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Math Help - Contravariant Vector (newbie in trouble)

  1. #1
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    Contravariant Vector (newbie in trouble)

    Hello everybody, great forum. This is my first post so any help you can provide would be great

    I have attached the problem with this post, the first part has been solved but I am confused for the second part.

    I know all the inverse differential coefficients but am unsure how to use them.
    Attached Thumbnails Attached Thumbnails Contravariant Vector (newbie in trouble)-revision-1.png  
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  2. #2
    GJA
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    Re: Contravariant Vector (newbie in trouble)

    Hi, AA23.

    I'm a little confused on the notation in the handout. It says the contravariant vector field is V^{\mu}=(x-y, x^{2}+y)^{T}. However, \mu is a summation index in the rule for contravariant transformation. I think the vector field should be just V, then V^{1}=1 at (3,2) and V^{2}=11 at (3,2). Am I understanding this right? If I'm correct does this help clear up what was confusing you?

    Good luck!
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  3. #3
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    Re: Contravariant Vector (newbie in trouble)

    Hi GJA, I see what u mean about it being a summation index in the rule for contravariant transformation, however that is the way the exam question has been written. I'm unclear as to how I should use the information established along with the inverse differential coefficients to establish the correct answer, please see attachment below for calculations.

    Thanks again Contravariant Vector (newbie in trouble)-revision-1.png
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  4. #4
    GJA
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    Re: Contravariant Vector (newbie in trouble)

    Hi again, AA23. I think I see what needs to be done here. The notation in the worksheet could use a little improving (mentioned last post). Also, a negative sign got dropped in \frac{\partial \theta}{\partial x}; it should be

    \frac{\partial\theta}{\partial x}=-\frac{\sin\theta}{r}.

    In fairness, typing all the sub/super-scripts and forumlas perfectly is never easy in geometry - particularly in MS Word.

    Now for your exercise the "barred coordinates" are the contravariant components in polar coordinates (r, \theta), and the "unbarred coordinates" are the contravariant components in Cartesian coordinates (x,y). So, in this example,

    1. u^{1}=x

    2. u^{2}=y

    3. z^{1}=r

    4. z^{2}=\theta

    Since your instructor is directing us to compute \frac{\partial\theta}{\partial x} and \frac{\partial\theta}{\partial y}, we are figuring out what \overline{V^{2}} is at the point (3,2). Expanding the summation in the contravariant transformation formula for \overline{V^{2}}, and using 1, 2 & 4 above, we have

    5. \overline{V^{2}}=\frac{\partial\theta}{\partial x}\cdot V^{1} +\frac{\partial\theta}{\partial y}\cdot V^{2}

    Now we know the following:

    6. V^{1}=1 at (3,2)

    7. V^{2}=11 at (3,2)

    8. \frac{\partial\theta}{\partial x}=-\frac{\sin\theta}{r}=-\frac{\sin\theta}{\sqrt{13}} at (3,2)

    9. \frac{\partial\theta}{\partial y}=\frac{\cos\theta}{r}=\frac{\cos\theta}{\sqrt{13  }} at (3,2).

    The thing that remains for you to determine is what are \cos\theta and \sin\theta at (3,2)? Once you figure this out you will know 8 & 9. After figuring out 8 & 9 you plug 6-9 in equation 5 and if everything was done correctly you should get your instructor's solution of 31/13.

    Let me know if there are any other questions. Good luck!
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  5. #5
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    Re: Contravariant Vector (newbie in trouble)

    Thank you so much for your detailed reply GJA

    After establishing values for sin theta and cos theta and substituting them back into equation 5 I do indeed get 31/13.

    However in the answer given it was, 25/ SQRT 13 , 31/13

    What is the 25/ SQRT 13 ?

    Thank you once again
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  6. #6
    GJA
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    Re: Contravariant Vector (newbie in trouble)

    25/\sqrt{13} is the contravariant component in the r-coordinate. Basically, what we just did was to get the theta portion of the coordinate, but you still need to determine the r part. You do the same thing we just did, only this time you will need to use \frac{\partial r}{\partial x} and \frac{\partial r}{\partial y} at (3,2). I will skip writing the details for now; I think you can do it if you follow the model of the previous part. Give it an honest effort, and if you're stuck I'll give more details. Good luck!
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  7. #7
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    Re: Contravariant Vector (newbie in trouble)

    Thank you once again GJA, I copied the previous method and obtained 25 / SQRT 13.
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  8. #8
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    Re: Contravariant Vector (newbie in trouble)

    Just some further questions regarding this topic, if for example I was asked to find the following but for a covariant vector, would this be correct?

    Please see attachments for my attempt
    Attached Thumbnails Attached Thumbnails Contravariant Vector (newbie in trouble)-revision-1-part-2.png   Contravariant Vector (newbie in trouble)-revision-1.png  
    Last edited by AA23; August 5th 2012 at 06:38 AM.
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  9. #9
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    Re: Contravariant Vector (newbie in trouble)

    Please guys any advice would be fantastic, is this answer correct I have posted?
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  10. #10
    GJA
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    Re: Contravariant Vector (newbie in trouble)

    Hi, AA23. The second part looks fine. The MHF hasn't been allowing me to respond after I post a couple of things on a thread for some reason. Don't know why it's letting me do it now.
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  11. #11
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    Re: Contravariant Vector (newbie in trouble)

    Hi GJA, thank you for responding. I actually believe I've made an error, I think it should be 23/SQRT17 , 7.

    I had the sin and cos values in the wrong place for the r coordinate calculation.

    I do have a follow up question, how do I go about establishing the following (see attachment). I imagine it is an identical method with just a few changes
    Attached Thumbnails Attached Thumbnails Contravariant Vector (newbie in trouble)-revision-5.png  
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  12. #12
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    Re: Contravariant Vector (newbie in trouble)

    Anyone have any advice on answering the question above, I've tried a similar technique to above but am sure is incorrect. Thank you in advance
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