Relativity and Cosmology

The 'Twin Paradox' of relativity is around a century old, but it still gets a lot of attention on many forums. I have also written on it a few times in this Relativity and Cosmology Blog, e.g. here and here. An interesting variant of the classical twin-paradox is to consider the situation at the halfway stage, which may be labeled the 'Half-Twin Puzzle' of relativity.

On the right is a graphic of what I call the 'RGB scenario', an all-inertial variant of the classical twin paradox. It differs only in that the 'away-twin' (Red) does not make a quick turn-around to come home again, but a third inertial observer (Blue) does a 'R/B flyby' in the opposite direction and performs the return leg. Both speeds are 0.866c relative to Green, giving a Lorentz factor γ = 1/√(1-0.866²) = 2 in both directions. This means that the 'moving' clock appears to tick at half the rate of the 'stationary' clock.

At the R/B flyby, Blue sets her clock to read the same as Red's clock (2 units) and since they are both inertial, presumably their clocks will tick at the same rate relative to Green. When Blue later passes Green, they compare clocks first-hand and conclude that Green has aged 8 units, double the sum of Red's and Blue's aging during the two halves of the test (a sum of 4 units). This is standard, verified relativity and not disputable.

The 'half twin puzzle' is where we simply ask the question: at the halfway (R/B) flyby event, which twin (Green or Red) has aged less since the original (G/R) flyby event? Here are some issues, lurking in very subtle, perhaps confusing logic. To understand them, we must look at the full GRB scenario, after which Green, Red and Blue can hold a teleconference and compare results. They will conclude that the proper time that has elapsed for Red until the R/B flyby (from 0 to 2) equals the proper time that has elapsed for Blue from the R/B to the B/G flybys (from 2 to 4). This is inevitable, since Red and Blue flew the same distance at the same speed relative to Green.

Green, Red and Blue will further agree that Green's elapsed proper time between the G/R and G/B flybys was 8 units, as measured by Green's own clock. It seems logical that between the G/R and R/B flybys, Green "really aged" 4 units while Red "really aged" only 2 units (Green's dotted line of simultaneity). After all, Red and Blue each recorded half of the 4 units, so how can Green not record half of the 8 units at the halfway point?

The first apparent paradox is this: since twins Green and Red were both stationary in their own inertial frames for the duration of the test, they should be equivalent and the one cannot age more (or less) rapidly than the other over the duration of this test. To argue that they age differently will mean giving preference to one inertial frame over another. Since the 2 units for Red is a given, it means that Green should also have aged only 2 units.

Secondly, according to special relativity, Red has the 'right' to view Green as moving at v = -0.866c, with a Lorentz factor of γ = 2. So per Red, Green should have aged only 2/γ = 1 unit between the G/R and R/B flybys (Red's dotted line of simultaneity). Thirdly, to make matters even worse, according to Blue's dotted line of simultaneity, Green's clock should have read 7 units when the R/B flyby occurred. After all, Green flew at 0.866c relative to Blue and while Blue aged 2 units, Green should have aged only 2/γ = 1 unit.

How do we reconcile these apparently paradoxical conclusions, i.e., did Green age 1, 2, 4 or 7 units during this 'Half-Twin test'? We will let interested readers puzzle a little, before offering attempted resolutions of the paradoxes.

-J

Imagine the cosmos some 200 million years after time zero. If the cosmos is presently in its prime (like a 30-something years old person), then in human terms the cosmos was about 6 months old at that stage. Very young, but probably a very violent place, with supernovae, not too far apart, going off everywhere. The stars of that time were apparently very hot and heavy and hence very short lived; also, their deaths were mostly violent supernovae. On the same human scale as above, few of them lived longer than 2 weeks - extreme infant mortality!

During those supernovae, lots of matter were thrown out, anything from elementary particles, up to atoms of some metals. Most of the matter were probably caught up in later generations of the star- and planet-formation, but some particles may have survived. For the purpose of this exercise, let's assume a few atoms made it out of that distant region and all the way to our region of the cosmos.^[1] Let's also suppose that they were ejected with a velocity of 0.999c relative to the overall cosmic frame of reference,^[2] the CMB - like high speed 'bullets' from afar.

What would the velocity of such bullets be when they eventually reach our neighborhood? And how long would they have taken to get here? If the cosmos was only 200 million years old and we can observe light from that era today, the light will be red-shifted to z = 19. Calculations show that when the light was emitted, the source had to be 1.8 billion light years from our region.^[3] Let's take this as the original proper distance of the bullets from our neighborhood.

At that time, the cosmos was expanding extremely rapidly, but slowing down under the dominant gravitational force of all the matter in the cosmos. At first, those bullets, traveling at 0.999c, could not make any headway towards us; the expansion took them farther from us for quite some time. However, after the expansion rate diminished somewhat, the bullets did indeed start to come towards us. So, how long would they have taken and at what speed would they arrive?

If there was no cosmic expansion, it would have been rather simple: 1.8 billion light years at a speed of 0.999c would have taken 1.802 billion years and they would have arrived at the original 0.999c. However, with the observed expansion profile of the cosmos, it becomes quite tricky, for two reasons: (i) the bullets were taken away at first and (ii), their speed relative to the CMB would have dropped (or decayed) to a lower value.^[4] In the end, the bullets would take almost 18 billion years to reach our neighborhood (~4 billion years from now) and they would arrive at the "sedate speed" of 0.64c.

Enough of these ramblings - let the graph on the right rather 'speak for itself'. The bullets were ejected at a proper distance of 1.8 Gly (the start of the dark blue D_proper curve) from where our Galaxy would form shortly thereafter (somewhere on the horizontal axis, D=0).

You can clearly see how the bullets first moved farther away from our quarters and then later started to get closer to us. They will take almost 18 billion years to reach us - where the D_proper graph intersects the horizontal axis, some 4 billion years into our future.

The green V_peculiar curve shows a bullet's momentum decay. V_peculiar starts at -0.999c and then gradually decreases in magnitude, until it passes us at around -0.64c. The reason for this decay is discussed in^[4], but in a nutshell, as the cosmic balloon expands (R increases), any free particle's angular momentum^[5] around the hyper-center remains constant; hence, it must lose some surface momentum. The light blue curve represents the Hubble velocity (V_Hubble = H x D_proper), which is the recession speed for the bullet's (then) proper distance. The red curve shows the proper velocity (V_proper = d(D_proper)/dt) of the bullet over time.

'Sexy' as these curves may appear, they represent some deep truths about the standard cosmological model. If you understand them, you could well be reckoned as an amateur 'cosmic ballistics detective'. If you don't, here's your chance to learn something more about that interesting trade - just ask...

Notes:

[1] Earth (and our galaxy) could not have been around yet, but that's a long story...

[2] It is known as the local rest frame w.r.t. the Cosmic Microwave Background (CMB), meaning an inertial frame in which the CMB has the same average temperature in all directions.

[3] These values are available from any good cosmological calculator. I have an up-to-date (2009) cosmo-calculator on my website Relativity-4-Engineers.

[4] See Blog entry Particle momentum decay. The equations can be found in my Blog entry on Tethered Galaxies.

[5] This angular momentum is equivalent to the so-called 4-momentum of the particle, which is constant during expansion.

After more than 2 years of sweat with two unexpected errors (misalignment torques and a varying polhode motion), the Stanford team now announced a result in reasonable agreement with Einstein's Theory of General Relativity. The 'geodetic precession' (space curvature near Earth) is relatively large and was clearly visible two years ago. The 'frame-dragging' (by Earth's rotation) is much smaller and was originally masked by the two errors mentioned.

They now wrote: "The accuracy of the GP-B experimental results has improved seventeen-fold since our preliminary results announcement at the American Physical Society annual meeting in April 2007. At that time, only the larger, geodetic effect was clearly visible in the data. Over the past two and one half years, we have made extraordinary progress in understanding, modeling and removing three Newtonian sources of error—all due to patch potentials on the gyroscope rotor and housing surfaces. The latest results, based upon treatment of 1) damped polhode motion, 2) misalignment torques and 3) roll-polhode resonance torques, now clearly show both frame-dragging and geodetic precession in all four gyroscopes (see figure at top right).

"The figure at lower right displays the science estimates as of September 2009, with the gyroscopes analyzed individually and combined. The estimates are indicated with colored "X"s, and the statistical uncertainty associated with each estimate is plotted with a corresponding colored ellipse.

"The combined four-gyro result in the figure gives a statistical uncertainty of 14% (~5 marcsec/yr) for the frame-dragging (EW). The gyroscope-to-gyroscope variation gives a measure of the current systematic uncertainty. The standard deviation of this variation for all four gyroscopes is 10% (~4 marcsec/yr) of the frame-dragging effect, suggesting that the systematic uncertainty is similar in size (or smaller) than the statistical uncertainty."

-J

The cosmic balloon analogy is normally used to illustrate a cosmos with a perfectly homogeneous mass distribution. In such a case the balloon is a perfect sphere. It is possible to loosely illustrate inhomogeneities by means of indentations on the balloon. For a given 'time slice' it is however possible to rigorously show the shape of the balloon for a single static black hole embedded into an otherwise homogeneous mass distribution (evenly spread 'cosmic dust').^[1]

A Monster

In order to picture the embedding on a cosmic scale, it is necessary to use a monstrously large black hole. Fig. 1 (right) shows the profile of a cosmic balloon, 'dented' by a single black hole with a mass of 0.01% of the mass of the otherwise homogeneously spread matter, the total mass of the hyperspherical cosmos. (The 'squiggles' on the curves are not inhomogeneities, but just artifacts of the image conversion processes.)

The black circle represents the original homogeneous cosmic balloon with hyperradius R = 100 Gly. The blue curves show the 'dent' caused by this monster black hole. The gravitational effect of such a huge mass concentration will essentially reach around the entire hyperspherical universe.^[2]

At the same time, the black hole's central singularity may reach all the way down to the original (hypothetical) singularity of the big bang. The 'may' and 'hypothetical' qualifiers are used because we do not know what happens at any singularity - just that the math of general relativity breaks down there. We require a quantum theory of gravity, which we do not (yet) have. It is nevertheless interesting to speculate that all black holes may be connected to the original big bang singularity...

Be that as it may, the hyperspace vectors of dust particles outside the singularity are a little bit easier to treat - we temporarily 'unfreeze' the situation now, so that the balloon can expand. A homogeneous expanding balloon would have driven free (static) particles outward on radial paths, i.e., with hyperspace vectors normal to the unperturbed (black) surface. It is reasonable to argue that spots on the dented (blue) surface of the expanding balloon will be driven outward with hyperspace vectors that are slanted towards the singularity. This means that free (static) particles at those spots will move closer to the singularity. Gravity 'explained'! Well, not quite. This is not the full picture of gravity - for that we must also consider the time dimension, which is outside of the scope of this article.^[3]

Event Horizons

Singularities do not exist without event horizons, those 'one-way valves' that allow things in but not out of black holes. The event horizon radius of this huge black hole would have been at least 300 million light years, which is almost 1% of the proper radius of our observable universe. This is comparable to the size of superclusters and quite unrealistic, but good for visualization.

Fig. 2 (right) shows the (almost) vertical 'throat' of the black hole's event horizon. 'Vertical' here actually means going inward radially, because this is still a segment of a hypersphere. The radials converge at the origin (0,0), way down below the segment shown here.

In Cartesian space, the event horizon radius is given by r_H = 2GM/c², where M is the conventional mass of the black hole. In spherical coordinates, the event horizon spans an angle θ_H = r_H/R, where r_H is measured along the unperturbed hypersurface and R is the radius of the unperturbed hypersphere (100 Gly for this time slice).^[4]

Does this mean that the event horizon radius of a black hole increases due to cosmic expansion? No, not unless matter is 'swallowed' by the hole. For constant black hole mass, r_H remains constant, while R increases. This means that θ_H = r_H/R deceases over time. Assuming a perpetually increasing expansion rate, the gravitational effect of a black hole on the global scheme of things will dissipate.

One final, intriguing thought: jumping into a black hole may take you 'back' to the original big bang singularity. Whether that also means being transported back in time is not quite clear. From Einstein's equations, it looks more like you will experience 'imaginary time' (whatever that may mean) until you are first stretched and then crushed...

Jorrie

Notes

[1] The 'given time slice' or 'snapshot' means that for simplicity, we ignore the complication that some of the 'homogeneous cosmic dust' will swirl into the black hole and so create a large void around the hole. We simply place the black hole and then 'freeze' the situation, with the homogeneous conditions still intact around the hole.

[2] If the minimum size of a hyperspherical universe is worked out with R_min = 100 Gly, the total mass of baryonic plus dark matter comes to M ≥ 10²⁵ Solar masses (Sols). This is just a minimum mass - we do not know how large the 'real cosmos' is. It means that the (0.01% of total mass) black hole must have a mass of at least 10²¹ Sols - improbable, but still...

[3] I have written a number of Blog articles on black holes, starting at this Blog entry. The effect of time is shown graphically there. More depth can be found in Chapters 4, 5, 6 and 7 of Relativity 4 Engineers. The present Blog article deals only with black holes in relation to expanding space.

[Update: I wrote above: "It is reasonable to argue that spots on the dented (blue) surface of the expanding balloon will be driven outward with hyperspace vectors that are slanted towards the singularity." This does not mean that any 'spot' on the balloon, outside the event horizon, moves closer to the singularity in proper space; it is only the angle θ (see [4] below) that becomes smaller, while spots may remain at the same proper distance, or move away from the BH singularity.]

[4] Some equations used in this article:

The radius of the 'dented' hypersphere (at angle θ from the singularity) and outside the event horizon, can be reworked from 'Gravitation' by Misner, Thorne and Wheeler (MTW), eq. 23.34b, where they give the 'embedding lift' for Schwarzschild space:

z(r) = [8M (r - 2M)]^½ + constant ---------- (a)

where MTW's M is a normalized mass, equating to the conventional (SI) mass by GM/c² (they use geometric units where c=G=1). If we convert this equation to polar coordinates and our usual units, we get:

R' = [8GM/c² (Rθ - 2GM/c²)]^½ + constant ---------- (b)

where R is the 'undented' hyper-radius and the constant is chosen to give R' = R when θ = Π, where ΠR gives the proper spatial radius of the present universe. From present observations, we can deduce that the present R ≥ 100 Gly, but we do not know the actual value.

It is clear from (b) that when Rθ < 2GM/c², the local hyper-radius R' goes imaginary, giving the event horizon angle in hyperspherical coordinates:

θ_H = 2GM / Rc² ---------- (c)

This event horizon angle is obviously relative to the center of the hypersphere, which seems to support the idea that, at least for a hyperspherical universe, the black hole's central singularity resides at the original BB singularity. In a 'flat' or 'open' cosmos, the original singularity was 'everywhere' (as infinite density) and the event horizon of black holes do not point to any specific hyperspace spot, but this is outside of the scope of the cosmic balloon analogy.

Erratum

While writing this article, I noticed that Figure 1.5 (page 28) of Relativity 4 Engineers contains a typo, as indicated on the right. That r inside the square root of z(r) must be above the line, i.e.:

z(r) = √[8 g_tt M_bar r]

where g_tt = 1-2M_bar/r and M_bar = GM/c² in units meters. It is fairly obvious that with r below the line, the curve cannot have a positive slope.

For those readers who have the eBook, you can download an annotated page to patch in, using Adobe Acrobat Professional 7.0 or later (or equivalent). The page contains notes to 'correct' the error (it is a little painful to regenerate the single page from scratch). Otherwise, drop me a CR4 email and I'll send you a link for re-downloading the complete annotated eBook (including the commented page and some extra notes on other pages, clarifying a few things a little better).

-J

In this final 'application' of the cosmic balloon, the effect of expansion on 'rigid bodies' will be investigated. Two buttons are separated by proper distance D on the surface of the cosmic balloon. We tether the two buttons to each other by means of a semi-rigid tether and then inflate the balloon.

Gravitational tidal forces^[1] attempt to either compress or stretch objects in a gravitational field. It has to do with the different spacetime curvatures at different parts of an extended object. Cosmic tidal forces have similar effects on extended objects, but it has to do with the expansion dynamics and not directly with gravity.

Cosmic tidal forces

Fig. 1 (right) shows the cosmic situation - two galaxies (red buttons) on the surface of the balloon, with a tether keeping them at a constant proper distance (D) apart. To determine the cosmic tidal force (if any), we measure the force on the tether, using strain gauges or some other practical method, taking into account that the forces can be either stretching or compressing.

It is reasonable to expect that cosmic tidal forces will depend on the inflation profile of the balloon. As it happens, it feels intuitively correct (and it is in fact easy to show) that a constant inflation rate (dR/dt = constant) causes no tidal force on the tether (apart form a transient when we start the inflation). With the balloon in uniform expansion, the buttons will essentially move inertially across the surface of the balloon. This is equivalent to a phase of the actual universe, around half of the present age, when the expansion rate was essentially constant (see this post).

If we blow up the balloon with a constant gas flow rate, the expansion rate will slow down over time and free buttons would tend to move towards each other, because they would have had initial momentum towards each other. The semi-rigid tether will prevent that and we should measure a compression force on the tether. This is equivalent to a phase in the actual universe before 7 Gy age, when the expansion rate was decreasing under the dominant matter density.

If we control the gas flow rate so that the expansion rate of the balloon increases over time, equivalent to the now dominant vacuum energy phase, one would expect the buttons to separate. Again, the tether will prevent the separation and we should measure a stretching force in the tether. This is all rather intuitive, so let's try to quantify these forces on a cosmological scale.

Our Local Group

Rather than working with billions of light years (as usual), let's keep it 'local' and make the tether 10 million light years long. This is about the diameter of our Local Group of galaxies (Andromeda, the Milky Way, Triangulum and some 27 dwarf galaxies). It will be interesting to find the magnitude of the cosmological tidal force on this scale (ignoring the local gravitational effects at first). Fig. 2 (right) plots the accelerative tidal force exerted on the 10 million light years long tether. The timescale starts some 10 billion years ago and ends far into the future. For the relevant equations, see note [3].

For the first 7 Gy, the tidal force was negative (compressing) and then it became stretching, as expected. At the present time (~13.6 Gy), the stretching tidal force over 10 million light years has an order of magnitude (OOM) 30 femto-g,^[2] or 300 femto-Newton per kg mass. This is small, really small. However, when compared to the gravitational acceleration at the edges of the Local Group, it may not be so negligible.

Gravitational acceleration

Let's get an idea of how much effect the cosmic expansion may have on the structure of our Local Group of galaxies. It has a total mass of ~ 1.3 × 10¹² Sols, or M ~ 3 x 10⁴² kg^[4] and a radius r ~ 5 x 10²² meter. If we include dark matter (at ratio 6:1), it puts the mass of the Local Group at M ~ 2 x 10⁴³ kg. The rough gravitational acceleration of a particle just outside of the Local Group is: a ~ -5 x 10^-13 m/s², (-50 femto-g).^[5]

If my calculations are correct,^[6],[7] this is of the same OOM as the 30 femto-g tidal stretching force at that distance. It may mean that this very feeble tidal stretching force has some influence on the structure of things like galactic clusters - and even more so on super-clusters. It probably only means slightly larger orbits for the galaxies right at the edge of clusters. Even so, never again 'sneeze' at an acceleration of a few tens of femto-g!

I have made a table of our family of gravitationally bound structures, using my equations and data from ref [7]:

I have included dark matter in the masses of the structures, although it does not make a huge difference. The last three columns are: g_acc is gravitational acceleration of a particle at radius r from the gravitational center of the structure (in femto-g), t_acc is the tidal acceleration at that same place; r/r_crit is the ratio of r to r_crit, the 'critical radius' where the two forces precisely balance each other.^{[3] eq. (6)}

It is clear that the Milky way is gravitationally bounded very well, our Local Group cluster is marginally bounded, the Virgo Cluster is bounded very well (it has a great density of galaxies) and the Virgo Supercluster is marginally unbounded. Above the size of superclusters, we get the 'filaments' and 'great walls', which are not gravitationally bounded at all.

[Edit: Also see reply to Roger below (effect of cosmic tidal forces on galaxy formation).]

Jorrie

Notes:

[1] Gravitational tidal forces are 'real', coordinate independent forces, measurable by means of strain gauges or accelerometers.

[2] It is convenient to work in nano-, pico- or femto-g, because my calculations output the acceleration in units of 1/Gy, which is roughly one nano-g. An acceleration of 1 ly/y² is roughly one g (9.8 m/s²), because the radius of spacetime curvature at Earth's surface is roughly one light year. One nano-g is equivalent to a gravitational radius of curvature of one Gly. (Can you think why this must be so?)

[3] Some relevant cosmological equations (read with the equations of the previous Blog):

The acceleration of expansion:

d²a/dt² = a H₀²(Ω_Λ - Ω_m/(2a³)) ---------------(1)

(From Peebles 1993, eq. 13.3, p 312, where one must read H₀ as implying the usual H₀/978 Gy^-1, in order to be compatible with our units convention. The factor Ω_m/(2a³) - Ω_Λ has historically been called the deceleration parameter q. It has the present value q ~ -0.6 (negative, because the deceleration is negative, i.e., it's an acceleration ).

The tidal acceleration over a fixed distance D follows exactly the same profile, but scaled to D. Any change in a causes the same % change in the proper distance between two free particles that were momentarily at rest relative to each other. Hence:

d²D/dt² = D H₀²(Ω_Λ - Ω_m/(2a³)) ---------------(2)

At the present time, with a=1, it means

d²D/dt² = D H₀²(Ω_Λ - Ω_m/2) ---------------(3)

Assuming the present values: H₀ = 72/978 Gy^-1_, Ω_Λ = 74% and Ω_m = 26% of the critical density, this works out to a present proper tidal acceleration of:

d²D/dt² = 0.00334 D Gy_^-1, ---------------(4)

or about 3.3 femto-g per Mly of proper distance (about 10 femto-g per Mpc).

If we work with radius r instead of with diameter D, we can find the critical radius of spherical structures (r_crit) where tidal and gravitational forces have the same magnitude. From equation (3) we can write:

GM/r_crit² = r_crit H'₀²(Ω_Λ - Ω_m/2) ---------------(5)

where H₀ must be expressed as inverse seconds (s^-1) in order to keep the (SI) units (m/s²) of the two sides compatible, i.e., H'₀ ~ H₀ /( 3 x 10¹⁹) s^-1. (Coming from 1 Gy ~ 3 x 10¹⁶ seconds, together with the usual H₀/978 conversion to Gy^-1). Hence:

r_crit = [GM / (H'₀² (Ω_Λ - Ω_m/2))]^1/3 meters ---------------(6)

We can 'cosmologize' it to (say) Mly: divide by 10²², roughly the meters in a million light years.

[4] http://en.wikipedia.org/wiki/Local_Group

[5] The gravitational acceleration is obtained as: a ~ (-6.67 x 10^-11 ) (2 x 10⁴³) / (5 x 10²²)² ~ -5 x 10^-13 m/s², or a ~ -50 femto-g. This is not strictly correct for a very non-homogeneous mass concentration like the Local Group, but the OOM should be about right.

[6] The only applicable reference that I managed to find so far is a quite technical paper by Gregory S. Adkins et. al (2006): 'Cosmological perturbations on local systems'. Their conclusion is that matter dominated phases do not perturb the orbits of local systems, but for the vacuum energy dominated phase they say: "However, the cosmological repulsion becomes more important for more extended clusters, and would make a contribution comparable to that of the gravitational term for clusters that are only marginally bound."

Unfortunately, they do not give quantified results (at least not in a form that I can understand).

[7] I have now found a very accessible paper: "On the influence of the global cosmological expansion on the local dynamics in the Solar System" by Matteo Carrera and Domenico Giulini (2006), which seems to confirm my approach. They use a different, more general approach than myself, but the results seem to be compatible.

-J