Time Dilation and Absolute Age of the Universe

[ramparts]

Cosmological time (and the age of the universe) is an approximation, based on a theoretical series of clocks in free-fall across time, under the least gravitational influence in their own epoch, co-moving with the expansion of the universe.

Beautifully put!

ah yes, and this reminded me how this is all related to the CMB!

When measured from Earth, the CMB (Cosmic Microwave Background) Radiation does look a tad warmer in one direction than it does in the other. This is because Earth is not perfectly comoving with the expansion of the Universe. It's in orbit around the Sun; The Sun "orbits" around in the Milky Way galaxy on one of the rings. And, the Milky way even has some overall motion towards the Hydra cluster.

The CMB exhibits the highest degree of homogeneity (and isotropy, really) in the frame that is comoving with the expansion of the Universe. Any measured change in this homogeneity would indicate that the frame from which you have measured it in, has some motion away from the expansion of the Universe.

The fact that the CMB is homogenous across the entire Universe, lets us be able to confidently define an age for the Universe that EVERYBODY, ANYWHERE can agree upon. Almost sounds "absolute" doesn't it!!

Actually, it almost sounds like the ether. But, not quite, thank goodness

Indeed - due to large-scale homogeneity and isotropy, we can talk about a CMB rest frame (in which the dipole you mentioned disappears). This is the closest we have to an "absolute" frame of reference in the universe - technically it's not actually privileged above any other reference frame, but we can treat it as such since it is unique in that the whole universe looks the same from it. </science>

[SpeedFreek]

The fact that the CMB is homogenous across the entire Universe, lets us be able to confidently define an age for the Universe that EVERYBODY, ANYWHERE can agree upon. Almost sounds "absolute" doesn't it!!

If only we knew exactly what had happened between the Big-Bang and the recombination epoch, when the CMBR was released at what we currently estimate to be 380,000 years later, eh?

[darkmatter4brains]

The fact that the CMB is homogenous across the entire Universe, lets us be able to confidently define an age for the Universe that EVERYBODY, ANYWHERE can agree upon. Almost sounds "absolute" doesn't it!!

If only we knew exactly what had happened between the Big-Bang and the recombination epoch, when the CMBR was released at what we currently estimate to be 380,000 years later, eh?

That's part of the reason I'm anxious for the LHC to finally turn on. If they find a Higg's particle, it will work towards confirming the Inflationary Big Bang Model too, since that uses a certain "type" of Higgs field. I think they're mainly looking for a different Higgs Particle/Field (the one for the SM, which fills in why particles have mass) with the LHC but, hey, if you find one, it makes the other a lot more plausible. That would get us back pretty darn far in time, provide a mechanism for (at least part of) an actual "bang", and help explain the shape and extent of the Universe quite a bit.

On the other hand, if they find out Higgs particles (and their associated fields) don't exist, well ......

[FlatEarth]

...The longer the path some one travels between 2 points of space-time, the shorter the time he measures.

But the earth loops around the sun, so there is a shorter path:
- Launch a rocket at 30 km/s (0.0001 c) in the opposite direction of earth translation to stop orbital speed.
- Keep it fixed relative to the sun compensating sun gravity with the rocket.
- Wait 1 year for earth returning to the same place.
One clock in the rocket will mark 1/sqrt(1-v/c) = 1.000000005 times what a clock in the earth marks
in 365.25 days of 24 hours of 3600 sec = 31557600 sec
the difference will be 0.000000005 * 31557600 = 0.157788 sec
it is small, less than 1/6 sec ...
but will be more than a clock on earth.

I think rpmath's post was lost in the flurry of posts. Does anyone care to comment?

Excellent post!

Here is another calculator:

http://www.astro.ucla.edu/~wright/CosmoCalc.html

Thanks SpeedFreek. I appreciate it.

[SpeedFreek]

...The longer the path some one travels between 2 points of space-time, the shorter the time he measures.

But the earth loops around the sun, so there is a shorter path:
- Launch a rocket at 30 km/s (0.0001 c) in the opposite direction of earth translation to stop orbital speed.
- Keep it fixed relative to the sun compensating sun gravity with the rocket.
- Wait 1 year for earth returning to the same place.
One clock in the rocket will mark 1/sqrt(1-v/c) = 1.000000005 times what a clock in the earth marks
in 365.25 days of 24 hours of 3600 sec = 31557600 sec
the difference will be 0.000000005 * 31557600 = 0.157788 sec
it is small, less than 1/6 sec ...
but will be more than a clock on earth.

I think rpmath's post was lost in the flurry of posts. Does anyone care to comment?

(my emphasis in bold)

To keep the rocket fixed, relative to the sun, compensating for the sun's gravity using the rocket, requires a constant acceleration for the whole year. Perhaps rpmath would like to do the math for this section too, and add it to his final total for the elapsed time on the rocket vs the elapsed time on the Earth?

If he does this he should find that the constant acceleration in order to fight against the suns gravity will cause a massive shift in simultaneity between the rocket and the Earth - the clock on the rocket will show less elapsed time at the end of the experiment as it has been constantly accelerating throughout the experiment.

[FlatEarth]

To keep the rocket fixed, relative to the sun, compensating for the sun's gravity using the rocket, requires a constant acceleration for the whole year. Perhaps rpmath would like to do the math for this section too, and add it to his final total for the elapsed time on the rocket vs the elapsed time on the Earth?

If he does this he should find that the constant acceleration in order to fight against the suns gravity will cause a massive shift in simultaneity between the rocket and the Earth - the clock on the rocket will show less elapsed time at the end of the experiment as it has been constantly accelerating throughout the experiment.

How would the deceleration of the rocket (leaving Earth to “park”) and the acceleration to land back on Earth influence time dilation?

[Jerromy]

I see an obvious debate started about the difference between relative velocity versus relative gravity. I am under the impression from what I know of Einstien's theories that time is distorted by velocity not gravity or acceleration (virtually identical).

Gravity being diverted into accretion or orbital velocity is far different from being held away from a gravitational pull. Flying in an airplane over the Earth is like our orbit of the sun. Being thrusted away from the sun's gravity is like standing on solid ground on Earth. There is always a trade-off in velocity versus acceleration. Fighting gravity is acceleration, PERIOD. Circumventing gravity is velocity, PERIOD. Find a balancing point between the milky way and andreomeda galaxies, send a probe to that location which sends clocks in various directions, and whichever clock records the shortest time relative to the others is going the slowest through space.

Seems alot more plausible than finding gravitons anyway.

[harper05]

It could accelerate in the opposite direction of earths orbit, then after a short period of time, slow down and stop.(I mean stop to the point on the orbital line where for 180 degrees, earth relative to the rocket would move away on the curve, then for the remainding 180, earth, relative to the rocket, would move back toward the rocket.) It would obviously then have to provide a small force to keep it at an earth orbit distance from the sun. But it would not then have to dec. or acc., it could simply be hit again by earth, on its way back around.

This would be an interesting way to test the speed of gravity. In that we would know exactly how much force, and time it would take to have an object in an exact fixed distance relative to the sun in its orbit of the galaxy. We could then deduce very accurately how fast the planets in a g-bound orbiting system interpret gravity?

Have we not done this already?

[FlatEarth]

I see an obvious debate started about the difference between relative velocity versus relative gravity. I am under the impression from what I know of Einstien's theories that time is distorted by velocity not gravity or acceleration (virtually identical).

Gravity being diverted into accretion or orbital velocity is far different from being held away from a gravitational pull. Flying in an airplane over the Earth is like our orbit of the sun. Being thrusted away from the sun's gravity is like standing on solid ground on Earth. There is always a trade-off in velocity versus acceleration. Fighting gravity is acceleration, PERIOD. Circumventing gravity is velocity, PERIOD. Find a balancing point between the milky way and andreomeda galaxies, send a probe to that location which sends clocks in various directions, and whichever clock records the shortest time relative to the others is going the slowest through space.

Seems alot more plausible than finding gravitons anyway.

Hi Jerromy. I originally had the same view as you, but after looking the Wiki article, I realized I was wrong. I can't say I understand it, but I know I was wrong. Apparently without acceleration both observers in the "twins paradox" would see the other slow down and there would be no time dilation. Please see below and see some of the earlier posts on page 3 for a better explanation.

For a more in depth discussion, I would recommend Wikipedia's article on the Twin Paradox.

[FlatEarth]

It could accelerate in the opposite direction of earths orbit, then after a short period of time, slow down and stop.(I mean stop to the point on the orbital line where for 180 degrees, earth relative to the rocket would move away on the curve, then for the remainding 180, earth, relative to the rocket, would move back toward the rocket.) It would obviously then have to provide a small force to keep it at an earth orbit distance from the sun. But it would not then have to dec. or acc., it could simply be hit again by earth, on its way back around.

Hello, harper05. Unless I'm missing the point completely, my understanding is that acceleration is key to measuring time dilation, so to make a final measurement of time the rocket must return to Earth. By impacting Earth, it would accelerate back to Earth's orbital velocity which should produce the same result. I think.

[harper05]

I agree flat earth, the rocket has to acc. away from earth, then stop at one point in the orbital path. then acc away from the sun in its current orbital path,(around the galaxy). For earth it will acc. away from the rocket(on its orbital path), then acc. back into it, all the while acc. with the sun on its orbital path. Thus, the clock on the rocket would read the same as the clock on the earth.

If I had to word it, it would be as follows (hence I haven't done the math, but i've worded it) know what i mean

rocket acc away from earth 30km/s, earth acc away from rocket 30 km/s. they cancel.
rocket stops. earth acc 180 away and 180 toward. they cancell.
both acc with the sun on its orbit. they cancell.

My point is that we would know the exact force required, to compensate for the propagation of gravity. Because we would know the exact force and time required to keep that rocket at a fixed point relative to the sun. Hence the time should be different if gravity propagates at c. It should technically take 8 mn or so for the suns gravitational info to get to our rocket.

[harper05]

We would have to take into consideration the gravitational time dialationl effect of earth of course. I mean calculate it in to the results.

[SpeedFreek]

I agree flat earth, the rocket has to acc. away from earth, then stop at one point in the orbital path. then acc away from the sun in its current orbital path,(around the galaxy). For earth it will acc. away from the rocket(on its orbital path), then acc. back into it, all the while acc. with the sun on its orbital path. Thus, the clock on the rocket would read the same as the clock on the earth.

If the rocket accelerates away from the Earth in the opposite direction to Earth's orbital path, and then turns off its engines, it will free-fall into a retrograde orbit around the Sun. It does not have to accelerate to keep up with the Suns path around the galaxy. It will never escape the gravity of the Sun whilst it remains in free-fall, unless it accelerates away from the Sun at over 42 km/s (the escape velocity of the Sun, for an object at Earth's orbital distance).