There is Nordstrom theory, which can be given as $$ C_{\mu \nu \alpha \beta} = 0. $$ The solution of Einstein equations for this case is conformally flat metric: $$ g^{\mu \nu} = e^{\epsilon \varphi (x)}\eta^{\mu \nu}. $$ How to show that this theory predicts no deflection of light?


1 Answer 1


In the solar system, there is only a weak gravitational field outside the sun. So for practical purpose, you can expand the metric to the first order in $\epsilon$(and I guess that why have this parameter in the definition),

$$g_{\mu\nu} = e^{-\Phi }\eta_{\mu\nu} = e^{-\epsilon\phi} \eta_{\mu\nu} \approx -(1-\epsilon\phi) dt^2 + (1-\epsilon\phi ) (dx^2 +dy^2 + dz^2 ) $$ Notice that spatial part of the perturbation metric has the opposite side to the linearized Einstein theory. That's exactly why you don't have light bending. It is a good exercise to do, since the derivation is completely parallel to the usual linearized theory, see for example, MTW, exercise 18.6.

However, the result is true even beyond linear order. You know the equation of motion of photon is geodesic equation, $$ \nabla_p p = 0 $$

where momentum $p$ is the tangent vector of the geodesic curve and for photon it is a null vector. In components, $$ \frac{d}{d\tau} p^{\mu} - p^{\mu} p^\nu \partial_{\nu} \Phi = 0 $$ where I have used the explicit form of connection and $p^{\mu}p_{\mu} =0$(please verify it).

This equation can be integrated out, $$ \frac{d}{d\tau} p^{\mu} - p^\mu \frac{d}{d\tau} \Phi = e^{\Phi} \frac{d}{d\tau}(e^{-\Phi} p^\mu) = 0 $$ so $e^{-\Phi}p^\mu$ is a constant alone the geodesic.

Following the standard procedure, we compare 4-momentum at emission and receiving; in both cases the photon are very far away from the star in the center, and thus $\Phi \sim 0 $. Therefore we can conclude that the 4 momentum doesn't change in the asymptotic flat space, i.e. no light bending.

You can read MTW exercise 7.1. That's a problem starting from an action of a scalar(Nordstorm theory) gravitational field, and there are also some useful hints and comments in the text.

Added: The conserved quantities are due to the four conformal Killing vectors: $\partial_{\mu} $. $$\mathcal{L}_{\partial_\mu} g = -\partial_{\mu}\Phi g $$ Let $\xi= \partial_\mu$, then $g(p,\xi)= p^\nu \xi_\nu$ is a conserved quantity. That's because $$\nabla_p g(p,\xi) =(\nabla_p g)( p, \xi ) +g( \nabla_p p, \xi) + g(p, \nabla_p \xi ) = g(p, \nabla_p \xi )\\= g(p, [p,\xi] ) + g(p, \nabla_{\xi} p ) = - g( p, \mathcal{L}_{\xi} p ) + \frac{1}{2} \nabla_{\xi}g(p,p ) \\= -\frac{1}{2} \mathcal{L}_{\xi} g(p,p) + \frac{1}{2} (\mathcal{L}_{\xi} g)( p, p ) = -\partial_\mu \Phi g( p, p ) = 0 $$

A quick check shows that $p^\mu \xi_\mu = p^\mu e^{-\Phi } $ are just what we have derived.

  • $\begingroup$ How did you get your form of geodesic equation? I only got $$ \frac{d p^{\mu}}{d \tau} - p^{\mu}p^{\alpha}\partial_{\alpha} \Phi = 0. $$ $\endgroup$ Dec 20, 2013 at 23:57
  • $\begingroup$ The connection coefficients are proportional to \Phi, as you can use Christoffel symbol to compute it. $\endgroup$
    – anecdote
    Dec 21, 2013 at 0:17
  • $\begingroup$ Sorry that's a mistake. Your equation is correct. $\endgroup$
    – anecdote
    Dec 21, 2013 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.