Question on Spacetime

krd

himnextdoor wrote: »

Such a cavity would need to be strongly reinforced to prevent collapse.

I'm imagining, that the gravity maximum of my water planet, is somewhere about midway through the radius. At the surface of the planet water is less dense, as you descend it gets denser, as the water bearing down from above - Approaching the gravity maximum, the water is at its' heaviest and densest, once you pass the gravity maximum, the water becomes less dense and less heavy, until you reach the centre, where for fun, I'm saying there is a cavity, of lose oxygen and water vapour, a sea that's in the shape of the inside of a sphere.

Gravity doesn't 'cancel out' any more than a million laser-beams focussed to a point cancel out. The energy is real but in all directions and this creates a 'crushing' force; the centre is maintained by this force causing is to resist motion relative to the surface of the planet.

Gravity is not light. And a light beam can cancel out another light beam through destructive interference. There are several experiments you can do with laser beams, mirrors and beam splitters, where you can see the laser cancel itself out.

I'm only familiar with Newton's gravitational equation. Which is straightforward. The resultant gravitational force between two masses at a distance. A child's marble, has it's own gravity. If you get 3,000 billion marbles and put them together. And say put your marbles orbiting earth - as well as having a moon made of marbles. The gravitational force between any mass at a distance (r in Newton's eq), and the marble moon, will be due to the resultant gravity of 3,000 billion marbles, and whatever the other mass is.

There's something apparent in Newton's equation, which is not completely obvious. Newton's equation, gives the resultant force. With two masses, this would be especially noticeable with large masses like planets - the gravity each body exerts on the other will have different intensities at different points in the masses. When the moon is directly overhead, the earth's surface directly beneath feels the moon's gravity at it's maximum....When it's above the other side of the earth, the earth's surface feels gravity the least.

Bill is right about one thing though; if you were at the centre of the earth, you would be in whatever free-fall the earth was in but... where on earth could you be in order not to be in free-fall with the earth?

You're being pedantic.

No cancellation as such; A and C accelerate B equally in opposite directions alright but the stresses caused by gravity are still acting on B which will be stretched in both directions. I the gravity of A and C were very large but equal then B could be ripped apart. Not exactly cancellation.

You have to consider counter stretching. Gravity is not equal at all points in the bodies. For the sake of the argument I'm saying the bodies are exerting perfectly symmetric force on each other - The symmetries counteract each other. So the body in the middle does not accelerate towards either at its' sides.

Again, no. The oceans simply experience the moon's gravity in addition to the earth's

.

The resultant gravity, the net gravity. You're not saying anything that I'm not. When the moon is directly over head. The water is lighter - it's experiencing the pull of the moon's gravity, which is partially cancelling the earth's gravity.

That's true anywhere but that 'net' gravity will act in one particular direction.

In Bryson's planet - in the cavity at the dead centre, the gravitational pull is equal from all directions. So, the 'net' gravity is zero.

If you're in a cavity below the surface of the earth then whatever it is you are stood on becomes the surface and your equation would need to be modified to reflect the new distance between you and the centre of mass.

Any mass, is composed of other masses. The gravity of any mass, has to be the resultant gravity of all other parts - and each will have it's own centre of gravity (think of the marble moon)

What if, the centre of mass of any object is actually fictious. The resultant gravity of an object at a distance, will imply a centre of gravity.

You seem to suggest that Bill Bryson thinks that the centre of mass is outside the centre of the planet; that would mean that the centre of mass is actually a spherical shell that encloses the central region of planets. This in turn would mean that not only does gravity pull away from the surface, it pulls away from the centre too. It would mean the points of highest density would be away from the centre and therefore density would decrease between the centre of gravity and the true centre of the planet.

Yep...that's what I am thinking.

Rather than thinking of the centre of mass in terms of balanced forces, think of it in terms of being the bottom.

But where is the bottom.....I know there's a balance at the centre of gravity......but from Brysonian model, there isn't a bottom, or the bottom has two sides, and it's at the gravity maximum.

The centre of mass is the point where acceleration due to gravity is equal in all directions; this point cannot be away from the centre.

Yeah....Let's call that the gravity minimum.

Let me ask you this; if the maximum force of gravity isn't dead centre (mass-wise) then how would fusion occur in stars? How could planets form?

Stars and planets would still form into the same spherical shape. And fusion will happen wherever in the star the pressure and temp is high enough - that being why big stars burn up quicker.

Yes, or rather, toward the centre. If you were able to fall through the centre of the earth, your momentum would take you a certain distance beyond the cantre and you would fall toward the centre again, oscillating, or bouncing, about a mean until you settled at the centre. (Not somewhere away from the centre.)

See, this is the whole thing. I could have it really arse ways. I probably have it arse ways.

But what caught me with the whole Bryson thing, is it made me wonder about the gravity distribution - and I am unaware of a simple equation that can confirm show me if I'm right or wrong. I'm am just so so rusty on my maths, I can imagine a series that could be used to model the resultant gravity at every point of a globe, if it were made of marbles.

I'm playing around with it in my head - and one minute it implies one thing, and the next the contrary.

Which is the hottest part of a kettle in the process of boiling water, the water or the heating element?

What does a kettle have to do with all this............no ..don't answer.
It is not 'less' at the centre it is the sum of the forces from all directions which is larger.

If you start at the edge of a gravitational field in freefall then your rate of acceleration will increase over time. It would a long time before you reached reached the viscinity of earth and achieved an acceleration of 9.8 m/s/s.

if I'm right. And you knew the masses, and the distance between them..And got the force F.....F/m should give you the acceleration. Jesus, the more I think about this the more mental it gets.........If you jump out of an airplane. Not only is the earth pulling you towards it, you are pulling the earth towards you.

As you fall below the surface of the planet, the rate of increase slows down as less mass exerts a pull on you and I would argue that the local mass behind (above?) you does not exert a decelerating force at all and if it did then the increasing density of the local mass in front of (below?) you would negate the effect.

It is not until you pass the centre until you begin to decelerate because the majority of the mass is then behind you after that point.

Is it?.....You're forgetting something - in Newton's equation r - the closer you are to a mass the greater the gravity. We're talking in terms of gravity.

Oh my god. I wish I hadn't forgotten all my maths, and had some kind of graphical modelling tool where I could see.

krd

LeighH wrote: »

The hottest part of the Earth is the inner core (about 5500 C, similar to the surface of the sun). It is solid because of the enormous pressure acting on it, not because it is cooler than the outer core.

Has anyone ever sent a temperature probe to the earth's core?

The inner core could be the hottest part of the planet, simply because it's the furtherest part away from the surface - and nothing to do with gravity.

It's been known for a long time beneath the earth's crust is very hot. I believe, early geologists believed it was left over primordial heat.

Morbert

krd wrote: »

This what I thought before - but when I start thinking about it. If the net gravity at the centre is zero. I wish I hadn't forgotten all my maths - even my basic maths, I could model this - I need a little practice. If the net gravity at the centre is zero - then there must be somewhere else, where the net gravity is at its' maximum. I'm guessing this is at mid way point of the radius.

Actually, I realise this becomes harder to model. I would expect the maximum pressure at the gravity maximum. Water, at the bottom of the ocean, is at its densest, and highest pressure. It's not simply there because it's there, it's there because gravity took it there. It the earth was made of liquid (and not spinning) I'm guessing, beyond the gravity maximum, as far as the water is concerned the world turns upside and it gets less dense as it approaches the centre - even allowing for a gas cavity at the centre.

I am I gone crazy?

If the gravity maximum, cannot be at the centre - why would the pressure maximum be there?

I know it's assumed that the pressure is at it's maximum at the centre of the sun - is it?

Pressures is not a measure of net gravitational force. It is the measure of the force per unit area of the surrounding water. When you are at the centre, all of the water is exerting a pushing force on you, even water at the very surface. To put it another way: As you submerge, the water above you is no longer gravitationally pulling you towards the centre, but it is still exerting pressure.

An analogy would be a line of people (with you at the front) pushing against you. The closer a person is, the less they push. But since there is a line of them, you feel the accumulated "push" of everyone in the line, not just the person directly behind you.

There is an assumption, that black holes start life as recognisable mass - collapsing stars etc.

If they're spherical - which they are - if the singularity happens at a critical point where - say a mass collapsing to its' Schwarschild radius, I'm wondering, if there are gravity maxima inside the mass, could they reach their critical gravity/density, before the outer radius - and you could have a singularity within a singularity.

No. Classically speaking, a black hole produces a singular point of great density. Rotating black holes can produce "ring" singularities, but a singularity in a singularity is physically meaningless.

krd

Morbert wrote: »

Pressures is not a measure of net gravitational force. It is the measure of the force per unit area of the surrounding water.

I know that. The high pressure at the bottom of the ocean is due to gravity. Indirectly.

When you are at the centre, all of the water is exerting a pushing force on you, even water at the very surface. To put it another way: As you submerge, the water above you is no longer gravitationally pulling you towards the centre, but it is still exerting pressure.

Yes, I know this. And this is the kind of thing that makes everything more complicated. When you're submerged, the water above you is no longer pulling you down, it's gravity is dragging you up. And it's mass is forcing you down. And it is gravity causing that mass to fall on you.

In terms of the planet, our oceans are just not that deep - but if you consider the telluric oceans of fire beneath the crust, it's a different story. And that's what's got my imagination going.

Maybe you could answer this?

From a distance, objects/masses have a resultant gravity -even in all directions. Internally in objects, is there gravity maxima, and minima. I have no way of calculating this but I had a crazy idea today, that the gravity density within any object would or could, look like a wave. Even have some weird fractal nature to it.

No. Classically speaking, a black hole produces a singular point of great density. Rotating black holes can produce "ring" singularities, but a singularity in a singularity is physically meaningless.

I know very little about blackholes. And I hate the term "singularity" - I assume it's some kind of mathematical paradox, and I don't have the maths to get there on it.

Maximus Alexander

krd wrote: »

Has anyone ever sent a temperature probe to the earth's core?

Is this question a joke? Has anyone sent a probe to a neighbouring star to verify that they are, in fact, like the sun? Perhaps they are just pictures on a giant LED screen surrounding the solar system?

Clearly they have not. There are many ways to approximate the core temperature. For example, seismic waves can be used to indicate the chemical and physical properties of materials they encounter at depths of thousands of miles.

krd

LeighH wrote: »

Is this question a joke? Has anyone sent a probe to a neighbouring star to verify that they are, in fact, like the sun? Perhaps they are just pictures on a giant LED screen surrounding the solar system?

Measuring the surface temperature of a star is very easy - using Stefan–Boltzmann law. A star can be seen. The sun can be seen. We know what the sun is made of, through spectrometry.

Clearly they have not. There are many ways to approximate the core temperature. For example, seismic waves can be used to indicate the chemical and physical properties of materials they encounter at depths of thousands of miles.

I'm not disagreeing with you........but tell me how you can tell temperature through seismology?

himnextdoor

krd wrote: »

I'm imagining, that the gravity maximum of my water planet, is somewhere about midway through the radius. At the surface of the planet water is less dense, as you descend it gets denser, as the water bearing down from above - Approaching the gravity maximum, the water is at its' heaviest and densest, once you pass the gravity maximum, the water becomes less dense and less heavy, until you reach the centre, where for fun, I'm saying there is a cavity, of lose oxygen and water vapour, a sea that's in the shape of the inside of a sphere.



Gravity is not light. And a light beam can cancel out another light beam through destructive interference. There are several experiments you can do with laser beams, mirrors and beam splitters, where you can see the laser cancel itself out.

I'm only familiar with Newton's gravitational equation. Which is straightforward. The resultant gravitational force between two masses at a distance. A child's marble, has it's own gravity. If you get 3,000 billion marbles and put them together. And say put your marbles orbiting earth - as well as having a moon made of marbles. The gravitational force between any mass at a distance (r in Newton's eq), and the marble moon, will be due to the resultant gravity of 3,000 billion marbles, and whatever the other mass is.

There's something apparent in Newton's equation, which is not completely obvious. Newton's equation, gives the resultant force. With two masses, this would be especially noticeable with large masses like planets - the gravity each body exerts on the other will have different intensities at different points in the masses. When the moon is directly overhead, the earth's surface directly beneath feels the moon's gravity at it's maximum....When it's above the other side of the earth, the earth's surface feels gravity the least.




You're being pedantic.




You have to consider counter stretching. Gravity is not equal at all points in the bodies. For the sake of the argument I'm saying the bodies are exerting perfectly symmetric force on each other - The symmetries counteract each other. So the body in the middle does not accelerate towards either at its' sides.

.

The resultant gravity, the net gravity. You're not saying anything that I'm not. When the moon is directly over head. The water is lighter - it's experiencing the pull of the moon's gravity, which is partially cancelling the earth's gravity.




In Bryson's planet - in the cavity at the dead centre, the gravitational pull is equal from all directions. So, the 'net' gravity is zero.




Any mass, is composed of other masses. The gravity of any mass, has to be the resultant gravity of all other parts - and each will have it's own centre of gravity (think of the marble moon)

What if, the centre of mass of any object is actually fictious. The resultant gravity of an object at a distance, will imply a centre of gravity.



Yep...that's what I am thinking.



But where is the bottom.....I know there's a balance at the centre of gravity......but from Brysonian model, there isn't a bottom, or the bottom has two sides, and it's at the gravity maximum.



Yeah....Let's call that the gravity minimum.



Stars and planets would still form into the same spherical shape. And fusion will happen wherever in the star the pressure and temp is high enough - that being why big stars burn up quicker.




See, this is the whole thing. I could have it really arse ways. I probably have it arse ways.

But what caught me with the whole Bryson thing, is it made me wonder about the gravity distribution - and I am unaware of a simple equation that can confirm show me if I'm right or wrong. I'm am just so so rusty on my maths, I can imagine a series that could be used to model the resultant gravity at every point of a globe, if it were made of marbles.

I'm playing around with it in my head - and one minute it implies one thing, and the next the contrary.




What does a kettle have to do with all this............no ..don't answer.
It is not 'less' at the centre it is the sum of the forces from all directions which is larger.




if I'm right. And you knew the masses, and the distance between them..And got the force F.....F/m should give you the acceleration. Jesus, the more I think about this the more mental it gets.........If you jump out of an airplane. Not only is the earth pulling you towards it, you are pulling the earth towards you.



Is it?.....You're forgetting something - in Newton's equation r - the closer you are to a mass the greater the gravity. We're talking in terms of gravity.

Oh my god. I wish I hadn't forgotten all my maths, and had some kind of graphical modelling tool where I could see.

Imagine that the centre of the earth is a giant beach-ball and that gravity is a large number of hose-pipes that are each pointing to equidistant points perpendicular to and distributed over the entire surface of the beach-ball. If the water-pressure from each hose is equal then where would the beach-ball settle?

If the pressure of the hoses was increased then the ball would feel a compression force which would cause it to 'contract' toward the centre. If the pressure were increased to a high enough level then the entire mass of the ball would occupy a single point; the centre of gravity.

On your water-world, if there was an internal ocean surface then the centre of gravity would have to be midway between the inner and outer surfaces of the sphere. (You have to be able to 'bob about' on both surfaces.) But this would mean that the inner surface would be exerting a gravitational force on the rest of the surface; opposite sides would be attracted to each other. In other words, the force that holds the moon in orbit around the earth would be equal to the force being exerted across the centre and in the opposite direction to the outer surface. Opposite sides would be attracted to each other and would meet violently at the centre and this would be the point of maximum density; the centre of gravity.

Are you sure that Bill Bryson thinks differently?

About the kettle. That was a reference to the fact that an increase in pressure causes an increase in melting point. The core is very hot but cannot melt because the heat generated is conducted away and provides energy that keeps the less dense matter above it molten.

The heating-element of a kettle doesn't melt because of the water but without the 'pressure' of the water, the kettle-element would indeed melt.

himnextdoor

krd wrote: »

Gravity is not light. And a light beam can cancel out another light beam through destructive interference. There are several experiments you can do with laser beams, mirrors and beam splitters, where you can see the laser cancel itself out.

This is somewhat off topic but I feel I should point out that destructive interference is not cancellation of energy. Photons are not plusses and minuses that can negate each other which, even if this were the case, would emit energy during recombination.

To put it another way, if you fired two beams from two one-kilowatt lasers, 180 degrees out of phase (to produce cancellation) and focussed to the same point, then if their combined energy is zero, no heating takes place and no other energy is produced, two-kilowatts of energy are leaving the Universe every hour; wouldn't the law of conservation have something to say about that?

Maximus Alexander

krd wrote: »

Measuring the surface temperature of a star is very easy - using Stefan–Boltzmann law. A star can be seen. The sun can be seen. We know what the sun is made of, through spectrometry.

While I understand your point, the suggestion is that you're comfortable accepting optical or near-optical evidence, because it is familiar to you. Seismic waves are no less reliable just because we don't have eyes to see them.

I'm not disagreeing with you........but tell me how you can tell temperature through seismology?

Seismology can tell us the density and stiffness of materials deep within the Earth. So we know that the inner and outer cores are both are both predominantly made of iron, that the inner core is solid, the outer core is not, and what the pressures are like.

So, it is quite possible to calculate what the temperature must be at the boundary of the inner and outer cores in orders for iron to become molten at that pressure.

The problem is, because those temperatures and pressures can not be recreated in a laboratory, it is difficult to confirm exact figures, so we are left with an approximation.

It is likely that it will never be possible to send a probe to the center of the Earth. Does that mean that efforts to examine it through other means should be abandoned and we should accept we will never know anything about it?

krd

himnextdoor wrote: »

This is somewhat off topic but I feel I should point out that destructive interference is not cancellation of energy. Photons are not plusses and minuses that can negate each other which, even if this were the case, would emit energy during recombination.

To put it another way, if you fired two beams from two one-kilowatt lasers, 180 degrees out of phase (to produce cancellation) and focussed to the same point, then if their combined energy is zero, no heating takes place and no other energy is produced, two-kilowatts of energy are leaving the Universe every hour; wouldn't the law of conservation have something to say about that?

Are you saying that when you do an interferometry experiment. Where you see dark lines, or circles appearing on the screen, the light recombines somewhere else, behind the screen, or somewhere within the screen. That all the energy is conserved even in the dark lines?

To tell you the truth, I don't know the answer.


If the two lasers are combined - does that energy leave the universe?

krd

LeighH wrote: »

It is likely that it will never be possible to send a probe to the center of the Earth. Does that mean that efforts to examine it through other means should be abandoned and we should accept we will never know anything about it?

I never said anything like that.

Seismology can get a very good picture of the interior of the earth. It would probably easier to send people to mars than send a temperature probe into the earth.

himnextdoor

krd wrote: »

Are you saying that when you do an interferometry experiment. Where you see dark lines, or circles appearing on the screen, the light recombines somewhere else, behind the screen, or somewhere within the screen. That all the energy is conserved even in the dark lines?

To tell you the truth, I don't know the answer.


If the two lasers are combined - does that energy leave the universe?

The interference is due to scattering, if I'm understanding you correctly, and they give clues as to energy dissipation.

The lasers; if the application of one causes heating then the application of the second would have to cause cooling in order for the net effect to be zero. Even if this were possible, the focus area would be absorbing and emitting photons. Imagine; one photon hits an electron raising its energy and on the other side of its orbit it his hit by another one which sends it back to its lower state - a photon is emitted. But equally, electrons could be liberated.

No, the energy does not leave the Universe, it powers it.

krd

himnextdoor wrote: »

The interference is due to scattering, if I'm understanding you correctly, and they give clues as to energy dissipation.

An experiment like Young's slits does not violate the law of the conservation of energy. The light is twice its' intensity at the maxima, appears not to be there at the minima.

I don't particularly understand it. Looking at the Young's slit experiment with electrons, the single electrons only seem to hit in the maxima. How they can be a wave one second, then a discrete particle the next, I do not know.

The lasers; if the application of one causes heating then the application of the second would have to cause cooling in order for the net effect to be zero. Even if this were possible, the focus area would be absorbing and emitting photons. Imagine; one photon hits an electron raising its energy and on the other side of its orbit it his hit by another one which sends it back to its lower state - a photon is emitted. But equally, electrons could be liberated.

I looked up that question. The energy is not lost. And it's impossible to do this in the real world, as the beams will diverge.

I don't know. In something like the Young's Slits experiment with individual electrons - do the electrons always decide to exist in the maxima. Or electrons deciding to cease to exist if they reach the screen in the minima.