's up on
the top then the height is plus R cosine
theta or just + sorry just plus R so the
minus sign is there to keep track of the
fact that when theta is zero the energy
is minimum it's down at the bottom
that's not you that's just the height of
the pendulum relative to the pivot
what's the potential energy of a
particle in a gravitational field mg
times the height so we have to multiply
this by minus M times the acceleration
of gravity this is the potential energy
let's write the Lagrangian the
Lagrangian is just M over 2r squared
theta dot squared minus the potential
energy minus the potential energy the
potential energy itself has a minus in
front of it so that means we should add
em G R cosine of theta again y plus
because we want - the potential energy
in the Lagrangian and the potential
energy has a minus sign two minus signs
make a plus sign that's a lot easier
than figuring out the acceleration and
using F equals MA for the two components
of motion and then juggling them
together Lagrangian all we really needed
to do was to calculate the velocity as a
function of theta dot and here it is all
right what about the equations of motion
let's write down the equations of motion
first of all we take the canonical
momentum conjugate to theta that's PI
sub theta that's equal to the derivative
of the Lagrangian with respect to theta
dot that's M R squared theta dot it's
actually the angular momentum but we
don't even need to call it that we just
call it M R squared theta dot and then
the equation of motion is that the time
derivative of this D by DT of M R
squared theta dot that's the usual D by
DT of the derivative of the Lagrangian
with respect to the velocity that's
equal to the derivative of the
Lagrangian with respect to theta what is
that that's minus M G R sine of theta or
we can just write this we can cancel out
the masses of course and then we can
write this as R squared remember R is
just a number that doesn't vary in the
problem times theta double dot the
angular acceleration is equal to minus G
R times sine theta this can be stuck on
a computer we just we either take it as
a differential equation to solve or we
break up the time axis into lots of
little intervals and do it numerically
I'm going to assume in this course that
every equation that we could write down
has a solution and that you know how to
find the solution if but by nothing else
by setting it up on a computer to
simulate the simulate simple
differential equations like this and
it's fun to do it if you haven't done it
do it for some simple equations of this
type and this simply tells you the
acceleration as a function of the angle
so it's a usual kind of F equals MA type
equation
um let's while we're at it let's just
calculate the energy we know what the
energy should be the energy should just
be the kinetic energy plus the potential
energy the Lagrangian is kinetic energy
minus potential energy but let's do it
by by the mindless method the mindless
method says that the Hamiltonian is
equal to the momentum what did I call it
the pi sub theta times theta dot this is
pi times Q dot we work that out last
time minus L what I'm writing is PI sub
I Q sub I dot minus L that's where you
sum over I but there's only one
coordinate in this problem so that's it
PI sub theta what's PI sub theta pi sub
theta is M R squared theta dot so this
becomes M R squared theta dot times
another theta dot theta dot squared
that's PI theta dot minus the Lagrangian
so it's minus M R squared over 2 theta
dot squared minus mg R cosine theta this
is these two things constitute minus L
okay so this is easy mr squared theta
dot squared M over 2 R squared theta dot
squared this cancels half of this and
just turns this into plus that's the
energy Hamiltonian or equivalently the
energy that's it okay it is the kinetic
energy plus the potential energy don't
get fooled by this minus sign here
potential energy had a minus sign
oh it's almost always equal to 2t you
know doesn't have to be but it's often
2t certainly if the potato connecticut
if the kinetic energy is of the form
one-half MV squared then it's always
going to be that way i can concoct some
in which it's not the case and i'll
remind me but for the simple examples
it's usually the case but don't trust it
don't trust a thing like that you know
you work out each example separately
alright that's the energy how does the
pendulum move well i'm not going to
solve the equations of motion but of
course we know how it moves we have a
pretty good idea if the energy is all
right let's first of all notice what's
the difference of energy at the bottom
of the orbit and the top the difference
of the energy at the top we would have
player we would have mg times cosine of
pi or minus that so at the top the
energy is plus MGR
and at the bottom it's minus MGR all
right so it takes some energy to get to
the top it from the bottom to the top
you have to pay a price twice MGR if
let's start let's imagine starting the
pendulum at the bottom here with a
certain velocity we give it a poke to
give it some velocity if the velocity is
not enough if the kinetic energy at the
bottom is not enough to overcome the
necessary energy to get to the top then
of course what will happen is the
pendulum will swing up run out of
kinetic energy it has converted some
kinetic energy to potential energy but
it didn't have enough to get to the top
and so it'll start back it'll swing to
the other side and it will just
endlessly oscillate back and forth
symmetrically on both sides that's
intuitively clear and it actually is a
feature of the equations
just endless oscillation almost like a
harmonic oscillator which is a system
we're gonna study shortly but a little
more complicated yeah what's that no
theta is PI at the top pi is 180 degrees
oh no oh I see what you mean no no I'm
sorry okay the top doesn't mean when
it's ER I'm imagine
I'm imagining a pivot point and the
pendulum swings around 360 degrees
around at that pivot point so the top
means up here good point okay yeah well
right yes now that that's right what I
meant by the top of the orbit was when
the pendulum is vertically up right and
above the ceiling okay what if it does
have enough energy to climb to the top
if it has enough kinetic energy at the
bottom so that by the time it gets to
the top it still has a little bit of
kinetic energy left over in other words
if its kinetic energy to begin with is a
little bit bigger than two mg R then it
will have kinetic energy to spare when
it gets to the top and it will just keep
going and it'll cycle around going
faster at the bottom slowing down going
over the top and just endlessly round
and round and round and round so there
are two kinds of orbits
two kinds of motions really strictly
speaking three there's oscillation back
and forth and there's what should we
call it
yeah okay
what kind of rotating motion to the left
and a rotating motion to the right of
course you can ask what happens if it
has exactly enough energy to get to the
top well then it'll just go to the top
and sit at the top but that's that's
highly fine-tuned and not the sort of
thing you're likely to encounter so
we'll ignore that possibility there's
nothing wrong with it just go up to the
top and just get there barely and you
stop okay those are that's the basic the
basic pendulum and as I said my main
reason for well two reasons for
illustrating it the first is just to
show you that it's easier to calculate
using lagrangian's
than it is would beat the F equals MA
for a system that's complicated as this
but really I want to show you another
system which is the double pendulum the
double pendulum is a far more
complicated system it has very very
complicated motions its motions are
chaotic we haven't learned what chaos
means yet we will but its motions are
chaotic and truly baffling if you watch
the double pendulum and if you were to
try to study it by using just F equals
MA without knowing the Lagrangian
formulation of it you would spend hours
working out its equation of motion it's
too complicated for certainly be too
complicated for me to do on the
blackboard but if you know how to
calculate lagrangian's it's just
mechanical here's what the double
pendulum is again a pivot point
pendulum hangs from it same deal as
before it has length R and mass M I
won't bother writing its mass but then
connected to this pendulum there's
another one
I'll also for simplicity give it radius
R and to keep it simple I'll give it the
same mass as the first one okay it's
also in a gravitational field this one
swings while the first one swings and
the whole thing is a really complicated
mechanical system and it's even
complicated just to work out its
equations of motion except yes yeah in
the same plane no we could do it more
generally no good big problem about
doing it more generally but I'll take
the case where they're in the same plane
you could have asked that up there too
there's a is it a planar pendulum or is
it a pendulum that can swing out in in
all directions I chose to do the planar
pendulum it might be fun to work out the
equations for the full three-dimensional
pendulum which can swing out of the
blackboard but I won't do that here in
fact it's easy enough we could do it
here but they won't all right let's give
it some coordinates the first thing that
you do when you see a problem like this
is to define some coordinates a little
bit of experience will teach you what
the best choice of coordinates is and if
you don't choose the best choice that's
okay too I mean
just pick a set of coordinates and a set
of coordinates means those things which
determine the position of the system a
minimal number of objects which
determine the position of the object all
right the first coordinate is of course
theta the angle that the first pendulum
makes relative to the vertical now you
kind of have a choice you could take for
the second pendulum you could take the
angle that the second pendulum makes
with respect to the location of the
first pendulum we could qualify
or we could take the angle of the second
pendulum with respect to the vertical
six of one half a dozen of the other
I chosen my notes to use the angle
relative to the vertical so this I call
Phi and that's it if you know theta and
you know Phi you know the location of
the of the pendulum completely now we
need to calculate the kinetic and
potential energy
so let's first calculate the kinetic
energy of the first bob over here it's
exactly the same no difference the
calculation is identical once again this
is our sine theta and so forth I can
just write down the answer for the first
piece of it it's just one-half M theta R
sorry one-half M R squared theta dot
squared that's the kinetic energy of the
first Bob well what about the second one
well the second one is moving for two
reasons
one of them is that the first Bob is
moving and if the first Bob is moving
even if the angle Phi were not changing
the second Bob would move so we have to
add two velocities together to get the
velocity of the components of velocity
of the second Bob to get the composer
yeah the components of velocity of the
second Bob are the sum of the velocity
as if this was standing still plus the
component of motion of the first Bob
I'll write down the answer and you'll
see why it is I think you won't have
much trouble of seeing why it is the
velocity of the second Bob let's give us
some components here it is velocity of
the second Bob it has two components
sorry be three plus three well I don't
know why you're calling interacting but
you can call it whatever you want
oh the fact that it's attached to this
one yeah we can also cool we can also
call these things whatever we want I
call them Bob you could call them Herman
it doesn't matter but we could call him
Bob and this one Herman I like that
this one's Bob this one's Herman okay so
here's Bob's kinetic energy and Herman's
kinetic energy has or Herman's velocity
has two components let's are of course
as two components but each component
itself consists of two parts alright so
the X component of motion the X
component of Herman's velocity will
consist of the X component the Bob's
velocity that is where is it our cosine
theta times theta dot and then the
component of Hermon relative to Bob
pretend Bob is standing still what would
be the X component of Herman's motion
well it'll be exactly the same as Bob's
except instead of theta put Phi if Bob
was standing still and Herman were
oscillating or swinging then we would
add to this plus R cosine of Phi times
Phi dot that's the X component of motion
two pieces to it one associated with the
fact that Bob moves that's this one and
the other associated with the fact that
Herman moves relative to Bob now what
about the Y component exactly the same
thing Bob has some Y motion and Herman
has some Y motion relative to Bob
exactly the same so they this is the X
component comma and the y component
would be R minus
r sine theta theta dot minus R sine of
Phi times Phi dot are there any
questions about what I've written down
all I've done was treat Herman as if he
were a pendulum of exactly the same kind
substituting the angle Phi wherever I
saw theta in the first calculation good
now we can calculate the kinetic energy
we know the components here's the
kinetic energy of Bob here's the
components of velocity of Herman let's
calculate Herman's kinetic energy let's
do it over here
Herman's kinetic energy is one-half M
same velocity
sorry same mass now we have to square
this and add it to the square of this
this is the X component here's the y
component what do we get let's let's
look at let's see if we can examine it
and see what we get we're going to get R
squared cosine theta squared theta dot
squared when we square the X component
we'll square this and we'll square and
we'll add to it the square of the Y
component so we're gonna get this
squared plus this squared what does that
give us that gives us R squared cosine
squared plus sine squared that's one
times theta dot squared okay so that's
just gonna be one-half M R squared theta
dot squared all I've done is said that
when I square VX and I square V Y and
add them together it gives me a term
with a cosine squared and a term with a
sine squared all right
those add together to just give theta
dot squared then we have the same thing
over here that's going to do exactly the
same thing plus one-half M R squared Phi
dot squared
that's coming from the square of this
piece plus the square of this piece
cosine Phi squared plus sine Phi squared
also adds to one what have I left out
I've left out the cross terms this times
this and this times this okay okay so
let's add them in we have plus of course
it's 1/2 M again 1/2 M R squared times 2
there's always a 2 when you a plus B
squared has to a be so there's a 2
that'll eat up this 2 in the denominator
it'll give us a theta dot times a 5 dot
that's something new later dot times Phi
dot this is the first time we've seen
velocities coming in multiplying each
other so stated that times Phi dot
there's M R squared and then there's
cosine theta cosine of Phi
but there's also a term from the cross
term from V Y from the y-component
that's going to be exactly the same
except that it will have sine theta
times sine Phi instead of cosine so
there's plus sine theta sine Phi this
looks to me like the kinetic energy of
Herman which I now add to the kinetic
energy of Bob to get the total kinetic
energy that just takes this factor of
1/2 away removes the factor of 1/2 just
doubles this term involving theta dot
squared the fact that it just doubled it
was because of a special choice of the
fact that I took the same radius vector
here the same radius here and the same
mass in general it would just do
something else wouldn't just double it
but I chose the numbers simply so that
it would just do something nice and
simple and that's it now what about
cosine theta times cosine Phi plus sine
theta times sine Phi anybody recognize
that yes so what is it huh cosine of
theta minus Phi ok this is cosine of
theta minus Phi cosine theta minus Phi I
hate putting brackets inside parentheses
inside parentheses I'll just put an
underline here to indicate that these
are grouped together oh no I guess we
don't know we can we can we can spare
our parentheses there that's it is it
surprising that it's cosine of theta
minus Phi not really and I'll tell you
why supposing we didn't have the
gravitational field here altogether
supposing there was no preferred
direction a preferred direction in this
problem is the vertical Direction due to
the the gravitational field if there was
no
preferred direction the problem would
have a symmetry the symmetry would be
rotation symmetry where you rotate both
theta and Phi simultaneously if you just
rotated the whole thing and so you kind
of expect that the Lagrangian should not
change if you added two theta and Phi
the same angle that would just be
rotating everything now that is more or
less obvious when you think about it if
you just took the whole configuration
and rotated it that can the Lagrangian
shouldn't change and that's what tells
you that only angle differences should
come in here an angle difference doesn't
change if you add to both fada and Phi
the sum of the angles would change if I
added two degrees to theta and I added
two degrees to Phi then theta plus Phi
would get added four degrees but theta
minus Phi wouldn't change at all so
there's a symmetry to this problem in
the absence of the potential energy if
we don't have a potential energy
associated with the gravitational field
of the earth then the problem has a
symmetry and the symmetry we'll work it
out later we're going to work out the
conserved quantity that's associated
with it it is of course some form of
angular momentum but before we do that
let's work out the potential energy yes
oh yes good all right so from Herman
we had a 1/2 mg sorry one half Bob and
Herman conspired to get rid of the half
what's that No
no no no no yeah
I added Bob together with Herman Herman
also had an mr squared theta dot squared
and so good good so that's that's the
kinetic energy what about the potential
energy well this is T kinetic energy
potential energy that's exactly the same
as it was for the single pendulum except
we have to add up a contribution for
both the Bob's
for Bob and Herman excuse me alright for
Bob it's minus mg cosine theta and then
for Herman that's another minus mg
cosine of Phi cosine ya cosine of Phi R
thank you mg R cosine theta right so the
total lagrangian we now have a
lagrangian and it is mr squared theta
dot squared that's more or less familiar
looking except for the factor of 1/2
which is usually their plus now we have
a truly familiar looking term M R
squared Phi dot squared then we have
this extremely odd cross term which
multiplies theta dot times Phi dot M R
squared theta dot Phi dot times cosine
of theta minus Phi then we have to
subtract you so that gives us plus mg R
cosine of theta plus cosine of Phi
what's that yeah right yeah no no yes
thank you
this one right yeah that one yeah
right okay now this is complicated
I'm not going to solve it what I will do
is write down some of the equations of
motion but before I do it let's work out
the conserved quantity why is there a
conserved quantity no sorry there is not
a conserved quantity because the special
direction associated with gravity means
that there's no rotational symmetry of
this problem this problem does not have
a conserved quantity altogether which
makes it about as complicated as a
simple system like this can be but if we
forgot gravity I don't mean if we forgot
it I mean if we did this way way out in
space where there was no gravitational
field it would still be a rather
complicated problem but it would have a
new feature of having a symmetry this
would not be there let's get rid of it
and then there would be a symmetry what
would the symmetry be the symmetry of
course would be rotation where you add
to each angle exactly the same amount so
it's theta goes to theta plus epsilon
where epsilon is a small angle and Phi
goes to Phi plus epsilon same angle you
can't shift one by one thing in the
other one by another thing you shift
them both by the same amount
remember the last time we define
symmetries so that Q went to Q plus
epsilon times F F of Q this just means
that the FS are both equal to one okay
the FS which I defined last time that go
along that tell you how a Q changes when
you perform the symmetry transformation
in this case they're both the same
okay can we see in fact that the
Lagrangian doesn't change when you do
this well if you add a constant to a
theta yeah if you add a constant to
something you don't change its time
derivative so you don't change the theta
dot squared you don't change the v dot
squared by adding a constant and you
don't change cosine of theta
- Phi because theta - fire doesn't
change at all so in fact this Lagrangian
has the symmetry I don't know would you
say we've discovered it or imposed it I
think we discovered it but of course we
imposed it by getting rid of the
gravitational field so it's a bit of
both
all right let's calculate the angular
momentum or the conserved quantity the
North achar jh
and see what it is which doesn't change
with time if we know that something
doesn't change with time that goes a
long way toward helping us solve a
problem it takes a problem involving two
variables and really replaces it with a
problem with only one variable it
diminishes the number of equations since
we know that something doesn't change
with time okay so let's see if we can
figure it out the rule is an earth the
charge is PI sub I Q sub I sorry PI sub
I times what I called F sub I summed
over I where PI sub I is partial of the
Lagrangian with respect to Q sub I dot
and what about F sub I in our case the
FS are just worn so this just becomes PI
sub theta plus PI sub Phi
not-so-surprising it's just the sum of
the canonical momentum of theta and Phi
why because the FS are both equal to 1
but now we have to calculate or PI sub
theta and PI sub Phi R you might expect
that PI theta is just related to theta
dot and you might expect that PI Phi is
just related to Phi dot but it's more
complicated because of this mixed term
here so let's calculate that PI sub
theta
is equal to the derivative of the
Lagrangian with respect to theta dot all
right what is that it has use theta dot
over here M R squared theta dot and a
factor of two when you differentiate
theta dot squared you get twice theta
dot partial of L with respect to theta
dot has twice M R squared theta dot but
when I finished there's this mixed term
here as a theta dot in it so there's a +
M R squared Phi dot times cosine of
theta minus Phi that's PI sub theta
what about Pais of Phi same game all
right Phi dot looks very similar the
first term except it doesn't have the
two in front of it because of the half
which appears in Lagrangian instead of
the two appearing in PI we just have M R
squared Phi dot this is partial of L
with respect to Phi dot but then we also
have a term from here plus M R squared
now it's theta dot times cosine of theta
minus Phi look how hard it's gotten PI
sub theta involves Phi dot and PI sub
Phi involves theta dot so things have
gotten a little mixed up not to worry we
just we just nevertheless follow our
nose mechanically add these things
together I won't bother I will write out
the answer the answer is 2m r squared
theta dot plus M R squared Phi dot I
think I'm missing an R M R squared
right plus M R squared theta dot plus
Phi dot times cosine of theta minus Phi
this is the thing which doesn't change
with time we can call it the angular
momentum it's the thing which is
conserved because of rotation symmetry
and it doesn't change with time it gives
us a piece of information if we know the
angular momentum at any given time let's
say at the start of the hour at the
start of the trajectory then it stays
fixed forever after and allows us to
figure out what theta dot is in terms of
Phi dot it allows us to eliminate one of
the variables of the problem okay only
one variable allows us to eliminate one
thing and simplifies the equations for
us but as you can see the main reason
for doing this was to show you how
mechanical it is to calculate the
Lagrangian to calculate the conserved
quantity and then to calculate the
equations of motion let's do the
equations of motion it's it's boring
it's it's quite boring anything which
has been reduced to being mechanical
like this is naturally going to be
boring and that's the goal of physics to
make everything really boring
oh this is this is not the Lagrangian
here this is the angular momentum let's
not confuse it with that L up there
okay so let's calculate the equations of
motion we've calculated pi so just do
theta I won't do fire you just do theta
the first thing is to calculate D by DT
of partial of L with respect to theta
dot okay that's already a little bit
complicated but it's straightforward D
by DT of what is it
- mr squared
theta dot that's just a fader second
derivative theta double dot M r-squared
they're constants then there's plus
what's the other term hmm question I
heard somebody say double dot this is
double dot D by DT of theta dot is Theta
double dot right all right but this is
not the full price of theta the rest of
Pi sub theta has this term so it has
another M R squared again M R squared is
a constant then it has Phi dot and then
cosine of theta minus Phi all right this
is a little bit complicated D by DT of
this term just gives us theta double dot
D by DT of this term is a little bit
complicated we have to differentiate Phi
dot times cosine of theta minus Phi you
know how to do that I'm not going to
waste your time it'll give us a five
double dot term and then it'll give us a
term where the time derivative hits the
cosine here I'll let you do that
straightforward calculus it'll give you
a few more terms than you might have
expected so on the left side we have D
by DT of a very definite quantity we can
feed this into a computer calculate five
out of each instant along a trajectory
differentiate it and all of this can be
worked out easily and that's equal to
what that's equal to the derivative of
the Lagrangian with respect to theta
itself that's equal to DL by D theta
that's Lagrange's equations where does
theta appear yep okay yeah
yes it is yes it is well you just follow
your nose why are you yeah the
derivative with respect to one of them
is - the derivative with respect to the
other it's true
whenever you have a function of a
difference of two variables the
derivative with respect to yeah you'll
pick up a minus sign when you
differentiate this with respect to Phi
you'll get a minus sign from over here
okay yeah yeah okay so I assume you know
how to do this derivative on the left on
the right the only place where theta
appears is in this cosine here okay so
yeah
yeah yeah yeah
you'll get one term with a theta dot and
another term with a Phi dot you know
have opposite sign right okay so I'll
leave that to you
there's nothing terribly interesting
straightforward calculus and the other
thing which is here is the right-hand
side which is the derivative of the
Lagrangian with respect to theta and
that is just equals mr squared Phi dot
and then the derivative of cosine theta
sorry the derivative of cosine with
respect to theta just becomes minus sine
of theta minus Phi can everybody read
this I can't laplacian laplacian yeah Oh
M R squared Phi dot times the do it
upstairs how am attire Oh
oh you're right thank you thank you
thank you thank you thank you very good
yes the derivative of Lagrangian with
respect to theta so that's gonna give me
what another 5 theta dot yeah
thank you good but its mechanical point
is its mechanical it's complicated the
final answer is complicated but
mechanical no no ambiguities in the
steps this is the equation of motion for
the double pendulum the theta equation
you can work out the fire equation in
the absence of a gravitational field
it's already complicated one of them
spinning around this exchanging energy
back and forth and angular momentum back
and forth with the other it's a rather
complicated motion that these things are
undergo even in the absence of the
gravitational field but the one thing
that you have going for you when there's
no gravitational field is you have a
conserved quantity that you can use to
replace theta dot by its expression in
terms of Phi dot you also have energy
conservation right right that
effectively eliminates one whole degree
of freedom and write your energy
conservation so you can fix the energy
and you can fix the angular momentum and
that there's effectively two equations
that allow you to eliminate let's say we
can eliminate Phi dot and Phi out of the
equation and just have an equation for
theta all right but we're not in the
business of solving equations here we're
in the business of studying the formal
structure of mechanics and as I said the
main reason for doing this tonight was
to show you that a problem which is very
very treacherously tricky when you try
to think about it in terms of F equals
MA and it would take us a lot longer to
do on the blackboard you simply set up
to what you have to do the rules are
figure out the velocity square it right
one-half MV squared figure out the
potential energy subtract it off and
then you're going your
on your way all right
I said I would take a simple problem
pendulum then a complicated problem the
double pendulum will come back to the
double pendulum when we talk about chaos
a little bit the double pendulum is a
very chaotic system which means it's
rather unpredictable despite the fact
that the equations of motion are in some
sense exactly predictable it does some
rather unusual and weird motions we'll
come back to it as an example of chaos
but for now it was just an example of
the canonical or the Lagrangian
formulation
what is the incidentally what is the
energy of the system well all the terms
of theta dots are kinetic energy the
term without the theta dots of potential
energy for a system like this the energy
is simply the kinetic energy plus the
potential energy instead of minus so you
would just change the sign of this term
I think that's right
so in that sense it's similar to the
ordinary pendulum all right the next
problem that I want to study is the most
basic problem of all of theoretical
physics it's the one that comes up over
and over and over again what is it a
harmonic oscillator there's one problem
which is a little bit simpler than the
harmonic oscillator a little more
fundamental of a fundamental a little
and it's of course just the free motion
of a free particle with no forces on it
that's really boring the harmonic
oscillator is a boring problem but not
quite as boring by boring I mean of
course that the motion is so simple and
so completely predictable that
it has no surprises but it does have a
structure a mathematical structure and
the which is which underpins so much of
physics that we need to get it right
from the beginning the pendulum is
almost a version of the harmonic
oscillator if you look at the pendulum
and you displace it just a little bit
from the origin that swings back and
forth very very much like an ordinary
harmonic oscillator if you displace it
far from the origin then it's not quite
true
the mathematical connection is as
follows the potential energy minus mg R
cosine theta what does that look like
let's let's draw a picture of it just
minus cosine theta cosine of theta looks
like this right from 0 to 2 pi swing all
the way around minus the cosine which is
what's in the potential energy looks
like this
so it's minimum at its minimum at theta
equals 0 in other words the energy is
minimum when it's hanging straight down
that's that's the expression for the
energy I could continue it back a little
bit to negative values of theta and it's
minimum right here any function which is
minimized at a point has a basic feature
that its first derivative is zero what
comes next is the second derivative any
function like this which as a minimum is
well approximated by a parabola at the
bottom in other words by something
proportional to theta squared this is
approximately equal to high-precision
when theta is small in fact to correct
the second-order in theta it's equal to
M G or let's get it straight it's minus
mg R which is just a constant
and not interesting plus 1/2 mg R theta
squared I'm simply using the fact that
near the bottom of the orbit like this
you can approximate this curve by a
parabola something proportional to theta
squared any function which has a minimum
a smooth minimum like this typically
will be approximated by a parabola well
that's not quite true
it could be flatter than a parabola but
that would be an exceptional situation
but typically when a function has a
minimum at the minimum you can
approximate it by something proportional
to theta squared if you did so and you
ignored the constant in the energy which
doesn't do which is of no particular
importance did I get this wrong did I
get the sign right here let's say U is U
as plus then we would have in the
Lagrangian minus 1/2 MGR theta squared
and in the energy we would have plus but
the important thing here is that it's
just proportional to theta squared it's
a very simple expression the simplest
expression you can imagine for the
kinetic energy about a minimum about a
point of minimum energy
something's quadratic that's what
defines the harmonic oscillator when the
energy or the potential energy is simply
quadratic in the coordinate that's
called a harmonic oscillator all right
so let's let's instead of calling the
coordinate theta let's call it X here's
another system that we might study an
object on a spring with a mass M if the
spring has some equilibrium position
some natural length when it's stretched
and not under any stress when it's not
stretched or compressed and if you
compress it or you stretch it the mass
will oscillate about the equilibrium
point
the if we call x equals zero the
equilibrium point let's call this the
x-axis and take X equal to zero at the
equilibrium point then we have to do
some work to stretch it we have to do
some work to compress it and the
potential energy is proportional to x
squared the potential energy in this
case is proportional to x squared with a
constant coefficient K and it's
conventional to put a 2 there this is
called the spring constant the spring
constant KX squared and this by
convention we put a 2 there yeah I just
dropped it because a constant doesn't do
anything but you're right it's there but
it wouldn't make any difference to
anything the reason is because it's a
constant and the only things which come
in two equations of motion are
derivatives of the Lagrangian that will
not contribute anything to the
derivatives of the Lagrangian so you're
right there is another term here it's a
constant term and just ignore it because
it doesn't it doesn't affect anything
okay so for a simple system like a
spring a good approximation is to say
that the potential energy is positive
you increase the energy by stretching
the spring you also increase the energy
by compressing the spring in either
direction you have to work to move the
spring KX squared divided by 2 the fact
that this is a numerical number there
that's not surprising the 2 is simply a
matter of definition K over 2 is call is
the coefficient in the in the energy the
Lagrangian is then 1/2 MV squared minus
K over 2 x squared what about the force
on the oscillator what about the force
remember the force
is equal to minus the derivative of
potential with respect to X and that's
equal to minus KX the derivative of this
with respect to X is just KX force is
equal to minus KX the minus sign means
that the force is a restoring force if
you stretch so that X becomes positive
the force is negative pulls you back if
you compress so that X is negative the
force is positive and again it's a
restoring force so this is the
Lagrangian the force being proportional
to X has a name that's called Hookes law
how do you spell hook
heito okay with it two O's II all right
I wasn't sure if it was any at the end I
didn't remember Hookes law this is
Hookes law Hooke was Newton's best
friend so this is your standard spring
mathematically idealize the real spring
of course would have a more complicated
formula in fact if you stretch the real
spring enough it would break so there's
all sorts of complicated nonlinearities
in it this is an idealization and it's
good for small excursions of the spring
no no of course not you can't The Rifles
law you can only experimentally measure
it but what you can say is that on
mathematical grounds
[Music]
a function which has a minimum has no
first derivative it has no first
derivative that means there's no term
proportional to X a constant is
irrelevant we don't care about a
constant in the energy next there could
be a term proportional to X but if we're
at the minimum then the derivative of a
function with respect to X is zero and
the first order correction the first
order term is zero the next term is
proportional to x squared in other words
if we expanded a function and a Taylor
series expansion and powers of X the
first term could be a constant which
we're not interested in the next term
would be X proportional to X but if
we're sitting at the minimum then
there's no slope X would mean a slope
there is no slope so that's not there
then there's x squared and then there's
all the higher terms that could be their
constant X cubed and other constants e
prime X to the fourth and so forth the
point is that if X is alright the main
assumption there's also a constant here
let's call it D I've already used C and
C Prime it's conceivable to think of a
function which begins with order X cubed
in other words where the constant
multiplying x squared is just plain zero
all right that would be rather special
it's usually deemed to be a fine-tuning
of some sort that there would be no
quadratic term it's not a fine-tuning to
say there's no linear term it's just a
statement that we started the bottom
line wherever the bottom is that there
would be no quadratic term would require
some careful tuning of a function if
there's a quadratic term that's the
first thing
next would be cubic quartic and so forth
these terms are not Hookes law but the
point is that if X is small enough that
if we only deviate a little bit from the
origin then X cubed that X fourth are
much much smaller than x squared so for
small displacements these terms can
typically be ignored unless the
first term is absent if the x squared
term is absent then you have to begin
with X cubed but if the first term is
not absent then for sufficiently small
oscillations it always begins with order
x squared so in that sense
Hookes law is derive abaut in the sense
of an approximation for small
oscillations under the circumstances
where nobody carefully fine-tuned the
way the first term here Hookes law is
very general this is the reason this is
the fundamental reason that the harmonic
oscillator shows up in so many places
it's always there when you're
oscillating about the minimum of a
potential energy unless somebody has
very carefully fine-tuned the way the
first the first order correction the
first order contribution second order
contribution yeah no no no no if there's
a term before to see cubed no of course
not no springs break if you stretch them
too much no no no it's right it's it's
always an approximation on the one hand
but it becomes a better and better
approximation as the oscillations are
smaller and smaller so you imagine
studying this very very small
oscillations very small oscillations it
becomes progressively more and more
exact it becomes a better and better
approximation as the oscillations become
smaller and smaller of course as the
oscillations get smaller and smaller
they get less and less interesting you
can't see them after a point they're too
small so you have a highly accurate
theory of something which is too small
to be measurable the question the real
question is where does this term come in
compared to this and what is the nature
of your
and how big are the oscillations and is
this term important relative to this
when the oscillations have a certain
size so it becomes a question of
estimating the relative importance of X
cubed
relative to x squared and that depends
on the size of these coefficients but
for almost all systems for sufficiently
small oscillations in other words
sufficiently small energy the x squared
is usually a good approximation the
linear term if we put a linear term in
let's put a linear term in okay give me
another letter okay okay I've already
used to K over here I don't want a ax
what would this do well it would put a
first derivative over here it would mean
that the function now has a first
derivative and it would look like this
well that would just mean that you
started at the wrong point instead of
starting at this point start at this
point shift X shift the definition of X
so that x equals zero at this point and
then there won't be a linear term yeah
okay so the X the constant coefficient
is of no importance or constant the
linear term can always be shifted by
going to the minimum by starting your
calculate by starting your coordinates
at the minimum that gets rid of the
first term and then you're left with the
x squared term right all right
so as I said in that sense it's a rather
general structure which is always a good
approximation for sufficiently small
oscillations it always breaks down when
the oscillations get big enough whatever
they are they are whether they are light
oscillations sound oscillations water
waves
Springs vibrating for very small
oscillations Hookes law is a good
approximation for large oscillations it
breaks down the pendulum is a perfectly
good example
what comes after theta squared if you
expand cosine theta theta to the fourth
is the theta it happens there's no theta
cubed term there's a theta to the fourth
term so if you take large swings the
large swings are sensitive to the higher
terms in the cosine but for small enough
swings only the theta squared term is
important all right yeah this one I've
lost track of it let's see um are you
all right I think that I think you're
right yeah sounds right no I was
yeah but cosine has 1 minus theta
squared over 2 yeah right yeah no idea I
think it's minus measures from theta
equals 0 the potential energy is
positive if the potential energy is
positive then Anna Lagrangian it should
be negative all right this the
importance of this system I've explained
why it's important why it recurs over
and over in physics now we have to
understand it we have to study its
properties its Hamiltonian its
Lagrangian its equations of motion and
all the things that that we've learned
up till now apply them to this system
yeah
say it again in other words if the
system is disturbed only a little bit
yeah yeah for the simple one-dimensional
system or systems more generally than
that its energy conservation which
prevents the oscillations from getting
too big you have only a certain amount
of energy let's write down the energy
the energy H is M over 2 X dot squared
plus K over 2 x squared a large
excursion large X will mean a lot of
energy if you only have a small amount
of energy this simply can't get too big
because it's positive if it was negative
if you study the system with a negative
term in the potential energy then X dot
can get big and X can get big and cancel
each other and keep the energy bounded
that's because both these terms are
positive x dot squared and x squared
neither one of them can get big if the
initial energy is small so that's a very
good question
and there's a very good answer energy
conservation it wouldn't work if the
potential was upside down potential
being upside down means you're standing
at the top of a hill if you displace the
thing a little bit it'll go flying off
the hill and your approximation of a
small oscillation will not be consistent
all right so this this is a system worth
exploring since it comes up so much here
it is I'll write it over again let's
work out its equation of motion we could
of course just write F equals MA for
this case I mean for this case it's
simple enough we can just write after
the F equals MA but let's not do it that
way the L by DX dot that's PI sub X the
L by DX dot is just M X dot just the
ordinary momentum D by DT of that that's
derivative of the Lagrangian with thats
Deeb I DT of DL by DX dot that's the
left-hand side of Lagrange's equations
the right-hand side is just DL by DX
that's this so that's minus K over 2
times 2x this of course is just good old
F equals MA M X double dot is equal to
minus KX Hookes law it's nice to divide
by the mass let's divide by the mass
here notice the only thing which comes
in to the motion is the combination K
over m K over m is the important
quantity if I were to change K and
change M keeping K over m fixed the
equation of motion would not change all
right this is a this is a an equation
whose solution is well known and
everybody know what it is anybody not
know what it is
that's cosines cosines of sines yeah
corresponding to just an oscillation
let's why is it cosines and sines
supposing I take a function like cosine
of T and I differentiate it once then I
get sine
what happens if I do sorry what do I get
if I differentiate cosine minus sine and
then if I differentiate it again minus
cosine so the second derivative of a
cosine is proportional to itself but
with a minus sign now that's not the
cosine T is not quite right
cosine Omega T is a better bet where
Omega is the angular frequency then um
just a number okay
it tells you how fast the oscillation is
oscillating Omega the bigger omega is
the faster the oscillation this is the
angular this is just called the
frequency of the oscillation what is the
first derivative of this with respect to
time then minus Omega sine Omega T
what's the second derivative minus Omega
squared cosine Omega T so if I call this
X then this is X double dot and notice
that it's just minus Omega squared times
X that's what this equation says it says
the second time derivative is
proportional to X itself with a
coefficient K over m same minus sign and
so we identify that the frequency of the
oscillator is K over m or frequency
squared Omega squared equals K over m or
Omega the frequency a square root of K
over m supposing instead of putting
cosine there I put sine same thing if I
differentiate sine I get cosine if I
differentiate cosine again I get minus
the sine so this will also work for sine
in fact it will work for any combination
any linear combination X is equal to a
times cosine Omega T plus B times sine
Omega T any function of this form will
satisfy this equation if Omega is equal
to the square root of K over m you can
check that each one separately does and
if you add them together they do
why are there two coefficients why are
they two any a and B will work
why are there two coefficients in the
solution right it's because you have to
specify two things to start a solution
out one way of saying it is you have to
say you have to specify a position and a
velocity so there are two parameters
that govern a solution
there have to be in every solution of a
simple equation second order equation
they have to be two constants for each
one of the coordinates the general
solution one to fix the initial
condition the initial position and the
other to fix the initial velocity you
can work out a and B if you know the
initial position and the initial
velocity so this is all the solutions
there are there are no other solutions
beyond this there's another way to write
it you can also write it as X is equal
to AE times cosine of Omega times t
minus t naught where t naught is just
some constant some constant starting
time again it has two parameters an
amplitude let's call it capital a an
amplitude of oscillation which tells you
how big the oscillations are and a
particular time which is the time when
the oscillation reaches its maximum in
other words you can either write the
general solution in this form with two
coefficients a and B or in this form
with a coefficient that's called the
amplitude
dude and this one is called the phase in
this form t naught is the time at which
the oscillation reaches its maximum so
or if you like the starting point if you
imagine stretching the spring and
letting it go stretching it to a maximum
and letting it go the time that you let
it go is this T naught over here when T
is equal to T naught the cosine is at
its maximum value so these are the two
forms that the general solution takes
and it's perfectly obvious why it's
called the harmonic oscillator a lot of
know why it's called harmonic but it's
called an oscillator because it
oscillates it vibrates it vibrates with
a cosine of Omega T alright that of
course is no doubt familiar to most of
you say it again the thing is is T 0
yeah
so it's two parameters two parameter
family of solutions which as I said you
can either take to be the theorem of
course is that if you add a sine and a
cosine like this it's always equal to a
cosine of a shifted of a shifted
argument all right that's the basic
setup for the harmonic oscillator but
now we want to think about its
Hamiltonian first and its canonical
momenta canonical momentum first of all
is P which is the same as PI sub X which
is the same as the derivative of the
Lagrangian with respect to X dot and
what is it it's just NX dot it's just
good old momentum nothing surprising
there after all it is just a particle on
the end of a spring so that's P P or PI
how about the Hamiltonian let's work out
the Hamiltonian the Hamiltonian is P X
dot a usual thing canonical momentum
times velocity minus the Lagrangian
minus L and that's let's see P what's is
MX dot so P X dot is MX dot squared
twice the kinetic energy but now minus
the Lagrangian so minus M over 2 X dot
squared plus K over 2 x squared
in the hamiltonian the KX squared term
has the opposite sign as it had in the
Lagrangian so this is exactly what you
would expect MX dot squared over two
plus K over two x squared and it's
conserved through all the reasons that
we worked out last time we worked out
that this quantity is generally
conserved so this is the conservation of
energy this is the energy well we can
write it in another form and now I'm
going to start to teach you about the
Hamiltonian formulation of classical
mechanics up till now we've dealt with
the Lagrangian form of the equations we
want to pass now to the Hamiltonian form
the Hamiltonian form works not with X's
and X dots or Q's and Q dots but it
works with the basic variables being
canonical momentum and coordinates Q's
and pies or Q's and peas why why are we
doing such a thing I have to admit I
have no idea why Hamilton started
thinking about things that way the
ultimate reasons boil down the quantum
mechanics we're not doing quantum
mechanics now we're gonna do some things
which may mystify you a little bit
you'll start to get familiar with them
they're very pretty they're elegant
they're simple but what the motivation
the Hamilton had for going along this
route is quite mysterious I don't know
what it was nevertheless Hamilton said
the important quantities well we'll come
to see why it's so important as we go
along but Hamilton said he wants to work
not with the quantities Q and Q dot but
rather with Q and P he saw some sort of
symmetry between Q's and PS he saw
something very special about the
canonical momentum that's not apparent
until you start to work with it
I will tell you what it is as we go
along but the goal is to rewrite
mechanics not in terms of positions and
velocities but in terms of positions and
canonical momenta this may be this at
this point this is probably a mystery to
you why wouldn't want to do that and it
really didn't achieve its full
importance until quantum mechanics so at
this point you have to hold off and say
ok he's doing some magical mystical or
mumbo-jumbo with P's and Q's instead of
Q's and Q dots I don't know why but
we'll start to see as we go along that
it has some in the form of the equations
is especially simple that's especially
simple in terms of P's and Q's yeah
that's right yeah some people think he
forced all quantum mechanics I don't
think he saw force or quantum mechanics
he just
say it again certainly
well again any system when disturbed a
little bit from equilibrium will behave
like an oscillator so a molecule I'm not
sure exactly what question you're asking
the reason that we study oscillators is
because small deviations from
equilibrium oscillate the reason is the
same in quantum mechanics as in
classical mechanics but there's a
feature in quantum mechanics which isn't
there in classical mechanics there's
zero-point energy it's impossible to
lower the energy to an arbitrarily small
amount so the idea of an arbitrarily
small oscillation is not sensible in
quantum mechanics that means that there
is no real limit in which the oscillator
becomes perfectly accurate but I didn't
want to get in that now I want to keep
away from quantum mechanics until until
we study quantum mechanics so let's
let's move on the let's write the
Hamiltonian in terms of P's and Q's
instead of Q's and Q dots here it is
here's the Hamiltonian or the energy
there's the expression for the energy
here's the expression for the canonical
momentum P is MX dot to write the
Hamiltonian in the form that Hamilton
would have wanted you to write it in he
would want you to get rid of all the X
dot s-- and replaced them by peas in
each case the peas will be functions of
X dots and X's in principle you can
solve for the X dots we'll work out some
examples you can solve for the X dots
and write them in terms of P for example
in this case it's very easy
x-dot is just P over m so let's write
that down
X dot is peel mm-hmm yeah yeah we're
gonna care of that oh really all right
so this this equation here tells us that
X dot is equal to P over m the momentum
divided by the mass is the velocity
let's plug that into here this now reads
then P squared over 2m why is it over 2m
x dot squared will be P squared over m
squared so when I write X dot squared I
will have two M's in the denominator two
M's in the denominator one M in the
numerator there will be p squared
divided by 2m so whereas when you write
the energy in terms of velocities the M
appears in the numerator when you write
it in terms of momentum the mass appears
in the denominator big deal but keep it
in mind that it's a it's a useful
mnemonic whenever you see something
multiplying an X dot squared in a
Lagrangian it usually winds up that
thing gets into the denominator in the
in the Hamiltonian er plus KX squared
over two
well notice is a kind of symmetry in
this problem between Pease and X between
P and X apart from the coefficients K
and M the Hamiltonian is perfectly
symmetric between K between X's and PS
it's of the form roughly speaking of X
square plus P squared well not exactly
it has some coefficients something times
P squared plus something corral times x
squared it has a very nice symmetry
between peas and and X's and that that
symmetry is one that will exploit over
and over again that the equations of
Hamiltonian mechanics are entirely
symmetric with respect to X's and peas
they look the same if you interchange
X's not not exactly but the structure of
the equations is symmetric between X and
PS but for now just recognize that the
energy has P squared plus x squared well
not quite
1 over 2m times P squared plus K over 2
times x squared let's make a diagram
representing on the horizontal axis X
and on the vertical axis P now you could
have used velocity on the vertical axis
after all velocity and momentum are
proportional to each other but we're
gonna use momentum now for now momentum
in the vertical axis horizontal axis is
is X how
a starting point for the motion consists
of a value of X and a value of P a
position and a momentum so if I tell you
the position and the momentum at the
start of the motion that corresponds to
a point on the X P planes incidentally
the X P plane has a name it's called
phase space and it has a symmetry
between X's and PS that it would not
have if you used velocities on the
vertical axis but that's something that
will come with time you start the system
out it starts with a velocity in a
position you can translate the velocity
into a momentum just by dividing by the
mass and so the starting point is some
point in phase space what happens next
the system starts to move it changes
both its X's and its Peas in this case
there's only one X and 1 P the velocity
changes the position changes the system
moves in the phase space how does it
move in the phase space can we track its
trajectory as it moves through phase
space well first of all we know that
energy is conserved that means that P
squared over 2m plus K over 2 x squared
which is equal to the energy call it the
Hamiltonian or the energy same thing
sometimes I'll call it energy sometimes
a core Hamiltonian we call it we tend to
call it Hamiltonian when we're talking
about its functional form in terms of X
and P we tend to call it energy when
we're just thinking about the numerical
value of the energy the numerical value
of the energy is conserved with time it
doesn't change and so that tells us
something about the motion can you see
from this what kind of motion it has
okay it would be a circle if M and K
were both equal to 1 then it would be P
squared plus x squared is equal to 2e or
something okay not quite a circle if you
put some coefficients in here
coefficient P squared plus another
coefficient times X
where is equal to a constant then it's
not quite a circle what is it an ellipse
so the motion alright in the phase space
is elliptical it's not a very good
ellipse is it let's figure out the
intercepts first of all oh oh if it were
a circle in other words if M and K were
adjusted so that the coefficients here
were the same then the radius of the
circle would be controlled by Eid in
fact e would be the square of the radius
roughly speaking P squared plus x
squared is equal to twice ie twice he
would be the square of the radius so if
we change the energy that will change
the orbit by changing its radius now the
ellipses are all of the same exact shape
there would be circles if K and M will 1
all right but they have different
distance from the origin they have
different radii same shape same
eccentricity different radius different
overall radius and the radius is
controlled by the energy so if we start
the system someplace there's an ellipse
passing through that point we take that
ellipse and the system will just
continue to move on that ellipse let's
find out where it intersects the x axis
and the P axis it intersects the x axis
at the point where P is 0 where P is 0 X
let's work it out X where P is equal to
0 for up here K over 2 x squared is
equal to e or X is equal to the square
root of e divided by twice e divided by
K I believe
twice a / K do I have that right yeah
okay
so the x-intercept here is square root
of 2 e over K notice that as a gets
bigger and bigger the intercept moves
out what about the Y or the p-intercept
here the p-intercept you do the same
thing the p-intercept is where x is
equal to 0 and we just solved for P
divided by 2m is equal to e and that's P
is equal to the square root of 2m e
square root of 2m e so the x-intercept
is 2 e over K or the square of square
root of 3 over K the Y or the
p-intercept is 2m e square root of that
so if we know the energy and we know the
mass and we know the spring constant we
can start drawing these ellipses in
phase space and wherever we start the
system just moves around on the ellipse
incidentally by squeezing the x axis and
expand or just by an appropriate the
redefinition of X and P we can turn
these into circles easily enough okay we
can easily turn them into circles by
rescaling either the x axis or the P
axis squeeze the x axis a little bit or
expand the P axis or some combination
and you turn them into circles I won't
do that I'll just leave it the way it is
but you can think in your head circles
circles in phase space or ellipses in
phase space that can be rescaled two
circles now how long does it take for
the phase point to go all the ways
around that depends on the frequency of
the oscillator Omega the larger Omega
the long the shorter the amount of time
for to go always around
all right so the Fae the the phase point
moves on ellipses it moves around in a
time which doesn't depend on where you
begin so in other words if you actually
thought of circles think of circles then
the whole phase space rotates around
with frequency Omega every point no
matter where you begin it takes exactly
the same amount of time to go around and
so wherever you start in this phase
space you just progress around with the
same angular frequency round and around
it around wherever you start
supposing you project the motion onto
the x axis in other words you don't look
at P you just look at what X is at any
given instant of time how does it move
well that's just projecting onto the x
axis and it just moves back and forth it
oscillates the oscillator oscillates and
it oscillates with exactly this kind of
motion it also oscillates if you
projected it on to the P axis and notice
when it's farthest out in X it's moving
fastest know when it's farthest out in X
when it's farthest out in X it's moving
slowest Y when it's way out here at
maximum X P is equal to zero when it's
up at its maximum X is equal to zero so
when the oscillator is far out it's
moving slow when it's close in it's
moving fast and you can think of the two
motions the X motion and the P motion as
the projection of the circular
phase-space motion onto the x axis and
onto the P axis
they space is quite fundamental notice
some other things now not all systems
move on circles but all systems do move
on surfaces of constant energy so when
you write the energy however you have to
write it in terms of X and P then the
trajectories in that XP space are
surfaces of constant energy their
contours of constant energy of total
energy the energy being expressed in
terms of X and P there's another fact
that's obvious for this system well if
you think about it in terms of circles
it's obvious for this system but it's
also very general it's going to play a
very very big role in the structure of
mechanics it's the following fact
supposing you take a little patch of
area in the phase space this patch could
stand for your uncertainty in the
knowledge of where you begin you know
that you begin somewheres in here and
that's all you know what happens to it
it just moves around and around and
around but the most important thing is
that it preserves its area area in phase
space is preserved with time I'm gonna
come back to that I'm a little getting
I'm reaching yeah I've reached 9 o'clock
and that's the end of my energy so I my
Hamiltonian has run out but yeah yeah
Yeah right we need to be able to solve
where's our equation we need to be able
to solve for X dots in terms of P in
order if if we want if our goal is to
write things in terms of P and X all
right
if that's our goal and we will see that
it will be alcohol but if that's our
goal to rewrite mechanics in terms of
X's and P's we've got to eliminate the X
dots out of the problem that means we've
got to solve for the X dots in terms of
the peas and plug back in we can't do
that unless unless X that to it that
would present some problems those
lagrangian's are deemed to be
fundamentally sick and the sickness I'll
tell you what we'll do an example to
remind me next week I'll work out an
example and you'll see what goes wrong
at the place where you can't solve the
equations for something happens
something happens it's bad the the the
motion doesn't make sense
[Music]
the preceding program is copyrighted by
Stanford University please visit us at
stanford.edu