Relativistic Projectile I

[00:00:00]Okay, so today I thought what we'd do is mix things up a little bit and try looking at our projectile motion problem relativistically.

[00:00:09]So imagine that we still have a uniform gravitational field, but now we're firing a projectile again with some initial launch angle theta, some initial velocity v kn, but we're lo using this uh relativistic speed. So this V knot could be very very high even approaching the speed of light not greater than the speed of light but it could be approaching that speed of light and we want to see how does the trajectory change in particular for this video anyway I'll change this to uh one we'll calculate the total flight time of this trae this projectile the range where it lands and then its maximum height now this might seem a little bit contrived at first blush Because if this was at the um surface of Earth for example, if you launched with any sort of relativistic speed, you're immediately going to be way outside of the approximation where G can be taken to be uniform. So we're not really thinking about this near the surface of Earth.

[00:01:09]That would definitely not be appropriate given um how quickly you would be away from the uniform field approximation.

[00:01:18]But there are plenty of places in the universe that do have relatively um uniform fields over a very large region.

[00:01:28]So for example, some stars uh super giants or even hyper giants can be so large they can be many light minutes across for example that if you were near the surface of that star imagining this sort of picture you could actually have a fairly uniform gravitational field even including pretty relativistic motion near that surface. It's just it's so large that the surface is approximately flat and you get approximately uniform fields. uh other places might be near super massive black holes. Again, even if the gravitational field is very strong, it's the uniformity that we're worried about here. And if you're even at the event horizon practically of a super massive black hole, the tidal forces, the the gradient in the gravitational field is so tiny, it's practically uniform. So, this could work in that situation, too.

[00:02:19]So, it might seem contrived at first, but it's actually not so bad. Let's also remember a few things from special relativity. First off, um I'm going to be imagining that I'm standing in some sort of laboratory frame. I'm I'm at the surface of this giant star or black hole surface. And so when I'm [clears throat] talking about things like how long is it in the air or looking at Newton's second law, I know I have to be really careful about what I mean by space and time.

[00:02:47]Everything is relative. So the frame of reference that I'm developing is um some sort of locally inertial frame where I mean GR aside some special relativistically inertial frame where t here is going to be the coordinate time.

[00:03:02]Okay, that's the time measured on my clock as I'm watching this thing fly around. I'm not talking about the proper time of the projectile itself. Maybe we'll do that in a different video.

[00:03:11]Let's also remember that Newton's second law still holds. the net force on this object still will be the rate of change of momentum. But I can't write that as m a anymore. Um instead I have to recognize that momentum relativistically is not just mass times velocity but it has this extra gamma factor. And this gamma is 1 over the<unk> of 1 - beta squ and beta itself is this non-dimensionalized velocity in terms of the speed of light c.

[00:03:38]So because of that extra piece there, we really can't write down MA quite the way you'd expect. And it does make things a lot more complicated because now gamma, which depends on V, will itself be changing with time in addition to V itself. So we've got kind of a product rule going on here. Okay. So what we want to do first is try to calculate how long is the particle in the air with respect to my uh coordinate time frame.

[00:04:04]And what we're going to do is apply Newton's second law. So let's go ahead and write that up.

[00:04:14]Okay, so there's Newton's second law written out. I've included in the uh net force component here. There's only one force. It's the gravitational force.

[00:04:23]We're assuming it's uniform. It's in the negative jhat direction. So I've I've implicitly chosen my coordinates here.

[00:04:29]Uh I guess explicitly y is going to be called up and x will be to the right.

[00:04:34]uh we do have to worry about this sort of product rule thing, but actually it's not quite so bad because we can immediately integrate this. This is a constant on this side. And I'll go ahead and separate this out into two components, the x component and the y component separately. We'll integrate both sides. And I'm actually not going to write this in terms of gamma mv just yet. I'm going to write it back in terms of p, the momentum. So let's do that.

[00:05:07]Okay, so let's quickly look at what we've got here. Uh, since this is just equal to a constant, I know that px equals a constant, meaning it's equal to its initial value. And px knot is just well this thing at time zero. I'll go ahead and write that as gamma kn as a shorthand for 1 over 1 minus beta kn.

[00:05:29]and beta KN is V KN / C. And then we need the cosine theta component because of course we're looking at just the x component in the y direction. Okay, the derivative isn't zero, but it it is a constant. So I can just integrate both sides. I'm still starting at time zero.

[00:05:44]So when I integrate that side, I'm going to get a minus mgt. And then my initial condition, my plus c, if you will, is that p y, which will also have the same form except it'll have a sin theta. So I can see that um px is a constant. PY decreases linearly with time. Okay. Now I'm going to invoke a little bit of symmetry here. Uh although I've drawn this like a parabola, it's not a parabola, but it is symmetric about some axis. I expect from the symmetry of the problem that when this thing is returning and it's about to hit the ground surface of the star, uh the y components of the momentum needs to be the same just in opposite direction. And we could check that more explicitly if you want with something like energy conservation in a relativistic setting.

[00:06:32]But I think from symmetry alone we can kind of see that py at time capital t that is this this flight time has to be negative of the initial py.

[00:06:46]I'm kind of playing fast and loose with the order of my uh my knots and my y's here. So by plugging in capital T here, we get this nice relation and that's very easy to immediately solve for t.

[00:07:10]Now notice all the M's cancel out. That kind of makes sense from a um equivalence principle perspective.

[00:07:18]And [clears throat] then we also see a very similar structure to what we saw with the classical case. In fact, I can just sort of pull out this gamma knot.

[00:07:27]That's the new bit and multiply it by the classical flight time, which remember we've derived many times before. It depends on that initial vertical velocity and of course inversely on G.

[00:07:43]So notice this is actually longer than the classical flight time because gamma kn is greater than or equal to 1. You can check that over here by noticing that beta has to be greater than or equal to zero.

[00:07:58]So the projectile will be in the air for longer and it'll depend entirely on u that initial launch speed. Gamma can be made arbitrarily large though we should note. So in addition to the v knot here making the overall flight time larger like it would classically we get an extra boost to the amount of time because um gamma can actually go to infinity as v approaches c. So if I did try to do this for a light beam or something the v kn here would be c. Of course I'm really not allowed to do that because then I can't have a massive particle moving at the speed of light.

[00:08:38]But you can kind of imagine what might happen here if gamma kn became infinite.

[00:08:43]The particle would never um return basically.

[00:08:48]Okay, so that's our first result. We've got our flight time. Now let's see if we can calculate what the range of this particle is. Okay, so to look at the range, we're going to need a couple extra facts. They come immediately from uh these definitions, but we can maybe just remind you what those are. First of all, uh I can always rewrite beta, this v over c in terms of the momentum directly and the total energy. I didn't write the total energy over here. Guess I could add that real quick. So the total energy of any particle here uh that's moving and we're not including by the way the um gravitational potential energy. So I should say this is the total energy excluding external potential that there is an external potential here. So let me rephrase it to mean the sort of kinematical energy and that's just going to be gamma mc^2. In fact an object at rest has gamma equal to 1 and we can see e= mc² that's Einstein's famous relation for the rest energy. Uh in instead of using things that involve these nasty gamas though with the square roots sometimes it's convenient to swap things out and try to write things without invoking this uh this lorren factor gamma. So by basically dividing these two equations by each other we can get beta in terms of the momentum and the energy and then by squaring these and adding them together we can get this nice invariance relationship that says the square of this m this let's call it kinetic energy or kinematic energy not including external potential is um this momentum p^ squ plus its rest energy squared and we'll need those to make things a little simpler in in just a second. The next thing we're going to do is to think about what velocity really is. So, we know that velocity is the rate of change of position. Nothing is new about relativity in that sense. And then I'm going to do something that might seem very weird. I'm going to use the chain rule, but I'm going to use it in a very strange way. I'm going to write this as dpy times dpyd.

[00:11:05]Now this is valid. This is just using the chain rule. But why am I using the y component of the momentum as my intermediate?

[00:11:14]Well, the answer is dpyd is this known thing. In fact, it is the force in the y direction which is minus mg. We can actually write that right now. Um, that's going to be minus mg. And then I'll have this this extra derivative.

[00:11:31]And that's a nice constant. I I could use dpx, but then I would just get zero, which is not terribly interesting. So, this is telling me some nice relationship between the velocity components. Um, oops.

[00:11:48]and something about how the position changes with y momentum. Okay, so now let's write this. Let's actually just look at the x component here. So if I write down dx dpy, there's going to be a 1 over mg times vx. Okay, so that's not so bad. Now, I'm going to turn back to this relation because V and beta are basically the same thing.

[00:12:19]V is nothing more than C * beta. So, I'll just stick an extra C on here.

[00:12:27]Okay, that doesn't seem so bad either.

[00:12:31]And then remember, we actually already figured out what PX is. It's a constant.

[00:12:35]The there's no force in the X direction.

[00:12:36]So, we know what that is.

[00:12:39]Maybe I'll just leave it in terms of its initial value.

[00:12:44]Uh we could always rewrite that if we want. And now E. Well, let's use this relation. Note that I'm going to need to take the positive square root here. And then my px^2 and my p y^2 that come from doing this sort of dotproduct squared are also known.

[00:13:02]And let's see what order I think I'll write the mass first.

[00:13:10]Okay, so that is writing in my energy relation.

[00:13:14]And let's go ahead and actually plug in or remind ourselves that this px is actually just px knot because again the x component of momentum is conserved.

[00:13:24]Now if we look at the far left and far right here, I see a whole lot of constants. the px knot the c's the m's the g's the mc squar more c's more c px knot again but there's a p y squared here but that is exactly the independent variable for this differential equation now it is a little ugly I'll admit uh all of this together is messy but it basically has the form of dxd something is equal to a big old constant one over the square root of a constant plus that thing squared.

[00:14:04]And that is a fairly standard integral.

[00:14:07]In fact, we could write it in terms of um what an inverse hyperbolic sign. But instead of actually doing all that work, I'm going to sort of jump down to what this answer is for x as a function of py. So I'm going to integrate both sides of this remembering that x starts at zero but py does not start at zero. py starts at py known thing and we'll just give the result of doing that integration.

[00:14:46]And just to simplify things a little bit here, I've gone ahead and defined this constant ex^2 plus this mass term.

[00:15:02]And the reason for that is it's just sort of a combo that comes up a lot in this problem because both of these are constant values. There is no such thing as a vector of energy. Please don't read this as like an x component of the energy. I'm just sort of splitting this up into ex squar and then the thing that actually varies which is the y component of the momentum.

[00:15:24]Now we could go ahead and plug in well py goes where p y is and then I could plug in py also and do a subtraction of these logarithms but I don't really need to in this case. Uh we're going to use this in a sec. Let's turn now to the other component. Remember we've only written down dx dpy. Let's also look at dy dpy.

[00:15:49]So from this same relation we'll get vy. From the same relation again we'll get something involving p y c ^2.

[00:16:03]But now this is a variable. This is something that depends on time. And so it's not just going to be this constant like it was here. However, we're still going to have an integral that's fairly standard. Something like an integral of x over<unk> 1 + x^2. That can be done with something like a u substitution.

[00:16:28]Okay, so this is kind of what we're looking at. We are going to want to integrate both sides of this with respect to py. And again, this integral on this side will be something that's fairly standard. If you want to do a sort of u sub to clean it up a bit first, you can. But I am going to again kind of cheat a little and skip right to the answer.

[00:16:50]And that is it's actually a little nicer than this one. It doesn't have any logarithms in it. Um it's just the square root. And you can kind of see, yeah, if you take a derivative of the square root, the square root will go into the denominator. And then the chain rule in here will give you a a two that cancels the 1/2 and a py in the numerator. So that's not too surprising. Again, I'm leaving it in this kind of unevaluated form. We would have to apply the fundamental theorem of calculus to to finish that off. But we have our two relations for x and y.

[00:17:25]Okay. Now let's see how we can use this to find the range. Okay. So I've just rewritten what the uh coordinates for the location of the projectile are. not in terms of time, not even in terms of velocity, but rather the y component of the momentum. Again, that might seem kind of weird. Usually when we do these problems, we write things as functions of time. We parameterize them that way.

[00:17:47]And we could do that. Um, if you go back and rewrite things in terms of time, you can get some differential equations that you can integrate. They're a little messy, but maybe in a bit we'll do that.

[00:17:58]Right now, however, I'm really interested in using that symmetry argument again that when the projectile actually hits the ground, I'm expecting the momentum in the y direction to just be the negative of the initial y momentum.

[00:18:11]In other words, the range which is the position of this particle when the momentum is negative the initial momentum can just be found directly from this.

[00:18:28]Now let's check very carefully what that's going to be.

[00:18:34]There will be this constant factor on everything. So I'll pull that out. All right. All right. So now all we have to do is plug in negative py kn for p y.

[00:18:44]And don't forget we have a fundamental theorem of calculus subtraction of positive py being plugged in for py. And then we have this natural log that's on all these terms.

[00:19:01]Uh did that fit? It did. Okay, good. So here we are. We have our subtraction here. The only thing that's really different is we plugged in a negative py knot for the end point and then we plugged in a positive py knot for the starting point and then we subtracted.

[00:19:17]Now, of course, logarithms have that nice property that we can combine them using um a log of a quotient.

[00:19:25][clears throat] Let's also while we're doing that go ahead. I don't want to have a negative value here and all these things are positive. So, I'm going to use that negative rule to flip the argument of the logarithm.

[00:19:38]That means this term will go on top and this term will go on bottom.

[00:19:47]And of course, squaring inside of this is going to leave that thing positive.

[00:19:53]Okay, it's not bad. But maybe you might recognize this as h it's a logarithm of some stuff involving these quadratic uh root relations between my my variables.

[00:20:09]This maybe might suggest a hyperbolic trig function. And in fact, this is actually the inverse hyperbolic tangent written in a particular way. And maybe we can see that a little bit better here by doing a couple simplifications. So for example, I'll go ahead and plug in the value for this uh px. Remember that that's going to be related to gamma m v cosine theta. The m's will cancel, so that's kind of nice.

[00:20:41]Next thing is notice that ex^2 which of course is a constant plus py c^2 all of that is just the total e knot maybe I'll call it e knot that's the total initial energy of again initial non-p potential energy of my projectile.

[00:21:06]That's kind of nice. Um, it makes this look a little cleaner. Again, it's just this thing evaluated at t=0.

[00:21:14]And now, like I said, this is probably best rewritten in terms of an inverse hyperbolic tangent. In fact, we could do one better here. Go ahead and use this relation again to divide everything by e kn and we'll have a py c over e. That's just beta y.

[00:21:40]And maybe now it's a little bit easier to see why this is going to be a hyperbolic tangent.

[00:21:47][snorts] Uh and in fact it's going to be a two because remember that hyperbolic tangent inverse is uh 1/2 of log the 1 plus something over 1 minus something.

[00:22:01]And just to put everything in sort of the same final form, since this involves beta not y, I'm going to go ahead and rewrite this uh v cosine theta with some c's in terms of beta x. That way it'll be a little more symmetric.

[00:22:17]That's actually not too bad of a final result here. So this is the range. It's the distance this projectile will land.

[00:22:24]Again we can see it does depend on the initial x velocity of course and the initial y velocity. But now unlike the previous case in the classical case where it was just the product of those two uh now it's got sort of a separate dependence for the y and x directions.

[00:22:44]And actually just to finish things off, maybe we can also take the limit as beta and beta beta x and beta not y are both relatively small compared to one.

[00:22:56]Remember that they're both dimensionless ratios with the speed of light. And in that case, if we take that limit inverse tangent um tailaylor series starts with the linear term. So as beta not x and beta not y are both much much less than one we do expect this limit to just become beta not y itself.

[00:23:24]I also expect gamma kn to approach one.

[00:23:27]Remember how that relates to these betas. And so this will just become 2 beta x beta y c^ 2 over g. But of course if I replace these uh things with v's we get exactly the classical result that we've seen. In fact we usually write this as v kn cosine v s. So v kn 2 and then the two cosine s is a double angle formula for sign.

[00:24:00]So it does result uh reduce to the classical case in the limit of low initial launch velocities.

[00:24:08]Okay. So that's kind of interesting. Not too hard to write this down. Almost always in special relativity when you deal with kinematical quantities. It's not surprising to end up with hyperbolic trig functions showing up. There is a deep reason for that baked into the um Laurenian signature of spacetime, but we'll talk about that later. All right, so we got our result for R. Let's finish this off by finding what is the maximum height that this projectile reaches above the ground.

[00:24:35]Okay, to find the maximum height, it's not so bad. U we can use this relationship here. Obviously, I want to find a ycoordinate. And I'm looking at the place where the y momentum goes to zero. It's kind of like what we've been doing in the past where the velocity in the y direction goes to zero. So h this maximal height is really just y evaluated at momentum y0.

[00:25:01]Okay. Well, we can just plug that in directly.

[00:25:04][snorts] Notice my first term here when I plug in zero is just going to be ex everything's positive. So the square root's fine. And then my second term I do have to plug in. I already factored out the minus sign here. So there will be a negative and then I'll plug in my initial y velocity uh momentum.

[00:25:28]All right. So I mean that is the answer but we can do a lot better and put it in terms of more recognizable things. For example, once again this whole thing together is just E not but we probably want to write it in terms of the parameters that are given in the problem. And to do that, let's see, I'll use this minus sign to flip the order. EN over MG.

[00:25:54]Uh, if we go back to this relationship, E over M is just going to be a gamma.

[00:26:00]And of course, since it's E KN, it's going to be gamma KN. So that first term will actually end up being a gamma.

[00:26:07]Don't forget there's also a C^ squ here.

[00:26:09]C^ squ gamma KN, sorry, gamma not C^ squ.

[00:26:13]Gamma look more gamma E. uh there's still the g in the denominator and then that first term has basically become a one. So that the e knot here over mg I factored that out. I'm going to factor an e not out of everything now.

[00:26:27]And then my next term this ex remember is just the piece that involves um the the x component of the velocity. And maybe h because it's probably going to be a little easier to write this involving uh the the total e rather than just the x component. I'll go ahead and write this as the square root of ex squared. Uh and when we do that, we're going to get what? I'm going to write it as e knus pyc ^2 kn all over e kn squared. Don't forget that I did kind of pull an e knot out of everything. So it's still in the root too. And then we can simplify that pretty easily.

[00:27:26]Notice that this first. Okay. So we have a one from on the outside. We got a root and then we got a one on the inside minus and now all of this together pyc over e kn is just beta y knot and then squared.

[00:27:42]Okay, so I think that's a pretty good relation here.

[00:27:46]Uh it does show us that it [clears throat] depends entirely on what's going on in the y direction with this beta knot here. Although unusually it also secretly depends on what's going on in the x direction because it involves this gamma knot. We could do the same check again as before. In the limit where beta y knot is relatively small compared to one, I can do a binomial expansion. Maybe we should do that.

[00:28:14]So in the limit where beta y knot is very small compared to one. Um gamma kn is still going to be approximately one.

[00:28:25]And then this we can use a binomial expansion and write this as since this is 1 minus something small to the 1/2 power I can pull down that 1/2 power and then I have my something small which is this thing squared and then sure enough 1 minus one cancels away the negative negative cancels away and I'm left with a thing that looks like this. And of course, if we make one more simplification because we know how beta depends on C, we get this relation that we've seen many times. So this is the classical result, which is good. Should always get the classical result in the non-relativistic case.

[00:29:12]Uh you can also go check that this function if you look at how it depends on beta this will be larger than the classical case kind of as you might expect given the uh relativistic nature of um how the momentum changes under this force and I think that is a pretty good finale here so let's go ahead and just add that to our final list and it feels Feel free if you want to replace all these beta knot x beta not y's and even the gamma kn with v knots and ss and cosiness of theta. U it'll just sort of clutter it up. I don't think it really adds much of interest.

[00:29:57]Okay, so those are the three relativistic versions of our typical projectile motion kinematical quantities. Uh in a future video we'll take a look at things like what is the shape of this trajectory? It's not parabolic, but what is the shape? What's the functional form? You might already be able to sort of see what that is. Or maybe you can just take what's given up here to eliminate py to get y as a function of x. You're going to have logs, you're going to have square roots.

[00:30:24]So something maybe logarithmic, something hyperbolic triggy maybe. And then um we might also want to look at uh things like the velocity as a function of time or the position as a function of time. Uh we might also want to look at things involving not just the coordinate time but the um proper time for someone that might be riding along this projectile. So, we'll look at those in a future video, but for now, we've got our first results for the relativistic projectile QED.