# How can gravity affect light?

I understand that a black hole bends the fabric of space time to a point that no object can escape.

I understand that light travels in a straight line along spacetime unless distorted by gravity. If spacetime is being curved by gravity then light should follow that bend in spacetime.

In Newton's Law of Universal Gravitation, the mass of both objects must be entered, but photon has no mass, why should a massless photon be affected by gravity in by Newton's equations? What am I missing?

Newton's law does predict the bending of light. However it predicts a value that is a factor of two smaller than actually observed.

The Newtonian equation for gravity produces a force:

$$F = \frac{GMm}{r^2}$$

so the acceleration of the smaller mass, $m$, is:

$$a = \frac{F}{m} = \frac{GM}{r^2}\frac{m}{m}$$

If the particle is massless then $m/m = 0/0$ and this is undefined, however if we take the limit of $m \rightarrow 0$ it's clear that the acceleration for a massless object is just the usual $a = GM/r^2$. That implies a photon will be deflected by Newtonian gravity, and you can use this result to calculate the deflection due to a massive object with the result:

$$\theta_{Newton} = \frac{2GM}{c^2r}$$

The calculation is described in detail in this paper. The relativistic calculation gives:

$$\theta_{GR} = \frac{4GM}{c^2r}$$

The point of Eddington's 1919 expedition was not to show that light was bent when no bending was expected, but rather to show that the bending was twice as great as expected.

• A downvote? Why was that? I didn't think I'd said anything controversial. If you'd like to say why you downvoted I'll have a look at editing my answer accordingly. Commented Aug 12, 2014 at 14:06
• Perfect! Thank you! I was reasonably certain that was the answer. I appreciate your help. Commented Aug 12, 2014 at 16:56
• Well... If a particle has zero mass then F is zero and you can't really use newtonian mechanics to begin with. (Im not the one who downvoted, just think that might be the complain) Commented Jul 17, 2017 at 18:32
• What John's answer left out is that when you take limit as m goes to zero, you have to apply L'Hopital's rule, which should come out to 1. Thus, F is not zero. Commented Jan 13, 2019 at 2:47
• No, the force would still be zero (without relativity) if m is zero. The LHopitals rule only applies to the acceleration. So I’m confused as well..if F is zero for a massless particle then what’s causing the acceleration? This leads me to question if this could really be explained purely classically to within a factor of 2. Commented Feb 25, 2019 at 9:56

One can in principle consider a Schwarzschild spacetime:

$ds^2 = -\left(1- \frac{2M}{r}\right)dt^2 + \frac{dr^2}{1- \frac{2M}{r}} + r^2 \left(d\theta^2 + \sin^2 \theta d \phi^2\right)$

The Lagrangian of geodesics is then given by:

$\mathcal{L} = \frac{1}{2} \left[- \left(1 - \frac{2M}{r}\right)\dot{t}^2 + \frac{\dot{r}^2}{1-\frac{2M}{r}} + r^2 \dot{\theta}^2 + r^2 \sin^2 \theta \dot{\phi}^2\right]$

After applying the Euler-Lagrange equations, and exploiting the fact that the S-metric is spherically symmetric and static, one obtains the orbital equation for light as (After defining $u = 1/r$) as:

$\frac{d^2 u}{d\phi^2} + u = 3 M u^2$.

It 's pretty difficult to solve this ODE. In fact, I don't think a closed-form solution exists. One can apply a perturbation approach. Defining an impact parameter $b$, one can obtain an ansatz solution to this ODE as:

$u = \frac{1}{b} \left[\cos \phi + \frac{M}{b} \left(1 + \sin^2 \phi\right)\right]$.

One can derive the following relationship:

$u\left(\frac{\pi}{2} + \frac{\delta \phi}{2}\right) = 0$,

where $\delta \phi$ is the deflection angle.

Now, finally Taylor expanding $u$ above around $\pi/2$, one can show that, in fact:

$\delta \phi = \frac{4M}{b}$,

which is the required result.

In this answer we derive the formula for the angle of deflection

$$\theta ~=~\frac{2GM}{b}\left(\frac{1}{v_0^2} + \frac{1}{c^2}\right) +{\cal O}(M^2) \tag{1}$$

of a (massive or massless) particle in a Schwarzschild spacetime. Here $$b$$ is the impact parameter and $$v_0$$ is the asymptotic speed (which for a massless particle is $$c$$). The Newtonian limit is $$c\to \infty$$.

Sketched proof:

1. By spherical symmetry, we can take the orbit plane = equatorial plane: $$\theta=\frac{\pi}{2}$$. We start from the geodesic equations \hspace{-5em}\begin{align} E~=~&n^{-1}\frac{dt}{d\lambda} , \cr n^{-1} ~:=~&1 - \frac{r_s}{r}, \cr r_s~:=~&\frac{2GM}{c^2}, \end{align}\tag{5.61/7.43/6.3.12} $$\hspace{-5em} L~=~r^2\frac{d\phi}{d\lambda}, \tag{5.62/7.44/6.3.13}$$ \hspace{-5em}\begin{align} \epsilon c^2~=~&n^{-1}c^2 \left(\frac{dt}{d\lambda}\right)^2 \cr ~-~&n\left(\frac{dr}{d\lambda}\right)^2 -r^2\left(\frac{d\phi}{d\lambda}\right)^2, \end{align}\tag{5.55/7.39/6.3.10} cf. Refs. 1-3. Here $$[\lambda]=\text{Time}, \qquad [E] ~=~\text{dimensionless}, \qquad [L] ~=~\frac{\text{Length}^2}{\text{Time}}. \tag{2}$$

• Massive particle: $$\epsilon=1$$ and $$\lambda=\tau$$ proper time.

• Massless particle: $$\epsilon=0$$ and $$\lambda$$ not proper time.

This leads to \hspace{-5em} \begin{align} \underbrace{ (cE)^2 }_{\text{"energy"}} ~=~& \underbrace{ \left(\frac{dr}{d\lambda}\right)^2 }_{\text{"kinetic energy"}}\cr ~+~& \underbrace{n^{-1}\left(\frac{L^2}{r^2}+\epsilon c^2 \right) }_{\text{"effective potential energy"}}. \end{align}\tag{5.64/7.46/6.3.14} The reader may ponder whether the $$\epsilon$$ parameter leads to a discontinuity in the angle $$\theta$$ of deflection (1) between the massive and massless case? We shall see below that it does not.

2. Q: How do we identify the constants of motion $$E$$ and $$L$$ with the observables $$b$$ and $$v_0$$ at spatial infinity $$r=\infty$$? A: Note that \begin{align}r v_{\phi}~:=~ r^2\frac{d\phi}{dt}\quad\longrightarrow&\quad bv_0~=~h~=~\frac{L}{E}\cr &\quad\text{for}\quad r~\to~\infty, \end{align}\tag{3} and \begin{align}\frac{dr}{dt} \quad\longrightarrow&\quad v_0~=~c\frac{\sqrt{E^2-\epsilon}}{E}\cr &\quad\text{for}\quad r~\to~\infty.\end{align}\tag{4} Eq. (4) means that the energy constant is $$E=\gamma_0=\left(1-\frac{v_0^2}{c^2}\right)^{-1/2}$$ in the massive case, and it is undetermined in the massless case.

3. If we define the reciprocal radial coordinate $$u~:=~\frac{1}{r},\tag{5}$$ we get a 3rd-order polynomial$$^1$$ \begin{align} \left(\frac{du}{d\phi}\right)^2~=~&P(u)\cr ~:=~&\left(\frac{cE}{L}\right)^2 - n^{-1}\left(u^2+\frac{\epsilon c^2}{L^2}\right) \cr ~=~&r_s(u-u_+)(u-u_-)(u-u_0),\end{align}\tag{6} with 3 roots \begin{align}u_{\pm}~=~&\pm\frac{c}{L}\sqrt{E^2-\epsilon} ~+~\frac{r_s}{2}\left(\frac{cE}{L}\right)^2~+~{\cal O}(r_s^2)\cr ~=~&\pm \frac{1}{b}~+~\frac{GM}{h^2}~+~{\cal O}(M^2) ,\end{align} \tag{7} and
$$u_0~=~\frac{1}{r_s} +{\cal O}(r_s) .\tag{8}$$

4. During the scattering process the reciprocal radial coordinate $$u$$ goes from 0 to the root $$u_+$$ and then back again to 0. The half-angle is then \begin{align}\phi&~~~\stackrel{(6)}{=}~\int_0^{u_+} \! \frac{du}{\sqrt{P(u)}} \cr &~~~=~\int_0^{u_+} \! \frac{du}{\sqrt{(u_+-u)(u-u_-)\left(1-r_su\right)}} \cr &\stackrel{u=u_+x}{=}~\int_0^1 \! \frac{dx}{\sqrt{(1-x)(x+\alpha)}}\left(1+\beta x\right) +{\cal O}(r_s^2) \cr &~~~=~\beta\sqrt{\alpha}+(\alpha\beta+\beta+2)\arctan\frac{1}{\sqrt{\alpha}}+{\cal O}(r_s^2) \cr &~~~=~\frac{r_s }{2b}+2\arctan\left(1+\frac{r_s }{2b}\frac{c^2}{v_0^2}\right)+{\cal O}(r_s^2)\cr &~~~=~\frac{r_s }{2b}+2\left(\frac{\pi}{4}+\frac{r_s }{4b}\frac{c^2}{v_0^2}\right)+{\cal O}(r_s^2), \end{align}\tag{9} where we have defined \begin{align} \alpha~:=~&-\frac{u_-}{u_+}\cr ~=~&1-\frac{r_s}{L}\frac{E^2}{\sqrt{E^2-\epsilon}} +{\cal O}(r_s^2) \cr ~=~&1-\frac{r_s }{b}\frac{c^2}{v_0^2} +{\cal O}(r_s^2)\end{align} \tag{10} and \begin{align} \beta~:=~&\frac{r_s}{2}u_+ ~=~\frac{r_s}{2}\frac{\sqrt{E^2-\epsilon}}{L}+{\cal O}(r_s^2)\cr ~=~&\frac{r_s}{2b} +{\cal O}(r_s^2). \end{align}\tag{11} The constant $$\beta$$ vanishes in the Newtonian limit $$c\to \infty$$.

5. Finally, we can calculate the angle of deflection \begin{align}\theta ~=~&2\left(\phi-\frac{\pi}{2}\right)\cr ~\stackrel{(9)}{=}~&\frac{r_s}{b}\left(1 + \frac{c^2}{v_0^2}\right) +{\cal O}(r_s^2),\end{align}\tag{12} which is the sought-for formula (1). $$\Box$$

References:

1. Sean Carroll, Spacetime and Geometry: An Introduction to General Relativity, 2003; Section 5.4.

2. Sean Carroll, Lecture Notes on General Relativity, Chapter 7. The pdf file is available here.

3. R. Wald, GR, 1984; Section 6.3.

--

$$^1$$ In the Newtonian limit $$c\to \infty\leftrightarrow r_s \to 0$$ the 3rd-order polynomial (6) is replaced with a 2nd-order polynomial $$\left(\frac{du}{d\phi}\right)^2 ~=~(u_+-u)(u-u_-).\tag{13}$$ Differentiation leads to the Binet equation $$\frac{d^2u}{d\phi^2}+u ~=~\frac{r_s}{2}\left(\frac{cE}{L}\right)^2~=~\frac{GM}{h^2} .\tag{14}$$ The solutions are hyperbolas $$u~=~\frac{GM}{h^2}(1+ e\cos(\phi\!-\!\phi_0)), \tag{15}$$ where $$e=\sqrt{\frac{b^2}{a^2}+1}>1$$ is eccentricity; $$\frac{GM}{h^2}=\frac{a}{b^2}$$ is the reciprocal semi-latus rectum; $$a$$ is the major semi axis; $$b$$ is the minor semi axis/impact parameter; and $$\phi_0$$ is a phase offset (which we put to 0 for simplicity).

In the limit of small angle of deflection (where $$a\ll b$$), the Newtonian angle of deflection becomes $$\frac{\theta}{2}~\approx~\sin\frac{\theta}{2}~\stackrel{(12)}{=}~-\cos\phi~\stackrel{(15)}{=}~\frac{1}{e}~\approx~\frac{a}{b}~\stackrel{(3)}{=}~\frac{GM}{bv_0^2}, \tag{16}$$ which agrees with eq. (1) in the $$c\to \infty$$ limit.

• Related discussion: physics.stackexchange.com/q/246853/2451 Commented Feb 24, 2019 at 22:39
• Notes for later: The scattering angle is minus the derivative of the radial action wrt. angular momentum. Commented Aug 30, 2021 at 14:16

I think what the OP was missing is the principle of the equivalence between mass and energy, as well as the fact that light rays do bend even in a vacuum, and even if that bending shows a curvature so subtle that it remains far beyond our perception without magnification equipment.

As pointed out by Viktor Toth at https://www.researchgate.net/post/Why_do_you_think_that_gravitational_lensing_is_due_to_time_dilation_Can_it_be_due_to_length_contraction , both gravitational time dilation and gravitational length contraction appear to have been equally involved in that deflection of light rays by gravitating objects, as verified during the 1919 eclipse mentioned by John Rennie: Some such deflection had been accepted as an effect of Newton's theory of gravity, and was exactly doubled by Einstein, leading to the 1st experimental confirmation of GR.

The most interesting thing about this, to me, is that the spatial aspect of the curvature involved appears to have been discovered more than a century after its temporal aspect: The shape of objects as familiar as the moon is suggestive of spatial curvature, but an understanding of time apparently held greater interest.

• My answer's among the least sophisticated, but I wanted to post it mainly to correct the misconception about light rays not bending, which could discourage interest in cosmological models whose local universes are closed. I'll be testing my preference for such models in subsequent questions.– Edouard Commented Jan 22, 2021 at 17:22

1) The bending of light rays is a general relativistic effect, not one due to Newton's law of gravity.

2) It's probably better to think about these things from a field perspective -- a distribution of mass-energy moves along, and it creates a gravitational field. Then, when things enter that field, they interact with it, and this changes their motion. These things might have their OWN gravitational field that can move the first things, or whatever else, but they are just interacting with the field, not the matter distribution that created the field.

• Light does bend in Newtonian gravity... Commented Jun 7 at 20:24

If the mass of light is assumed to be strictly zero, Newton gravity would produce zero force. However the orbit of light is determined by acceleration, not force. For zero mass, acceleration is undefined. In the limit of photon mass going to zero the force goes to zero, but acceleration is of course independent of photon mass. One can therefore simply apply Newton acceleration to an object moving at speed c. This results in the value of Einstein's paper of 1911, which is half the GRT prediction and the experimental value. See https://en.m.wikipedia.org/wiki/Gravitational_lens #History.

• @PM2Ring Einstein indeed. Commented Jan 22, 2021 at 3:20

How can gravity affect light?

According to Einstein famous formula $$E=mc^2$$, everything which has energy,- must have inertial properties defined by mass $$m$$ (not necessary a rest mass $$m_0$$, it can be relativistic mass as well, which is due to motion). And photon has own energy defined in quantum mechanics as $$E=h \nu$$. So if we will put these things together, we can express Newton gravity law for a photon passing near-by to a massive body $$M$$ :

$$F_G = G~r^{-2} M \left( \frac {h}{\lambda c} \right)_{m_{\gamma}} \tag 1$$

Hence, gravity also affects photon.

To apply Newton gravitation with the maintain of angular momentum and mechanical energy, you must assume that the photon has a mass, even infinitesimal.

Then you get the results of the trajectories of a photon deflected by a gravitational field: hyperbola, parabola, ellipse or even circle, depending on the initial conditions.

Note that the mass of the particle (the photon) appears at the beginning of the calculations but is reduced later on.

However, Newton gravitation leads to a variable photon speed which can be less or more than $$c$$ speed of the light in vacuum, which does not fit with the actual observations.

This shows that classical mechanics can not be applied to particle with no mass, and you have to use general relativity to find the deflection of a photon by a massive object.

A stated in the previous answers, for a photon coming from infinity, the GR deflection is twice the (wrong) deflection calculated with classical mechanics.