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.

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?

thank you

amazing 😉

Really neat and clear explanation….

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

How is he writing on the screen? Thats so cool!

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?

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?

Thanks MIT.. It's the best explanation of transformation of inertial frame….

But why this videos have this small like?…

really thank u