In Newtonian gravity, an infinite planar mass of uniform density σ produces a uniform gravitational field: at every point in space, test particles experience an acceleration of magnitude 2 π σ (in units where G=1), directed towards the plane and orthogonal to its surface.
Is there an analogous situation in General Relativity, and if so, in what ways does it depart from the Newtonian case? In fact there are at least two exact solutions of Einstein’s equation that deserve to be called “infinite plane” solutions [1], but they both have exotic aspects (a cosmological constant, and relativistic pressures) that make them less similar to the Newtonian case than we might wish, if we’re not interested in a strictly infinite plane for its own sake, but only in the mathematical simplicity that it yields.
So, instead of a literally infinite plane, we will consider the General-Relativistic version of the approximately uniform field near the centre of a finite planar mass. We will employ the weak-field approximation, where we use the linearised version of Einstein’s equation. This is appropriate if σ is small enough for us to neglect terms of second or higher order in σ. We will also simplify matters by assuming that the pressure within the planar mass, maintaining its shape and its uniform density against self-gravitation, is itself utterly negligible as a source of gravity. This will be true if, as well as being light, the planar mass is not too large. Finally, we will limit most of our analysis to the region of space that is both close to the plane itself, and close to its centre; this makes it “seem” arbitrarily large to test particles moving within that region, without requiring us to confront the messy problems that arise if we let the plane’s size go to infinity.
What we will find is that test particles behave in almost the same manner as in the Newtonian case, with one major difference: for particles with relativistic velocities, their rate of acceleration towards the plane (measured in a frame in which the plane is fixed) will be velocity-dependent.
Of course, this does not violate the requirement of the Equivalence Principle: that two free-falling observers with different velocities whose trajectories cross will see each other’s motion (near the time and place where they cross paths) as inertial. That is, each will describe the other, near the crossing, as moving on a straight line with a constant velocity. The fact that the second derivatives of these observers’ coordinates will be different — in a coordinate system tied to the plane — does not contradict this.
We are going to be using the framework of linearised gravity [2], in which the metric is assumed to depart from that of flat space-time by a small perturbation, and we discard all terms that are second-order or higher in the perturbation. We choose a coordinate system (t,x,y,z) in which the metric gab is equal to the standard Minkowski metric, diag(–1,1,1,1), plus a small perturbation hab. Suppose the planar mass is a square of half-side-length H, arranged symmetrically around the origin of our coordinate system at z=0. In other words, it occupies the region
S = {(x,y,0)|–H≤x≤H, –H≤y≤H}.
Note that because the metric is only approximately equal to the standard metric, we are not stating that the proper length of one side of this square is exactly 2H. However, that level of detail is a second-order effect that we can safely neglect (and in any case we’ll be mostly interested in results that are independent of H).
In linearised gravity, the perturbation h can be found by summing (or integrating), for all sources, the perturbation that results from each single point source. For a point source of mass M located (and at rest) at (xs,ys,zs), this is (see Exercise 18.3 of [2]):
hab = 2 M / √((x–xs)2 + (y–ys)2 + (z–zs)2) diag(1,1,1,1)
Because our planar mass is stationary in the coordinates we’re using, there’s no need to worry about the retarded integrals of Equation 18.14 in [2] that take account of the light-speed time lag in communicating any movement of the source; everything is time-independent, so we can simply integrate the formula above over S, where each infinitesimal source in the integral is given the mass σ dxs dys. Again, we’re neglecting the second-order issue of any correction due to the metric assigning a slightly different area than dxs dys to this infinitesimal region.
The exact result of this integral is a messy expression involving logs and arctans that we’ll relegate to a footnote [3]. Because we’re interested in the region close to the origin of our coordinate system, we will take a Taylor expansion to second order in x, y and z, and define the scalar quantity h as:
h = 2 σ (8 H ArcSinh[1] – 2 π z – √2 (x2 + y2 – 2 z2) / H)
We then have:
hab = h diag(1,1,1,1)
gtt = –(1–h)
gxx = gyy = gzz = 1+h
Note that to first order, the metric g is independent of x and y. However, we need to know the metric to second order to be able to compute the Riemann curvature tensor and determine tidal effects.
We compute the connection coefficients Γabc in the usual way, from derivatives of the metric components, then we raise the first index using the flat Minkowski metric rather than gab, because the distinction is a second-order effect (see Equation 18.2 of [2]). This yields:
Γttz = Γtzt = Γztt = Γzxx = Γzyy = σ (2 π – 4 √2 z / H)
Γxxz = Γxzx = Γyyz = Γyzy = Γzzz = – σ (2 π – 4 √2 z / H)
Γtty = Γtyt = Γytt = Γyxx = Γyzz = σ 2 √2 y / H
Γxxy = Γxyx = Γyyy = Γzyz = Γzzy = – σ 2 √2 y / H
Γttx = Γtxt = Γxtt = Γxyy = Γxzz = σ 2 √2 x / H
Γxxx = Γyxy = Γyyx = Γzxz = Γzzx = – σ 2 √2 x / H
with all other components zero. For x, y, z << H, we treat x/H, y/H and z/H as zero, yielding the only non-zero components:
Γttz = Γtzt = Γztt = Γzxx = Γzyy = 2 π σ
Γxxz = Γxzx = Γyyz = Γyzy = Γzzz = – 2 π σ
The world line of a test particle, parameterised by its proper time τ, obeys the geodesic equation:
d2xa / dτ2 = – Γabc (dxb/dτ) (dxc/dτ)
which, for the four individual coordinates, yields:
d2t / dτ2 = – 4 π σ (dz/dτ) (dt/dτ)
d2x / dτ2 = 4 π σ (dz/dτ) (dx/dτ)
d2y / dτ2 = 4 π σ (dz/dτ) (dy/dτ)
d2z / dτ2 = 2 π σ ((dz/dτ)2 – (dt/dτ)2 – (dx/dτ)2 – (dy/dτ)2)
Using the chain rule to convert all the derivatives with respect to τ into derivatives with respect to t, we obtain:
d2x / dt2 = 8 π σ (dx/dt) (dz/dt)
d2y / dt2 = 8 π σ (dy/dt) (dz/dt)
d2z / dt2 = – 2 π σ (1 + (dx/dt)2 + (dy/dt)2 – 3 (dz/dt)2)
These results demonstrate that the second derivatives of a free-falling particle’s coordinates in the frame in which the planar mass is fixed depart from the Newtonian result by relativistic modifications that depend on the particle’s velocity in that frame.
Note, though, that dx/dt, dy/dt and dz/dt are not quite equal to vx, vy and vz, the components of the ordinary velocity in a local orthornormal frame that a stationary observer would measure for the test particle, because x, y and z do not quite measure proper distance, nor t proper time for such an observer. The relationship between the two is:
dx/dt = √[(1–h)/(1+h)] vx
dy/dt = √[(1–h)/(1+h)] vy
dz/dt = √[(1–h)/(1+h)] vz
From this, we can find the rates of change of the velocity components:
dvx/dt = 4 π σ vx vz
dvy/dt = 4 π σ vy vz
dvz/dt = –2 π σ (1 + vx2 + vy2 – vz2)
To obtain these results, we used an approximation for h as a linear function of z alone (i.e. discarding the quadratic terms in our series). After substituting for dx/dt in terms of vx, etc., and solving for the time derivatives of the velocity components, h has been discarded from terms such as (1+h), leaving a result to first order that is independent of h, and hence of z.
For the case of purely vertical motion we can integrate the second-order differential equation for the z coordinate to obtain an explicit trajectory:
z = z0 – log[cosh[(√3) a t] – (√3) d0 sinh[(√3) a t]] / (3 a)
dz/dt = ((√3) d0 cosh[(√3) a t] – sinh[(√3) a t]) / ((√3) cosh[(√3) a t] – 3 d0 sinh[(√3) a t])
d2z/dt2 = a (3d02 – 1) / (cosh[(√3) a t] – (√3) d0 sinh[(√3) a t])2
where a = 2 π σ is the acceleration unmodified by relativistic effects, and z0 and d0 are the values of z and dz/dt respectively at t=0. For the case d0=0, fall from rest at z0, we have:
z = z0 – log[cosh[(√3) a t]] / (3 a)
dz/dt = – tanh[(√3) a t] / (√3)
d2z/dt2 = –a sech2[(√3) a t]
The second-order equation that we integrated shows that d2z / dt2 becomes positive when |dz/dt| > 1/(√3), which at first glance might seem to suggest that gravity becomes repulsive for vertical velocities at around 57% of lightspeed! The asymptotic behaviour of dz/dt, which converges on –1/(√3) so long as d0 < +1/(√3), suggests the same conclusion. However, while these statements about dz/dt are true, this does not reflect what happens with vz, the vertical velocity that would actually be measured by a static observer. As our equation for dvz/dt shows, this will always be negative, so gravity will always remain attractive. The subtle point here is that even though dz/dt and vz are related by a factor close to 1, they can still have time derivatives of opposite sign so long as those derivatives are small — and the fact that our planar mass is light means that σ, and hence a = 2 π σ, will be small.
The equation for vz for purely vertical motion is:
vz = (vz,0 cosh[a t] – sinh[a t]) / (cosh[a t] – vz,0 sinh[a t])
where vz,0 is the value of vz at t=0. For vz,0=0:
vz = – tanh[a t]
From the connection coefficients we found before discarding all first-order terms in x, y and z, we can compute the Riemann curvature tensor. Discarding terms that are second-order in σ, we get:
Rtztz = Rxyxy = Ryxyx = Rzttz = 4 √2 σ / H
Rtzzt = Rxyyx = Ryxxy = Rztzt = – 4 √2 σ / H
Rtxxt = Rtyyt = Rxtxt = Rxzzx = Rytyt = Ryzzy = Rzxxz = Rzyyz = 2 √2 σ / H
Rtxtx = Rtyty = Rxttx = Rxzxz = Rytty = Ryzyz = Rzxzx = Rzyzy = –2 √2 σ / H
From these values, it follows that a cloud of test particles that are initially at rest will be subject to tidal squeezing of 2 √2 σ / H in the x and y directions, and tidal stretching of 4 √2 σ / H in the z direction. While these effects are inversely proportional to H, and hence diminish as the planar mass is made larger, for the solutions we are considering H is always finite, so the tidal effects will always be present. What’s more, the two solutions for infinite planes given in [1] also imply tidal effects. So the idealised Newtonian situation, where tidal effects are absent for an infinite plane, can never be realised in General Relativity.
[1] “The general relativistic infinite plane” by Jones, Muñoz, Ragsdale & Singleton, http://arxiv.org/abs/0708.2906.
[2] Gravitation by Charles Misner, Kip Thorne and John Wheeler, W.H. Freeman, San Francisco, 1973. Chapter 18: Weak Gravitational Fields.
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