# Change of vector in different frames

1. Aug 31, 2010

### aim1732

I just read this in a text:
The rate of change of a vector with respect to a fixed frame is same as that with respect to a frame in translation.
But supposing it were a position vector then rate of change(velocity that is) is different in different frames translating with respect to each other(that is why we have relative velocities). Now I am obviously missing a point but can not pinpoint the error in my logic.

2. Aug 31, 2010

### CompuChip

But do you agree that if you, standing by the side of the road, see the velocity of a car change by 10 m/s, then so will I, sitting in another moving car ?

We will not agree on the relative velocity, but we will agree on the change in velocity. This continues to hold if you look at instantaneous changes.

Have you have any calculus yet? If so: suppose that there is a relative (constant) velocity u. Then you can easily see mathematically that the two velocity vectors v and v + u have the same time derivative.

3. Aug 31, 2010

### aim1732

We do not agree on relative velocity,right? But velocity itself is rate of change of position vector so rate of change in position vector is different in different frames?

4. Aug 31, 2010

### hikaru1221

I think the text is wrong.

5. Sep 1, 2010

### aim1732

In that case the source happens to be Beer and Johnston.

6. Sep 1, 2010

### vela

Staff Emeritus
Say you have two frames, S and S', with S' moving with constant velocity relative to S, and a point P. The position vectors (to point P) you're talking about are two different vectors. One goes from the origin of S to point P, and the other goes from the origin of S' to point P. Obviously, the rate of change for those two vectors, though related, aren't going to be the same.

The text is talking about just one vector, say the one going from the origin of S to point P. If you were in S, you'd call the rate of change of this vector the velocity of point P. If you were in S', you'd say it's the velocity of P relative to S. The text just means that those two quantities are the same.

Say instead you had two frames R and R' whose origins coincide but one rotates relative to the other, and say point P is at rest in R. The same vector r connects the origin to the point P in either frame, but in frame R', the point P appears to move. So you have dr/dt=0 in R while dr/dt≠0 in R'. Non-translational motion of R' relative to R causes the rates of change to differ.

7. Sep 1, 2010

### aim1732

Thanks.That makes a lot of sense.