1 Attachment(s)

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.

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 . However, is a summation index in the rule for contravariant transformation. I think the vector field should be just , then at (3,2) and at (3,2). Am I understanding this right? If I'm correct does this help clear up what was confusing you?

Good luck!

1 Attachment(s)

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

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 ; it should be

.

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 , and the "unbarred coordinates" are the contravariant components in Cartesian coordinates . So, in this example,

1.

2.

3.

4.

Since your instructor is directing us to compute and , we are figuring out what is at the point (3,2). Expanding the summation in the contravariant transformation formula for , and using 1, 2 & 4 above, we have

5.

Now we know the following:

6. at (3,2)

7. at (3,2)

8. at (3,2)

9. at (3,2).

The thing that remains for you to determine is what are and 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!

Re: Contravariant Vector (newbie in trouble)

Thank you so much for your detailed reply GJA (Clapping)

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

Re: Contravariant Vector (newbie in trouble)

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 and 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!

Re: Contravariant Vector (newbie in trouble)

Thank you once again GJA, I copied the previous method and obtained 25 / SQRT 13. (Rock)

2 Attachment(s)

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)

Re: Contravariant Vector (newbie in trouble)

Please guys any advice would be fantastic, is this answer correct I have posted?

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.

1 Attachment(s)

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)

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