I've been studying gravity with a dynamical preferred frame, namely one in which the preferred frame is set by the existence of a unit time-like vector field called the "aether". I've been using this link:


When trying to look for spherically symmetric and static solutions of this theory, the aether can be written as \begin{equation} u = a(r) \partial_t + b(r) \partial_r \end{equation}

(equation (11) of the link). Does anyone know why the vector can be written this way? All the papers I find on similar subjects don't explain why as well...

Thank you!

  • $\begingroup$ What about it confuses you? The field is independent of the angular spherical coordinates and of time, and so it is static and spherically symmetric. Is this what you're asking? $\endgroup$ Jan 26, 2018 at 18:01
  • $\begingroup$ It's confusing me that this vector is written in terms of the derivatives with respect to t and r, which are operators. When we are talking about vector fields we are talking about assigning a vector to a point in space (in this case, space-time). But this expression for the vector depends on d/dt and d/dr, which makes me think it needs to act on something in order to retrieve a vector. This is what is confusing me, I don't know if this was clearer? $\endgroup$
    – Sth99
    Jan 26, 2018 at 18:34
  • $\begingroup$ This is a standard notation for vectors in differential geometry. You may want to glance at Frankel's "The Geometry of Physics" or another book that deals with geometry itself to help you understand how vectors are related to differential operators. $\endgroup$ Jan 26, 2018 at 20:35
  • $\begingroup$ Just to be a bit more specific, in this notation $\partial_t$ is just a way of writing the basis vector with only a $t$ component, i.e. $(\partial_t)^\mu = \delta^\mu_t$. The reason it uses the same symbol as a partial derivative is that vector fields act naturally on functions as directional derivatives by contracting with the gradient. So the action of the vector $\partial_t$ on a function $f$ is $\delta^\mu_t \partial_\mu f = \partial_t f$. However, for the above reference, it suffices to know that $\partial_t$ is just a basis vector pointing in the $t$-direction $\endgroup$
    – asperanz
    Jan 27, 2018 at 5:25

1 Answer 1


If you impose spherical symmetry + staticity to the spacetime you want ($g_{ab}$) you should impose it as well to the energy momentum tensor ($T_{ab}$) in order to find consistent or at least easy solutions to Einstein’s equations. The energy-momentum tensor of what they call aether (which is like a gravitating fluid with some equation of state) is given in terms of the velocity vector field $u_a$, so it is impossed by hand that this vector has spherical symmetry and staticity.

This means that if you take the commutator of $u_a$ with the $4$ Killing vector fields that generate time translations and rotations respectivelly, the result should be zero. You can solve these 4 equations, but the result that you will get is that the components of $u_a$ do not depend on $t$ nor on any spatial angle, and that (in spherical coordinates, which are adequate to study spherically symmetric problems) you do only have a time component and a radial component, so that $u_a$ is what you wrote.

  • $\begingroup$ I understand why a(r) and b(r) can only be functions of r. What I don't understand is why the vector field is written in terms of d/dt and d/dr $\endgroup$
    – Sth99
    Jan 26, 2018 at 18:24

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