Gravity between two Photons (I searched for an answer online already but I couldn't quite find what I was looking for...)
I thought about this for a long time now. If two Photons fly in the same direction, one behind the other one, for my understanding the one behind the other one should be pulled towards the photon in front of it due to it's gravity, and because it cant get faster it should increase it's frequency and therefore gain energy. The one in front cant be pulled backwards though because gravity travels with the speed of light itself(?) and therefore the gravity of the rear photon cant reach the one infront of it, which would therefore not lose energy. 
But that would break the law of conservation of energy, wouldn't it? So I'm confused... 
Am I thinking something wrong? Or how does it work/what would actually happen in this scenario?
Thanks for answers in advance!
 A: Each photon will be at rest relative to the other, since they both travel in the same direction at the same speed.  A photon at rest has zero frequency, hence zero energy.  So, at least to first order, there should be no gravitational interaction between the two photons within their mutual rest frame.
I'm not sure what an outside observer would see.  It seems to be accepted that gravitational waves interact with each other very very weakly, which suggests that the same should be true of electromagnetic waves.
A: The speed of gravity is also only c, so the photon in front never gets overtaken by the gravity of the photon in the back. The photon in the back travels through the gravitational field of the one in the front, but since the direction of pull is to the front while the back photon is already travelling with c in the direction on the pull it already has the maximum velocity which it can not exceed. Since the distance stays constant so does the energy.
A: They do not interact because their rest masses are zero. 
You can not attach to the photon neither an inertial nor a gravitational "mass" as
$$ m \neq\frac{\hbar \omega}{c^2}$$ 
See for example on this topic https://arxiv.org/abs/physics/9907017.
This part is pure kinetic part and is related to the momentum of a photon in the energy–momentum relation (with $m_{rest}=0$ )
$$E=p c $$ 
As example, for a moving particle with non-zero rest mass, the momentum part $E=p c $ does not influence its gravitational interaction (i.e. the gravitational potential that it creates). Because you can always consider the case in a comoving reference frame where $p=0$.
A: From my understanding:. 
Photons have a rest mass of zero, but the electromagnetic waves do carry energy which causes gravitational pull. Since light isn't affected by gravity because its rest mass is zero, but the space light is moving on gets curved. This is what causes light beams to curve around massive objects. The two light beams wont be pulled into one another, but will start 'bending' if they are not moving parrelel to each other. This could get them closer to each other without changing speed. 
A better explanation can be found in this Physics.SE post
A: I don't know enough relativistic E&M to answer Henning Markholm's salvage of the question...but, for the OP:
Photons are massless, so they exert no gravitational force.  So neither will experience a force.  
Furthermore, your notion of the "speed" of a force is fundamentally wrong.  The "speed" of a force is the rate at which changes in the force field propagate.  So even if two massive, yet lightspeed-traveling particles did exert a force on each other, the first particle would still feel the force exerted by the second particle.  There are confusing pieces of physics associated here; see https://en.wikipedia.org/wiki/Wheeler%E2%80%93Feynman_absorber_theory (sorry no link; I'm writing this on a phone).  If you really want to understand want the "speed" of a force means, look into getting yourself a textbook; any E&M textbook at the level of Purcell or better will do.  
