Mutual gravitational acceleration (or deflection) of light beams as a function of the angle between them Given Einstein's famous equivalence between energy and mass, $E=mc^2$, a light beam with power, $P$, has an energy per unit length of $P/c$ and an equivalent mass per unit length of $P/c^3$.  A short element, $\delta l$, of the beam would therefore be expected to produce a gravitational field of $G(\delta l P/c^3)/r^2$.
However, I gather the actual deflection of a light beam in a gravitational field is twice the amount expected from Newtonian mechanics. From conservation of momentum, it follows that the mutual force between the light beam and the deflecting mass must be twice that prescribed by Newtonian mechanics. I think this means the gravitational field produced by an element of a light-beam is actually $2G(\delta l P/c^3)/r^2$. (?)
The question is about the gravitational interaction between two light beams. There seems to be another complication. I gather that two parallel light beams traveling in the same direction see zero mutual gravity. I also gather that parallel light beams traveling in exactly opposite directions see four times the Newtonian gravitational interaction.
This is pure guesswork, but can I take the field (acceleration) of an element of one light beam due to an identical element of another identical light beam to be simply $4Gsin^2(\theta /2))(\delta l P / c^3)/r^2$, where $\theta$ is the angle between the two beams and r is the distance between the two elements?  Or do I need a more complicated expression?
 A: Newton's formula $F=\frac{GMm}{r^2}$ is only an approximation, valid in non-relativistic contexts. Light is highly (!) relativistic, and so it's not surprising that the formula doesn't hold for it. To find the actual deflection of light due to gravity one must use the field equations of general relativity. In these equations the source of gravity is not mass or energy, but rather the stress-energy tensor. That tensor includes energy as one term, but includes many other terms as well, some of which are significant for light.
A: The question was not posed well. A single angle isn't sufficient to define the configuration.
Two line elements each defined by in position (3 variables) and orientation (2 variables) gives a total of ten variables. Without loss of generality we can assume one element is at the origin, 0,0,0, and oriented along, 0,0. We can also extract the separation, |r|, as a scalar. However, there remain four angles to define the configuration: two describing which direction the second element is to be found and two decribing the orientation of the second element.
An element, $\delta l$, of a light beam, power $P$, produces a static gravitational field, $2G(\delta l P/c^3)/r^2$, as above. But it also produces a gravitomagnetic field. This is the effect that causes parallel beams to experience no attraction when they are traveling in the same direction but twice the attraction when they are traveling in opposite directions. The total force on an element $\delta l_1$ from an element $\delta l_2$ of two light beams will be the sum of force from the static field and force from the gravitomagnetic field.
$\delta^2 F = {4GP_1P_2 \over c^6r^3}[\delta l_1.\delta l_2.\vec r + \vec{\delta l_1} \times (\vec{\delta l_2}\times \vec r)]$
The first term inside the square braces corresponds to the static gravitational force and is directed along $\vec r$. The second term is the the gravitomagnetic force derived by analogy with the force between two elements of current. The gravitomagnetic force is pependicular to $\vec {\delta l_1}$ and in the plane defined by $\vec {\delta l_2}$ and $\vec r$.
If we simplify things and say that the two line elements are in the same plane, then there are only two angles to worry about: the angle of the second element as seen from the first, $\theta$, and the orientation of the second element, $\phi$ (see diagram).

If we consider only the force perpendicular to $\delta l_1$, then
$\delta^2 F = {4GP_1P_2\delta l_1\delta l_2 \over c^6r^2}[sin(\theta) - sin(\theta-\phi)]$
If the two elements are parallel and side by side, $\phi = 0^o, \theta = 90^o$, then there is no attraction between the beams.  If the beams are antiparallel, $\phi = 180^o$, then there is twice the expected attraction.
