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ozone
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Given the metric ds^2 = \frac{dx^2 + dy^2}{y^2} find a set of coordinates which yield local flat space. i.e. (g_{\mu\nu} = \delta_{\mu\nu} + second order terms).
My text outlined a process to go through to find the flat space coordinates, but the actual execution is slightly beyond me. I would really like to know how to complete this problem for my own understanding.. Let's dive into the math then:
Our given metric is g_{yy} = \frac{1}{y^2} , g_{xx}= \frac{1}{x^2}. Then without any loss of generality we may expand this metric about the point (0,y_{*}) the resulting expansion will leave us with new metric terms given as
g_{yy} = \frac{1}{y_{*}^2} - \frac{2}{y_*^3} (y-y_*) + \frac{6}{y^4_*} (y-y_*)^2, g_{xx} = \frac{1}{y_{*}^2}
we might write this in the form g_{\mu\nu}(x) = g_{\mu\nu}(0) + A_{\mu\nu,\lambda} x^\lambda +B_{\mu\nu,\lambda \sigma} x^\lambda x^\sigma
However we really don't need to worry about our second order coefficient, as our only aim right now is to take our constant term and turn it into the identity matrix, and to find a coordinate system which will remove our first order matrix coefficient A.
Then doing a coordinate transformation to x^{\mu} = k^\mu_\nu x^{'\nu} + L^\mu_{\nu\lambda} x^{\nu} x^{\lambda} + ...
We are now ready to solve for k^\mu_\nu and for reasons which aren't 100% clear to me (I believe it has to do with the definition of the metric tensor being the first order derivatives of x and the way it transforms) my book states that we wish to solve the equation g'_{\rho\sigma}(0) = k^\mu_\lambda k^\nu_\sigma g_{\mu\nu}(0). This seems logical to me and I can generally follow up to this point, I can solve that with ease(we want our primed 0th order term to be an identity matrix), but I did not know how to deduce the relationship we want to solve for L^\mu_{\nu\lambda}, furthermore my book made the cryptic statement that solving for k will change the resulting value of our coefficient L. Any help on solving for the with which we can L eliminate our coefficient A would be much appreciated
Homework Statement
Given the metric ds^2 = \frac{dx^2 + dy^2}{y^2} find a set of coordinates which yield local flat space. i.e. (g_{\mu\nu} = \delta_{\mu\nu} + second order terms).
My text outlined a process to go through to find the flat space coordinates, but the actual execution is slightly beyond me. I would really like to know how to complete this problem for my own understanding.. Let's dive into the math then:
Our given metric is g_{yy} = \frac{1}{y^2} , g_{xx}= \frac{1}{x^2}. Then without any loss of generality we may expand this metric about the point (0,y_{*}) the resulting expansion will leave us with new metric terms given as
g_{yy} = \frac{1}{y_{*}^2} - \frac{2}{y_*^3} (y-y_*) + \frac{6}{y^4_*} (y-y_*)^2, g_{xx} = \frac{1}{y_{*}^2}
we might write this in the form g_{\mu\nu}(x) = g_{\mu\nu}(0) + A_{\mu\nu,\lambda} x^\lambda +B_{\mu\nu,\lambda \sigma} x^\lambda x^\sigma
However we really don't need to worry about our second order coefficient, as our only aim right now is to take our constant term and turn it into the identity matrix, and to find a coordinate system which will remove our first order matrix coefficient A.
Then doing a coordinate transformation to x^{\mu} = k^\mu_\nu x^{'\nu} + L^\mu_{\nu\lambda} x^{\nu} x^{\lambda} + ...
We are now ready to solve for k^\mu_\nu and for reasons which aren't 100% clear to me (I believe it has to do with the definition of the metric tensor being the first order derivatives of x and the way it transforms) my book states that we wish to solve the equation g'_{\rho\sigma}(0) = k^\mu_\lambda k^\nu_\sigma g_{\mu\nu}(0). This seems logical to me and I can generally follow up to this point, I can solve that with ease(we want our primed 0th order term to be an identity matrix), but I did not know how to deduce the relationship we want to solve for L^\mu_{\nu\lambda}, furthermore my book made the cryptic statement that solving for k will change the resulting value of our coefficient L. Any help on solving for the with which we can L eliminate our coefficient A would be much appreciated
