MARKUS KLUTE: Welcome back to 8.20, special relativity.
In this section, we want to discuss Lorentz transformation.
Or, in other words, given an event observed by Bob,
we want to express that event as observed by Alice.
We want to find the translation between the observations
in Bob's reference frames to the observation
in Alice's reference frames.
We have already done this for the classic case
as Galilean transformation.
Now, we want to do this in the framework
of special relativity.
In order to simplify the discussion,
we don't worry about the y- and z-component here.
Those dimensions can be neglected
if we assume that the relative motion between the two
reference frames only in x-direction.
We also know from the previous discussion
that you can use the invariant interval.
ct squared minus x squared is the same observed in Bob's
and in Alice's reference frame.
We'll make use of this fact.
And, lastly, we can assume that this transformation
has to be linear.
Why?
Because we transform something like a measurement of distance
into a measurement of distance.
It has to be linear.
If not, we find something like a length squared
or the same for time.
And we might end up on time squared
if we don't do this correctly.
All right, so we can write this down as a linear equation,
which is a multiplication of a matrix with a vector, ct, x,
into a vector, ct [? x ?] [? prime, ?] x prime.
OK, so the goal here now is to find
the parameters or the coefficients of this matrix,
OK?
I invite you to stop the video here and try to work it out.
It's an interesting exercise.
It tests your algebra knowledge.
There's not much physics in here,
but it's still useful to go along and try to work this out.
So the first thing we want to do is
assume that the origins coincide at t equals 0.
And then we can follow along the trajectory of the origin
of S prime in the S frame.
So this is just ct, vt.
OK, great.
This already gives us a constraint on the coefficients
a1,0 over a1,1, which is equal to minus v/c, OK?
And then we can use the invariant interval,
which is another constraint.
And we can use this to obtain the set of equations here.
I will not read this for you.
And that's already enough in order
to solve the set of equations.
So, if you do this and follow along,
you find answers for all four coefficients
given gamma and beta as we defined them before.
This then simplifies to our Lorentz transformation.
So the only thing we did here is we simplified a little bit.
We assumed that this is a linear transformation.
We used the invariant interval in order
to set the constraints.
And we find Lorentz transformation.
If I summarize this, we find this matrix
here with coefficients gamma, minus gamma beta, minus gamma
beta, and gamma.
Great.
Or, if you want, you can write this
as an equation for the spatial component and the time
component.
So does this make sense?
There's always a chance that we make a mistake
in this kind of calculation.
So we want to make sure that the answers we developed
in previous sections actually are reflected
by this transformation.
So let's go one by one.
The first thing we can do is check units.
If we do that, we see that this first equation here
is of unit meter, and then we can analyze the second part
of the equation.
OK, so gamma is unitless.
x is of unit meter.
And then we have beta ct.
Beta is unitless.
c is meter per second times second, also of unit meter.
So this checks out.
The second equation is very similar.
c times t is of unit meter.
Meter per second times second is of unit meter.
Gamma is unitless.
Beta is unitless.
And then we have an x, unit [? meter, ?] plus ct,
c, meter per second times second, also meter.
So this checks out.
So this is great.
At least we find that we have a linear transformation
by design, and the units work out.
So now we can see, what happens now
if we use this for velocities which
are much, much smaller than the speed of light?
In this case, gamma is equal to 0, and beta is very close to 0.
If we put this in our equations, you find x prime
is equal x minus vt.
And t prime is equal to t.
OK, this checks out because this is our Galilean transformation.
So, for systems which move relative
with very low difference in velocities,
we can use Galilean transformation
as an approximation of Lorentz transformation.
OK, at a third part, now we can investigate a little bit
further.
For example, what happens now to a distance,
just a measure of distance or a measure
of length, which we obtain by making
this measurement simultaneously at t2 equal to t2?
We find delta x prime is equal to gamma delta x.
All right, that's length contraction.
If we do the same thing for delta t,
for doing the measurement of time
at x equals-- x2 equals x1, we find time dilation.
All right, this is exactly what we expect.
And then we can look at two events which
happen at the same time in frame S
and see what happens to the time,
as measured in system S prime.
Delta t prime is equal to gamma delta t.
Well, in this example, we set this to 0.
And then we have the second term, which is minus
beta over c gamma delta x.
So we find that, while this event happened simultaneously
in our frame S1--
or in S, it does not happen simultaneously
in our frame S prime.
There's an extra term, which is not 0
unless you actually measure at the very same point,
l is equal to 0.
So this is the relativity of simultaneity.
Again, this checks out.
And I think we're good with our Lorentz transformation.