This is the last and final
chapter of this subject, and I think it's,
again--we're starting by backing off a little bit towards
the end of last class. I think that there are maybe
parts of it were maybe difficult for you to follow.
Let me remind you of what we were trying to do.
We had learned that space-time is described by four
coordinates, so that if you're following some particle or some
event, you label it with four numbers
c, t, x, y and z,
which I want you to let me combine into the single vector
r. They are written as
x_0, x_1,
x_2, x_3.
Now, I'm of two minds about x_2 and
x_3; sometimes I carry them around,
sometimes I don't. x_1 is just
our familiar x; x_2 is
y and x_3 is
z. It's possible to arrange all
the motion to take place in the x direction,
in which case y and z don't change when you
go from one frame to the other, so I don't usually bother with
that. But sometimes I put it back
because if you have only two components, x_0
and x_1, you might say,
"Hey, the world is two dimensional."
So, I want to keep reminding you now we think of it terms of
four dimensions. So, at least for that reason
sometimes they carry all the three components of position,
sometimes only one. But I'm sure you guys can
figure out from the context which is which.
So, going back to the simpler situation of only x and
time, the point of those coordinates is that if you went
to a moving frame, x_1 prime will
be x_1 - βx_0,
divided by the square root. And x_0 prime
will be x_0 - β times
x_1, divided by that square root,
where β is v/c.
Sometimes it's useful to introduce γ,
which is [reciprocal of square root of ]
1 - u^(2)/c^(2). Also to prevent all confusion,
u will be the velocity of your frame relative to mine.
v will stand for the velocity of some particle that
we are both studying. So, let's reserve the symbol
v for motion of actual particles, which is why we are
there. We are all observers but we're
observing something. That something has a velocity
that I'll call v and you will call w,
and u is the speed relative to your frame and my
frame. All right, so this is the
notion of space-time because space and time are getting mixed
up when you go to a new frame of reference.
But even in this world where nothing is sacred,
the following combination is very important,
which we denote as X dot X.
There's a new dot product now in space-time.
The new dot product will be the square of the time-like part
minus the square of the space-like part.
If you wanted to bring in all the components of space-time,
then you can modify it to x_0^(2) - r dot
r. r dot r will be
x^(2) + y^(2) + z^(2). So, you can use either notation.
Whenever possible I just keep two, but you've got to remember,
this is why it's a four-dimensional dot product.
And this is the same for all people.
So, I can calculate it, you can calculate it,
our x_0 and x_1 will
differ, but it's the property of the
Lorentz transformation, that if I square this guy and
subtract it from the square of this guy,
on the right-hand side some magical things will happen and
I'll get the same combination without the primes.
That's like the length of the vector.
And that's called the space-time interval.
Let's call that s^(2), the space-time interval.
If it took two events separated in space by Δx and
separated in time by Δt, then we like to use the symbol
Δs^(2) for the space-time interval.
That's the same thing applied to the differences rather than
to the coordinates themselves. And this is invariant.
Invariant is a term mathematicians use for saying
same for all people, or unaffected by Lorentz
transformation. It's the analogue of how far
you are from the origin in the old days, when I took two
frames, one which is the standard
xy, the other is the rotated xy.
If you go to the rotated frame, the components x and
y change, but how far you are from the
origin, or what's the length of the vector is the same for all
people. More generally,
the dot product of two vectors is invariant under rotations,
and this is the new dot product.
I use the symbol dot product, but I don't draw arrows,
because if I don't draw any arrows and use capital letters,
it means it's a four-vector. That's the X vector.
Then I said, "Let's try to reason by analogy
and try to invent a momentum vector."
How did we invent the momentum vector in the old days?
We said, let the particle move a distance Δr,
vector Δr, in a time Δt,
and a divide by Δt, and take some limit.
But I know this is a vector because Δr is a
difference in two vectors. Where the particle is now,
and where it is later, the difference of two vectors
is also a tiny vector going from here to there.
Dividing by a scalar, by a scalar I mean something
that is not affected by rotations, means taking a vector
and multiplying it by some number.
Therefore, that's also a vector, so this ratio is also a
vector, and the limit of the ratio is also a vector.
I'm sorry, I forgot the mass here, and we call it the
momentum vector. So, you can manufacture vectors
starting with a position vector by taking derivatives with
respect to something like time that is the same for all
observers, where observers,
I meant in the old days, people with an axis rotated
anyway you like. I wanted the same thing now in
this present problem, but I now want to define
obviously a four-dimensional vector.
So, to do the four-dimensional vector, my plan is to define
something called momentum, which is now going to be the
change in this fourth position divided by something.
The question is, "What do I divide it by?"
I cannot divide by Δt. In the old days Δt was
a sacred number, same for all people.
Now Δt is like Δx or Δy,
it's one of the components, and you don't divide the change
in the vector components by the change in one of the components
to get a velocity. You've got to divide by
something that's going to play the role of time,
and have the same value for all people.
Well, we sort of know what that quantity is.
Obviously, if a particle is moving from here to here in
space-time, this is x_0 = ct and
this is x_1. It moves the distance
Δx_1 in a time Δt, and we know that for
this piece of its motion the following quantity,
Δs, which is [root of] cΔt^(2) - Δx^(2),
is the same for all people. So, I should really be dividing
by this. Let me write it slightly
differently as cΔt times the square root of 1 minus
Δx over Δt^(2) times 1/c^(2).
So, we are asked to divide by this crazy thing.
Well, let me write this crazy thing as c times
something I want to call dτ,
is dt times the square root of 1 minus Δx over
Δt^(2) times 1/c^(2).
Then I said, "We can give a meaning to this
thing." In fact, let's take the
derivative with respect to τ not s.
They differ only by this factor c, which is a constant so
it doesn't matter. τ has units of time,
and you can ask, "What does that time stand
for?" It's not the time according to
me, but I argued that it's the time according to the particle.
Why? Because you can evaluate this
quantity in any frame you like. So, go to the frame moving with
the particle. If you're moving with the
particle, and if you're here now and there later,
later is certainly later, but here remains here for the
particle. It doesn't move in its own
frame. So, Δx will vanish,
therefore Δs simply reduces to c times the
time elapsed according to the particle.
So, dτ is the time elapsed according to the
particle itself. According to the particle.
That's what we're going to divide by.
So, it's a very clever way to circumvent the question of what
should I take the derivative respect to,
my time and your time all are variable, we don't agree on it.
But let's ask how much time elapsed according to the one
thing we are studying, and let's divide by that time.
That answer will be the same because you and I may not agree
on how long it took to go from here to there,
but we can all ask how long did the particle say it took to go
from here to there. We're talking about the same
thing, and we'll get the same answer.
To get the particle time you've got to take the time according
to you and me and multiply it by this factor.
This factor, by the way, is nothing but
usual time dilatation. Because the particle says it
took time dτ because the clock has slowed exactly this
amount, I will say it took longer by this amount.
So, let's rewrite this in a way that's going to be more helpful
to us. Take the ratio and write
dt over dτ is 1 over the square root of 1 minus
v^(2)/c^(2). I'm going to use that in a
minute. That just says the time
according to me and the time according to the particle are
related by the time dilatation factor that's controlled by the
velocity of the particle. Now, we are ready to define the
four-momentum p. The four-momentum p is
going to be m times dx_0/dτ times
mdx_1/dτ. I'm not going to bother with
the other components. You can always put them back. But x_0 you
remember was ct. So, this really becomes
mdt divided by dτ, mdx divided
dτ. This is guaranteed to be a
four-vector. That means, if you know the
components of this in one frame of reference,
and you call one of them as p_0,
and you call this one as p_1.
Then, in a frame that's moving relative to your x speed
u, our p_0 prime will
be p_0 minus β times
p_1 over the square root,
and p_1 prime will be p_1
minus β times p_0 over the
square root. It will transform just the
components of x. Yes?
Student: For p_0,
where does the c go? Professor Ramamurti
Shankar: Where did the c go?
No, it did not go, you're quite right.
The c, x_0 the
ct, you're quite right. It is c times dt.
You need that; then, this is correct.
So, we've got a four-vector, but we don't know what it
means. Because we don't know what
dτ means. Of course it's the time
according to the particle, but I don't care about the
particle. I'm studying this particle,
I've got my own clock, I've got my own meter sticks,
I want to think in terms of time according to me.
Period. Well, we can do that.
We can do that by going to anything where you take a
derivative such as dx_1/dτ and
you can write is as dx_1/dt times
dt/dτ. dx_1dt is
something we understand, it's just the velocity
according to you and me. If you use that rule here,
you will find the momentum has got components m times
c, dt/dτ is dt/dt,
which is 1, times dt divided by dτ,
which is this factor here. And the second thing will be
mv divided by the same thing.
In other words, you can trade τ
derivatives for t derivatives everywhere,
provided you attach this factor.
That's the ratio of the τ derivative to the
t derivative. So, now I have a four-vector.
I still don't know what it means, but at least it is
written in terms of things with a direct operational
significance for me. I see a particle moving,
it's got a velocity, put that in and I get this
thing. Apparently, the relativity
theory says this should play the role of momentum,
but a four dimensional momentum in your new world.
If you want to bring the other components, we'll put an arrow,
you can keep it or leave it; it doesn't matter. Now, notice once again
that--Okay, I'll continue some more.
I've already done this last time, but let me remind you.
We looked at this guy and said, "Look, at smaller velocities,
when v/c is very small, the denominator is just 1.
Numerator is mv. And then I said,
"Uh-huh, that's the momentum." So, it must be that with all
these factors, this is the momentum in the new
relativistic theory. In the new relativistic theory
momentum is not just mass times velocity.
Mass times velocity divided by all this stuff.
The theory says, "Use that as your momentum."
That's good. Then, we say,
"Okay, that's my momentum, who is this?"
I expanded this out for small velocity.
If you put velocity equal to 0, you get mc.
That looks like just the mass and we don't know what that
corresponds to. It's the mass,
nothing kinematical. But if you expand the square
root a little further out to one more term, we found that the 0
component, p_0,
looks like mc times 1 plus v^(2) over
2c^(2) plus dot dot dot. That becomes mc +
1/2mv^(2) times 1 over c plus dot dot dot.
And now I finally have something I understand,
because ½ mv^(2) is familiar kinetic energy.
But I don't like this annoying c here.
If I multiply both sides by c, I learn that c
times p_0 is mc^(2) plus ½
mv^(2), plus more stuff that's more and
more powers of v over c.
But this I argue. If that's the kinetic energy,
everything is an energy formula, so the whole thing
stands for energy. So, I learned that the first
component, of P is energy divided by c.
So, I come back here and I write this as E/c. So, for all of you,
or for many of you who found some of this mathematics
difficult to follow, it depends.
For me it was love at first sight.
This is really when I said, "I want to go into business."
For some reason this worked for me.
But it doesn't matter. What you have to know is what
I'm going to tell you now. Operationally,
I'm not going to hold you responsible for this part of the
material at all. For me, it's part of explaining
how logically starting with the very first axiom--first two
axioms you can deduce everything.
But now, I want to tell you operationally what you guys are
supposed to know. You are supposed to learn the
following two things. In the old days,
we had time and we had space; they have been united into a
single quantity; (ct,r),
which we write as x_0,
x_1 etcetera. In other words,
the three components of a vector joined with one quantity
that looked like a scalar together to form a
four-dimensional object. Now, we are learning that there
is another four-dimensional object, one of which is just the
momentum, and the other is energy divided by c.
And we want to call that as p_0,
p_1 etcetera. Again, the familiar momentum,
which is a vector, joined with another quantity,
which the best way to think about it is kinetic energy,
plus rest energy, plus other corrections.
But it's a scalar; it doesn't depend on
orientation of axis in the old days.
But now under Lorentz transformation,
namely, when you go from one frame to another,
E and p will all mix with each other just like
x and t will mix with each other,
but this is the new four-vector. It's called the energy-momentum
four-vector. Just like that's called the
space-time four-vector. This is the energy-momentum
four-vector. So, all over relativity you
will find three objects, which used to be part of a
vector, will combine with the fourth one, which used to be a
scalar; together, they'll form a new
four-dimensional vector. So, we should know the
following. When we are studying particle
dynamics, we have to know in more detail--in fact,
let me dedicate a whole board to that, it's very important.
The new definition is that it's E over c and
p. Yes?
Student: [inaudible]
Professor Ramamurti Shankar: Well,
it looks like mc^(2) plus ½ mv^(2) plus other
stuff, right?
So, I recognize ½ mv^(2) as an energy;
then, the other powers v^(4)/c^(4),
they're all corrections obviously to the energy at
higher velocity. Just like momentum,
even in relativistic theory was not mv.
It started out like mv, but when you expand the
denominators and powers to v/c you get more and more
corrections. I identify this as an energy
because if I go to very low velocities, it reduces the rest
mass, which is never going to change,
plus a ½ mv^(2), which played the role of
kinetic energy. That's why it has units of
energy, and one term in it looks like energy.
So, the whole thing, the theory is telling you,
it's all just the energy of the particle.
In the old days when velocities were very small,
we saw only that two terms are written: ½ mv^(2) and
mc^(2). mc^(2) we didn't bother
with because it doesn't change in any collision.
You add the mc^(2) before, each particle has got
its own mc^(2) before and after;
they simply cancel. So, the ½ mv squares
before and after that would equal, and we use that to solve
collisions. Now we will be told you should
use the conservation of this energy and the conservation of
this momentum in a collision. Okay, so the first thing is
this is E and this is p, and if anyone says,
"What is E?" It is mc^(2) divided by
1 minus v^(2)/c^(2), and p in one dimension
is mv divided by 1 minus v^(2)/c^(2). This you must know.
So, momentum and relativity looks a bit like the old
momentum but it's got this ubiquitous factor.
And energy looks really weird; doesn't look like anything on
Earth, but if you expand this in powers of v,
you will see the terms we wrote down.
Rest energy plus everything else you can call energy of
motion, and the first term of the energy of motion is the
familiar ½ mv^(2). In other words,
this is how the world always was.
It was revealed to us in the old days when we only looked at
low energy particles as mc^(2) + ½
mv^(2). And we couldn't see the dot dot
dot terms because v/c was so small, the fourth power and
sixth power and eighth power of v/c were too small for us
to detect. Likewise, the momentum of a
particle was never mv; it was mv divided by all
this. But for small values the
v/c looked like that, and they thought,
"Hey, that's what is really happening."
So, we were looking at an approximation to these
quantities in the old days when velocities were small,
and as velocities go up, those quantities aren't useful,
this is what the theory says should play the role of the new
four-momentum. So, the significance of this
being the forward momentum, first of all is what is written
in the top right corner there; namely, if you know the energy
and momentum in one frame, you can find the energy and
momentum in another frame of reference moving relative to me.
For that purpose, you should take the energy and
divide by c, you should call it
p_0, the momentum can be left as it
is, and this will behave like x_0,
that will behave like x_1,
under Lorentz transformation. But there are other
consequences of the fact that P is a four-vector.
The first consequence it this: you remember the dot product of
a vector with itself is the same for all observers.
So, what's the dot product of P with itself?
Well, before I do that, let me remind you in general,
in relativity you will have four-vectors:
(A) you'll have one part that looks like time,
and one part that looks like an ordinary vector.
And (B), is another four-vector, it's got a part
that looks time and another part that looks like an ordinary
vector. It's a union up of former
vector with a former scalar. And they will transform under
Lorentz transformation just like x_0 and
t. But the following quantity
A dot B, which is A_0
B_0 minus the usual dot product of the spatial
components, is invariant.
That means all people will agree on this quantity. And a special case is A
dot A, which equals
A_0^(2) minus the length of the vector part,
which for our purpose will be just A_0^(2) -
A_1^(2). This is all reasoning by
analogy. Yes?
Student: [inaudible]
Professor Ramamurti Shankar: No.
Quaternion are a slightly different object.
You can form with these four numbers something that you may
call a quaternion. Right.
They were invented before relativity by Hamilton,
and they are of great mathematical beauty.
They're a generalization of numbers.
You know, we generalize real numbers to complex numbers;
then, we went from complex numbers to things called
quaternions. They are also numbers;
you can multiply them by each other, but the product AB
differs from the product BA.
They're called hyper-complex numbers where the order in which
you multiply them matters. Okay.
So, now let's ask what is the invariant length square of the
momentum four-vector with itself.
That means, you've got to do p_0^(2) -
p^(2). That means you are trying to
take m_0--I'm sorry, you're going to take
mc over this factor square,
minus mv over this factor square.
Well, let's try to do this in our head, guys.
m^(2) is going to come out of the whole thing,
you agree? On the top, I have got 1 over 1
minus v^(2) over c^(2),
yet I have minus v^(2) over 1 minus v^(2) over
c^(2), so they will all cancel giving
you mc^(2). Uh, m^(2)c^(2).
[Note: Here, the Professor has started
dropping the subscript on m_0,].
So, the four-vector of any particle has a length that does
not at all depend on how fast it's moving.
It depends only on this. When you take a velocity vector
for a particle, it doesn't have a fixed length.
The particle can be moving slow, it can be moving fast.
So v_x^(2) + v_y^(2) could be
whatever you like depending on how it's moving.
But this relativistic length square of a four-vector is
independent of velocity because the velocity here when combined
with this number just cancels out.
Well, that's just the way it is. Now, if you're not good in
algebra, there's a shortcut for finding this.
Let me tell you what the shortcut is, and we love the
shortcuts because they're very helpful.
And the moral of the story is also illustrated by the
shortcut. I want to find P dot
P. It's an invariant because it's
like a dot product, so anybody can calculate it.
How about somebody riding with the particle?
If you're riding with the particle your 0 component is
m_0c, is mc,
there is no velocity for you, and you have no momentum.
As the particle has no momentum as seen by the particle so the
components of the vector P become mc and 0.
What's the length square of this vector?
Well, it's the square of this guy and case closed;
there's nothing here to subtract from here.
That's a quick way to verify that this length of vector with
itself is in fact m^(2)c^(2).
In other words, the trick is ride with the
particle and calculate the vector.
Then, the vector has only one component, namely mc.
It doesn't have a spatial part because it's not moving
according to itself. That's where P dot
P is m^(2)c^(2). So, people write it also as
follows: remember p_0 was
E over c, this is just P^(2)
is m^(2)c^(2). This is one way to write it
explicitly, but here is how people write it:
multiply both sides by c^(2) and write
E^(2) is equal to c^(2)p^(2) -
m^(2)c^(4); that deserves a box.
That's how we remember it. I have just rearranged this
equation. P dot P is
m^(2)c^(2). By definition P dot
P is the time-like component square minus the
spatial component square. We agree the time-like part is
E/c; the spatial part is just the
momentum. So, I've just rearranged it to
get this. And it says the energy of the
particle depends on its momentum, if it comes complete
to rest, you kill this term, that's E = mc^(2);
that's the rest energy of a particle. Now, I am almost finished with
the kinematics except for one thing.
There are strange particles in the world called photons.
A photon has no rest mass; therefore, if you naively went
by any of these formulas, if you killed the mass of the
photon, you might think energy and
momentum vanish because m is going to 0,
but we all know enough mathematics to know that if it's
0 on the top, you could beat it if there's a
0 in the bottom. That will occur only for v =
c. So, particles moving at the
speed of light do have energy and momentum,
but they don't come from a mass divided by something,
or mass times velocity divided by something.
They simply have energy and momentum that obeys this
equation without an m on the right-hand side.
For a photon, E^(2) equal to (c
times p)^(2) is what a photon obeys.
So, photons can have energy, and photons can have momentum,
but their energy will be c times their momentum so
that if you find the invariant square of the photon momentum
you will get 0. So, photons are particles whose
momentum square is 0. But the square is the
four-dimensional square with the time square minus the spatial
square. So, let me summarize once more
for people who may not want to follow every detail.
The practical part of what you must understand from what I've
been teaching you is that when you study a single material
particle, which has a mass,
so that it can be brought to rest and made to sit on the
table, we associate with that at
four-vector. The first part of that is the
energy over c; the second part of that is
called the momentum, and the explicit formula for
these are m_0, or mc divided by this
and mv divided by this. And one can explicitly verify
that P dot P is m^(2)c^(2).
This is a summary of all that I've said.
If you don't know anything at all, know this. Yes?
Student: [inaudible] Professor Ramamurti
Shankar: Yes. A capital P without any
arrows on it is supposed to be a four-vector,
and the dot product of it is defined as the square of the
first component minus the square of the second one.
That is correct. And for photons,
we always like to use the symbol K for the momentum
of a photon rather than P.
Luckily, people agreed more or less by convention,
photon momentum is called K,
and it's called Q by some people, but we call it
K. And the energy of the photon is
when we call , the momentum of the photon is
going to be called k. And the condition is =
kc. And there's' the condition
K dot K = 0. K^(2) if you like,
shorthand for K dot K, is 0 for photons.
A photon is a particle whose square of the four-momentum is
0. Material particles that can
come to rest have a square of the fourth momentum,
which is m^(2)c^(2). This is something you have to
know. Another thing you have to know
is that when I write them in this notation as four-vectors,
you should know the following: if you know the components,
p_0 and p_1 in one
frame, and you move relative to
S with speed u, p_0 prime and
p_1 prime will be related to
p_0 and p_1 by the
Lorentz transformation. Just like x_0
prime and x_1 prime are related to
x_1 and x by that formula on the
top left. I'm going to do a lot of
problems for you so you get used to doing relativistic
kinematics. Last thing I want to tell you
is why do we take this definition of momentum.
Okay, you agree with the theory naturally gave out something
that it said deserves to be called momentum.
But what's so nice about this? Why is it important that the
momentum-energy form a vector? I'm going to give you a reason
for it, and try to follow the reason.
The whole advantage of energy and momentum--What is the big
deal about energy and momentum, let me ask you that.
Yes? Student:
[inaudible] Professor Ramamurti
Shankar: They are conserved. They were not useful unless the
special utility came from the fact they're conserved.
So, you can put a restriction on possible collisions.
You add the momentum before, it's got to agree with momentum
after. You add the energy before,
it's got to agree with the energy after.
Of course, in relativity, energy and momentum have been
assembled into four-vector, so conservation of momentum
will mean really four equations. The zero part will be the
energy before, will be the energy after,
and the three special parts will be x component and
momentum. Before will be x
component after and so on. That's all very good,
but you've got to make sure that if a momentum is conserved
for me, forward momentum, it must also be conserved for
you. There's no use having a
quantity that's conserved for me and not conserved for you.
So, by the time you bring up something and say,
"We're going to use it as a conserved quantity,"
you've got to make sure the conservational law is not
limited to one observer. And that's what I want to show
you. I want to argue without writing
any questions that if energy and momentum are conserved for me,
they are going to be conserved for you, because they are
vectors. Let's understand why.
Suppose you've got a collision; a lot of particles come in;
one particle brings momentum-energy
P_1, other one brings in energy and
momentum P_2. And at the end of the collision
maybe they become particle 3 and particle 4 and particle 5.
In relativistic theories, if two particles come in,
you can have 36 particles coming out.
In fact, that's why they built accelerators.
If two went in and two came out, it's a waste of time.
You can smash any two things and with that energy you can
produce matter. So, it's not a mistake I've
made. I do mean that sometimes two
can go into three, or four or five.
Let's take an example where this is true.
Each P is a four-component object like this.
Suppose I say incoming equals outgoing.
Now, if you see the whole collision from your frame of
reference, then I maintain that the numbers you get,
P_1 prime and P_2 prime,
will also equal P_3 prime plus
P_4 prime plus P_5 prime.
That's guaranteed, and I want to tell you why.
Suppose these were ordinary vectors, and let me make life
simple by saying that P_1 +
P_2 became P_3. What that means is one vector,
P_1, another vector,
P_2, added up to one vector,
P_3. Except these are not ordinary
arrows, these are arrows in four dimensions.
The fact that two arrows add up to a third one,
are equal to a third one, means they form a triangle.
Now, if they can form a triangle for me,
and you can stand on your head and do what you like,
it will still be true. Even though it looks like this,
it is still true that this guy plus this guy is equal to that
guy. If two vectors add up to a
third one; namely, if a sum of two vectors
is equal to another vector; namely, the left-hand side
vector is equal to the right-hand side vector,
going to a rotated frame will continue to say that the rotated
vector is equal to the rotated vector.
That's because equal vectors upon rotation will remain equal
vectors. That's the whole point.
The total incoming momentum four-vector, call it P
incoming, and this is the outgoing momentum,
call it the P outgoing. If P incoming is equal
to P outgoing for me, so two vectors that equal in
space-time, for the rotated observer they will be equal.
Do you understand why? Because my total incoming
vector will undergo Lorentz transformation to become your
incoming vector. My outgoing vector will undergo
Lorentz transformation to find your outgoing.
But if the input of the Lorentz transformation is the same,
the output will also be the same because you will be putting
the very same numbers to get out the output numbers.
That's why it's very important that this momentum that we have
be used, because it has a chance that if it's conserved in one
frame, it's conserved in all frames.
If you went and said, "You know what?
I don't like your formulas. I'm really wedded to this
formula for momentum," what you will find is that even if it's
conserved in your frame of reference,
in a different frame of reference, that combination,
sum over mv, will not be conserved.
Okay, that's the final property of four-vectors you've got to
use. Again, if you don't follow the
details, you must know that in collisions that we are going to
study we will add up all the [energy-momentum]
four-vectors coming in, we'll equal them to all the
four-vectors going out. That's it.
So, now I'm going to do sample problems to give you a feeling
for that. So, as usual,
they will start out with the simplest one and then try to
make it more complicated. So, the first problem is this.
I'm standing here, I've got some mass m,
and a photon is coming towards me with some momentum P.
And this hits me on the face and I absorb the photon.
It's completely possible for the atoms forming the body to
absorb a photon. They absorb it;
so in the end there is no photon.
There is me, but possibly with a new mass,
m prime. In relativistic collision,
there is nothing called inelastic collisions.
Remember the old days when one thing came and stuck to another
thing, and they moved together? Momentum is conserved,
but energy is not. There is no such thing.
Energy will always be conserved in relativistic theories.
In other words, even though this will become a
single unit, you might say, "What happened to the energy?
Previously, I had a moving photon, I got nothing moving
here." Well, first of all,
I will be moving; secondly, I will have more mass.
The best thing is to consider two particles coming in opposite
directions. They collide and form a new
mass. In relativity,
there is no momentum in the final particle,
it is not moving at all. But it will be more massive
than before, and that extra mass times c^(2) will be the
extra energy brought by the particles that are formed.
So anyway, the question here is, "What is m prime?"
That's what we have to figure out.
So, the equation we will use is, I will call my momentum as
P. I will call
P_1 as the initial one, and
P_2 as the final one.
Let's write down all the equations we can;
the only equation in town is K + P_1 =
P_2. Everybody is a four-vector,
so don't be fooled. It's four equations.
But in this problem--;luckily, since there is no motion
perpendicular to the x axis, we can have this x
momentum; there's no need for y
and z. Because the photon comes this
way, and I can start moving this way, but all the action is in
the x direction. So, here it will be actually
two-dimensional vectors and not four.
There's nothing vertically before or after,
it will be 0 = 0. So, what are those equations?
Before you do that, let's write down the momentum
of the photon first. Momentum of the photon has got
some energy [should have said p_0]
/c, and it's got some momentum
k, where = kc, you've got to remember that.
P_1 has got an energy which is mc,
and no spatial momentum because I'm not moving. At the end of the day
P_2, I am moving,
but I also have a new mass, and I have a new velocity,
which we don't know, and I have m prime
v divided by the square root of 1 minus
v^(2)/c^(2). Now,all of you should be able
to write this part down. If you're not,
you should take a minute, you should think about it,
you can stop me, you can question me.
But if you cannot do this, you cannot do any of the
problems. You've got to think,
"Would I have done this? Does it make sense to me?"
The bookkeeping is all in the top.
When you've got a material particle, it's first term,
0 component, will be m times c
divided by the square root of all those things.
But if the particle is at rest, you only have mc.
There is no momentum. The final object,
me--After I got hit by the photon, I am recoiling;
so, I will have a velocity. I don't know what it is,
but my energy in terms of that, the first term,
p_0 will be this;
this will be P_1.
Now, the Law of Conservation of Energy and Momentum,
namely K_1 + P_1 = K_2 +
P_2, will really be saying all the
energy parts of the 0 components will add up before and after,
and the first components will add before and after.
In other words, /c plus mc,
will be equal to m prime c over this.
Then, k plus 0 will be equal to m prime v
divided by this. This is not any different from
the old days. You had Law of Conservation of
Momentum, you had Law of Conservation of Energy,
except energy is not ½ mv^(2);
energy is this crazy thing. Plus, even if a particle is not
moving, don't forget to give it mc as the energy;
that's the new part. So, what is the question I
asked you? I said, "What is m
prime." So, your job is to take these
two simultaneous equations, this one and this one,
and solve for m prime. So, what should I do?
Can you think of a strategy that will give me m
prime? Student:
[inaudible] Professor Ramamurti
Shankar: Into this one here? Yes, you still have to solve
for m prime, and you also don't know the
velocity here. Remember, you have to solve for
the velocity, and you have to solve for the
m prime. If you say what are the
unknowns and what are the knowns, let's be very clear.
The photon has a known momentum and known energy--that's given
to you. Initial me has a mass m,
that's given to you. Your job is to find the unknown
mass m prime, and the unknown velocity.
Two equations for two unknowns, but they're not simple linear
equations where you can add and subtract.
The simple linear equations are multiply this by three,
multiply that by four and subtract, and you'll get what
you want. But it's more complicated.
The trick that you really want to use is the fact that for any
particle, if you took the square of the 0 component,
subtract from it the square of the spatial component,
the answer will be m prime square c^(2).
So, I'm saying: this square minus this square
will be m prime times c^(2),
and that you get by taking the square of this minus the square
of this. You can juggle that,
but I'm not going to do it that way, I'm going to show you a
much more efficient and rapid way to do this.
So, let's do it the more efficient way,
so let me come down here so you can see.
Here is a shortcut. The shortcut is to work
entirely with four-vectors and four-dimensional dot products,
and don't get down to the ugly business of writing each
component. We know that
P_2 is P_1 plus
K, right? And who are we after?
We're after m prime; then we remember that
P_2^(2), namely P_2 dot
P_2 is equal to m prime square
c^(2). Yes?
So, the question you are asked is--No one asked you how fast I
am moving after I got hit by the photon;
you're only asking me what is my new rest mass.
See, I've swallowed a photon, so I weigh more;
it's really that. How much more do I weigh is the
question; namely;
what is my rest mass. Namely, I swallowed it,
and I recoiled, and if you travel with me and
find me at rest, what mass m prime will
you measure? So, no one's asking me my
velocity of recoil v. So, you shouldn't bother to
calculate that. You could if you wanted to,
but no one is asking you that. So, go directly for
P_2^(2). So, the quantity we're looking
for is m prime square c^(2).
That's got to be P_1 + K,
the whole thing square. Well, do the old algebra.
That's equal to P_1^(2) + K^(2)
plus twice P_1 dot K. Okay, now let's see if I can do
what we do next. Would you like to guess?
Yes? No?
Do we know any of these quantities?
Do we know anything about the three terms I've written here?
Yes? Student: [inaudible]
Professor Ramamurti Shankar: K^(2) = 0
because of what? Student:
[inaudible] Professor Ramamurti
Shankar: For a photon I told--For any particle the
square root of its momentum is really the square root of its
mass times c^(2). For photons,
I told you long back, K dot K is 0.
You don't have to worry about that.
How about the square of the momentum of me?
Well, that was my rest mass before, times c^(2) plus
0. Now, P_1 dot
K you've got to be careful.
That dot product means--let me write down the vector of
P_1 and K for you.
P_1 was mc and 0,
and K was /c,
and the spatial momentum, where was equal to
kc. Remember that?
For a photon, once you pick the energy,
you have no choice with the momentum.
The magnitude of the momentum has to be the magnitude of
energy divided by c. So, let's take
P_1 dot K;
let's try to do P_1 dot
K. Remember the dot product of two
vectors is time times time. mc times over
c minus space dot space, that was 0. So, I cancel that,
and I find it's really m. And that's equal to m
prime c^(2). Yes?
Student: Can you just remind me what the
stands for again? Professor Ramamurti
Shankar: Yeah, the photon, remember the
four-vector's first component is energy over c.
Second component is the momentum.
It's traditional to call the energy of the photon by the
symbol Ω. And it's traditional to call
the momentum of the photon, the usual momentum,
by the symbol K. And that's very helpful;
otherwise, you're going to have P_1 and
P_2 and P_1 sub-0 and
K_1 sub-0, too many indices.
So, what we're trying to do is whenever we write a K
whose components are and little k,
we know we're talking about a photon.
And whenever I write a P, I know I'm talking
about the electron. So, is just the
energy of the photon. So now, if I cancel all the
c squares, then I get m prime
square plus m^(2) plus 2m/c^(2).
Now, you can take the square root, if you like,
and that's the answer. So, my mass would have gone up.
You might think, "How much would the mass have
gone up?" First of all,
this guy swallowed a photon of energy ,
so maybe that energy over c^(2) will be the change
in mass. That is not correct,
and you've got to think about why that is not the answer.
Why is that wrong? Why doesn't the photon energy
turn into my extra mass? Yes?
Student: [inaudible]
Professor Ramamurti Shankar: In other words,
when I absorb the photon, I also have to recoil to
conserve the momentum of the photon.
Consequently, my final energy will equal old
energy plus photon energy, but that does not mean all of
it is rest energy because there's some energy of motion.
So, you've got to make sure that in the end you calculate it
so that the total energy is balanced, but it's not contained
in rest mass. Because you produce the [extra]
rest mass, and you also move the rest mass.
And for all of that you've got only the energy of the photon to
work with. Okay, so this is one problem.
I was going to do one or two more, but I want to make sure
you guys are following; I just don't know.
See, if I ask you what you would do next,
if you have some idea, then we can proceed,
but if you don't know, maybe we should slow down.
And you've got to talk about which part of it you find more
difficult. So, let me say a few things.
And the first time you hear something that you find
difficult, you've got to put your hand up,
okay? Otherwise, you're not going to
make progress. The first thing I say is this:
particles of mass have an energy and momentum that depend
on their velocity and rest mass as given by the first equation.
And the energy and momentum are assembled into a four-vector,
the first component of which is E/c, the second component
is just the momentum. That much you should know.
Secondly, if you took the dot product of that vector with
itself, and the dot product is defined to be Lorentz invariant,
so that it is the same for all people, that's
p_0^(2) - p_1^(2),
you'll always get m^(2)c^(2) for any
particle of mass independent of its state of motion.
And the simplest way to remember that is to evaluate the
dot product in the frame moving with the particle.
In that frame, it only has a
p_0, which is mc,
and doesn't have a spatial momentum;
so that's got to be the answer. Second thing,
when you deal with photons, they're a little tricky.
Photons have energy, photons have momentum,
but energy is not mass times some velocity square over 2
etcetera. They have no mass,
but they have energy and they have momentum,
and the condition of the photon is that the energy must be
momentum times c. That will make sure that
K dot K is 0. Because for a particle of mass
m, I showed you the relation, E^(2) =
(c times P)^(2) + m^(2) c^(4).
For a photon, there is no m.
That means E = pc. So, photon energy and photon
momentum have relative magnitude.
Yes? Student: [inaudible]
Professor Ramamurti Shankar: That's what I'm
telling you. For a photon,
it's not defined as mv over something.
It is not defined that way because you don't even get off
first base because m vanishes, and v = 0,
you're in real trouble. So, you've got to realize that
for a photon there's an exception of how you think about
it. You don't think of it as an
object of some mass that moves and has some energy and some
momentum. It always moves at one speed.
You might say, if it always moves at one
speed, it must have only one unique momentum.
That's not true either. So, they are a little tricky.
For photons, you have to accept the fact
that photons have energy and photons have momentum,
with E = pc. And apart form that,
there is no restriction, you can give a photon any
energy you like as long as the momentum you give it is such
that E^(2) = c^(2)p^(2). And you cannot go to the frame
in which the photon is at rest; there is no such thing,
it has no rest mass. But particles with no rest mass
can have energy, and can have momentum.
When enough photons hit you, you can feel the force,
that's called radiation pressure.
So, they carry momentum, they carry energy,
but the formula for momentum energy is not written in terms
of mass. So, that's the second thing
I've said. Yes?
Student: [inaudible]
Professor Ramamurti Shankar: For the photon?
Yes. k_0 =
k_1 for the photon.
And if I wrote k_0 as
/c, then it means = kc.
That's the second statement. That's not obvious from
anything I've taught you. So, that's some new stuff.
Because suddenly, I've got an object with no mass
and the whole derivation for momentum started with m
times some velocity and so on, so that fails for photons.
But they exist and they carry energy, they carry momentum.
You should think of them as particles that obey E^(2) =
(cp)^(2) + m^(2)c^(4), with m = 0,
so they obey E = pc. That's a second statement.
Third statement is, in a collision you add up all
the Ps and the Ks for everybody on the left,
and that equals the sum of all the Ps and the Ks
on the right. And you do that so you've got
to make sure you can do that part.
For particles, I will tell you this particle
has this rest mass, it's going at this velocity.
And you should be able to build up its energy and momentum.
For photons, I won't tell you this is the
velocity, velocity is always c.
For photons, I'll have to tell you this is
the of the photon; this is the k of the
photon. So, you form a little a
vector with /c as the first component,
and k as the second one. Then it's very easy.
Add all the first column, second column,
third column, fourth column,
and those four things are separately conserved.
That's it. That's the trick to collisions,
balancing energy and momentum in one mega equation,
which is four-momentum before equals four-momentum after.
And the beauty is any observer can look at this collision from
any frame of reference; different people will disagree
on what's the energy and what's the momentum of the particle.
For example, in this example I gave you,
I said I am at rest, and I got hit by a photon,
then I'm moving. Well, that's the point of view
of some observers for whom I was at rest initially.
You can see the whole thing from a rocket.
In that case, my initial momentum will have a
spatial part because on the rocket, I'll be moving
backwards. This photon will also have a
new and a new k, which you can find
from Lorentz transformation; this final object will be
moving at some speed, so everything will look
different. Nonetheless,
the momentum and energy will balance on the two sides for you
as well. Yes?
Student: [inaudible]
Professor Ramamurti Shankar: No.
You've got to be very careful about what you mean by mass.
Some people think of mass as m divided by square root
of 1 minus v square root. I never want to take that point
of view. The only mass is the rest mass,
and nobody will disagree on what the rest mass of a particle
is. Do you understand that?
I may think the particle is at rest, you may think it's moving,
that's just fine. But if we both ask the
following question, "What is its mass in a frame in
which the particle is at rest?" the answer to that question is
a unique answer. To find that,
you go sit with the particle and you find its mass.
So, the rest mass, in this case m and
m prime, are the same for all people.
You can see the whole collision on the rocket,
the m prime is the m prime.
E and p are transforming as Lorentz
variables, but m is the length squared of a four-vector;
therefore, it's the same for all people.
Now I want to do a second very interesting problem,
which is this. The year is 1953 or '54.
You are trying to build a collider in Berkeley.
It's going to have a proton target. By the way, I'm going to show
you one more trick that's very useful.
Put c = 1 everywhere. You can always put c = 1
by choice of units. For example,
if your unit of length instead of being a meter stick was 3
times 10^(8) meters, and that was your new unit of
length, let's call it a light second.
For you, the velocity of light is 1 because it goes 1 light
second in 1 second, the distance over time will be
1. Even if that's not your unit,
I say forget about c, do the math without c;
in the end you can always put back the c.
We'll always know in the end how to put back the c.
Maybe I have an example here. Yeah, in this problem,
if I've forgotten c everywhere, I get an answer that
looks like this: m prime square is
m^(2) + 2m. I already know that if that's
the mass square and that's the mass squared.
Well, that's the mass, and that's an energy,
so I know that should be a c^(2) here.
You can always fudge the c^(2); at the end of the
day, there's no reason to carry it from the beginning.
Just from dimensional analysis, when you're looking for energy
and you see a mass, put a c^(2) there.
When you're looking for a mass and you see an energy and it
should be a mass, divide by c^(2).
So, that's what I'm going to do. So, here is a proton at rest.
Here is a proton that I've created from my Bevatron,
which is going to come and smash into this.
And the job of this reaction is the following:
where this notation is very unfortunate [p + p],
this is not a momentum, this stands for proton.
Proton plus proton should form two protons and a new proton and
anti-proton created from the vacuum.
Created from nowhere. This is what happens in a
collider. Particles and anti-particles
are created in pairs using the incoming energy.
So, it's the opposite of the atomic bomb.
In an atomic bomb, mass is destroyed and energy is
created. In an accelerator,
energy is converted into mass. Which is why two particles can
go in, and four particles, or ten, or 14,000 can come out.
We want to look at the experiment where people were
trying to make anti-protons. The theory says the anti-proton
mass is the same as the proton mass.
It's a very general symmetry of the theory.
But you want to make it, and the question is,
"What is the minimum energy of this incoming proton that will
make this reaction possible?" And this guy starts out at rest. So, what energy or what
momentum should the incoming projectile have?
Yes? Student:
[inaudible] Professor Ramamurti
Shankar: Let's see. So, here are four protons,
and in the cheapest collision what will they be doing in the
end? Student:
[inaudible] Professor Ramamurti
Shankar: Sitting in a clump. And what will be the energy of
that clump? Student:
[inaudible] Professor Ramamurti
Shankar: Do you agree that's 4m?
Student: Yes. Professor Ramamurti
Shankar: Okay, this guy brings energy into the
collision, being the static target.
So, you would conclude the energy of this should be
3m. By that we'll mean
mc^(2) divided by 1 minus v^(2) over c^(2)
should be 3 times mc^(2). That's the energy.
Because that 3m plus this m will make
4m. Yes?
Student: [inaudible]
Professor Ramamurti Shankar: No.
The mass of these particles is always m.
I'm saying the energy of this guy is 3m.
Yes? If this is 3m and that
guys brings an m just by sitting there,
and we need 4m to make four particles,
and we want the cheapest deal, so nobody is doing anything.
But this wrong. This is an underestimate for
what you need. You need more.
You've got to think about why do we need more energy than
this? Yes?
Student: [inaudible] Professor Ramamurti
Shankar: No. Let's say they don't annihilate;
they form and they just--they don't annihilate that fast,
so they can be produced. Anti-protons can be produced.
That's not the reason why. I want you to visualize the
question we're asking. Obviously, if we send a
super-duper proton beam, it can knock the hell out of
this proton and produce however many particles you want.
We want the minimum energy, and the picture we have is
what's the minimum energy of four particles?
Well, it's four particles at rest.
That's the one for which we computed the energy,
but there's something wrong with our argument.
Yes? Student:
[inaudible] Professor Ramamurti
Shankar: Yes. Let me repeat what he said.
The Law of Conservation of Momentum will not allow you to
take a situation where the proton comes in with momentum.
This guy is doing nothing, and you cannot have four
particles at rest, because their momentum is 0.
Therefore, the four guys who are produced must be drifting
forward with some minimum momentum, so that that momentum
equals incoming momentum. That means, you've got to pay
the price to make them, you've got to pay the price to
move them at a minimum momentum to balance the incoming
momentum. So, that should tell you that
the real energy of the incoming particle has got to be more than
just whatever I said, 3m.
So, we've got to find out the real momentum.
So, the trick is again going to be the same.
So, this is the initial state, this is the final state,
but this final state has got some motion.
It cannot be at rest. So, what's the initial momentum?
Initial momentum for this guy is some energy E;
for this guy is some mass m.
And the momentum of this is some number P.
And the momentum of this guy is 0;
that's the initial momentum. And that's got to be equal to
the final momentum. Very good.
Final momentum you can write down in detail by taking these
four masses, giving them some velocities, but I'm not going to
do any of that. Yes?
Student: [inaudible]
Professor Ramamurti Shankar: This m,
this particle has an energy m.
It's really mc. Sorry, it's E/c,
mc and 0. This guy has got E/c and
some momentum P. I've forgotten the c.
Remember, we agreed that we're not going to worry about
c in this problem? So, this is just m. So, this is the initial
momentum, that's going to be the final momentum.
That also means initial momentum square is final
momentum squared. If two vectors are equal,
the squares are equal. Now, the final momentum
squared, I say we can evaluate not in the lab frame,
but in another frame called the center of mass frame.
And here is a center of mass frame.
Suppose you start moving to the right at some speed,
what do you think will happen? Imagine in your head.
You move to the right, this proton will start moving
to the left, this proton will appear slower.
Is that clear to you? Pick up your speed until
they're coming in exactly opposite momenta.
That's a frame of reference you can imagine.
In that frame, the collision looks rather
different. In that frame,
this guy and this guy are approaching each other with
opposite velocities, and therefore equal and
opposite momenta; the total momentum is 0.
In that frame, these two guys are allowed to
go into four sitting ducks, because the total momentum is
0. So, that is the frame of
reference called the center of mass frame in which incoming
momentum adds up to 0, so the final particles are
allowed to be at rest if they want to, and the minimum energy
collision, they will be produced at rest,
because that's the least expensive way.
So, I found a frame in which this is possible.
That's not the lab frame, that's the fame in which they
are moving in opposite directions.
But what's the total momentum in this frame?
Total momentum in this frame, P total,
is 4m, 0. Now, no one says the total
momentum in that frame has to be the total momentum in my frame,
but--sorry, say P final. No one says that P final
should be my P final, but the square of that P
final is the square of my P final because the
square of any vector is the same for all observers.
So, what is P final square?
You guys should know by now. This dot product of this vector
with itself is the square of this number minus 0.
There is nothing here. Therefore, I argue here,
that P initial square is equal to P final square.
That I calculate in the center of mass frame,
and I get 16m^(2). Yes?
Student: [inaudible]
Professor Ramamurti Shankar: Right.
Why is the square of the four-momentum the same in two
frames of reference? Anybody answer that?
Why is the momentum not equal, but the square is equal?
Why is that? Because it's a dot product of a
vector with itself, and the whole idea of defining
these squares is so that they are the same for all observers.
The whole idea of defining four-vectors is that when you
take the length square of a four-vector defined in this
crazy way, 0 component squared minus the
first component squared, that's the same for all people,
right? That's the beauty.
That's why, if you really look at the total momentum vector in
the lab frame, it looks like this.
And that's got to be the final total momentum too.
These numbers are not--they are both non-0.
So, no one says the total momentum vector is the same.
In the center of mass frame it looks like this,
but the square of the vector must be the square of the vector
for all people. Now, let me finish this and
maybe we'll have a discussion. So, P initial square is
16m^(2), so this is the top most line.
Find the square of this vector, is (E + m)^(2) -
P^(2) = 16m^(2). Now, so, let's see where to
write this; maybe I'll write it here. That means E^(2) + m^(2)
+ 2Em - p^(2) = 16m^(2). But you must know that E^(2)
- p^(2) is an m^(2). I'm saying E^(2) = p^(2) +
m^(2), if you banish all factors of c^(2).
So, this E^(2) - p^(2) is an m^(2);
there is already an m^(2) here,
so I get 2m^(2) + 2Em = 16m^(2).
You cancel the 2 and make it 14m^(2),
and you divide by 2m, you find E = 7m.
By the way, this is posted on the web, on the internets.
So, you guys can read it whenever you like.
So, don't worry about this last detail;
I've given you every possible detail here.
If you didn't get every factor, it doesn't matter.
What I want you to understand is, I've also given you the
whole argument on how you go to this frame and that frame.
The point is you can answer E = 7m,
now we realize it's time to put back the velocity of light,
and everybody knows, even man and woman on the
street, that this should be really 7mc^(2).
If we carried factors of c throughout the
calculation, it would have been very messy,
but here is where c^(2) would have to turn up.
So, if you're looking for an energy and you get a mass,
you put back the c^(2). So, what we are told is,
the incoming proton, as it moves,
its energy goes from mc^(2) to bigger and
bigger numbers, because the denominator of 1
over v^(2) over c^(2) is vanishing.
Wait until the energy is 7 times the rest energy,
and when it's 7 times the rest energy,
that's when it's ready to hit the static proton and produce
proton/anti-proton pairs from nowhere.
So, some of the 7m^(2) is going to give a momentum to
the fragment of the collision, and some is to create them.
And the way we found out the minimum energy is we did
energy-momentum calculation. But the minimum energy
configuration is easily imagined in the center of mass frame,
that's the whole point. Not in the lab frame,
the center of mass frame. That's why accelerators,
nowadays, are taking one particle that can move faster
and faster and hitting a target they produce intersecting
storage rings where the particles are going in opposite
directions. And every time they meet in the
racetrack, they're allowed to collide.
What's the beauty of that? The beauty of that is,
since they come with opposite momenta, the final fragments
don't have to move at all. So, you can make more and more
mass. So, don't put your energy all
into one beam and a static target;
put them into two oppositely moving beams.
Then, all the energy is available for producing mass.
All right, so I'll stop now. I think this is a difficult
chapter, so you have to go and read, and you have to talk to
each other and try to do as many problems as you can.
