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Particle Colliders and Collision Energy

  1. Apr 26, 2010 #1
    Sorry for taking your time with elementary problems, but this question seems to be so simple that the only people who take it seriously are the ones who don't know what they are talking about.

    Consider the question of whether it is worse to be in a car hitting a concrete barrier of nearly infinite mass and negligible give, compared to a car that hits an oncoming car of the same mass and equal but opposite velocity.

    Now, in a particle collider, the reason for arranging collisions of similarly-massed particles travelling in opposite directions is that the energy levels attained can be many thousands of times greater than bombarding a solid target with the same types of particles at the same velocities. I want to discuss why such colliders can achieve significant gains at all, and I beg someone who really knows what he is talking about, to come down from the levels of sophistication at which he (or of course, she) normally works and kindly make matters clear.

    First: a matter of faith, dogma, confidence, logic, etc etc: The total amount of energy the collision yields cannot exceed the amount put into accelerating the particles, no matte what, no matter whether the particles are relativistic or not. In fact, there always will be losses to side effects of various types.

    Consider a case first, of suspended steel balls, such as you could find in a Newton's cradle. Next to a single ball we put a smooth, near-as-dammit rigid steel wall. We raise the ball to a standard height and let it swing against the wall. In a perfect world it would bounce back to the starting position. As nothing really is perfect, It would only recoil most of the way till it bounces to a stop. We observe that a certain amount of energy is expended in the process, energy in the form of momentum, kinetic energy, potential energy etc.

    Well then, we remove the steel wall and replace it with a second steel ball. We repeat the experiment, but this time we raise and drop BOTH balls, thereby priming the system with precisely twice the amount of energy that we started with before. And we see...

    Both balls recoil to the same height as before, right? and bounce the same number of times, yes? In short, each ball experienced the same impacts and accelerations as before, with the same energy as before, no?

    Near as dammit, of course.

    But when we do the analogous thing in a particle collider, although we still get out no more energy than we put in, we find the process thousands of times more profitable than hitting even the most rigid of stationary targets. How much of that is because of relativistic effects, and how much of a role does relativity play? I assume that part of the problem is that at such energies there is no such thing as a "rigid" stationary target? That for such purposes atoms within say a diamond crystal do not differ significantly from atoms in liquid methane?

    There is another thing of course. Our steel balls approximate to elastic rigid objects. Suppose that we replace them with crushable material, possibly something like putty. I would expect that the amount of deformation of the putty PER BALL would be effectively the same in both experiments.

    Am I wrong? Why? Given the equation for kinetic energy of a given mass at a given velocity: e=0.5*m*v*v it is true that the amount of energy for a fixed mass rises quadratically with velocity; it then makes sense that having put twice as much energy into the two balls as into one ball, I should get out twice as much energy as when I dropped just one ball against the rigid barrier. And yet, I could argue that I am entitled to regard the two moving balls as one ball stationary and the other at twice the velocity bearing four times as much energy, instead of twice as much. But I did not put in four times as much; only the same old twice as much!

    Why is this not true? What am I missing?

    And for the two cars; will the drivers be worse off than the driver of the single car hitting a rigid barrier? And if not, who would like to demonstrate, using swinging eggs on strings -- eggs that he had paid for personally?

    And why should matters be different for relativistic particles in collision? I realise that their velocities will not change much when their energies do, but even so...?

    Anyone, please? Real, crushing, simple arguments? Equations permitted, even welcomed, but not as substitutes for sense please! Links to direct answers will do, if they are clear and pertinent, say, if they were in reply to essentially the same questions in the past.

    Thanks for your patience and attention,

    Jon
     
    Last edited: Apr 26, 2010
  2. jcsd
  3. Apr 26, 2010 #2
    So in essence you're partially asking for a derivation for Kinetic Energy? In of which case try wiki, depending on your level of mathematics you may understand or not, but it seems fairly straight forward. I would link you, but Im in a bit of a hurry. Hope that helps.
     
  4. Apr 27, 2010 #3
    A more appropriate analogy would be, for example, a car hitting a stationary car that is sitting there in neutral, or a steel ball swinging into a freely-hanging steel ball. Having a totally rigid barrier that is completely fixed to the earth's reference frame does not really describe what happens in a "fixed" target accelerator experiment.

    For the car analogy, you would get essentially the same collision for two cars colliding, each traveling toward the other with speed v, or with a single car of speed 2v hitting an initially stationary car (but one that is allowed to move backwards once it gets hit). Because of relativity, once the velocities get near the speed of light, this advantage becomes much greater---a collision of particles with some center-of-mass energy E is equivalent to a collision where one of the particles is at rest and the other has an energy that is much, much greater than E.

    Your analogies would be like taking a collision with equal mass particles traveling toward each other and comparing it to a really, really, really massive particle at rest being hit by a much smaller particle. It's not really a good comparison for the question at hand.
     
  5. Apr 27, 2010 #4
    I appreciate your difficulty and your kindness in your hurry. (Not sarcastic! I really do.)
    No, I do understand e=mvv/2. I even knew it offhand from very rusty Phys I. My problem is slightly subtler and more humiliating than that. I trapped myself into making untenable statements (at least, I could not logically support them and now think I was wrong) concerning the collisions of cars with each other, as compared to with solid barriers.
    Now, "So what," you say, "correct your statement and get over it!"

    Well, that is not so simple. Firstly, if what I had said really was nonsense, then where does the extra energy yield in particle colliders come from (factors of thousands in extreme cases, I read) or alternatively, where does the energy loss come in when accelerated particles hit a rigid barrier instead of head-on collision with particles of equal mass and opposite velocity?

    Secondly, in checking up on discussions of the "paradox" which surely is only an apparent paradox in such a simple matter, I find such confused mutual argument and contradiction that it is hard to believe. "So, why worry your curly little head about that," you ask? "Just check on the logic of the argument, discard the crud, and go your way! Why bother us?"

    The trouble is that having put my foot in it with one thoughtless response, I now want to get it right; really watertight, copper-bottomed right, without tripping over my simplistic tongue this time! I would have thought that I had it right already, but then there are those minor problems:
    1: A lot of apparently educated, intelligent folk get into really persistent argument without anyone being able to provide a clincher. Why?
    2: I don't understand this collider business. A doubling of the energy yield I could understand; where does this factor of thousands come in?
    3: If the effect of head-on collision between two matched cars at equal speed, is about the same as hitting a solid barrier, then why is there any energetic advantage to a collider rather than an accelerator using a rigid barrier.

    I hope that shows up the difficulty more clearly?

    Thanks again,

    Jon
     
  6. Apr 27, 2010 #5
    Hi House,
    I take that to mean that the reason it is so important in the collider that the target is not stationary, is that I was right in suggesting that the rigidity of any "rigid" practical barrier is negligible from the point of view of a relativistic particle, right? The accelerated particle might as well have hit a free, more or less stationary electron, hadron or the like, right? Essentially the molecular bonds of matter (probably even nuclear forces) are tiny compared to the forces exerted by a near-c particle?

    Am I with you so far?

    Let's see whether I can get this one straight. Please bear with me and stand by to correct.

    Suppose I take a relativistic particle and I collide it with a matching particle such that practically all the energy E is released at the centre of mass. So far so good. Had the particles been at Newtonian velocities, I could have got numerically the same yield by doubling the speed of one particle and leaving the other one stationary. (And if they had been cars, the effect on the drivers and cars would have been the same too, right? And a "rigid" Newtonian barrier would have the same effect on one car travelling at half the closing speed of the two cars, right? If not, why not?)

    However, getting back to our relativistic particles, we cannot double the velocity, because that would demand trans-c speeds. No problem; we don't care about the speed as such; we are thinking about energy, OK? So all we have to do is increase the velocity by a (carefully calculated and very expensive, energetically huge) smidgen, and that would have the same effect? yes?

    Only, it is hellish hard and correspondingly expensive and takes a lot of energy to increase relativistic velocities by smidgens, what with bremsstrahlung and so on. It is energetically and technologically cheaper to produce a colliding beam in the opposite direction at the same (opposing) velocity. Yes?

    If yes, then the reason for the thousands of times increase in energy yield, rather than just doubling it, is that the effective increase in relative kinetic energy is far greater than from a Newtonian doubling of speeds, yes?

    Note: I am not telling you, but asking!



    Um... But it would seem to me that I have two not-too-similar analogies. I think I am now reasonably happy with the collider mystery, but for the car, it still seems to me that a good approach to an immovable, rigid barrier is indeed practical (ask any heavy duty, steel-shod reinforced concrete bridge pier if you won't accept my word for it). It also seems to me that for cars crawling along at say less than Mach 1, it would make little difference whether they hit that pier, or each other at the same velocity. For each car the energy would be the same. For hitting a stationary car at twice...

    Now-just-a-ruddy-second...!

    My car ramps up to a terajoule at N kph, right? It hits the twin car at the same opposing kinetic energy, right? Yield in one form or another; 2 terajoules, OK? After all, my fuel bill shows how much energy went into my acceleration. But my twin driver is stingy and wants to save energy. He refuses to waste fuel,so he stays stationary. So to get the same effect I need to go at 2N kph, right? Root 2 won't work because the closing velocity would no longer be the same. But if I speed at 2N kph, then my kinetic energy will be 4 terajoules, not 2!!! And saying that our speeds are relative don't cut the mustard; I still have the fuel bill to prove it. And again, what about hitting that solid bridge pier at N kph and 1 terajoule?

    I don't get that part one tiny bit!!!

    Sorry!

    Help!!!

    Jon
     
  7. Apr 27, 2010 #6
    I think you're OK up to here.

    There are a few issues here, since relativistic systems can be unintuitive if you're not familiar with them. What we want to do here is to compare a collision to the exact same collision, except viewed in a different (moving) reference frame. The simplest is to compare the center-of-mass frame (where two identical objects move with equal speed/energy toward each other), with a frame where one object is at rest.

    In a relativistic system these two situations will NOT have the same closing speed, as you mistakenly assume. In fact, if the relative motion is extremely relativistic, it is easy to estimate the closing speed, which depends very little on the actual energies involved. In the center-of-mass frame the two objects are moving toward each other at very close to the speed of light, for a closing speed of ~2c. In the frame where one is at rest, the other is again moving close to the speed of light (a little bit closer than before, but both values are very close to c), and so the closing speed is actually half. Note that this is different from the non-relativistic case, where the closing speeds are indeed the same in both reference frames.

    OK, I just realized that your 1 terajoule car isn't really relativistic. I'm not sure if that's intentional and I'm not sure I follow what exactly you're getting at in this paragraph anyway. And since I have to go now, hopefully this will give you a start at least. Ask again if things still don't make sense and I should be able to reply later (or I'm sure smarter people will probably reply as well).
     
  8. Apr 27, 2010 #7
    Apologies if this sounds patronising, but as I am unaware of your knowledge of physics, I've tried to say this in as simple and as broken down way as possible.

    In the scenario which you described with the cars, or even metal balls, energy is conserved quite rightly. But the thing is, they are rigid objects which have an inertial mass that stays there and does so for the entire experiment. But when we look at particle collisions we can no longer talk about just the conservation of energy. We now have to talk about the conservation of Mass-Energy. As everyone is these days, I'm sure you're familiar with Einstein’s famous equation E=MCC. Well, this equation comes from the notion that mass can be destroyed, by converting the "matter" into energy via the speed of light twice. Or in other words, when two particles collide and they annihilate energy is released in the form of photons (light). 2 to be precise, hence c squared. So when you ask how do we get more energy out then what we put in, then the answer is say in proton-proton collisions, the same mass that went in doesn’t always equal the mass that comes out. This is because as the protons split and each subatomic particle decays or various other scenarios some annihilate, the mass gets destroyed, and is emitted as energy. Also you may think, well why is there any mass at all? And this is because the converse is also true; two photons can collide to create particles. If you wish, I can show some equation and mathematics later on, but effectively the energy of the photons released is equal to the energy due to the mass that is destroyed. This results in our conservation of mass-energy law.

    To follow on from before, the difference in particle collisions rather then accelerator-barrier collisions, is that there are two mass which can annihilate, meaning there is twice the possible energy to be released, rather then just having the 1 particle and 1rigid barrier, which only allows the proton to split into subsidual energy parts, whilst leaving the barrier as it was.

    To the first: This is because alot of intelligent folk haven’t studied it far enough along the chain, and so there understanding of what truly happens, peters out before they can find said clincher.

    Finally to the second question: The factor of a thousand can be seen quite happily because of the energies and masses involved. And the fact that c squared= 9x10^16.



    I hope this helps

    Si
     
  9. Apr 27, 2010 #8
    OK, Between You and Si, you had clarified that point rather nicely.

    Hm. I take it you mean "~2c" metaphorically?

    No. The Newtonian reference was in fact deliberate. Sorry I didn't make that clear enough. What had happened was that in my readings from the early years of collider technology, the fact that relativistic considerations played a major part in the improvement of energy yield from accelerators was not made clear, and I had been thinking in terms that amounted to Newtonian mechanics.

    However, I now feel vaguely but more comfortably confident about one major factor, namely where the large discrepancy comes from in relativistic collisions. So I was thereafter concentrating on the Newtonian case, which I had just decided that I understood very nicely thank you, until a nasty accountancy problem raised its head, which once again recalled my original source of confusion.

    Again.

    You certainly have helped. Many thanks.

    Jon
     
  10. Apr 27, 2010 #9
    No. In reality for an ultrarelativistic collision the relative velocity in the center-of-mass frame might be something more like 1.9999c, but that's pretty close to 2c, which is what the "~" is supposed to represent

    This is correct. Non-relativistically, you get the same collision with two identical opposing objects, each with energy E, or with one object stationary and the other with energy 4E (twice the original velocity). Going to a relativistic system makes that energy discrepancy much, much larger.

    Simon's response was pretty good (which is a good thing, since I somehow missed the post he replied to, asking about things like energy yield), although I would like to correct two minor things:

    1) There is no relationship between the power of 2 in E=mc^2 and the number of photons produced in some annihilation reaction. I'm not sure where that idea might have come from.

    2) Since proton-proton collisions are mentioned specifically, I should point out that it's impossible to "release" energy in a proton-proton collision. Baryon number conservation ensures that the total final mass will be greater than or equal to the mass of two protons, and the total kinetic energy will correspondingly be less than or equal to the initial kinetic energy. Of course this is not the case for something like proton-antiproton collisions, in which case what you describe is correct--you could in principle end up with more total mass or with less total mass than you started with.
     
  11. Apr 27, 2010 #10
    You did exactly right, and I did not read what you said as patronising at all.

    Mmm... Fine mostly, but I seem to have misunderstood something. I certainly understand that conversion of fermions to photons would yield energy, sort of, and might affect the accounting perhaps, but I had not thought that that was the source of a major part of what I have been calling the energy yield (meaning the amount of energy useful to the experimenters). In fact I would have expected that particle interconversions would generally consume more energy than they yielded. Otherwise we could look thither for nuclear power, which afaik, we can not. I had thought that practically all the usable energy we got in the colliders came from what we had pumped into the acceleration. I accept that "matter"/photon interconversion would complicate the accounting, but would not have thought that that would have figured in the order of magnitude of difference between collider energies and the energies from accelerators with stationary targets. I thought that was essentially a matter of the implications of working in the very high relativistic regimes rather than at predominantly Newtonian velocities.

    Am I wrong there?

    Without knowing the details, I am happy with that item. No problem. Good Ol' Feynman diagrams & so on. Exactly what kind of disturbance might promote say, the splitting of a MeV gamma photon, potentially immortal, into an electron and positron, and how it could do it, I couldn't guess, but that would be a side issue here.


    Uh... Yeah? Proton/anti-proton I can see. But what about lead atoms? And anyway, a proton is only about a GeV energy equivalent; 2 GeV shouldn't figure very impressively in the workings of modern colliders, surely, never mind in the LHC surely? Please clarify.

    This I was beginning to appreciate. It seemed to me to amount to the nature of the "softness" of the nearest we could get to a rigid target, namely anything made of atoms, no matter how firmly bound.

    That part I had appreciated vividly! :wink: It was to avoid joining them in their futility that I was pestering you folks.

    As always, yes, but what I am embarrassed to say is that my main problem was, and still is, that the relativistic examples were inadvertent digressions. I have learned a lot from them that I find very valuable, even if what I learned is intuitive and qualitative.

    My still operative problem however, is trivial in comparison. I am not comfortable with why (and therefore whether) Newtonian collisions are eluding my insights, irrespective of the figures.

    If your patience will stand it, let me put the problem again, in the light of my Newton's cradles and steel or concrete barriers, and my projectiles, variously "rigid"-elastic or deformable.

    Let us accept as a unit of energy and its implication of velocity 1 Gigajoule. (F) Never mind how appropriate that might be for speeding bullets, cars or the like. Let us accept that in all cases our masses are gravitationally insignificant, grammes or tonnes, not petatonnes or the like. Furthermore, our velocities are thoroughly Newtonian, M/s or k/s, not thousands of k/s.

    OK? Ordinary Newton's cradle or jet plane stuff, neither snails nor cosmic ray particles.

    Now. I crash my ball or my car into the rigid barrier. Assume that the barrier has negligible "give". I have invested a resultant 1F in acceleration. The effect dissipates 1F, whether in rebound or in deformation of the projectile.

    So far, so simple. Right?

    Now, I get lonely and call on my friend who has a matching projectile. Here we need to watch the accountancy more carefully.

    When we bounce our balls they rebound or deform exactly as they did from the barrier, whether we bounce them directly off each other, or whether we bounce them simultaneously from opposite sides of the barrier. Our closing velocities still are thoroughly Newtonian.

    Right? If this is not right, then where did the energy discrepancy go, or where did it come from?

    =================First checkpoint===========

    So now, as before, my friend gets lazy or stingy. "Energy is energy," he says, "I'll park here with my brakes and gears disengaged, a frictionless mass. You accelerate to 1.4142 of the velocity that took you 1F of energy to attain last time. Unlike you I am an advanced physicist and I understand all about e=mvv/2. That way you will have a kinetic energy of 2F, which is the sum of the amount of energy we put into our acceleration before."

    "That sounds wonderful" I agree. "How lucky I was to find a physicist like you! This way we can save a lot of energy, can't we? I would have thought that to keep our outcome consistent I would have had to expend 2F in acceleration. But tell me, how come we can get the same effect with a smaller closing velocity? Surely Newtonian relativity should demand the same closing velocity?"

    "Don't be silly," he says, "If you doubled your velocity you would quadruple your kinetic energy! Do you want to kill us both with a 4F collision?"

    "But, that was the closing velocity we had last time. Why did we not have a 4F collision last time?"

    Now am I putting the question more comprehensibly?

    Thanks to both for your help.

    Jon





    Si[/QUOTE]
     
  12. Apr 27, 2010 #11
    Phew! Some overlap in this thread! :smile:
    Yes, I knew that, I just was not used to thinking in terms of beyond-c relative velocities. In fact I still am uncomfortable with the idea.

    Now there I fall off the bus. What happens to the energy discrepancy? All sorts of things begin to ring alarm bells.

    Suppose the projectiles are effectively rigid-elastic balls, too strong to be damaged or visibly deformed in the collision. Each of us spends F in fuel to accelerate his ball. They meet in the centre and rebound with about the same separation velocity as their final closing velocity. Total fuel bill 2F.

    Right?

    That was fun, so now we try a variation. You leave your ball at rest and I accelerate mine to twice the original velocity. Not surprisingly this costs me 4F in fuel. Yes? After the collision, if I have been waiting next to the stationary ball, I see the moving ball stop dead, and the other one take off at the closing velocity.

    Great stuff. If instead I had taken off after the originally static ball at half the closing velocity, it would have looked to me exactly like the first collision, with the same closing velocity, the same directions, and the same parting velocity. Beautiful. Indistinguishable.

    Except that my (total) fuel bill is now 4F instead of 2F.

    Suppose that instead of steely balls we had had cars with crunch zones, or putty balls. Now, when the balls collide, they stick and undergo 2F-equivalent distortion, forming a new single lump of a given shape. Energy bill 2F.

    Happy?

    You are ahead of me, right?

    So we leave one ball stationary and ramp the other up to twice the velocity, at a cost of 4F in fuel.

    What happens? Well, one reasonable expectation seems to me that the balls collide and stick, achieving exactly the same shape from 2F worth of distortion, then move off at half the closing velocity, which would make it look the same to the observer who moves off at such a velocity that it all looks the same to him as the event when the two moved at the same velocity around the same centre of mass.

    But what about my fuel bill??? Sorry to sound so obsessionally mercenary in a scientific matter, but after all, 4F... It didn't go into the distortion, nor into the velocity, did it? Where did it go into?

    And, the question that started this off: does the driver in the crumple-zone car experience anything worse for more fuel than for less? Does it take 4F hitting a pier to make him feel as bad as 2F hitting an oncoming car?

    Whether yes or no, why?:confused:


    Thanks, but still importunate!

    Jon
     
  13. Apr 27, 2010 #12
    I guess I'm still not entirely sure what you're getting at (probably because I haven't had time to read everything as thoroughly as I should have), but lets see if this helps:

    First consider the case of two identical objects colliding. In that case, all we need to do is talk about different reference frames, but the same actual physical collision. It's important to note that the total energy in the system is NOT the same in each reference frame. In fact, the center-of-mass frame will always be the one for which the total energy of the system is the smallest. Any frame of reference that is moving with respect to the center-of-mass frame will measure a larger total energy, since there's both the relative motion, but also motion of the center-of-mass. At the risk of myself sounding patronizing, you can think of just the simple case of a single particle. In one frame it's at rest, but in a different frame of reference it is moving and so will have a larger energy.

    In the non-relativistic case for a colliding system, as you mention, the frame with one object at rest has twice the total energy as the center-of-mass frame. (In the relativistic case, the two frames are boosted compared to each other at close to the speed of light, so it happens to be a factor larger than 2, but that appears to be irrelevant to your question).

    So if I'm reading your question correctly "where did the energy go?", you would say that the extra energy went into the motion of the entire system as a whole (the center-of-mass of the two cars), which was at rest in the lower-energy frame/case.

    I have no idea if that helps, but hopefully it can't hurt.

    I'm not sure what to say about the case of the car hitting a rigid fixed object. From the perspective of the car, it's the same as hitting another car with opposite velocity, since the acceleration of the car is the same in both cases. Other than that, I'm not sure what there is to say about it. You'll just have to keep asking questions until they've finally been answered in a satisfactory manner, I guess.
     
  14. Apr 27, 2010 #13
    Patronising doesn't bother me, especially when people are trying to help.

    What does bother me is my total inability to imagine how changing the frame of reference could alter the fuel bill in a problem such as this. Never mind the money; what bothers me is that more (or less) actual fuel had to get burned to achieve the same results while all the observers in all the frames of reference were watching the whole thing. They agreed on all sorts of things except for trivial differences in timing and direction, but above all, they all would agree on how much fuel got burnt, and to what effect.

    There I am afraid you lose me utterly.

    Doesn't something about that discrepancy bother you?

    Or why is there no discrepancy except in my own mind? Where have i mis-applied good ol' e=mvv/2???


    But, as always, thanks for your patience and courtesy,

    Jon
     
  15. Apr 27, 2010 #14
    That's good. Something like the "fuel bill" should be the same for every frame of reference, if you're looking at the same process. I think perhaps you're not thinking of equivalent situations.

    If you take two cars at rest and accelerate them until they are moving toward each other, each with speed v, the initial and final state (just before they collide) differ by 2E = mv^2. If you start with two cars at rest and accelerate one of them until it's moving at 2v, the initial and final state have an energy difference of 4E. Both collisions are equivalent (what I'm calling the "final state" in each case differs only by the frame of reference), but the "initial" state is not the same if you look at it compared to the center-of-mass reference frame of the cars just before the collision: in the first case the two cars were initially at rest in the center-of-mass frame, while in the second case they were both initially moving (together) in the center-of-mass frame (note that I'm talking about the center-of-mass frame after the car or cars have been accelerated). It should be no surprise that it takes a different amount of energy to accelerate the cars to their final state.

    Does that make sense?
     
  16. Apr 28, 2010 #15
    That is very much like what I was groping for. I knew I had to be looking at the wrong distinction, but I could not objectively identify where I was erring. I am working on some clarificatory model examples.

    I deduce that in fact, running into another (matching) car at the same speed is as good as running into a concrete wall. A bigger or faster car is worse, and a smaller or slower car is better.

    Would you agree?

    Thanks for your trouble so far!

    Jon
     
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