# Velocity of a photon according to a stationary observer in Schwarzschild metric

Consider the Schwarzschild metric

$$\left(1-\frac{2M}{r}\right)dt^2-\left(1-\frac{2M}{r}\right)^{-1}dr^2-r^2\left(d\theta^2+\sin^2\theta\, d\phi^2\right)$$

Final goal: calculate the observed velocity of a photon on a radial geodesic according to an observer that's stationary at some $$r>2M$$ above the singularity.

Such a stationary observer must have a wordline $$\gamma(\tau)=\left(t,r,0,0\right)$$

such that $$\dot r=0$$ and $$\dot\gamma^\mu\dot\gamma_\mu=1$$, this yields

$$\frac{\mathrm{d}t}{\mathrm{d}\tau}=\frac{1}{\sqrt{1-\frac{2M}{r}}}$$

The wordline for an observer stationary at some height $$h$$ is thus

$$\gamma(\tau)=\left(\frac{\tau}{\sqrt{1-\frac{2M}{h}}},h,0,0\right)$$

Now, a photon geodesic must obey

$$\frac{\mathrm{d}r}{\mathrm{d}\lambda}=\left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\lambda}\implies\frac{\mathrm{d}r}{\mathrm{d}t}=1-\frac{2M}{r}$$

The velocity of the photon according to the observer is

$$\frac{\mathrm{d}r}{\mathrm{d}\tau}=\frac{\mathrm{d}r}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}\tau}=\frac{1-\frac{2M}{r}}{\sqrt{1-\frac{2M}{h}}}$$

The problem: if I plug in $$r=h$$ I would expect to obtain $$1$$, as the photon is supposed to move at the speed of light near the observer. What happened?

If we want to be precise, your calculations seem correct but it is just the coordinates you are using that are not locally inertial and that is why you are not getting what you would like. Just a comment, do not set $$\theta = 0$$ that is a coordinate singularity of spherical coordinates and there fore of the metric in these coordinates. Instead, take $$\theta = \frac{\pi}{2}$$ which is the equator.

In general relativity, locally inertial coordinates means the following: given some coordinates in a neighborhood of a world-line $$\gamma(\lambda)$$, these coordinates are called locally inertial if the metric written in these coordinates at each point $$\gamma(\lambda)$$ of the world-line is the Minkowski metric. However, away from the world-line the metric in these coordinates is not Minkowski!

Your coordinates are $$(t, r, \phi, \theta)$$ and the world-line (a non-geodesic one by the way) is let's say $$\gamma(\tau) = \left(\, \frac{\tau}{\sqrt{1-\frac{2M}{h}\,}},\, \, h,\,\, 0,\,\, \frac{\pi}{2} \,\right)$$ parametrized with respect to proper time. The tangent coordinate vectors along the coordinate $$t-$$lines, $$r-$$lines, $$\phi-$$lines and $$\theta-$$lines are $$\frac{\partial}{\partial t} = \begin{bmatrix}1\\0\\0\\0 \end{bmatrix}, \,\,\, \frac{\partial}{\partial r} = \begin{bmatrix}0\\1\\0\\0 \end{bmatrix}, \,\,\, \frac{\partial}{\partial \phi} = \begin{bmatrix}0\\0\\1\\0 \end{bmatrix} \,\,\, \text{ and } \,\,\, \frac{\partial}{\partial \theta} = \begin{bmatrix}0\\0\\0\\1 \end{bmatrix} \,\,\,$$ respectively. They are Schwarzschild orthogonal, but their Schwarzschild length is not unit ($$+1$$ for the time-like vector and $$-1$$ for the three space-like vectors) along the world line $$\gamma(\tau)$$. If you rescale these vectors, however, you can achieve normalization $$\frac{\partial}{\partial \tau} = \frac{1}{\sqrt{1-\frac{2M}{h}}}\, \frac{\partial}{\partial t} = \begin{bmatrix}\frac{1}{\sqrt{1-\frac{2M}{h}}}\\0\\0\\0\end{bmatrix} , \,\,\, \frac{\partial}{\partial \rho} = {\sqrt{1-\frac{2M}{h}}}\, \frac{\partial}{\partial r} = \begin{bmatrix}0\\ \sqrt{1-\frac{2M}{h}} \\0\\0 \end{bmatrix}, \,\,\,$$ $$\frac{\partial}{\partial u} = \frac{1}{h\,\sin(\frac{\pi}{2})}\,\frac{\partial}{\partial \phi} =\frac{1}{h}\,\frac{\partial}{\partial \phi} = \begin{bmatrix}0\\0\\\frac{1}{h}\\0 \end{bmatrix}, \,\,\,\,\, \frac{\partial}{\partial w} = \frac{1}{h}\,\frac{\partial}{\partial \theta} = \begin{bmatrix}0\\0\\0\\\frac{1}{h} \end{bmatrix} \,\,\,$$ Now, you can perform the linear change of variables from $$(t, r, \phi, \theta)$$ to $$(\tau, \rho, u, w)$$ as follows \begin{align} &t = \frac{\tau}{\sqrt{\, 1 - \frac{2M}{h}\,}}\\ &r = \left(\sqrt{\, 1 - \frac{2M}{h}\,} \right)\, \rho \,+ \, h\\ &\phi = \frac{u}{h}\\ &\theta = \frac{w}{h} + \frac{\pi}{2} \end{align}
and the the inverse relations are \begin{align} &\tau = \left({\sqrt{\, 1 - \frac{2M}{h}\,}}\right)\, t \\ &\rho = \frac{r \, - \, h}{\sqrt{1 - \frac{2M}{h}}} \\ &u = h\,\phi\\ &w = h\,\theta \, - \, \frac{h\,\pi}{2} \end{align}
The world line $$\gamma(\tau)$$ in the new coordinates is $$\gamma(\tau) = \big(\tau,\, 0, \, 0, 0 \big)$$ The light-like radial geodesic in the original $$(t, r, \phi, \theta)$$ coordinates is $$g(t) = \Big(\,t, \, r(t), \, 0, \, \frac{\pi}{2}\,\Big)$$ where $$\frac{dr}{dt} = 1 - \frac{2M}{r}$$ In the new coordinates it is $$g(\tau) = \Big(\,\tau, \, \rho(\tau), \, 0, \, 0\,\Big)$$ where we can find the function $$\rho(\tau)$$ by changing the coordinates in the differential equation for $$r(t)$$, i.e. \begin{align} \frac{d\rho}{d\tau} &= \frac{\frac{dr}{\sqrt{1-\frac{2M}{h}}}}{\sqrt{1-\frac{2M}{h}}\, dt} = \left(1 - \frac{2M}{h}\right)^{-1}\, \frac{dr}{dt} = \left(1 - \frac{2M}{h}\right)^{-1}\, \left(1 - \frac{2M}{r}\right)\\ & = \left(1 - \frac{2M}{h}\right)^{-1}\, \left(1 - \frac{2M}{h + \rho \, \sqrt{1-\frac{2M}{h}}}\right) \end{align} Observe that in the new coordinates the Schwarzschild metric is the Minkowski metric for each point on the world-line $$\gamma(\tau)$$. Now, when the light-like geodesic $$g(\tau)$$ intersects the world-line $$\gamma(\tau)$$, i.e. the photon reaches the observer, then $$\rho = 0$$ and from the differential equation for $$\rho(\tau)$$ we see that $$\frac{d\rho}{d\tau} = 1$$ i.e. in this locally inertial coordinate system, the speed of light is still $$1$$ as expected.

What you've calculated is called a coordinate velocity. Coordinates don't mean much in general relativity, because GR allows you to use any coordinates you like. For example, I could change to a new coordinate $$u=7r$$ rather than $$r$$, and then the coordinate velocity $$du/dt$$ would be 7 times bigger than what you calculated.

Final goal: calculate the observed velocity of a photon on a radial geodesic according to an observer that's stationary at some r>2M above the singularity.

Coordinates don't have anything to do with observers. GR doesn't have global frames of reference, only local ones.

• So what I calculated doesn't really have physical meaning, does the question itself have physical sense? – user2723984 Jan 28 at 15:18
• An inertial observer (i.e., one who is free-falling) can measure the speed of light locally, and the result will always be $c$. – Ben Crowell Jan 28 at 17:08