This is lecture 9.
Remember that there are 10 in special activity for over 5
where I talk about the beginnings of general activity.
So the aim is to get this the end of this chapter and Chapter
7 on dynamics done in the next two lectures and we can move
straight on. At that point I might drift slightly into
slightly intellectual 11, but that's that's not a big problem.
If it happens,
I don't think there are announcements other than to
mention and to remind you that the paddle there exists and is
useful. What previous years, I found it very useful by this
stage. In previous years, the party has been full of
questions,
ask questions. I mean, obviously I'm not
terror systematic about looking at that every day, but I do aim
to look at it fairly regularly and questions that you put there
can be answered and other folk can see those excellent
questions. What you thought about during the process of
mulling things over in your head. After the lectures
I shall move on. Are there any questions?
We're about where we are
and
where we are now,
emotionally or otherwise.
Any questions?
OK. Well, you're being you'll, you'll be excellent asking
questions in the lecture. So that's good.
Where we got to last time
was I had introduced the idea of four vectors
as the fairly direct analogue of three vectors in ordinary clean
space. There's an XYZ and ATT, and the point I was hammering
home
was that these lectures, these vectors in 4 dimensions have
basically the same properties as the vector you're familiar with,
except that the definition of length
in for these vectors
had those extra minus signs.
So the the definition, I don't think I have a slide which has
that on it.
Well yes, I I talked to the the inner product or the scale of
product or the dot product for everyone to call it. You know
magician would distinguish those various things. We're not going
to bother. The dot product of 2 four vectors
has this these remaining signs
which and and. And the matrix of the ETA which has those lines in
it. The diagonal matrix plus minus, minus, minus
is. Wild maintenance come from. That's called the metric. It's a
constant for special activity it becomes non constant and of
physical interest for general activity. I noted that the
length of the displacement vector that's just dot X, dot Y
delta Z is
recovers this invariant interval that I talked about several
times in previous lectures. So that's that's interesting and
and and and that's that shows where doing something familiar.
I talked about the way we did a quick quick question in which I
looked at the lengths and inner product of 2 vectors which we
discover that those two vectors A&B were orthogonal to each
other in the specific sense that their inner product product,
scalar product we want to call it
is 0.
So that so that they are as orthogonal in Minkowski space as
two right angles to the right angles would be in three space
and they both have different magnitudes.
The
if A has a squared length, squared magnitude of which is
negative, so it's a space lake
vector.
It points roughly along the X axis.
Vector B had a positive squared magnitude. It's time like it
points roughly in a sort of time like along the same axis
direction.
And I went on to talk about
the
derivatives of the displacement vector
and how we can define the
relativistic velocity and just the component by component
derivative of the position
DX0 editor DX1 by detour DX and so on. We're dividing by the
Tor, not DT, because DT Tor the is the Lorentz scalar that
that's the framing variant thing.
The the the time between two events is A-frame dependent
thing, so that we can't divide by that is not that that changes
from frame to frame, but tall. The proper time between two
events is frame invariant, so we can divide by that. It's a
scalar under the red transformation. We can do that
twice and get an acceleration.
So it's nothing complicated, just it is just the component
component derivative.
And I walked through the the brief calculation that shows
that this is another another manifestation of time violation.
The the the rate of change of of of the time coordinate with
proper time
is
non. It is not one that they don't vary in the same in the
same way.
I committed at that a little a little more. I also described
mentioned the little notation for velocity rather than
the sanitation. Where the the 0 component of of the velocity U0
is gamma,
the ex important is gamma VX, gamma vy, gamma VZ and so on and
write that in that particular way and I showed you. I hope I
convinced you. I hope that the dot product of you with itself
the the length of the vector or the velocity vector was
over 1.
And because that's true in one frame and the nice easy frame to
calculate, which is the case for is not moving, so the velocity
vector points along the T axis, we can easily calculate what U
is in that frame. And since the frame independent because
there's reports all frame independent, that's true always.
So the velocity vector is strange
in that its length is always one
that unexpectedly
so. The
the point being that as you move, as something moves at
rustic speed, its velocity vector doesn't get longer,
doesn't change length, you just change direction in the
Minkowski space,
right? So that's where we got to last time. Where we're going to
go on is
and
coming up appropriately.
OK,
I mentioned that you know it was equal to DX naughty by D Tau or
more generally that you MU is DX UB yd
reminding you that the this index MU the Greek ones go over
0123 Latin one goal 123.
So we have the velocity,
how do we get the the next thing we do? Look at the acceleration,
the acceleration vector. That's straight forward. So a naughty
is going to be
the
you not by detour
before,
which is
the tea by detour
at the you not by
DT.
I'm gonna write this.
Which
yes and remembering that you you not
but you is equal to gamma. One VU naughty is equal to gamma.
So this is going to be gamma
the gamma by DT
gamma gamma dot.
I'm not doing anything complicated that that looks very
fiddly, but I'm not doing anything more complicated than
just
between rule for differentiation.
Already you can see this is looking a little bit a little
little messier than the than the. I've lost you
vector
the special components AI. That's DU
UI by D Tau
equals DT by D Tau
DUI by
ET,
which is Gamma
D by DT
Gamma
VI.
So the the the the the 8th component of you is just Gamma
VI
from the thing on the right.
Differentiate that
and we get
gamma.
Gamma dot
VI plus gamma
would be
I dot
which is Gamma
Gamma dot
the I plus gamma
EI. Defining the
EI as the just straightforward derivative with respect to the
claim coordinate of T and you can see that looks a bit
messier.
The the relationship between position, velocity and speed is
nice and simple. In Newtonian mechanics, it ends up being a
bit messier
in this case, but it's just because there's a gamma in there
making things up.
So OK, that's that's messy. We don't want we don't want to do
calculations with that as much to the extent we can avoid it.
But
a point worth making is that
there is no
there are no frame in which a particle is not. If a particle
is accelerating
and there's no frame in which it's not accelerate,
the particles moving at a constant speed and there's a
frame in which it is stationary,
there's no freedom which is not accelerating.
It's exaggerating one after the other.
Admit that it's complicated.
If a particle is moving to constant speed, then there is a
frame in which that particle is not moving
the the frame that's cool moving with the particle.
A particle is accelerating, then there is no frame in which it is
not accelerating.
It's just accelerating from a different.
But
at any point there is a frame in which the particle, though
accelerating, is briefly at rest.
And if you if for example, if you throw something up,
then all the way through its motion it's accelerating, but at
the very top of its motion it's it's vertical speed is 0, so
it's more entirely at rest,
but it was accelerating.
OK,
that's called the momentarily cool moving reference frame or
instantaneously cool moving reference.
I'll come back to that. Yeah. It's it's it's what do you, you
know, lodge that thought in your head. And now in that frame.
The
velocity acceleration vectors are nice and simple
in that frame.
The moment you can move reference frame,
the
velocity is
nice and simple. It's directed entirely along the T axis. It's
the special component of. Its velocity is 0. It's. It's
temporarily at rest
in that frame. Also
the
you look at look at this. The velocity is 0 and gamma is equal
to 1
in in that frame.
So that means the acceleration is 0. Velocity is 0
and the gamma is
is 1
so that the
acceleration in that frame is just is
camera 1 and gamma dot is
2nd
gamma dot. It turns out continuous factor of of of TV.
So dot in that reference stream is 0
and V0 gamma dots 0, so gamma is 1.
So in that moment actually that's more convenient in that
more entire local movement. The point. The point is, in that
moment helical movement reference frame, the
international symbol,
the velocity has zero spatial component and gamma is 1. The
acceleration
of the gamma, gamma dot which tends to be 0 in that frame and
justice acceleration. But look what we found in that frame.
You A
is equal to 0,
it's 1 * 0 minus
zero times dot east.
So in that frame
you and A are orthogonal.
But
that sort of recent variant is just a number,
so it's not frame dependent.
So it's true in every frame.
So in any frame,
the velocity and its aeration are
orthogonal.
And that's that's that's a technique that
this is another variant of choose the right coordinates,
which you you you, you. Well, we've drummed into you again and
again over the course of your physical mathematical career.
Pick the right coordinates and and problems are simple
or problems can be simple and if you have an invariance then the
answer you get in that simple frame is the answer you get and
what more generally. And that sort of makes sense, because if
you
remember that the velocity vector is always
the same length, you don't use always one
and change the velocity correspond to changing the
direction of that vector in Minkowski space,
then something is accelerating. What's happening is not it's
being.
This vector is being extended, it's just being
moved,
and that sort of makes sense that the acceleration acting on
the velocity would rotate it if you like, rather than extend it.
That's not a detailed mathematical
a code, but it it intended to to make this feel a little less
strange,
right? Also in that frame,
we can look at
what
the length of the of the acceleration vector is.
It's just the timing component squared minus the space
component squared 0 ^2 -, a ^2.
So the acceleration vector
is space lake. It's negative. A space like vector
the length of which is referred is called the proper
acceleration,
and that's true in this realm, and therefore it's true in every
stream. Even though in in another frame,
the components of U would change in that other frame, the
components of a would change in that other frame, but in such a
way that the
the dot product of the two would have been the same. The
components of the acceleration 4 vector would be different in a
different frame,
but the components would change systematically in a way that the
dot product would remain at this value. That's what it means,
that the dot product is frame invariant.
The components of the vectors change the dot product, doesn't
it? Our family variant quantity. The components are frame
dependent. The dot product is framed, independent
and her very important distinction to have in mind.
Um,
and and and. I'm not going to go through the details, but but
if we have two velocity vectors then
we can find a quantity like that and go through the details if
you're interested. It's not terribly important, but if you
look at the notes
and the point here is that
this
these expressions for the acceleration for Vector are
looking messy.
Messi, isn't good
looking messy. But if we instead fall back on the geometrical
properties of these vectors,
we can find these we clearly very important relations really
quite promptly. We didn't have to do a lot of fiddly
differentiation here. We didn't have to explicitly change frames
here. We just had to be smart, do the calculation in the right
frame,
get the answer very straightforwardly
and we've, we've discovered a couple of very important,
clearly very important properties of these two vectors
just from the geometry,
OK. I think that's quite a good example of where the geometry,
our general approach, has power.
Umm,
on the stage do we have?
That's what I've just
soon
quick question.
If I threw a ball vertically into the air at the top of its
path, what is it? What is the instantaneously cool moving
reference frame?
Who is it with the family room?
Who would see it with a free moving upwards with the same
speed as I threw the ball?
Who's? Who's? Who's able to frame moving downwards with the
balls terminal velocity?
Who had put the hands up yet?
OK, then we'll start again. Guess. I mean, I don't care if
you get wildly because in the moment you're gonna talk to each
other and explain why? Why? Why? Why why you, right. Who said the
frame of the room?
Who said it was a frame of upwards with same speed
free moving downward of the terminal velocity?
I still have to eat all those hands. If if you have to guess,
I don't mind. I just want to see all those hands up. I want to
make some sort of commitment at some point.
Who would see with the frame of The
Who has stayed with the free movie upwards? Who say with the
frame moving downwards?
That's enough. You talk to your neighbours. Tell them why you're
right.
Look what it should be.
Well,
OK, having I thought. I remember. I say again, I I don't
keep track of who answers what. I don't care. All I'm all I'm
interested is aggregate numbers to see if I'm going about the
right speed.
Who's here with the name of the room?
Who is it with this free move upwards with the speed of the
ball?
Who is it with the free moving downwards the terminal velocity.
OK. It's the fame of the room and and and and and and and I
think I'm going to high speed three of the room because at
that at the point where the balls at the top of its of of
its trajectory or the pens at the top of its trajectory there
the frame in which it's not which is not moving is the frame
of the room. So at the very end, for instantaneously at the top
of its of its of its trajectory is not moving vertically.
OK,
it's that's the frame. So the frame which is not moving
vertically is instantaneously called moving frame and that
frame is the frame of the room. So does that make sense?
OK.
And
I mean in every other instant the ball is moving with respect
to the room. Although all the all the pain or whatever is
moved respects the room, but only that that that instant is
the is instantaneously
key points
right now we're going to go on to an exciting different thing.
So that's all very nice and somewhat abstract.
Ohh, I'll, I'll just mention that.
Yep.
And in in your experience of Newtonian mechanics, you're
you're you're you're familiar with acceleration and force and
so on, being things that you use all the time to solve problems
and so on. They turn out not to be slightly unexpectedly they
turn out not to be terribly useful in relativity context,
partly because they're a bit messy like that.
But there is a sort of constant acceleration equations for
interactivity, which I quote from the notes just so you've
seen them. But they turn out not to be as as useful because in
relative to context we do the sort of puzzle we have to solve,
like how fast is this electron moving in the accelerator? How
fast is this accretion disc moving around
a black hole? They tend not to be constant acceleration type
problems, so that that that that tends not to be an important
technique. But they do exist and they're in the notes,
OK. And we're moving on
applications of this.
I've mentioned the
velocity acceleration for vectors and I said we can derive
them fairly straightforwardly just differentiating the
displacement vector component by component.
This is straight forward, but I I have just admitted they're not
terrifically useful too for talking problem.
But there is another important four vector which we can look at
quite closely and discover
useful things immediately.
We're going to do is.
So imagine
ah
ohh it's a wave fronts.
But let's talk about these being realistic. The in water or
whatever right now would be completely generic, but these
are waves off
the way France
and there's a
wavelength. Of course
there's a
a direction,
a special direction, and that's
out of LMN type vector. Go write that in.
Nothing special with that of that. At that age,
I would imagine 2 successive events
on this wave front.
And then
a little while later when the we've moved on a bit
on
an event
on another wave front
and we're going to add what are the displacement
between those two events.
We could call that Delta R
and that Delta R
it's going to be Delta T, Delta X.
But why
Doctor Z in the expected fashion?
OK,
no. In the team, there's way front is moving at a speed You
I've lost to you.
So in a time delta T
each wave front will have moved forward
a distance.
You just
so the length
between
one and two
is going to be.
You
Delta T plus
in Lambda with
any of the number of away fronts we're talking about.
So we're building up this construction step by step.
But that
special separation
between 1:00 and 2:00 is also
just
and dot dot R
which is equal to L delta T
plus m.m dot X + M dot Y + N
doctor Z.
Keep that with that. Just the, the, the, the, the length. Or
two different ways of of getting the length of the spatial
distance between these two events, and we rearrange that a
bit. We get
you just a T minus L X -, m, Delta Y minus N, Delta, Z
equals minus
and Lambda. And you from the pattern of saying you could
probably see where this is going
and we can divide through by Lambda
and write F is equal to U for Lambda that F dot at t -, L over
Lambda, delta X -, m over Lambda, delta Y minus N over
Lambda Z. You're going to make N
and we can write that
as
L
dot R where delta R is this thing here
before
and L squeeze it in. The bottom here
is
F,
L over Lambda,
over Lambda, and over Lambda.
So if I the point is all right, *****.
If I define
this thing L in this way,
then and and recall the definition of delta R as above,
then L to R is equal to this thing here.
OK, that's
suggestive.
This looks a bit like a a bit like a vector. Is it a vector?
Or is it just some some numbers in a row
and
I'll look at? Let's look at that, just to help to suggest
that a bit.
And
images of water waves move along the X axis at speed 10 metres a
second with wavelength 5 metres.
What's its frequency? For Vector?
We're going to, well, let. Let's all just stare at that for a
moment and then I'll get you to talk to each other. And
what I'll do is I'll put the other thing on one
the other thing
you're looking at this.
So what's the frequency for vector of that in that
situation?
And
OK,
who would say it was the first one?
Hoosier was the second one.
Who's the third one?
OK,
I'm just going to. I don't trust my own arithmetics, I'm just
going straight to answer.
Move along the X axis so So N is equal to 1.
The frequency is 2/2 Hertz, 10 metres second divided by 5
metres
and so that means that the
you you over Lambda
the you know the this
first component over Lambda in the 5th so it would be two 5th
00. I mean
don't worry if you got that I I got I was doing that but doing
that with you I got it wrong as well because I've got my
arithmetic rubbish. But the point is, there's nothing
particularly exotic about this frequency 4 vector.
It's just there's not really obvious thing.
No. So so where we got to
is
is this. I'm going to write that down again.
L is equal to F&L over Lambda, M over Lambda, N over
Lambda and delta R is equal to delta T, Delta X, Delta Y, Delta
Z
and we worked out the L... R is equal to minus N and that which
is just a number.
But look back to the beginning of this chapter and one of the
things that I showed you
was that the components of a vector
in one frame
turn into
component of the same vector in another frame
when the when the actually upon by the transformation matrix big
Lambda.
And so the components of delta R,
delta X, Delta Y, delta Z
in terms of the components of the same vector in another frame
are going to be
add primed plus
gamma, delta X primed plus V delta T prime primed
data. Why prime data
their frame? Now we know that simply because we R is our
prototype
full vector displacement vector is a prototype 4 vector and we
deduced that its components transformed in this particular
way.
I'm falling behind time here. I'm talking too much
now. We can substitute that into this expression here
and they would rearrange it
and we end up with
dot dot R
equal to
camera
and
F -, V one over Lambda,
but elbow Lander
camera
over Lambda minus
PDF.
Aim of a Lambda
and Orlando.
Daughter
Print,
so all of them there is. Substitute that back into L dot
r = -, n and we get the same thing. Is this expression here
after a bit of rearrangement. I encourage you to go through the
steps of that just to reassure yourself I'm not pulling a fast
one.
But what we've got here
is the
the L primed that we would get if we took L prime equal to
Lambda L where Lambda is the transformation matrix
that gets us this
out of that
is L primed dot
R print.
So just to repeat, what I'm doing there is if we took the
components of L up at the top there and acted upon them with
the same matrix that get got is that that that transformed delta
R what we would get is L primed with these components here.
So what we discovered
is that this vector L
would find at the top,
transformed in the scene. We have our four vector.
It behaves the same way as the displacement vector. It's a four
vector in other words.
OK, now is that not sort of inevitable?
I seem to have been trivial here.
No I'm not. Because if I were to say I don't know what what S
doing in that in that in in that in that definition of. Well, OK,
let it seems pointless. Let's just put define L because I want
to
to be 0 over Lambda, M over Lambda over Lambda. Why
shouldn't I do that? Why should I do that? Because if I do the
same process here and operate on that with the transformation
matrix L Lambda rather
what I get
is what we'll have a a non 0 component
in the in the time slot it'll have changed its form. And
moreover this property of the dot product which does R
wouldn't remain frame invariant.
In other words, it's only when we include that F at non zero
with frequency in the frequency 4 vector
that we get these correct properties of a vector. So this
is behaving like a four vector is the point
by by his behaviour. Shall you know it?
OK. And that's called the frequency for vector
and it's an important thing.
Uh, no. Next thing is
imagine we had our
away free
moving right before
and in our
a prime frame is moving at speed at at at an angle Theta with
respect to the X prime axis
in that frame. Then
the
4 vector would be
it. Frequency in that frame
cost you frame over Lambda.
Something strange over Lambda that's going to be its
the the four vector describing the the motion in the prime
frame when it's moving with an angle Theta primed to the X
axis. So it's moving in that direction, but like that, like
that.
OK.
What then
would that look like in the unframed frame? In the frame
with respect to which that frame is moving?
Uh,
I'm
that L is equal to
inverse transformation.
Hell
transformation on
L primed
equals
what equals,
but that
it also the case that in
this
frame, in the frame with respect to which the frame is moving,
the the same wave train is going to have our frequency,
it's going to be moving at an angle.
So these are all primed as well
if we move at an angle Theta not Theta primed
over Lambda
0.
And we can compare these two expressions for L component by
component
to get
the frequency
in the
thank you thank you. The frequency in the prime frame as
our function
frequency in the unframed frame as a function of the frequency
in the frame frame.
This factory Gamma
1 + V
Cos Theta over
the frequency of the wave train in that frame,
and a similar expression for the the change in the angle. In
other words, simply by virtue of
the of regarding the four vector as our the the frequency 4
vector as a four vector,
that tells us we're allowed to relate the components of that
frequency of that vector in the prime frame L prime
and deduce the bit of missed out the... Deduce what the frequency
pulling point of that same vector R in the unprimed frame
and compare them to what we decide that we're going to call
those components in this frame and so get a way of transforming
the frequency from one frame to another.
And then we're we're, we're getting towards the, the, the.
At the point here,
umm,
and because I'm running out of time, I'm not going to go
through the the the, the blow by blow details, but we can
and discover that the the length of the of the four vector
is
OK
Yeah
and what? What do I miss out?
OK. I'll I'll, I'll I'll skip a bit about and
talk about the important case
where the the the the way free in question is
a light wave moving at the speed of light.
OK, so the velocity U is equal to to one and the bitter missed
out. It's a question.
ITV Cos Theta primed over U Primed
so but but that you is the is the speed of the of the wave
train. But we're going to but what we're interested in really
is the special case where the wave train is moving. The speed
of light is a is a light wave,
and in that case
the
this expression changes. It simplifies to.
Is equal to
if primed gamma 1 plus
the Cos, Theta
and Costa
equals.
You're playing plus V
over 1 plus
Cos Theta
praying where V is the open question. Is it gamma times that
or is it gamma off that? Ohh, gamma 10 times that? Yes. Good
point. Yes. So it's just gamma of V where V is the relative
speed between the two frames. Thank you. Yes,
and and that is a nice way. There are other ways of doing
it. If you look at Carol and Astley, the recommended book
they have, they're a different way of deriving this, which I
think is is much messier. But this way we have deduced this
relativistic Doppler shift directly from the requirement
that the the frequency for vector is a full vector.
So it's it. So the first equation we wrote in this in
this in this chapter the beginning of chapter 8 about the
transformation of vectors from one frame to another.
It's what lets this pop out.
And you are familiar with the Doppler shift of for sound
waves. You're familiar with Doppler shift for light waves,
for example. You can tell how fast a Galaxy is is moving back
at the Doppler shift from the light emitted from things on the
edge of the Galaxy, and so on. And that there's a bit, These
are the staples of observational astronomy.
That Doppler shift is just the non realistic limit of this
expression here.
With the, the, the, the, the feature in question being a nice
straightforward line of sight. Zero 100 degrees, whatever you
want.
And so I'm gonna write the remaining 5 minutes
having
rushed there. And this is this is the double shift. It's it's a
very important and and directly observable
feature
of the change of the change of frequency and wavelength and and
and the direction of light we move from one frame to to to to
another. It's directly observable.
You can see that book you you you learned in in in
last year in astronomy last year, how to use the Doppler
shift to
measure the speeds of things at astronomical distances and the
fact that the direction changes. The direction of Starlight
changes depending on what direction the Earth is moving
in.
Because these changes depending on on on on on whether the Earth
is moving in that part of it, Robert, or on this part of its
orbit,
The change of direction is not much that that that. That
particular aberration is not much, but it was detected in the
18th century.
People were were were able to make precise enough measurements
of the procession of astrometric measurements of the direction of
stars at different times of the year to detect that there was a
an anomaly over the course of the year which with this rustic
aberration
and you can detect this in a laboratory, it's nice. It's very
easy to detect the non relativistic Doppler shift that
that that that the longitudinal Doppler shift where where Theta
is 0 or or π. They used to take that thing just changed speed
change colour very obviously. It's much less less easy to
detect the transverse Doppler shift because as something is
moving past you
perfect you perpendicular state. There's no there's there's zero
classical Doppler shift there because it's not moving moving
towards you.
So the only Doppler shift you get is because of time dilation.
If it
effectively could have obtained
and that's a very small effect
and it's very hard to measure.
But it it was detected in 1932, I think the two Eyes and
Stillwell at Bell Labs. In fact there were
there were industrial researchers and they managed to
detect this effect. And that was one of the one of the early
straightforward observational
confirmations of of lots of defects in in in a laboratory.
So this is very, very practical, very, very, very
straightforward, detectable and an immediate consequence of
relativity in the laboratory.
And I am going to stop there because I've run out of time. I
had hoped to take on to Chapter 9 in this in this lecture, and I
think it would be ambitious to you all the way through. Chapter
Definitely. I think we're basically get all the way
through that next time, so I think we will drift eat into
lecture 11 a bit, but I aim to get through Chapter 7 fairly
briskly, so it would be a good idea to look at that.