# Null geodesics in uniform gravitational field metric

I'm trying to understand the null geodesics in the metric:

$$\mathrm{d}s^2 = -(1+gz)^2 \mathrm{d}t^2 + \mathrm{d}z^2 + \mathrm{d}x^2$$

In particular I'm wondering if the following intuition is valid: If a photon is emitted from the origin at an angle $\theta$ to the horizon, then could it be that the photon is "dragged" by the gravitational field back down onto the surface $z=0$ at some later time?

To answer this we can write the Lagrangian (for a null geodesic): $$L = (1+gz)^2 \dot{t}^2 - \dot{x}^2 - \dot{z}^2 = 0$$ And the conserved quantities: $$E = (1+gz)^2 \dot{t},\quad u = \dot{x}$$

This allows us to reduce to a 1-dimensional problem for $z$ with potential:

$$V(z) = -\frac{-E^2}{(1+gz)^2}$$

I'm struggling to match this with my above intuition and would appreciate any help in doing that.

• These are called Rindler coordinates. Although the title says "uniform gravitational field," actually particles with constant z have different proper accelerations depending on z. The metric is flat, so the null geodesics are simply the null geodesics in Minkowski space, transformed into Rindler coordinates. The WP article on Rindler coordinates has a discussion of the null geodesics and their geometric interpretation. – Ben Crowell May 15 '16 at 15:03
• Thanks for your comment - to use that fact then surely we'd need to find the transformation taking this metric to Minkowski coordinates, is that easy to do here? – Wooster May 15 '16 at 16:24

The metric reads, restoring $c$, $$\mathrm{ds}^2 = -(1+gz/c^2)^2 c^2\mathrm{dt}^2 + \mathrm{dz}^2 + \mathrm{dx}^2\:.$$ The metric can be used to determine the geodesics by means of the associated quadratic Lagrangian $$L = -(1+gz/c^2)^2 c^2\dot{t}^2 + \dot{z}^2 +\dot{x}^2\tag{0}$$ where the dot denotes the derivative with respect to the affine parameter $s$.

N.B. The more usual Lagrangian $$L = \sqrt{|-(1+gz/c^2)^2 c^2\dot{t}^2 + \dot{z}^2 +\dot{x}^2|}$$ is not suitable here as it does not determine the light-like geodesics whereas (0) determines all types of geodesics, already written in function of an affine parameter as well.

The case of a vertical photon (and vertical massive object)

I first consider the case of a photon whose story is described in the plane $t,z$. In other words, these are vertical photons, the vertical direction being the one of the gravitational field, $z$.

Euler-Lagrange equations of Lagrangian (0) produce, $$-\frac{d}{ds} (1+gz/c^2)^2 \frac{dt}{ds}=0\tag{1}$$ and $$\frac{d}{ds} \frac{dz}{ds}=-(1+gz/c^2)g\left(\frac{dt}{ds}\right)^2\:.\tag{2}$$ The former implies $$(1+gz/c^2)^2\frac{dt}{ds}=T\:,\tag{2'}$$ so that, assuming $T > 0$ as it must be for future-directed causal geodesics, $$\frac{d}{ds}= \frac{T}{(1+gz/c^2)^2}\frac{d}{dt}\:.$$ Using it in (2) we have $$\frac{T}{(1+gz/c^2)^2}\frac{d}{dt} \frac{T}{(1+gz/c^2)^2 }\frac{d}{dt} z = -(1+gz/c^2) g\left(\frac{T}{(1+gz/c^2)^2}\right)^2$$ which simplifies to $$\frac{d^2z}{dt^2} = 2 \frac{g}{c^2(1+gz/c^2)}\left(\frac{dz}{dt}\right)^2- (1+gz/c^2) g\:.$$ You see that in non-relativistic regime, i.e., $|gz|<< c^2$ and $\left(\frac{dz}{dt}\right)^2<< c^2$, the equation for $z=z(t)$ can be approximated with the Newtonian equation $$\frac{d^2z}{dt^2} =-g$$ as expected for massive objects evolving along time-like geodesics.

Light-like geodesics can be explicitly determined following a shorter way.

First of all we observe that the Lagrangian (0) does not depend on $s$ explicitly. Therefore Jacobi's theorem says that the Hamiltonian function is conserved along the solution of Euler-Lagrange equations. Due to the absence of any potential term in the Lagrangian, the Hamiltonian function coincides with the Lagrangian itself. In other words $$-(1+gz/c^2)^2 c^2\dot{t}(s)^2 + \dot{z}(s)^2 +\dot{x}(s)^2= constant\tag{3}$$ along geodesics. For light-like geodesics in the plane $t,z$ we have in particular $$-(1+gz/c^2)^2 c^2\dot{t}(s)^2 + \dot{z}(s)^2 = 0$$ because $L$ is nothing but the squared Lorentzian norm of the tangent vector, so that $$(1+gz/c^2)^2 c^2\dot{t}(s)^2 = \dot{z}(s)^2\:.$$ Since, as observed above, causal geodesics can be parametrized by $t$ (though it is not an affine parameter), the found equality yields $$\frac{dz}{dt} = \pm c(1+gz/c^2)$$ and thus $$\frac{dz}{1+gz/c^2} = \pm cdt \tag{4}$$ Notice that (3) also holds for time-like geodesics but the consequent first order differential equation is not as easy to solve as for light-like geodesics. Instead, the equation for light-like geodesics (4) can be integrated immediately producing $$z(t) = \left(z(0) + \frac{c^2}{g}\right) e^{\pm \frac{g}{c}t}- \frac{c^2}{g}\:.$$ This is the general formula describing light-like geodesics in the plane $t,z$ of our accelerated system of coordinates. It is worth remarking that there are two types of geodesics emitted at $z(0)$ for $t=0$, the ones with the sign $-$ in the exponent propagating (for $t>0$) towards the Killing horizon, situated at $z_0 = -\frac{c^2}{g}$ (where $g_{tt}=0$), and the ones propagating (for $t>0$) towards $z= +\infty$. The geodesics of the first class reach the horizon spending an infinite amount of Killing time $t$. The other type of curves escape to infinity with exponential rapidily.

Summing up, for massive bodies evolving along time-like geodesics the action of the gravitational field is similar to the classical one in the non relativistic regime. These particles are "dragged" by the gravitational field back down onto the surface $z(0)$ where they were emitted.

The picture turns out completely different for light-like geodesics describing particles of light. These particles are not "dragged" by the gravitational field back down onto the surface $z(0)$ where they were emitted. However this analysis is valid for particles of light emitted along the direction of the gravitational field, i.e., the vertical direction $z$.

The case of a non-vertical photon

Let us finally check if any dragging effect exists for light particles if considering motions with non-vertical initial direction. As the problem is rotationally symmetric around $z$ we can consider the case of $y=0$ constantly but non-constant $x$. For the sake of simplicity I henceforth assume $c=1$.

The equation for the coordinate $x$ arising from the Lagrangian (0) is trivial, $\frac{d^2x}{ds^2}=0$, so that $$x= Xs\tag{5}$$
for some constant $X>0$. There is another additive constant I can always suppose to be $0$ by redefining the origin of the $x$ axis because the problem is invariant under translations in the $xy$ plane. Now (3) for light-like geodesics produces, if taking (5) into account, $$\left(\frac{dz}{ds}\right)^2 + X^2 = (1+ gz)^2 \left(\frac{dt}{ds}\right)^2\:.$$ Eventually, (2') yields $$\left(\frac{dz}{ds}\right)^2 = \frac{T^2}{(1+ gz)^2}-X^2\:,$$ That is $$\frac{1}{g^2}\left(\frac{d(1+gz)}{ds}\right)^2 = \frac{T^2}{(1+ gz)^2}-X^2\:.$$ Defining $\zeta := (1+ gz)^2$, this equation can be re-written as $$\left(\frac{d\zeta}{ds}\right)^2 = 4g^2( T^2 - X^2\zeta)$$ and its solution reads $$\sqrt{T^2 - X^2\zeta(s)} = gX^2s +C$$ for an arbitrary constant $C$. In other words, re-defining $C$ $$\zeta(s) = \frac{T^2}{X^2} - g^2X^2(s+C)^2\:.\tag{6}$$ Making use of the definition of $\zeta$ we finally get $$z(s) = -\frac{1}{g}\pm \sqrt{\frac{T^2}{g^2X^2} - X^2(s+C)^2}\:.$$ Actually, since the coordinates we are employing are defined for $z> -1/g$ only the solution
$$z(s) = -\frac{1}{g}+ \sqrt{\frac{T^2}{g^2X^2} - X^2(s+C)^2}\tag{7}$$ is permitted.

It is possible to parametrise this curve with the Killing time $t$ integrating (2') since $(1+gz)^2 = \zeta$ is now explicitly given by (6). However it is not necessary since we are only interested in the shape of the trajectory.

The curve (7) is an ellipse in the plane $z,s$ centered in $z= -1/g$, $s=-C$ and with axes parallel to the Cartesian axes. We can always assume $C=0$ since the origin of the affine parameter is arbitrary.

The physical interpretation is now easy. If we emit our photon from the surface at $z=z_0> -1/g$ along the direction $z>0$ but also with a horizontal component of its affine velocity $\dot{x}=X>0$ at some initial affine time $-s_0<0$, after a finite amount of affine time, more precisely at $s= s_0$, the photon comes back at $z_0$.

Your intuition was correct: The photon is, in fact, dragged by the gravitational field back down onto the emission surface at $z=z_0$.

• Thank you for this detailed response! Is it not the case though that when these light-like geodesics have some $x$ component (I understand you've ignored the $x$ component here), then there will be some dragging? It's not clear to me how you've used the condition that the geodesics are emitted at an angle $\theta$ in the $xz$ plane? – Wooster May 19 '16 at 11:54
• Well actually there is no angle here! Or better it is $\pm \pi/2$ the "vertical direction" , i.e., the one along the direction of the gravitational field, $z$ in our case. Perhaps some dragging effect takes place including the coordinates $x, y$. Let me think about it. – Valter Moretti May 19 '16 at 14:08
• Yes, it makes sense to me that there should be no dragging in the vertical direction but I would expect some if the light was emitted at some angle (c.f. gravitational lensing) so I would be very curious to see how it works out in that case - I haven't been able to solve it! – Wooster May 19 '16 at 14:12
• @Wooster I obtained the result of the trajectory of the photon with non-vertical initial direction. Your intuition is exact, the photon is, in fact, dragged by the gravitational field back down onto the initial surface! – Valter Moretti May 19 '16 at 17:05
• you're welcome :) Indeed, it was the first time I considered that problem. I might confess that I though you were wrong! You have a deep physical intuition... I will use this computation as an exercise for my students. – Valter Moretti May 21 '16 at 13:20