Classical Mechanics - Tranformations

[teme92]

Homework Statement

Consider a transformation to a relatively uniformly moving frame of reference, where each position vector r_i is replaced by r^l_i = r_i − vt. (Here v is a constant, the relative velocity of the two frames.) How does a relative position vector r_ij transform? How do momenta and forces transform? Show explicitly that if equations (1.1) to (1.4) hold in the original frame, then they also hold in the new one.

Homework Equations

(1.1) p^l_i = m_ia_i = F_i

(1.2) F_i = F_i1 + F_i2 + · · · + F_iN = ∑F_ij

(1.3) F_ji = -F_ij

(1.4) F_ij = r^_ijf(r_ij)

The r^ is supposed to be the unit vector but I can't get r hat to work.
p=mv
F=ma

The Attempt at a Solution

So I said r_ij = r_i - r_j

As r^l_i = r_i − vt, rearranged and got r_i on its own and then subbed into r_ij = r_i - r_j, giving:

r_ij = r^l_i - r_j +vt

I'm confused about the next part.

v is relative velocity so: v = v_i - v_j

What does this mean for the momenta and the forces? Any help would be much appreciated.

 

[mfb]

The r^ is supposed to be the unit vector but I can't get r hat to work.

$\hat{r}#[/color]# ->$\hat{r}##

So I said r_ij = r_i - r_j

Fine so far.
Now the question asks you to find r'_ij.

r_ij = r^l_i - r_j +vt

That statement is true, but not helpful - it mixes coordinates from one frame with coordinates from the other frame.

v is relative velocity

No, you can (and should) look at the absolute velocity in this frame. The position is the only part of the question where relative quantities are looked at.

 

[teme92]

Hey mfb thanks for the help.

So r'_ij = r'_i - r'_j?

 

[mfb]

= ... ?

Can you express that in terms of the old r_ij?

 

[teme92]

r'_ij = r'_i - r'_j = r_i - vt -(r_j - vt)

I'm not sure about the r'_j as it doesn't specifically say what it is in the question.

 

[mfb]

i or j as index does not matter, that is just an arbitrary letter.
You can simplify the right side now. The answer will get really short.

 

[teme92]

So vt cancels and I'm left with:

r'_ij = r_i - r_j

which is the same as r_ij. So there's no change?

 

[mfb]

Right.

 

[teme92]

So does this mean there is no change in the momenta or forces also then?

 

[mfb]

What are the equations that brought you to this conclusion? Guessing does not count.

 

[teme92]

Well p=mv: m is constant and seeing as r doesn't change the derivative which is velocity will be the same.

F=ma: once again m is constant and a is the second derivative of r, so there should be no change

 

[mfb]

r (the position coordinate) does change, you directly see this in the problem statement.
What you calculated before was the relative position of two objects.

 

[teme92]

Ok so going on p_i = m_iv_i

v'_i= v_i - v

but v = v_i - v_j

so v'_i= v_i - (v_i - v_j)

v'_i = v_j

so p=mv_j

Is this correct? And if it is do I just differentiate again to get acceleration for F=ma?

 

[mfb]

but v = v_i - v_j

What is v_j?

And where is p'?

And if it is do I just differentiate again to get acceleration for F=ma?

That will work, but you have to find the correct p' first.