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.

  • $\begingroup$ 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. $\endgroup$ – John Rennie Aug 12 '14 at 14:06
  • $\begingroup$ Perfect! Thank you! I was reasonably certain that was the answer. I appreciate your help. $\endgroup$ – math and mountains Aug 12 '14 at 16:56
  • 1
    $\begingroup$ 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) $\endgroup$ – P. C. Spaniel Jul 17 '17 at 18:32
  • $\begingroup$ 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. $\endgroup$ – Sergiy Kolodyazhnyy Jan 13 at 2:47
  • $\begingroup$ 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. $\endgroup$ – InertialObserver Feb 25 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.


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.


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. We start from the geodesic equations $$ E~=~cn^{-1}\frac{dt}{d\lambda} , \qquad n^{-1} ~:=~1 - \frac{r_s}{r}, \qquad r_s~:=~\frac{2GM}{c^2}, \tag{5.61/7.43/6.3.12} $$ $$ L~=~r^2\frac{d\phi}{d\lambda}, \tag{5.62/7.44/6.3.13} $$ $$ \epsilon~=~n^{-1}c^2 \left(\frac{dt}{d\lambda}\right)^2 -n\left(\frac{dr}{d\lambda}\right)^2-r^2\left(\frac{d\phi}{d\lambda}\right)^2, \tag{5.55/7.39/6.3.10} $$ where $\epsilon=1$ for a massive particle and $\epsilon=0$ for a massless particle, cf. Refs. 1-3. Here the specific energy constant $E$ is taken to be dimensionless, and the specific angular momentum constant $L$ is taken to have dimension of length. This leads to $$ E^2 ~=~ \left(\frac{dr}{d\lambda}\right)^2 + n^{-1}\left(\frac{L^2}{r^2}+\epsilon \right). \tag{5.64/7.46/6.3.14} $$ 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 observables $b$ and $v_0$ at spatial infinity $r=\infty$? A: Note that $$r^2\frac{d\phi}{dt}\quad\longrightarrow\quad bv_0~=~h~=~\frac{cL}{E}\quad\text{for}\quad r~\to~\infty, \tag{2}$$ and $$\frac{dr}{dt}\quad\longrightarrow\quad v_0~=~c\frac{\sqrt{E^2+\epsilon}}{E}\quad\text{for}\quad r~\to~\infty.\tag{3}$$ Eq. (3) 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. In both cases, we can write $$ v_0~=~c\frac{\sqrt{E^2-\epsilon}}{E} .\tag{4}$$

  3. If we define the reciprocal radial coordinate $$ u~:=~\frac{1}{r},\tag{5}$$ we get a 3rd-order polynomial$^1$ $$ \left(\frac{du}{d\phi}\right)^2~=~P(u)~:=~\frac{E^2}{L^2} - n^{-1}\left(u^2+\frac{\epsilon}{L^2}\right)~=~r_s(u-u_+)(u-u_-)(u-u_0),\tag{6}$$ with 3 roots $$u_{\pm}~=~\pm\frac{\sqrt{E^2-\epsilon}}{L}+\frac{r_s}{2}\frac{E^2}{L^2}+{\cal O}(r_s^2) ~=~\pm \frac{1}{b}+\frac{GM}{h^2}+{\cal O}(M^2), \tag{7}$$ 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 $$\begin{align}\phi&~~~=~\int_0^{u_+} \! \frac{du}{\sqrt{P(u)}} \cr &~~~=~\int_0^{u_+} \! \frac{du}{\sqrt{(u_+-u)(u-u_-)\left(1-r_su\right)}} +{\cal O}(r_s^2) \cr &\stackrel{u=u_+x}{=}~\int_0^1 \! \frac{dx}{\sqrt{(1-x)(x+\alpha)}}\left(1+\beta x\right) +{\cal O}(r_s^2) \cr &~~~=~\beta\sqrt{\alpha}+(\alpha\beta+\beta+2)\arctan\frac{1}{\sqrt{\alpha}}+{\cal O}(r_s^2) \cr &~~~=~\frac{r_s }{2b}+2\arctan\left(1+\frac{r_s }{2b}\frac{c^2}{v_0^2}\right)+{\cal O}(r_s^2)\cr &~~~=~\frac{r_s }{2b}+2\left(\frac{\pi}{4}+\frac{r_s }{4b}\frac{c^2}{v_0^2}\right)+{\cal O}(r_s^2), \end{align}\tag{9}$$ where we have defined $$\alpha~:=~-\frac{u_-}{u_+} ~=~1-\frac{r_s}{L}\frac{E^2}{\sqrt{E^2-\epsilon}} +{\cal O}(r_s^2) ~=~1-\frac{r_s }{b}\frac{c^2}{v_0^2} +{\cal O}(r_s^2) \tag{10}$$ and $$\beta~:=~\frac{r_s}{2}u_+ ~=~\frac{r_s}{2}\frac{\sqrt{E^2-\epsilon}}{L}+{\cal O}(r_s^2) ~=~\frac{r_s}{2b} +{\cal O}(r_s^2). \tag{11}$$ The constant $\beta$ vanishes in the Newtonian limit $c\to \infty$.

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


  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 $c\to \infty$ 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 $$ \frac{d^2u}{d\phi^2}+u~=~\frac{r_s}{2} \frac{E^2}{L^2}~=~\frac{GM}{h^2} .\tag{14}$$


It's just a simple concept according to Einstein's relativity when light travels through high gravation field or high mass containing object (i.e. same when a object has high mass means it can attract other objects of lesser mass) the photons present in the light gets attract towards the other object and we see light bending in the universe, but there is one thing is to note that photons are massless matters but in this case the object with higher mass attracts the object with lesser mass weather it is 0.


protected by ACuriousMind Jul 17 '17 at 9:23

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