Physics Asked on March 15, 2021
I am reading the book ‘Gravity’ by Hartle and presently I am at the section discussing orthonormal and coordinate bases. I am confused about a few points I had read previously and can’t exactly correlate with the above mentioned section.
Here the author says that at every point in spacetime I can define a basis that is orthonormal in the tangent space of that point. This would not necessarily be the coordinate basis which I would get from the line element. This would mean that the metric in the orthonormal basis becomes the flat spacetime metric at the point (from the definition of the components of the metric in terms of the dot product of basis vectors and the requirement of one timelike and three spacelike components). Now, I know that the way to locally transform the metric to the flat spacetime metric is to do a coordinate transformation on the line element that does the above, i.e to go to a local inertial frame for that point.
My question is – Is there any correlation between a coordinate transformation that transforms my line element to that of flat spacetime in the vicinity of a point to that of choosing an orthonormal coordinate system at that point? Is the latter something similar to choosing a coordinate system where my coordinate basis would align with the orthonormal basis at the point? So, I guess my question boils down to the two definitions of the metric – one from the dot product of basis vectors and the other from the line element. Are they completely equivalent?
My last question is – Is the coordinate transformation that takes me to a local inertial frame of a point unique? Can there be more than one coordinate transformations that take me to the local inertial frame of a point? Does this mean that at every point I can choose an orthonormal basis at every point in one and only one unique way?
I am sorry for not writing equations. I am very poor with Tex. I hope I have been able to put my question clearly.
First thing of note is that every frame (basis) at a single point is a coordinate frame. To see this, let $mathrm dx^mu$ be a coordinate cobasis, and let $x$ be a point of the manifold $M$. An arbitrary frame at $x$ can be expressed as $$ theta^a|_x=L^a_{ mu}mathrm dx^mu|_x, $$ where $L^a_{ mu}$ is a constant invertible matrix. But then if we define the new coordinates $u^a=L^a_{ mu}x^mu$, then $$ mathrm du^a=L^a_{ mu}mathrm dx^mu=theta^a, $$ where I was lazy in denoting it, but every expression is evaluated at $xin M$.
The problem starts if one is considering frame fields, which are defined on an open domain. If $theta^a$ is a coframe field on an open domain, then by Poincaré's lemma the necessary and sufficient condition for it to be a coordinate frame is that $mathrm dtheta^a=mathrm dtheta^a_nuwedgemathrm dx^nu=partial_mutheta^a_numathrm dx^muwedgemathrm dx^nu=frac{1}{2}left( partial_mutheta^a_nu-partial_nutheta^a_muright)mathrm dx^muwedgemathrm dx^nu=0$, which is a nontrivial integrability condition.
Even if this integrability condition is not satisfied, one can still write $$ theta^a|_x=theta^a_mu(x)mathrm dx^mu|_x $$ for some invertible matrix function $theta^a_mu(x)$, however in general there will be no such coordinate functions $u^a(x)$ such that $theta^a_mu(x)=partial_mu u^a(x)$.
Nontheless, for any $x_0in M$ you may try to see $theta^a|_{x_0}$ as a fixed cobasis at a point, and try the coordinate transformation $u^a_{(x_0)}(x)=theta^a_mu(x_0)x^mu$, which will give at $x_0$, and only at $x_0$ $$ theta^a|_{x_0}=mathrm du^a_{(x_0)}(x_0). $$
Which means one can take an orthonormal frame to arise from coordinate transformations, but it will not come from a single coordinate transformation that fits together nicely everywhere, but rather one separate coordinate transformation at each point, which do not fit together nicely.
Answered by Bence Racskó on March 15, 2021
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