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
add comment

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.