Gravitational lenses magnifies the appearance of themselves, so in the case of a galaxy it looks bigger than it is. Thus the outer stars in a galaxy seems to have a higher tangential velocity.

For example, the Sun deflects light passing from minus infinity to us 1.75 seconds of arc. Say that half of that value is the amount it deflects its own light at its edge, then the Sun looks 200 km wider.

Does this effect noticeably contribute to the galaxy rotation curve (dark energy/matter) problem?

  • $\begingroup$ How do you arrive at 200 km? At 1 a.u. 0.875" extends about 630 km. (Without claim that this is the correct answer). $\endgroup$
    – my2cts
    Feb 20, 2021 at 11:35

1 Answer 1


For a spherically symmetric potential, the apparent radius of an object seen from far away is $$R_{\infty} = R \left(1 - \frac{2GM}{Rc^2}\right)^{-1/2}\, $$ where $M$ is the total mass-energy and $R$ is the radius in Schwarzschild coordinates.

For a big spiral galaxy (like the Milky Way), and ignoring any contribution from dark matter, then approximate numbers would be $M \sim 5\times 10^{10} M_{\odot}$ and $R \sim 20$ kpc.

Using the equation above, $R_{\infty} = 20.0000002$ kpc.

If one includes dark matter, then you might lose one of the zeroes.

So no, the effect you ask about has no significant influence on the measurement of galactic sizes or velocities.

  • 1
    $\begingroup$ Do you have a reference for this result? $\endgroup$
    – my2cts
    Feb 20, 2021 at 11:49
  • $\begingroup$ @my2cts what, the $R_{\infty}$ equation? It's used all the time in neutron star research (e.g. p.7 in aip.scitation.org/doi/pdf/10.1063/1.5117791). A simple "derivation" is to consider a spherical blackbody radiator in the Schwarzschild metric. The number of photons per second received by a distant observer is reduced by the GR time dilation factor of $(1 - 2GM/rc^2)^{1/2}$, but the photon frequency is reduced by the same factor, so the luminosity decreases by a factor $(1 - 2GM/rc^2)$. $\endgroup$
    – ProfRob
    Feb 20, 2021 at 12:02
  • $\begingroup$ But the $T_{\rm eff}$ of the blackbody is also redshifted and reduced by the factor $(1-2GM/rc^2)^{1/2}$. i.e. $L_{\infty}/L = (1-2GM/rc^2)$, $T_{\rm eff,\infty}/T_{\rm eff} = (1-2GM/rc^2)^{1/2}$, then using the Stefan-Boltzmann relation $R_{\infty}/R =(1-2GM/rc^2)^{-1/2}$ $\endgroup$
    – ProfRob
    Feb 20, 2021 at 12:11

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