How much $G$ is required to bend light?

Question

How much $G$ is required to bend light?. Also, please explain what happens if a powerful light beam in a vacuum is intercepted by electromagnetic fields on either sides.

In that case, since G is a constant, the answer is the value of G. — jinawee, Jan 19 '15 at 22:48
I think he means $g$. As in $g=9.81 ms^{-2}$. And to answer the OP, as soon as you have gravity you have gravitational lensing, a.k.a. bending of light. Now, the question could be rephrased as "how does the bending of light depend on gravity", in which case look at @HDE 226868's answer below. — Demosthene, Jan 19 '15 at 23:13

score 3 · Accepted Answer · edited Apr 13 '17 at 12:39

The equation for gravitational lensing is $$\theta=\frac{4GM}{rc^2}$$ where $M$ is the mass of the object doing to bending and $r$ is the distance of the light to the object. But $G$ is a constant - it doesn't change. The literal response, therefore, to your question is "Any positive $G$." In a universe where $G>0$, gravitational lensing is possible.

Also explain what happens if a powerful light beam in vacuum is intercepted by electromagnetic fields on either sides.

So light interacting with light? In that case, this and this should help. But this is a completely different question than your first one.

If Demosthene is correct and you mean $g$, the question sort of stays the same. We've established what $G$ must be. $M$ can also be any non-zero value - even your body bends light! It's just such a tiny influence that it's really, really, really hard to observe. And any value of $r$ will bend light, although of course as $r \to \infty$, $\theta \to 0$.

score 1 · Answer 2 · edited Jan 19 '15 at 23:05

I'll answer the question in your headline. I suppose by "G" you mean "Gravity". If so, the answer is that any non-zero mass will cause a gravitational potential, hence deflecting the path of light. To "bend light" enough that it can be measured, depends upon 1) how close the light beam is to the mass, and 2) how precise your detector is (i.e. what is its angular resolution; for instance Hubble's resolution is 0.05 arcsec.).

In 1919, Arthur Eddington measured the deflection of the light from stars passing close to the Sun during a solar eclipse (of around 1 arcsec, I think), but you would be able to measure the effect for lighter objects if they were smaller so that the light could pass nearer the object. The angle $\theta$ of deflection is given by

$$\theta = \frac{4 G M}{c^2b}$$

where $G$ is the gravitational constant, $M$ is the mass of the object, $c$ is the speed of light, and $b$ is the distance at which the light passes by (the "impact parameter"). You see that as long as $M > 0$ (and $b < \infty$), there is a deflection, however small.

How much $G$ is required to bend light?

2 Answers2