[timey (OP)]

Are clocks ticking faster where gravity is stronger?


https://en.wikipedia.org/wiki/Gravity_of_Earth

As one can see on this wiki gravity map from GRACE (NASA), red shows the areas where gravity is stronger than the smooth, standard value, and blue reveals areas where gravity is weaker.

Clocks are supposed to tick slower where there is stronger gravity, yet we observe clocks to tick faster at the top of mountains.

A clock on the top of a mountain will be moving faster relative to a clock at the bottom of a mountain due to centripetal motion. A clock moving faster will tick slower relative to a slower moving clock...

So why are we observing (relative to bottom of the mountain clocks) clocks ticking faster on the top of mountains where gravity is stronger and the centripetal speed is faster?

Anyone got any answers?

[Petrochemicals]

]Gravity gets weaker as you increace height above the surface and decend beneath. I believe you weigh less downa mine the same as you do up a mountain. The difference in clock speeds in satelites is due to the relative position AND velocity to gravitational fields, in satellites cases the gravity of earth being by far the largest factor it experiences. Satelites in low earth (fast orbit) will differ from those in fixed position, which will only suffer from gravitational effects.


From
https://www.animatedscience.co.uk/ks5_physics/general/Mechanics/Gravitational Fields.htm

The airplane clock experiment is a good one to look at following gravitational and velocity differences.
https://en.m.wikipedia.org/wiki/Hafele–Keating_experiment

[Janus]

Are clocks ticking faster where gravity is stronger?


https://en.wikipedia.org/wiki/Gravity_of_Earth

As one can see on this wiki gravity map from GRACE (NASA), red shows the areas where gravity is stronger than the smooth, standard value, and blue reveals areas where gravity is weaker.

Clocks are supposed to tick slower where there is stronger gravity,

No they aren't.   Clocks at a lower gravitational potential will tick slower than those at higher potential, but this does have to related to where gravity is stronger.   It is related to specific  potential energy difference between the clocks or how much energy per unit mass it takes to lift some thing from one point to another.
Take three scenarios,  Each with one clock higher than the other.
1. Clocks are in a gravity field that decreases in strength with height so that the lower clock is in stronger gravity
2. Clocks are in a gravity field that remains constant so both clocks are in the same strength of gravity
3. Gravity increases with height  so the upper clock is in the stronger gravity.
In all three of these cases, the lower clock runs slower than the upper clock, because it would take work to move from the position of the lower clock to that of the upper one.

yet we observe clocks to tick faster at the top of mountains.

A clock on the top of a mountain will be moving faster relative to a clock at the bottom of a mountain due to centripetal motion. A clock moving faster will tick slower relative to a slower moving clock...

So why are we observing (relative to bottom of the mountain clocks) clocks ticking faster on the top of mountains where gravity is stronger and the centripetal speed is faster?

Anyone got any answers?

As stated above it gravitational potential that counts and not local gravity strength.  The tops of mountains are at higher potential, so clocks run faster there for that reason.  This is slightly offset by the higher tangential velocity, but you would need an extremely high mountain ( much higher than practical) for the two effects to cancel each other out.

[timey (OP)]

Gravity gets weaker as you increace height above the surface

The point being that the data from GRACE shows that gravity is strongest at the top of mountains, and clocks are observed to tick faster at top than at bottom where GRACE shows that gravity is weaker.
(I have studied physics for 10 years and am no stranger to any of the time related experiments)

As stated above it gravitational potential that counts and not local gravity strength

If that that were the case then the argument for time running slower on the bigger mass (black holes) is surely compromised. Unless you can give me a reason for time running slower on the bigger mass that isn't associated with stronger gravity.
(yes - of course there is a specific orbit/orbital speed above earth (that no mountain reaches) where the effects of gravity potential time dilation are cancelled out by relative motion time dilation. If you are the same Janus I talked to elsewhere, then we talked about this 5 years ago. In anycase, the same effect occurs at sea level earth, at every longitude of the equatorial bulge, doesn't it?)

[chiralSPO]

The point being that the data from GRACE shows that gravity is strongest at the top of mountains, and clocks are observed to tick faster at top than at bottom where GRACE shows that gravity is weaker.

My understanding is that GRACE is measuring the strength of the gravitational field at a fixed altitude that is well above the surface of the Earth. It detects an ever-so-slightly stronger gravitational acceleration at this altitude when over mountains (think of it as the earth having an ever-so-slightly greater radius immediately below the sensor, and therefore ever-so-slightly greater effective mass).

However, this effect is not as significant as what is experienced due to varying altitude between the mountains and the valleys. Being higher up on the mountain places one further out of the gravitational well than being in the valley.

[timey (OP)]

Ok - I get where you are coming from @chiralSPO

But:

Being higher up on the mountain places one further out of the gravitational well than being in the valley.

Being placed further outside of a gravity well is a gravity potential consideration, rather than a strength of gravity consideration. If GRACE feels more acceleration (stonger gravity) over a mountain, compared to a valley, then a clock on that mountain, or in that valley will be feeling the same difference.

[Janus]

Gravity gets weaker as you increace height above the surface

The point being that the data from GRACE shows that gravity is strongest at the top of mountains, and clocks are observed to tick faster at top than at bottom where GRACE shows that gravity is weaker.
(I have studied physics for 10 years and am no stranger to any of the time related experiments)

As stated above it gravitational potential that counts and not local gravity strength

If that that were the case then the argument for time running slower on the bigger mass (black holes) is surely compromised. Unless you can give me a reason for time running slower on the bigger mass that isn't associated with stronger gravity.
(yes - of course there is a specific orbit/orbital speed above earth (that no mountain reaches) where the effects of gravity potential time dilation are cancelled out by relative motion time dilation. If you are the same Janus I talked to elsewhere, then we talked about this 5 years ago. In anycase, the same effect occurs at sea level earth, at every longitude of the equatorial bulge, doesn't it?)

Time dilation factor as viewed by an distant observer:

T = T0/(sqrt(1-2GM/rc^2) 
For the Earth, with a mass of 6e24 kg and radius of 6378,000 m,  this works out to be a factor of 0.9999999993
 And a value of g of 9.842 m/s^2

For Uranus, with a mass of 8.68e25 kg and radius of 25559000 m,  we get a time dilation factor of
0.9999999974
but a g value of 8.866 m/sec^2
Time dilation is greater at the surface of Uranus than on the surface of the Earth, yet Uranus' surface gravity is the weaker of the two.

1 solar mass black hole:
At a distance of  3,682,424,742 m. you get a gravity of 1 Earth g and A time dilation of
0.999999799,  even more than the at the surface of Uranus. 

Clocks run slow near a black hole due to the amount of energy needed to lift a mass way from the black hole and the rate at which they tick is not directly related to the strength of gravity at that point.  It is more directly related to the escape velocity from that point.  The escape velocity from Uranus is 21.29 km/sec ( compared to Earth's 11 km/sec), even though the surface gravity is less, and the escape velocity from the above distance from the black hole is 269.2 km/sec even though the local force of gravity is the same as that for the surface of the Earth. This can be seen in the fact that  escape velocity is found by v= sqrt(2GM/r),  and if put this instead of v into  sqrt (1-v^2/c^2) from the gamma function  you reproduce the gravitational time dilation equation.

[Janus]

Ok - I get where you are coming from @chiralSPO

But:

Being higher up on the mountain places one further out of the gravitational well than being in the valley.

Being placed further outside of a gravity well is a gravity potential consideration, rather than a strength of gravity consideration.

Right, and gravitational time dilation depends on a difference of gravitational potential. Lower potential, slower clock, higher potential, faster clock.

If GRACE feels more acceleration (stonger gravity) over a mountain, compared to a valley, then a clock on that mountain, or in that valley will be feeling the same difference.

[Petrochemicals]

T = T0/(sqrt(1-2GM/rc^2) 
For the Earth, with a mass of 6e24 kg and radius of 6378,000 m,  this works out to be a factor of 0.9999999993
 And a value of g of 9.842 m/s^2

For Uranus, with a mass of 8.68e25 kg and radius of 25559000 m,  we get a time dilation factor of
0.9999999974
but a g value of 8.866 m/sec^2
Time dilation is greater at the surface of Uranus than on the surface of the Earth, yet Uranus' surface gravity is the weaker of the two.

Err, is that not what you would expect, weaker gravity, faster clocks ? Or am i reading that wrong ?

Edit.

And whats wrong with you all ? Its 3 in the morning !

[timey (OP)]

Time dilation factor as viewed by an distant observer:

T = T0/(sqrt(1-2GM/rc^2) 
For the Earth, with a mass of 6e24 kg and radius of 6378,000 m,  this works out to be a factor of 0.9999999993
 And a value of g of 9.842 m/s^2

For Uranus, with a mass of 8.68e25 kg and radius of 25559000 m,  we get a time dilation factor of
0.9999999974
but a g value of 8.866 m/sec^2
Time dilation is greater at the surface of Uranus than on the surface of the Earth, yet Uranus' surface gravity is the weaker of the two.

1 solar mass black hole:
At a distance of  3,682,424,742 m. you get a gravity of 1 Earth g and A time dilation of
0.999999799,  even more than the at the surface of Uranus. 

Clocks run slow near a black hole due to the amount of energy needed to lift a mass way from the black hole and the rate at which they tick is not directly related to the strength of gravity at that point.  It is more directly related to the escape velocity from that point.  The escape velocity from Uranus is 21.29 km/sec ( compared to Earth's 11 km/sec), even though the surface gravity is less, and the escape velocity from the above distance from the black hole is 269.2 km/sec even though the local force of gravity is the same as that for the surface of the Earth. This can be seen in the fact that  escape velocity is found by v= sqrt(2GM/r),  and if put this instead of v into  sqrt (1-v^2/c^2) from the gamma function  you reproduce the gravitational time dilation equation.

As we have no clock on Uranus or black hole to check that with, then this data you provide can only have been derived via GR equations...as what we expect from theory.
The difference being the degree of compression of mass as per radius size.

Ok - so the point is that GRACE is not showing a uniform scenario of escape velocity. (albiet we are talking small differences here) The data sugggests that the escape velocity at the top of the Andies will be greater than the escape velocity somewhere in Western Australia.

And clocks will tick 'faster' at top of Andies than they will in Western Australia, not slower.

Given that there is a uniform time at sea level at every height of the equatorial bulge, where each increase in height of the bulge from poles to equator will cause an increase in time via gravity potential, that is entirely cancelled out by the decrease in time caused by the increase in centripetal motion, so that time at sea level is the same from poles to equator - could the cancelling out of the gravity potential increase in time (from poles to equator) be partly to do with there being more mass as the bulge increases towards the equator (slowing time down) as well as being to do with the increase in centrpetal speed (slowing time down) as the bulge increases towards the equator?

[Janus]

Time dilation factor as viewed by an distant observer:

T = T0/(sqrt(1-2GM/rc^2) 
For the Earth, with a mass of 6e24 kg and radius of 6378,000 m,  this works out to be a factor of 0.9999999993
 And a value of g of 9.842 m/s^2

For Uranus, with a mass of 8.68e25 kg and radius of 25559000 m,  we get a time dilation factor of
0.9999999974
but a g value of 8.866 m/sec^2
Time dilation is greater at the surface of Uranus than on the surface of the Earth, yet Uranus' surface gravity is the weaker of the two.

1 solar mass black hole:
At a distance of  3,682,424,742 m. you get a gravity of 1 Earth g and A time dilation of
0.999999799,  even more than the at the surface of Uranus. 

Clocks run slow near a black hole due to the amount of energy needed to lift a mass way from the black hole and the rate at which they tick is not directly related to the strength of gravity at that point.  It is more directly related to the escape velocity from that point.  The escape velocity from Uranus is 21.29 km/sec ( compared to Earth's 11 km/sec), even though the surface gravity is less, and the escape velocity from the above distance from the black hole is 269.2 km/sec even though the local force of gravity is the same as that for the surface of the Earth. This can be seen in the fact that  escape velocity is found by v= sqrt(2GM/r),  and if put this instead of v into  sqrt (1-v^2/c^2) from the gamma function  you reproduce the gravitational time dilation equation.

As we have no clock on Uranus or black hole to check that with, then this data you provide can only have been derived via GR equations...as what we expect from theory.
The difference being the degree of compression of mass as per radius size.

Ok - so the point is that GRACE is not showing a uniform scenario of escape velocity. (albiet we are talking small differences here) The data sugggests that the escape velocity at the top of the Andies will be greater than the escape velocity somewhere in Western Australia.

No. it doesn't.  The slight increase in local g at the Andes is more than offset by the increase in R from the center the the Earth.   The Andes are gravitationally  "uphill" from Western Australia, regardless of the fact that g might be slightly higher in the Andes.  Extreme case in point.  If you were able to drill a hole to the center of the Earth, at the bottom g would be 0, yet the escape velocity from the bottom of the hole would be greater than that at the surface. A ballistic object would lose velocity climbing out of the hole, but would still need to be moving at surface escape velocity when it got the surface in order to escape Earth's gravity. And a clock at the center of the Earth would run slower than one at the surface, despite being at zero g.

[jeffreyH]

Compared to the whole earth a mountain has an insignificant amount of mass. This would make an insignificant difference. As Janus pointed out, it is escape velocity and potential that matter.

[timey (OP)]

The slight increase in local g at the Andes is more than offset by the increase in R from the center the the Earth.   The Andes are gravitationally  "uphill" from Western Australia, regardless of the fact that g might be slightly higher in the Andes.  Extreme case in point.  If you were able to drill a hole to the center of the Earth, at the bottom g would be 0, yet the escape velocity from the bottom of the hole would be greater than that at the surface. A ballistic object would lose velocity climbing out of the hole, but would still need to be moving at surface escape velocity when it got the surface in order to escape Earth's gravity. And a clock at the center of the Earth would run slower than one at the surface, despite being at zero g.

It is actually 'the offset' that I'm particularly interested in. Could you by any chance give me some maths for the equatorial bulge showing this offset?

OK as a matter of curiosity - so you say that escape velocity would have to be greater from centre of earth is g=0, than it would be at surface of earth.
But if we look at how much thrust would be needed to achieve that velocity from a 0g start point, surely the 'work done' will be the same?
If same amount of thrust is applied at 0g centre of earth as is applied at surface of earth to achieve escape velocity, then isn't the rocket traveling at a speed that escape velocity is still possible from by the time it reaches the surface? ie: same amount of work.

Ignoring the fact that the difference in air density between the Andie's and Western Australia will cause a rocket engine to perform differently, and ignoring that a rocket doesn't maintain a trajectory of flight that is directly above its launch site - a rocket taking off with same thrust at the Andie's, compared to the rocket taking off with same thrust at Western Australia, will not reach the 'same speed' due to the Andie's having more downwards pull than Western Australia.
In order for the rockets to achieve the same amount of 'distance' (not height), in the same amount of time (as per a distant observer), won't the Andie's rocket have to 'work' harder?

[Janus]

The slight increase in local g at the Andes is more than offset by the increase in R from the center the the Earth.   The Andes are gravitationally  "uphill" from Western Australia, regardless of the fact that g might be slightly higher in the Andes.  Extreme case in point.  If you were able to drill a hole to the center of the Earth, at the bottom g would be 0, yet the escape velocity from the bottom of the hole would be greater than that at the surface. A ballistic object would lose velocity climbing out of the hole, but would still need to be moving at surface escape velocity when it got the surface in order to escape Earth's gravity. And a clock at the center of the Earth would run slower than one at the surface, despite being at zero g.

It is actually 'the offset' that I'm particularly interested in. Could you by any chance give me some maths for the equatorial bulge showing this offset?

OK as a matter of curiosity - so you say that escape velocity would have to be greater from centre of earth is g=0, than it would be at surface of earth.
But if we look at how much thrust would be needed to achieve that velocity from a 0g start point, surely the 'work done' will be the same?

If same amount of thrust is applied at 0g centre of earth as is applied at surface of earth to achieve escape velocity, then isn't the rocket traveling at a speed that escape velocity is still possible from by the time it reaches the surface? ie: same amount of work.

No.  Imagine you are standing at the bottom of the hole and are launching a 1 kg mass to escape velocity.  You've allowed yourself 1 meter to reach that speed.   The  specific gravitational potential At the bottom is - 3/2(GM/r) = -93767638.76 J/kg  where M is the mass of the Earth and R its radius.    Escape velocity is the velocity needed to  raise the  specific KE of the object to a value such that  sum of  the two equal 0.   This gives an answer of 13.69 km/sec.
We know that the escape velocity at the surface is 11.19 km/sec. which, for our 1 kg object gives a KE of 62608050 J
If we allow ourselves that same 1 meter distance to gain that velocity, the "extra" energy needed to accelerate up to this speed is that amount needed to lift 1 kg at the surface or ~ 9.8 J  This is nothing compared to the KE difference between a 1 Kg object moving at 13.69 km/sec compared to one moving at 11.19 km/sec.  And for that fact, you don't need any extra thrust to get a rocket up to escape velocity at the surface. Just fire it horizontally instead of vertically.  The direction o fthe vector doesn't matter (as long as you aren't aiming at the ground).  Sure, you can arrange things so that the rocket at the surface climbs very slowly, and thus has to lift itself a large distance against gravity in order to reach escape velocity, and thus is required to expend a lot more energy, but this is just being inefficient.  The point is that when you compare the theoretical minimum work needed to reach escape velocity, it is more launching from the center of the Earth than from the surface.   

Ignoring the fact that the difference in air density between the Andie's and Western Australia will cause a rocket engine to perform differently, and ignoring that a rocket doesn't maintain a trajectory of flight that is directly above its launch site - a rocket taking off with same thrust at the Andie's, compared to the rocket taking off with same thrust at Western Australia, will not reach the 'same speed' due to the Andie's having more downwards pull than Western Australia.
In order for the rockets to achieve the same amount of 'distance' (not height), in the same amount of time (as per a distant observer), won't the Andie's rocket have to 'work' harder?

The same argument as above applies.  You could launch both rockets horizontally such that the local g has no effect on the thrust needed to attain escape velocity.  The Western Australia rocket would have to reach a higher speed in order to attain escape velocity. 
If you were able to put the Andes in the Middle of Western Australia, while maintaining the g value difference ( so that, at the base of the mountains g was lower than at the peaks).  Starting at the base, you would still have to lift yourself against gravity and expend work to climb to the top.  It is this work you would have to perform that adds to the escape velocity you would need to achieve from the base compared to the peak.

GR predicts that, despite the fact that g is slightly higher in the Andes vs. Western Australia, a clock in the Andes runs slower. And this is what we measure with real clocks. 

[timey (OP)]

Ok, well these rocket considerations are fun to think about, but while launching a rocket from Western Australia is possible, launching a rocket from top of Andes or centre of earth are improbable.

I'm more interested in how GR time dilation and relative motion time dilation cancel at sea level of every longitude of the equatorial bulge, and how that relates to the increase of mass of the bulge, bc this will have some bearing on the question I am asking.

you have said:

GR predicts that, despite the fact that g is slightly higher in the Andes vs. Western Australia, a clock in the Andes runs slower. And this is what we measure with real clocks.

I'm quite sure that this part of your post is a slip of the tongue, because a clock in the Ande's is higher than a clock in Western Australia, and the clock in the Ande's ticks faster. This is what we measure with real clocks.

[Janus]

Ok - I get where you are coming from @chiralSPO

But:

Being higher up on the mountain places one further out of the gravitational well than being in the valley.

Being placed further outside of a gravity well is a gravity potential consideration, rather than a strength of gravity consideration. If GRACE feels more acceleration (stonger gravity) over a mountain, compared to a valley, then a clock on that mountain, or in that valley will be feeling the same difference.

No, it doesn't.   GRACE is measuring the local field strength at a fixed altitude.  From this, you could calculate the field strength at any other fixed altitude (say, Mean Sea Level).   This means that the map does not represent the value of g at the actual physical surface,  But the difference in measured g at the same altitude for different parts of the globe. 
So let's say that our reference altitude is Mean Sea level.   This means that an object sitting in  valley floor at MSL in the Andes would feel more gravity and weigh more than an object sitting at MSL in Western Australia.   But it does not mean that an object sitting on a peak of the Andes would feel a stronger gravity and weigh more than it would in Western Australia, which is at a lower altitude.   If you look at the image from GRACE, you will note that the scale at the bottom is marked in milligals. A "gal" ( which named after Galileo) is 1cm/sec^2.  a milligal is 1/1000 of that.   Between The region of the Andes and Western Australia, we get a difference of ~50 milligals or ~5e-7g  (with 1g being standard Earth surface gravity) 
That's a really small variation in g.     Compare this to the actual difference in surface gravity when you take the altitude of the surface into account.   Kumarina, in Western Australia is 610m above MSL, and the tops of the Andes are over 6000 m above MSL.   Going from MSL to 610 m above causes a decrease in local gravity of ~0.00019g. Going from MSL to 6000 m, causes a decrease of ~0.00188g  or about 10 times more a decrease.   And both of these completely swamp out the 5e-7 g difference caused by local variations in the Earth gravitational field.  So a clock on a mountain peak of the Andes still weighs less than it would in Kumarina, Australia, despite the local variations in the Earth's gravitational field.
But again, even if this were not the case, As I already pointed out, GR would still predict that a clock would run faster on the mountain top ( It would just change the exact amount of the differential. For example, if we assume a 1g standard at MSL, and compare a clock at MSL to one on top of a mountain, If gravity strength lower at the top of the mountain than it is at MSL, a clock there will run faster than the one at MSL by a smaller factor, than if the gravity was stronger at the top than at MSL. It would take more work to climb from MSL to the peak against gravity that gets stronger as you climb, than it would to do so against gravity that gets weaker as you climb.)

[timey (OP)]

No, it doesn't.   GRACE is measuring the local field strength at a fixed altitude.  From this, you could calculate the field strength at any other fixed altitude (say, Mean Sea Level).   This means that the map does not represent the value of g at the actual physical surface,  But the difference in measured g at the same altitude for different parts of the globe. 

So are you saying that when GRACE takes these measurements over the top of the mountain (or valley), that gravity is not reducing by the inverse square law within the distance between the mountain top (or valley) and GRACE's position of altitude?
That gravity is somehow weaker at the mountain top than GRACE actually reads, and stronger in the valley than GRACE actually reads?

[jeffreyH]

@Janus I don't know why you bother. Timey just won't get it.

[timey (OP)]

If anyone would care to relate this question to the observation of time at sea level at the longitudes of the equatorial bulge?

[Janus]

Ok, well these rocket considerations are fun to think about, but while launching a rocket from Western Australia is possible, launching a rocket from top of Andes or centre of earth are improbable.

I'm more interested in how GR time dilation and relative motion time dilation cancel at sea level of every longitude of the equatorial bulge, and how that relates to the increase of mass of the bulge, bc this will have some bearing on the question I am asking.

you have said:

GR predicts that, despite the fact that g is slightly higher in the Andes vs. Western Australia, a clock in the Andes runs slower. And this is what we measure with real clocks.

I'm quite sure that this part of your post is a slip of the tongue, because a clock in the Ande's is higher than a clock in Western Australia, and the clock in the Ande's ticks faster. This is what we measure with real clocks.

Definitely a mistype. 

As far as time running the same a at sea level World round.   Water will run from high potential to low potential.   You can treat the effects of gravity and those of the rotating Earth as a "battle" between two opposing "potentials".  There is the gravitational potential which increases as you move away from the center of the Earth, and there is the centripetal potential caused by the Earth rotation which increases as you move towards the axis of rotation.  The surface defined by where the Sun of these two potentials gives the same result is a surface of equipotential.  As a result, this is the also the surface over which the oceans will distribute their surface.  MSL follows this equipotential surface.  In the same way,  a Clock on the this surface is subject to these two potentials by virtue of the equivalence principle.  Without the Earth's gravity, pulling down, you would have to expend energy to move from the surface of the rotating Earth to the axis of rotation.  This has just an much an effect on a clock in the rotating frame as a difference in gravitational potential does.   
Since sea level automatically follows the surface of equipotential, clocks at MSL would be subject to the same equipotential, and would run at the same rate.  So the reason why sea level settles where it does, and why clocks at this level all run at an equal rate is one and the same.