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If time measured by one observer moving at a greater velocity than another observer is observed to be passing more slowly, does this imply that there's such a thing as "absolute time" or "base time" which would be the passage of time as measured by a completely static observer of the universe? Basically time as measured from the universe's inertial frame of reference.

I mean given the speed with which the milky way is hurtling through space, doesn't that mean we on earth are experiencing a dramatically slowed down version of this "true time" or "base time" or whatever you would call it?

Qmechanic
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temporary_user_name
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  • You seem to be describing "comoving time": https://en.wikipedia.org/wiki/Comoving_and_proper_distances. Strictly speaking, this is a question about General Relativity. – m4r35n357 Aug 24 '23 at 08:48
  • You might be interested: https://physics.stackexchange.com/questions/516661/if-there-is-no-absolute-time-how-can-we-say-the-big-bang-was-13-8-billion-years There is indeed such a time (or at least you can argue there is such a time, since every observer's time is equally valid). – Allure Aug 24 '23 at 08:54
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    The convention of working with comoving time has good experimental motivation. But the answer to whether the OP has interpreted special relativity correctly is an emphatic no. The so called "completely static observer" is moving relative to the Milky Way and can therefore also be described as having "dramatically slowed down" time. Nothing intrinsically makes one frame more of a base than the other. – Connor Behan Aug 24 '23 at 10:44
  • Yes, I agree in SR that is the case. In GR, however, a comoving observer in the middle of "deep space", or one at Schwartzschild infinite R, absolutely is such a reference, and proper time elsewhere (nearer to a centre of mass/energy) passes more quickly. The OP may have heard a discussion like that, and mis-applied it to speed in SR. – m4r35n357 Aug 24 '23 at 13:23

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If time measured by one observer moving at a greater velocity than another observer is observed to be passing more slowly, does this imply that there's such a thing as "absolute time" or "base time" which would be the passage of time as measured by a completely static observer of the universe?

No. This is not implied because the same is true for each observer’s frame. Observer A’s clock is time dilated in B’s frame and observer B’s clock is time dilated in A’s frame. The situation is symmetrical and does not identify or distinguish between the two observers or their frames.

Dale
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  • You would need to come up with a satisfactory solution to the twin paradox to hold this as correct. – foolishmuse Aug 24 '23 at 15:33
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    @foolishmuse My favorite is the spacetime diagram, but they are all satisfactory https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html see also https://www.physicsforums.com/insights/when-discussing-the-twin-paradox-read-this-first/ If you have a question about the twin paradox that is not a duplicate then please ask it – Dale Aug 24 '23 at 15:56
  • I just posted my question. https://physics.stackexchange.com/questions/777470/twin-paradox-problem-with-acceleration-turning-issues-removed The references you gave do not cover this situation. – foolishmuse Aug 24 '23 at 17:21
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Your question is based on a misunderstanding about the meaning of time dilation. My watch ticks every second. In the frame of a passing muon, the interval between each tick of my watch might be a minute, say, but that effect is entirely symmetrical- a watch in the muon frame would tick every second, just as my watch does, but the interval between each tick on such a watch would be a minute in my frame.

Marco Ocram
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