How the Big Bang’s alternatives died (Synopsis)

“We were marching down the street, & we were at the head of the troops. We went on marching, & the troops went off to the left.” -Geoffrey Burbidge

It's such a part of our cosmic and scientific history, that it's difficult to remember that it's only been for the past 50 years that the Big Bang has been the leading theory-and-model that describes our Universe.

Image credit: Edwin Hubble, 1929. Image credit: Edwin Hubble, 1929.

Ever since the 1920s, when Edwin Hubble discovered the apparent expansion of our Universe, we've recognized that it's a much bigger place than simply what's in the Milky Way. But the Big Bang was hardly the only game in town. Yet the discovery of not only the Cosmic Microwave Background, but the detailed measurement of its temperature and spectrum, was able to rule out every single alternative as a non-viable model.

Image credit: Bell Labs, circa 1963, of Penzias and Wilson with the Horn Antenna, via Image credit: Bell Labs, circa 1963, of Penzias and Wilson with the Horn Antenna, via

Come find out the whole story on exactly what it was that killed the Big Bang's alternatives, and why they still fail today.


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@Ethan: Very nice précis of the data! I've got one question about the old COBE/FIRAS plots you show. They all label the Y-axis in units of [MJy/sr] (note the uppercase 'M'). I read that as "megajanskys per steradian." That can't be right, can it?

I mean, really bright radio sources are just a few janskys (hence the choice of units, eh? :-), so how can the CMB, which is much dimmer, for example, than the Milky Way, be megajanskys? Is it that steradians are just really gigantic units of area? Or am I missing something about the CMB?

By Michael Kelsey (not verified) on 22 Oct 2014 #permalink

The entire celestial sphere is just 4 pi steradians, so your inference of MegaJanskys is, in fact, correct. Integration over the continuum of the sky at peak wavelengths will get you up into the MJys without much of a problem, but what's amazing is that the fluctuations are only on the order of a few kJy!

Thanks, Ethan! I usually trust my intuition about units, but it was nice to get a reality check.

By Michael Kelsey (not verified) on 23 Oct 2014 #permalink

@Michael Kelsey

Jansky / megaJansky

Thanks. Unbeknownst new unit. I was surprised. Looked it up.

Read the comments while printing out a table of Orders of Magnitude to give to someone.

By MandoZink (not verified) on 24 Oct 2014 #permalink

@MandoZink #4: There are an amazing number of non-SI units out there in science. My favorite is the unit of smell, which is equivalent to torque (since bad smells make you turn your head). Normalized to that of cows, horses and pigs crammed together in a small space, it's the barn-yard-atmosphere ;->

P.S. for you and Ethan: when spelled out, units are always in lower case (hertz, angstroms, janskys), even if their abbreviations are capitalized (Hz, Å, Jy), with degrees Celsius being the exception. And the capitalized units, at least in SI, are only the ones named after people.

By Michael Kelsey (not verified) on 24 Oct 2014 #permalink

A question related to "Tired light" and its "lost energy":

Where does that "lost" energy go to?
Does it reflect?
Does it evenly distribute/scatter?

I compare this to black-body radiation generation, where in principle a superposition of fitting multiples of wavelengths result in the Planck-curve: here we superpose all "lost" energy and only those waves with integer multiples of wavelengths will not eliminate, thus producing an exact Planck-curve...

Is this a bad "Ansatz"?


I don't know from mathematic side, but in general tired light models require some "resistance" within spacetime. So that loss of energy is transferred to whatever is causing the light to interact with it. Be it some particles, or intergalactic "stuff". So some form of scattering takes place. One of the reasons tired light was ruled was because you don't see distant galaxies as blurred or other optical effects if there was some scattering that's unaccounted for.

By Sinisa Lazarek (not verified) on 27 Oct 2014 #permalink

@Sinisa Lazarek #7: In electrical environments "resistance" generates reflection of energy and thus delay of signals, somehow "redshifting" them on their forward path. _That_ energy will not disappear "somehow": it simply flows in reverse direction.

Can't that happen in the same way within the universe, having a "resistance" of let's say 10^35 Ohm or so?

@SCHWAR_A #6, #8: Since the "tired light" hypothesis has been fairly well discredited (that is, our observational evidence not only does not support it, but directly contradicts its predictions), your questions don't really have any basis in fact. We could engage in idle speculation, of course, but that's beer-swilling philosophy, not science.

I'm not sure you understand black-body radiation very well. The black-body spectrum is not "fitted" from some set of special wavelengths ("integer," you called them). It is a continuous emission due to thermal motion. The primary feature of the black-body spectrum is that it depends *ONLY* on temperature, and not in any way on the materials which create the emission.

By Michael Kelsey (not verified) on 27 Oct 2014 #permalink


Am just an amateur, but from what I have gathered, in principal you could propose something like space having a resistance as a reason for redshift. I think that from physics standpoint it is a reasonable hypothesis. In order for it to work, there should be a description of that "resistance" of space is and a process by which it interacts with light as to only manifest in z-shift.

According to wiki, Zwicky spent a lot of time pondering different scenarios, some sort of Compton effect or similar, but no model worked or agreed with observation.

But that aside, since there is no real model or explanation for redshift other than it's just lambda and that of simplest form from de-sitter space, if one would explain the redshift that agrees with observation, through some light interaction with something instead of spacetime expansion, it would certainly be of great significance.

By Sinisa Lazarek (not verified) on 27 Oct 2014 #permalink

@Sinisa #10: Keep in mind that "redshift" is not some isolated bit of data. If you (Schwar_A, Duffield, whomever) want to propose some alternative to the current cosmology, then you'd better be prepared to reproduce *ALL* of the existing results: redshifts, including their detail patterns of structure; the CMB, both the overall value and the fluctuations on many different scales; gravitational lensing; the time dilation of distant cosmic events (such as supernova light curves) and the correlation of that time dilation with redshift; and many more.

And if you have some alternative cosmological model which does reproduce all that existing data, then it better also make predictions for new observations, such that it can be distinguished from the current model.

By Michael Kelsey (not verified) on 27 Oct 2014 #permalink

@ Michael

I have no intention of making alternative cosmology models, nor did I say that current models are wrong. What I did say, is that if someone was able to explain cosmological redshift through some "transfer" of energy model, instead of spacetime expansion, that fits all observations, that it would be really something. I also did say, that all such models were eventually discarded as failing to agree with observations.

On the other hand, again as an amateur, IMO you linked some things that are not part of same topic. Mainly, tired light models have little or nothing to do with CMB and it's statistics. Redshift/distance is just one small set of problems/observations in the whole of cosmology. Tired light is only about weather or not EM radiation transfers part of it energy to something else, while it travels large distances. nothing else. It has nothing really to do with CMB, inflation, gravitational lensing and so on. One needn't produce new theory of gravity or of BB as a whole if one ponders how light might loose energy. . i.e. what does pattern of CMB structure have to do with redshift and talk of tired light hypothesis? That's inflation and statistics distribution... different theory trying to address different problems.

By Sinisa Lazarek (not verified) on 27 Oct 2014 #permalink

of course, just so there is no misunderstanding, you are absolutely right that looking as a whole - what we call the standard cosmological model.. with FRW metric + inflation + DM + DE really ties everything together and different pieces collaborate each other. That's why it's so hard to change one without affecting the rest.

But Schwar's question is a valid question, and looking to see how something is working and looking to find answers how it might work if certain things are different, is a natural part of learning. It's not about dissing established cosmology, it's about learning the finer points of it :)

By Sinisa Lazarek (not verified) on 27 Oct 2014 #permalink

We're searching for a different source for the same kind of Planck radiation distribution.

We need something similar to the statistically identical vibrating of atoms and molecules (described by temperature). The idea is that this is nothing else than oscillators, which statistically have the same base frequency with arbitrary phases within a certain part of volume. Superposition results in black-body radiation.

If there actually is some kind of backward-reflection of radiation energy on its way forward, and the producing wavelength-delta is constant per distance, Δλ=2πKr, then I think we have such condition, because then all this backward flowing energy can be seen as oscillators, all statistically generating the same base frequency ("vibration" and thus "temperature") with arbitrary phases within a certain volume, giving the Planck-spectrum.

Is this a correct flow of thoughts or do I miss something important? Or is it totally stupid?

@Sinisa #12 and #13: Sorry for any misunderstanding! In my comment #11, I was using "you" in the generic sense of "someone."

At the same time, you (anyone) _can't_ take the redshift data out of the context of cosmology, and try to "explain" it in isolation. The redshift data were the first thing which revealed the large, expanding Universe we currently understand. That understanding _includes_ all of the other more recent observations we have: of the CMB temperature, of gamma-ray bursts with correlated redshifts and time-dilated light curves, of the CMB fluctuations, etc. If you (anyone!) build a model that "explains" redshift in some non-cosmological way, then that model _must_, at the same time, explain all the other observational evidence which is correlated with those redshift data. In particular, such a model must reproduce all of the correlations and consilience we currently have in cosmology.

By Michael Kelsey (not verified) on 28 Oct 2014 #permalink

@SCHWAR_A #14: I suggest you read up on the Sunayev-Zeldovich effect. It is basically what you are describing, is a well-understood and well-measured physical effect, and is NOT sufficient, in our universe, to substitute for cosmological redshift.

By Michael Kelsey (not verified) on 28 Oct 2014 #permalink

@ Michael #15

Of course, no argument about the redshift needing to coincide with ALL other measurements. Personally, I don't think anything was missed back in the 70's and 80's when tired light models were ruled out by observations. But at the same time, I think it's great if someone wants to re-visit the problems and maybe try things differently.

Even in the best case scenario... one could only add minor adjustments to the now established picture. You could still have everything as is now, but the age of the universe might be couple of million years younger or so... the "tired" effect might be small... Those error bars on supernova aren't small, so someone could get away with small effects in theory.

By Sinisa Lazarek (not verified) on 28 Oct 2014 #permalink

@Michael Kelsey #16: My idea was totally free of atoms, thus no Compton scattering, no SZ-Effect.
Just pure self-interacting radiation...


If you dismiss any scattering and return to your original idea about "resistance" from electricity, then that only leaves the medium or spacetime.

You do have things like electric and magnetic constants for vacuum, tied to fine structure constant etc.. from Maxwell's equations for EM.. the equation that defines "c" in units. So there is "resistance" to EM propagation in vacuum, thus it's not infinite but has a certain speed. I don't know of any other "self-interaction" that EM could have that could "cost" energy... and is not some form of scattering.

By Sinisa Lazarek (not verified) on 29 Oct 2014 #permalink

@Sinisa Lazarek #19:
I know that and used that already: the constant K in #14 actually is
K = Z_0 / R_0 = µ_0 · c / R_0,
with R_0 about 10^35 Ohm. This produces the very small phase-turning like in telegraph equation, redshifting the signal.

@Sinisa Lazarek #19:
Another “self-interaction” could be caused by gravity, I think...

@Schwat #21

but that is already accounted for. gravitational redshift that is. You are trying to find something not accounted for :)

By Sinisa Lazarek (not verified) on 29 Oct 2014 #permalink

@Sinisa Lazarek #22:

Don't photons, escaping from their previous position, suffer from time dilation due to their own energy, i.e. mass?

I'm not sure I understand the setup... or the escaping part. and previous position.. in regular words.. do you mean a photon moving from point A to point B?

By Sinisa Lazarek (not verified) on 30 Oct 2014 #permalink

I mean a chain of photons on its way. A photon has assigned a mass depending on its energy. There should be a gravitational influence within the chain...

Speaking in broad terms, since I'm not a physicist, in domain of relativity.. any energy will plug into stress energy tensor and thus cause some effect. What the particular solution of einstein's equation looks like for a single photon, I don't know. I can maybe imagine it as a single gravitational wave on top of the photon..or the grav. potential following the photon, both moving at c.

What the quantum picture would look like.. I have no clue.. since that requires a description of graviton and quantum gravity interaction which we don't have :)

By Sinisa Lazarek (not verified) on 30 Oct 2014 #permalink

How do you feel about this:

Imagine a photon as a sphere: in moving direction its front end together with its following part will generate an extremely distorted, thus one-sided potential "funnel" limited exactly to the cylinder in the opposite direction.

The immediately following photon passes through this zone of time dilation, thus it is a very tiny bit slower than c relative to its predecessor, and thus suffers from redshift.
Let's say v=c/(1+K), and because photons always move with c instead its wavelength increases by
Δλ = 2πKr.

@SCHWAR_A: In #23 you asked about whether photons experience time dilation. They do, in the sense that, if you ask what a "photon rest frame" looks like (which is actually a nonsense question, as Einstein realized when he came up with relativity!), you discover that a photon experiences "infinite" time dilation. In proper terminology, a null world line has zero time elapsed.

In #23 and #25, you show a basic misunderstanding of what "mass" means and how it is used in relativity. When computing momentum and energy, or equivalently the Poincare transformations, the _rest-mass_ is the only useful quantity. It is a frame independent scalar, and is not at all what non-physics-based popularizations call "relativistic mass."

Light has a rest mass of exactly zero. This is what leads to the equality of energy and momentum for light (E = pc for light, but E = sqrt(p^2c^2 + m^2c^4) for massive objects).

In #27, you show some basic misunderstanding of how the Poincare transformations work. They are NOT, in any way, changes to the shape of an underlying frame of reference. Objects do not, in any way, "pass through a region of time dilation."

The Poincare transforms are mathematical tools, or tricks, which allow you to take data which is recorded in one frame of reference, and compute what that data would look like in a different (uniformly moving) frame of reference. Period. That's it.

By Michael Kelsey (not verified) on 30 Oct 2014 #permalink

@Michal Kelsey #28:

thank you.

"When computing momentum and energy,... the _rest-mass_ is the only useful quantity."

Do you mean, a photon never produces any form of "funnel" in the medium it rushes through?
There is no gravitation at all between photons?

Depends what you mean by "some sort of funnel". There's no medium to "funnel".

The problem here (and the reason why I wonder about self-gravitating photons and gravitons interacting in the frame of reference of the moving photon) is that you get different qualitative results thinking about them separately.

It's one reason why we know QM and GR are incompatible, at least at the level of the graduate student.

Speaking in broad terms, since I’m not a physicist, in domain of relativity.. any energy will plug into stress energy tensor and thus cause some effect.

Yup, It's one reason why you get a different "shape" of event horizon on a rotating massive black hole, where it *may* be possible (at least according to the maths used) to have an exposed singularity.

It may also be why spin is quantised and invariant in the same particle group: if it were different spin, then the energy would be different, and therefore either unstable or a different particle if stable.

That, though, was a thought that just occurred to me whilst typing this...

A question related to “Tired light” and its “lost energy”:

Where does that “lost” energy go to?

Problem: there's nowhwere FOR it to go! In standard cosmology where we have redshift, the energy goes into the gravitational potential of the photon coming out of the parent body's gravity well.

QM it goes into "breaking the graviton link" between it and the parent.

And, further to Michael's comment #11, and one I think some who've been slapped down for "silly ideas" will recognise in my lecturing of everyone here (including me), I REITERATE that if you want an idea SERIOUSLY discussed as plausible, rather than "just mentally doodling here", YOU MUST include EVERY piece of evidence and result that supports the idea you wish to replace.

Just coming along and saying "this one thing is sorted out if it's $TIREDLIGHTTHEORY, hence an alternative way to look at it!" but if you want to say "$TIREDLIGHTTHEORY is right, BECAUSE it explains this one thing!" you may think you're saying the same thing, but you're not.

A replacement theory TO BE A REPLACEMENT must solve more than just the "problem" it was constructed to explain. It must fit ALL THE REST of science, much of which will be unaffected, but not everything.

If you're not going to think how your theory ALSO fits with, for example, the quantisation of the HII transition lines, then you're just flinging dung.

@ SCHWAR #29

couple of things. you can't think of EM radiation ONLY as particles and try to model everything like that. they are waves with interferences phases and all that, just as well.
secondly... as for gravitation. EM radiation (light, heat, xrays etc...) does contribute to the tensor.. in other words local curvature. i.e. the curvature sun exerts is the sum of it's energy.. all of it.. not only from elements but also from all radiation. so yes.. light, like any other form of energy does bend spacetime.

But on the other hand, if you talk about gravitation between 2 photons.. that's what we don't have.. i don't know i.e. of feyman diagram with photon and graviton. that is the missing link.

By Sinisa Lazarek (not verified) on 31 Oct 2014 #permalink


Hi. Could you imagine that, just as a thought experiment for the moment, a very small thing in the complete physics might need a tiny correction at the right place, and one person alone is actually able to roll out this alone, "prooving" against ALL known items at the same time without help of other interested persons?

I believe, if there is a "curiosity" or "strange" thing, asking very special questions and trying to peek the very essence of something should be allowed without being inpolitely titled to do #"@§ (censored ;-). I would wish you to be interested and not immediately crying "shut up", as if there were something to be "defended"...

@Sinisa Lazarek #33:

Thank you, I understand that there seems to be a difference between "radiation" and "photon": photons do not produce gravity, while radiation does.

For me this sound schizophrenic and needs some deeper understanding. Would you help me, please?

How does the model look like for radiation producing gravity resp. bending spacetime?


", I understand that there seems to be a difference between “radiation” and “photon”: photons do not produce gravity, while radiation does."

no no no... you got it backwards :) ..there is no difference. what I'm saying is that you can't treat light as buckshot and assign macroscopic effects like funnels and treat them as spheres moving through something. nor are they "elastic" or any of those things. it's radiation, yes, it moves in packets/photons, but it's not a paintball.

By Sinisa Lazarek (not verified) on 01 Nov 2014 #permalink

"Would you help me, please?
How does the model look like for radiation producing gravity resp. bending spacetime?"

I wrote before that I'm not a physicist. Am an amateur astronomer and am genuinely interested in cosmology, but can't give help you with finer points of math and different branches and laws etc. But just google photon and GR and you can get what is there... i.e.

By Sinisa Lazarek (not verified) on 01 Nov 2014 #permalink