4.3 Reference Frames
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4.3 Reference Frames

October 14, 2019


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.

Only registered users can comment.

  1. what does it mean if "the coordinate axes in frame S' are parallel to, but displaced away from the coordinate axes in frame S"? Also, is the object moving from the origin of Frame S to where it is shown in the video?

  2. MIT is improving the world by providing free education. Meanwhile my school doesn't even give students soft copy of our lecture notes and tutorials and doesn't give access to lecture videos of higher order subjects to everyone

  3. Sir in 4:43 you said that r prime will appear to be in motion because the distance between the observers changes. What if the S prime observer is moving in a circular path?

  4. I didn't understand the part where it says the a and a' are equal. isn't the r' moving while the r is resting? how come they have the same acceleration?

  5. Thanks MIT.. It's the best explanation of transformation of inertial frame….
    But why this videos have this small like?…

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