Basis vectors and the metric tensor 

Thanks for your comment. Much appreciated!

Thank you for your comment.

Hello Logan and thank you for your comment. Much appreciated!

@@TensorCalculusRobertDavie Hi, you are very welcome. I browsed through some of your other titles just now, and I am excited to see a rich source of mathematics of my most favorite type. Would you mind if I cite you as a source in the text I am writing on general relativity (with a rigour in tensor calculus and differential/Riemannian geometry) -- especially for any instances that I am inspired to add to my work because of your content? If you would like, this is a hyperlink to my document. drive.google.com/open?id=1-MU7daeZ0Q8TefNOwzImcGD2uZIhktvZl3FO13R5UkQ Thank you for the content!

You are welcome to cite my material andgood luck with your efforts.

Hello Marina and thank you for your comment. Much appreciated.

Yes, a bit ironic. I hope there aren't too many ads?

@@TensorCalculusRobertDavie No, it's not too bad. That's the price of posting stuff on youtube. They can put ads in your stuff and there's nothing you can do about it except not post videos. But I think it's a small price to pay for the freedom of being able to post mathematical content. I'm pretty grateful for youtube both as a viewer and as a poster.

Thank you again!

Please have a look at the first few minutes of this video.

You're welcome!

Thank you David.

Hello Sjaak, the content covered here does assume some prior knowledge of vector calculus. The main point of the video are the two forms of basis vectors that can be formed so could I suggest that a good starting point would be to focus on the meaning of the diagrams before moving on to deal with the notation and what it is trying to express. Hope that helps?

Hello and thank you for your comment. The answer is no because this video provides you with a basic introduction to basis vectors and one forms (the objects with raised indices). However, the more you learn the better so do continue to study linear algebra if you can. Thank you for the feedback and good luck with your studies.

@@TensorCalculusRobertDavie Thankyou!

Hello Benedek and thank you for your question. Wikipedia discusses this issue in the quote below and further in the link below that. "The vector is called covariant or contravariant depending on how the transformation of the vector's components is related to the transformation of coordinates. Contravariant vectors are "regular vectors" with units of distance (such as a displacement) or distance times some other unit (such as velocity or acceleration). For example, in changing units from meters to millimeters, a displacement of 1 m becomes 1000 mm. Covariant vectors, on the other hand, have units of one-over-distance (typically such as gradient). For example, in changing again from meters to millimeters, a gradient of 1 K/m becomes 0.001 K/mm." www.wikiwand.com/en/Covariance_and_contravariance_of_vectors

@@TensorCalculusRobertDavie So it doesnt matter if you use contravariant or covariant they are just used whenever most conveniently for transforms?

Thank you.

The two bases are distinct, hence the upper and lower indexes, and behave in different ways unlike in Euclidean space where they really are just the same thing hence no reason to raise or lower indices. Sorry for the short answer. Have a look at this article; en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors and this video; ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-CliW7kSxxWU.html In General Relativity we use a metric to raise and lower these indices that is not the same as the Euclidean metric.

@@TensorCalculusRobertDavie Thank you for the reply, will take a look! :)

You are right. Thank you for spotting that.

@@TensorCalculusRobertDavie Thank you for your all efforts, highly appreciated 👍👏👏

Vicente Matricardi Muchos gracias.

Gracias a usted por generar y divulgar tan buena calidad de informacion , le escribo en español por que me agrada que sepa que a mucha gente le interesan estos temas , un saludo !!!!

Vicente Matricardi Thanks Vicente.

Thanks, Robert Davie

Hello Gary and thank you for your question. The inverse metric is the g^ij part and the nabla u is the derivative giving us the maximum direction of increase in the scalar u in each of the directions j. The inverse metric raises the j index on the resultant of nabla u so that we obey the Einstein summation convention and don't end up with two j's down below. We CANNOT have (nabla u)j e_j but we can and must have (nabla u)^j e_j. Hope that helps?

You're welcome.

Hello Ron, thank you for your comment and you are correct however, in this case, we have u(covariant)v(contravariant) = u(contravariant)v(covariant) which was the point I was trying to show across lines 3 and 4. The point here is that there are four different looking ways to get the same result. At the time I did um and arrgh about whether I should write it in the form you have pointed out but my goal took precedence in the end.

Thank you Zoltan, that is a good point about differentiability, I should have mentioned it at the beginning.

Thank you Anthony.

Which slide?