The Laboratory FrameThe nucleon-nucleon center-of-mass frame is ideal for deriving the relations between spin observables and the interaction matrix, but it is not always a useful way to conduct experiments. Intermediate energy nuclear physics experiments are typically fixed target experiments, so it is necessary to convert the expressions for to be in terms of laboratory frame polarization transfer observables.

The laboratory frame will be defined as:

and

The axes are typically referred to as longitudinal , normal , and sideways . and represent right-handed Cartesian coordinate frames In terms of the laboratory frame coordinates equation can be expressed as

The analyzing power associated with the out-of-plane component is still referred to as by convention. Therefore, the laboratory frame polarization observables can be obtained by measuring both the initial and final polarization.

In order to determine the Bleszynski, et al. observables, , [BBW82] it is necessary to express the laboratory frame observables (i.e. ) in terms of the center-of-mass spin observables (i.e. ). This requires a transformation of the reference frame which includes a relativistic effect on the spin rotation. This transformation is presented in detail by [IcK92].

Figure: Scattering kinematics in the center-of-mass and laboratory reference frames including the relativistic effect on the spin rotation explicitly [IcK92].

First of all, the unit vector normal to the scattering plane will remain unchanged , so . For the in-plane components the final state coordinates, and , will be rotated by an angle, , which is the relativistic spin rotation as is shown in figure . is given by

where is the velocity of the particle, is the velocity of the c.m. frame with respect to the laboratory frame, , and and are the scattering angles in the c.m. and laboratory frames, respectively. If we take to be the angle between the incident beam direction, , and the unit vector as defined in equation (see fig. ) then the transformations look like

Given these expressions between the coordinates the relations between in-plane spin observables is

Response Functions

To this point in the formalism only nucleon-nucleon (nn) scattering has been considered. In order to relate this to the nucleon-nucleus (nA) scattering of the experiment it is necessary to introduce the idea of nuclear response functions. The cross section for the nA collision can be expressed in terms of the nn cross section [KMT59] [BeS82]:

where is the free nn cross section, is the effective number of nucleons, which for reactions would be neutrons, that can participate in the reaction. This is also sometimes called the distortion factor [Che93]. The effective number of nucleons will, in general, be less than the full number of nucleons in the nucleus because the projectile nucleon will be strongly attenuated as it travels through the bulk of the nucleus--exactly how much is a function of energy--and therefore a portion of the possible target nucleons will be shadowed.

The Response Functions, , to a given projectile scattering operator, , are defined as [IcK92] [Bro95]:

where is the nuclear wavefunction. The operators, , can be defined as corresponding to the spin operators , where j = 0,n,p,q:

The operators, , are then defined as

where sum is over the number of neutrons, N, and the normalization means that the response is given as the response per neutron. In the limit of no Pauli Blocking ( orthogonal to all ) [Che93] [BeS82]

Just as the total cross section for nn scattering can be defined as the sum of the partial cross sections defined in equations to

one can use the response functions to define the nA scattering cross section. As can be inferred from equation the nA cross section is given by

where the 's are still the partial nn cross sections.

The Optimal Frame

In order to factor out the expression for total nA cross section into the product of several independent terms such as the nn cross section, response function, and other corrections (eqn. ) the assumption must be made that the nn amplitudes depend only on the incident energy and the momentum transfer, q. However, in general, the stuck nucleon will have its own momentum, , due to Fermi motion which enters the problem and can break down that simple factorization. Instead of integrating over the struck nucleon's Fermi momentum the process becomes much simpler if the nn amplitudes can be factored out by evaluating them in some ``optimal'' frame where the struck nucleon has a constant value, . This procedure has been discussed in [Gur86], [Smi88], and in detail in [IcK92]. For inelastic scattering, in the non-relativistic case, the optimal momentum was worked out to be [Gur86]

Notice, the optimal momentum will be zero for free scattering where .

The scattering matrix in the optimal frame, , is related to the nn scattering matrix, M (eqn. ) by [Che93]

where is the incident proton energy, is the Möller (kinematic) factor, and is the matrix which performs the frame rotation as defined in [IcK92]. The new scattering matrix can be written as

with the spin-dependent pieces being defined

where the superscript 0 indicates the incoming proton and the superscript i denotes a sum over A nucleons. The amplitudes ( , , etc.) are related to those in equation by equation . From the equations above it can be seen that ( ) does not excite the purely longitudinal(transverse) mode in the optimal frame. Therefore, and are no longer related directly to the spin-longitudinal and spin-transverse nuclear response functions. Instead, the nA center-of-mass partial cross sections, 's, given by

will be

where is a kinematic factor defined in [Che93] and is the distortion factor.

Obviously the choice of the optimal reference frame somewhat distorts the simple relationship between the center-of-mass spin observables, 's, and the nuclear response functions. However, it is the only way to deal with amplitudes consistently over a range of energy transfers, such as in the quasifree region, without violating energy conservation in each nn (nucleon-nucleon) interaction.

EXPERIMENTAL PROCEDURE

Research is the process of going up alleys to see if they are blind.

Marston Bates

The results presented in this work are based on data acquired during experiments E385 ( and other heavy targets) and E387 ( ) at the Indiana University Cyclotron Facility (IUCF) using the beam swinger charge-exchange experimental area and both the INPOL and KSU neutron polarimeters. These experiments took place in three parts: July 1995, December 1995, and February 1996. This chapter will discuss the facility, equipment, and procedures used to gather data during these experiments.

Center of mass reference frame

What we'll do is change to a moving coordinate system, which goes with the velocity of the center of mass! That's sounds weird. It makes the problem sound evem harder. But actually you'll see how simple things look in this frame.

Remember we said that if momentum is conserved, the center of mass velocity of the system is also. As the collision is taking place, it doesn't alter the motion of the center of mass a bit. It just plods along at a constant velocity. If we were coasting along on a bike at this center of mass velocity, watching the collision, what would we see?

Well in this reference frame, the center of mass velocity, by definition, is zero.

the total momentum is also zero. I'll notate all variables in this new system the same as the old, but just to remind ourselves that we're in this new frame I'll also add " ' " to them. So for example, the initial momentum of thr first particle is denoted .

So let's play before and after again, but this time in the center of mass reference frame.

Before:

The total momentum . The total energy is .

After:

The total momentum . The total energy is .

This looks a lot simpler. The momentum equations say that the particles have equal and opposite momenta, and Using this, equating energy is almost as easy

Factoring the masses and cancelling gives . There are two solutions to this. One is kind of boring, . It means that before and after, nothing changes. This certainly obeys conservation of energy and momentum, but means that the particles haven't bounced off each other. So what's the other more interesting solution? It's . From conservation of momentum, that means . In terms of velocity this gives

This says that after the collision, the two balls have reversed their initial velocities. That's it. This satisfies both momentum and energy conservation.

Back to the old frame

What did going to the center of mass velocity frame do for us? It told us the two balls reverse their velocties in that particular frame. But we want to know what happens in our original frame. Can we express what we just learned in a way that'll help us solve our original problem?

If we ask what happens to the difference in velocities before and after the collision, that'll tell us something independent of reference frame. The difference in the velocity of the two balls is always the same irrespective of frame. So what happens to the difference in velocities?

In the center of mass frame we see from eqn. 1.57 that . But as was just stated, this is true in our original frame

That is, the difference in velocities just reverse sign after the collision. You could also say that for any one dimensional elastic collision relative velocity of approach equals the relative velocity of recession. This is true in any reference frame. This equation is nice because it's linear. Its key to figuring out problems with elastic collisions.

So we now have two linear equations that we have to solve

Multiply the second equation by , and add it to the first equation to eliminate

Solving for

We can get most easily by interchanging the indices 1 and 2

Let's do a couple of examples now.

Collisions

Collisions are when two objects bang into each other. An example is when you play pool. If you're really good, you'll be able to actually aim the stick well enough so that the white ball hits another ball on the table. When that happens, you get a collision. There are two extreme versions of collisions. Totally inelastic collions, and elastic collisions.

Totally inelastic collisions are when the two objects the hit each other stay in contact. Think about two pieces of chewing gum being thrown at each other and sticking. On the other hand don't. It's too disgusting.

When a dropped superball hits the ground, it bounces up to almost it's original height. That's the opposite case of an elastic collision. In both cases the actual collision process is quite short. Often of the order of milliseconds. We'll try to understand the process of the collision now in a little more detail.

The Impulse

Suppose we look at a superball hitting a hard table with some high speed camera. What will we see? Well, we'll see the ball getting kind of smushed against the table. The more smushed, the greater the force the table exerts on the ball. So like a brontosaurus, the normal force of the table starts off small gets bigger, and then gets smaller again.

So the external force acting on the superball, namely the normal force of the table is some function of time pictured below.

Now how do we relate this to the momentum? Let's use eqn. 1.12.

The time derivative of the momentum of the ball is equal to the normal force of the table

So as you can think of this as saying that normal force of the table is decelerating thethe ball, causing it to slow down, so the momentum decreases. Eventually the ball completely changes direction because the force continues to act on the ball even after it has become maximally squashed. So the ball reaccelerates away from the wall until it looses contact with the table.

So if the normal force is responsible for the change in momentum, how is the total change in momentum related to this force?

Well we just integrate eqn. 1.12 from the initial time of contact to the final time of contact . This gives an expression for the change in momentum

This is just the area under the force curve pictured in figure 1.4.1. We can get an idea of the magnitude of the force by calculating the forced averaged over the interval from to .

From the last equation this means

OK, the derivation of this seemed like a lot of work, but the final result is quite interesting. The average force felt during a collision is the ratio of the change in momentum to the change in time.

So what happens if a collision is really fast? The average force during that time will be really large. Let's think about this in everyday life. When you drop a glass on a tile surface it breaks, where as on a shag carpet it won't even chip. What's the difference? About $10 if was a fancy glass like the kind my mom likes to keep around (but never uses except for when they have ``sophisticated" company). But seriously, the difference is that the carpet ``gives" more than the rug. It's softer. This means that the collision last a lot longer than on tile. So the average force will be quite a bit large, causing the glass to break.

With a lot of collisions, baseballs against bats, pool balls against each other, glass against tile, and countless other situations, the time of the collision is very short, of order milliseconds and the change in momentum is quite large. In this case you get quite humongous forces during a collisions. Let's do an estimate. Assume a time of contact of a mass of 1kg, and a change in velocity of 1m/s. You get an average force of which is . The weight of a 1kg mass is about 10N, so that the force during the collision is 100 times greater than the force of gravity. No wonder people try to avoid getting hit by baseball bats, they can pack quite a punch!

What that means is that during one of these fast collisions, you can ignore the effect of gravity, friction, springs, or any other force acting. The collisional force will exceed these other forces.

With this in mind, let's discuss collisions one dimensional collisions in more detail.

Totally inelastic collisions

Suppose Bobby is cycling east at 5m/s, and Lisa is cycling north at 10m/s happen to collide. They and their bikes get all tangled up and start moving together. What's the velocity of the Bobby-Lisa-bicycle glob right after the collision? Let's take the combined mass of Bobby and his bike to be and Lisa and her bike to be . (Don't worry, Bobby and Lisa are fine. This is not a real examaple.)

To figure this out, we'll use conservation of momentum, as we claimed that to a good approximation, we can ignore other forces during the collision.

Before:

The total momentum

After:

The . where denotes the final velocity, right after the collision.

Equating these two expressions (since momentum is conserved) we have

Now and so we get

The combined mass goes shooting off in the north-east direction.

This is an example of a totally inelastic collision. When the two masses hit, they stick together. The final velocity is just the center of mass velocity of the system, since the center of mass velocity is constant for any process obeying conservation of momentum.

Momentum is conserved but in general, energy is not. You could calculate the change in kinetic energy during this collision and would find that it is negative. What happens to this energy? It mostly goes into heat, some of it goes into sound waves. The scrunching sound that you hear is powered by the initial kinetic energy of our two unfortunate bicyclists.

Elastic collisions

With elastic collisions, energy is also conserved. The only really good examples of elastic collisions involve atomic particles and the like. But you can get pretty close in ordinary life, say with pool balls or a superball.

Lets say we start with two balls with masses and having initial velocities and respectively. They collide and end up with velocities and

Before:

The total momentum . The total energy is .

After:

The total momentum . The total energy is .

Equating the total momentum and energy before and after the collision

Let's try to analyze the collision of two balls in one dimension.

elastic collisions in one dimension

So now we just have to solve for the final velocities and and we're done. But how do you do that? One equation is linear and the other quadratic? If you try to substitute one into the other, you get a big mess. You can get around this by being a little clever, but there is a better way to think about all this which gives you more physical insight into what's going on with the physics of this problem.

example

A 1000kg lead ball traveling at 1m/s hits a .01 kg superball that is at rest. Calculate the final velocity of the two objects after the collision.

solution

In this example we have , , and . Plugging these into eqns. 1.62 and 1.63 we have to a good approximation

and

So what does this say? The heavy ball keeps on going pretty much with its same initial velocity. That's not too surprising. However the lighter ball gets propelled to twice the initial velocity of the heavy ball. That might seem a but unexpected, but you can understand this by going to the reference frame of the heavy ball.

In this frame the heavy ball sees this piddling little ball coming towards it and 1m/s and then bounce off it. The lighter ball will reverse it's velocity traveling away at 1m/s. But the heavy ball will just keep on truckin, its velocity hardly altering due to the collision. Now let's look back in the reference frame of the ground. If you add up the 1m/s of the heavy ball to the 1m/s of the lighter ball relative to the heavy ball, you get 2m/s.

solution

In this example we have , , and . Plugging these into eqns. 1.62 and 1.63 we have to a good approximation

and

So what does this say? The heavy ball keeps on going pretty much with its same initial velocity. That's not too surprising. However the lighter ball gets propelled to twice the initial velocity of the heavy ball. That might seem a but unexpected, but you can understand this by going to the reference frame of the heavy ball.

In this frame the heavy ball sees this piddling little ball coming towards it and 1m/s and then bounce off it. The lighter ball will reverse it's velocity traveling away at 1m/s. But the heavy ball will just keep on truckin, its velocity hardly altering due to the collision. Now let's look back in the reference frame of the ground. If you add up the 1m/s of the heavy ball to the 1m/s of the lighter ball relative to the heavy ball, you get 2m/s.

example

Consider a block going towards a mobile hump. Mobile hump? Is this getting esoteric or what! Well, look why not? This could be an oddly shaped piece of ice that slides friction-free on a surface. See the figure below.

{

Lets suppose the block and the hump have the same mass m and that the block, pictured in red, starts off with a velocity towards the hump which is initially stationary. The hump is of height 1m.

a

How far will the block rise?

b

At that point, what is the velocity of the hump and the block?

c

What is velocity of the hump and block after the block slides down again?

solution

First of all, it might rise all the way to the top and back down the other side! We don't know that yet. Let's assume that it doesn't get all the way to the top and solve for what height we get. If this is greater than 1m, then we know that it get's to the other side.

So what is the physics at the point where the block reaches it's maximum height. Well in the reference frame of the hump, the block rises up it and comes to a maximum as pictured above. At this point the block is stationary relative to the hump. That means that at this point, the block and the hump have the same velocity. We already considered this type of problem earlier! It's identical to the problem of inelastic collisions. The velocity in this case is just the center of mass velocity of the system which by eqn. 1.37 is

This is the answer to part b of the problem.

At that point the total kinetic energy of the system is

The inital kinetic energy was just , so the difference in these two kinetic energies must be the potential energy mgh (by conservation of energy).

Therefore

or

This is less than a meter so the block doesn't go over the top. This is the answer to part a of the problem.

Now how do we figure out c? We know that momentum is conserved and energy is conserved, so we this is an example of an elastic collision. Even though it is a little bizarre, it is a collision nevertheless.

So we can use our painfully derived formulae 1.62 and 1.63 to calculate the final velocities of the objects. In this case , and . Pluggin these in, we see that , and . So the block stops completely and the hump moves off with a velocity of 1m/s.

Notice how fiendishly difficult this problem would have been to solve using F=ma!

We could have solved this whole problem starting from Newton's three laws of motion. It would have been much much harder. Although conservation of momentum and energy are a consequence of Newton's laws, for many questions, they provide much insight to what's going on. You shouldn't forget however that they are a consequence of Newton's laws. They contain no extra information.

elastic collisions in two dimension

Now let's figure out what happens when objects collide elastically in higher dimension. Two should be enough for us don't you think? Let's ask what we can learn from eqns. 1.54, which are the equations for energy and momentum conservation . If we're given the initial velocities of the two objects before impact, we'd like to know what they'r velocities are after the collision. That's like we'd like but I'm afraid we're not going to be able to get those completely. What??? How can that be? In one dimension, we saw that applying these equations gave you the final velocities, so what's going wrong?

Well we applied Newton's laws directly, we're certain to get the complete trajectory of the particles, before, during, and after the collision. But that's quite tough. Instead we're using conservation laws. However even though conservation laws are derived from Newton's laws, they aren't equivalent. In general they contain less information so we shouldn't be surprised if they give an incomplete description of what's going on.

So back to the problem at hand. If we want the final velocities of the objects, that means we want to know four quantities since each velocity has both a magnitude and a direction that we wish to know. The problem is that we have only three equations. (It's three equations because the momentum equation is a vector equation, so each component is an equation in its own right. Two components means two equations.) So we have four unknowns and three equations. So we don't have enough equations to find the final velocities.

Does that make sense physically? Sure! If two balls collide with each other, they go off in different directions depending on how they collide. Think about playing pool. Depending on where the moving ball hits the stationary one, it goes off in a different direction. So without more detailed knowledge of the situation, it makes sense that we aren't going to know what the outcome of the collision will be.

Let's consider a particular case, where the masses of the balls are equal . Let's also take mass to be initially stationary, that is .

Then cancelling out the m's eqns. 1.54 become

The first equation says the vector sum of the final velocities is the initial veloicity. If you represent the two final velocity vectors and as the sides of a triangle, then will be the hypotenuse. The second equation looks kind of like the Pythagorean theorem. What kind of triangle obeys this theorem? A right triangle. So this says that and are perpendicular!

If you don't follow this reasoning you can get the same result by taking and squaring it

Subtracting gives . So the final velocities are perpendicular.

So you can see the momentum and energy conservation imply an interesting result. Of course we still don't know what final direction the balls will be traveling in. However we know they'll be going in perpendicular directions.

Rockets

We've all seen majestic pictures of big rockets blasting off from the earch, burning millions of dollars a second, and carrying interesting payloads into space, such as experiments on how spiders build webs in space. Fascinating.

But there are some general things about this picture that might strike you as odd. One is that payloads are pretty small. Also the size of these big rockets is absolutely enormous. Is there a reason for this? Yes I know what you're thinking, it's boys playing with their toys. But no, that's not the only reason. They're big because they have to be. We're now in a poistion to understand rockets, and you'll see why this happens.

What is a rocket? It's this thing that spews out lots of fuel at high velocity. The expulsion of fuel speeds up the rocket. The more fuel expelled, the faster the thing goes. We want to know how fast the rocket will be going after total mass of the rocket has been reduced from it's initial mass to its final mass . This will cleary have to depend on the velocity that fuel is expelled at. Let's call that U.

Let's try to understand this with a simple low tech example. A small child called Cindy sits on top of her little red go-cart loaded up with a huge pile of bricks. She's a strong girl and can hurl a large number of these out in one go.

Suppose she starts at rest with 128 bricks each weighing 1kg. We'll ignore Cindy and the go cart's mass. She hurls out 64 of these bricks, with velocity of . What will be the final velocity of the cart?

Well the initial momentum was zero, so assuming a frictionless go-cart, the final momentum must be the same. So if half the brickes are travelling to the left at 1m/s the other half must be travelling to the right at the same speed, 1m/s.

OK, now they're 64 bricks left in the cart. Cindy hurls out half of these, again with the same velocity relative to her. Now what will the final velocity of the cart be? To figure this out, go to the reference frame of the cart, which was initially travelling at 1m/s. In this frame, it looks like the initial problem, where the cart was stationary, but with 64 bricks inside it. After 32 bricks are hurled, by the same reasoning as before, the cart will be going an additional 1m/s, so that relative to the ground the cart is going 2m/s.

Now let's ask what happens when Cindy again hurls out half of these 32 bricks, at the same velocity she did before. Going to to a reference frame of 2m/s. we see as we did before, that the cart will be going 1m/s faster than before.

You get the idea. Every time half the mass is thrown out, the velocity increases by 1m/s. So if the mass decreases by , the velocity will increase by . In this case there is a logarimthic dependence of mass on velocity. You can check that the final velocity

Here U is 1m/s, and is the ratio of initial to final masses. What does this say? It's pretty hard to speed up a rocket. If you want the speed to go up by U, you have to expell half half your mass. We can rewrite the above equation as

So if you have a payload of mass , the amount of mass you have to start with, depends exponentially on the final velocity. Exponentials are very rapidly increasing functions. So you're initial mass normally has to be very large.

Now that we understand this simplified problem, we can ask what happens in a real rocket. Now we don't have a little girl hurling bricks, but a rocket called Cindy. Cindy expells fuel at a velocity -U (to the left), but does so continuously, not in the punctuated way we just analyzed. So if initially, we choose a reference frame where the rocket is at rest. What happens after it expells a little fuel of mass dM? The initial mass of the rocket is M. After expulsion, the mass will have gone down to M-dM. Let's call the increase in velocity dv. So conservation of momentum says that

This says that the velocty after this is

Since dM << M (formally its an infinitessimal, we can safely ignore it. So we get the equation

This is the increase in velocity after a very short time. If we now add up all these velocity increases, we'd formally be integrating the left hand side:

which says that

This is almost the same form as our previous analysis except now we get a logarithm base e instead of base 2. Qualitatively it acts the same way.