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24 Mar 18
3 edits
Originally posted by @lemon-limeYeah, your probably correct. I believe the Earths oceans bulge at the equator, meaning a stronger gravitational field, and the equatorial velocity is greater. These effects independently and taken together point toward a slower running clock at the equator. So your catch phrase "The colder one will become the older one." is correct (I believe).
Meanwhile, back at the ranch (here on earth)...
As yet no one has jumped in to prove me (the layman) wrong. This tells me I'm probably right about which twin ages faster than the other.
"The colder one will become the older one."
EDIT: After reading a bit into it, I retract my statement. The judgement should not be made until some careful physics are preformed. The Earth is an oblate spheroid ( as Deep Thought mentioned ). Flattened at the poles bulging at the equator. The more mass under your feet at the equator is apparently more than offset by the increase in in gravitation at the poles from a decreased radius. So gravitation will be stronger at the poles ( slowing clock speed ) and the velocity of the equator will slow clock speed. The effects may cancel, or they may not. The analysis (General Relativity) is great deal beyond my level of training, so its back to waiting!
24 Mar 18
Originally posted by @joe-shmoEvery diagram I've seen of earth as an oblate spheroid has been exaggerated in order to make the oblation more apparent. It's because of this I'm guesstimating the gravitational difference between equator and pole to be negligible... the actual oblation of earth is so slight it's hardly noticeable.
Yeah, your probably correct. I believe the Earths oceans bulge at the equator, meaning a stronger gravitational field, and the equatorial velocity is greater. These effects independently and taken together point toward a slower running clock at the equator. So your catch phrase "The colder one will become the older one." is correct (I believe).
EDIT: ...[text shortened]... analysis (General Relativity) is great deal beyond my level of training, so its back to waiting!
25 Mar 18
1 edit
Originally posted by @joe-shmoThe more I think about this the less sure I am.
Yeah, your probably correct. I believe the Earths oceans bulge at the equator, meaning a stronger gravitational field, and the equatorial velocity is greater. These effects independently and taken together point toward a slower running clock at the equator. So your catch phrase "The colder one will become the older one." is correct (I believe).
EDIT: ...[text shortened]... analysis (General Relativity) is great deal beyond my level of training, so its back to waiting!
It just occured to me that 1040 mph (rotation speed at equator) is not very fast... it might not be fast enough to overcome even a very small gravitational differece between the equator and North pole.
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25 Mar 18
Originally posted by @lemon-limeI did some Equation grabbing ( a practice frowned upon in physics ) just to see if I there were orders of magnitude difference.
The more I think about this the less sure I am.
It just occured to me that 1040 mph (rotation speed at equator) is not very fast... it might not be fast enough to overcome even a very small gravitational differece between the equator and North pole.
I used the following equation for gravitational time dilation: https://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere
Assumptions/simplifications ( and probably killers of my argument )
1) The earth is spherically symmetric ( directly contradicts, but since the eccentricity of Earth is very slight..meeehhh)
2) The earth is inertial ( non-rotaing - usually a decent approximation in classical mechanics - don't know how well it holds here )
3) Since these time dilations are tiny; superposition would be reasonable approximation. That is t = t_s + t_g. Where t_s is the time dilation from special relativity and t_g is the time dilation from gravitation.
4) The Polar and Equatorial radii of 6357 km and 6378 km respectively. See:https://en.wikipedia.org/wiki/Figure_of_the_Earth#Sphere
5) Mass Earth = 5.98 x10^24 kg
6) G = 6.67 x 10 ^(-11) N m^2/kg^2
7) c = 3.00x10^8 m/s
8) Equatorial speed is 465 m/s (1040 mph)
Then compute the ratio:
t_g ( poles) / ( t_s (equator) + t_g (equator) ) = 0.5
This ( which could be complete nonsense ) seems to indicate the clock is slower at the equator and in fact the time dilation at the poles is half the time dilation at the equator. "Meaning the colder one is NOT the older one"
I look forward to have holes torn into this argument by those who know what they are doing!
25 Mar 18
Originally posted by @joe-shmoNow it's my turn to say you're probably right, or at least close enough to tip the scale toward the warmer one being the older one.
I did some Equation grabbing ( a practice frowned upon in physics ) just to see if I there were orders of magnitude difference.
I used the following equation for gravitational time dilation: https://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere
Assumptions/simplifications ( and probably killers of my argument )
1) ...[text shortened]... "
I look forward to have holes torn into this argument by those who know what they are doing!
I intuitively went with rotation speed, and forgot all about gravity as a possible factor. If the earth was a perfect sphere and gravity was uniform, this would have been a no- brainer.
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25 Mar 18
5 edits
Originally posted by @lemon-limeNow I feel bad, I'm getting this end bit all screwed up. I have to make a correction to this statement.
Now it's my turn to say you're probably right, or at least close enough to tip the scale toward the warmer one being the older one.
I intuitively went with rotation speed, and forgot all about gravity as a possible factor. If the earth was a perfect sphere and gravity was uniform, this would have been a no- brainer.
"This ( which could be complete nonsense ) seems to indicate the clock speed is slower at the equator and in fact the time dilation at the poles is half the time dilation at the equator. "Meaning the colder one IS the older one"
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25 Mar 18
Originally posted by @joe-shmoHere is a PhD's quick answer.
Now I feel bad, I'm getting this end bit all screwed up. I have to make a correction to this statement.
"This ( which could be complete nonsense ) seems to indicate the clock speed is slower at the equator and in fact the time dilation at the poles is half the time dilation at the equator. "Meaning the colder one [b]IS the older one"[/b]
https://www.quora.com/Does-time-run-slower-on-the-equator-than-at-the-poles
He says the effects cancel.
I was also wondering about whether the centrifugal force felt in our rotating frame effects the clocks as far as relativity is concerned, it does...
One question:
He goes on to say, "And the earth, being only semi-solid, sloshes around until the combined effective gravitational+centrifugal potential is even over the surface"
So if the earth has an effective equipotential gravitational surface, than why is the time dilation at the equator not discernable by placing a clock at each location?
He says the effect of rotational velocity is detectable:
"There is a effect from things whizzing in circles because of the earth's rotation, and although it's quite small, of order 1.4E-11, it's well within the precision of modern atomic clocks."
25 Mar 18
Originally posted by @joe-shmoThat's what I suspected. Wonder how close to true his analysis is?
Here is a PhD's quick answer.
https://www.quora.com/Does-time-run-slower-on-the-equator-than-at-the-poles
He says the effects cancel.
I was also wondering about whether the centrifugal force felt in our rotating frame effects the clocks as far as relativity is concerned, it does...
One question:
He goes on to say, "And the earth, being o ...[text shortened]... ugh it's quite small, of order 1.4E-11, it's well within the precision of modern atomic clocks."
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25 Mar 18
Originally posted by @sonhouseI can't answer that ( because I can't derive it for myself ), but I suspect is much closer to true than my own.
That's what I suspected. Wonder how close to true his analysis is?
26 Mar 18
Originally posted by @joe-shmoI made the assumption they may cancel just visualizing the two effects realizing they were opposing.
I can't answer that ( because I can't derive it for myself ), but I suspect is much closer to true than my own.
03 Apr 18
Originally posted by @joe-shmoI think he's wrong. Sorry to be slow about this, but I had to get around to actually doing it, look up all the numbers and then feed them into a spread sheet. The result is that a clock at the pole will gain about 3 nanoseconds per second relative to one at the equator (assuming I haven't goofed up somewhere).
Here is a PhD's quick answer.
https://www.quora.com/Does-time-run-slower-on-the-equator-than-at-the-poles
He says the effects cancel.
I was also wondering about whether the centrifugal force felt in our rotating frame effects the clocks as far as relativity is concerned, it does...
One question:
He goes on to say, "And the earth, being o ...[text shortened]... ugh it's quite small, of order 1.4E-11, it's well within the precision of modern atomic clocks."
The way to calculate this is to start with the Schwartzschild metric (I'll check the Kerr Metric as well, but I really doubt it will make any difference).
The metric is:
c²dτ² = (1 - rₛ/r)c²dt² - dr² / (1 - rₛ/r) - r² (dθ² + sin²θ dφ² )
t is the elapsed time as measured by an asymptotic observer (one far away from the Earth and stationary with respect to the Earth's centre).
r, θ, and φ are the normal polar coordinates.
dx where x is one of t, r, θ, or φ refers to a small change in one of these coordinates along the worldline of either of the two twins.
τ is the elapsed proper time along the worldline of one or other of the twins. In other words the time shown on their clocks.
rₛ = GM/c² is the Schwartzschild radius.
Neither twins world lines involve r or θ changing so we have that dr = dθ = 0. At the pole θ = 0 and sin(0) = 0. So the change in proper time dτ for the twin at the pole is given by:
c²dτₚ² = (1 - rₛ/r)c²dt²
and we have
dτₚ/dt = sqrt(1 - rₛ/r)
This is the rate that the twin at the poles clock ticks relative to the asymptotic observers.
For the twin at the equator θ = π/2 and sin(π/2) = 1. So in this case we have:
c²dτₑ² = (1 - rₛ/r)c²dt² - r²dφ²
dφ/dt is just the angular velocity ω and the linear velocity v = ωr. So we can write:
dτₑ/dt = sqrt( 1 - rₛ/r - ωr²/c² ) = sqrt (1 - rₛ/r - v²/c² )
So, were the Earth a perfect sphere the observer at the pole would have the faster clock. When I took the differing radii into account and put the numbers into a spreadsheet I got:
dτₚ/dt = 1 - 3.448E-10 for the twin at the pole and
dτₑ/dt = 1 - 6.977E-10 for the twin at the equator.
Dividing these two gives us the rate at which the polar twin's clock gains on the equatorial twin's clock:
dτₚ/dτₑ = 1 + 3.489E-10 > 1
However this calculation does not take the effect of rotating frames into account (I'd need to use the Kerr Metric and have some sort of estimate to the moment of inertia of the Earth, or at least its angular momentum). It might be that my results converge with your PhD's answer if I did that (which would be interesting as I wouldn't have expected the Kerr metric to be needed for this).
03 Apr 18
Originally posted by @deepthoughtSo with the numbers you generated, if the twins lived to be 300 years old the equatorial one would be about one second younger....
It's just occurred to me that the PhD might mean that the effect due to the equator being further out than the pole is canceled by the effect due to the earth's velocity. They are in opposite directions, but the equatorial twin's clock still runs slower.