the following content is provided under a Creative Commons license your support will help MIT OpenCourseWare continue to offer high quality educational resources for free to make a donation or to view additional materials from hundreds of MIT courses visit MIT opencourseware at ocw.mit.edu okay guys welcome back as you can see we're not using the screen today this is gonna 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 will 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 gonna 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 one that's firing at a large nucleus two and afterwards a different small nucleus three and a different large nucleus four 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 three went off at angle theta and particle four went off at angle fee so hopefully those are differentiable enough and if we were to write the overall Q equation showing the diff 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 to C squared plus t2 has to equal mass 3 C squared plus t3 plus mass 4 C squared plus t4 we're just four symbols M refers to a mass t refers to a kinetic energy and so this conservation of total mass or total energies got to be conserved and we'll use it again because again we can describe the queue 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 SI Squared's together for easier writing if we take the initial masses - 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 and keep with one 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 T - that's just a simplification that will 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've 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 probably know that we probably know what things were 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 we didn't actually even know the masses yet we 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 three and four is going to depend on what angles they fire off at let me give you a limiting case let's say theta was zero what would that mean physically what would be happening to particles one two three and four if theta and fie were zero if they kept on moving in the exact same path yeah we don't know well let's see if 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 one scatters off particle two at an angle of zero degrees that's what's known as forward scattering which is to say that theta equals zero so this is another quantity that we might not know we might not know what theta and phiy are and the problem here is we've got like three or four unknowns and only one equation to relate them so what other question when you say 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 & 2 were to fuse like a capture event for example or a capture and then a readmission let's say of a neutron yeah if it was readmitted in the forward direction then that could be an elastic scattering event but still in the same direction or an elastic scatter at an angle of theta equals zero could be like there wasn't any scattering at all because really in the end it can matter let's say if you have a neutron firing at a nucleus depends on what angle it bounces off of in like the billiard ball sense if it bounces off at an angle of zero that means it missed we would consider that theta equals zero 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 momentum right so let's start writing some of the momentum conservation equations so we can try and nail these things down so let's 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 equation separately so what's the momentum of particle one how do we express that yep so it would be like m1 v1 so we'll have a little box right here from momentum we could say mass times velocity or how do we express that in terms of 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 to MT because in this case we would have root two let's see root two times m1 times one-half m1 v1 squared the twos here cancel the let's see I 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 one just what I've got up there to em 1 T 1 what's the X momentum in this frame of particle 2 0 it's not we're assuming that it's at rest and now what's the X momentum of particle 3 I heard a couple of things can say him louder it's ready - yep root 2m 3 T 3 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 to em 44 cosine fee now let's do the same thing for the Y momentum what's the Y momentum of particles 1 & 2 0 there's not moving in the Y direction to start and how about particle 3 I hear whispers but nothing vocalized yep same thing but root 2 M 3 T 3 sine theta and M 4 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 zero so must the final momentum in the Y direction so these two momentum has to be equal and opposite and that's time sine fee 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 the DS in terms of some of the others so let's get started with that because we angles are kind of messy and theta will use theta should uniquely define fee let's try and get things in terms of just one angle so I'm going to start by separating the Thetas and the fees on either side of the equal 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 it will separate angles so we'll have root 2 m 1 t 1 and minus root 2 M 3 t 3 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 m 4 T 4 cosine fee 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 equal sign so that would give us minus root 2 M 3 t 3 sine theta equals minus root 2 M 44 sine fee 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 everything here has a square root of 2 so we'll just get rid of all of the square root of twos to simplify as much as possible and now we look a little stuck but now's the time to remember those trigonometric identities back from high school that I don't think does anyone use these since in 1801 or 1802 anyone used a trig identity a little bit okay um 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 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 unangan the 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 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 fee plus a sine squared of fee which also equals one so we can destroy this fee 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 M 1 T 1 this so we're gonna take that stuff squared and that square it's not too hard neither are those so we'll have root M 1 t 1 squared which just gives us M 1 t 1 minus root M 1 T 1 times root M t-33 cosine theta let's just lump those terms together as root M 1 and 3 t 1 t 3 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 classes for okay and then we've got another one let's just stick a two in front of there and Plus that term squared so we'll have em 3t three let's see yeah looks like cosine squared of theta yep equals this one's easier m44 cosine squared of fee okay now we'll do the same thing for the Y momentum equation much easier because there's no addition anywhere and we have m3t three sine squared theta here equals M for T for sine squared feet 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 and equations and let's try and group all the terms together so we have M 1 T 1 minus 2 root M 1 M 3 it's getting hard to write over the lip of the chalkboard here cosine theta and we have M 3 t 3 cosine squared theta plus M 3 t 3 sine theta so we're sine squared theta equals M 44 cosine squared theta plus sine squared theta or fee I'm sorry cosine squared fee plus sine squared fee okay 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 6 instead of just the 4 visible but I think we'll finish this derivation in 4 boards so let's write what we've got left see your remaining so we have M 1 T 1 minus 2 root M 1 M 3 so much easier to write standing up cosine theta equals M 44 quite a bit simpler did I miss a term ah thank you right you're right and we had a plus M 3 T 3 yep that would be important thank you equals M 44 so we now have a relation between the math 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 yeah probably because like you guys have been doing on problem sets 1 & 2 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 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 gonna copy it up from there t3 plus t4 minus t1 so we can isolate t4 say t4 equals Q plus T 1 minus T 3 and continue substituting 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 M 1 t1 minus 2 root m1 m3 t1 t3 cosine theta plus M 3 T 3 equals M 4 times Q plus t1 minus t3 I anticipate us needing to see this side of the board soon 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 don't feel so old we had I was in 1802 3 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 like 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 M for 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 over or sorry t1 times M 1 over m 4 minus 2 over m 4 times root of all that stuff plus T 3 times M 3 over m 4 equals Q plus T 1 minus T 3 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 t ones together and group all the t threes together so if I subtract T 1 I get T 1 times M 1 over m 4 minus 1 minus 2 over m 4 root M 1 M 3 T 1 T 3 cosine theta plus and if I add T 3 then I would get M 3 over m 4 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 T 1 is probably how about the masses M 1 and M 4 if we know what particles are reacting we can just look those up or measure them or whatever we know M 4 we know our masses we know T 1 what about T 3 we don't necessarily know yet so T 3 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 queue 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 queue 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 oh 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 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 it is it's a quadratic hoop 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 T 3 and you have something with t3 better known as root t 3 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 yep yep over to a 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 is it ends up with root T 3 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 a squared plus T or s let's see if I can remember this without looking it up now 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 B 1 cosine theta over m 3 plus M 4 and T equals M 4 Q is it a - it is a + / M 3 plus M 4 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 II 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 EES 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 okay 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 zero 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 is exothermic which means it'll 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 okay there we go so that's kind of the analogy so t1 has to be greater than or equal to zero it's pretty much not a condition right it happens all the time so if we were to say t1 were to equal zero let me get my crossing out color again if t1 were to equal zero then s could equal 0 and t10 here and then you just get that's an S t equals m 4q over M 3 plus M 4 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 III plus e 4 has to be greater than or equal to e 1 all this mmm thank you cuz ease will be used in a different point of this class so we'll stick with T for kinetic energy thank you so although that's how 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 for 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 a mission 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 equal to greater than or equal to 0 or else the roots of the our 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 our nuclear reactions not or are they allowed let me rephrase that when our 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 as everyone clear on the implications here if not let me know because that's what this class is for yes so for exothermic reactions where Q is greater than zero all that says from our initial part of the Q equation if Q is greater than zero 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 zero which makes s equal to zero because there's a 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 three right here yeah up there particle three 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 zero that just means that t3 equals lower case T which equals that so then you uniquely define 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 zero it's like the it's like the excess you need to get a chemical reaction going does anyone here ever played with let's that's the one a striking one here well has anyone ever led 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 anyone think of one yeah it's a hydrogen generator I think that happened 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 - is this great that it's on video you do that plus put minging these dioxide and 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 I'm paying 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 out of them it's a pretty sweet job so if you guys are interested in academia you know these these things happen in life it's pretty cool yeah all right so 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 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 that's the general explanation I'd give so I forget who had asked the question but does that help explain it a bit cool okay I'll take five minutes and let's do a severely reduced case of this the case of elastic neutron scattering is 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 QED equation everything else that we're gonna study is just a reduction of that and this is about as reduced as it gets so inelastic 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 one so we can say that m1 equals one 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 and an arbitrary nucleus what's m2 yep 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 gonna use these two we're gonna 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 it write 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 m 1 equals M 3 M 2 equals M 4 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 T one times M 1 is 1 over M 4 is a minus 1 minus 2 over m 4 is a this is where it gets nice and easy M 1 and M 3 are just 1 so 1 times 1 times T 1 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 and what do we have left Plus T out and let's make T 1 into an in right there x + 3 / M 4 + 3 was one M 4 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 one a of Oreille here we can call one a over a there then lets us combine our denominators and stick the sign 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 A's go away there we have a much simpler equation 0 equals T in of let's see that's one minus a over one okay we'll just call it one minus a minus two root T in T out cosine theta plus T out a plus one and okay it's ten minutes of or it's five minutes of five minutes of so I'm gonna 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 form which hopefully you all know now exists yep so let's let's look at elastic scattering as an example so inelastic scattering two particles bounce off each other like billiard balls in forward elastic scattering the neutron after interacting somehow with particle two keeps moving forward unscathed so in the last excatly sense forward scattering is also known as missing you can have other reactions let's say we have a particle at rest another particle slams into it and the center of 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 interests mass or inelastically or 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 it's just technically a case that can be treated by this because all you have to do is plug in theta equals zero and you have the case for how much how much energy do you think the neutron would lose if it misses particle two 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 neutron slowing down or moderation in reactors yeah T&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 two at a different angle it will have bounced off a particle two moving particle two at some other angle fee and giving it some of its energy elastically the total amount of that kinetic energy will be conserved so let's say what would we call it what is it yeah so T three T one would have to be the same as T 3 and T 4 together for this Q equation where Q equals zero to be satisfied so that what you said can happen but it's only the case for forward scattering any other questions yeah so we the question was at an exothermic reaction why did we say t1 equals zero 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 zero it can also happen when t1 is greater than zero so we're not putting any restrictions on that but in the case that t1 equals 0 s is destroyed and the comment 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 to 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 court 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 and center of mass coordinates but for most people and I'm gonna 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 zero and it has to remain zero 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 a 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 ax 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 like 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 v of I would actually make sure to get you to your next class on time and I'll see you guys on Thursday