MIRA 47
Christmas 1998





Editors Bit

The subject of cosmology is one in which all amateur astronomers can have an interest, how did the universe start and where will it end? How big is it and is anybody out there? These are questions on which many amateurs have their own ideas based on the amount of knowledge they have on this fascinating subject. So it was with some interest that when Dennis offered to send me a full answer to his Cosmological Conundrum, I decided to print it in full. This is the first time an issue of MIRA has been devoted to a single subject. A subject that is also one of the hardest to grapple with and one of the easiest to get one's brain in knots! The conundrum sounds simple at first reading, but upon reflection requires an in-depth knowledge of basic cosmology to answer correctly (I know because I got the answer right but with the wrong reason, which makes it wrong). The shear size of the subject overwhelms the human scale of existent in every way, but we can, now for the first time in our history, get a small clue to what the universe is actually like.
The author, Dennis Spratley, has spent a great deal of time researching this subject and has studied many of the research papers on the latest findings in cosmology. Many of these findings are contradictory to older work (and indeed contradictory to contemporary work also), so much more research is needed to clarify the issues. In many fields, specialists are backing one or other findings and many stake their professional reputation on a few sets of observations of a subject. Large telescope time is at a premium and does not allow astronomers to spend as long as they would wish for on their pet projects and researching subjects. With the new large Earth based telescopes built in the last 10 years, equipped with state of the art detectors and the Hubble Space Telescope we can see back to the earliest moments in the history of our universe. Back to the earliest times soon after its formation when the first stars formed. This is an incredible achievement which we can all feel proud of, and we are now nearly ready to be able to state with some certainty how it all started.
What all this new data will tell us about the formation of the stars and galaxies is now only just coming into the public domains, it is always a year or two behind as the research papers need to be written and printed. Of cause the arguments will get more involved and the subject will get more complicated as more results come in. But that shouldn't stop us from enjoying the wonder of discovery and the new ideas about the universe which will be forthcoming in the future.
I enjoyed Dennis's "Cosmological Conundrum" and answer and kicked myself for not seeing the reason why I was wrong until I read the answer. Oh well, I will have to think a little deeper next time. If anyone as any thoughts on this subject or disagrees with the answer please write in to the Editor so that I can print your letter in the next issue of MIRA. I look foreword to your replies.

Ivor Clarke, Editor





A Cosmological Conundrum 

By Dennis W. Spratley


Suppose that the Universe we inhabit is indeed one that will continue to expand for ever.  That is, we consider the situation of a model universe that expands without limit unlike the model that collapses back into a final singularity.  There are two classes of objects that we may have to account for in such a universe: those which are now "visible" to the astronomer and those which can never be "visible", the latter class being further away.  These two classes of objects are separated by a boundary, which may be moving: it can be static depending on the model we choose.  This boundary, called the event horizon, is spherically symmetric about us, but every observer in this universe has his own horizon.  For any observer in this universe staying say, in his own galaxy, events occurring beyond this horizon are for ever beyond his powers of observation.  Due to the cosmological expansion away from any observer objects cross his horizon at the speed of light.


Consider a quasar, Q, emitting, as is to be expected, its vast outpouring of radiation, a proportion of which, regarded as a stream of photons, is heading in the direction of the Earth situated at O, which we regard as the origin.  Due to the universal cosmic expansion, when this radiation is received at O it has a high red-shift.  Suppose that somewhere between O and Q is a galaxy G, which may be taken as a member of a cluster that is receding from O.  The quasar Q and galaxy G are seen with very small angular separation in the night sky.  Radiation from G also arrives at O with red-shift but this red-shift is less than that of the radiation from Q.  From the view obtained from G, however, the radiation arriving at G from Q will have less red-shift than that of the radiation arriving at O from Q.


This is the general picture.  Suppose now that Q is just on the event horizon of O which will be taken as static for simplicity.  That is, the event horizon for any observer in the space is at a fixed distance from him: consequently this is so for O.  No information about Q can be received by O: the light from the quasar has infinite red-shift on its arrival at O.  Now a photon PQ ejected from Q towards O leaves Q with no red-shift.  On its journey to O the photon PQ passes the galaxy G where it has a finite red-shift, and we may imagine that at this time a photon PG is emitted by G at exactly the same frequency as that of PQ and that both PQ and PG (an identical pair) head towards O.  Photons PQ and PG arrive at O together but we have arranged that on arrival PQ has infinite red-shift.  Since photons PQ and PG can be regarded as starting from G together with identical frequencies should not PG also have infinite red-shift on arrival at O and thus the galaxy G like Q, cannot be seen from O?  What is the answer and why?.


The trap into which the unwary fall thereby making this conundrum difficult is one of their own making.  The result is that they either fail to arrive at the solution or, if they succeed, they do so by false reasoning.  Being right by means of an erroneous argument is hardly satisfactory!


What is of the utmost importance is to begin by considering why the objects such as G and Q have red-shifted light in the first place.  It is, of course, due to the expansion of the universe.  Now the nature of the cosmological expansion is not the situation where both the galaxy G and the quasar Q are moving through a static space, but where all space itself is expanding or stretching outwards from any observer in the universe.  Consequently this includes the space between O and G, O and Q and G and Q.  This gives rise to the recessional speeds.


At O's event horizon the quasar Q is sitting in its "local" space. By this is meant, roughly speaking, the space around the quasar, a region of space where the expansion is very small, even negligible.  But between O and Q the accumulative expansion is such that Q is receding relative to O at the speed of light.  The same can be said for the galaxy G: it too sits in its "local" space where expansion is negligible but the accumulative expansion between O and G will give G a substantial recessional speed relative to O.  One more point of importance is that O itself sits in its own "local" space.


Now at O, G and Q within their "local" spaces the "local" speed of light is the same.  For example, the photon PQ will leave Q relative to Q at what we refer to as the speed of light: likewise Pwill leave G relative to G at the speed of light.  The same is true for O, of course: a photon created there will leave O at the speed of light relative to O's "local" space.


Just as important is the fact that photons entering the "local" spaces of O, G and Q from whatever source traverse the "local" spaces at the speed of light.  The speed of light in a "local" space or local reference frame is a universal constant.


However, when we consider the speed of PQ relative to G at the time when PQ is leaving Q the situation is different: the relative speed here is not the speed of light; it is less.  The reason is that between G and Q space is stretching and PQ is on an expanding "track". As PQ approaches G its speed relative to G increases and when PQ arrives at G it passes G... through G's "local" space... at the speed of light having finally overcome the expansion effect along its path from Q. So the travel-time of a photon is not determined by the distance between its initial and final points and the speed of light in any simple manner.


At this point we may consider what is happening to the frequency of the photon.  Since light possesses the so-called wave/particle duality it is easier to think of light as a wave motion.  Without recourse to mathematical analysis it is not difficult to appreciate that the stretching of space due to the universal expansion also stretches the wave along its path thereby increasing its wavelength.  This produces the red-shift.


This will relate total red-shift on arrival with travel-time.  If a photon starts out from O's event horizon heading towards O and leaving Q relative to Q's "local" space at the speed of light, with Q crossing O's event horizon relative to O at the speed of light it is not difficult to see that it will take the photon an infinite time to reach O! It has no speed relative to O.  In fact it is debatable as to whether one should discuss infinite red-shift if infinite time is interpreted as meaning that the photon never reaches O. But if PQ starts out just inside O's event horizon it can reach O after a tremendous time span and corresponding huge red-shift.


So the physics is thus. Q is on the edge of O's event horizon... just inside... and Q is "about" to cross this boundary at a speed relative to O of that of light. Q emits the photon PQ which leaves Q relative to Q... Q's "local" space... at the speed of light.  From O's point of view Q is just about to recede beyond O's horizon and O can just detect Q but by means of light arriving at O that began its journey in the distant past.  The photon PQ is just starting out: O cannot see it, of course!  This photon is almost "hovering" on the inside edge of O's event horizon, virtually held back but not quite, by the inexorable cosmic expansion, the accumulative space-stretching between O and Q.  Eventually the photon PQ will begin to make progress and increase speed relative to O and will finally arrive in O's "local" space at the speed of light.  But the journey will take a long, long time and PQ will have high red-shift on arrival.


Now, finally, we have to consider galaxy G which has yet to play its part.  This galaxy is also visible to O by photons that set out from G in the distant past but later than the photons that make Q visible to O. Galaxy G is receding from O and increasing its recessional speed relative to O, a speed which until G arrives at O's event horizon is always less than the speed of light. G is going to meet the photon PQ that is almost "hovering" relative to O.  Note that G is making an "outward" journey, i.e. away from O, and is swept along by the cosmological expansion.  On the other hand PQ is making an "inward" journey, i.e. towards O, and is hampered by the cosmological expansion.  The proportion of the total travel-time between Q and O that PQ spends in this "struggling" position close to O's event horizon, and therefore relative to O, is large and this results in G having ample time to finally reach the fringe of O's event horizon.  This is the crux of the conundrum.  When G and PQ meet it is at a point just a "little further" inside O's event horizon than was Q when it emitted PQ.  And, by the time G and PQ meet, Q has gone beyond O's event horizon.  The photon PQ has now entered G's "local" space which it crosses at the speed of light, relative, of course, to G.  The photon PQ is now joined by PG, the photon emitted by G and with identical frequency to that which PQ has in G's "local" space: these photons are a matched pair.  Soon G too slips over O's event horizon leaving the pair of identical photons PQ and PG continuing their struggle against the cosmological expansion and en route to O.  They do make the journey together after a tremendous time-span and with very large red-shift which is, of course, identical for both of the photons.  In the limit of Q being on O's event horizon when PQ is emitted, then the red-shift is infinite.


This is the complete answer to the conundrum: the "what is the answer", together with the "why of the answer".


The pitfall that the unwary dig for themselves is invariably the one caused by a simple and naive application of the Hubble Law.  Now in itself this well-known law says nothing about cosmological expansion: it is simply red-shift v. distance.  This can draw the "student" into attempting a solution to the problem whilst inadvertently assuming a static space where the red-shift is caused, not by the expansion of the space, but by the expansion of the material system through space radially outwards from any observer.  This leads to a further serious error which is the taking of the speed of light relative to an observer as being independent of the distance of the photons from the observer.  This applies only in the case where the "fabric" of space is unchanging, for example in the Special Theory of Relativity which is a "local" theory...the observer's neighbourhood... where space (and space-time) is static.  The Special Theory is thus a limiting case: hence a student applying this theory to the whole cosmos will make grave errors.


An incomplete answer to the conundrum can be obtained in the sense of merely solving the "what of the answer".  It is possible to show by application of a now quite well-known physical principle derived from the General Theory of Relativity that the photons PQ and PG must arrive at O with infinite red-shift (the "why of the answer" remains unresolved).


To do this we consider the result from the General Theory of Relativity that no information can "escape" from on and beyond an event horizon.  Now the photon PQ may be regarded as being a signal sent by Q to O.  This is the general situation.  But no signal carrying information can be sent to O if Q is on O's event horizon.  However, G's event horizon is beyond O's, so Q can send a signal to G: this is photon PQ, or rather one like it.  Also, G is inside O's event horizon so G can send a signal to O.  If the transmission of a signal from Q to G is made, i.e. PQ, or one like it, and on receipt of this signal G transmits a signal, PG, to O so that PQ and PG are now travelling to O together, it appears that although PQ will arrive at O with infinite red-shift PG will not, and we seem to have extracted information from the event horizon by using G to relay the signal.  This cannot be true because of the very nature of an event horizon.  Hence photon PG must lose its information content which is to be interpreted as PG having infinite red-shift at O.  Unfortunately, this does not lead to an understanding of why this is so because the above argument is made in static space. The expansion is the key to the complete solution.


This concludes the reply to the mathematical timid readers.  For those with a mathematical bent the following "model of a model" is offered in the hope that it gives further understanding.  The mathematics used is no more than A-level: well, no more than A-level 1955 vintage, for it may well be that this is now not part of any curriculum until first year undergraduate level.  One trusts not! I have called the "model of a model" the Photons in an "Elastic" Space.  It may appear to be somewhat frivolous but it does contain the complete solution to the conundrum.



The Mathematical Solution: The Photons in an "Elastic" Space.


The model here is simplistic.  Firstly, no gravitation effects will be considered and the stretching of space will be driven, effectively by a cosmological constant.  Secondly, no space curvature is included and the size of this model is indefinite.  And finally, time will be what is termed in the General Theory of Relativity: "cosmic time".  This will not be discussed but it means that the problem may be treated using a reference time that is the same at all points moving with the expansion.  Effectively this means that the problem may be examined non-relativistically as is required in this "model of a model". In the text by "time" is meant "cosmic time".  Like Newton we can imagine for convenience... but this is certainly not true!... a "cosmic-clock" defining time.


The only property that space has is that at any point within it, say at the location of an hypothetical observer, O, all objects surrounding this point... galaxies, quasars etc... are receding from it and that the further the object is from the observer, the faster is the velocity of recession.  We may speak of recessional velocity because it is directed radially outwards: no transverse motion is considered.  Now this expansion is caused by the stretching of space as if it possesses "elastic" properties: the objects do not move through space but with it.  The exception to this is that light, because of its very nature, does move through space but at the same time moves with the expanding space.


Suppose that we select any point, P, at a radial distance denoted by r, from the observer O.  There does not have to be an object at this point.  The point will recede so r will vary with time: we write r = r(t).  At the point there will be a recessional velocity which increases as r increases.  We will choose a very simple law relating this velocity, v, to the radial distance, r.

We have



where H can indeed be regarded as the Hubble Constant.  Actually, current literature is now referring to H as the Hubble Parameter which applies to most cosmological models and is correct usage because, in general, H does vary with time.  However, in equation (1) H is a constant throughout all space and time.


We can eliminate the velocity v by writing v = dr/dt where dr is now to be regarded as the amount by which r is stretched in a time dt. Consequently,



This simple equation may now be used to derive some interesting and important consequences. Consider another observer, O', who is situated along the line joining O and P, and whose radial distance from O is, at a particular time... nothing escapes the expansion..., given by r' = r'(t).  Thus O sees O' receding at a velocity v' = dr'/dt caused by the expansion, where


as is to be expected. And it follows, quite trivially, that O' will observe the point P to be receding from him at a velocity given by H(r - r')... assume r > r' for all time... which may be obtained by subtracting equation (3) from equation (2)



If the observer O' is at the point P then r - r' = O and as is to be expected he does not see the point P receding.


The next step is to consider a photon travelling radially inwards from P towards O' and O.  The photon's radial distance at the instant of time considered and as measured from O is say r.  Now if there was no expansion with all observers in fixed positions the velocity of the photon moving through space towards O would be - c. We may write



the minus sign arising because dr/dt has to be negative as r = r(t) is decreasing with time.  The symbol c is the speed of light which, as is now very well known, appears in the most famous equation, E = me2.  If now the cosmological expansion is taken into account equation (5) is modified to



We regard r in equation (6) as the distance of the photon from O, not the distance of a point.  This change is important.  Also, dr/dt is the velocity of the photon relative to O.  If we put r = 0, that is, the photon is located "with" the observer O, then the photon's velocity is simply - c.  This is what is meant above by a photon in a "local" space moving at what we call the speed of light.  Since O is arbitrary this is true throughout all space including galaxy G and quasar Q.


Another immediate and very important consequence of equation (6) is that as r is increased from zero there comes a value for r, say rH, for which -c+ HrH = 0. This gives



which is the distance, rH, of O's event horizon. Any increase in r above rH will make dr/dt always positive for O, so photons beyond the observer's event horizon always travel away from him even though emitted towards him from any source!  It should be noted that the distance to an observer's event horizon is the same for all observers because in equation (7) the constants c and H are universal constants.


In the conundrum we have a quasar Q and a galaxy G whose motions relative to O are governed by the differential equation, equation (2).  We have for the quasar and galaxy, the respective equations:



and


where rQ and rG are the radial distances of Q and G, respectively.  There are also a couple of photons, PQ and PG to consider and whose motions are governed by the differential equation as given in equation (6).  We have for the photon PQ and photon PG, the respective equations:



and


where rq and rg are the radial distances of PQ and PG, respectively.  For the initial condition on time we take as "time-zero" by our "cosmic clock" the time when Q emits PQ, i.e. when this event occurs time, t = 0.  At time zero, t = 0, Q is located at rQ = rQ0, and G is located at rG = rG0: these are the initial conditions.  The solutions to equations (8) and (9) are then



and


To solve equation (6) for general initial conditions r = r0 at t = t0 we have, separating the variables:



which yields



that is,



giving as the photon radial trajectory equation:



In the case of equation (10) and (11) we have as solutions



and



Note that the initial radial position of PQ is that of Q at time t = 0. At time t* the radial position of PG is rG*, the point in space-time where G and PQ meet, as yet not calculated.


It is very convenient to replace the initial radial distance, rQ0, of the quasar from O at t = 0 by its initial distance, at t = 0, from O's event horizon.  To do this let ε > 0 and let rQ0 = (c/H} - ε. Then Q can be made arbitrarily close to O's event horizon by letting ε 0.


If rQ0 = (c/H) - ε is substituted into equation (15) we obtain



which is a great simplification.  Note from equation (17) if ε 0 then rq → c/H meaning that the photon stays on the event horizon.  Or, looking at the situation from the view-point of the time of travel of PQ to the origin, O, we may set rq = 0 in equation (17) to obtain:



and



where Tq is the travel-time of PQ to the observer O.  Now, if we allow ε 0 then Tq → ∞, meaning that the closer Q is to O's event horizon at t = 0 the greater is the travel time of PQ to O and in the limit the time is infinite.


The photon does, however, pass G in a finite time even if ε 0. Let PQ and G meet after PQ has travelled from t = 0 (when PQ was created) to t = t*.  At this time, t = t* we have as the radial positions of G and PQ, rG* and rq* respectively, where


and



These may be equated: rG* ≡ rq* because G and PQ have met. This gives:



from which we obtain



which is finite even if ε 0.  We may now find the radial distance of G at the event of G meeting PQ. This was denoted above (equation (13)) as rG*( ≡ rq* ). We have




which, since ε is small, is very close to the event horizon.  If Q is on the event horizon when PQ is emitted then G does not meet PQ until the galaxy G itself reaches O's event horizon because then from equation (20)



Since ε is very small, when G and PQ meet they are close to O's event horizon and PQ has made little progress.  G now emits a photon PG of its own with frequency identical to that of PQ... as seen by G... and in the same direction as PQ i.e. towards O.  Note: the time t* is the time of creation of PG.  We may now substitute into equation (16) the values of



to find the trajectory of the photon PG.  The result is



which matches equation (17), the equation for the radial photon trajectory obtained for photon PQ.  This shows what was expected: that PQ and PG set off on identical radial trajectories.  We may repeat the calculation leading to the time after t = 0 that PQ arrives at O: the result is that Tg, the time recorded by the cosmic clock when photon PG arrives at O, where rg = 0, is given by



All that remains now is to evaluate the red-shift.  This is yet another trap for those "students"... including the self-taught... who have studied some aspects of the Special Theory of Relativity, in particular the Doppler effect but nothing of the General Theory of Relativity.  It must be emphasised once more that the cosmological red-shift is an expansion effect whereas the Doppler red-shift... and blue-shift... is a velocity effect.


The standard result will be quoted first, followed by a proof which is similar to that which led to equation (14).


Suppose that we have two points in our expanding space separated at a cosmic time t0 by a cosmological distance r0: these points will be referred to as G and G' as if they were two galaxies but it is important to realise that, since galaxies have finite size, strictly speaking G and G' are points within the galaxies and have no peculiar motion relative to the cosmological expansion. In the model considered which has an event horizon, we must have r0 < c/H, otherwise no photon could travel between the galaxies.


At the cosmic time t0 a photon is emitted from G (or G') with wavelength λ0 in G's (or G''s) local space. The photon travels through the expanding space and at a cosmic time t1 arrives at G' (or G) with wavelength λ1. At this cosmic time G and G' are separated by a distance r1, and r1 could now place G and G' beyond each others event horizons.


The red-shift is defined as z = ( λ1 - λ0 ) / λ0 and the standard equation is



We write r1 = r(t1) and r0 = r(t0) to emphasise that the distances are functions of cosmic time.


Since we are now discussing the wave-nature of light rather than the property of discrete particles, one of the most instructive approaches to deriving equation (23) is to consider the wave at radio frequencies. Suppose we have a radiating di-pole aerial and that the electrical oscillations in the conductor are periodic with frequency v. The time for one cycle is thus v-1 = r. The electromagnetic waves travelling away from G, where we take the aerial to be located, although it could be G', are such that the particular wave-crest emitted at t0 will be at a distance r from the aerial at time t, where



an equation obtained in a similar fashion to that yielding equation (14). At t = t0, r = 0 as is easy to show. To derive equation (24) consider the speed of a photon distant r from G' at time t. This is similar to equation (6) but with a change of sign on c.



To solve equation (25) for initial conditions r = 0 and t = t0, we have, separating the variables,


which yields


that is



giving the photon radial trajectory equation



The wave-crest that was emitted at a time t0 + r, i.e. one oscillator cycle-time later is, after a time t, at a distance r' from the aerial where



and the two successive waves crests are separated by a distance r - r' where



which yields



If we now define the radiation wavelength at time t and distance r from the aerial as λ = r - r', or since the wavelength varies with time, write λ = λ(t), we have for the red-shift, z, the equation



Hence 1 + z = eH(t-t0).


This equation gives the red-shift of the radiation that left G and under went a travel time of t - t0. Suppose that the radiation arrives at G' at a time t1. Then the red-shift at G' is given by



If G and G' are separated at time t = t1 by a distance r1 and were, at a time t = t0 separated by a distance r0 then r1 = r0eH(t1-t0), hence



The above proof lies within the model discussed, i.e. the "elastic" space model: nothing outside this model is used. However, an erroneous result is obtained by means of the misuse of the Special Theory of Relativity equation



and taking λ0 as the wavelength of the radiation transmitted from G and λ as the received wavelength at G', with v = Hr. This is quite wrong!


Equation (27) can be applied only between two infinitesimally close inertial frames on the line joining G and G': i.e. we have to set the relative velocity of the two frames as dv = H dr. If the wavelength changes from λ to λ +  we may use equation (27) to write



which gives dλ/λ = Hdr/c = Hdt where dt is the infinitesimal time for a photon to traverse dr i.e. leave one inertial frame and enter the other. Then


which yields


which gives



or


as before.


At first sight this proof seems easier: it does, however, assume equation (27) and Special Theory of Relativity effects. It is included solely to emphasise that equation (27) must not be misused. This equation does not apply over cosmological distances but only between "local" i.e. infinitesimally close reference frames which possess the property of being inertial, which in the context of the model means that they are fixed in the expansion.


These equations and results verify the description given above. Or, correctly, the description is a word picture of the result of the mathematical analysis. It is important to note that the result is in no way determined by gravitating mass and that the cosmological event horizon is radically different from the more familiar one surrounding a black hole. This should be no surprise because the space (or rather the space-time) surrounding a black hole is static: it is not contraction of the space that draws objects in to the black hole: the objects fall through the space. Also, due to the mass of the black hole, the space (and space-time) has curvature.


It might be asked if the mathematical analysis given above is valid. It is but it is not a proof: this has to be based on the General Theory of Relativity. However, the analysis does reveal some essential features of one relativity-based cosmological model known as the de Sitter space.


If gravitating mass is introduced then the analysis is obviously made more complicated. This is because the effect of gravity is to oppose the expansion. However, if, as always, by starting the problem with suitable initial conditions it can be ensured that the expansion dominates. After all, the basic premiss of the conundrum is that this is so. It can then be shown that the modified mathematical solution gives the same general result for the behaviour of the photons PQ and PG.


However, it has to be said that the equations used in the Standard Model (1998) of the Big Bang contain no repulsion effect: no cosmological constant. In that case, should the universe be as one of the two possible solutions within the framework of the Standard Model that expand forever, then there is no event horizon! The Standard Model has a particle horizon in these two cases.... but that is another story!


Acknowledgement:   My sincere thanks are given to my nephew, Andrew P. Spratley of The Queen's College, Oxford. I am indebted to him for reviewing this article and independently working out the mathematical solution as a check. His assistance has been invaluable.


©Dennis W. Spratley 1998

Desford, Leciestershire.






This, the full answer to the "Cosmological Conundrum" was sent to me by Mr. Dennis Spratley, and printed from his supplied sheets. As you can see the format is not the normal MIRA format but this has been done so no mistakes occur in the mathematics of this article. The equations are not the sort to type in without carefully checking of each symbol and these pages are reproduced from his corrected proofs. Hopefully containing no errors.


Ivor Clarke, Ed