I said that we had finished chapter 8 but it chapter 6, But
it's got to me that it would be very useful for me to show a
sort of worked example there of using things like the
relativistic Doppler shift just to so you can see the the the
having been done once, we'll then moved roughly on to chapter
seven. I think we won't get all the way through Chapter 7 today,
but I we will encroach, I think on Lecture 11.
But I I I have high hopes that we'll get into Chapter
8
next time.
Any questions about organisation, material and all
of that like that?
OK,
what I want to do, there's an example I want to work through
is the case of Imagine
are
a relativistic ambulance is going past you
and you you see or it's going to pass at A at A at a
a lot of speed and you see it's blue light. You see it's blue
light as it's passing you.
What frequency is the light?
OK,
so the way we work that out is using the centre that the the
recipe that I approximation of the recipe that I mentioned for
in Chapter 5 talking with the restaurant
what we do is we talk about
the
2 frames.
So let's have a
primed
white frame frame that's in the frame of the
I'm gonna
and that's moving
at some some speed V
and the light
in that frame being emitted
at some angle
Teacher prime.
So this is the frame
with the ambulance.
OK
and we want. The question is what is the and the? The light
has frequency F primed
OK.
The question is what frequency do we see that light in offering
and our. So what is
F in
frame
in which
the
and be you Lance
moving at
speech?
And that would of course be neater if I were handing this
in.
At that point, Once you've set the question up
and being clear and explicit about what the frames are, what
we have been given, Theta what we been given, we will give
ourselves an F prime and prime. Once we've set up correctly, the
answer
almost pops out. It's almost automatic
is that
the expression for the relativistic Doppler shift F
equals framed.
Umm,
Camera
or gamma F prime 1 + v Cos Theta prime.
So the question is this. For a question in an exercise or a
question in a in a text or an exam, the question would then be
if if the light were emitted perpendicular but to to the
ambulance, what frequency would you see it as
and what that is, is that's telling you what Theta prime to
use? What what what angle do you use
If if I see the light is emitted perpendicular to that that the
light you see the flash you see that you're looking at the the
frequency of it emitted perpendicular to the ambulance.
What is that telling me?
Anyone.
What
bit of data that we know we need is that telling me
is telling me F
is it not
it will it will be upshifted, yes. But the light that that
that that's emitted perfectly perpendicular to the ambulance
is the ambulance driver just shines out the window.
Is is telling us? Is it telling us Theta or Theta primed
the depraved? Exactly.
So it's telling us Theta framed, not Theta.
So the light
committed perpendicularly
to ambulance
at
or -π do whatever
future
depriving by two. In other words, cost to prime
is 0
and so that F equals gamma F prime.
In other words, even though this light is emitted
perpendicular to the direction of motion
to where classically there would be no Doppler shift,
it's
blue shifted.
OK,
that now that might seem slightly surprising.
Why is that slightly surprising
that that that fits in with you? You know that Doppler shift blue
blue shift things, but why is that slightly surprising that F
equals gamma times F prime?
It's surprising because
the train violation
the
I'm really moving to relativistic speed.
Therefore you'd expect time to slow down frequency to to to to
to be lower. So you'd expect F
to be lower than F prime.
Any idea why that why That seems to be the wrong way around?
Let's ask a different quest.
What would I? How about what is the light of the frequency of
what? The frequency of the light that I see when I look directly
across the path of the of the embrace of motion and see those
when I see the light come in
directly in this direction. What? What different? How is
that different? How is that a different question?
Different.
They're the same 2 frames, so it's the frame of the ambulance
is moving in this direction with positive V and I'm in this
frame.
The same 2 frames, same speed.
But how have I changed the question?
In the first case I talked about the
that that not F&F primed but
Theta and Theta brain.
So in the first case I was asking what is the the frequency
that I see of the light that's emitted perpendicular to that to
to its motion by the by the ambulance.
Now I'm asking what are the light, the frequency of light
that I see arriving perpendicular to the direction
of motion of of the ambulance.
And they are not the same thing
because in the second case
the light
or
they've they've
particular
feature equals
π by 2.
And if I look at similar like F prime equals gamma F
1 -, v Cos Theta
if Theta equals root of the counterpart to this. Going in
another direction, if Theta is equal to 0, is a π by two, then
Cos Theta is equal to 0 and F primed
framed
is F
if you could
yeah F primed over
gamma.
And
in the first case,
if Theta primed is equal to 0,
then cost teacher will be equal to well, V some number which is
less than one.
So the light will come in at a different direction from π by 2
and if looking at that we're owned. If Costa was 0 because
I'm looking at the light I'm observing from the from this
direction, then the counterpart of that the that light was
emitted in a different direction by by the ambulance
and that's why there's so the light that I'm seeing
arriving perpendicular to me that's nicely red shifted
because obtained election
as opposed to the light which is emitted perpendicularly and
making a fuss. That's not because you have to care very
much about observing the colour changes of of ambulance lights,
but because
in I have asked two different questions in that example
which are different questions.
They were carefully worded and I would word them more carefully
if we were saying that as a an exam answer. But in both cases
the thing you had to work out was what does he mean? Which
does he mean Theta or Theta? Primed I was talking about π by
Theta prime is π by two or Theta is π by two. If I'm talking
about light being emitted forwards or backwards is that's
Theta equals 0 or Theta equals equals π and vice versa. So
several of the exercises
essentially
questions of interpretation. How do you read the the the question
and turn it into which you know. So that's why it's it's like
the recipe that I talked about for doing Minkowski for doing
our transformation things in Chapter 5.
Step one is writing down, being clear and explicit what the
frames are.
Step 2 is writing down what you know, what you've been told. And
that's that's an the non trivial step
going from what the question says to that means it's that
that that that, that that it's that, that is that that figure
that picture frame would ever be told, not theatre. And toward
the end you're turning the handle, you just plugging things
into expressions.
So that's a a rather handy question. The diagram at the
top, yeah
time is there's like a street lane and then there's the wave
coming in. Yeah, the angle there is not shown to be π / 2. Ohh no
no no, not that example no no. So that's more general. So in
the, in the, in the case that we're talking about
and that would be in the X prime,
my prime. So there it's, it's.
Is that is that the?
So yes, I drew the the the general sort of general diagram
1st and then particularised it given the information in the in
the question.
OK. And again, not very pretty, not very neat, but
good enough for no talking over
any other questions about that.
There are half a dozen, I think, questions in the exercises in
part in in chapter 6 which are variants of that.
OK, right. Then let us move on
and
and we can talk about
dynamics,
OK.
Kinematics, I said, was about describing motion.
Let's stick with that for more. Can immatics is about describing
motion, dynamics is about explaining motion. So dynamics
is where we talk about forces and momentum and and so on. The
the stuff that you learned that you're thoroughly schooled in,
in terms of Newtonian mechanics.
In the in the Kinematics chapter we talked about relativistic
velocity.
There's all we'll go back on to this.
We talked about rustic velocity and it's slightly unusual
properties.
Now
we are very familiar with how we get we get non relativistic
momentum from velocity, we just multiply it by a mass.
So let's do the same.
Let's define a thing P
where? Where
we just the the the momentum of an object is its mass times the
speed of the object. So the just as a momentum I've I've lost
before vector. There's a momentum 4 vector
and the mass here just the massive. It's nothing
complicated, but the matters. It's how much?
Yeah. How much material the race will you hold in your hand.
OK, And the math doesn't change. The mass is just a number. It's
it. It it's a, it's a a variant
that's too far so good. That is nice and simple
in the kit, but we remember that in the case where the particle
is not moving,
it's velocity
will be
A4 vector pointing along the time axis. So it has it's time,
It's time component is 1 and it's special components are 0.
In that frame,
the
P will be equal to
M
1
0,
so P dot P
will be equal to
m ^2,
which is our Lorentz invariant, so that that's that. The length
of the momentum 4 vector is just the mass squared. The length
squared is just the mass squared, and that's true in
every frame. So because this is a four vector, in different
frames this momentum will have different components. They're
nice and simple in this, in the frame of which The thing is not
moving
been other frames, you know the the the tank component will be
bigger, the spatial components will be bigger and and so on.
But when you work out the length squared of the object, it'll be
m ^2 again as before,
because that's the range and variety. The the the length
length squared of the vector is extremely dependent.
OK, now imagine we've got some.
Umm,
a couple of questions I'm not going to talk about. Imagine
we've got.
Uh,
a collision like this?
So there's a
2.2 particles come in and two particles come out. So it's not
it's not a very exciting collision from the point of view
of particle theory, particle physics, but it's nice and
simple from the point of view of our analysis of it. Two parts
come in P1 and P2.
Something happened and they and two parts come out, P3 and P4.
Now we're going to make a wild guess
and suppose
that
P1 plus P2
equals
P 3 + P
4.
We're going to suppose
that maintenance formulation is conserved. We know that three
momentum is conserved in non relativistic collisions
and you're doing with that I trust
and we're just going to suppose that the same is true for this
thing. What we've got here, we've no evidence for that. Is
that just a guess? Because we like doing that with momentum.
At the moment, it's just a guess.
OK, So what is that? Let us
do
that and equation
and equality between an equation between 4 vectors.
So it's true component by component,
and what that means is that. So P here
and will be.
He won, for example, will be
M1
gamma
1
V if I. If I look back at the
slightly strange notation that I mentioned in the in the other
notes, just to show you that you have seen that before,
it's just this
with an aim in front of it.
OK,
so it's it's and written down component by component. What
that means is that gamma 1
M1
gamma V1 plus M2, gamma V2 equals M3 gamma V3 plus M4 gamma
3/4.
And all I'm doing there just rating out long hand
the let's see the the the the X component
of
the expression above.
So that's just the the the see the X component of that vector
equation above it.
So as you can see,
sorry
and
yeah, so, so, so, so that is, it is.
And
I've written that wrong incorrectly.
There should be a V
and
in each of these V1.
Me too
P3
before my note. The equation 7.2 B in the notes should have V's
in
throughout and that is just P.
It's actually gamma
1P1 plus gamma 2P2 equals gamma 3P3 plus gamma 4P4. In other
words, that is just
the spatial components of this recover the normal conservation
of momentum just for some extra gammas in here, which of course
in the non relativistic case are all approximately 1.
So this this relativistic expression conservation of
relativistic momentum is consistent with the
with the
conservation of
ordinary special momentum.
In the limit when gamma is
it won. You know there's a a slow speeds so. So this isn't
seeing anything different from what you're really familiar
with.
OK,
OK
mumble mumble positive Mumbles
and
so much for the spatial part.
Now if we look at the the time component of that,
what we see
is that the. If we look back at
this expression here
we see that the time component
is
and
yeah I'm I'm, I'm gonna I'm gonna switch away briefly from
the
conservation of momentum and look just at the the the the the
time component of our four vector. So if if the overall
thing is
gamma one
the then the time component, the 0 component of this former
mentor of this 4 vector is just gamma north,
right?
That's that's not surprising at all. I'm just all I'm seeing is
is that's fairly obviously the the, the, the time component,
the 0 component of that expression.
But
yeah,
like anything to ask what's the low speed limit of that? Can we
find an interpretation for this zeroth component of this
momentum?
Yes we can
because if we
look at the
low V expansion of VS that using the a Taylor series
then gamma which you could do 1 -, v ^2 to the power minus 1/2
is is going to be a 1 + V ^2 / 2 plus things of order
V to the 4th.
So that is going to be
M plus
half
MV squared plus terms of order
the 4th.
And that is, I hope, a rather suggestive expression.
You've got something in there which is which goes like 1/2 MV
squared,
which looks a lot like the kinetic energy of the particle.
And what that hint to us
is that
if
momentum is conserved, momentum is conserved,
then each of the components are conserved. And we saw that with
the spatial momentum there's a hinting that this there's
indicating, indicating also that this zeroth component is
conserved
in collisions and something that looks like energy
to that, prompting us to interpret this 0 component
of the four momentum as the energy
of the of the moving particle. So the spatial components
correspond to the spatial momentum that we're familiar
with.
The youth component corresponds to the energy of the particle,
sort of, but it's not. It's clearly not the engine familiar
with because there's a half MV grid in there, but there's also
this M.
In other words, this is telling us that this thing that we that
is a bit like the energy of the particle,
because it's conserving collisions,
isn't zero when the particle isn't moving
to review is equal to 0. That reduces to just P not equals M
and if we.
Look at this expression here. P not.
And we're going to write that as E.
But we're going to jump to the conclusion there. If you go to
Gamma M
this remember is in the units where
she is equal to 1.
So we can ask what is the
the The version of that expression in physical units for
CC is not equal to 1,
so we can do that conversion. We can put C back in or
and make sure that with with the right power so that the
dimensions work in physical units
and we get
gamma, MC squared
and the zero speed limit of that
we're gonna be equal to 1.
The
is
MC squared,
which has been called perhaps the most famous equation of the
20th century.
So what we've done here so, so this seems, well, equals MC
squared. We've got that
as you, as I'm sure you you hope we would at some point.
All we've done here
is guess
that by putting an arm in front of the relative velocity we've
got, we've got, we've got I think which we're calling the
formentor which is physically meaningful.
We are reassured that it's physically meaningful because
when we decide to say
let's suppose that the formentor was conserved,
then we get first of all the
something which looks like the conservation of three momentum
or which which reduces the conservation theory momentum in
the rugby limit.
And we've also got something which looks a bit like energy,
which we said called energy. So this this P naughty is what we
have now on calling the energy of the particle
and the fact that we we're giving it that name is is
plausible because we we spot that express the kinetic energy
in there and we discover that it's non 0
with the things isn't moving
and recover. This equals Gamma MC squared.
In other words,
as far as relativity is concerned, as far as the
dynamics of relativity is concerned,
the energy that you that that's important is not the kinetic
energy, the 1st order term in in that expression, but the whole
thing which is non 0 even when the particle is stationary. So
there is energy in mass.
And by the way, this so, so, and all this reassures us that we
are right to say
this former mentum.
It's conserved in collisions. And that is the third example of
a physical statement that I've made In this course.
The first two physical statements were the 1st 2
axioms,
the expression of Dalian rotor. You can't tell you. Moving the
2nd axiom, everything moves C.
Those are things that could be otherwise, but I'm saying that
in our universe they appear to be that way.
The other things we've done the last 10 weeks have been logical
consequences of that,
so that you know they can't be otherwise. If you take those
axioms as true, the other things just follow.
This conservation of four momentum is another physical
statement.
You can imagine that being otherwise, and you know it would
be mathematically wrong for it to be otherwise. But in our
universe it appears that's just true.
OK. And that's a statement about physics
or astronomy over you.
And it's important to be clear with the sanctions, those two
things
and more things I want to see on that section.
So, So what that also means.
As you can see here that this is the zeros component of the.
This kind of empty squared is 0 component of momentum and that
changes
at a freedom dependent thing.
So the
spatial components of the four momentum and the time component
of the four momentum are different in different frames.
So the the the the energy, the spatial momentum of a particle
is different different frames. That's not surprising because I
mean that's true in
non rustic physics too, but also the energy of a particle is
different in different frames.
The energy being the 0 component of all for momentum,
but always
so so so so so energy isn't it? I think you can stripe to
article anymore but what you can notice is that what will still
be true is that P dot P
will be m ^2
because that that because as I've said repeatedly the
dot product of 24244 vectors and I thought you were either doctor
4 vector with itself is framing variant. So the length squared
of of the four momentum vector
is always going to be the the mass of the particle square,
even though it have different components and different
umm.
I will also just quickly write down that UM
with
rating equals Gamma M&P.
Lower case
P is equal to
gamma M
V
and
yeah the former momentum P is going to be.
It's 10 component is the energy,
that's just P not And the special components are what I'm
writing out. The the the the the spatial momentum P
to P
dot P
which is equal to m ^2
is going to be
e ^2.
Nice
P ^2
and the instant physical unit is it is also eastward equals
P ^2 C ^2 + m ^2 C
4th. I'm not going to go through the details of that, but is it
good to see that expression written down at some point? The
point is that there's a a nice relationship between the mass,
the energy and the the rest of the energy and the relative
momentum.
That's a key point I want to mention before moving on. But
before I go on and talk about photons,
are there questions about that?
Over what question are you excellent questions? From the
working there was this
gamma,
what it was, and it was to the minus 1/2. That's right. Yes.
OK.
So the definition of gamma
is it's 1 / 1, one of us square root 1 -, v ^2.
So that's one of most V ^2 to the power minus 1/2.
Now if you're if you catch me back you have you done Taylor
Series. I'm McLaren Series and all that stuff so you will
remember possibly or or you may not remember but you can go back
and and and just confirm. You seen before that 1 + X to the
power N
The Taylor expansion of that is 1 plus N, X + 1/2 N
X ^2 plus
I and so on.
Adequate, general
for general thing and and and doesn't have to be integer.
I think that's also called binomial theorem, blah blah
blah. There's got a couple of right. It's very special cases,
have a couple, have a variety of different names, but the
expression just above there is just that
applied to
the one most V ^2 to the power minus 1/2.
I think it's raising it.
One, one, one 1 -, -, 1/2 * V ^2
question,
four term and that just never matter because it would just be
such a small number. Yes, for vehicle small. For very small,
yeah. So so this is is the sort of thing that you you will see
again and again in in
in this sort of context and and a bit of bits of physics. And
when something is small,
you're also often interested in the leading, so-called the
leading order behaviour.
So what? What is the? The the the behaviour of something when
the
the the the next leading order terms had you know could be
ignored because they're small. So yeah, so this the the
expression for the Taylor series is
Yeah,
it's valid for v ^2 less than one.
Thanks
and and the thing that that that's telling us is is fairly
obviously tells us that that that gamma goes to one when v =
0. But also is telling us that it goes to one quite quickly as
V goes to zero it goes to one that the the deviations from
from one goes as v ^2. So when V gets small, gamma gets very
close to 1.
Yeah. Thank you for
anything else.
OK.
When we were talking about.
Driving the the the full velocity, you may remember I did
it by just differentiating the displacement 4 vector delta R
term by term, Dr nought by D Tau DR1 by tour Dr 2 by 2 and so on.
Does that work? Ohh And the momentum of of particle was just
the mass times that that full velocity.
Does that work for photons?
That particular plan doesn't work for photons because if you
remember, photons always move to be right and that means that the
displacement
they will, which is is that they'll move as much through
time. The the the the the time component of the displacement of
a photon is always going to be the same as the spatial
component of the photon. In other words, which means that
the times training component squared minus the spatial
component squared will be 0 or. Photons always move along a
light like or null
well
paths,
so the full velocity. So by that definition the full velocity of
our
of of a photon is always null.
So if we were to naively talk about the momentum of a photon
in the same way as here, we discovered that photo 0 momentum
because it's mass times a null vector. So we don't do that.
So what that is saying is that the the prescription we had for
deriving a physically meaningful quantity didn't work in the in
the extreme case where the the, the the the the the vector in
question was null.
Uh, what? We can instead do
what was so that for a photon.
I'm right P gamma. As for that that is always null.
But if we look back at.
This expression Here
we see that for a massive object we simply need to have the
energy and the momentum
equal to yeah equal to each other.
So for a Masters particle
eastward
will be equal to P ^2
as this article. And what that means in practise is something
like the form. The formatum of our photon will be
something like HF.
Where I have you remembered a little bit quantum connect
perhaps to be told that the former meant that the energy of
a photon is Planck's constant frequency
that to the. If we suppose that is the energy of this of a
photon with this momentum, then we know that the
we will only X axis. Then the X component must be the same value
in order that the
in a product of this
4 vector
with itself end up being dull.
So we can by this by this means talk sensibly about the four
momentum of
a massless particle moving through a light.
Uh,
in the way that the that we can't, we we couldn't with a
prescription before.
There's not a lot I could read here, but that that is I think.
Thank you.
We are making good progress here. That's that's
before moving to the next section. Other things wrapping
up things to say about before
OK,
So what what what other things we have does that
Ohh yes so so so that should have a.
Have I have I just? If that's wrong, it's wrong for a very
long time.
Yeah, those those those shoots, there should be V's in there
gamma gamma V1M1V1. I think I've written the one is not one to
write the second time whatever. So that's that's incorrect and
as as expression seven point
277.2 B.
OK,
now let's look at the
simplest situation where 2 particles.
Collide and form
a single outbound particle, perhaps that. Perhaps the
Collider that and it bounced off into the beyonder. Perhaps it
stayed at rest or something.
Two particles arrive and collide
into one particle
and as before P1 plus P2 will be equal to P3. Just form into
conservation, but form into conservation rather 3 minute
conservation.
OK,
so let us step through this.
In each of those cases,
right, Pi will be
Gamma I MI one
VI where Gamma I is just shorthand for Gamma
of the eye. So that's for I123. So for the three different
particles,
OK,
so let's go through the numbers here
and
and what we're gonna do is go to
assume that these particles are travelling only in the along the
X axis, so that the the incoming
in this direction One Direction and the resulting particle will
end up also moving all in the X direction. So I'm going to miss
out why and Z, just to make things easier to write.
So let's suppose that M1
and M2
are 8 units of mass. I'm not gonna, I'm just not gonna worry
about what the what the unit of the mass are. They're not very
big. OK for for particles and the first one,
you're travelling at speed
15 seventeens.
Will this be late?
And the second one,
it's stationary.
OK, so this is a particle particle sitting there. Particle
one comes in at 5017, super light hits it, they join
together and we're interested in what the speed of the particle
is going along the X axis afterwards.
OK, you've got the picture.
I trust
that means
that we can start to look at the
and
let's see
if I ask what is the?
Zeros component of part of of the of the energy of particle
one. That's going to be gamma 1
M1.
If V1
is equal to
15 or 17, then gamma, 1
gamma or V1 will be 17 / 8,
one of the Pythagorean triples. So why did I pick that odd
fraction? Because 8 ^2 + 15 ^2 is 17 ^2 1 of the Pythagorean
triples, right? Nice and easy to do.
That means that P that's used component of particle one is 17
eighteen 17817 / 8 times the mass of particle one which is
17
the
X component of particle particle 1 momentum is gamma 1M1V1
which is that times 15 seventeens
2/1
and we can therefore build up the table that's in
in the news. I'm not going to write it out.
So there I I worked out what
the serious component of Article 1
momentum for vector was.
The one component X component,
the velocity, the
gamma factor, and
notice that the
mass, the length squared of that four vector is 17 ^2 - 15 ^2
with the youth component minus the spatial component which is 8
^2. The square root of that is
through. The mass of this particle is still 8, as you
would expect,
which is the same thing. Same calculation I'm going through.
Go through the steps
Tim Cook calculation. For the particle that's at rest
then it's
V is 0 so it's gamma is 1
so equals gamma M
for the for particle 2 is m = H * 1 + 8.
The the velocity of that particle is 0 so the the spatial
component of the former momentum is 0. So the four mentum T and
the components are 8 zero plus 20. Gamma is 1 and H ^2 - 0 is 8
^2.
Whether that is 8, so again the particles you know the mass is
correct.
What about the particle particle 3? We can know what code what
the energy momentum of the particle
are. Going particle is just by conserving momentum.
So here the 0 component of particle 3 is 17 + 8. It's just
the sum of the
of 0 components of the incoming particles. 70 + 8 is 25
15 plus 08/15
so the outgoing particle has four momentum 2515
and then
getting a no to get the these from that.
We know that
looking at this expression here that VI is equal to P.
This is the
get me through this front. Yeah, the one component
divided by zero component,
but for each particle.
So the velocity of this particle is 15 / 25
3/5 which gives us the gamma factor.
And to find out the length of this four of this 4 vector,
we have 25 ^2 -, 15 ^2 which is 20 ^2
and we were filled out the rest of the table.
No,
there are a couple of points to make here in the I'll make these
points again at the beginning of our next lecture,
just because quite important.
So the the two incoming particles
have this have have
former mentor
which have the same length,
so the the same
moment energy momentum
in both cases.
Now that may seem surprising
since one of them is moving in, other one isn't.
But if but since the length of the
and you mentioned Victor
is Freeman variant
issue with the the Case No matter what frames we pick,
so it has to be the same in all frames. So in the frame with the
particle of all particle one, it's particle 2 that's moving.
So in in three of particle one it's particle 2 that is moving
fast.
So therefore it should not be surprising
that in both cases the argumentum of the two particles
is the same. Of two incoming particles is the same even
though in the frame in our in our lab frame only one of them
is moving.
So the energy momentum is that the total length instrumental
vector squared length is the mass of the particle.
It doesn't reflect how fast the particles moving the direction
of the four momentum vector
reflect how fast the particle moving in your frame, but that's
because the components are framed dependent.
You change change frames, you have different components.
The mass
the length of this of the outgoing particle is not the sum
of the ingoing particles
is not 16.
It's the because it's not because because mass or the
length of the of the four vector is not a thing that's that's
conserved in collisions. The length of the four vector, the
the total amount of energy momentum of the particle has is
framed independent,
meaning the same in all frames, but it's not conserved, meaning
it's it's the same through a collision.
Whereas the components of the forum mentum are conserved,
meaning they stay the same through a collision, but the
note frame independent, there are different different frames.
OK, so 20 is not eighteen 8 + 8.
Ohh yeah, and that's a good point. So M ^2,
the length of the four vector, is frame invariant but not
conserved.
The individual points are conserved but not being negative
and I'll pick up here.