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I wanted to do a little more with the
idea of a particle in an electromagnetic
field I wanted to work out in more
detail the theory of a particle in an
electromagnetic field a charged particle
we're not going to worry in this course
about the dynamics of a field itself of
course the dynamics of the field itself
the field is a dynamical system itself
has equations of motion it will follow
them for our purposes now the electric
and magnetic fields are simply given to
us and in fact to make things simple
let's suppose that the electric and
magnetic fields don't depend on time
although they may depend on space I've
worked out for you the theory of a
charged particle moving in a in a
magnetic field I thought we would redo
that talk about it a little bit because
it is it is unintuitive some things
about it are unintuitive in particular
the canonical formulation of it the
Hamiltonian formulation the notion of
momenta is a little bit surprising and
there are some unintuitive things about
the way a particle moves in a magnetic
field in particular when you add an
electric field so it's called the Hall
effect for those who wonder well may
know what I'm talking about all right so
let's sin we have various and in fact we
have various different formulations of
mechanics that we've done up till now
let's just lift list them list the
various forms we're going to learn one
new form but the first form is just
equations of motion F equals MA all
right ah let's call it equations of
motion and those are typically
differential equations which come down
from above Newton or somebody tells us
what those equations of motion are and
they're of the form of second-order
differential equations for the
coordinates of a system
next was the principle of least action
not any equation that we can write down
will be of the form that it can be
derived from a principle of least action
but all the fundamental equations of
physics are derived from the principle
of least action including incidentally
the field equations governing the
electromagnetic field but we haven't
gotten to that yet least action from the
principle of least action we can derive
another form namely the Lagrange
equations of motion Lagrangian form
those are derived by applying the method
of the calculus of variations to the
principle of least action in other words
the Euler Lagrange equations and that
gives us the Lagrangian form
and then finally by concentrating on
phase space the peas and the Q's we
ultimately wound up with a hamiltonian
form they each have their advantages and
disadvantages
the Hamiltonian form is perhaps the
least intuitive on the other hand it has
the most to do first of all with quantum
mechanics but second of all with
information flow it has this lovely
picture of a flow of points through
phase space which if you follow them you
not only see how a single system behaves
you see how the whole ensemble of
systems with different initial
conditions will behave and allows you to
compare them it allows you to discuss
for example information conservation and
other things that will come true so
that's the advantage of the Hamiltonian
form the disadvantage is in some ways
it's less intuitive in particular we'll
see that tonight when we talk about the
electromagnetic particle Lagrangian form
very very general very very powerful uh
I can't think of very many disadvantages
of it it's a powerful except that it
doesn't do some of the nice things that
the Hamiltonian form does namely allow
you to plot a flow through the phase
space the principle of least action that
underlies everything extremely deep but
in its form in the form in which you
write it down it is not terribly useful
the way you use it is to derive
equations of motion from it namely go to
the Lagrangian form so the principle of
least action underlies the whole shebang
but by itself it doesn't tell you how to
solve problems so easily minimize the
action usually you do that by solving a
differential equation an equations of
motion well those of course are the
lowest common denominator in the sense
that all systems have equations of
motion
we learn about them in elementary
physics F equals MA but just knowing
that there are equations of motion
differential equations is not sufficient
to tell you about energy momentum and
all the special things that go with the
special form of the equations that
physics makes use of so they're all
interesting in all useful we can apply
each of these to the motion of a
particle in an electromagnetic field now
for tonight I'm mostly going to be
interested in a particle moving in an
electromagnetic field in two dimensions
just because it's easy to visualize in
other words a particle free to move
around on the blackboard in an
electromagnetic field where for
simplicity the magnetic field will be
into the board no components along the
surface of the board and the electric
field will be parallel to the board
we'll come to that but before we do let
me write down the general equation of
motion for a non relativistic particle
for a non relativistic particle is one
this means one moving slowly by
comparison with the speed of light and
it's just F equals MA
force is equal to mass times
acceleration but now I have to tell you
what the forces the force is given in
terms of the first of all the charge of
the particle called Q then is a term
proportional to the electric field the
electric field is a vector that makes
sense
acceleration is a vector and magnetic
term which is velocity dependent depends
on the velocity again it's proportional
to the electric charge it contains the
velocity vector cross the magnetic field
are depending on your choice of units
some familiar units put a see down here
where C is the speed of light I'm going
to leave it out you'll find in some
choices of units in fact on a lot of
choices of units the speed of light is
included down here but this doesn't
matter you can either absorb the speed
of light into a new definition of the
charge or a new definition of the
magnetic field or you can remember that
it's always V divided by the speed of
light when I write V you can think to
yourself he really means V divided by
the speed of light
either way the main arm structure of the
equation is mass times acceleration is
equal to an electric field term plus a
magnetic field term the magnetic field
term has the property that the force
depends on the velocity but unlike a
friction force the force is
perpendicular to the velocity cross
products are perpendicular to the
Constituent vectors which make them up
so R and in fact what that means is that
if a particle is moving the magnetic
force does not accelerate it along the
trajectory it accelerates it
perpendicular to the trajectory and for
that reason it doesn't change the speed
of the particle not changing the speed
of the particle means that it doesn't
change the kinetic energy magnetic
fields do no work that's the that's the
catch phrase all right so this is one
form arm as I said I will mainly be
interested in magnetic fields or motion
in the plane in the plane of a
blackboard the whiteboard where the
magnetic field is perpendicular to the
plane the velocity of course is in the
plane so the velocity is in the plane
the magnetic field is perpendicular the
cross
dr. V cross B is also in the plane it's
perpendicular to B if B is perpendicular
to the plane the force lies in the plane
so the whole motion takes place in the
plane or can be constrained to take
place in the plane and if the electric
field also lies in the plane that just
gives another contribution to the
acceleration in the plane itself all
right so this is this is all set up now
let's go to the principle of least
action oh let me write down one other
equation which doesn't come in in any
useful way into into the equation of
motion in this form but it is necessary
to go either to make the step from
equation of motion to any of these other
formulations you need another idea you
need the idea of the vector potential
all right
the vector potential the vector
potential is just defined by saying the
magnetic field is equal to the curl of
some vector that's called the vector
potential now there's something special
about vectors which are curls of other
vectors anybody know what it is they
have zero divergence we talked about
what a divergence is but basically it's
the spread of the vector field away from
a point anything which is a curl has
zero divergence the magnetic field has
zero divergence
what about an electric field as an
electric field have zero divergence
what's if we write the divergence of the
electric field del dot e what's on the
right-hand side the charge density the
charge density what is it telling us
that del dot B is equal to zero
yes it's telling us that there's no
analog in nature of electric charge
which creates sources of the magnetic
field no point magnetic sources which
create magnetic field now is it true
that they really are no magnetic
monopoles probably not but nobody's ever
discovered one so until somebody
discovers one we can at least assume
that for the purpose of the physics that
we normally do that there are no
magnetic charges and the curl of the
magnetic the divergence of the magnetic
field is zero any field whose divergence
is zero can always be expressed as the
curl of something else but not in a
unique way not in a unique way we'll
come back to that non uniqueness there's
a non uniqueness in the vector potential
here that the vector potential can be
changed without changing the magnetic
field the equations of motion only
depend on the magnetic field we could
write that the magnetic field is the
curl of a vector potential but the
equations are such that changing the
vector potential keeping the magnetic
field the same doesn't change the motion
of the particle that's called gauge
invariance we will come back to gauge
invariance in fact we'll probably come
back to it tonight but for the moment
just keep in mind that this construction
of a vector potential is ambiguous I'll
give you some examples as we go along
what about the electric field
I don't usually represent the electric
field is just a more ordinary non
velocity dependent force it's very much
like the ordinary forces that we studied
up till now it doesn't depend on
velocity it does depend on position it's
also conservative conservative means
that it respects the conservation of
energy and what do we know about
conservative forces we know that they're
generated from potential energy
functions from potential energy
functions force is equal to minus the
gradient of a potential energy function
to write it in terms of components it's
the equation that we've written the ithe
component of the force is minus the
derivative with respect to X I of a
potential energy function now in
electricity and magnetism it's normal
oh all right first of all the electric
force has this this particular behavior
and to ensure it what one writes is that
the electric field is equal to minus the
gradient of a quantity called v v is
essentially the potential energy but not
quite what's it missing well we have
mass times acceleration is equal to
charge times e so e is not quite force
charge times e is force and therefore
potential energy is charge times V
that's all the the V what does V stand
for well it's measured in volts and a
volt is an energy per unit charge is
what it is that's all the vault is is
tuned energy per unit charge and this is
a convention to some extent that we
instead of writing you we write V but
then multiply it by Q to find the
potential energy of a charge but it also
tells us that all things being equal the
force and the potential energy on a
charge is proportional to the charge
double the charge you double the
potential energy and so forth
okay so we can rewrite the
these equations here in terms of the
potential v sometimes just called the
electric potential and the vector
potential
AE by just plugging in two here and here
let's come next to the principle of
least action the only known formula for
an action which gives rise to this
equation requires you to use the vector
potential you can't get away with it
nobody knows and in fact you can prove
that you can't write a principle of
least action which deals directly with
the magnetic field you have to use the
vector potential well that's a little
bit odd because at the end of the day
the the behavior of a particle doesn't
depend on the vector potential it
depends on the magnetic field you can
change the vector potential without
changing the magnetic field and the
motion of the particle doesn't depend on
the other vector potential but
nevertheless to write the principle of
least action you must introduce the
vector potential here it is the action
for particle isn't as always is an
integral it has the usual 1/2 MV squared
the kinetic energy one-half M let's
write it out explicitly X dot squared
plus y dot squared I'm leaving out the Z
component of the motion because as I
said let's concentrate on two dimensions
if we wanted to put the third dimension
in we would and X dot squared or just MV
squared it then has the good-old are
potential energy which would just be
minus U for an ordinary potential energy
we would write minus U of X and Y now we
write minus charge times V of X and why
and we integrate this DT what else would
we integrate it with DT but then that
does not capture the physics of the
magnetic field we have to add plus the
integral again the charge the vector
potential let's write it in components
first a sub I D X sub I we take the path
and we break it up into little
infinitesimal segments each segment is a
DX and dy we call the combination DX and
dy we call it the X sub I and we take
the component of the vector potential
multiply it by the corresponding
component of DX and integrate it we
don't need any DT in here we have a DX
so that's just along the trajectory in
each little interval there we take we
can call it DX dot a if we think of DX
as a little vector and a is a vector
then what appears here is a dot DX so
this is also a line and integral along
the trajectory from the beginning I
won't bother putting in the boundaries
the endpoints of the integration is from
the beginning of the trajectory to the
end of the trajectory and that is the
action for a particle and an
electromagnetic field now let me take
just two to show you some of the
symmetry of this formula let me take
this term and put it over here and write
it as minus Q V DT let's remove it from
here
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this is kind of interesting cue comes
out cue goes along with all of the terms
what do we have here we have for each
each direction of space we have an a sub
I times a DX sub I and then for the time
component we have a V times DT it's as
if if we plotted the motion in space
time now that we're not talking about
relativity for the moment but you can
see that what's being set up here is
going to be particularly simple when we
think about the special theory of
relativity where X and T form space-time
but let's put in T here x and y and
imagine a trajectory we now break up the
trajectory into little bits and now what
we have from the electromagnetic field
is a term a sub ID X sub I minus V DT
it's almost as if not almost as if it is
as if V was the fourth component of a
vector potential and that this is a
combination that involves all four
directions of space or sorry all four
directions of space and time dotted into
a three dimensional vector potential and
its fourth component now if you're not
familiar with special relativity don't
worry about it
right now this is the form that the
action takes this a DX looks a little
bit unfamiliar up till now our action
has involved the integral of a
Lagrangian and a Lagrangian is a
function of velocities and positions a
is a function of position a is a vector
field in space it depends on position
all three components of it depend on
position V also depends on position in
general a could also depend on time and
so could V but it does not depend on the
velocities
so this is something which depends on
position but it has this odd form of
being multiplied by the X sub I well to
get it to look more ordinary just divide
by DT and multiply by DT we can also
divide by DT over here and now we see
that the lagrangian for the particle in
the electromagnetic field has the usual
form ah yeah we could write this VI here
no V is a scalar 8 oh yes oh yes a dot
DX is a sum no that's right but we're
calling tooth yeah we're going to wind
up calling as usual two things by the
same symbol is velocity and this
potential I'll try to use if I ever here
here it is right here let's be careful
let's call a small V small V for
velocity a sub i DX by DX by DT can also
be written as the dot product of the
vector potential with the velocity all
right so let's do that let's just call
this a say dot V
sorry a dot small V and these are both
vectors so that's our Lagrangian this is
our principle of least action the true
trajectory minimizes the action and it
immediately tells us what the Lagrangian
is to get the Lagrangian all we have to
do is erase the integration signs and
the Lagrangian is plus Q times a dot V a
dot little V minus V of X
this is now on the wrong line it should
be next to Lagrangian here we can go
through the construction of the
Hamiltonian and we'll do that now it's
the standard construction let's leave
that for either later maybe we won't
even need it very much altogether but
let's concentrate now on the gauge
invariance what kind of things can you
do to the vector potential that don't
change the magnetic field and therefore
do not change the physics remember the
real basic differential equations here
do not involve the vector potential that
means there's some sort of redundancy of
the description because we can change
the vector potential without changing
the physics and we can also think of it
as a kind of symmetry a kind of symmetry
of the system we're changing the vector
potential in a certain way has no
influence on the motion it's kind of
symmetry it's called gauge invariance
and here's the way it works let's take
as a let's just there to understand it
let's focus on a particular component of
del cross a e let's say the Z component
but all the other components work the
same way what is the Z component of del
cross a it is derivative with respect to
X that means D by DX of a Y minus the
derivative with respect to Y of a X
that's its definition now supposing I
add I'll write it in in vector notation
and then just explicitly in components
supposing I add to the vector potential
something which is itself a gradient we
call it lambda lambda of X what is a
gradient a gradient simply means the
ayth component we add to the ayth
component of a we add derivative with
respect
exei of lambda now lambda can be
anything the point is that changing a in
this way doesn't change the curl of a e
let's see why that is
supposing I now recalculate the curl of
a after having introduced this extra
term here for example let's take the x
and y component a sub X will now have
the new term d by DX of lambda and a sub
y will now have the new term d by dy of
lambda lambda of X means lambda of X and
Y in general x and y let's just leave it
that way okay what happens to the curl
the curl gets another piece it gets a
plus derivative with respect to X of the
new term in a y the new term in a Y is d
by dy of lambda but then we have to
correct this term that's minus D by dy
of the new term in ax the new term in ax
is derivative with respect to X of
lambda well these are exactly the same
thing d by DX D by dy of lambda is
exactly the same thing as d by dy D by
DX the order of operations of
derivatives is immaterial and so by
adding the gradient of a scalar to a
vector it does not change its curl
another way to say it is the gradient of
a scalar doesn't have a curl is curl
free so by adding to the vector
potential the gradient of a scalar we
change the vector potential we change
the Lagrangian but we don't change the
equation of motion
I will remind you in some ways in your
notes I worked out well maybe we should
do it let's do it over again let's just
to keep the lecture self-contained let's
work out the equation of motion and
starting with the Lagrangian and see
that it really does have the form
incidentally this equation up here has a
name
the right-hand side is the Lorentz force
law Lorentz re ntz
the same gentleman who first wrote down
the Lorentz transformations ok so this
term is called the Lorentz force law
let's check that this Lagrangian gives
rise to Lorentz this equation of motion
for the motion of a particle we just do
the usual thing let's just do the X
component that will just do the X
component let's not bother with the y
component we start with the X component
and we first calculate the canonical
momentum conjugate to X that's the
derivative of the Lagrangian with
respect to X dot so P sub X P sub X is
equal to the derivative of Lagrangian
with respect to X dot that's MX dot well
that's not surprising
that's just mV good old momentum but
then there's another term the X dot also
appears in a dot V this is X dot times a
sub X so there's another term Q a sub X
and that's it Q times a sub X so we see
something odd that the canonical
momentum is equal to the thing we
usually call momentum MV plus an extra
term which depends on the vector
potential but this is strange
because the extra term is itself not
gauge invariant it depends on the vector
potential if I change the vector
potential by adding something to it I
will
change the momentum well that's okay we
follow the mathematics if the
mathematics tells us to do it we do it
the canonical momentum of a point
particle in a magnetic field is not
gauge invariant means it's not a thing
you really measure it depends on your
choice of the very we're going to give
some examples of different choices of
vector potentials and see how the whole
thing works but this is the canonical
momentum now what's the equation of
motion the equation of motion is that
the time derivative of this the
derivative of P X with respect to T is
equal to the derivative mathis just
Lagrange's equations d by dl let's write
it down derivative of Lagrangian with
respect to X dot d by dt e that's just
derivative of P is equal to partial of L
with respect to X well the left-hand
side is just a time derivative of this
so we can write MX double dot that's the
first term of D by DT the L DX dot and
then the second term is plus Q times the
time derivative of a sub X I told you
that a sub X was not a function of time
does that mean that it doesn't depend on
time no what it means is it doesn't
depend explicitly on time here's a point
you may have a field in space which
varies in space but does not vary in
time all right doesn't vary in time at
all it's just something some spatially
dependent thing but if a particle moves
through it then by virtue of the motion
of the particle from point to point the
field at the particle does change with
time right and to work out the change in
time we just write that this is equal to
the derivative of a sub X with respect
to X times the time derivative of X plus
the derivative of a sub X with respect
to Y times the time derivative of Y
alright a at the position a sub X of the
position of the particle changes because
the particle is moving and here's the
time derivative of a that's the left
side what about the right side equals
armed now what he it depends on X let's
write this out this is a Q a sub X times
X dot plus a sub y times y dot minus V
of x + y oh this should be a Q here yeah
Q alright
what depends on X a X depends on X in
general a Y depends on X and V depends
on X the X dots don't depend on X just
ax and a Y and V depend on X so on the
right hand side let's write down what we
get we get Q times the derivative of a
sub X with respect to X times X dot what
I'm writing is partial of L with respect
to X so we get one term by
differentiating a sub X over here with
respect to X another term from
differentiating a sub Y with respect to
X let's
plus Q derivative of a sub y with
respect to x times what x times y dot
that's the derivative of this term with
respect to X and then we get minus Q DV
by DX
first of all q DV by DX what is that
that's Q times the electric field so
that's this term in Lawrence's force law
we can forget about it for now that's in
good shape on the left hand side we have
some terms each of which involves the
derivative of a with respect to x times
some sort of X dot on the right hand
side we have similar things perhaps some
of them cancel let's see what cancels
both the left and the right we have the
ax sub X with respect to X times X dot
here and here so the whole right-hand
side then what do we have here sorry
we can now transpose this term over here
to the right hand side and let's see
what we have we have first of all from
here Q then da Y by the x times y dot
and then from here we have minus da X by
D Y times y dot the a y by DX minus the
ax by dy what is that that's B sub Z so
this is Q times B sub Z times y dot
that's exactly this term here it has a
velocity it has a magnetic field and if
we're interested in the X component of V
cross B then it has B Z and Y component
of velocity so that's this so indeed we
do reproduce the Lorentz equation of
motion but even more than that we see
that the vector potential only appears
in the form of a magnetic field despite
the fact that we had to use it to write
down the Lagrangian the vector potential
appears in a gauge invariant way in
other words what appears here the B
field does not change when you make a
gauge transformation so that's that's
one of the oddities of physics that it
requires you quite often to introduce
some redundant descriptions of things
it's not that you have to introduce a
redundant description of things you
could have worked just with this but in
order to make the equations look like
principle of least action or grungy ins
hamiltonians and so forth there are
times when you have to introduce
redundant descriptions of things and
those redundant
descriptions are called gauge invariance
--is the transformations which
correspond to these redundancies the
changes that you make that are really no
changes at all in the underlying physics
those are typically called gauge
transformations so whenever you hear
about gauge invariance one is always
talking about some kind of redundancy
that's necessary to work out the
canonical formulation of a theory but
where the redundancy doesn't enter into
the actual motion of a system okay so
there we are we have the Lagrangian
nervous now I want to give you some
explicit examples of vector potentials
in particular I want to concentrate on
the problem of a uniform magnetic field
one which doesn't vary from place to
place that's the simplest situation
magnetic field pointing into the
blackboard which has no variation just a
uniform constant magnetic field how do
we construct one from a vector potential
we want B Z to be independent of
position so BZ is equal to derivative
with respect to X of a Y plus minus
derivative with respect to Y of AX I'll
give you two examples of vector
potentials which do the job you're on B
Z to be a constant let's say a constant
let's take B let's assume that B is into
the blackboard in other words it is
positive and it's and it's magnitude
I'll just call a little B the magnitude
of the magnetic field alright then we
can take a sub is an example take a y
equal to the number B times X and ax
equals zero in that case the derivative
with respect to X of a Y is just be the
X goes away when we take the derivative
on the other hand ax is equal to zero
and this doesn't change the answer so
this is one possible choice of vector
potential which will give rise to the
uniform magnetic field here's another
one a y equals zero and a X anybody want
to guess minus B times y just that just
check it this time the X of a y is equal
to zero but the Y by of the ax is equal
to minus B and gives you the same answer
here I'll give you another example it
has to be minus because there's a minus
here Y on if we want to get if we want
to get the same answer yeah here's
another example all giving the same
magnetic field a y equals B over 2 X a x
equals minus B over 2 Y also will give
the same answer these different vector
potentials are related by gauge
transformations it's not too hard to
work out that there's some scalar I
won't but write it down there's some
scalar that takes you from one to the
other by taking by taking its gradient
all right so these are three different
and there's an infinite number of other
ones an infinite number of other
formulations for the vector potential
which give exactly the same magnetic
field one can use any one of them it
doesn't matter since the net final
equations of motion don't depend on the
vector potential they only depend on the
net magnetic field let's take these two
examples let's forget this one here and
concentrate on
these two armed let's let me rewrite it
ax equals B Y a y equals 0 or and the
other possibility is a y equals 0 a X is
equal to know our ax is equal to 0 a y
is equal to minus B times X so in the
lingo these would be two different
gauges two different gauges giving rise
to the same to the same field okay let's
see what we learn by from conservation
laws let's start with this gauge over
here okay
let's forget the second gauge let's
start with the gauge in which ax is
equal to B Y and a Y is equal to 0 let
me rewrite the Lagrangian in detail now
and I'll tell you what for simplicity
for the moment let's forget the electric
field we'll come back to the influence
of the electric field in a little while
okay so a times V means now ax times X
dot I've set a y equal to 0 in this
first choice of gauge and furthermore ax
is equal to B times y so that's my
lagrangian and now let's calculate the
canonical momentum conjugate to x + 2 y
let's start with PX PX is the derivative
of the Lagrangian with respect to X dot
that's M X dot plus Q B why is this
conserved or isn't it conserved well
let's write down the other one py
is equal to my dot plus nothing
which one do you think if any are
conserved I will take a vote is this
conserved is this conserved any of them
conserve yes no good what yeah okay but
still are any of them conserved it's
that's one possible answer
anybody say yes come on I want to
embarrass you okay this one is conserved
why is it conserved this is the X
component of momentum and notice the
Lagrangian doesn't depend on X it
depends on X dot but X dot is not X and
it depends on Y if I make a
transformation X goes to X plus epsilon
the X component of the velocity doesn't
change
I've just translated the system a little
bit in the X direction
that doesn't change X dot it doesn't
change y dot doesn't change X dot over
here and it doesn't change Y why because
I haven't changed Y that's why that
means that the X component the momentum
is conserved there's a symmetry X goes
to X plus a constant what about Y goes
to y plus a constant mm-hmm this term
changes if Y goes to if you shift Y a
little bit this term changes so the Y
component of momentum is not conserved
all right let's forget it since it's not
concerned for the moment what we've
learned is that in this gauge this
quantity is conserved well I don't need
to say in this gauge I can just say n X
dot plus Q B Y
is conserved for example if it starts
out being zero let's take that case
suppose it starts out being zero then it
will stay equal to zero let's take that
case for simplicity if it starts out
zero it stays zero if it's not zero
it'll stay whatever it is okay and we'll
understand a little while what that
means okay that tells me that
immediately solves for me and tells me
that X dot the X component of velocity
is minus Q B over m times y that's what
I learn from conservation laws it's very
odd first of all it's odd because the X
component the momentum is not what you
might have thought namely just X star
and X dot and second of all in the
particular gauge that were writing since
there's an X translation invariance we
can immediately conclude that the X
component of momentum is conserved and
that tells us if it starts zero it stays
zero
later on we'll worry about what if it's
not zero okay now let's go to the other
gauge oh and incidentally this equation
here that X dot this doesn't have
anything to do this is completely gauge
invariant that just tells us that the
velocity X component of velocity is
proportional to Y let's find out another
piece of information by working out the
theory in the other gauge in the other
gauge the Lagrangian becomes minus QB x
times y dot
the y-component the momentum which is
now the one I'm going to be interested
in py is my dot minus QB x in this gauge
py is conserved oops sorry
y dot yeah in this gauge py is conserved
why because the Lagrangian doesn't
depend on Y seems to me I said that
before do I have a Lagrangian doesn't
depend on Y so py is conserved okay a
shift of Y doesn't change Lagrangian in
this gauge a shift of X changes
Lagrangian but a shift in y doesn't
change the Lagrangian and so this
quantity here is also conserved in the
motion of a charged particle for example
if it starts 0 it will stay 0 in which
case y dot is equal to this time it's
plus Q B over m times X I haven't had to
solve any equations all I did was write
down the conservation laws and I found
out that the X component of velocity is
proportional to minus Y and the y
component of velocity is proportional to
X that should ring a bell to some of you
maybe not to all but we'll work it out
I'll ring your bell for you what kind of
motion is this that's circular motion so
let's prove that it's circular motion
and see what we can find out about the
circular motion
let's consider a particle moving in a
circle about the origin for simplicity
what is what is the what is its equation
well let's call this angle theta
let's call its distance from the origin
R then X is equal to R times cosine of
theta but we're going to allow theta to
evolve and swing around here with an
angular velocity Omega we can write that
this is equal to R cosine of Omega T and
y is equal to R sine Omega T this is
uniform motion with an angular velocity
Omega number of radians per second is
Omega let's check that this solves this
equation here so what is X dot X dot is
just differentiating this with respect
to time that's equal to R Omega and then
the derivative of cosine is minus sine
minus sine Omega T and y dot is equal to
R Omega cosine Omega T X dot is supposed
to be minus something times Y X dot is
minus something times y and y dot is
supposed to be that same something times
X Y dot is that same something times X
in order to find out what Omega is we
just have to match this equation so here
we have X dot is equal to minus Q B over
m times y here we have that X dot is
minus Omega times y X dot is minus Omega
times y the RS cancel out all right so I
can write just from this here that X dot
is equal to minus Omega
x y and also a similar equation for y
dot well if I compare that with this
equation here I immediately find that
Omega the angular frequency it's called
the cyclotron frequency is equal to
minus Q B over m yes and it's Omega what
am i running q Omega Omega is QP over m
the charge of the particle divided by
its mass notice that combination ratio
of charge to mass times the magnitude of
the magnetic field that's the angular
frequency the larger the charge the
larger the angular frequency the larger
the mass the slower the angular
frequency this combination Q over m was
what made it hard about about the turn
of the 20th century to determine the
charge and mass of the electron because
there always comes into most things in
the ratio Q over m and so measurements
of the trajectory of electrons and so
forth and magnetic fields yield the
ratio of Q to M but it doesn't give Q&M
separately something else had to be done
to do that okay
now this missed the whole thing they
strike you as odd that we have this
redundancy of description we can shift
the vector potential change the whole
formalism but nevertheless extract out
interesting and useful information from
the conservation laws in fact I used two
different gauges to get two different
pieces of conservation laws what happens
if I did not send if I did not require P
X let's write this correctly yeah I
think we had P X is equal to n X dot
plus Q B Y but that was engage number
one gauge number one means this one over
here and an entirely different gauge
number two we had P y equals M Y dot
minus QB x gauge two because the real
motion doesn't depend on the gauge I
could use gauge one to calculate what X
does I could use gauge two to calculate
what why does so it's an odd kind of
logic the circular motion about the
origin here corresponding to both P this
px and this P y equal to zero if I set
this equal to zero that's setting my
equal to QB X what if I have a circular
motion since after all the magnetic
field is completely uniform one place is
exactly the same as another I should be
able to consider a circular motion
moving around some other point let's
consider that and then calculate whether
these quantities are still conserved
to do that all we have to do is add an X
naught here and a Y naught here X naught
and y naught are just the center
positions the central coordinates of the
circle X Norton why not I've shifted the
circle by amount X naught and y naught
no you're absolutely right it's a good
thing you caught me before I did it here
plus X naught plus y naught doesn't
change X naught and y dot at all okay
let's calculate what M X dot plus QB y
is okay MX dot is M R Omega minus R
Omega sine Omega t + QB Y what is y R
sine Omega t + y naught right
that's a that's px
now assuming again that Omega is related
to Q and B where is it where is the
connection between oh here it is
q b over m then these cancel each other
q b over m is omega so these again
cancel each other and we simply find
that PX is conserved it's just equal to
Y not the central position of the
circular motion the central position of
the circular motion of course does not
change with time because we're talking
about a circle moving about a fixed
central position likewise this py over
here will be MA I don't quite have that
right the way it's QB x why not QB x why
not the QB comes from this QB over here
and the py is minus QB X naught so you
see what the physical meaning of the
momenta is it's something entirely
unexpected the two momenta in the two
gauges the X component in the Y
component which are conserved quantities
simply correspond to the coordinate
locations of the circle about which the
particle is revolving and what the
conservation tells you is that the
central location of the circle doesn't
move to stays there entirely unexpected
way to think about momentum conservation
but this is exactly the way a charged
particle moves in a magnetic field so
you'll have to trust the grunge and
Hamilton and so forth that they knew
what they were doing and you have to
trust gauge invariance but the
mathematics gives you gives you the
right answer and in fact gives you the
right answer and probably
the easiest possible way just to use the
conservation laws okay let's stop for
questions for the moment yeah what's
that
this one you mean it's units yeah I
don't know what's the units of Q do you
know yeah exactly yeah if you choose
consistent units the units will be
consistent all right
so let's go one step further and do one
other thing let's put back the the
electric field let's put back the
electric field if I put back the
electric field then and let's take this
px over here let's put let's put an
electric field in the X direction all
right then it will not be the case
anymore
that px will be conserved why if there's
an electric field in the X direction
let's so let's go backwards
okay yeah let's put an electric field in
the X direction that means a potential
which looks like Q sorry a potential
which looks like some electric field
magnitude of the electric field times x
times minus X I believe if I want the
electric field to point in the X
Direction here E is just a magnitude of
the electric field and remember the
derivative of V sub X the derivative of
V with respect to X is e sub X
derivative of V with respect to Y in
this case would be zero so the
derivative of V with respect to X would
just be the minus sign here would be e
equals zero okay how does that affect
these equations well it doesn't affect
this equation the canonical momentum
conjugate to X is still we it affects
this equation here puts in minus V which
is plus e times
xqe times x q e times X now is the
electric field force or is the electric
field energy if you like now the
canonical momentum in this gauge is
still just P sub X it's just MX dot plus
QB why it hasn't changed because this
doesn't depend on the velocity but it's
no longer conserved its time derivative
is equal to the X derivative of this Y
that's Lagrange's equations Lagrange's
equations d by dt of partial of L with
respect to X dot but that's just piece
of X is equal to DL by DX and now we now
we do have an X dependence we just have
our QE x / QE okay so now instead of
having a conserved quantity we have the
new equation that let's write it out the
time derivative of px is MX double dot
plus Q B Y dot I've taken the time
derivative of P sub X and that's equal
to Q times E what about the other
equation what about the equation for P
sub y in the other gauge that's
unchanged because I haven't put any
force in in the Y direction so in the Y
direction there's been no change M y
double dot plus or minus Q B X dot is
still equal to zero alright now I want
to look for a particular kind of a
solution of the equations just for
simplicity only for simplicity I want to
look for a solution with no acceleration
let's just look for a solution with no
acceleration that means the particle is
somehow moving with uniform velocity
ordinary Newton's equations of course
have such solutions in any velocity as a
possible velocity well that's not true
anymore let's look for a solution of the
equations with no acceleration that
means that the velocity has to be
constant but it can't be any constant
let's throw these terms away and see
what we get from here we get that X dot
has to be equal to zero so if we plot x
+ y x + y X dot has to be equal to zero
the particle doesn't move to the left to
the right but from here
we see that y dot y dot is equal accuse
cancel let's see Y dot is equal to e
over B alright is the ratio of the
electric to the magnetic field so how
does the particle move it moves with a
constant velocity oh this is constant
notice that Y dot is constant and
therefore we check there really is no
acceleration I was correct in looking
for a solution with no acceleration but
what I find is a very special kind of
velocity namely a velocity in the X
direction with uniform motion equal to
the ratio of the electric to the
magnetic field it can be anywhere it can
move over here it can move over here but
it moves with a very specific velocity
in the Y direction now that's kind of
interesting because I put the force in
in the X direction I put the electric
field in the X direction II was in this
direction and it causes the particle to
move along the Y direction a bit of a
surprise that's called the Hall effect
if you have a magnetic field if you
first of all have a magnetic field in no
electric field then first question is
there any solution with no acceleration
motion in the circle has acceleration
but nevertheless there are solutions
with no acceleration namely a circle of
zero radius where the particle just sit
still that is a solution of the Lorentz
force law no acceleration no velocity we
have no electric field now no velocity
no acceleration that's a solution but
when we turn on the electric field in
the X direction the particle moves off
along the y axis that's called the Hall
effect
in fact if you take a distribution of
charge in the plane and you impose a
magnetic field into the blackboard and
you don't stir up the particles to give
them any special kinetic energy they
will just sit there but then if you turn
on an electric field in the X direction
the whole flow of charge will be along
the y axis and the velocity will not
depend on the charge but it will depend
on the ratio of electric to magnetic
field so that's called the Hall effect a
Hall effect what spin up the what Oh is
all right he I don't know about it I see
hmm it's not nor Denari computers is it
yes oh no okay
why it spins up anything I don't really
know but I can guess why it might put
looks like this okay that's the Hall
effect it's very similar and the
mathematics is very close but I won't do
it to the gyroscope effect the gyroscope
effect is if you have a spinning top
spinning along the x axis and it's in a
gravitational field pulling it down it
doesn't go down it processes in the
perpendicular direction same mathematics
very similar mathematics okay that's our
that's the motion of a charged particle
as described at least in these three
ways I will write down for you what that
I won't I'll assign it as a as a problem
I think it's one that I did in class but
to work out the Hamiltonian of the
charged particle and work out Hamilton's
equations and see that you get the same
answers of course all right so we've
been some through some good stuff
electric particles magnetic particles
gauge invariance now I want to come to
another form of mechanics it's really
very closely related to Hamilton's form
it's a kind of abstraction out of
Hamilton's forms and mechanics it's
called the form in terms of pour sone
brackets
now this is highly abstract there is not
something that I would bother teaching
you if it were not for the crack that
it's essentially related to quantum
mechanics it is much easier to
understand the quantum mechanical origin
of this than it is to understand the
classical mechanical origin of it one
usually thinks that classical mechanics
is something much easier to understand
in quantum mechanics well at some point
it becomes quite clear that the
classical mechanics is really some
overly complicated way of describing how
the motion of wave packets behaves in
quantum mechanics and in quantum
mechanics the whole idea Poisson
brackets is really very much easier but
nevertheless since we've come this way
without quantum mechanics and since it
is considered a part of classical
mechanics I'm going to teach you about
Poisson brackets I think we've already
talked about a little bit but there's an
entirely is it there's a entire
formulation of quantum mechanics of
classical mechanics which is based on
Poisson brackets Poisson brackets and
canonical transformations we will not
get the canonical transformations
tonight but we will get the Poisson
brackets right now let's begin with
Hamilton's equations of motion I know I
did this before but it never hurts to do
a thing again
is it now I always forget where the sign
goes P squared over 2m da out that's
equal to Q sub I that I'm now using P
and Q rather than P and X and the H by
the Q sub I is equal to minus P sub I
got these are the two Hamilton equations
of motion one for each component I or
rather two for each component I double
as many equations as the Lagrange form
but the equations of first order instead
of second order they contain only first
time derivatives alright let's take some
function of P and Q any function P
squared plus Q squared us is any
function of momenta and velocities some
things that we might measure I'm having
a hard time thinking of something that
would be interesting to measure but all
the interesting quantities that one
would do experiments on in a classical
mechanical situation of functions of
velocity and position or equivalently
functions of momentum and velocity and
position so let's take some arbitrary
function a of all of the peas and all of
the Q's all right so here's our phase
space and we have some function which
varies from place to place in the phase
space well-defined function when the
fluid moves through the phase space the
value of this function changes with time
not because the function changes with
time but because the point in the fluid
is moving through the fluid and changing
with time how does a vary with time
along a trajectory well that's not too
hard to figure out let's just
differentiate it a dot this is the time
derivative of a along the trajectory
assuming that a has no explicit time
dependence well that's going to be the
derivative
a with respect to P P sub I times P sub
I dot and when I write a thing like this
you can assume that I mean sum over I
I'm not going to explicitly write all
the summation signs plus the derivative
of a with respect to Q sub I times Q sub
I dot but now let's use Hamilton's
equations this becomes partial of a with
respect to P sub I times looks like
there's a minus sign here I should I
wanted to start with the plus sign so
let's start with the plus sign the a by
DQ sub I and Q sub I dot is the H by D P
sub I and then we have the a by the P
sub I times P sub I got but P sub i dot
first of all has a minus sign and then
it has the H by D Q sub I this
combination given two functions in this
case a and H but they could be any two
functions of positions in momenta
there's an object called the Poisson
bracket and this is it so let's write it
generally
for a general pair of functions let's
call them a and B now then by definition
the Poisson bracket of a and B first of
all represented by by an A and a B with
a comma in between sandwiched between
two squiggly brackets like that that's
called the Poisson bracket and it's by
definition summation I'll now write the
summation sign just just when writing
definitions I'll just to be very clear
summation over ie derivative of a with
respect to Q sub I the rivet of of B
with respect to P sub I minus derivative
of a with respect to P sub I derivative
of B with respect to Q sub I the Poisson
bracket of a with B is not quite the
same thing as the persone bracket of B
with a what's the relationship
not quite this is not true - as a minus
sign if I interchange the a and B these
two terms instead well if I interchange
they and B they switch and this would be
- so the Poisson bracket is an anti
symmetric expression involving any two
functions
that's number one so after you have to
remember which one you wrote down first
let's keep track of it but this is it
that's its definition what do we prove a
moment ago we proved that the time
derivative D by DT along a trajectory of
any function of a and P and Q is equal
now it's either plus or minus plus the
Poisson bracket of a with the
Hamiltonian so this of course includes
the to the two hamilton equations for
example supposing we chose for AE what's
the simplest thing we could choose for a
either P or Q I'm going to stop writing
sub I let's just do this for one p in
one queue but it's the same for many of
them just because I don't feel like
writing a subscript all over the place
let's concentrate on one p in one queue
but nothing I do is special to that case
you can figure out where to put the
summation signs and the indices um
what's the simplest thing that a could
be P well P is pretty simple
what else anything simpler
how about zero zero is very simple one
is pretty simple just a function one the
function which is everywhere is the same
on the entire phase space doesn't vary
on the phase space what's the time
derivative of the function one as the
particle moves through the phase space
zero because 1 is just one everywhere is
let's just check that supposing I plug
in for a I plug in just warning well the
derivative of 1 with respect to either
appear a Q is zero so quite obviously
the Poisson bracket of one with anything
is zero ok so the function 1 does not
change let's let me make you snow of
your brilliant suggestions of a moment
to go let's let a just be p and see what
we get
alright the left-hand side will just be
DP by DT the right hand side will be the
P by BQ the H by V P minus DP by DP the
H by DQ I have just substituted in for a
P the H by DQ sorry the P by the Q the H
by D P minus DP by DP the H by DQ now VP
by DQ is 0 P and Q are independent
variables and the derivative of one
independent variable with respect to
another is just playing 0 so this is 0
what about DP by DP that's just 1 and so
we find the 8 the P by DT is just the H
by DQ one of Hamilton's equations the
other Hamilton equation is the
e q by DT is equal to now it's
substituting DQ by DQ D H by D PE and
now we're going to have another term
where we'll have DQ by D P but that's 0
DQ by D P is 0 the Q by DQ is 1 so the
other Hamilton equation is also of the
form that the time derivative of an
arbitrary function of P and Q is in fact
equal to the Poisson bracket of the
function with the Hamiltonian so we're
sort of reducing now the physics all of
classical physics to a kind of algebraic
structure involving Poisson brackets let
me tell you a few properties that
Poisson brackets have they're actually
sufficient to recover any Poisson
bracket that you might be interested in
I'm just going to write down some
properties first of all we've already
encountered one property a with B is
minus B with a you can think of this as
a kind of axiomatization of mechanics
based on Poisson brackets why is it
interesting to do so it isn't until you
come to quantum mechanics but it is very
very centrally well it may be
interesting but it really achieved its
sum its full glory shall we say when it
was realized that how closely it is
connected to the algebraic structure of
quantum mechanics a with B Poisson
bracket of a with B is minus the Poisson
bracket of B with a
all right how about the Poisson bracket
of P with P now we could write I'm going
to do P with P but then I'll tell you
what the answer is if there are any
number of variables what is P with P
well that's equal to partial of P with
respect to P partial of Q with respect
to Q sorry partial of P with respect to
Q partial of Q with respect to PE hmm
yeah you're right
good I don't even need to do it the
Poisson bracket of p with p is equal to
minus the Poisson bracket of p with p
from the above rule here but anything
anything which is equal to its own
negative is zero so Poisson bracket of P
with P is equal to the Poisson bracket
of Q with Q and is equal to zero now we
can go a little bit further if we have a
system of in coordinates then we can
take the Poisson bracket of one p with
another Priya with another one I with J
will still get 0 it's easy to prove and
we'll also get 0 for the cubes you can
see there's a symmetry well not an exact
symmetry but some symmetrical
relationships between P's and Q's
they're on the same footing
how about P with Q or let's go to Q with
P plus own bracket of Q with P
that's equal to the partial of Q with
respect to Q partial of P with respect
to PE minus partial of Q with respect to
P partial of P with respect to Q well
the derivative of a Q with a P or a P
with a Q is zero so this is zero the
derivative of a Q with a Q or a P with a
P is one and so the Poisson bracket of Q
with P is just one more generally if I
have Q sub I and P sub J then that's 0
if I is not equal to J for example if we
take the Poisson bracket of the x
coordinate with the Y momentum that's 0
so this is 0 if I is not equal to J and
it's 1 if I is equal to J the symbol for
that is the Kronecker Delta symbol IJ
delta IJ Delta IJ is by definition 0 if
I equals J not equal to J excuse me 1 if
I equals J
you
now notice that there is a little
asymmetry Q with P is minus P with Q so
if I were to put P with Q I would get
minus 1 instead of one so you have to
remember Q if Q is first and P is second
then the contour in another commutator
the the Poisson bracket is 1 but
otherwise there's a great deal of
symmetry in fact the general structure
of classical mechanics is that whenever
you interchange Q and P s is always a
minus sign it's called the symplectic
structure of classical mechanics okay
all right some other rules here's
something how about the Poisson bracket
of Q sub I well I'll do it without
indices and then I'll tell you the
general answer the Poisson bracket of Q
with any function of Q and P let's do P
P with any function of Q and P
[Music]
taking the Poisson bracket of a function
with respect to P does something to that
function let's see what it does it gives
you first of all DP by D q DF by D P
minus DP by D q DF sorry DP by D q DF by
DP - right the p by DQ the f by DP
that's right
the PDP
thank you DF DF by DQ now the first time
is zero the P by DQ is zero the second
term the P by BT P is one and so it just
gives - DF by DQ in other words role
taking the Poisson bracket of an
arbitrary function with respect to P is
the same as taking its derivative with
respect to Q with the minus sign if we
wanted to do this with many variables it
would be the Poisson bracket of a
function with P sub I gives you the
derivative of the function with respect
to Q sub I so taking Poisson brackets
with P's and Q's is basically the act of
taking the derivative with respect to P
and Q likewise
is equal to plus DF by D P sub I so
taking the Poisson bracket with a Q
differentiates with respect to P taking
the Poisson bracket with P
differentiates with respect to Q but you
have to keep track of the minus signs so
um actually I think there's that there's
one more ingredient that you need to add
in that would allow you to have a full
algebra of Poisson brackets that could
be used to oxidize classical mechanics I
don't particularly like ax Emmett izing
anything but knowing these rules here is
not quite enough to tell you what the
Poisson bracket just knowing these rules
without knowing supposing nobody told
you what a Poisson bracket was they told
you that Poisson brackets satisfy these
relationships here hmm no they it is
something like that um yeah that that is
there is a linearity I'll write that
down in a minute they told you the such
thing called Poisson brackets they told
you a little bit about the Poisson
brackets these things here and then also
told you that mechanics can be
summarized by saying that the time
derivative of anything is the Poisson
bracket of that thing with H how much do
you need to know about Poisson brackets
that you could start operating and
completely forget where everything else
came from well you need to know I think
I think whoever said linearity I think
you do need that all right so here's
here's linearity it hadn't even occurred
to me but yes a plus B all right
linearity means the following it means
that if you take any number times a and
take its plus own bracket with
to be it's just the same as multiplying
the Poisson bracket by alpha so that
means if you multiplied a by two its
plus home brackets with everything would
just be twice as big no surprise there
another thing if you take the Poisson
bracket of a plus C with B that just
gives you the plus own bracket of a with
B you can check these things now check
it from the definition a with B plus the
plus own bracket of C with B these two
things together are called linearity the
linearity of the Poisson bracket with
respect to one of the entries if you
double it you double the Poisson bracket
if you add two things together and take
that plus some bracket with B it gives
you the sum of the two plus on brackets
that's almost enough to determine the
Poisson bracket of anything with
anything else any function of P and Q
with any other function of P and Q there
is one more element what happens if I
take a times B times now not plus a
times B with the Plus on or with with C
well let's work it out and then abstract
from what I worked out the general
answer what is this this is equal to
derivative with respect to Q of a B
times derivative with respect to P of C
minus the derivative with respect to P
of a B times the derivative of C with
respect to Q but now all we have to use
is the product rule for derivatives the
product rule for derivatives says that
this is a times the derivative of B with
respect to Q times the derivative of C
with respect to P now let me not finish
this expression I'll come back to the
other half of it but let me write it let
me combine it with a times the
derivative of B with respect to P times
derivative of C with respect to Q that's
the term that I get when I do the
derivatives by pulling out a and
differentiating B then there's another
term where the derivative hits B and not
a that will be very similar except a and
B will be interchanged so that will be B
times the derivative of a with respect
to Q derivative of C with respect to P
minus B derivative of a with respect to
P derivative of C with respect to Q
right if you summarize it all and put it
all together this piece here is just a
times the Poisson bracket of B with C a
times the derivative of times the plus
own bracket of B with C and this one is
just B times a plus on bracket of a with
C
all right so the Poisson bracket of a B
with C is a times it persone of B with C
plus B times the persona of a with C
that believe it or not Plus knowing some
a few fundamental Poisson brackets
between P and Q enables you to calculate
any Poisson bracket that's enough to get
started
calculate sums of things peas plus Q's
calculate products and eventually to
calculate any function with any other
function I'll let you experiment around
with that and see that that you can
calculate or at least any functions that
can be written as sums of products of
functions of P and Q even infinite sums
and products can be computed so these
basic rules linearity the product rule
the anti symmetry and the fact that
Poisson brackets of P with Q are 0 or 1
and so forth those completely determine
the entire family of plus own brackets
of anything with anything else and if
you then add that the rules of the game
are at the time derivative weird like
way that I put it I think I've erased it
that the time derivative of anything is
plus on bracket with a Hamiltonian that
is another formulation of kind of
axiomatic formulation of classical
mechanics that of course is not
independent of the others that follows
from the others but it is another
another elegant the French were great at
this kind of super elegance it wasn't a
friend to invented quantum mechanics so
they were very very good what's that
yeah although was a German Germans well
not as good at it's super elegance yeah
this was before Bourbaki Bourbaki didn't
exist of course this was before Bourbaki
but there's no question the French were
extremely good at finding these very
elegant and Hamilton was another
Frenchman and Newton was another
Frenchman and but the people who really
well a person who really the two people
who really put poisson's and all this
stuff on the map or Heisenberg and Dirac
and not in the context of classical
mechanics so this would be next quarter
not this quarter
[Music]
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