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MARKUS KLUTE: Welcome back
to 8.20, special relativity.

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In this section, we want to
discuss Lorentz transformation.

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Or, in other words, given
an event observed by Bob,

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we want to express that
event as observed by Alice.

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We want to find the translation
between the observations

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in Bob's reference
frames to the observation

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in Alice's reference frames.

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We have already done
this for the classic case

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as Galilean transformation.

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Now, we want to do
this in the framework

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of special relativity.

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In order to simplify
the discussion,

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we don't worry about the
y- and z-component here.

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Those dimensions
can be neglected

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if we assume that the relative
motion between the two

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reference frames
only in x-direction.

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We also know from the
previous discussion

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that you can use the
invariant interval.

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ct squared minus x squared
is the same observed in Bob's

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and in Alice's reference frame.

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We'll make use of this fact.

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And, lastly, we can assume
that this transformation

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has to be linear.

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Why?

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Because we transform something
like a measurement of distance

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into a measurement of distance.

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It has to be linear.

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If not, we find something
like a length squared

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or the same for time.

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And we might end
up on time squared

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if we don't do this correctly.

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All right, so we can write
this down as a linear equation,

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which is a multiplication of
a matrix with a vector, ct, x,

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into a vector, ct [? x ?]
[? prime, ?] x prime.

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OK, so the goal
here now is to find

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the parameters or the
coefficients of this matrix,

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OK?

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I invite you to stop the video
here and try to work it out.

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It's an interesting exercise.

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It tests your algebra knowledge.

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There's not much
physics in here,

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but it's still useful to go
along and try to work this out.

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So the first thing
we want to do is

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assume that the origins
coincide at t equals 0.

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And then we can follow along
the trajectory of the origin

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of S prime in the S frame.

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So this is just ct, vt.

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OK, great.

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This already gives us a
constraint on the coefficients

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a1,0 over a1,1, which is
equal to minus v/c, OK?

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And then we can use
the invariant interval,

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which is another constraint.

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And we can use this to obtain
the set of equations here.

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I will not read this for you.

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And that's already
enough in order

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to solve the set of equations.

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So, if you do this
and follow along,

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you find answers for
all four coefficients

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given gamma and beta as
we defined them before.

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This then simplifies to
our Lorentz transformation.

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So the only thing we did here
is we simplified a little bit.

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We assumed that this is
a linear transformation.

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We used the invariant
interval in order

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to set the constraints.

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And we find Lorentz
transformation.

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If I summarize this,
we find this matrix

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here with coefficients gamma,
minus gamma beta, minus gamma

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beta, and gamma.

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Great.

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Or, if you want,
you can write this

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as an equation for the
spatial component and the time

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component.

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So does this make sense?

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There's always a chance
that we make a mistake

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in this kind of calculation.

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So we want to make sure that
the answers we developed

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in previous sections
actually are reflected

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by this transformation.

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So let's go one by one.

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The first thing we
can do is check units.

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If we do that, we see that
this first equation here

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is of unit meter, and then we
can analyze the second part

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of the equation.

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OK, so gamma is unitless.

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x is of unit meter.

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And then we have beta ct.

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Beta is unitless.

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c is meter per second times
second, also of unit meter.

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So this checks out.

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The second equation
is very similar.

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c times t is of unit meter.

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Meter per second times
second is of unit meter.

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Gamma is unitless.

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Beta is unitless.

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And then we have an x,
unit [? meter, ?] plus ct,

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c, meter per second
times second, also meter.

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So this checks out.

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So this is great.

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At least we find that we
have a linear transformation

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by design, and the
units work out.

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So now we can see,
what happens now

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if we use this for
velocities which

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are much, much smaller
than the speed of light?

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In this case, gamma is equal to
0, and beta is very close to 0.

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If we put this in our
equations, you find x prime

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is equal x minus vt.

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And t prime is equal to t.

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OK, this checks out because this
is our Galilean transformation.

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So, for systems
which move relative

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with very low difference
in velocities,

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we can use Galilean
transformation

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as an approximation of
Lorentz transformation.

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OK, at a third part, now we
can investigate a little bit

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further.

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For example, what happens
now to a distance,

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just a measure of
distance or a measure

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of length, which
we obtain by making

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this measurement simultaneously
at t2 equal to t2?

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We find delta x prime is
equal to gamma delta x.

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All right, that's
length contraction.

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If we do the same
thing for delta t,

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for doing the
measurement of time

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at x equals-- x2 equals
x1, we find time dilation.

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All right, this is
exactly what we expect.

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And then we can look
at two events which

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happen at the same
time in frame S

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and see what
happens to the time,

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as measured in system S prime.

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Delta t prime is equal
to gamma delta t.

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Well, in this example,
we set this to 0.

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And then we have the
second term, which is minus

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beta over c gamma delta x.

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So we find that, while this
event happened simultaneously

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in our frame S1--

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or in S, it does not
happen simultaneously

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in our frame S prime.

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There's an extra
term, which is not 0

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unless you actually measure
at the very same point,

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l is equal to 0.

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So this is the relativity
of simultaneity.

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Again, this checks out.

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And I think we're good with
our Lorentz transformation.