Does red shift approach infinite near time zero?

  • #1

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Hi. I'm trying to get an idea how to look at the beginning, before the cosmic background radiation (CBR) and what we can detect with our eyes, and what we can assume about earlier times than light was around.

I was looking into the red shift effect effect, as not only a measure of relative velocity, but also a measure of relative time shift.

Such that, for the greatest red shifted locations in the universe, and therefor the most distant, we see events there unfolding at a slower rate of time also, the amount depending on the distance to the galaxy. The time shifts can be mapped out just as the velocities. Is this correct? If incorrect, which in cosmology is likely, no probs. :smile:

So if so, then, if we imagine continuing back to before the first cosmic light is released, we can imagine the red shift and time shift continuing on back, without our need seeing it, by using theory and extrapolation. Is that correct? Is that what the math tells us?

If so, let's take it to 1ms. second after the beginning. Assuming 'red shift' and time shift doesn't end at the CBR event, but continues back beyond CBR, can we calculate the red shift at time 0 +1ms.?
Will it be near infinite? Some other value?

How about the time shift? Near infinite, frozen in time from our perspective? If so, is this what gives the universe a similar look to a black hole, where both the boundaries of the accessible universe is analogous to the event horizon of a black hole?
 

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  • #2
BvU
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Hello,DMH, :welcome: ( a bit late ... ?:) )

cosmic background radiation (CBR)
Where did you find that term ? I know of CMBR, and suppose that's what you are referring to ? That goes back until ( and I quote from here ) ' about 378,000 years after the Big Bang (at a redshift of z = 1100). '
Earlier than that there's no radiation and therefore no redshift, but phystorians go back to ##t_0 + ## about a picosecond or even less
 
  • #3
Orodruin
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Earlier than that there's no radiation and therefore no redshift
There was radiation, but the Universe was opaque to it. It was continuously being emitted and absorbed.
 
  • #4
kimbyd
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To the question in the title: in a way.

The mathematical definition of redshift is ##z = 1/a - 1## (note: in this notation, the scale factor ##a## is always less than one). At ##t=0##, ##a=0##, so the redshift is infinite.

But nobody should take this answer seriously. What this is really saying is that when you get somewhere close to ##t=0## (or ##a=0##), the equations we're using can no longer work.

Also, the other posters are correct to point out that even though the redshift can be mathematically defined as far back as you like, before our universe was a few hundred thousand years old, it was opaque. So there are no photons out there with redshifts much greater than ##z=1090##.

Higher redshifts could be viable for gravitons, however.
 
  • #5
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Yes, sorry: there was no radiation that could reach us. Wiki recombination lemma compares the opaque plasma state to the current state of the interior of the sun. By the time it could get out it was alreaady 378000 years o'clock ?
 
  • #6
JMz
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To the question in the title: in a way.

The mathematical definition of redshift is ##z = 1/a - 1## (note: in this notation, the scale factor ##a## is always less than one). At ##t=0##, ##a=0##, so the redshift is infinite.

But nobody should take this answer seriously. What this is really saying is that when you get somewhere close to ##t=0## (or ##a=0##), the equations we're using can no longer work.

Also, the other posters are correct to point out that even though the redshift can be mathematically defined as far back as you like, before our universe was a few hundred thousand years old, it was opaque. So there are no photons out there with redshifts much greater than ##z=1090##.

Higher redshifts could be viable for gravitons, however.
Also for neutrinos and, I imagine, certain (rare) transitions inside He nuclei, right?
 
  • #7
Orodruin
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Also for neutrinos
Yes, but not that much. The early Universe was also opaque to neutrinos. Just as there is a last scattering surface for light, there is a last scattering surface for neutrinos. Theoretically there should exist a cosmic neutrino background, but detecting it experimentally would be extremely challenging.
 
  • #8
Chronos
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It's not much different than Sally watching brave [?] Bob fall into a black hole, he just redshifts into oblivion from Sally's POV. That is the way of the universe. We will never 'see' the BB. It would be so reshifted it would be smeared out beyond detectability. It would be like trying to view a single electron, the' light' would have to be such high energy it would blind any conceivable detector and annihilate the electron you were attempting to view.
 
  • #9
JMz
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Yes, but not that much. The early Universe was also opaque to neutrinos. Just as there is a last scattering surface for light, there is a last scattering surface for neutrinos.
Yes: t ~ 1 sec, as I recall, about 10^13 times earlier than recombination: pre-nucleosynthesis. That would greatly expand our view into the past. I suppose there would be a neutrino Dark Age following that, lasting until nucleosynthesis at a few minutes, a factor of a few hundred in age.
Theoretically there should exist a cosmic neutrino background, but detecting it experimentally would be extremely challenging.
You have a gift for understatement. ;-)
 
  • #10
kimbyd
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Yes, but not that much. The early Universe was also opaque to neutrinos. Just as there is a last scattering surface for light, there is a last scattering surface for neutrinos. Theoretically there should exist a cosmic neutrino background, but detecting it experimentally would be extremely challenging.
There's also the problem that the neutrinos are (as I understand it), non-relativistic now, which means that they'll be impacted much more strongly by gravitational interactions than photons, which may wash out any directional signal we might hope to glean.

Plus they're virtually impossible to detect at their current energies (the cosmic neutrino background is something like half the temperature of the cosmic microwave background).
 
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  • #11
Orodruin
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There's also the problem that the neutrinos are (as I understand it), non-relativistic now, which means that they'll be impacted much more strongly by gravitational interactions than photons, which may wash out any directional signal we might hope to glean.
This depends on the absolute neutrino mass, which we do not know. There are some papers discussing the gravitational clustering of neutrinos depending on their mass, for example https://inspirehep.net/record/657114

Plus they're virtually impossible to detect at their current energies (the cosmic neutrino background is something like half the temperature of the cosmic microwave background).
Indeed. There are some ideas to measure it but, as we are already aware, even high-energy neutrinos are notoriously difficult to detect. Actually, gravitational clustering would be good for the detection prospects (but not for background neutrino astronomy of course) as it would increase the local number of background neutrinos.
 
  • #12
JMz
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the cosmic neutrino background is something like half the temperature of the cosmic microwave background.
Yes: more precisely, 1.9 K, as I recall. Of course, that's a prediction, not a measurement, but the physics is well understood.
 
  • #13
kimbyd
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Yes: more precisely, 1.9 K, as I recall. Of course, that's a prediction, not a measurement, but the physics is well understood.
I think the physics at work is extremely well-understood, such that it would be hard for that temperature to be off by any significant amount. Even the specific physics of neutrinos isn't that important: the temperature difference between the cosmic neutrino background and the cosmic microwave background draws from how much matter went non-relativistic between the emission of the CNB and the CMB. When a form of matter goes non-relativistic (e.g. electrons/positrons), those particles annihilate with one another, dumping extra energy into the photon field (but not the neutrino field after neutrinos have decoupled). That extra energy is why the CMB has a higher temperature.
 
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  • #14
JMz
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I think the physics at work is extremely well-understood, such that it would be hard for that temperature to be off by any significant amount. Even the specific physics of neutrinos isn't that important: the temperature difference between the cosmic neutrino background and the cosmic microwave background draws from how much matter when non-relativistic between the emission of the CNB and the CMB. When a form of matter goes non-relativistic (e.g. electrons/positrons), those particles annihilate with one another, dumping extra energy into the photon field (but not the neutrino field after neutrinos have decoupled). That extra energy is why the CMB has a higher temperature.
Nice. I had forgotten the origin of the temperature difference. Thanks.
 
  • #15
Orodruin
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I think the physics at work is extremely well-understood, such that it would be hard for that temperature to be off by any significant amount. Even the specific physics of neutrinos isn't that important: the temperature difference between the cosmic neutrino background and the cosmic microwave background draws from how much matter went non-relativistic between the emission of the CNB and the CMB. When a form of matter goes non-relativistic (e.g. electrons/positrons), those particles annihilate with one another, dumping extra energy into the photon field (but not the neutrino field after neutrinos have decoupled). That extra energy is why the CMB has a higher temperature.
To be a little more specific, when it comes to neutrinos they are not completely decoupled during ##e^\pm## annihilation. We typically define a number ##N_{\rm eff}## as the effective number of neutrinos acting as radiation during the CMB epoch. The expected number using three completely decoupled neutrinos would be 3, but the expectation using what we know of neutrino interactions is 3.04 [1], which means that neutrinos would be slightly hotter than what you would expect if they were completely decoupled. The measurement of Planck is 3.15±0.23 [2].


[1] Mangano et al, 2002
[2] Planck Collaboration, 2015
 

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