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We can consider a
given coordinate system

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as a reference frame within
which we can describe

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the kinematics of an object.

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By "the kinematics," I mean
the position, the velocity,

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and the acceleration
as a function of time,

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basically a geometric
description of the motion.

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Some aspects of these
kinematics will look different

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in different reference
frames and I'd

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like to examine that now.

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First, I want to define what I
mean by an "inertial reference

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frame."

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An inertial reference
frame is one

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in which an isolated body, one
with no net force acting on it,

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moves at constant velocity,
where that constant velocity

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might be zero.

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Another way of saying this
is that an inertial reference

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frame is one in which
Newton's laws of motion apply.

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Recall that Newton's
first law of motion

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states that an isolated object
with no forces acting on it

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moves at constant velocity.

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So let's begin by
considering an observer

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in a particular reference frame.

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We'll call that reference
frame S and denote it

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by coordinate axes x and y.

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And let's consider an object
that in that reference frame

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is at a position vector small r.

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We can then consider a
second reference frame, which

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I'll call the frame
S prime, and I'll

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denote that with coordinate
axes x prime and y prime.

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In my example here,
I'm going to assume

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that the coordinate axes in
frame S prime are parallel to

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but displaced away from the
coordinate axes in frame

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S. More generally, we could
have the S prime coordinate axes

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rotated with respect
to the frame S axes.

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That's a complication
I'm not going to add now

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but conceptually, it's
not really different.

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For simplicity, we'll stick to
parallel axes in this example.

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So imagine we have
two observers, one

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in frame S at the origin and one
at the origin of frame S prime,

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both looking at the same object.

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Observer S will measure a
position vector little r.

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Observer S prime will measure a
position vector little r prime.

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The two observers have a
relative position vector,

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capital R, which is the
position of S prime relative

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to the origin of frame S.

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So what we'd like to see is how
are these different position

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vectors related.

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Well, from the geometry
of the diagram,

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we can see that the position
measured by observer S, which

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is just little r, is equal
to the position of observer S

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prime relative to
observer S, which

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is capital R, plus the
position vector measured

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by the observer at S prime,
which is little r prime.

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I can rewrite that if I'd like
to write the position measured

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by the observer at S
prime in terms of what's

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measured by the observer
at S. I could just

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rearrange this and
write that little r

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prime is equal to little
r minus capital R.

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Now let's add a
further complication

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and assume that the
observer S prime is not

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just at a different location
from the observer of S

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but is moving at constant
velocity relative to frame S.

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So we'll assume that
frame S prime is moving

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at constant velocity
with respect

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to frame S at a constant
velocity vector V.

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So V vector is a constant.

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And in that case, my
offset of observer

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S prime relative to observer S,
which is the vector capital R,

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is a function of time.

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So capital R is a
function of time

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and it's given by the
offset at time 0, which

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I will call capital R0,
plus the elapsed motion

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due to the constant velocity,
which is capital V times time.

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So since capital R is a
function of time, that tells me

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that in this equation, little
r prime, the position vector

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measured by the observer
in frame S prime,

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is also going to be
a function of time,

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even if capital-- sorry-- even
if little r is a constant.

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So notice what that means.

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If the object is
at rest in frame S,

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the object will
appear to be moving.

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Its position vector will be
time-dependent in frame S prime

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because capital R, the location
of S prime relative to S

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is changing.

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So this relation tells us how
the position vectors in the two

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frames are related.

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What about the velocities?

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Well, to compute how the
velocities are related,

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we can just take the time
derivative of the relation

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of the position vectors.

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So in this particular
case, we have

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that the time derivative
of the S prime position,

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d little r prime dt,
is equal to the time

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derivative of the position in
frame S, which is d little r

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dt, minus the time derivative of
the offset of S prime relative

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to S. So that's
minus d capital R dt.

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And I can rewrite that in terms
of symbols for the velocity.

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So I have here the velocity
little v prime, which

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is the velocity measured by
an observer in frame S prime,

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and that's equal to the
velocity of little v, which

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is the velocity
measured by frame S,

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minus d capital R
dt, which we see

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is just capital V vector,
which is the velocity of S

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prime relative to S. So this is
how the velocities are related.

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And again, notice that if the
object is stationary in one

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frame-- so if it's
0 in one frame,

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it will be nonzero
in the other frame.

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So in general, you will
measure different velocities

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in the different frames, even
if one of those velocities is 0.

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Now, how are the
accelerations related?

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Well, again, we can just
take the time derivative

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of the velocities
to figure out what

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the relationship of
the accelerations is.

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So differentiating
this equation,

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I have that d
little v prime dt is

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equal to d little v dt
minus d capital V dt.

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But here, something interesting
happens because remember,

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we said that capital V
is a constant vector.

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And remember, capital
V is the velocity

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that frame S prime has
relative to frame S.

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So since capital V
is a constant, that

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means that this term goes to 0.

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And so we see that the
acceleration in frame S prime

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is equal to the
acceleration in frame a.

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So if I have two reference
frames, one moving

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at a constant velocity relative
to the other, in general,

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I will measure
different positions

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and different velocities
for an object as

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measured by the two frames.

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However, the accelerations
measured in both frames

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will be identical.

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Because the accelerations
are identical,

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we'll see that Newton's
laws will look identical

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in the two frames.

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And we can see that
in the following way.

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In frame S, we
have that the force

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is equal to the mass
times the acceleration.

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In frame S prime,
the force F prime

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is equal to the
mass times a prime.

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But a prime is equal to
a, as we calculated here.

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So we see that the
forces in the two frames

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are identical, even though
the positions and velocities

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in general will be different,
as measured in the two

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frames for the same object.

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But the accelerations
will be identical

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and so the forces
will be identical.

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So if one of these
reference frames

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is an inertial
frame, one in which

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an isolated body moves
at constant velocity,

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then any other frame moving at
constant velocity with respect

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to the first frame will
also be an inertial frame.

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What this means is
that you're always

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free to transform from one
inertial frame to another.

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And what that means
is that you can always

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transform to another
frame that is moving

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at constant velocity
with respect

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to an original inertial frame.