# Why is $p\cdot u_{\mathrm{obs}} = -E$ valid in curved spacetime for a stationary observer?

An observer stationed at a fixed Schwarzschild radial coordinate R near a spherical star of mass M observes a photon moving radially away from the star and measures its energy to be E. What are the components of this photon's four-momentum $p$ in the Schwarzschild coordiante basis? (Hint:$p*p=0$ for a photon.)

I wasn't sure if $p^t=E$ in a curved spacetime. So I it looked here, which gives out the answer: $$p=\left[E\left(1-\frac{2GM}{R}\right)^{-\frac 12},E\sqrt{1-\frac{2GM}R},0,0\right]$$

My question is: Why is $p*u_{obs}=-E$ valid all the time, not only in special relativity but also in this case? The article on that website simply says because the euqation is a tensor equation, but I still don't get it.

• I've changed the title of the question to reflect your specific issue with the calculation. – JamalS Dec 4 '14 at 8:02

If $(M,g_{\alpha\beta})$ is the space-time and $p$ is some event on the worldline of an observer then at $p$ this observer can construct an orthonormal basis,specifically a local Lorentz frame $e_{\hat{\alpha}}$ (where hatted index indicates orthonormal frame), so that $g_{\mu\nu}e^{\mu}{}{}_{\hat{\alpha}}e^{\nu}{}{}_{\hat{\beta}} = \eta_{\hat{\alpha}\hat{\beta}}$ i.e. at $p$ the metric just reduces to the Minkowski metric; this is due to a very standard result in linear algebra.
What this means for this observer then is, at $p$, all the laws of physics that can be expressed entirely in terms of quantities depending on $p$ and $p$ alone will reduce to the laws of physics in SR when using the local Lorentz frame. Since $p^{\alpha}$ and $u^{\alpha}_{\text{obs}}$ are both just 4-vectors living in $T_pM$ this applies and so the SR relation $p_{\hat{\alpha}}u^{\hat{\alpha}}_{\text{obs}} = -E$ holds at $p$ in $e_{\hat{\alpha}}$ for this observer. But this is clearly a tensor equation so it must hold generally in any basis, be it an orthonormal one or a coordinate one.