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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK, guys. Welcome back. As you can see, we're not
using the screen today. This is going to be one of
those fill-the-board lectures. But I am going to work you
through every single step. We're going to go
through the Q equation and derive its most
general form together, which, for the
rest of this class, we'll be using
simplified or reduced forms to explain
a lot of the ion or electron-nuclear
interactions as well as things like
neutron scattering and all sorts of other stuff. We'll do one example. For any of you that have
looked at neutrons slowing down before, how much
energy can a neutron lose when it hits something? We'll be answering that
question today in a generally mathematical form. And then a few
lectures later, we'll be going over some of the
more intuitive aspects to help explain
it for everybody. So I'm going to show you the
same situation that we've been describing sort
of intuitively so far, but we're going to hit
it mathematically today. Let's say there's
a small nucleus, 1, that's firing at a large
nucleus, 2, and afterwards, a different small nucleus, 3,
and a different large nucleus, 4, come flying out. And so we're going to keep
this as general as possible. So let's say if we draw angles
from their original paths, particle 3 went
off at angle theta and particle 4 went
off at angle phi. So hopefully those are
differentiable enough. And if we were to write the
overall Q equation showing the balance between
mass and energy here, we would simply have
the mass 1 c squared plus kinetic energy of 1. So in this case,
we're just saying that the mass and
the kinetic energy of all particles on the
left side and the right side has to be conserved. So let's add mass 2 c squared
plus T2 has to equal mass 3 c squared plus T3 plus mass
4 c squared plus T4, where, just for symbols,
M refers to a mass, T refers to a kinetic energy. And so this conservation of
total mass or total energy has got to be conserved. And we'll use it again. Because, again, we can describe
the Q, or the energy consumed or released by the reaction,
as either the change in masses or the change in energies. So in this case, we
can write that Q-- let's just group all of
the c squareds together for easier writing. If we take the initial masses
minus the final masses, then we get a picture of
how much mass was converted to energy, therefore,
how much energy is available for
the reaction, or Q, to turn it into kinetic energy. So in this case, we can
put the kinetic energy of the final products minus
the kinetic energies-- I'm going to keep with 1-- of the initial products. And so we'll use this
a little later on. One simplification
that we'll make now is we'll assume that if we're
firing particles at anything, that anything
starts off at rest. So we can start by
saying there's no T2. That's just a simplification
that we'll make right now. And so then the
question is, what quantities of this situation
are we likely to know, which ones are we
not likely to know, and which ones are left
to relate together? So let's just go
through one by one. Would we typically know the mass
of the initial particle coming in? We probably know what we're
shooting at stuff, right? So we'd know M1. What about T1, the
initial kinetic energy? Sure. Let's say we have a reactor
whose energy we know, or an accelerator, or
something that we're controlling the energy,
like in problem set one. We'd probably know that. We'd probably know what
things we're firing at. And we would probably
know what the masses of the final products are,
because you guys have been doing nuclear reaction
analysis and calculating binding energies and everything
for the last couple of weeks. But we might not know
the kinetic energies of what's coming out. Let's say we didn't actually
even know the masses yet. We'd have to figure out a way to
get both the kinetic energies. And what about
these angles here? This is the new variable
that we're introducing, is the kinetic energy
of particles 3 and 4 is going to depend on what
angles they fire off at. Let me give you a limiting case. Let's say theta was 0. What would that
mean, physically? What would be happening
to particles 1, 2, 3, and 4 if theta and phi were
0, if they kept on moving in the exact same path? Yeah? AUDIENCE: Is it a fusion
event, or [INAUDIBLE] PROFESSOR: We don't know. Well, let's see. Yeah. If it was a fusion event-- let's
say there was one here and one standing still-- then the whole center
of mass of the system would have to move that way. So one example could
be a fusion event. A second example could
be absolutely nothing. It's perfectly valid to say if,
let's say, particle 1 scatters off particle 2 at an
angle of 0 degrees, that's what's known as
forward scattering, which is to say that theta equals 0. So this is another quantity
that we might not know. We might not know what
theta and phi are. And the problem here is we've
got, like, three or four unknowns and only one
equation to relate them. So what other-- yeah? Question? AUDIENCE: For
forward scattering, when you say theta equals 0, do
you mean they just sort of move together forward, kind of
like an inelastic collision, and they just keep moving
in the same direction? PROFESSOR: An inelastic
collision would be one. And since we haven't gone
through what inelastic means, that would mean some
sort of collision where-- let's see. How would I explain this? I'd say an inelastic
collision would be like if particles 1 and 2 were to
fuse, like a capture event, for example, or a capture and
then a re-emission, let's say, of a neutron. Yeah. If it was re-emitted in
the forward direction, then that could be an
inelastic scattering event-- AUDIENCE: Oh, OK. PROFESSOR: --but still
in the same direction. Or an elastic scatter
at an angle of theta equals 0 could be like there
wasn't any scattering at all. Because really in
the end, can matter-- let's say if you have a
neutron firing at a nucleus, depends on what angle
it bounces off of, in the billiard ball sense. If it bounces off at an angle
of 0, that means it missed. We would consider
that theta equals 0. But the point here
is that we now have more quantities
unknown than we have equations to define them. So how else can
we start relating some of these quantities? What else can we
conserve, since we've already got mass and energy? What's that third quantity
I always yell out? AUDIENCE: Momentum. PROFESSOR: Momentum. Right. So let's start writing some
of the momentum conservation equations so we can try
and nail these things down. So I'm going to write
each step one at a time. We'll start by
conserving momentum. That's what we'll do right here. And we can write the x and
the y equations separately. So what's the momentum
of particle 1? How do we express that? AUDIENCE: Mass times velocity. PROFESSOR: Yep. So it would be like M1 V1. So we'll have a little box
right here for momentum. We could say mass
times velocity-- or, how do we express that
in terms of the variables that we have here,
like we did last week? What about in terms
of kinetic energies? Well, another way of
writing mass times velocity would be root 2MT. Because in this case, we
would have root 2 times M1 times 1/2 M1 V1 squared. The 2s here cancel. Let's see. You have M1 squared. You have a V squared. And the square root of M
squared V squared is just MV. So this is an equivalent
way of writing the momentum in the variables that
we're working in already. And so since that doesn't
introduce another variable like velocity--
which we do know, but it's kind of confusing
to add more symbols-- let's keep as few as possible. So what's the x
momentum of particle 1? Just what I've got up there. 2 M1 T1. What's the x momentum in
this frame of particle 2? 0. We're assuming
that it's at rest. And now, what's the x
momentum of particle 3? AUDIENCE: [INAUDIBLE] PROFESSOR: I heard
a couple of things. Can you say them louder? AUDIENCE: Square
root of 2 M3 T3? PROFESSOR: Yep. Root 2 M3 T3. But in this case,
if we're defining, let's say, our
x-axis here, it also matters what this angle is. So you've got to multiply by
cosine theta in this case. And that's the x
momentum of particle 3. And we've also got to
account for particle 4. So we'll say add 2 cosine phi. Now let's do the same
thing for the y momentum. What's the y momentum
of particles 1 and 2? 0. They're not moving in
the y direction to start. And how about particle 3? I hear whispers, but
nothing vocalized. AUDIENCE: Sine? PROFESSOR: Yep. Same thing, but root
2 M3 T3 sine theta. And M4-- I almost
wrote the wrong sign there-- has got to be a minus. If the momentum of the initial
particle system in the y direction is 0, so
must the final momentum in the y direction. So these two momenta have
to be equal and opposite. And that's times sine phi. So now we actually have sets
of equations that relate all of our unknown quantities. We have the mass
conservation equation, we have the Q equation,
we have the x momentum, and we have the y momentum. And from this point on,
it's a matter of algebra to express some of
these quantities in terms of some of the others. So let's get started with that. Because angles
are kind of messy, and theta should
uniquely define phi, let's try and get things
in terms of just one angle. So I'm going to
start by separating the thetas and the phis on
either side of the equals sign, so that hopefully later
on we can eliminate one in a system of equations. So all I'm going to do is
I'm going to subtract or add the theta terms to the
other side of the equation. So let's say we'll
separate angles. So we'll have root 2 M1 T1
and minus root 2 M3 T3 cosine theta. I'll be depending on you
guys to check for sign errors here because those
will be messy. I do have notes in
case, but I'm hoping I won't have to look at them. And all we have
left on this side is root 2 M4 T4 cosine phi. So that's the x
momentum equation. Let's do the same thing with
the y momentum equation. So all we'll do is
take the theta term and stick it to the
left of the equals sign. So that would give us minus
root 2 M3 T3 sine theta equals minus root 2 M4 T4 sine phi. Right away, we can see that
the minus signs can cancel out, just for simplicity. And what else is common to
these that we can get rid of? Yep? AUDIENCE: Square root of 2. PROFESSOR: Everything here
has a square root of 2. So we'll just get rid of
all of the square root of 2s to simplify as much as possible. And now we look a little stuck. But now is the time to remember
those trigonometric identities back from high school
that I don't think-- has anyone used these since? In 1801 or 1802, anyone
used a trig identity? A little bit? OK. I would hope so. But I don't know what other
people are teaching nowadays. At least this way I'll make sure
you remember the high school stuff. We're going to rely on
the fact that we already have got a cosine and a sine. We have a set of
simultaneous equations. If we can add them together
and destroy the angles somehow, that will make
things a lot easier. So for the thetas, we
have a cosine, a sine, and an unangled term
that looks kind of messy. Here we have a
cosine and a sine. Anyone have any
idea where we could go next to destroy
one of these angles? Anyone remember any handy
cosine or sine trig identities? AUDIENCE: If you
squared both terms, you could get square root of
cosine squared, square root of-- sorry. You get cosine squared
and sine squared and then you factor out
the square root of M4T4 and then cosine squared
plus sine squared equals 1. PROFESSOR: Exactly. So we can rely on
the fact that if we can square both sides of both
equations and add them up, we would have a
cosine squared of phi plus a sine squared of
phi, which also equals 1. So we can destroy this phi angle
and make things a lot simpler. So we'll start by
squaring both sides. Let's start with the
x momentum equation. So if we have-- let's see-- root M1T1. So we're going to take
that stuff squared. And that squared
is not too hard. Neither are those. So we'll have root M1 T1
squared, which just gives us M1 T1, minus root M1 T1 times
root M3 T3 cosine theta. Let's just lump those terms
together as root M1 M3 T1 T3 cosine theta. Also, anyone, raise
your hand or let me know if I'm going too fast. I'm trying to hit
every single step. But in case I skip one,
please slow me down. That's what class is for. OK. And then we've got another one. Let's just stick a
2 in front of there and plus that term squared. So we'll have M3 T3. Let's see. Yeah. Looks like cosine
squared of theta. Yep. Equals-- this one's easier-- M4 T4 cosine squared of phi. OK. Now we'll do the same thing
for the y momentum equation. Much easier because there's
no addition anywhere. And we have M3 sine
squared theta-- over here-- equals M4
T4 sine squared phi. So this is quite nice. Now if we add these
equations together, we get rid of all of the
cosine and sine squared terms. So let's add them up. Let's see. We'll add the two equations. Add equations. And let's try and group
all the terms together. So we have M1 T1
minus 2 root M1 M3-- it's getting hard to write
over the lip of the chalkboard here-- cosine theta. And we have M3 T3 cosine
squared theta plus M3 T3 sine squared theta. Equals M4 T4 cosine squared
theta plus sine squared theta-- or phi. I'm sorry. Cosine squared phi
plus sine squared phi. OK. Hopefully that's as low
as I'll have to write. And like we saw
before, cosine squared plus sine squared equals 1. So that goes away. That goes away. And let's keep going over
on this side of the board. I told you this would
be a fill-the-board day. Let's see if we
actually get all six instead of just
the four visible. But I think we'll finish this
derivation in four boards. So let's write what
we've got left. Let's see. Remaining. So we have M1 T1
minus 2 root M1 M3-- so much easier to
write standing up-- cosine theta equals M4 T4. Quite a bit simpler. AUDIENCE: [INAUDIBLE] PROFESSOR: Did I miss a term? AUDIENCE: The M3 T3. PROFESSOR: Ah. Thank you. You're right. You're right. And we had a plus M3 T3. Yeah, that would be important. Thank you. Equals M4 T4. So we now have a relation
between the masses, the energies, and one angle,
which is getting a lot better. We still have one more
variable than we can deal with. So let's say if we're-- let's see. Which of these
variables do you think we can eliminate using any
of the equations you see, let's go with, on that
top board over there? Well, what other
quantities are we likely to know about
this nuclear reaction? Let's bring this back down. Are we likely to
know the Q value? AUDIENCE: Yeah. PROFESSOR: Probably. Because like you
guys have been doing on problem sets one and
two, if you know, let's say, the binding energies, or the
masses, or the excess masses, or the kinetic energies
of all your products, any combination of
those can get you the Q value of that reaction. And if you just look up those
reactions like, let's say, radioactive decay reactions,
on the table of nuclides, it just gives you the Q value. So chances are we can express
some of these kinetic energies in terms of Q. And all we've got left
is T1, T3, and T4. So which of these are
we most likely to be able to know or measure? T1, we probably fixed it
by cranking up our particle accelerator to a certain energy. T3 or T4, what do
you guys think? Let's say we had a
very small nucleus firing at a very big one. Which one do you
think would be more likely to escape
this system and get detected by us standing a couple
feet away with a detector? Yep? AUDIENCE: T3. PROFESSOR: Probably T3,
the smaller particle. We've just arbitrarily
chosen that. But for intuitive
sake, let's say, yeah. Why don't we try and get
T4 in terms of Q T1 and T3? That's not too hard,
since it's addition. So our next step
will be substitute. And we'll say that Q equals-- I'm just going to copy
it up from there-- T3 plus T4 minus T1. So we can isolate T4 and say
T4 equals Q plus T1 minus T3. And continue substituting. I usually don't like to have
my back to the class this much. But when you're writing this
much, it can be a little hard. So let's stick this T4 in right
here and rewrite the equation as we've got it. M1 T1 minus 2 root M1 M3 T1 T3
cosine theta plus M3 T3 equals M4 times Q plus T1 minus T3. I anticipate us needing to see
this side of the board soon. I also apologize for
the amount of time it takes to write these things. There's another
strategy one can use at the board which is
defining intermediate symbols. And here's why I'm
not doing that. When I was a freshman, back in-- whoa-- 2001. Who here was born after 2001? Nobody. OK. Thank god. I don't feel so old. I was in 18023, which was
math with applications, which was better known as
math with extra theory. And in one class, not only
did we fill nine boards, but we ran out of
English letters-- symbols-- and we ran out
of Greek letter symbols, and we moved on to Hebrew. Because they were distinct
enough from English and Greek. And being, I think, the only
Hebrew speaker in the class, I was the only one that
could follow the symbols, but I couldn't follow
the math anymore. So I am not going to define
intermediate symbols for this and just keep it
understandable, even if it takes longer to write. OK. So let's start off
by dividing by M4. Our goal now is to
try to isolate Q. Because this is something
that we would know or measure. And it will relate all of the
other quantities, only one of which we won't
really know yet. So let's divide
everything by M4. So we have T1 times M1 over
M4 minus 2 over M4 times root of all that stuff plus
T3 times M3 over M4 equals Q plus T1 minus T3. And we've almost isolated
Q. I'll call this step just add and subtract. And I'm going to group
the terms together. So let's, for example,
group all the T1s together and group all the T3s together. So if I subtract T1, I get
T1 times M1 over M4 minus 1, minus 2 over M4 root M1 M3
T1 T3 cosine theta, plus-- and if I add T3, then I would
get M3 over M4 plus 1 equals Q. So this is a good place to stop,
turn around, and see you guys, and now ask you, which of
the remaining quantities do we probably not know? So let's just go
through them one by one, just to remind ourselves. Are we likely to
know what T1 is? Probably. How about the masses M1 and M4? If we know what
particles are reacting, we can just look those up,
or measure them, or whatever. We know M4. We know our masses. We know T1. What about T3? We don't necessarily know yet. So T3 is a question mark. How about cosine theta or theta? We haven't said yet. And T3 we don't know. And the masses we know. And the Q we know. So finally, to solve for-- well, we only have two
variables left, T3 and theta. So this here-- this is actually
called the Q equation in its most complete form-- describes the relationship
between the kinetic energy of the outgoing particle and
the angle at which it comes off. How do we solve this? How do we get one in
terms of the other? Anyone recognize what kind
of equation we have here? It's a little obscure. Well, it's not obscure. But it's a little bit hiding. But it should be a
very familiar one. Think back to high school again. Yes. AUDIENCE: Is it the cosine angle
for the triangle [INAUDIBLE] PROFESSOR: Let's see. Certainly, there's probably
some trig involved in here, in terms of, yeah, if
you know the cosine, then you know, let's
say, the x or the y component of the momentum. But there's something
simpler, something that doesn't require trigonometry. Yep. AUDIENCE: Is it quadratic? PROFESSOR: It is. It's a quadratic--
so who saw that? It's actually a quadratic
equation, where the variable is the square root of T3. That's the trick here, is you
have something without T3, you have something
with square root T3, and you have something
with T3, better known as root T3 squared. And there. So this is actually
a quadratic equation. Despite the fact
that it may not have looked that way in the
first place, there we go. So now, someone who
remembers from high school, tell me, what are the roots
of a quadratic equation? Let's say if we have the form
y equals ax squared plus bx plus c, what does x equal? Just call it out. AUDIENCE: Negative b-- PROFESSOR: Yeah. AUDIENCE: [INAUDIBLE]
square root-- PROFESSOR: Yep. AUDIENCE: --b
squared minus 4ac-- PROFESSOR: Over-- AUDIENCE: 2a. PROFESSOR: 2a. And in this case,
a is that stuff. b is that stuff without the T3. And c is that stuff. Because we have,
like, 15 minutes before I want to open
it up to questions and I don't think we have to
repeat the quadratic formula stuff, I will skip ahead. Skip ahead. This is when I'd normally say
it's an exercise to the reader. But no. It's not the phrase
I like to use. It's boring. And I can just tell you
guys what it ends up as. It ends up with root T3 equals--
and this is the one time I am going to define new symbols
because it's just easier to parse-- ends up being, we'll call
it s, plus or minus root s squared plus t, where s-- let's see if I can remember
this without looking it up. No. I have to look at my notes. I don't want to get it wrong
and have you all write it down incorrectly because of me. There we go. The remaining stuff
in the square root, E1 cosine theta over M3 plus M4. And t equals M4 Q-- is it a minus? It is a plus. Over M3 plus M4. So these are the roots
of this equation. This is how you
can actually relate the kinetic energy of
the outgoing particle directly to the angle. So I want to let that
sink in just for a minute, stop here, and check
to see if there's any questions on the
derivation before we start to use it to do something
a little more concrete. Yep. AUDIENCE: Where did
the E come from? PROFESSOR: The E. Oh, I'm sorry. That's a T. Thank you. Kinetic energy. Again, we should be
consistent with symbols. And I think-- I don't see any
other hanging Es. Good. Thank you. So any other questions on the
derivation as we've done it? We managed to do it in
less than four boards. There we go. OK. Since I don't see
any questions, let's get into a couple of the
implications of this. So let's now look at what
defines an exothermic reaction where we say if Q
is greater than 0-- which is to say that some of the
mass becomes kinetic energy-- if an exothermic reaction is
energetically possible, then what is the minimum T1? Ah. That's why I brought it. What's the minimum T1 up here
to make that exothermic reaction happen? We'll put a condition on T1. So if the reaction's
exothermic, which means it will happen
spontaneously, how much extra kinetic energy do
you have to give to the system to make the reaction happen? Let's think of it in
the chemical sense. If you have an exothermic
chemical reaction, is it spontaneous or is it not? It is spontaneous. Same thing in the nuclear world. If you have an exothermic
nuclear reaction, do you need any kinetic
energy to start with to make it happen? No. OK. There we go. So that's kind of the analogy. So T1 has to be greater
than or equal to 0. It's pretty much not
a condition, right? It happens all the time. So if we were to say
T1 were to equal 0-- let me get my crossing
out color again. If T1 were to equal 0,
then s could equals 0. And T1 is 0 here. And then you just get-- that's an s-- t equals
M4 Q over M3 plus M4. And this just kind of
gives you a relation between the relative kinetic
energies of the two particles. Another way of
writing this relation would just be that
E3 plus E4 has to be greater than or equal to E1. AUDIENCE: T? PROFESSOR: All this-- hmm? AUDIENCE: T? PROFESSOR: Ah. Thank you. Because Es will be used in a
different point of this class. So we'll stick with
T for kinetic energy. Thank you. So all that this
condition says is that if mass has been
converted to energy, then that kinetic
energy at the end has to be greater
than at the beginning. And that's all it is. So it makes this equation
quite a lot easier to solve for an
exothermic reaction. You can also start to look
to say, well, what happens as we vary this angle theta? What does the kinetic energy do? Let's take the case of
an endothermic reaction. Now we are running out of space. For an endothermic reaction
where Q is less than 0, you would have to have
T1 to be greater than 0. Otherwise the
reaction can't occur. So you have to impart additional
energy into the system to get it going. And it also means that not every
angle of emission is possible. You might wonder, why do
we care about the angle, because the reaction
still happens anyway? Well, it doesn't
happen at every angle. And reactions have
different probabilities of occurring depending on the
angle at which the things come out. So you could see here
that as you vary T1 and as you vary cosine
theta, you still have to make sure that this
quantity on the inside here-- so, s squared plus t-- always has to be greater
than or equal to zero or else the roots of this are imaginary
and you don't have a solution. So it's kind of nice that
this came out quadratic. Because it lets you take some of
the knowledge you already know and now apply it to say, when or
when are nuclear reactions not or are they allowed? Wait. Let me rephrase that. When are nuclear reactions
allowed or not allowed? You can now tell, depending
on the angle of emission and the incoming energy and the
masses, which are all things that you would tend to know. So is everyone clear on
the implications here? If not, let me know. Because that's what
this class is for. AUDIENCE: Yeah. Can you just go over
it one more time? PROFESSOR: Yes. So, for exothermic reactions
where Q is greater than 0, all that says from our initial
part of the Q equation, if Q is greater than 0, then
we have this thing right here, where the final kinetic
energies have to be larger than the initial one. Which is to say that
some mass has turned into extra kinetic energy. And the solution to
these is pretty easy because you don't need
any kinetic energy to make an exothermic reaction happen. So you can just set T1 equal
to 0, which makes s equal to 0, because they're all
multiplied here. And then it
simplifies lowercase t as just a ratio of those
masses times the Q equation, which will tell you pretty
much how much kinetic energy is going to be sent off to
particle 3 right here. Up there. Particle 3. Because then we have this
condition, if root T3 equals s plus square root of
s squared plus t, and we've decided
that s equals 0, that just means that T3 equals
lowercase t, which equals that. So then you've uniquely
defined the kinetic energy for an exothermic
reaction, as long as you have no incoming
kinetic energy. For the case of an endothermic
reaction, first of all, we know that the
incoming kinetic energy has to be greater than 0. It's like the excess
energy that you need to get a chemical
reaction going. Has anyone here
ever played with-- what's the one, a
striking one here? Well, has anyone ever
lit anything on-- no, that's-- yeah. Of course you have. And that's not a
good explanation. Hmm. What's a good, striking
endothermic chemical reaction? Can anyone think of one? Yeah? AUDIENCE: When you put tin
foil in Liquid-Plumr and it releases-- PROFESSOR: And it's
a hydrogen generator? AUDIENCE: Let's see. I guess that's an explosion. PROFESSOR: I think
that happen-- yeah. That's more like an explosion. That's, like, the intuitive
definition of exothermic. Yeah. Actually, there's a fun
one you can do, too. This is great that
it's on video. You do that plus put manganese
dioxide in hydrogen peroxide and you have an
oxygen generator. And then you have the purest,
beyond glacially pure, spring water. You just mix H and O directly. Just don't get near it. Because it tends
to be pretty loud. We do this for our RTC
or reactor technology course, where I've got to
teach a bunch of CEOs enough basic high school chemistry
so they can understand reactor water chemistry. And the way I make sure that
they're paying attention is with a tremendous explosion. So folks come here, pay
about $25 grand apiece for me to fire water-powered
bottle rockets at them. It's a pretty sweet job. So if you guys are
interested in academia, you know, these
things happen in life. It's pretty cool. Yeah. All right. Since I can't think of any
endothermic chemical reactions off the top of my
head, I'll have to keep it general
and abstract and say, if you have an
endothermic reaction, you have to add energy
in the form of heat to get the reaction going. In an endothermic
nuclear reaction, heating up the material
does not impart very much kinetic energy. You might raise it from
a fraction of an electron volt to maybe a couple
of electron volts if things are so
hot that they're glowing in the ultraviolet. That doesn't cut it for nuclear. So you have to impart kinetic
energy to the incoming particle such that the kinetic
energy plus the rest masses is enough to create the rest
masses of the final particles. And that's the general
explanation I'd give. I forget who had
asked the question. But does that help
explain it a bit? AUDIENCE: [INAUDIBLE] PROFESSOR: Cool. OK. I'll take five minutes. And let's do a
severely reduced case of this, the case of
elastic neutron scattering. It's kind of a flash
forward to what we'll be doing in the next month or so. Does everyone have what's
behind this board here? I know that was, like,
three boards ago. So I hope so. So let's take the case of
elastic neutron scattering. Remember I told you that
after we developed this highly general solution
to the Q equation, everything else that
we're going to study is just a reduction of that. And this is about as
reduced as it gets. So in elastic
neutron scattering, we can say that M1-- well, what's the mass
of a neutron in AMU? And let's forgive our six
decimal points' precision for now. What's it about? AUDIENCE: 1. PROFESSOR: 1. So we can say that M1 equals 1. And in the case of
elastic scattering, the particles bounce
into each other and leave with their
original identities. So that also equals M3. If we're shooting neutrons at
an arbitrary nucleus, what's M2? Yep? AUDIENCE: A? PROFESSOR: Just A,
the mass number. Same as M4. Now, we don't have
M2 in this equation. Whatever. But the point is, yeah. We're going to use these two. We're going to use these two. So let's substitute that in. Oh, and one last other
thing I mentioned. What is the Q value
for elastic scattering? AUDIENCE: [INAUDIBLE] PROFESSOR: Right. 0. Because the Q value is
the difference in the rest masses of the ingoing
and outgoing particles. If the ingoing and outgoing
particles are the same, M1 equals M3, M2 equals
M4, that sum equals 0. Therefore, Q equals 0. So let's use these
three things right here and rewrite the general Q
equation in those terms. Which board is it on? Right there. So let's copy that down. So let's say we have T1 times
M1 is 1 over M4 is A minus 1 minus 2 over M4 is A. This is
where it gets nice and easy. M1 and M3 are just 1. So 1 times 1 times T1. We don't know what that is yet. So let's call it the Tn, T
of the neutron coming in. How about this? We'll call it T in and T out
for ease of understanding. Cosine theta. What do we have left? Plus T out. And let's make T1 into
an in right there. Times M3 over M4. M3 was 1. M4 is A. Plus 1
equals Q, equals 0. This is quite a simpler
equation to solve. So let's group
this all together. There's a couple
of tricks that I'm going to apply right
now to make sure that everything has
A in the denominator to make stuff easier. We can call 1 A over A here. We can call 1 A over A there. That lets us combine
our denominators and stick the sine right there. That becomes an A.
Same thing here. I'll just connect the dashes
and stick the minus sign there, leaving an A right there. Now we can just
multiply everything by A, both sides
of the equation. So the As go away there. We have a much simpler equation. 0 equals T in of-- let's see-- 1 minus A over 1. OK. We'll just call it 1 minus
A. Minus 2 root T in T out cosine theta
plus T out A plus 1. And, OK, it's 10 minutes
of, or it's five minutes of five minutes of. So I'm going to
stop this right here at a fairly simple equation. We'll pick it up on Thursday. And I want to open the last
five minutes to any questions you guys may have. Since that request came
in on the anonymous rant forum, which hopefully
you all know now exists. Yep. AUDIENCE: So what exactly
is forward scattering? I didn't really get that before. PROFESSOR: So let's look
at elastic scattering as an example. So in elastic
scattering, two particles bounce off each other
like billiard balls. In forward elastic
scattering, the neutron, after interacting
somehow with particle 2, keeps moving forward unscathed. So in the elastic
scattering sense, forward scattering is
also known as missing. AUDIENCE: [INAUDIBLE] PROFESSOR: You can
have other reactions, let's say, where you
have a particle at rest, another particle slams into it,
and the whole center of mass moves together. I don't know if you'd
call that forward scattering as much
as, let's say, capture or fusion or something. But in this case, scattering
means that two particles go in, two particles leave. Whether it's
elastically, which means with no transfer of
energy into rest mass, or inelastically,
where, let's say, a neutron is absorbed
and then re-emitted from a different energy level. And that's something we'll
get into in, like, a month. So you can have forward elastic
or inelastic scattering. In this case, I'm talking
about elastic scattering, which is the simple case of, like, the
billiard balls miss each other. Which is technically a case
that can be treated by this. Because all you have to do
is plug in theta equals 0 and you have the case
for how much energy do you think the neutron would
lose if it misses particle 2. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah. It wouldn't lose any energy. Right? It would have the same energy. So that's the case for
forward scattering. A neutron, when it interacts
somehow with another particle, can lose as little as
none of its energy. If it misses, no one said
it had to lose any energy. And by solving
this equation here, which we'll do on
Thursday, we'll see what the maximum amount of
energy that neutron can lose is, which is the basis
for neutrons slowing down or moderation in reactors. Yeah. AUDIENCE: Are T in and T out
equal there, in which case that equation is used
to solve for theta? PROFESSOR: T in and T
out are not always equal. But in the case of forward
elastic scattering, they would be. Because the neutron
comes in with energy T in and it leaves with energy T in. For any other case in
which the neutron comes off of particle 2 at
a different angle, it will have bounced off of
particle 2, moving particle 2 at some other angle
phi, and giving it some of its energy elastically. The total amount of that kinetic
energy will be conserved. So let's say-- what
did we call it? What is it? Yeah. So T1 would have to be
the same as T3 and T4 together for this
Q equation where Q equals 0 to be satisfied. So what you said can happen. But it's only the case
for forward scattering. Any other questions? Yep. AUDIENCE: In the case of
an exothermic reaction, we assume that T1 equals 0. Can you re-explain why
we made that assumption? PROFESSOR: So the question
was, in an exothermic reaction, why did we say T1 equals 0? It's not always the case. But it provides the simplest
case for us to analyze. So an exothermic reaction
can happen when T1 equals 0. It can also happen when
T1 is greater than 0. So we're not putting any
restrictions on that. But in the case that T1
equals 0, s is destroyed and the harder part
of T is destroyed, making the solution
to this equation very simple and intuitive. Which is to say that if you
just have two particles that are kind of at rest and
they just merge and fire off two different pieces
in opposite directions, their energies are
proportional to the ratio of their single mass
to the total mass. So that's like a
center of mass problem. You'll notice also I'm not using
center of mass coordinates. Center of mass coor-- who here
has used those in 801 or 802? And who here enjoyed
the experience? Oh. Wow. No hands whatsoever. So center of mass coordinates
and laboratory coordinates are different ways of
expressing the same thing. Usually you can write
simpler equations in center of mass coordinates. But for most people-- and I'm
going to go with all of you, since none of you raised your
hand-- it's not that intuitive. That's the same way for me. So that's why I've
made a decision to show things in
laboratory coordinates, so you have a fixed
frame of reference and not a moving frame of
reference of the center of mass of the two particles. But the center of mass idea
does kind of make sense here. If you have two particles
that are almost touching and then they touch and they
break into pieces and fly off, the total amount of momentum
of that center of mass was 0. And it has to remain 0. And so each of these particles
will take a differing ratio of their masses away. We already looked at this
for the case of alpha decay, where if you have one nucleus
just sitting here-- let's say there was no T1. There was just some unstable
T2 that was about to explode and then it did. Remember how we
talked about how the Q value of an alpha reaction is
not the same energy that you see the alpha decay at? Same thing right here. So this Q equation describes
that same situation. Notice there's no hint of M1. There was really
no M1 in the end. We don't care what the initial
mass of the particle that made alpha decay is. All we care about is what are
the mass ratios and energy ratios of the alpha particle
and its recoil nucleus. So it all does tie together. That's the neat thing,
is this universal Q equation can be used to
describe almost everything we're going to talk about. So this is as
complex as it gets. And from now on, we'll be
looking at simpler reductions and specific cases of each one. So it's five of. I want to actually make sure
to get you to your next class on time. And I'll see you
guys on Thursday.
