2

We know that through GR, in a gravitational field light bends twice as much as predicted by strict Newtonian physics. In researching this I've found explanations like light bends on both the space and time components of spacetime on page 116. And in the PBS Spacetime at 10:50, he says that it is caused by the wave front of light refracting in the gravitational field.

In either case, if light is bending twice as far as predicted, is it falling twice as fast as say a bullet shot horizontally across the planet? So on the Earth, would light fall at 18.4 m/s^2?

Edit: This question on the effect of gravity on light and bagels has just popped up on my sidebar.

foolishmuse
  • 4,551
  • 1
    No, because of the equivalence principle: gravitational fields locally act like uniform accelerations. The doubled bending of light is because of the spatial curvature, which is a global feature of Schwarzschild spacetime. – knzhou Apr 06 '23 at 20:59
  • For more on how light bends in a strong gravitational field, see the Veritasium videos How to Understand the Black Hole Image and First Image of a Black Hole! – mmesser314 Apr 06 '23 at 21:04
  • @knzhou your comment reinforces my question. If there is BOTH time dilation falling AND space curvature falling, don't they add up to something faster than normal? – foolishmuse Apr 06 '23 at 21:32
  • @mmesser314 both good videos, but they don't discuss my particular question. – foolishmuse Apr 06 '23 at 21:33
  • 3
    No, because in a lab the spatial curvature is totally negligible. Also, it's not like massive matter is exempt from feeling spatial curvature. If you shot an extremely light, ultrarelativistic massive particle around a star, it would bend twice as much as the Newtonian prediction, just like light does. People emphasize light because that's what we use in the experimental test but the rules are the same. – knzhou Apr 06 '23 at 21:58
  • @knzhou so it's the velocity that matters more than the mass of the particle? Can you point me to any references on this? – foolishmuse Apr 06 '23 at 22:04
  • BTW, the equation that predicts double the light deflection in GR compared to Newtonian gravity is only an approximation. It's quite accurate when the impact parameter of the ray is a large multiple of the Schwarzschild radius, eg for a ray passing near a planet or star. But it doesn't work for rays passing near a black hole. Such rays can make multiple loops around the photon sphere of the BH, which is impossible in Newtonian gravity. I have a diagram & some graphs here: https://astronomy.stackexchange.com/a/49874/16685 – PM 2Ring Apr 06 '23 at 23:49
  • 2
    @knzhou Please convert your comments to an answer; please don't use comments for brief answers. – rob Apr 07 '23 at 00:20

1 Answers1

2

foolishmuse asked: "In either case, if light is bending twice as far as predicted, is it falling twice as fast as say a bullet shot horizontally across the planet?"

No, at the vertex (at $\rm y=r$ with the initial motion along the $\rm x$ direction) it would fall (vertically accelerate) by a factor of $\rm 2 \ v^2/c^2+1=3 \times$ faster:

acceleration of a photon

The $\rm d^2 y/dT^2$ at the direct flyby is $3 \times$ higher, but if you integrate that over a large $\rm \Delta x$ the resulting $\rm dy/dT$ (the area instead of the curve) is only $2 \times $ higher than under Newton since the radial difference is smaller than the transversal and farther away the radial component dominates.

The factor $2$ is also not exact, that is an approximation for almost straight paths with small deflections where $\rm \tan \phi \approx \phi$. In the strong field, for example if the photon grazes the photon sphere at $\rm r=1.5 \ r_s$, the deflection angle is even stronger than $2 \times$ Newton.

The $\rm x$ axis on the plot is actually the local shell time $\rm T$, but at a large distance like the initial $\rm y=1000 \ GM/c^2$ that is almost equal to the actual $\rm x$ direction if we don't count half pixels.

A very good explanation about how that fits with the equivalence principle is here:

Einstein Online wrote: "There is a fundamental restriction implicit in our use of the equivalence principle: Only in a small region of space, and over a brief time period, can we use the laws of special relativity and end up with a good approximation. Only in a small falling elevator can we assume that light propagates at constant speed, along a straight line. On the other hand, if we actually want to measure the deflection of light, we will have to look at the big picture. We must consider not just local, but global deflection"

If the earth was not round but infinite and flat the equivalence principle would even hold over long distances and times, and the vertical acceleration would be the same regardless of the initial velocity (in terms of the local time $\rm d^2 y/ d T^2$, while in terms of the proper time $\rm d^2 y/ d \tau^2$ would be higher by $\gamma^2$, as would be the $\rm g$ force if the particle travels on the ground in a straight line).

Yukterez
  • 11,553
  • What I'm getting from this is that the light follows the circle of the Earth, much like astronauts in the international space station are constantly falling. So this is not a gravity effect, so much as a refraction of light as it passes by the circle of the planet. – foolishmuse Apr 07 '23 at 17:04