Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does the curvature of space-time cause objects to look smaller than they really are? What is the relationship between the optical distortion and the mass of the objects?

share|cite|improve this question

If, as I understand, your question is "Does the mass of an object make the object itself appear smaller?", then wow! it is a very good one.

I believe the answer is the opposite: its own gravitational potential makes an object seem bigger than it is. We see the Sun slightly bigger than it really is. It is essentially the same effect that makes an olive inside a Martini cup appear bigger. Instead of martini and glass slowing down light around the olive, there is spacetime curvature slowing down light around the Sun. You may think about it that way, although the shape of the glass plays an important role in the analogy.

Consider a light ray that departs from a point quite near north pole of the Sun (A) and arrives later to the Earth (B). If the Sun were massless, AB would be a straight line.

How do we find mathematically that AB is a straight line? by Fermat's Principle, that tells us that light travel time along the trajectory followed by a light ray is a stationary quantity:

$$\delta \int_{t_A}^{t_B}dt=0$$

Stationary here means that you cannot modify the shape of the trajectory even a slight little bit without getting an increase in the total travel time, it is therefore a minimum travel time path. In the absence of gravity, and under the assumption that the space between A and B is a perfect vacuum, the speed of light is constant and that minimum time is achieved by the straight line. Since [velocity]=[space]/[time], then the latter integral translates into

$$\delta\int_{A}^{B} \frac{dl}{c} = 0$$

Now, since the Sun has a mass, it warps spacetime. The nearer a clock to a mass, the slower it goes, you know that. The speed of light, being a quotient of space and time as any other speed, goes therefore slower near the Sun. Therefore, $c$ is not a constant quantity between A and B, and the result of the integral in Fermat's principle is no longer a straight line.

For a qualitative answer, you may imagine the gravitational potential of the Sun as a disposition of concentric layers. In the inner layers, the speed of light is smaller. In order to arrive from A to B in the least travel time, photons try to minimize the path through the slower, inner layers, therefore bending the trajectory. In this drawing, the segment $AI^1$ across the inner region is clearly shorter than $AI'^1$, therefore light chooses the path $A I^1 I^2 B$. Observers at the Earth (B) believe that light travels in straight lines, and therefore they perceive the diameter of the Sun as being bigger (see the small drawing in the upper right corner)

enter image description here

(I might upload a better drawing when I am back from the weekend)

More formally, we can get a position-dependent expression of the speed of light to plug into the integral, by considering that the metric around the Sun is nearly flat, therefore using that of Special Relativity plus a small perturbation: $$g_{\mu\nu} \approx \eta_{\mu\nu} + A\delta_{\mu\nu}$$ $$|A|\ll 1$$ In order to get Newtonian gravity asymptotically from this, A turns out to be the Newtonian gravitational potential scaled, $$A=\frac{2\phi}{c^2}$$ therefore, the infinitesimal interval (with +---) is: $$ds^2=(1+\frac{2\phi}{c^2})c^2dt^2-(1-\frac{2\phi}{c^2})(dx^2+dy^2+dz^2)$$ But photons evolve in phase space along null geodesiscs, therefore $ds^2=0$ and you can get an expression for the speed of light (squared) around the Sun as $$c'^2=\frac{(dx^2+dy^2+dz^2)}{dt^2} = c^2\frac{1+\frac{2\phi}{c^2}}{1-\frac{2\phi}{c^2}}$$ Since $\frac{2\phi}{c^2} \ll 1$, $(\sqrt{\frac{1+x}{1-x}})^{-1} \approx 1-x$, and then you have that: $$\frac{1}{c'}=\frac{1}{c}(1-\frac{2\phi}{c^2})$$ Therefore, Fermat's principle between the Sun and us translates into:

$$\delta\int_{A}^{B} (1-\frac{2\phi}{c^2}) dl = 0$$ where $\phi = -\frac{M_{Sun}}{\sqrt{x(l)^2+y(l)^2+z(l)^2}}$ is the Newtonian potential. Solving for the path $x(l), y(l), z(l)$ is another question, involving the Euler-Lagrange equations and a more or less interesting crash against reality, in which you will probably have to do this or that assumption in order to arrive at an analytical solution...

In the standard case of photons from background stars arriving from infinity and just grazing the surface of the Sun, Einstein did the calculations in 1916 and found the deflection angle to be 1.8 arcseconds. Since we are interested here in photons that originate at the Sun, then the potential has been deflecting them half the way along, and so the angle must be about 0.9 arcseconds.

That is, we see the Sun nearly 1.8 arcseconds bigger than we "should". With an apparent diameter of 32 arcminutes, who cares?, but you will be glad to know that extragalactic astrophysicists count with a not yet well quantified amplification bias in the universe, which roughly means that we must probably be overestimating the size and light from distant sources due to the amplification caused by the masses in between (not exactly your claim, but closely related).

Related to your question too, is the well known fact (predicted by Einstein himself) that frequencies in spectral lines from the Sun photosphere are slightly redder than that measured in Earth laboratories, another consequence of the slower rate of clocks near massive objects.

share|cite|improve this answer
The rest of the calculations similar to those Einstein did for the Sun, involve the Euler-Lagrange equations and then integrating along a tangent, unperturbed path from -inf to +inf (similar to the Born approximation to scattering) to find the deflection angle. I am lazy and that is in the literature. The point is, that massive objects curve the path of its own light inwards as seen from a distant observer, and so they appear bigger than they are, for that distant observer. – Eduardo Guerras Valera Mar 17 '13 at 18:42

"The true diameters of the Sun and Earth are 4.1 km and 4.4 mm greater, respectively, than one would expect from applying Euclidean geometry (C = pi * d) to the observed surface of these bodies." -

share|cite|improve this answer
The OP's question is specifically about optics. – Ben Crowell Mar 31 '13 at 20:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.