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Just to clarify, I am not asking about clock accuracy per se at all. The reason I am asking this question stems from the fact that on this site, most questions about the age of the universe answer with using the CMB as a method for the calculations. Now my question is simply put, when we use the CMB for calculating the age of the universe, how "accurate" is this?

I have read this question:

The "age of the Universe" of about 14Gyr you frequently hear about is a good approximation for any observer whose peculiar velocity is non-relativistic at all times.

The age of the universe

There is no special reference frame, as per relativity (time is relative), except all this is not completely true in practice, because there is a special frame, the CMB. All these questions about on this site about "13.8Bn years in what frame?" answer, that it is the age of the universe in the CMB frame. Up to date, as far as I understand, the most accurate calculations about the age of the universe (and so time since it was ticking) use the CMB. This should mean, that we measure time (since it began) most accurately using the CMB.

So shouldn't this mean, that the most "accurate" calculation we can use is the CMB frame itself (for determining the age of the universe)?

Question:

  1. How "accurately" does the CMB tell the age of the universe?
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  • $\begingroup$ This is nothing to do with 'accuracy'. You could pick a different frame and calculate a different $t$, what matters is how intuitive and useful that definition is. $\endgroup$
    – Eletie
    Commented Mar 2, 2022 at 23:00
  • $\begingroup$ There is not one clock that keeps time perfectly. One clock can never keep time perfectly. $\endgroup$ Commented Mar 2, 2022 at 23:00
  • $\begingroup$ Why should you choose a special frame? All clocks in the universe tick differently, though the intergalactic ones tick rather uniform. A clock that has resided in an inertial frame all its life will show the latest time. $\endgroup$ Commented Mar 2, 2022 at 23:05
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    $\begingroup$ A frame is not a clock, and vice-versa, so this question seems hopelessly confused. Are you really trying to ask whether there's something special about the CMB frame that makes clocks at rest with respect to it "more accurate"? If so, then the answer is "no". $\endgroup$
    – Eric Smith
    Commented Mar 2, 2022 at 23:52
  • $\begingroup$ @EricSmith I think this could be an answer if you could elaborate on why "CMB frame that makes clocks at rest with respect to it "more accurate"? If so, then the answer is "no".". $\endgroup$ Commented Mar 3, 2022 at 4:02

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Is the CMB frame the most accurate "clock"?

Not even remotely. The current estimate of the age of the universe is 13.787±0.020 billion years. So it is only known to a little better than one part in a thousand.

Modern atomic clocks are accurate to one part in $10^{18}$ which is about a thousand trillion times better.

The other thing to consider is the time scale that can be measured with a clock. The CMB might give you one part in a thousand accuracy over a few billion years, but closer to one part in one over centuries or less.

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  • $\begingroup$ Thank you so much! $\endgroup$ Commented Mar 3, 2022 at 4:03
  • $\begingroup$ Though I must ask, does this in any way mean, that this "clock" has "lost" 0.020Bn years through 13.787Bn years of "ticking"? $\endgroup$ Commented Mar 3, 2022 at 4:05
  • $\begingroup$ What does it mean to “lose” time? You asked about accuracy. That is the appropriate measure of accuracy if you were to use the age of the universe as a clock. $\endgroup$
    – Dale
    Commented Mar 3, 2022 at 4:32
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"Is the CMB frame the most accurate "clock"?"

No. It's no clock at all. You can't see the time by looking at it. Most clocks that were put in the state of origin of the CMBR, will these days show the latest time. Only clocks that would end up in gravitationally condensed states of matter would show earlier time. There is nothing special about the CMBR that can turn it into a "clock".

You ask in a comment:

does this in any way mean, that this "clock" has "lost" 0.020Bn years through 13.787Bn years of "ticking"?

No. The CMBR isn't a ticking clock. If it lost that 0.020Bn years it would be very inaccurate compared with one second in a billion years. It just means you don't know how long clocks have been ticking. It means that if a clock had been placed at the start of the big bang it would show about 13.787Bn years, but it could be more or less, about 0.020Bn years.

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  • $\begingroup$ This is actually a nice answer. Please see the edit because the question was misunderstood, as I am not asking about clock accuracy. If you like the question please upvote. $\endgroup$ Commented Mar 3, 2022 at 16:15
  • $\begingroup$ I undid my downvote. I understand what you mean now. $\endgroup$ Commented Mar 3, 2022 at 17:04
  • $\begingroup$ thank you so much! $\endgroup$ Commented Mar 3, 2022 at 17:08
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I think you are conflating two issues, one being accuracy of clocks and the other being the fact that elapsed times are frame dependent.

If you had two perfectly accurate clocks on your desk, one sat on top of the other, they would record different times because they are each subject to slightly different gravitational effects and they are moving at slightly different speeds. The difference is not a limitation of accuracy- it reflects the fact that each clock is, with perfect accuracy, measuring a different duration.

Considered in that context, your questions seems to be asking whether the passage of time in the frame of the CMB is somehow more valid or true than the passage of time in any other frame. In an absolute sense, the answer is no. Whether one measure of time is more relevant than another depends upon context and one's interests and objectives. To estimate the age of the Universe, the frame of the CMB is a convenient one to consider- to judge the effects of time dilation on the lifetime of muons, the CMB frame might be utterly irrelevant.

You might also want to consider the point that the CMB frame is not flat space. Depending upon where you are in the CMB frame, you will be subject to different gravitational field strengths, which means that elapsed times will be subject to local variations.

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  • $\begingroup$ Thank you so much! $\endgroup$ Commented Mar 3, 2022 at 16:07
  • $\begingroup$ I've edited based on these comments about the conflating issues, because I believe my goal was misunderstood. If you like the question please upvote. $\endgroup$ Commented Mar 3, 2022 at 16:16

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