Anderson Video - Relativistic Momentum

Professor Anderson
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<font color="#ffffff">So, how does this play into things that we can measure like momentum?</font> <font color="#ffffff">Well we remember what momentum was, according to Newton's model, which we can</font> <font color="#ffffff">call classical, P is just M times V.</font> <font color="#ffffff">Okay, if you double V, you double the amount of momentum,</font> <font color="#ffffff">but you can't just keep doubling V forever and ever</font> <font color="#ffffff">because there's this universal speed limit to the world that we live in,</font> <font color="#ffffff">which is C. All right, you can't take V up above C, something bad happens when you do that.</font> <font color="#ffffff">And so we need the relativistic version of momentum,</font> <font color="#ffffff">which looks almost the same except we're gonna stick a gamma right in front of it.</font> <font color="#ffffff">Alright, and again, gamma is equal to 1 over the square root,</font> <font color="#ffffff">1 minus V squared over C squared.</font> <font color="#ffffff">Okay, it's a number bigger than 1. So relativistic momentum has more momentum in it</font> <font color="#ffffff">than classical momentum. All right, so let's think about this now in terms of one of</font> <font color="#ffffff">your homework problems and this is --</font> <font color="#ffffff">let's see if I can figure out what number it is.</font> <font color="#ffffff">Okay, this is one that we looked at earlier in office hours, it is</font> <font color="#ffffff">26.18 and it says the following. Let particle of mass M</font> <font color="#ffffff">move at some V, we'll call it V1 of 0.37 C.</font> <font color="#ffffff">How fast does V2 have to be to have twice the momentum?</font> <font color="#ffffff">Okay, so particle moving at V1, it's got some momentum,</font> <font color="#ffffff">we want to go faster to V2 and we want to hit twice the momentum.</font> <font color="#ffffff">So, if you were in the classical world obviously you would just double it, right?</font> <font color="#ffffff">Double V, it doubles the momentum.</font> <font color="#ffffff">But now, in this relativistic world we have this factor gamma which has V in it.</font> <font color="#ffffff">All right, so we have two relationships, we have P1 is equal to gamma 1 mV1,</font> <font color="#ffffff">and then we have P2, which is equal to gamma 2mV2,</font> <font color="#ffffff">and we know that that is going to be twice the initial momentum P1.</font> <font color="#ffffff">All right, so now we have this relationship here</font> <font color="#ffffff">and we need to solve for V2. All right, so let's see if we can do that.</font> <font color="#ffffff">Alright, so let's write it over here we've got two times P1</font> <font color="#ffffff">is equal to gamma 2mV2, and we know what P1 is it's gamma 1mV1.</font> <font color="#ffffff">Alright, right off the bat we can get rid of m.</font> <font color="#ffffff">Okay, m is the rest mass, they were calling it M naught, it's the</font> <font color="#ffffff">same thing, M is the rest mass. Those cancel out -- those cancel out and now</font> <font color="#ffffff">let's do a little trick, let's divide by gamma 2 and let's divide by gamma 1 on</font> <font color="#ffffff">the right side, okay? So, I took gamma 1 and put it under there,</font> <font color="#ffffff">I took gamma 2 and put it under there, and now let's square both sides.</font> <font color="#ffffff">So if I square this and I square this,</font> <font color="#ffffff">what do we get? We get 4 V1 squared over gamma 2 squared,</font> <font color="#ffffff">equals V2 squared over gamma 1 squared.</font> <font color="#ffffff">Okay, but we know that gamma 2 squared</font> <font color="#ffffff">is 1 over the square root of 1 minus V squared over C squared, so 1 over gamma 2 squared</font> <font color="#ffffff">has to be 1 minus V2 squared over C squared. 1 over gamma 1 squared is</font> <font color="#ffffff">going to be 1 minus V1 squared over C squared. And so now this whole thing here</font> <font color="#ffffff">becomes the following, we get 4 V1 squared times 1 over gamma 2 squared,</font> <font color="#ffffff">which is 1 minus V2 squared over C squared, and on the right side we have</font> <font color="#ffffff">V2 squared times 1 over gamma 1 squared, which is 1 minus V1 squared</font> <font color="#ffffff">over C squared. And now we can just simplify this and solve for V2.</font> <font color="#ffffff">All right, let's just do it real quick. Multiply everything by C squared we get</font> <font color="#ffffff">4 V1 squared times C squared</font> <font color="#ffffff">minus 4 V1 squared V2 squared,</font> <font color="#ffffff">equals V2 squared C squared minus V2 squared V1 squared,</font> <font color="#ffffff">and now I have a V1 squared, a 2 V2 squared --</font> <font color="#ffffff">I have a V1 squared, a 2 V2 squared and so I can put those together,</font> <font color="#ffffff">and then I have 4 V1 squared C squared that's left.</font> <font color="#ffffff">When I add 4 to that negative 1, I get 3 V1 squared V2 squared</font> <font color="#ffffff">and now we can put the V2's together,</font> <font color="#ffffff">V2 squared times C squared plus 3 V1 squared and then finally we</font> <font color="#ffffff">can divide by that quantity and you end up with the following. V2 is equal to --</font> <font color="#ffffff">and I'm going to take the square root, so I can do that right now,</font> <font color="#ffffff">2 V1 C, all over the square root of C squared plus 3 V1 squared.</font> <font color="#ffffff">Okay, I went through that slightly quickly but you can replay the</font> <font color="#ffffff">video later on and double-check your algebra and make sure you get the same thing,</font> <font color="#ffffff">and now you have all these numbers, right? We have V1 we have C obviously</font> <font color="#ffffff">and if you plug in all those numbers, you can double check with mine,</font> <font color="#ffffff">you should get 0.62, and it wants it in the units of speed of light, and so you</font> <font color="#ffffff">write 0.62 and then the units are C.</font> <font color="#ffffff">Okay, so those were for my numbers, you of</font> <font color="#ffffff">course will have different numbers for your V1. That one look familiar Jessica?</font> <font color="#ffffff">Yeah. That's the one we went over earlier in office hours,</font> <font color="#ffffff">painfully, I'm a add. All right, questions about this one before --</font> <font color="#ffffff">before we move on?</font>
<font color="#ffffff">So, how does this play into things that we can measure like momentum?</font> <font color="#ffffff">Well we remember what momentum was, according to Newton's model, which we can</font> <font color="#ffffff">call classical, P is just M times V.</font> <font color="#ffffff">Okay, if you double V, you double the amount of momentum,</font> <font color="#ffffff">but you can't just keep doubling V forever and ever</font> <font color="#ffffff">because there's this universal speed limit to the world that we live in,</font> <font color="#ffffff">which is C. All right, you can't take V up above C, something bad happens when you do that.</font> <font color="#ffffff">And so we need the relativistic version of momentum,</font> <font color="#ffffff">which looks almost the same except we're gonna stick a gamma right in front of it.</font> <font color="#ffffff">Alright, and again, gamma is equal to 1 over the square root,</font> <font color="#ffffff">1 minus V squared over C squared.</font> <font color="#ffffff">Okay, it's a number bigger than 1. So relativistic momentum has more momentum in it</font> <font color="#ffffff">than classical momentum. All right, so let's think about this now in terms of one of</font> <font color="#ffffff">your homework problems and this is --</font> <font color="#ffffff">let's see if I can figure out what number it is.</font> <font color="#ffffff">Okay, this is one that we looked at earlier in office hours, it is</font> <font color="#ffffff">26.18 and it says the following. Let particle of mass M</font> <font color="#ffffff">move at some V, we'll call it V1 of 0.37 C.</font> <font color="#ffffff">How fast does V2 have to be to have twice the momentum?</font> <font color="#ffffff">Okay, so particle moving at V1, it's got some momentum,</font> <font color="#ffffff">we want to go faster to V2 and we want to hit twice the momentum.</font> <font color="#ffffff">So, if you were in the classical world obviously you would just double it, right?</font> <font color="#ffffff">Double V, it doubles the momentum.</font> <font color="#ffffff">But now, in this relativistic world we have this factor gamma which has V in it.</font> <font color="#ffffff">All right, so we have two relationships, we have P1 is equal to gamma 1 mV1,</font> <font color="#ffffff">and then we have P2, which is equal to gamma 2mV2,</font> <font color="#ffffff">and we know that that is going to be twice the initial momentum P1.</font> <font color="#ffffff">All right, so now we have this relationship here</font> <font color="#ffffff">and we need to solve for V2. All right, so let's see if we can do that.</font> <font color="#ffffff">Alright, so let's write it over here we've got two times P1</font> <font color="#ffffff">is equal to gamma 2mV2, and we know what P1 is it's gamma 1mV1.</font> <font color="#ffffff">Alright, right off the bat we can get rid of m.</font> <font color="#ffffff">Okay, m is the rest mass, they were calling it M naught, it's the</font> <font color="#ffffff">same thing, M is the rest mass. Those cancel out -- those cancel out and now</font> <font color="#ffffff">let's do a little trick, let's divide by gamma 2 and let's divide by gamma 1 on</font> <font color="#ffffff">the right side, okay? So, I took gamma 1 and put it under there,</font> <font color="#ffffff">I took gamma 2 and put it under there, and now let's square both sides.</font> <font color="#ffffff">So if I square this and I square this,</font> <font color="#ffffff">what do we get? We get 4 V1 squared over gamma 2 squared,</font> <font color="#ffffff">equals V2 squared over gamma 1 squared.</font> <font color="#ffffff">Okay, but we know that gamma 2 squared</font> <font color="#ffffff">is 1 over the square root of 1 minus V squared over C squared, so 1 over gamma 2 squared</font> <font color="#ffffff">has to be 1 minus V2 squared over C squared. 1 over gamma 1 squared is</font> <font color="#ffffff">going to be 1 minus V1 squared over C squared. And so now this whole thing here</font> <font color="#ffffff">becomes the following, we get 4 V1 squared times 1 over gamma 2 squared,</font> <font color="#ffffff">which is 1 minus V2 squared over C squared, and on the right side we have</font> <font color="#ffffff">V2 squared times 1 over gamma 1 squared, which is 1 minus V1 squared</font> <font color="#ffffff">over C squared. And now we can just simplify this and solve for V2.</font> <font color="#ffffff">All right, let's just do it real quick. Multiply everything by C squared we get</font> <font color="#ffffff">4 V1 squared times C squared</font> <font color="#ffffff">minus 4 V1 squared V2 squared,</font> <font color="#ffffff">equals V2 squared C squared minus V2 squared V1 squared,</font> <font color="#ffffff">and now I have a V1 squared, a 2 V2 squared --</font> <font color="#ffffff">I have a V1 squared, a 2 V2 squared and so I can put those together,</font> <font color="#ffffff">and then I have 4 V1 squared C squared that's left.</font> <font color="#ffffff">When I add 4 to that negative 1, I get 3 V1 squared V2 squared</font> <font color="#ffffff">and now we can put the V2's together,</font> <font color="#ffffff">V2 squared times C squared plus 3 V1 squared and then finally we</font> <font color="#ffffff">can divide by that quantity and you end up with the following. V2 is equal to --</font> <font color="#ffffff">and I'm going to take the square root, so I can do that right now,</font> <font color="#ffffff">2 V1 C, all over the square root of C squared plus 3 V1 squared.</font> <font color="#ffffff">Okay, I went through that slightly quickly but you can replay the</font> <font color="#ffffff">video later on and double-check your algebra and make sure you get the same thing,</font> <font color="#ffffff">and now you have all these numbers, right? We have V1 we have C obviously</font> <font color="#ffffff">and if you plug in all those numbers, you can double check with mine,</font> <font color="#ffffff">you should get 0.62, and it wants it in the units of speed of light, and so you</font> <font color="#ffffff">write 0.62 and then the units are C.</font> <font color="#ffffff">Okay, so those were for my numbers, you of</font> <font color="#ffffff">course will have different numbers for your V1. That one look familiar Jessica?</font> <font color="#ffffff">Yeah. That's the one we went over earlier in office hours,</font> <font color="#ffffff">painfully, I'm a add. All right, questions about this one before --</font> <font color="#ffffff">before we move on?</font>