# Contravariant Vector (newbie in trouble)

• Aug 3rd 2012, 10:27 AM
AA23
Contravariant Vector (newbie in trouble)
Hello everybody, great forum. This is my first post so any help you can provide would be great (Hi)

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.
• Aug 3rd 2012, 11:35 AM
GJA
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 $\displaystyle V^{\mu}=(x-y, x^{2}+y)^{T}$. However, $\displaystyle \mu$ is a summation index in the rule for contravariant transformation. I think the vector field should be just $\displaystyle V$, then $\displaystyle V^{1}=1$ at (3,2) and $\displaystyle V^{2}=11$ at (3,2). Am I understanding this right? If I'm correct does this help clear up what was confusing you?

Good luck!
• Aug 3rd 2012, 11:45 AM
AA23
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 Attachment 24414
• Aug 3rd 2012, 12:53 PM
GJA
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 $\displaystyle \frac{\partial \theta}{\partial x}$; it should be

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

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 $\displaystyle (r, \theta)$, and the "unbarred coordinates" are the contravariant components in Cartesian coordinates $\displaystyle (x,y)$. So, in this example,

1. $\displaystyle u^{1}=x$

2. $\displaystyle u^{2}=y$

3. $\displaystyle z^{1}=r$

4. $\displaystyle z^{2}=\theta$

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

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

Now we know the following:

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

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

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

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

The thing that remains for you to determine is what are $\displaystyle \cos\theta$ and $\displaystyle \sin\theta$ 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!
• Aug 3rd 2012, 01:19 PM
AA23
Re: Contravariant Vector (newbie in trouble)

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
• Aug 3rd 2012, 01:27 PM
GJA
Re: Contravariant Vector (newbie in trouble)
$\displaystyle 25/\sqrt{13}$ 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 $\displaystyle \frac{\partial r}{\partial x}$ and $\displaystyle \frac{\partial r}{\partial y}$ 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!
• Aug 3rd 2012, 01:38 PM
AA23
Re: Contravariant Vector (newbie in trouble)
Thank you once again GJA, I copied the previous method and obtained 25 / SQRT 13. (Rock)
• Aug 5th 2012, 06:23 AM
AA23
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 (Nod)
• Aug 6th 2012, 04:19 AM
AA23
Re: Contravariant Vector (newbie in trouble)
• Aug 6th 2012, 05:17 AM
GJA
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.
• Aug 6th 2012, 05:38 AM
AA23
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 (Nod)
• Aug 8th 2012, 07:20 AM
AA23
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