# GR, time-like energy, and photon-path bending?

1. Dec 7, 2005

### cefarix

Does time-like curvature of spacetime cause light to bend, for example, around the sun? Or is the gravitational bending of light only due to the space-like curvature? So far, I understand the concepts as: Time-like curvature is caused by the mc^2 part of E, while space-like curvature is caused by the pc part of E. So the sun should have a huge curvature time-wise, but very little space-wise (perhaps only due to its rotation?). Since photons have zero rest mass, they only have the space-like component of their energy (just the pc part), and so should only be bent by space-like curvature. So a perfectly still mass should not bend light? I think I'm wrong here somewhere. Do photons still have a time-like component to their energy despite having zero rest mass? Need some help here...confused...

2. Dec 7, 2005

From reading the "sun Bending Light" thread I gather that photons do have a time like componet to there energy, but I can not explain why sorry =/.

3. Dec 7, 2005

Staff Emeritus
The momentum energy tensor $$T_{\mu\nu}$$ is responsible for spacetime curvature. Bothe time and space are curved together.

The tensor has 16 components and is symmetric so that its top row and first column are the same. These carry the time ($$T_{00}$$) and space ($$T_{0i} = T_{i0}, i = 1,2,3$$) components of the momentum enrgy four-vector. The other components describe spatial stress of energy-momentum.

Einstein's field equations start out with one nonlinear partial differential equation for each component of T. Because of the symmetry of T and other symmetries in Riemannian geometry, the 16 equations contain only 10 independent ones. These 10 equations are simultaneous; you have to solve them as a set.

You cannot say this piece or that is responsible for space curvature or time curvature. Spacetime curvature is caused by the whole tensor.

4. Dec 7, 2005

### pervect

Staff Emeritus
When people talk about the deflection of light being twice that of the Newtonian value due to spatial curvature, they are not talking about either the Einstein or the Riemann curvature tensors.

They are talking about something else, the Christoffel symbols.

If you write the equation for a body falling directly into a massive body in a Scwarzschild metric, you get a differential equation (the geodesic equation of motion) that looks like this.

$$\frac{d^2 r}{d \tau^2} + \Gamma^r{}_{tt} \left( \frac{dt}{d\tau} \right) ^2 + \Gamma^r{}_{rr} \left( \frac{dr}{d\tau} \right)^2 = 0$$

(This is the simplest case, there are similar terms due to $\Gamma^r{}_{\theta \theta}$ and $\Gamma^r{}_{\phi \phi}$ in the more general expression.

If the velocity is much lower than 'c', $dt/d\tau$ is essentially one, and $dr/d\tau$ is << 1. In this case, the acceleration of an object is essentially constant and independent of its velocity.

We can therfore identify the Christoffel symbol $\Gamma^r{}_{tt}$ with radial gravitational acceleration in the Newtonian limit.

This is no longer the case when $dr/d\tau$ becomes of an order of magnitude near unity - the second term in the differential equation of motion becomes important.

The first Christoffel symbol involves only two time subscripts, the second Christoffel symbol involves only spatial subscripts. Because the name "Christoffel symbol" scares people, sometimes they are talked about as curvatures, though strictly speaking they are not. Note that the first symbol has only time-like subscripts, which is why it is sometimes very losely called a "time curvature", and the second symbol has only spatial subscripts.

Last edited: Dec 7, 2005