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MARKUS KLUTE: Welcome back
to 8.20, special relativity.

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In this video, we talk about
Galilean transformation.

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So what is it
you're going to do?

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We want to describe
our event P, maybe

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Professor Klute exploding, with
two different reference frames.

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We can call one of the reference
frames our laboratory frame.

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Maybe that's the frame in which
Professor Klute was stationary.

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It has an origin and
has axis x, y, and z.

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And then we have a
moving frame, which

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is moving with a constant
velocity with respect

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to the laboratory frame.

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The origin is o prime.

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The axes are x prime,
y prime, and z prime.

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All right, so now what
do we learn from this?

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Let's think about an example.

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An example most of
you have experienced

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before is the one where
you sit in a train car.

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And, if it's not a train car,
it can be a car or a plane,

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something which is moving
with respect to Earth.

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If you look out of the
window and the acceleration

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is very, very minor,
it's often not clear

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whether or not the Earth, the
train station, or the train car

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is moving.

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And so you have this kind of
weird feeling that, you know,

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I don't know if maybe the
neighboring train started

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moving or I'm moving.

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But what we're
going to do here is

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describe you sitting
in the train car,

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reading a newspaper, once
within your laboratory frame,

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within the frame of
the train, and then we

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want to describe the very
same events or sequence

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of events in the frame of
the stationary train station.

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All right, let's look
at a specific example.

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Here again our professor
is exploding at a time tP

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at xP, yP, and zP.

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To make this a little bit
easier, we define, at time t

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equals 0, the origin of
the two frames coincide.

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That just means
that, at the origin,

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we have two clocks, two watches.

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And we make sure that
they're synchronized.

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And then my watch stays with
me, and then the second watch

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may be with you,
which moves along.

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And those are great watches.

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They are synchronized.

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We also want to
simplify-- we can always

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define the direction of
our coordinate system

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such that the velocity,
the relative velocity

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between the two
reference frames,

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is in one specific direction.

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And here I decided to use x.

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I could have used y
and z, and I could

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rotate the coordinate systems
or the relative movement

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of the coordinate
systems in any way.

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It's just a simplification here.

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When I do that, I can
rewrite this event P

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in the S prime frame through the
S frame in the following way.

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So, for x, we find that xP is
given by-- x prime P is given

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by xP minus v, the velocity,
relative velocity--

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I could put a little
label x here--

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times tP.

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And then, for the y-coordinates
and the z-coordinates,

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there's no change.

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For the time, intuitively,
you say that those two watches

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are run with the
same speed, meaning

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that the time in both frames for
the same events are the same.

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So now I'm asking you, if
you're watching this video,

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to find the velocity
and the acceleration.

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It might be good to stop
and just write this down.

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So find the velocity and
acceleration of S prime,

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of S prime expressed
by the S-coordinates.

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So let's try to do that.

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So, first, we build the
derivative dx prime dt prime,

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which is our velocity.

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That's the velocity of an
object in my prime frame.

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And that's given by d dt--

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I can just do that here because
the times are the same--

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times x minus vt.

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That's [? x, ?] ux, the
velocity in my S frame,

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minus v. That's really
what you expect.

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You just subtract or
add the velocities.

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If I then build
the acceleration,

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I have to build the
derivative of u x prime, which

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is our acceleration,
in the prime frame.

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Here again I just do
this in x-direction

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because the solution for the
y-direction and z-direction

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are trivial.

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So now I find d
ux dt minus dv dt.

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Now, the velocity,
as we defined,

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between the two reference
frames is constant.

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Therefore, this is 0,
meaning that the velocity--

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the accelerations in the
two frames are the same.

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If the accelerations
are the same,

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that means that the forces in
the two frames are the same.

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And that means that the
forces or the accelerations

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are invariant.

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They do not change based on
the reference frame that I use.

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So now, coming
back to the example

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we discussed in our
very first lecture,

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you are experiment
in the train car.

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S, the velocity between
those two frames,

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the frame in your car or a
second frame, are constant.

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There's no way to tell
whether or not your train

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car is moving or not.

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That is only true as
long as the velocities

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are constant and
unchanged, constant.

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So, in summary, in
Newton's mechanics,

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time and accelerations are
invariant and, therefore,

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also the forces.

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There is no inertial frame
which is above another one.

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So you can pick one or
can pick another one.

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There's no difference in the
descriptions of the physics

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between those two frames.