# Is this classic or general relativity?

1. Jun 2, 2013

### birulami

Suppose a photon with speed $|\vec{v}(t)|=c$. With its frequency $f(t)$ it has the momentum $h/c^2 f(t)\vec{v}(t)$.

Further suppose a (classical) gravitational field $\vec{g}(x,y,z)\in R^3$.

Putting the photon at position $s(t)\in R^3$ into the field and equating

a) the force as the product of field and mass $h/c^2 f(t)$
b) with the first derivative of the momentum, we get

$$\frac{h}{c^2}f(t)\vec{g}(\vec{s}(t)) = \frac{h}{c^2} f'(t)\vec{v}(t) + \frac{h}{c^2} f(t)\dot{\vec{v}}(t) .$$

Dividing by the constant $h/c^2$ leaves the slightly simpler form

$$f(t)\vec{g}(\vec{s}(t)) = f'(t)\vec{v}(t) + f(t)\dot{\vec{v}}(t) .$$

As an additional equation we can throw in $\vec{v}\dot{\vec{v}}=0$, which follows from the constancy of the speed of light.

What I would like to know is, whether this description is still classic, or whether it is possibly equivalent with the general theory of relativity. I don't think it is completely classic, since I use the equivalence of mass and energy and allow the photon's energy to change in the field. Yet I would not be surprised to hear that from these equations the gravitational deflection of light is half of what general relativity correctly predicts, meaning this is still equivalent with Newton's mechanics? But I would not know how to solve the differential equation $fg = f'v + f\dot{v}$ to find out myself, even if I use the specific form $g(s(t))=-\vec{s}(t)/\vec{s}(t)^3$.

Any hints appreciated.

Harald.

2. Jun 2, 2013