We can consider a given coordinate system
as a reference frame within which we can describe
the kinematics of an object.
By "the kinematics," I mean the position, the velocity,
and the acceleration as a function of time,
basically a geometric description of the motion.
Some aspects of these kinematics will look different
in different reference frames and I'd
like to examine that now.
First, I want to define what I mean by an "inertial reference
frame."
An inertial reference frame is one
in which an isolated body, one with no net force acting on it,
moves at constant velocity, where that constant velocity
might be zero.
Another way of saying this is that an inertial reference
frame is one in which Newton's laws of motion apply.
Recall that Newton's first law of motion
states that an isolated object with no forces acting on it
moves at constant velocity.
So let's begin by considering an observer
in a particular reference frame.
We'll call that reference frame S and denote it
by coordinate axes x and y.
And let's consider an object that in that reference frame
is at a position vector small r.
We can then consider a second reference frame, which
I'll call the frame S prime, and I'll
denote that with coordinate axes x prime and y prime.
In my example here, I'm going to assume
that the coordinate axes in frame S prime are parallel to
but displaced away from the coordinate axes in frame
S. More generally, we could have the S prime coordinate axes
rotated with respect to the frame S axes.
That's a complication I'm not going to add now
but conceptually, it's not really different.
For simplicity, we'll stick to parallel axes in this example.
So imagine we have two observers, one
in frame S at the origin and one at the origin of frame S prime,
both looking at the same object.
Observer S will measure a position vector little r.
Observer S prime will measure a position vector little r prime.
The two observers have a relative position vector,
capital R, which is the position of S prime relative
to the origin of frame S.
So what we'd like to see is how are these different position
vectors related.
Well, from the geometry of the diagram,
we can see that the position measured by observer S, which
is just little r, is equal to the position of observer S
prime relative to observer S, which
is capital R, plus the position vector measured
by the observer at S prime, which is little r prime.
I can rewrite that if I'd like to write the position measured
by the observer at S prime in terms of what's
measured by the observer at S. I could just
rearrange this and write that little r
prime is equal to little r minus capital R.
Now let's add a further complication
and assume that the observer S prime is not
just at a different location from the observer of S
but is moving at constant velocity relative to frame S.
So we'll assume that frame S prime is moving
at constant velocity with respect
to frame S at a constant velocity vector V.
So V vector is a constant.
And in that case, my offset of observer
S prime relative to observer S, which is the vector capital R,
is a function of time.
So capital R is a function of time
and it's given by the offset at time 0, which
I will call capital R0, plus the elapsed motion
due to the constant velocity, which is capital V times time.
So since capital R is a function of time, that tells me
that in this equation, little r prime, the position vector
measured by the observer in frame S prime,
is also going to be a function of time,
even if capital-- sorry-- even if little r is a constant.
So notice what that means.
If the object is at rest in frame S,
the object will appear to be moving.
Its position vector will be time-dependent in frame S prime
because capital R, the location of S prime relative to S
is changing.
So this relation tells us how the position vectors in the two
frames are related.
What about the velocities?
Well, to compute how the velocities are related,
we can just take the time derivative of the relation
of the position vectors.
So in this particular case, we have
that the time derivative of the S prime position,
d little r prime dt, is equal to the time
derivative of the position in frame S, which is d little r
dt, minus the time derivative of the offset of S prime relative
to S. So that's minus d capital R dt.
And I can rewrite that in terms of symbols for the velocity.
So I have here the velocity little v prime, which
is the velocity measured by an observer in frame S prime,
and that's equal to the velocity of little v, which
is the velocity measured by frame S,
minus d capital R dt, which we see
is just capital V vector, which is the velocity of S
prime relative to S. So this is how the velocities are related.
And again, notice that if the object is stationary in one
frame-- so if it's 0 in one frame,
it will be nonzero in the other frame.
So in general, you will measure different velocities
in the different frames, even if one of those velocities is 0.
Now, how are the accelerations related?
Well, again, we can just take the time derivative
of the velocities to figure out what
the relationship of the accelerations is.
So differentiating this equation,
I have that d little v prime dt is
equal to d little v dt minus d capital V dt.
But here, something interesting happens because remember,
we said that capital V is a constant vector.
And remember, capital V is the velocity
that frame S prime has relative to frame S.
So since capital V is a constant, that
means that this term goes to 0.
And so we see that the acceleration in frame S prime
is equal to the acceleration in frame a.
So if I have two reference frames, one moving
at a constant velocity relative to the other, in general,
I will measure different positions
and different velocities for an object as
measured by the two frames.
However, the accelerations measured in both frames
will be identical.
Because the accelerations are identical,
we'll see that Newton's laws will look identical
in the two frames.
And we can see that in the following way.
In frame S, we have that the force
is equal to the mass times the acceleration.
In frame S prime, the force F prime
is equal to the mass times a prime.
But a prime is equal to a, as we calculated here.
So we see that the forces in the two frames
are identical, even though the positions and velocities
in general will be different, as measured in the two
frames for the same object.
But the accelerations will be identical
and so the forces will be identical.
So if one of these reference frames
is an inertial frame, one in which
an isolated body moves at constant velocity,
then any other frame moving at constant velocity with respect
to the first frame will also be an inertial frame.
What this means is that you're always
free to transform from one inertial frame to another.
And what that means is that you can always
transform to another frame that is moving
at constant velocity with respect
to an original inertial frame.