DIY Experement: Time Dilation

The article in Science does. Believe it or not, I don't really care. My opinion is that their analysis is a little bit sketchy but they've done enough to show their point I think.

The article in Science does. Believe it or not, I don't really care. My opinion is that their analysis is a little bit sketchy but they've done enough to show their point I think.

I should have explained better.

A massive flat Earth would produce exactly the same results in time dilation as a light fast spinning Earth. Without proving the mass of the Earth, there is no way to separate out the GR and SR components.

Its been a while since i've done GR but I've never been convinced by this. But as I have no intention of working it out so I'm happy to let people shout loudly about it.

I should have explained better.

A massive flat Earth would produce exactly the same results in time dilation as a light fast spinning Earth. Without proving the mass of the Earth, there is no way to separate out the GR and SR components.

Is the mass of the earth unknown?

Quote from: zeroply on February 20, 2009, 08:24:06 AM

I should have explained better.

A massive flat Earth would produce exactly the same results in time dilation as a light fast spinning Earth. Without proving the mass of the Earth, there is no way to separate out the GR and SR components.

Is the mass of the earth unknown?

That would take a different experiment. FE and RE models might have quite different masses.

It's an interesting question - I'm genuinely not sure which would have a bigger effect at that altitude.  I suppose you can argue that they both scale linearly with radius (instantaneous linear velocity and gravitational potential) so if you're making a differential measurement you'll see them both change by the same amount from ground to top of skyscraper.  Trig seems to like doing maths so I'll let him figure out which is dominant, but my suspicion is that the GR signal will dominate until you're talking about geostationary orbits substantially higher. Meh, I dunno!

I am too lazy to explain the maths here, but my result is that in the 100 story building on the Equator the time dilation due to the mass of Earth is around 1e-14, so it should be detectable with the 50,000 dollar atomic clocks. My results confirm that a much better clock would require recalibration when moved several meters up or down.

It is true that this effect is dominant by three orders of magnitude and this experiment does not permit differentiating between the two effects.

You could argue that a mass of 5.97e+24 kilograms is necessary to have such a big time dilation, so Earth has to be spherical. But simpler arguments are considered dubious by FE theorists, so this one will only be treated with the same selective acceptance that all of Relativity receives.

Quote from: Pongo on February 18, 2009, 03:16:00 AM

Also, Midgard is very correct.  This experiment will not conclusively prove the shape of the earth, nor does it prove the earth is spinning.  All it proves is that a clock in the top of a tower will appear to tick slower than one at the base.  The next question though, is why does it do this?  I cited Einstein's theory of relativity as the explanation for the phenomenon.  If someone could demonstrate and prove another explanation for why the top clock would appear to tick slower then the impact that person would make on the scientific community would by truly staggering.  However, what this experiment will indicate, assuming your clocks tick differently, is that the world isn?t moving uniformly upwards.

I am NOT a relativity expert, but your understanding of this phenomenon is way off.

The reason that clocks slow down as they get further away from the Earth is not a consequence of special relativity, it's a consequence of general relativity. It has little to do with the relative speeds they are moving at, it's because gravity is different at different elevations.

If you constructed a massive flat lead plate (say the size of Asia), and floated it somewhere in space motionless with respect to the Sun, you would obtain exactly the same results when doing your experiment on a tower somewhere on that plate.

Matrix and some of the other physics guys on here should be able to corroborate (or correct if I'm wrong).

There is a way to prove your hypothesis without constructing grand space-continents.  Find a low point on the earth, somewhere below sea level, and build a round track.  Send the clock around the track really fast for a predetermined amount of time.  Compare the clock to one you left motionless at the same height.  Repeat the experiment at a much higher altitude.  Compare both sets of data to a clock you left motionless at a height between the two experiments.  If the clocks at a lower height match each other, and the clocks at a higher height match each other then you can conclude that proximity to the earth is what is causing the time dilation and not speed of movement.  Interestingly, if you note the clock at the higher altitude, both the one at rest and the one moving, are respectively slower than the clocks at a lower altitude, you have effectively tested the tower experiment as well.

If you constructed a massive flat lead plate (say the size of Asia), and floated it somewhere in space motionless with respect to the Sun, you would obtain exactly the same results when doing your experiment on a tower somewhere on that plate.

The problem with all these ingenious uses of gravitational pull is that they immediately require the huge masses to be spherical, since every part of those masses is attracted towards the centre of the mass by the same gravitational pull. Anything bigger than a few kilometres wide will crumble into a somewhat spherical shape unless it is solid rock, and even solid rock will crumble if greater than a few hundred kilometres wide. In fact, on Earth (the real one, described by modern science), not even a mountain like the Everest can maintain its height unless it is permanently pushed upwards by plate tectonics.

So your lead plate would have to have a nucleus of solid steel to be even close to the size required, and you would have to explain why Earth itself has not crumbled into a spheroid.

Quote from: zeroply on February 19, 2009, 12:22:19 PM

If you constructed a massive flat lead plate (say the size of Asia), and floated it somewhere in space motionless with respect to the Sun, you would obtain exactly the same results when doing your experiment on a tower somewhere on that plate.

The problem with all these ingenious uses of gravitational pull is that they immediately require the huge masses to be spherical, since every part of those masses is attracted towards the centre of the mass by the same gravitational pull. Anything bigger than a few kilometres wide will crumble into a somewhat spherical shape unless it is solid rock, and even solid rock will crumble if greater than a few hundred kilometres wide. In fact, on Earth (the real one, described by modern science), not even a mountain like the Everest can maintain its height unless it is permanently pushed upwards by plate tectonics.

So your lead plate would have to have a nucleus of solid steel to be even close to the size required, and you would have to explain why Earth itself has not crumbled into a spheroid.

Do you have any math to back up these wild assertions? You're saying that a metal block say 6000km x 6000km x 1000km could not exist for a few days without crumbling into a sphere? Even if made of solid lead?

*especially made of solid lead.

Ideally, it would form an object that gravity could pull on equally from all directions. 

Quote from: trig on February 21, 2009, 08:29:59 AM

Quote from: zeroply on February 19, 2009, 12:22:19 PM

If you constructed a massive flat lead plate (say the size of Asia), and floated it somewhere in space motionless with respect to the Sun, you would obtain exactly the same results when doing your experiment on a tower somewhere on that plate.

The problem with all these ingenious uses of gravitational pull is that they immediately require the huge masses to be spherical, since every part of those masses is attracted towards the centre of the mass by the same gravitational pull. Anything bigger than a few kilometres wide will crumble into a somewhat spherical shape unless it is solid rock, and even solid rock will crumble if greater than a few hundred kilometres wide. In fact, on Earth (the real one, described by modern science), not even a mountain like the Everest can maintain its height unless it is permanently pushed upwards by plate tectonics.

So your lead plate would have to have a nucleus of solid steel to be even close to the size required, and you would have to explain why Earth itself has not crumbled into a spheroid.

Do you have any math to back up these wild assertions? You're saying that a metal block say 6000km x 6000km x 1000km could not exist for a few days without crumbling into a sphere? Even if made of solid lead?

Why do you require that the crumbling has to take only a few days? Earth as we know it, with the same annoying similarities to a sphere, has been that way for at least a few millennia, and if it is or ever was a sphere it never was flat for more than a few thousand years.

I will work on the maths, and tell you if your chunk of lead can stay that shape, so don't feel well for much, it will only take a day or two.

OK, I have some numbers crunched:

Suppose, as you do in static mechanics, that your 6000 by 6000 by 1000 kilometer block is really a 6000x5000x1000 and a 6000x1000x1000 kilometer block, just touching to make the original block. Then you just have to calculate the attraction between the blocks:

• Density of lead: 11340 kg/m³
• Mass of block1: 3.4x10²³kg
• Mass of block 2: 6.8x10²²kg
• Distance between centers of gravity: 3000000 m
• Big G: 6.673x10^-11
• Gravitational pull: G*m₁*m₂/r²=1.72x10²³ Newtons
• Area of contact: 6x10¹⁶ cm²
• Pressure: 2,800,000 N/cm²

For comparison, the pressure you can produce with a hammer is of the order of 10,000 N/cm

². A block of lead is not capable of resisting a blow with a hammer, so it would not be able to keep its shape with any protruding feature of more than 1000 kilometers of height. If you refine the maths you will get to the conclusion that any protruding feature of a few kilometers in height will collapse under its own weight.

In fact, since a low grade steel block does not resist a blow with a hammer either without loosing its shape, and steel has about half the density of lead, it follows that any feature of a few tens of kilometers in height will collapse if the original block is made of low grade steel.[/list]

OK, I have some numbers crunched:

Suppose, as you do in static mechanics, that your 6000 by 6000 by 1000 kilometer block is really a 6000x5000x1000 and a 6000x1000x1000 kilometer block, just touching to make the original block. Then you just have to calculate the attraction between the blocks:

• Density of lead: 11340 kg/m³
• Mass of block1: 3.4x10²³kg
• Mass of block 2: 6.8x10²²kg
• Distance between centers of gravity: 3000000 m
• Big G: 6.673x10^-11
• Gravitational pull: G*m₁*m₂/r²=1.72x10²³ Newtons
• Area of contact: 6x10¹⁶ cm²
• Pressure: 2,800,000 N/cm²

For comparison, the pressure you can produce with a hammer is of the order of 10,000 N/cm

². A block of lead is not capable of resisting a blow with a hammer, so it would not be able to keep its shape with any protruding feature of more than 1000 kilometers of height. If you refine the maths you will get to the conclusion that any protruding feature of a few kilometers in height will collapse under its own weight.

In fact, since a low grade steel block does not resist a blow with a hammer either without loosing its shape, and steel has about half the density of lead, it follows that any feature of a few tens of kilometers in height will collapse if the original block is made of low grade steel.[/list]

Is this the preferred way to do this type of calculation? I would think it would make more sense to talk about a gravitational vector field and look at the maximum values (at the corners). Why did you pick these particular block sizes instead of say two identically sized blocks?

I would imagine that the internal pressure would be very great as your calculations are suggesting, but how does that affect the surface?[/list]

Gravity (oooo no he didn't, oh yes he did) is not a vector field it is a tensor field though for these situations we approximate to Newtons equations quite adequately. This seems a perfectly sensible way of doing the calculation to me.

You know a vector is just a special case of tensor, right?  Same goes for scalars?

I think I heard that somewhere, maths was never my forte.

Yeah ditto.  Basically, most of the time you can get the job done with tensors of rank 2 or less, which are vectors and scalars.  GR has some crazy rank 4 tensors of course, but yeah, generally not needed

I was being a little glib. I haven't used 4th rank tensors, but QFT has plenty of rank 2 tensors.

Is this the preferred way to do this type of calculation? I would think it would make more sense to talk about a gravitational vector field and look at the maximum values (at the corners). Why did you pick these particular block sizes instead of say two identically sized blocks?

I would imagine that the internal pressure would be very great as your calculations are suggesting, but how does that affect the surface?[/list]

You are, of course, enthusiastically cheered to do the calculation in any way you like, and we will compare results, which will be within 30% of mine unless one of us makes a big mistake.

The calculation at the edges proves nothing, since there is almost no pressure. I can tell you off my head that the pressure is not very different from the one felt by the floor when a kilogram of lead is left on the floor on Earth.

The calculation of the pressure at the center of the block is quite elemental and you can do it yourself. The pressure there is even higher, and further proves that a block of 6000x6000x1000 kilometers will crumble under its own gravitational pull, even if it is made of solid steel.

Finally, the surface is not very relevant. The central 90% of the block (at least) will end up making a sphere and the rest will create mountains of at most a few tens of kilometers high if made of steel or a few kilometers if made of lead. And that is a tremendous under-calculation, since the gravitational pull will get stronger as the block crumbles into a sphere.

bowler, I'm thinking just Newtonian physics for the purposes of force calculations, so a vector field should work fine.

Also, in Trig's static model, you're computing pressure between two internal points (the centers of gravity), but you're ignoring all the mass on the outside of those two points, which should be exerting gravitational force away from the center of the block. That's why I was leery of the calculation. The two massive blocks might have a net attraction as calculated, but that would not be what the actual surfaces in contact experience.

bowler, I'm thinking just Newtonian physics for the purposes of force calculations, so a vector field should work fine.

Also, in Trig's static model, you're computing pressure between two internal points (the centers of gravity), but you're ignoring all the mass on the outside of those two points, which should be exerting gravitational force away from the center of the block. That's why I was leery of the calculation. The two massive blocks might have a net attraction as calculated, but that would not be what the actual surfaces in contact experience.

Truly, zeroply, you should get some help with the maths. The calculation is correct and you can check in several ways. What I did not say is that the pressure along the axis perpendicular to the line between the two centers of gravity is much less than the one I calculated, so the metal will move out.

You can do this the hard way, of course, doing the calculation with differential equations, or you can trust mine. But, before having to do all that work and taking into account your abilities for mathematics, think first: are you going to get a radically different result, or are you searching for the minuscule error that might be hidden somewhere?

Just accept that the forces you are looking at are humongous, enough to twist the best metal structure or block like butter.

Quote from: zeroply on February 25, 2009, 11:33:43 AM

bowler, I'm thinking just Newtonian physics for the purposes of force calculations, so a vector field should work fine.

Also, in Trig's static model, you're computing pressure between two internal points (the centers of gravity), but you're ignoring all the mass on the outside of those two points, which should be exerting gravitational force away from the center of the block. That's why I was leery of the calculation. The two massive blocks might have a net attraction as calculated, but that would not be what the actual surfaces in contact experience.

Truly, zeroply, you should get some help with the maths. The calculation is correct and you can check in several ways. What I did not say is that the pressure along the axis perpendicular to the line between the two centers of gravity is much less than the one I calculated, so the metal will move out.

You can do this the hard way, of course, doing the calculation with differential equations, or you can trust mine. But, before having to do all that work and taking into account your abilities for mathematics, think first: are you going to get a radically different result, or are you searching for the minuscule error that might be hidden somewhere?

Just accept that the forces you are looking at are humongous, enough to twist the best metal structure or block like butter.

Can you point me to some reference that this is the correct way to calculate gravitational forces on the inside of a solid body?

Here's my problem - consider two points that are 1m each from the center of the Earth, diametrically opposed, so that the total distance between them is 2m. If you consider the Earth as two hemispheres and calculate the force between the COGs, that does NOT give you the force between those two points. By doing the calculation that way, you would be off by orders of magnitude.

Similarly, while the total force between the two blocks would be what you have calculated, that does not equate to having that force spread equally across a plane, as you are doing. You need to consider the gravitational force from every point, not just the COGs.

Can you point me to some reference that this is the correct way to calculate gravitational forces on the inside of a solid body?

Here's my problem - consider two points that are 1m each from the center of the Earth, diametrically opposed, so that the total distance between them is 2m. If you consider the Earth as two hemispheres and calculate the force between the COGs, that does NOT give you the force between those two points. By doing the calculation that way, you would be off by orders of magnitude.

Similarly, while the total force between the two blocks would be what you have calculated, that does not equate to having that force spread equally across a plane, as you are doing. You need to consider the gravitational force from every point, not just the COGs.

You are endlessly talking about forces, when the real problem you have to solve is either pressure or, better yet, stress and strain. The sum of forces on every particle of a sphere when there are no external forces is zero.

My textbook is "Statics and Dynamics" by J.L. Meriam, ISBN 0-471-07862. But any statics textbook will do. On page 71 (chapter 3, Equilibrium) you are instructed on how to use the basic constraint of the problem (that the system is in equilibrium) and use that fact to isolate the part of the system where you want to know the forces.

Since we are assuming that our 6000x6000x1000 km block is in equilibrium, we can cut it in any place we want and we know that the sum of the forces in that part is zero. In this case we know that the force that the first block exerts against the second block is the same as in the opposite direction. Furthermore, we know that the only force between the two blocks is the gravitational pull, and we know the exact amount. Therefore, the average force between the blocks is known, and therefore the average pressure made perpendicular to the plane is known, and I already gave you the numbers.

Now, if you calculate the pressure on another plane, for example the one that divides the block in two 6000x6000x500 kilometer blocks, you will see that the pressure is not the same in every direction, so the metal will not keep its shape. In your example, inside a sphere the pressure is the same in every direction and that is why the forces in two points a meter from the center are the same.

Please do your homework, do not tell me "I don't like your maths". You are free to use any method you like, but not to do as you did, arguing against the calculation of pressures but talking only about forces.

P.S. Please explain to me what this means:

that does NOT give you the force between those two points.

Force is something that acts upon an object in every definition I have found. Is there a definition where a force is between two points that I am not aware of?

P.S. Please explain to me what this means:

Quote from: zeroply on February 25, 2009, 02:05:55 PM

that does NOT give you the force between those two points.

Force is something that acts upon an object in every definition I have found. Is there a definition where a force is between two points that I am not aware of?

I believe that's what a vector field is. Since you're calculating based on a point mass, there's no actual "object" in the sense you would have with mechanical engineering.

I thought about it further, and your calculation just doesn't work. Given a block R, you must integrate across all the points to get an accurate calculation of the force acting at a certain internal point. There is no way I can think of to do this without integral calculus. The force you are calculating is a net force between two center of gravities. It does not imply that that force would be spread across the surfaces.

I did some napkin calculations. Figuring out the gravitational vector field on a cuboid is either a lot harder than it appears at first glance, or my calculus is extremely rusty.

Here is my plan:

1. Calculate gravitational vector field over entire cuboid.

2. Consider a corner of the cuboid. See if the force is enough to deform it.

If anyone has some formulas please spit them out, otherwise I will try to calculate it.

Quote from: trig on February 25, 2009, 08:16:45 PM

P.S. Please explain to me what this means:

Quote from: zeroply on February 25, 2009, 02:05:55 PM

that does NOT give you the force between those two points.

Force is something that acts upon an object in every definition I have found. Is there a definition where a force is between two points that I am not aware of?

I believe that's what a vector field is. Since you're calculating based on a point mass, there's no actual "object" in the sense you would have with mechanical engineering.

I thought about it further, and your calculation just doesn't work. Given a block R, you must integrate across all the points to get an accurate calculation of the force acting at a certain internal point. There is no way I can think of to do this without integral calculus. The force you are calculating is a net force between two center of gravities. It does not imply that that force would be spread across the surfaces.

I did some napkin calculations. Figuring out the gravitational vector field on a cuboid is either a lot harder than it appears at first glance, or my calculus is extremely rusty.

Here is my plan:

1. Calculate gravitational vector field over entire cuboid.

2. Consider a corner of the cuboid. See if the force is enough to deform it.

If anyone has some formulas please spit them out, otherwise I will try to calculate it.

So, now you have a plan. That is a very good step in the right direction.

But again you are trying to find the forces on the top of the block instead of looking at the pressures inside. What you are doing is the same as looking at the forces on a brick on top of the Empire State Building and trying to determine whether the building will crumble. Right now I can tell you that the forces on the edge of your 6000x6000x1000 kilometer block are of about 3 Newtons for every kilogram of edge you decide to observe and that the block will not crumble from the edges downwards but from the center upwards.

And it is becoming tiresome already. What you need to calculate is the pressure on the most vulnerable parts and whether this pressure is the same in every direction.or whether those most vulnerable parts will change shape because of the uneven pressure. The sum of forces is close to zero in every part of the block, just as the sum of forces in every part of a building that is not falling is zero.

And another P.S.:

Please study vectors again. Vectors are not lines between two points. And neither are the vector fields.

And another P.S.:

Please study vectors again. Vectors are not lines between two points. And neither are the vector fields.

I've actually got a degree in math, so I think I've stumbled upon the whole "vectors" thing before.

Are you actually familiar with calculus, or are is high school mechanical engineering (based on textbook) as far as you've gotten so far?

You still haven't shown any examples of people calculating gravitational forces using a similar model. I'm suspecting that you have a hammer and everything is starting to look like a nail. You can't use the "truck weighing 4000lbs is 80ft from one end of the bridge" approach to solving all physics problems, you know?

So, now you have a plan. That is a very good step in the right direction.

But again you are trying to find the forces on the top of the block instead of looking at the pressures inside. What you are doing is the same as looking at the forces on a brick on top of the Empire State Building and trying to determine whether the building will crumble. Right now I can tell you that the forces on the edge of your 6000x6000x1000 kilometer block are of about 3 Newtons for every kilogram of edge you decide to observe and that the block will not crumble from the edges downwards but from the center upwards.

And it is becoming tiresome already. What you need to calculate is the pressure on the most vulnerable parts and whether this pressure is the same in every direction.or whether those most vulnerable parts will change shape because of the uneven pressure. The sum of forces is close to zero in every part of the block, just as the sum of forces in every part of a building that is not falling is zero.

Ok, here's a relevant section from Wikipedia. As I had guessed, your entire calculation is junk. You can't just consider the centers of gravity, you have to consider each point in both bodies. You are assuming that you can consider a block as a point mass, but this is not true for non-spherical bodies.

Don't design any bridges until you've mastered the maths further... 

Bodies with spatial extent

If the bodies of question have spatial extent (rather than being theoretical point masses), then the gravitational force between them is calculated by summing the contributions of the notional point masses which constitute the bodies. In the limit, as the component point masses become "infinitely small", this entails integrating the force (in vector form, see below) over the extents of the two bodies.

In this way it can be shown that an object with a spherically-symmetric distribution of mass exerts the same gravitational attraction on external bodies as if all the object's mass were concentrated at a point at its centre.[2] (This is not generally true for non-spherically-symmetrical bodies.)

Quote from: trig on February 26, 2009, 01:53:58 PM

And another P.S.:

Please study vectors again. Vectors are not lines between two points. And neither are the vector fields.

I've actually got a degree in math, so I think I've stumbled upon the whole "vectors" thing before.

Are you actually familiar with calculus, or are is high school mechanical engineering (based on textbook) as far as you've gotten so far?

You still haven't shown any examples of people calculating gravitational forces using a similar model. I'm suspecting that you have a hammer and everything is starting to look like a nail. You can't use the "truck weighing 4000lbs is 80ft from one end of the bridge" approach to solving all physics problems, you know?

If you have a degree in maths, please, please, please just write the integrals and stop whining. It will take just 5 minutes and you will not look stupid when I decide to work on them and beat you to it.

I have been trying to understand the "vector between the two points" you described and it makes no sense. A force vector is the combination of a 3-dimensional position and a 3-dimensional direction and strength. There is no two points.

And you have not shown any maths at all, not even a single number. Why does a mathematician elude every occasion to use maths to make his point?