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Please HELLLP

  1. Feb 17, 2007 #1
    Please HELLLP!!!!!

    This is driving me NUTS!

    Can someone please explain to me whats going on here, because for the life of me this makes no sense! It appears, at least to me, that this 'proof' has a lot of flaws in it.

    [​IMG]

    [​IMG]


    Notice, on the next page the author says explicitly:

    "It should be reiterated that r is a vector whose components are fixed in the relative frame [tex]x_1'x_2'x_3'[/tex]."

    HUHHHHHHH? But in this very derivation, we had xi and xf being two different points on the transformation coordinate system!

    How can that be a vector with "fixed components" in the relative frame??????????????????????????????!!!!!:mad: :mad: :mad:

    [/frustration!]

    :frown: :mad: :grumpy: :cry:

    I have been trying to understand this for a week now. Its actually kinda sad......
     
  2. jcsd
  3. Feb 18, 2007 #2

    AlephZero

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    The point x is rotating in the same way as the frame.

    In a small time interval the frame rotates from the XYZ to X'Y'Z' axes. The point also rotates from x_i to x_f.

    You can describe the position of x_i and x_f in both the XYZ and the X'Y'Z' frames.

    Because the point and the frame rotate the same way, the coordinates of the point in the intial frame before rotation are the same three numbers as the coordinates in the rotated frame after rotation. That's not the same as saying the point doesn't move.

    Maybe a picture in two dimensions is easier to understand. And keep in mind that the pictures are all drawn as seen by an observer "fixed in space" and "not rotating" (whatever those two statements mean!)
     

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    Last edited: Feb 18, 2007
  4. Feb 18, 2007 #3
    Then why would the author relate the initial point [tex]\{x_i\}[/tex], in the XYZ axis, to a point [tex]\{x'_i\}[/tex], in the X'Y'Z' axis before a change in time delta{t}, where the axis makes a small rotation? This is what does not make sense.

    The way I read this proof, I see one inertial frame, and one rotated frame. But I do not see the rotated frame as moving, I just see it as nonparallel to the original frame (but fixed). And it is the point x that is moving, (not the frames!).

    I dont see the author saying anywhere that the frame is moving.
     
    Last edited: Feb 18, 2007
  5. Feb 18, 2007 #4

    AlephZero

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    Because you already know how to transform between two different coordinate systems that are defined at the same instant in time, and you know how to rotate a point in a fixed coordinate system. He's building on those two things to define "angular velocity".

    If you like, imagine the process happens in two steps: first you rotate the point with the coordinate system fixed, then you rotate the coordinate system by the same amount. Or, you can rotate the coordinate system first, then rotate the point.

    Of course it gets more interesting later, when the point IS moving relative to the rotating system...
     
  6. Feb 18, 2007 #5
    Ok, now your picture has been approved.

    Observe your picture. I get what your saying. That is xi and xf.

    But that is not what the book is saying. The book has an xi and an xf and and x'i and an x'f in the same frame (Both relative and inertial).

    You, on the other hand, have one point in each of the two frames. I.e. you ONLY have one xi and ONE x'f.

    One point in EACH respective frame.

    See what Im saying?

    As I wait for your reply, I am going to look over the book again taking what you said into account.
     
    Last edited: Feb 18, 2007
  7. Feb 19, 2007 #6
    Ahhhhhhhhhhhhhhh, I need more help with this badly. Can you please show me using the equations in my book how this reflects what you are saying?

    I see what your saying, I just dont see it reflected in the equations. We spent two days doing the entire chapter on this stuff (up to relative velocity and acceleration) and we are going to fly through another chapter on tuesday. I simply cant afford to spend more time trying to figure this out anymore. :frown:
     
    Last edited: Feb 19, 2007
  8. Feb 19, 2007 #7

    AlephZero

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    As I read it, the book has only one point x that rotates.

    You can decribe its position before rotation in two ways:

    xi in the frame before the frame rotates, and x'i after the frame has rotated

    You can also describe its position after rotation in two ways:

    xf in the frame before the frame rotates, and x'f after the frame has rotated.

    And since the frame and the point are rotating in the same way, the coordinates of xi are the same as the coordinates of x'f.

    That's what 2.5.5 is saying IMO. If that doesn't help, sorry, I dunno how else to explain it.

    Agreed, this stuff is hard to get your head round, if you are not used to "rigorous mathematical proofs" of things that look "obvious".

    The point of doing it this apparently tortuous way is that what is true about rotations is not so "obvious" as what is true for translation vectors. For example, you can't just "add up" large rotations like you can add up vectors: In a fixed coordinate system , a rotation about X followed by a rotation about Y is not the same as a rotation about Y followed by a rotation about X (try it for 90 degree rotations with a physical object).

    Hope that helps
     
  9. Feb 19, 2007 #8
    Yes, agreed.


    Yes, I agree.

    Ok, fine.

    Yes, I agree with that.


    ....But now look at eq. 2.5.6. He is taking the change in {x'f-x'i}.

    Instead of saying something, I am going to ask that you explain your take on what that "means" and see if I am thinking along the same lines. This, to me, is where big problems start to occur.



    Well, not really. That is shown earlier. The 'point' of this proof is to show not only that dr/dr = wxr, but (more importantly), that omega is a vector which, when crossed into any vector in the moving frame gives the rate of change of that vector in the inertial frame. That is the core of this proof.


    Thats the thing, I am used to rigorous proofs, which is why this is driving me crazy!
     
    Last edited: Feb 19, 2007
  10. Mar 6, 2007 #9
    AlephZero my friend, looks like you and I were both right on this on!

    You were right in the fact that the coordinate frames are, but not necessarily, coincident.

    Here's the major flaw I hope everyone sees. We are talking about a fixed vector in the body frame, yet we are writing equations for the rate of change of a fixed vector in a rotating frame!

    Thats wrong.

    Heres another book I found that had a very interesting quote.....
    Now, this is not EXACTLY the full story of whats going on above. But it certainly is a problem. According to what was just said, ANY derivation in the body frame is total JUNK. The correct way to derive this equation is to do it relative to an INERTIAL frame, and then, *and only then*, to relate that solution to components in the body frame. Anything otherwise is a *physically meaningless proof*

    A very very good proof that is stupidly simple and gets rid of all this nonsense can be found in "Analytical Dynamics-Udwadia\Kalaba.


    What the author above is doing is holding the primed axis fixed and letting the vector r'_i rotate to r'_f, within that fixed primed frame. But then he later goes on to say r is fixed in the body frame. Wrong, wrong wrong. What hes REALLY doing is looking at the solution from different angles without telling you. In one case, he keeps the primed axis fixed and lets r'_i and r'_f move, then he looks at r being fixed in the primed axis but observing the axis rotation. Its a really bad mess where he goes back and forth without saying why or how...this proof is total crap.
     
    Last edited: Mar 6, 2007
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