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Classical Mechanics - Tranformations

  1. Sep 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider a transformation to a relatively uniformly moving frame of reference, where each position vector ri is replaced by rli = ri − vt. (Here v is a constant, the relative velocity of the two frames.) How does a relative position vector rij 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.


    2. Relevant equations

    (1.1) pli = miai = Fi

    (1.2) Fi = Fi1 + Fi2 + · · · + FiN = ∑Fij

    (1.3) Fji = -Fij

    (1.4) Fij = r^ijf(rij)

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

    3. The attempt at a solution

    So I said rij = ri - rj

    As rli = ri − vt, rearranged and got ri on its own and then subbed into rij = ri - rj, giving:

    rij = rli - rj +vt

    I'm confused about the next part.

    v is relative velocity so: v = vi - vj

    What does this mean for the momenta and the forces? Any help would be much appreciated.
     
  2. jcsd
  3. Sep 18, 2014 #2

    mfb

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    ##\hat{r}## -> ##\hat{r}##

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

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

    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.
     
  4. Sep 18, 2014 #3
    Hey mfb thanks for the help.

    So r'ij = r'i - r'j?
     
  5. Sep 18, 2014 #4

    mfb

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    = ... ?

    Can you express that in terms of the old rij?
     
  6. Sep 18, 2014 #5
    r'ij = r'i - r'j = ri - vt -(rj - vt)

    I'm not sure about the r'j as it doesn't specifically say what it is in the question.
     
  7. Sep 18, 2014 #6

    mfb

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    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.
     
  8. Sep 18, 2014 #7
    So vt cancels and I'm left with:

    r'ij = ri - rj

    which is the same as rij. So there's no change?
     
  9. Sep 18, 2014 #8

    mfb

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    Right.
     
  10. Sep 18, 2014 #9
    So does this mean there is no change in the momenta or forces also then?
     
  11. Sep 18, 2014 #10

    mfb

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    What are the equations that brought you to this conclusion? Guessing does not count.
     
  12. Sep 18, 2014 #11
    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
     
  13. Sep 18, 2014 #12

    mfb

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    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.
     
  14. Sep 18, 2014 #13
    Ok so going on pi = mivi

    v'i= vi - v

    but v = vi - vj

    so v'i= vi - (vi - vj)

    v'i = vj

    so p=mvj

    Is this correct? And if it is do I just differentiate again to get acceleration for F=ma?
     
  15. Sep 18, 2014 #14

    mfb

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    What is vj?

    And where is p'?

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