A Taylor expansion metric tensor

mertcan · Jul 1, 2016

hi, when I dug up something about metric tensors, I found a equation in my attached file. Could you provide me with how the derivation of this ensured???? What is the logic of that expansion in terms of metric tensor???? I really need your valuable responses. I really wonder it. Thanks in advance.....

mertcan · Jul 1, 2016

Actually I want to ask why do we have $$x^k x^l$$ instead of $$\partial x^k$$ $$\partial x^k$$ ??? In taylor series, I know we always write the infinite smalls..

for instance in this link (http://mathworld.wolfram.com/TaylorSeries.html), they have (x-a) and x approaches to a. In short (x-a) becomes $$\partial x$$...

mertcan · Jul 1, 2016

I hope my question is clear and I really would like to ask Is there anyone who is capable of responding to my question???

mertcan · Jul 2, 2016

Guys, I do not know why you do not give answer. İs there a situation that bothers you in my question ????? Uncertainity really makes me bad.....

strangerep · Jul 2, 2016

mertcan said: ↑

Guys, I do not know why you do not give answer. İs there a situation that bothers you in my question ????? Uncertainity really makes me bad.....

You are being too impatient. Sometimes it takes a while for someone to answer such a basic question.

Have a look at this Wikipedia page about Taylor series for several variables. The metric taylor series in your image just shows expansion around a point "p". I.e., x is like a displacement from the point p.

mertcan · Jul 3, 2016

strangerep said: ↑

You are being too impatient. Sometimes it takes a while for someone to answer such a basic question.

Have a look at this Wikipedia page about Taylor series for several variables. The metric taylor series in your image just shows expansion around a point "p". I.e., x is like a displacement from the point p.

thank you strangerep, I consider that your answer is close to my first thought. You mean If we look at or make taylor expansion around so close points, "x" in the image becomes infinite small distance as I thought before.

Lucas SV · Aug 2, 2016

Given you accept the result in calculus of several variables (as given in the wikipedia page linked by strangerep), here is the explanation, for a smooth manifold of dimension ##d## with metric ##g##. Take a point ##p## in your manifold and take a chart ##(\mathcal{U},x)##, such that ##x(p)=0##, i.e. we set the coordinates of point ##p## in this chart, to be the zero vector in ##\mathbb{R}^d##.

Then the components of the metric in that chart, denoted by ##g_{ij}## are just mappings from ##x(\mathcal{U})## to ##\mathbb{R}##, i.e. they are functions of several variables. Note that ##x(\mathcal{U})## is a subset of ##\mathbb{R}^d##, which contains zero, so you can apply the formula from wikipedia, by setting ##a=0##. Just use contravariant indices for the coordinates, Einstein summation convention and note that to evaluate at ##p## means essentially to evaluate at ##0## in this chart.

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A Taylor expansion metric tensor

mertcan

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