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Why can't gravity affect light directly without any reference to space-time?

How does light bending show that space-time is curved?

I would have thought that it shows that massless particles are subject to classical gravity just like massive particles. The curvature or lack of curvature of space-time seems irrelevant and unnecessary to an understanding of gravity.

What have I missed?

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  • $\begingroup$ There is nice explanation in Gravitation by Misner, Wheeler, Thorne, chapter 7.3 "Gravitational redshift implies spacetime is curved" $\endgroup$
    – Umaxo
    Sep 7, 2021 at 9:54
  • $\begingroup$ If you do not have the book, it is partially described here physics.stackexchange.com/questions/413910/… $\endgroup$
    – Umaxo
    Sep 7, 2021 at 10:04
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    $\begingroup$ The general theory of relativity has been consistent with all experimental observations. This indirectly implies that the idea of light bending should be correctly described by curvature of space-time. It is due to the agreement with observations that we have accepted this picture $\endgroup$
    – KP99
    Sep 7, 2021 at 12:30
  • $\begingroup$ Photon trajectories can be altered by gravity directly proportional to their proximity to the mass they are passing. $\endgroup$ Sep 7, 2021 at 15:48

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Why can't gravity affect light directly without any reference to space-time? How does light bending show that space-time is curved?

This is a good question, and like many good questions the answer is not straightforward.

First, strictly speaking, the observation of light bending does not support spacetime curvature per se, what it supports is specifically the theory of general relativity (GR) which is most commonly expressed in terms of curved spacetime. It turns out that Newtonian gravity can also be expressed in terms of curved spacetime. This is called Newton Cartan gravity, which uses curved spacetime but is experimentally identical to Newtonian gravity.

Secondly, it is not exactly clear how to treat light under Newtonian gravity. With the Newton Cartan formulation you would get a bending of light rays, but the value of the bend would be 1/2 the value of the bend predicted by GR. With the regular formulation of Newtonian gravity, if you treated light as a massless particle then since the acceleration is independent of the mass you would get a gravitational deflection. Again, the amount of deflection is half the GR prediction. Finally, if you use the standard Newtonian formulation and Maxwell’s equations in a Newtonian inertial frame then you would get no gravitational deflection.

Now, the observed bending matches the GR prediction and excludes either the 0 bending or 1/2 bending predicted by Newtonian physics. So the observation supports GR over either the standard formulation of Newtonian gravity without spacetime curvature or even the Newton Cartan formulation that does use curved spacetime.

Overall, I would say that the observed bending supports GR, not specifically spacetime curvature. With the understanding that spacetime curvature is a big part of GR.

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General relativity has not been created to explain the bending of light, which was observed years afterwards only.

And indeed, the idea of gravity having an effect on light was not absent from Newtonian mechanics (see for example https://en.wikipedia.org/wiki/Gravitational_lens#History).

So it definitely cannot be said that light bending entails the curvature of spacetime. It is the other way around: general relativity describes the whole of gravity via the curvature of spacetime and, in that framework, light bending becomes a (confirmed) prediction of the theory.

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In support of Dale's excellent answer I will show in the following how the deflection of light can be computed in Newton's theory.

For this purpose we consider a light ray usually moving along the x-axis passing along the sun which is located below the ray (and therefore below the x-axis). The light ray gets deviated in the direction of the y-axis according to the well-known law (may be here Special Relativity is needed to explain the attraction that is exerted on the light ray via $m =\frac{E}{c^2}$):

$$y = \frac{g}{2} t^2$$

where $g$ corresponds to the gravitational acceleration which can be quantified by $g = \frac{GM}{R^2}$ with $M$ mass of the sun, $R$ as radius of the sun (we assume that the light ray passes really very close along the sun's surface). We can relate the time it takes with the position of the ray along the x-axis: $x=ct$. Thus we get an relationship between the y-coordinate of the ray with its x-coordinate:

$$ y(x) = \frac{g}{2c^2}x^2 = \frac{GM}{2R^2 c^2}x^2$$

Of course the value of the y-coordinate is very small, grosso modo the approximation that the ray moves along the x-axis is still very good what justifies $x=ct$. In order to obtain the angle we take the derivative of the y-coordinate of the ray:

$$y'(x) = \frac{GM}{R^2c^2}x$$

So we get the deviation angle as a function of the x-coordinate of the ray. In order to get the ray's full deviation the light needs to travel a distance along $x$ that corresponds to the diameter of the sun (the distance the light ray is under the impact of the gravitational force of the sun):

$$\delta = y'(2R) = \frac{2GM}{Rc^2} = \frac{1}{2} \delta_{GR}$$

Yes, Newton's theory indeed predicts the deviation of light, but only half the value of the deviation of General Relativity (GR). So in order to explain the measured light deviation of $\delta_{GR}=1.75''$ General Relativity --- i.e. curvature of space-time --- is needed. But it is actually only a test of the linear part of GR. By the way, even Einstein originally predicted this value 1911 (wrongly) 4 years before the full accomplishment of his theory of General Relativity.

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  • $\begingroup$ Very nice walk through the math. +1 $\endgroup$
    – Dale
    Sep 7, 2021 at 22:03
  • $\begingroup$ @Dale actually the explanation should come along with a picture, without picture I tried to be more explicit ... in case of interest I could still upload a drawing. $\endgroup$ Sep 7, 2021 at 22:05
  • $\begingroup$ Well, if you have a picture handy that would be great, but if it is a lot of effort you probably needn’t bother. Answers with pictures tend to get more upvotes $\endgroup$
    – Dale
    Sep 7, 2021 at 22:08
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    $\begingroup$ @Dale may be tomorrow. Anyway, it 'd be very schematic only. $\endgroup$ Sep 7, 2021 at 22:36
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Understanding spacetime curvature is very relevant to understanding gravity. Gravity is not a true force. One can tell this by the fact that under free fall the proper-acceleration of the body is zero. Whereas an observer standing on earth would experience a net force or will have a proper acceleration.

Light travels in a straight line when the observer is in inertial frame. But when we look from a non-inertial frame, that straight path gets curved (called a lightlike geodesic). This concept tells us there has to be some curvature that is responsible for gravity.

Why at all the spacetime is curved, cause we try to relate it with the curved path that light follow. This suggest that spacetime is curved rather than being flat, and it uniquely describes all the events (in space and time).

Gravity cannot influence light if we consider Newtonian gravity (as light has no mass). But to explain the phenomenon of light bending. Einstein devised the concept of spacetime curvature.

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    $\begingroup$ I am sorry, but I think this answer repeats confusing things about GR you hear everywhere else. Gravity is of course a force, and a true one. It is 'just' described geometrically (by the way: All the other forces are described geometrically, too, if you ask a mathematician or a mathematical physicist). $\endgroup$
    – Koschi
    Sep 7, 2021 at 6:34
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    $\begingroup$ Gravity is different from other forces. $\endgroup$ Sep 7, 2021 at 6:35
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    $\begingroup$ That is true... but it still is a force. $\endgroup$
    – Koschi
    Sep 7, 2021 at 6:36
  • $\begingroup$ When other forces like electromagnetic, mechanical act on a body then the body definitely has proper acceleration. But in case of gravity under free fall, body's proper acceleration is zero. $\endgroup$ Sep 7, 2021 at 6:37
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    $\begingroup$ @Koschi The answer states that gravity is not a "true force" in the traditional sense of force. The answer is stating that it is force that emerges from spacetime curvature. The answer does not say "gravity is not a force". $\endgroup$
    – joseph h
    Sep 7, 2021 at 6:39

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