Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Let $x^\mu$ be the coordinates of a reference frame, $K$, where all bodies feel the same constant and uniform acceleration $\textbf{a}=\textbf{g}=-\nabla\varphi$; let $\xi^\mu$ be the coordinates of a Locally Inertial Frame, $LIF$. Using the Principle of Equivalence, show that the linear part in $\textbf{g}$ of the interval for $K$ is $$ ds^2=(1+2\varphi/c^2)c^2dt^2-d\textbf{x}^2+o(1/c^2). $$

My attempt:

P.E states the metric tensor, $g_{\mu\nu}$, respect to a generic system of coordinates its related to metric tensor of a flat Minkowskian $\eta_{\alpha\beta}=\text{diag(1,-1,-1,-1)}$ via the Jacobian of the diffeomorphism $x\mapsto\xi$. Obviously such a diffeo can be of the form $$\xi^0=ct,\xi^i=x^i-1/2g^it^2,$$ and the P.E states that $$ g_{\mu\nu}=\frac{\partial\xi^\alpha}{\partial x^\mu}\frac{\partial\xi^\beta}{\partial x^\nu}\eta_{\alpha\beta}, $$ so the 00 component is $$ g_{00}=\frac{\partial\xi^\alpha}{\partial x^0}\frac{\partial\xi^\beta}{\partial x^0}\eta_{\alpha\beta}=\frac{\partial\xi^0}{\partial x^0}\frac{\partial\xi^0}{\partial x^0}-\frac{\partial\xi^i}{\partial x^0}\frac{\partial\xi^j}{\partial x^0}\delta_{ij}=1-\Big(\frac{v^i-g^it}{c}\Big)^2\simeq1-2\varphi/c^2++o(1/c^2,\textbf{g}^2), $$ here i take $2v^itg^i/c^2=2x^ig^i/c^2=-2\varphi/c^2.$ Other problem arises becouse my $g_{0j}\neq0$. What i wrong?

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.