Massive stars ending their lives in supernova explosions...

Key Concepts

Mass-Energy Equivalence

This principle, encapsulated by Einstein's famous equation E = mc², states that mass and energy are interchangeable. In high-energy processes like relativistic collisions, part of the kinetic energy is converted into mass, meaning that the rest mass of a system can change when energy is added or removed. This concept is fundamental in understanding phenomena such as nuclear fusion in stars, where energy and mass play interchangeable roles.

Relativistic Inelastic Collisions

In an inelastic collision under relativistic conditions, colliding bodies combine and some of their kinetic energy is transformed into additional rest mass. Unlike elastic collisions where kinetic energy is conserved, relativistic inelastic collisions require the use of both energy and momentum conservation laws to determine the properties of the resultant particle. This effect is particularly important when the speeds involved are a significant fraction of the speed of light.

Lorentz Factor

The Lorentz factor, denoted as ?, arises in the equations of special relativity to account for effects such as time dilation and length contraction at high velocities. It modifies the classical expressions of momentum and energy, with the energy of a moving particle given by E = ?mc². The Lorentz factor is essential in accurately describing the dynamics of particles in high-speed collisions.

Conservation Laws

Conservation of energy and momentum remain fundamental even in the realm of special relativity. In relativistic collisions, although the expressions for energy and momentum differ from their classical forms, the total energy (including rest mass energy) and momentum of the system must be conserved. These conservation laws are used to analyze and solve for unknown quantities in collision problems, such as the mass and velocity of a combined particle after impact.

Relativistic Kinematics

Relativistic kinematics deals with the motion of objects moving at speeds close to the speed of light. It provides the framework and equations necessary to understand how energy, momentum, and time are interrelated in such contexts. This field is crucial for correctly modeling and predicting the outcomes of high-speed collisions and interactions, which differ significantly from those predicted by classical mechanics.

Transcript

00:01 So the purpose of this problem is to show that particles moving at very high fractions of the speed of light.

00:08 Colliding with each other can create new particles, composite particles, with higher mass.

00:14 And the implication is this is a rough model of how supernova creates the heavy metal, the heavier elements.

00:26 So this is a relativistic inelastic collision.

00:30 In this particular case, the masses in the inclusion are not equal.

00:38 The frame of reference has been chosen to be between the mass 1 and mass 2 and so that their incoming speeds are it's all purely on the x -axis, so we can talk about speed instead of velocity.

01:00 Their incoming speeds are equal and opposite.

01:03 One half of the speed of light.

01:06 So they collide.

01:07 Their masses are not equal.

01:09 So the composite particle has a speed v in the direction that the heavier mass was moving.

01:20 So we're going to end up having to find the value of that speed v.

01:25 And we'll use that to find the mass of the resulting particle.

01:35 Because in a relativistic collision, it's not simply the summation of the two masses.

01:40 There's energy, net energy, leads to a higher rest mass in the composite particle.

01:53 So we're given that mass one is 1 .99 times 10 to negative 26 kilograms.

02:05 So that's analogous to the weight of an atom.

02:10 The second mass is one -third of that mass.

02:20 And so basically we're talking about atoms absorbing smaller atoms becoming bigger atoms.

02:29 So we'll end up needing two equations to solve this problem.

02:37 We need the conservation of momentum equation and we'll need the conservation of energy equation.

02:47 Conservation so relativistic momentum has this form so for each particle there's a gamma term a mass term and a speed term so p1 plus p2 equals p total so the total momentum so this becomes explicitly written now gamma 1 m1 u1 plus gamma 2 m2 m2 equal sorry mv okay so we know that m1 is just m you know m 2 is m divided by 3 so in non so non -relativistic collisions you might expect you might expect the final mass to be something like this or equivalent to square root of four.

05:00 But this is wrong.

05:03 We're going to find out that it's actually going to be two times square root of five.

05:12 And i'm going to show you how to get to that.

05:20 So another important thing to notice is that since the speeds, since the magnitude of u1 equals the magnitude of u2, gamma 1 equals gamma 2, gamma 1 equals gamma 2, because the gammas are, they're not independent terms.

06:05 They're actually determined by the speed terms, if you actually explicitly write out gamma in general is like that.

06:30 Those u determined so we can write this as right because we can just substitute these equivalences into these terms and then we can combine them so that works out to 3 over 3 minus 1 over 3 so that's 2 over 3 gamma 1 m1u1 equals gamma mv so another point that's very, very important to notice otherwise nothing's going to add up right and you'll spin in circles trying to do this different ways.

08:05 So i'm going to add a subscript to this one.

08:09 I'm going to call it gamma v because this is, actually the gamma corresponding to the v speed.

08:23 So the question here is how many unknowns do we have? these two are unknown.

08:41 But gamma v is determined by v.

08:43 So we actually only have two unknowns because every term on the left hand side is known or can be calculated for something that's known.

08:52 So we have one equation and two unknowns.

08:54 So we need two equations to solve this problem.

09:00 So therefore use e equation, substitute gamma m.

09:25 So then we'll be left with the v term.

09:28 If we can re -express this gamma -v, m term, in terms of stuff on the left -hand side and substitute in, then we can solve for v and then calculate gamma -v.

09:38 Then once we have that, we can go back and calculate the value of m.

09:47 So you need two equations to solve two unknowns.

09:49 So we'll use the conservation energy equation.

09:53 There are more complicated equations, like sometimes use e squared, and you have a plus 2e1, e2 term, and so on.

10:09 These collision problems can get pretty complicated, but for this one all we need is e1 plus e2 equals e total.

10:18 So then they just have the form of gamma -m, c squared.

10:25 So this becomes, let's see, gamma 1, m1, c squared plus gamma 2, m2, c squared equals gamma v, mc squared.

10:50 So the, so v doesn't actually appear in this equation, but it doesn't need to because it's the same gamma term.

11:04 It's determined by the speed of the composite particle.

11:12 If you're working with this equation, you would need to use the momentum equation to determine this gamma turn.

11:22 These kinds of problems are usually a lot easier to a lot simpler to solve using four vectors notation, but i don't believe the textbook wants you to do it that way.

11:34 So let's see.

11:39 So we do the same kind of substitution.

11:48 So gamma 1 equals gamma 2, just not equal gamma v.

11:56 M2 equals m1 divided by 3.

12:03 So that becomes c squared divides out.

12:21 So we can write this as gamma 1, m1, m1, scama 2, 1 over 3, scm.

12:41 Because remember, we're trying to isolate this turn...