# Metric tensor definition

1. Jun 5, 2009

### muzialis

Hi All,

let me preface I am an engineer, in classical terms, a trivial techician.

Still I am putting some effort in improving my tensorial calculus.

I am struggling with the definition of a metric tensor, as found for example in

[PLAIN]http://en.wikipedia.org/wiki/Metric_tensor[/URL] [Broken]

in particular I am not grasping how the length of the curve is expressed, as I obtain a different expression. I have posted my line of thought in the attached file, as I do not know how to use mathematical notation within the post.

Your help would be the most appreciated, as I am relaly looking forward to understanding where the mistake is.

Have a nice weekend

Muzialis

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2. Jun 5, 2009

### xepma

The final question in your document: yes, that is the definition of r_u.

Your expression for the length of the arc is missing a square root. Furthermore, you should throw all terms under one integral. I haven't checked to see if what you have matches the wiki though...

P.S. the math used here on the forum is build using Latex. You might want to check into it - it's quite funky ;)

3. Jun 6, 2009

### Fredrik

Staff Emeritus
I don't know why you get three identical terms in your integral, but the fact that the derivatives of y don't appear anywhere in your expression should give you a hint about what your mistake was.

This might confuse you even more, but I would like to show you a notation that I think makes this sort of calculation much easier. I would define z1=u and z2=v, let ",j" mean differentiation with respect to variable number j, and use the convention that if an index appears exactly twice, we're supposed to do a sum over all of its values (and since we know that, we don't need to write a summation symbol). Then the calculation looks like this

$$\|\frac{d}{dt}\vec r\|^2=\frac{dr_i}{dt}\frac{dr_i}{dt}=r_{i,j}z'_j r_{i,k}z'_k$$

That's actually the whole thing (except that we still haven't taken the square root of both sides). If we want the result to look like what Wikipedia got, we rewrite the right-hand side as

$$\vec r_{,j}\cdot\vec r_{,k} z'_j z'_k$$

and then do the sums explicitly to get

$$\frac{\partial\vec r}{\partial u}\cdot\frac{\partial\vec r}{\partial u}u'^2+\frac{\partial\vec r}{\partial u}\cdot\frac{\partial\vec r}{\partial v}u'v'+\frac{\partial\vec r}{\partial v}\cdot\frac{\partial\vec r}{\partial u}v'u'+\frac{\partial\vec r}{\partial v}\cdot\frac{\partial\vec r}{\partial v}v'^2=\vec r_u^2 u'+2\vec r_u\cdot\vec r_v+\vec r_v^2$$

By the way, if you use the "quote" button, you see how I did the math symbols.

I should also add that I don't think it's necessary to go through these things to understand the metric tensor, but it is necessary to learn some differential geometry (at least if you want to understand it in the context of GR). "Modern differential geometry for physicists" by Chris Isham looks like a good place to start. (I recently bought that book myself because I still don't know the stuff in the last few chapters, but I haven't read it yet).

Last edited: Jun 6, 2009
4. Jun 8, 2009

### muzialis

Xepma, Fredrik,

many thanks for your time, everything is clear now

Best Regards

Muzialis

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