Relativity and Cosmology

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

The so-called 'tethered galaxy' thought experiment has created a lot of cosmic interest in the past. It is a rather complex issue with significant pedagogical value, but it is surprisingly easily and simply handled by the cosmic balloon analogy.

Button on a Balloon

The best balloon variant of the 'tethered galaxy' thought experiment has been suggested by StandardsGuy before. I use it with a slight modification here. Pick any spot on a partially inflated cosmic balloon's surface and attach one end of a tether (string) there. Call this spot the origin of the coordinate system. At the other end of the tether, attach a button lying frictionless on the surface - the button is now 'tethered' to the origin at a distance D from it.

Increase the inflation of the balloon from radius R₀ to R in time Δt, as in Fig. 1 (right). Assuming that the tether does not stretch with the skin of the balloon, the button will follow the red space-hyperspace vector, being pulled across the surface of the expanding balloon. Without a tether, the expanding balloon would have taken the button along the blue space-hyperspace vector and hence increased the distance from the origin at the Hubble velocity (H D). The starting angle between the blue and the red vectors represents a negative peculiar velocity (relative to the skin), although the button stays at a constant proper distance (D) from the origin - in other words, the tethered button has zero proper velocity. (These are important terms and I shall be asking questions later.)

The Curves

Now cut the tether, so that the button (galaxy) becomes 'untethered'. Wait some time and observe what happens to the proper distance of the button. We need no more than the 'decay of particle momentum' of the previous Blog entry to predict what will happen to the button relative to the origin.

It depends on how the balloon is being blown up. For simplicity, let's first control the gas input rate so that it keeps the expansion rate (dR/dt) of the balloon constant. Starting the simulation at the present age of the cosmos, the curves in Fig. 2 (right) result. For clarity, let's take things step by step, starting with the graph title.

(i) Case (0,0,0) means zero matter, zero radiation and zero vacuum energy density. This gives a constant expansion rate (dR/dt = constant) over all of time. Although not a realistic case, it is a good, simple starting point for understanding the various terms and dynamics.

(ii) The blue D_proper curve is essentially constant at a distance 0.5 Gly (note the y-scale Gly/10). It means that the button's peculiar velocity (across the balloon surface) towards the origin is canceled by the balloon's expansion that is trying to carry it away from the origin. The slight apparent drop over time is just a numerical integration error (only 1000 steps over the 500 billion year time span were used).

(iii) The V_Hubble curve drops down in a typical inverse of time fashion, because for a constant expansion rate the Hubble constant (H) changes proportional to 1/t. Think of a very large balloon that expands at the same rate (dR/dt) than a very small balloon. Since the Hubble constant is recession rate divided by distance, the large balloon will have a much smaller H.

(iv) V_proper is zero in this case, because the proper distance is constant. Proper distance and proper velocity are measures of the instantaneous distance and recession speed of an object. It is as if we use a tape measure and synchronized clocks to find the distance and separation rate between two points on the balloon's surface over a very small time interval.

(v) The green V_peculiar curve represents the (negative) velocity of the button relative to the local balloon surface. The conservation of angular momentum (discussed in the previous Blog) causes the curve to approach the time axis asymptotically. Because the expansion rate is constant in this case, V_peculiar also changes proportional to 1/t. V_peculiar + V_Hubble = V_proper = 0 in this case, so V_peculiar is a mirror image of V_Hubble around the time axis. This is not generally true for more realistic cases, though, as will become clear later.

I suggest that interested readers first 'digest' this information and ask questions as required, before we move on to slightly more complex cases. These Blog posts are necessarily compact, cryptic issues for discussion, so do not be afraid to ask questions. There is no thing like a 'dumb question' - the only 'dumb thing' is not to ask!

Due to some questions asked, here is the (presently) realistic case.

Realistic Expansion

For the more realistic case with present matter energy 26% and vacuum energy 74% of the critical energy density, the balloon is blown up with an accelerated expansion (dR/dt gets larger with time at present). In Fig. 1 above, the distance D remains constant. Here it is to be expected that the faster-growing expansion will drive the button farther from the origin, despite its peculiar velocity towards the origin. V_peculiar now decreases, while V_Hubble quickly starts to increase.

Fig. 3 (right) shows the same curves as for Fig. 2 above, but for the realistic scenario, again starting at the present cosmic age. What is surprising is how "quickly" (in cosmological terms) the proper velocity (red) of the button starts to 'follow the Hubble flow', i.e., how quickly the peculiar velocity of the button decays to near zero.

Some other salient points on this chart:

(i) The V_Hubble line first dips a little, because the Hubble constant initially decays faster with time than the away movement of the button happens. In this scenario, the Hubble constant drops from ~ 74 km/s/Mpc and it settles to a constant ~ 63 km/s/Mpc in another 13 billion years or so. Hence, the Hubble constant eventually becomes a true constant.

(ii) V_peculiar of the button (relative to the balloon skin) remains negative, i.e., towards the origin. The increasing expansion rate does however quickly carry the button farther from the origin. Conservation of angular momentum relative to the center of the ever larger-growing balloon decays the peculiar velocity to zero over time (approaching zero as time tends to infinity).

The above case untethers the galaxy at the present cosmic age. The curves become a bit more complex if the untethering is done much earlier, as shown below.

Multiple epoch case

Here we postulate some 'early universe astronomer' that untethered the galaxy when the cosmos was less than a billion years old. Matter density still largely dominated the cosmic expansion (until vacuum energy took over) and the expansion rate first decreased and later increased. It results in the quite complex (but very interesting) curves of Fig. 4 (right).

At cosmic time t = 0.2 Gy after the BB, the tethered galaxy was at D = –0.1 Gly.^[1] At that stage, the expansion rate was decreasing under the dominant (99.96%) influence of matter density of the time.

The Hubble velocity at D=-0.1 Gly was a whopping –0.335c and hence the peculiar velocity of the tethered galaxy at that time was 0.335c towards the origin. But, due to the decreasing expansion rate, the Hubble velocity quickly diminished for that proper distance and hence when untethered, the galaxy started to 'fall' rapidly towards the origin. It 'fell through' the origin at some 1.5 billion years and then continued to move in the positive D direction.

The proper velocity (red) started at zero (because the galaxy was tethered). When untethered proper velocity increased rapidly until the galaxy passed through the origin and then the proper velocity started to decrease – that is for as long as matter density dominated and the expansion rate slowed down.

At around 7 Gy, the acceleration of expansion caused by the constant vacuum energy density more or less balanced out the deceleration of expansion caused by the decreasing matter density. During this time the expansion rate remained more or less constant and the proper velocity of the galaxy also remained constant.

After 8 Gy age the vacuum energy started to win the 'tug-of-war' and the expansion rate started to increase (and so did the proper velocity of the galaxy). During all this, the peculiar velocity (green) continuously decayed and will keep on doing that. The proper velocity eventually joins the Hubble velocity, as before.

Jorrie

Notes:

[1] The tether here had to be a solid rod, not a string, because for a decreasing expansion rate, the tether must actually have keep the galaxy from 'falling' towards the origin. This is essentially a 'cosmic tidal force' at work. More about that in a future Blog post.

[2] Here are some of the equations used (for those who just cannot live without them). Actually, it's good to have a compact set of reference equations for the curves (for my own selfish purposes).

V₀ = -H₀D₀ ---------------(1)^[a]

ψ₀ = D₀/R₀ ---------------(2)

L = R₀ V₀/√[1-V_H₀²/c²] = constant ---------------(3)

V_peculiar = L/√[L² + R²] ---------------(4)

D = D + R ψ + V_peculiar Δt ---------------(5)

ψ = ψ + V_peculiar Δt/R ---------------(6)

H = H₀ √[Ω_m/a³ + Ω_v] ---------------(7)^[b]

da/dt = a H ---------------(8)^[c]

V_Hubble = H D ---------------(9)

V_proper = V_Hubble + V_peculiar ---------------(10)

Footnotes:

[a] V₀ of Eq. (1) is the initial peculiar velocity of the button and is specific to the tethered galaxy scenario.

[b] H is the time varying Hubble constant, also sometimes denoted H(t).

[c] a is the expansion factor, R/R₀.

-J

We have seen how cosmic redshift can be understood as photons conserving angular momentum by 'shedding energy' in terms of frequency as the cosmic balloon expands. Is it reasonable to think that particles simply shed speed in order to also maintain a constant angular momentum? So it seems.

Intro

Fig. 1 (right) shows a single massive particle (pink arrow) being shot along the balloon surface at a time when R=25 units. As the balloon expands, the particle maintains its angular momentum relative to the balloon center by traveling more slowly along the balloon surface. This is visible as the shrinking of the angle between the arrow and the radial lines, representing static observers on the surface. Instead of continuously spiraling out, as a photon would have done, the particle's angle to a local radial line eventually approaches zero. This means that over time the particle's surface velocity (momentum) decays to zero.

This effect is qualitatively independent from the expansion profile, be it decelerating, constant or accelerating expansion, as long as there is expansion and not contraction. This is a rather surprising result of cosmic expansion, because it looks like a 'drag' or 'friction' that the cosmic balloon surface (space) exerts on particle movement. However, when there is no expansion (a static universe), the effect disappears, ruling out that it is a friction. What is more, should the cosmic balloon shrink, the particle will gain speed relative to balloon surface, as it has to in order to maintain a constant angular momentum.

Analytics

In Galilean dynamics, it is the kinetic energy (½mv²) and the momentum (mv) of a particle that decay due to an expanding universe. The velocity v is also known as the 'peculiar velocity' in order to distinguish it from the 'Hubble velocity', which is the apparent recession velocity (relative to us) of a distant object that is static relative to the balloon surface.

As the radius R of the balloon increases, the v of the particle decreases in order to keep angular momentum L=mRv constant - at least in Galilean (low speed) dynamics. In relativistic (high speed) dynamics, things are not much different, just with the special relativistic Lorentz factor gamma (γ) entering the equation.^[1]

L = m R v γ = constant -------- (1)

where γ = (1-v²/c²)^-0.5. It is relatively easy to plot the curve of Fig. 1 from this equation, together with the expansion equations of Friedman, of course. Depending on the speed of the particle and the expansion profile, the particle path may be virtually straight, as in Fig. 1, or it may be curved. In the end, it always ends up going more and more radially, meaning with less and less momentum along the surface.

Figure 3.5 above is from a doctoral thesis by Tamara M. Davis, under supervision of prof. Charles Lineweaver, titled "Fundamental Aspects of the Expansion of the Universe and Cosmic Horizons".^[2] It shows the decay of the velocity of particles for various initial particle speeds, with light (v=c) the top curve. It is clear that the larger the kinetic (momentum) part of the total energy of a particle (see Eq. 2 below), the more it behaves like a photon.

Particle Redshift?

It is possible to analyze this case in terms of redshift of de Broglie waves.^[3] Following the thoughts in the previous Blog entry ("The linear momentum of a photon is given by p = h/λ, where h is the Planck constant") and Roger's suggestions on de Broglie waves, the following interpretation is very interesting.

The relativistic energy of a moving particle is given by

E = √[p² + m²] c² = √[(h/λ)² + m²] c² -------- (2)

where p is the momentum, m is the rest mass and λ the de Broglie wavelength of the particle. It seems like a "de Broglie momentum" (h/λ) is added vector-wise to the rest mass (m), giving the particle its wave-particle duality. Particle mass is presumed constant, so in order to keep angular momentum relative to the balloon center constant, the de Broglie wavelength of the moving particle must redshift more or less like for a photon.

Angular momentum of a particle moving on the balloon is given in special relativity by eq. (1) above as: L = m R v γ = constant.

But, according to ref. note [3]:

γ = h / (λ m v) -------- (3)

so we can also write:

L = R h / λ = constant -------- (4)

exactly as for a photon. If R is increasing, λ must be increasing by the same ratio in order to keep L constant. This is same as a velocity decrease in order to keep the relativistic angular momentum of the particle constant on the expanding balloon. The velocity equivalent to a specific λ can be extracted from Eq. 3, taking into account that γ is also a function of v.

Caution

One must always remember that the balloon analogy is just that, an analogy. It may help to get 'handles' on some of the puzzles of the cosmos, but it does not represent the real thing. As an example: the cosmos might be precisely flat, with no curvature, yet the behavior discussed here is apparently still present. How and why, I don't know.

Jorrie

Notes

[1] This is essentially the special relativistic form of Kepler's second law of planetary motion: "a planetary orbit sweeps out equal areas around the Sun in equal time intervals", where the equal time intervals are now measured in proper time of the moving particle. We do not need the general relativistic form here, because particles on the balloon feel zero net gravitational acceleration (the whole surface is at the same gravitational potential).

[2] http://arxiv.org/abs/astro-ph/0402278, a doctoral thesis by Tamara Davis, under supervision of Charles Lineweaver, page 50.

[3] http://en.wikipedia.org/wiki/De_Broglie_Wave

-J

Despite having apparently simple explanations, cosmological redshift remains one of the mysteries of the universe. Neither 'expanding space', nor Doppler shift offer a completely satisfactory answer. Here is a better one (perhaps).

What is cosmological redshift?

It is the observed phenomenon that distant galaxies glow at a redder and redder frequencies the farther out they are. Astronomers measure galactic redshift by comparing the absorption lines of certain elements in the light spectrum of the galaxy to the same light spectrum in the laboratory (or to the spectrum of our Sun). The redshift is then defined as the change in wavelength (Δλ) divided by the laboratory wavelength (λ₀) of the absorption line in question, i.e.,

z = Δλ/λ₀ = (λ-λ₀)/λ₀ = λ/λ₀ - 1 ----------- (Eq. 1)

where λ is the observed wavelength.

In the expanding cosmic balloon, this is very easily pictured in terms of the ratio of the radius of the balloon now (R₀) to its radius (R) at a certain time in the past, i.e.,

z = R₀/R - 1 ----------- (Eq. 2)

as shown in Figure 1 (right). Here R₀=100 and the red circles represent earlier values of R. It comes directly from Eqs. 1 and 2. It is not difficult to see why wavelength is inversely proportional to the radius. Or is it?

Figure 1 shows the balloon size from around last scattering of photons (the red dot at the origin, R=R₀/1089, representing the CMB) up to today (R=R₀=100). The red rings are not time based, but size based, at 25%, 50%, 75% of the radius (or circumference) of today. The picture is valid for any expansion profile, provided that there is at least some expansion (also called a perpetual expansion scenario).

Explanations

In the time that the CMB photons were in flight, the balloon expanded by a factor 1089 and the photon wavelengths were 'stretched' by a factor 1089, giving their redshift as: z = λ/λ₀ - 1 = 1088. Quite reasonable, it seems at first sight,^[1] but how can a photon's wavelength be stretched? A single photon does not even have a defined size, so stretching it is not conceptually very palatable!

Another reasonable explanation may be that it is just different frames of reference between transmission and reception of the photon and that the redshift is a coordinate transformation issue, resulting in Doppler shift. However, the two frames of reference may be moving away from each other at greater than the speed of light in vacuum (c), yet we still measure a real, finite redshift. How do we reconcile this fact with the Doppler shift equations of Einstein, which do not work for recession speeds equal to or larger than c?

So, where does cosmological redshift come from?

The balloon analogy offers a neat 'crutch' that makes the phenomenon a little more palatable. It appears as if photons conserve angular momentum as they travel along the surface of the expanding balloon, almost as if they go into a larger orbit around the center of the balloon.^[2] Since photons cannot shed speed in order to keep angular momentum constant in the larger 'orbit', they shed linear momentum in another way - by reducing frequency.

The linear momentum of a photon is given by p = h/λ, where h is the Planck constant. Angular momentum magnitude of a photon relative to the balloon center is given by: R p = R h/λ. If R is increasing, λ must be increasing by the same ratio in order to keep angular momentum constant. Increased λ is the same as cosmological redshift.

I suppose in the end it is not too important which 'crutch' you use - stretching of wavelengths, conservation of angular momentum, or even Doppler shift, as long as it is accepted that the received to emitted photon wavelength ratio (λ/λ₀) is the same as the expansion ratio (R₀/R) since emission.

Jorrie

[1] See the animation on the Webb Space Telescope site.

[2] This is equivalent to Kepler's second law of planetary motion, stating that a planetary orbit sweeps out equal areas around the Sun in equal time intervals. This is the same as the conservation of orbital angular momentum. The same considerations cause any (massive) particle with a velocity relative to the skin of the balloon to conserve angular momentum and hence slow down, provided that the balloon is expanding. The opposite (speed up) happens if the balloon is shrinking. More about that in a follow-on Blog post.

-J