# Incandescence

## Deriving Newtonian Spacetime Geometry

In Chapter 12 of Incandescence, the Splinterites succeed in deriving a possible geometry for the spacetime they inhabit. They come up with the simplest possible geometry that conforms to Zak’s principle (that the sum of the three perpendicular “weights”, or tidal accelerations, is zero, after the effects of spin have been removed) while explaining the fact that the ratio of the garm-sard weight to the shomal-junub weight is less than three.

Before reaching that solution, though, they make a couple of false starts. Initially, they assume that there is a notion of universal time, and when they look for a connection that is consistent with that assumption, they end up deriving a spacetime geometry where the ratio of weights is always precisely three.

That first solution — which of course is empirically untenable for the Splinterites — is in fact a description of Newtonian gravitational physics in terms of spacetime geometry. In our world, the redescription of Newtonian gravity in the same terms as Einstein’s general relativity was carried out by the brilliant French mathematician Élie Cartan in the early 1920s.

On this page, we present the detailed calculations that lead to a spherically symmetrical, Newtonian spacetime geometry. In order to make it easier to compare these calculations with those on the main page, where we derived the Schwarzschild geometry, we will use equation numbers that match up exactly with the corresponding equations on that page.

### The Connection for a Spherically Symmetrical Newtonian Spacetime

In a spacetime that conforms to the Newtonian notion of universal time, in what ways can two observers differ in their descriptions of the relationship between two nearby events? Suppose Alice uses time and space coordinates such that two events close to her are separated by ut, ux, uy, and uz, while Bob — who is also near to the events, but who might be moving relative to Alice, and whose coordinates might be oriented differently — measures them to be separated by u't, u'x, u'y, and u'z. Since we’re assuming the existence of a universal time coordinate, we must have u't = ut; there is no difference at all between the separations in time that the two obervers measure. What about the separation in space? If Bob’s coordinates are oriented differently from Alice’s, then we’ll need to apply an appropriate rotation to the spatial separations ux, uy, and uz; also, if Bob is moving relative to Alice, we will need to multiply Bob’s relative velocity by ut and subtract that, to get the final result for u'x, u'y, and u'z.

When it comes to constructing a spacetime connection that respects these results, we’re led to three considerations. Firstly, when we parallel transport any vector along any path, it can’t experience a change in its time component. But, secondly, that’s not to say that a vector with a non-zero time component can’t change at all. If Bob is moving relative to Alice, although they agree on the amount of elapsed time between any two events, they still have different personal time vectors: Alice’s idea of “staying still and moving only in time” is definitely not the same as Bob’s idea of “staying still and moving only in time”. To get from Alice’s time vector to Bob’s, you can’t mess with the time coordinate of her vector, but you must add whatever you need to the spatial part of it to reflect how much Bob is moving, in Alice’s eyes, when he thinks he’s staying still.

Thirdly, any changes in the spatial vectors under parallel transport will be due to rotations, and hence will obey the symmetry that spatial rotations require:

If changeab = F c,
then changeac = – F b.

We will use the same four vectors east, north, out and time as in our original calculations, and the same coordinates t, r, θ and φ, where a surface of constant r and t has area 4 π r2, and θ and φ are lattitude and longitude. When we take into consideration spherical symmetry and the restrictions on the connection discussed above, we have a general form for the connection of:

 changetimetime = G(r) out (2a) changetimeout = 0 (2b) changenorthout = H(r) north (2c) changenorthnorth = – H(r) out (2d) changeeastout = H(r) east (2e) changeeasteast = – H(r) out + (tan θ) / r north (2f) changeeastnorth = – (tan θ) / r east (2g)

The only difference from our original connection is in equation (2b), which originally had the out vector picking up a time component, G(r) time, when transported forward in time.

### Imposing Zak’s principle

As before, we will impose Zak’s principle on two different moving bodies, giving us two equations that we can combine to solve for the unknown functions G(r) and H(r). Again, the first moving body we will consider is the Splinter itself. Each point on the Splinter that lies in the plane of the orbit will move on the equator of one of our coordinate spheres, remaining at some fixed r coordinate, and at the fixed latitude θ=0. The world line of each such point will form a helix in spacetime: advancing in time while circling the Hub. The 4-velocity vector tangent to the world line will take the form:

 u(r) = C(r) east + time (3)

where C(r) is a function of r that we need to determine. The fact that the coefficient of the time vector here is 1 is just another way of saying that there is no difference between Splinter time and time measured by an observer who is stationary in our coordinate system.

If we use our connection to compute the acceleration of any point of the Splinter in the plane of the orbit as a function of its r coordinate, we find:

 acc(r) = changeu(r)u(r) = changeC(r) east + time[ C(r) east + time ] = C(r)2 changeeasteast + changetimetime = [ G(r) – C(r)2 H(r) ] out (4)

where we’ve made use of the fact that we’re on the equator, θ=0, to discard the (tan θ) / r term from changeeasteast.

We will call the r coordinate of the Splinter’s centre of mass r0; the centre of mass will be in free fall, so acc(r0)=0. This will only be true if:

 C(r0) = √[G(r0) / H(r0)] (5)

which in turn gives us the tangent to the world line of the Splinter’s centre of mass:

 u(r0) = √[G(r0) / H(r0)] east + time (6)

From the east component of this tangent vector, we can deduce the rate at which the Splinter is orbiting the Hub. Because east is defined to be a unit vector in terms of distances measured by a stationary observer, we only have to multiply its coefficient by (1 / r0) to get an angular velocity; by definition of the r coordinate the full circumference of the equator is 2 π r0, with all angles and distances related by the same factor of r0. So the angular velocity of the Splinter — by any clock — will be:

 dφ/dt = (1 / r0) √[G(r0) / H(r0)] (7)

We are assuming a spherically symmetrical spacetime geometry, so this equation will also hold for an object orbiting at a slight incline to the Splinter’s orbit, such as a stone displaced a small distance from the Null Line in the shomal or junub direction. Such a stone will cycle back and forth across the plane of the Splinter’s orbit, completing one cycle with each orbit. As in our original analysis, the tidal acceleration in this direction, which we will call ashomal, will be equal to minus the square of the angular frequency of the motion, so we have a value for ashomal in terms of G and H:

 ashomal = – (1 / r02) [G(r0) / H(r0)] (8)

Next, we will determine the rate at which the Splinter is spinning. As we move along the Splinter’s world line, how does the out direction (to which the Splinter is fixed, because it’s tidally locked to keep the same face towards the Hub) change with respect to parallel transport?

 changeu(r0)out = √[G(r0) / H(r0)] changeeastout + changetimeout = √[G(r0) / H(r0)] H(r0) east = √[G(r0) H(r0)] east

It follows that the tidal acceleration ararb that precisely balances the centrifugal force due to the Splinter’s rotation must be:

 ararb = – G(r0) H(r0) (9)

The third tidal acceleration we need is in the garm/sard, or out direction. We start by noting that every point on the Splinter is orbiting the Hub with the same angular velocity, and the argument we used to link that angular velocity to the east component of u(r) works just as well for values of r other than r0. It then follows from equation (3) that:

 C(r) / r = C(r0) / r0 C(r) = C(r0) r / r0 = (r / r0) √[G(r0) / H(r0)]

Substituting this into equation (4) gives us:

 acc(r) = [ G(r) – (r / r0)2 H(r) (G(r0) / H(r0) ] out (10)

We want to know how this acceleration (which is zero at r=r0) changes as we move further from, or closer to, the Hub than the Splinter’s centre of mass. As in our previous analysis, we need to know the ratio between a change in r and an actual distance in the out direction; once again, this turns out to be given by:

 dr/ds = r H(r) (11)

We then compute the total acceleration per unit distance in the garm/sard direction. We take the opposite of the derivative of equation (10) with respect to r, multiply by dr/ds, and set r equal to r0. The result is:

 Asard = G(r0) (r0 H'(r0) + 2 H(r0)) – r0 G'(r0) H(r0) (12)

We have used the notation Asard here as a reminder that this acceleration is a total which comprises both the pure tidal acceleration, asard, and an extra acceleration due to centrifugal force. Because the centrifugal force is balanced exactly by the tidal acceleration along the direction of the orbit, ararb, we have asard = Asard + ararb, and hence Zak’s principle tells us:

 ashomal + ararb + asard = ashomal + 2 ararb + Asard = 0 (13)

With a little rearrangement, and putting r in place of r0 for simplicity, equations (8), (9), (12) and (13) yield:

 r3 H(r) [ H(r) G'(r) – H'(r) G(r) ] + G(r) = 0 (14)

Surprisingly, this is exactly the same as our original equation (14)!

To obtain a second equation, as before we will use a cloud of dust that starts from rest and free-falls towards the Hub. The whole analysis of the transverse tidal acceleration — the squeezing together of the dust cloud as it falls — proceeds identically, once again giving us:

 atransverse = – G(r) H(r) (15)

We find the radial tidal acceleration as before, using the equation:

 tidal(a,b) = changea(changeba) – changeb(changeaa) + change[b, a]a

for the squeezing together of geodesics that point in direction a and are separated by direction b. To find aradial we put a=time and b=out. The difference now is that [out, time] is zero thanks to (2b), leading to a simpler result:

 aradial out = changetime(changeouttime) – changeout(changetimetime) = – changeout(G(r) out) = – dG(r)/ds out = – (dr/ds) dG(r)/dr out = – r H(r) G'(r) out (16)

Because the transverse acceleration applies in two of the perpendicular directions, Zak’s principle gives us:

 2 atransverse + aradial = 0 (17)

which in turns means that:

 H(r) (2 G(r) + r G'(r)) = 0 (18)

It makes no sense for H(r) to be zero, so we assume the second term is zero and solve immediately for G(r):

 G(r) = M / r2 (26)

This is just the Newtonian inverse-square formula for gravitational acceleration. Substituting this into (14) gives us:

 r2 H(r) [ 2 H(r) + r H'(r) ] = 1 (14')

This equation is solved by:

 H(r) = 1 / r (27)

This is the value for H(r) that makes space (though not spacetime) perfectly flat, because (2cg) now describe the way the directions east, north, and out change in ordinary Euclidean geometry.

Using (26) and (27) to evaluate the tidal accelerations in the Splinter, we have:

 ashomal = – M / r3 (29a) ararb = – M / r3 (29b) Asard = 3M / r3 (29c) Asard / ashomal = – 3 (29d)

This all works out nicely ... but we’ve simplified things by ignoring the fact that there’s actually a more general solution to (14'):

 H(r) = (1 / r) √(1 + D / r2) (27')

Using a non-zero value for D here would mean that space would no longer be flat, so this isn’t what would normally be described as a Newtonian solution, but the spacetime does still possess a universal time coordinate.

With this pseudo-Newtonian solution, the formula for the tidal ratio turns out to be:

 Asard / ashomal = – (3 + 2 D / r2) (29d')

With a negative value for D, this could be used to describe a spacetime where the magnitude of the tidal ratio is less than three. Had the Splinterites considered this family of solutions, they might have taken even longer to find the true geometry. But one conceptual point in favour of the relativistic Schwarzschild geometry (and shared by the Kerr geometry) is that it forces the tidal ratio to have a magnitude less than three, whereas this family of solutions introduces a parameter, D, that could as easily be chosen to give a ratio greater than or equal to three.

In other words, we need to “fine tune” the pseudo-Newtonian model to get the correct kind of tidal ratio, whereas in the relativistic models this property emerges automatically.  Incandescence / Deriving Newtonian Spacetime Geometry / created Friday, 3 April 2009 / revised Sunday, 5 April 2009