Exploring Geometric Shapes of Gravity

In summary, the conversation is discussing the concept of gravity from a geometric perspective, rather than a Newtonian perspective. The main question is about why matter follows certain paths, or geodesics, in spacetime. These geodesics can be either local, determined by the starting point and direction, or global, determined by the shortest distance between two points. The conversation also touches on the concept of escape velocity and how it is affected by the curvature of spacetime. The analogy of water pressure is used to explain the uniformity of spacetime's curvature.
  • #36
Hoku said:
I want to be sure too many people don't get hung up on the use of the word "care" as it relates to space or time. It's used metaphorically. I don't think time or space has "emotions". These kinds of metaphors are helpful for me and I hope people can find the grace to let me use them.
I understand the spirit in which it was meant and I used it in the same way. Before we go on I think it is worthwhile re-asking the question I posed. Do you have an intuitive feel for why geodesics get deflected outwards in polar coordinates? Do you understand why it is always an outward deflection and not an inward deflection, why it "cares" that you always get deflected away?

Hoku said:
Because space and time are inseparable, following a geodesic in time necessitates movement in space in some way.
Only in the case of curved spacetime. In a flat spacetime if you are going purely in the time direction then you will continue going in the time direction. Is that clear?

Hoku said:
Part 2: It seems like there are two geodesics that are simultaneoulsly in effect when stationary on the Earth's surface. The first is a local geodesic and it's the one I described in "Part 1". The other is more "global-like".

In the global-like geodesic, we have a beginning point and and ending destination. The beginning point is the spot on which we are standing. The destination is the center of the Earth. The radius from Earth's surface to Earth's center is NOT curved and it's the fastest way.
OK, what you are describing here is not a geodesic in spacetime, but a geodesic in space. Remember spacetime is 4 dimensional and includes time as well as space. So, if you want to specify a spacetime geodesic similar to what you are describing you would have to say something along these lines: "The beginning point is the spot on which we are standing at t = 0. The destination is the center of the Earth at t = 1 min. A parabolic path along the radius from Earth's surface to Earth's center is a geodesic and it's the extremal way." Do you see the difference? Now that time is explicitly included you can talk about spacetime geodesics between the beginning and the destination.

Now may be a good time to talk about how to measure the length of a path in spacetime and what it means for it to be maximized or minimized.
 
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  • #37
Hoku said:
Part 1: Even when we're stationary on the Earth's surface, we're still SUCCESSFULLY moving through space. We're stationary relative to the Earth, but not the sun, or stars. Our actual movement through space is congruent with the spinning of the Earth and because the Earth is spherical, it's also in the "downward" direction. Therefore, we have an outlet for movement in space. This makes "pressing" into the Earth seem redundant.

Part 2: It seems like there are two geodesics that are simultaneoulsly in effect when stationary on the Earth's surface. The first is a local geodesic and it's the one I described in "Part 1". The other is more "global-like".

In the global-like geodesic, we have a beginning point and and ending destination. The beginning point is the spot on which we are standing. The destination is the center of the Earth. The radius from Earth's surface to Earth's center is NOT curved and it's the fastest way. Part of what I've been trying to ask is why this radius can't be extended through and past the atmosphere. This would be a "straight" geodesic that shouldn't require our return to Earth if we're looking at it from a purely gemetric viewpoint. Of course, because the Earth is spinning, it wouldn't actually be straight but I'd think it could be simulated.

As Dalespam has pointed out, we normally consider the time dimension when considering geodesics but it is OK to talk about paths in space as long as it is clear that that is what we are doing (i.e. considering only the spatial components of the geodesic). If we drop a particle from 6ft above the ground it follows a short path terminating at the surface, but this geodesic is artificially cut short by the ground. If we create a tunnel right throught the Earth (and stop it spinning for a while) then if you release a particle from your hand at the surface of the Earth, it falls to the centre and then continues right through to briefly appear at the surface on the other side of the world before falling back to the centre and continuing back up until it eventually returns to your hand. I am of course ignoring friction due the air which would exert a damping force on the particle. If you do not catch the particle it will continue to fall up and down like this for a long time. The point is that the path does not end at the centre of the Earth unless the path is artificially terminated by placing an obstruction at the centre of the Earth. This spatial geodesic is a straight line that extends form one side of the world to the other, but it does not extend far out into space. If another particle was released from high up in the atmosphere then it would have its own geodesic in space that is longer than the first one, but extends no further on either side of the world than its original release height. On a larger scale these straight line geodesics through space would appear like a spirograph path though space orbiting around the Sun and the Galaxy to an observer at rest with the distant stars, so that the appearance of the path is dependent on the location and state of motion of the observer. To an observer that falls into the tunnel at the same time as the particle, the particle appears stationary and from their point of view the particle is only moving in the time dimension.
 
  • #38
This weekend we've got a date with the Easter Bunny, missing egg mysteries to solve and other springtime festivities. That's why you may not see me again 'till Monday.
 
  • #39
Hoku said:
This weekend we've got a date with the Easter Bunny, missing egg mysteries to solve and other springtime festivities. That's why you may not see me again 'till Monday.

Have a fun Easter Hoku! When you return, why not post again here so we all know to fire the thread back up again, and it doesn't get lost in the shuffle?
 
  • #40
There are only a few weekends where I will never tell my daughter, "Sorry, I have to work". This was one of them. Thanks for the friendly place-holder, FD. BTW, I liked how you applied timelike geodesics to biology!

DaleSpam said:
Do you have an intuitive feel for why geodesics get deflected outwards in polar coordinates?
If the space axis only refers to the "up/down" direction then I can understand how the deflection is outwards only. What's NOT intuitive for me is why it curves back in the first place. I can imagine an angled line that is straight and does not return to 0m. It's not intuitive as to why that shouldn't be an option (from a strictly geometrical perspective).
DaleSpam said:
Only in the case of curved spacetime. In a flat spacetime if you are going purely in the time direction then you will continue going in the time direction. Is that clear?
Here's how I understand it: One body may curve spacetime, but that curvature is meaningless without another body to interact with. Therefore, effective curvature requires two bodies to manifest, which means flat spacetime geodesics don't require movement in space.
DaleSpam said:
[...] a simple change in direction can radically change the shape of the geodesic because the geodesic then crosses different regions with different curvatures at different angles than it would have crossed had another tangent been chosen.
I completely understand how inital position and velocity affect different geodesics in a given geometry of space. I'm beginning to realize that it's all the "givens" of the geometric view that I take issue with.
DaleSpam said:
It is more than just the energies involved. The stress-energy tensor is what relates the distribution of matter and energy to the curvature of space. It has 1 term for energy, and 15 terms for other things including pressure, momentum, energy and momentum flux, stress, etc. All of these things together determine the spacetime curvature which is, in turn, one of the initial conditions.
There may only be one "official" term for energy, but the other terms, like pressure, momentum and stress, are still products of energy. They're just specifically defined expressions of it. So I still come back to the idea that energy is what drives all of these terms, which collectively define the force of gravity, which is the underlying cause of gravity's geometry.

Over the weekend I remembered something that I thought would make a good analogy for all of this. I think what I'm comparing it to is a type of suction, although I'm sure you can correct me if that isn't right. Many years ago I was at a pool party. There was a ball in the pool that one of the guys was playing with. He held the ball at the surface but his hand was underwater. He then spun the ball very quickly. When he let go, he was able to pull the ball underwater a small distance by the "suction" that was created between the ball and his hand by the moving water. The water was certainly moving in a way that is analogous to the curvature of spacetime, yet, without the force spinning the ball, neither this "curvature" not the "gravitational suction" effect would've have happened.

kev said:
If the tunnel is blocked the particle is no longer following a geodesic and it feels a force acting on it [...] and so it no longer has inertial motion, which is defined as the motion of a particle with no forces acting on it.
I know that what you're saying is true. It is verified by Wikipedia's "gravitaton" entry:
[general relativity] equates free fall with inertial motion, and describes free-falling inertial objects as being accelerated relative to non-inertial observers on the ground. In Newtonian physics, however, no such acceleration can occur unless at least one of the objects is being operated on by a force.
But this seems silly. When an object is on Earth's surface it's affected by a force but when it's falling through the center of the Earth it isn't? Is this one of those "frame of reference" problems?
 
  • #41
Hoku said:
There are only a few weekends where I will never tell my daughter, "Sorry, I have to work". This was one of them. Thanks for the friendly place-holder, FD. BTW, I liked how you applied timelike geodesics to biology!

If the space axis only refers to the "up/down" direction then I can understand how the deflection is outwards only. What's NOT intuitive for me is why it curves back in the first place. I can imagine an angled line that is straight and does not return to 0m. It's not intuitive as to why that shouldn't be an option (from a strictly geometrical perspective).Here's how I understand it: One body may curve spacetime, but that curvature is meaningless without another body to interact with. Therefore, effective curvature requires two bodies to manifest, which means flat spacetime geodesics don't require movement in space.I completely understand how inital position and velocity affect different geodesics in a given geometry of space. I'm beginning to realize that it's all the "givens" of the geometric view that I take issue with. There may only be one "official" term for energy, but the other terms, like pressure, momentum and stress, are still products of energy. They're just specifically defined expressions of it. So I still come back to the idea that energy is what drives all of these terms, which collectively define the force of gravity, which is the underlying cause of gravity's geometry.

Over the weekend I remembered something that I thought would make a good analogy for all of this. I think what I'm comparing it to is a type of suction, although I'm sure you can correct me if that isn't right. Many years ago I was at a pool party. There was a ball in the pool that one of the guys was playing with. He held the ball at the surface but his hand was underwater. He then spun the ball very quickly. When he let go, he was able to pull the ball underwater a small distance by the "suction" that was created between the ball and his hand by the moving water. The water was certainly moving in a way that is analogous to the curvature of spacetime, yet, without the force spinning the ball, neither this "curvature" not the "gravitational suction" effect would've have happened.

I know that what you're saying is true. It is verified by Wikipedia's "gravitaton" entry:But this seems silly. When an object is on Earth's surface it's affected by a force but when it's falling through the center of the Earth it isn't? Is this one of those "frame of reference" problems?

Welcome back Hoku, and thanks. :smile: For your last question, it's just a matter of: what exerts a force against you falling? Earth... the ground. In the absence of a planet to "stand on", there's nothing to keep you on the SURFACE of the planet. You're still constantly being "sucked" at a rate of 9.8m/s^2, but now you're experiencing the resistance of the ground.
 
  • #42
Before we continue, we should look at the notion of "distance" in spacetime. This is also called the spacetime interval or sometimes the Minkowski norm. It is defined as follows:

ds² = -c²dt² + dx² + dy² + dz²

As you can see, this is similar to the usual Euclidean norm, but with an extra term for time (scaled by c to get units of length) which has the opposite sign from the other terms. This is the notion of "distance" that is used in spacetime.

There are a few important things to notice about this definition. First, it is Lorentz invariant, meaning that all observers in any reference frame will agree on this quantity regardless of effects like time dilation and length contraction. Second, it divides spacetime up into 3 regions: these are the region where ds²<0 which is called timelike, the region where ds²>0 which is called spacelike, and the region where ds²=0 which is called lightlike or null. Light pulses follow lightlike world lines where v=c in all reference frames, and massive objects follow timelike worldlines, meaning that v<c in all reference frames. Third, for any arbitrary timelike worldline, the spacetime interval is proportional to the amount of time elapsed on a clock traveling along that worldline which is called the "proper time" and is a Lorentz-invariant quantity. Fourth, for timelike worldlines the paths which locally "minimize" the spacetime interval (in the sense of being a geodesic) are the paths which locally maximize the proper time.

Now, let's think about the global definition of a geodesic and how this applies to throwing a baseball given this information about the spacetime interval. First, just applying the formula you can see that a good baseball pitch goes about 60 feet in the space direction, but about 400 million feet in the time direction. That is the reason why the small curvature in the time direction becomes so important. Now, we are claiming that the spacetime path from the release of the pitch to the contact of the ball with the bat is a geodesic (neglecting air resistance) meaning that it is the shortest path through spacetime connecting the release and the contact, or equivalently that it maximizes the amount of time that a clock attached to the baseball would record. So, from what we know of time dilation, we can make a "rule of thumb" or two: the ball will tend to go as slow as possible on average since that will minimize the velocity time dilation, similarly the ball will tend to go as high as possible on average since that will minimize the gravitational time dilation. However, there is a trade-off between these two, the higher it goes the faster it needs to go on average in order to get up and back in time, and the slower it goes the lower it needs to travel on average in order to reduce the distance. The best trade-off is the parabolic path, this path spends most of its time going as high and slow as it can without "overdoing" it and requiring overcompensation on some other part of the path. This is the path which minimizes the spacetime "distance" by maximizing the time measured on a clock between the release and the contact. It is in this sense that it is a geodesic "straight line" in a curved spacetime.
 
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  • #43
I need to pick this apart.
DaleSpam said:
Third, for any arbitrary timelike worldline, the spacetime interval is proportional to the amount of time elapsed on a clock traveling along that worldline which is called the "proper time" and is a Lorentz-invariant quantity.
Are you talking about two different times here? One that makes up the interval and one on the clock we are traveling with? Is this relevant to the twins paradox?
DaleSpam said:
Fourth, for timelike worldlines the paths which locally "minimize" the spacetime interval (in the sense of being a geodesic) are the paths which locally maximize the proper time.
So this is just like time efficiency during ones day? To minimize the space we use in a day we need to maximize the time we use. Is this right? When talking about "time" the shortest path is "maximizing" and when talking about space/spacetime intervals the shortest path is "minimizing"?
DaleSpam said:
First, just applying the formula you can see that a good baseball pitch goes about 60 feet in the space direction, but about 400 million feet in the time direction.
Is there some ratio equivalence between feet in time and feet in space? I think there must be. What is the common ground that I can compare them to? It seems there must be a net inequivalence between them that necessitates distortion.
 
  • #44
Frame Dragger said:
In the absence of a planet to "stand on", there's nothing to keep you on the SURFACE of the planet. You're still constantly being "sucked" at a rate of 9.8m/s^2, but now you're experiencing the resistance of the ground.
Whether it's how you intended it or not, this description reinforces the absurdity that falling objects have no forces acting on them (are "inertial"). If the electromagnetic force is constantly and steadily "preventing" us from continuing our fall, then there must be a "force" that it is resisting. The geometry of Earth's surface is constant. This certainly means that the electromagnetic force keeping us on the surface is a "resistant" force and not an "offensive" one (I'm sure there's a better word than "offensive", but I can't think of one right now). How could it be offensive if the Earth's surface is not moving?

It seems like relativity calibrates "0-force" at the force of gravity. In other words, the "force" of gravity is not acknowledged because it's set as the "0" point. Does that makes sense? In post #4 of the thread "Are photons affected differently by gravity", bcrowell said,
In GR, the "lines" are geodesics, which are interpreted as the world-lines of particles that aren't subjected to any nongravitational forces.
This is different from the standard definitions I see because he adds the word "nongravitational". Other definitions would just say "aren't subjected to any forces."
 
  • #45
Hoku said:
I need to pick this apart.Are you talking about two different times here? One that makes up the interval and one on the clock we are traveling with? Is this relevant to the twins paradox? So this is just like time efficiency during ones day? To minimize the space we use in a day we need to maximize the time we use. Is this right? When talking about "time" the shortest path is "maximizing" and when talking about space/spacetime intervals the shortest path is "minimizing"?

Consider two observers, each carrying their own clock. Initially both are together at some point in time in space and then leave at the same time and go on slightly different paths before coming together at another point in time and space. The one that free falls from event 1 to event 2 will show the greatest elapsed time on their own clock (their proper time). This is an example of inertial motion maximising the proper time. The observer that does not free fall, has to be artificially accelerated at some point in order to leave the geodesic and this decreases the proper time for him. In terms of the twin's paradox, the twin that follows the shortest path through spacetime (as drawn on a time and space diagram) always shows the most elapsed time on their own clock, which seems counterintuitive, because we are used to the shortest route in everyday life being the one that takes the least time.

Hoku said:
Is there some ratio equivalence between feet in time and feet in space? I think there must be. What is the common ground that I can compare them to? It seems there must be a net inequivalence between them that necessitates distortion.

The speed of light is about 1,000,000,000 feet per second, so if you remain stationary in the spatial sense for one second, you travel 1,000,000,000 feet in the time direction. The ration factor between feet in space terms and feet in time terms is the speed of light c.

Hoku said:
Whether it's how you intended it or not, this description reinforces the absurdity that falling objects have no forces acting on them (are "inertial"). If the electromagnetic force is constantly and steadily "preventing" us from continuing our fall, then there must be a "force" that it is resisting.

I think the important point here is that when you are standing on the Earth, there IS a force of gravity acting on you and there IS also a reactive force of the ground pressing back on you. If the surface you are standing on collapses and assuming no air friction, then you fall freely and no forces are acting on you, so you follow a geodesic.
 
  • #46
Hoku said:
Whether it's how you intended it or not, this description reinforces the absurdity that falling objects have no forces acting on them (are "inertial"). If the electromagnetic force is constantly and steadily "preventing" us from continuing our fall, then there must be a "force" that it is resisting. The geometry of Earth's surface is constant. This certainly means that the electromagnetic force keeping us on the surface is a "resistant" force and not an "offensive" one (I'm sure there's a better word than "offensive", but I can't think of one right now). How could it be offensive if the Earth's surface is not moving?

Well, there goes any hope of my being the next "Great Communicator"! :smile: No... I didn't intend that, but I can see how that would be misleading. Perhaps I should join you on the learning end of this thread, and leave the teaching for sometime when I'm not killing my own point! :wink:

EDIT: "If the surface you are standing on collapses and assuming no air friction, then you fall freely and no forces are acting on you, so you follow a geodesic." Forgive me, but the image in my mind is of some hapless man in a business suit, suddenly plummeting into a hellish oscillation along his geodesic. That's a damned effective way of making a point! :-p
 
  • #47
Hoku said:
I need to pick this apart.Are you talking about two different times here? One that makes up the interval and one on the clock we are traveling with?
Yes, the dt is "coordinate time" which is the time measured in an inertial coordinate system (related to each other via the Lorentz transform). That is only equal to the "proper time" in the rest frame of an inertially-moving clock.

Hoku said:
So this is just like time efficiency during ones day? To minimize the space we use in a day we need to maximize the time we use. Is this right? When talking about "time" the shortest path is "maximizing" and when talking about space/spacetime intervals the shortest path is "minimizing"?
Technically the real term is "extremizing" but that just sounds awkward, so I usually say "minimizing" and hope I don't confuse anyone. But yes, you are maximizing proper time and minimizing the square of the spacetime interval (ds² up above).

Hoku said:
Is there some ratio equivalence between feet in time and feet in space? I think there must be. What is the common ground that I can compare them to? It seems there must be a net inequivalence between them that necessitates distortion.
Yes, as you can see from the formula (ds² = -c²dt² + dx² + dy² + dz²) the ratio is the speed of light. In the case of a baseball pitch, the distance from the mound to the batter is 60 feet, and a good pitch takes about 0.4 s (100 mph). 0.4 s times c is 400 million feet. So, a good pitch goes 60 feet through space and 400 million feet through time (0.4 s times c).
 
  • #48
kev said:
The observer that does not free fall, has to be artificially accelerated at some point in order to leave the geodesic and this decreases the proper time for him. In terms of the twin's paradox, the twin that follows the shortest path through spacetime (as drawn on a time and space diagram) always shows the most elapsed time on their own clock, which seems counterintuitive, because we are used to the shortest route in everyday life being the one that takes the least time.
Wouldn't this depend on two things, one being the path the artifically accelerated observer took and the second being the point in spacetime where they met up again? Let's use a whirlpool to represent curved spacetime. Let's say the inertial observer (A) and the accelerated observer (B) intend to meet up at 90 deg. from where they started. The inertial observer travels around the whirlpool along his geodesic but (B) has two travel options; he can either go out from the center of the whirlpool, making a big loop or take a "shortcut" by cutting through the inside of the whirlpool. If (B) takes the shortcut, he would have to go slower than (A) and his clock would show a lesser elapsed time because he was closer to the center of the whirlpool. However, if (B) took a loop away from the center of whirlpool he'd have to go faster and his clock would show a greater elapsed time because he was farther from the center.
 
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  • #49
kev said:
If we create a tunnel right throught the Earth (and stop it spinning for a while) then if you release a particle from your hand at the surface of the Earth, it falls to the centre and then continues right through to briefly appear at the surface on the other side of the world before falling back to the centre and continuing back up until it eventually returns to your hand. If you do not catch the particle it will continue to fall up and down like this for a long time. This spatial geodesic is a straight line that extends form one side of the world to the other, but it does not extend far out into space.
So it is agreed that falling up and down through the center of the Earth is a "straight" geodesic. I will use this in my argument that gravity is a "force". What is happening at the crutial moment when an object stops it's momentum and returns to the center? It is "change in the motion of a particle". It is a change in velocity. These things are the result of a force.
kev said:
I think the important point here is that when you are standing on the Earth, there IS a force of gravity acting on you and there IS also a reactive force of the ground pressing back on you. If the surface you are standing on collapses and assuming no air friction, then you fall freely and no forces are acting on you, so you follow a geodesic.
I'm going to use another diving analogy. Divers have a pressure gauge to determine how many atmospheres of pressure they are under. When at the surface, the pressure gauge says "0". When at 66 feet the gauge says, "2". Does this mean there is no pressure at the surface? No. There is a pressure of "1". "1" is the absolute pressure at the surface and "3" is the absolute pressure at 66 feet, even though the gauge says "2". But the pressure gauge isn't concerned with pressure at the surface so it doesn't factor it in. I believe this is exactly the same with the force of gravity. Once the ground "falls away", gravity IS a force in action. Relativity theory says there isn't but I'm certain the reason why is that relativity theory is going on "gauge pressure", or "gauge force". I don't mind speaking in gauge terms, but I think it is important that we acknowlege that the "absolute force" is a different value.
 
  • #50
Hoku said:
So it is agreed that falling up and down through the center of the Earth is a "straight" geodesic. I will use this in my argument that gravity is a "force". What is happening at the crutial moment when an object stops it's momentum and returns to the center? It is "change in the motion of a particle". It is a change in velocity. These things are the result of a force...

It's the result of the initial condition the falling body had. Think of a simple rock-on-a-rope Pendulum; All of the 'force' involved was the person drawing it back, the rest is simple oscillation.

As the falling sucker races towards the center of the Earth, there is in fact no force to STOP him! The return trip (and all subsequent miserable experineces for the faller) would diminish in terms of the space-like magnitude, just as a pendulum does. It's still a matter of following a geodesic (and ignoring friction, etc).
 
  • #51
Hoku said:
Wouldn't this depend on two things, one being the path the artifically accelerated observer took and the second being the point in spacetime where they met up again? Let's use a whirlpool to represent curved spacetime. Let's say the inertial observer (A) and the accelerated observer (B) intend to meet up at 90 deg. from where they started. The inertial observer travels around the whirlpool along his geodesic but (B) has two travel options; he can either go out from the center of the whirlpool, making a big loop or take a "shortcut" by cutting through the inside of the whirlpool. If (B) takes the shortcut, he would have to go slower than (A) and his clock would show a lesser elapsed time because he was closer to the center of the whirlpool. However, if (B) took a loop away from the center of whirlpool he'd have to go faster and his clock would show a greater elapsed time because he was farther from the center.
Be careful here. A geodesic is a local minimum, meaning that it is shorter than any possible path that deviates by an infinitesimal amount. When you are comparing paths that deviate by large amounts then it is possible for a non-geodesic path to be shorter than a geodesic path.

E.g. consider geodesics between two points on a sphere, there are generally two geodesics which connect the two points, both are great arcs but one is much longer than the other. There are non-geodesic paths which are shorter than the larger of the two geodesics (but no non-geodesic path is shorter than the short geodesic).
 
  • #52
Frame Dragger said:
It's the result of the initial condition the falling body had.
This is exactly the diverging point between the force view and the geometric view. The geometric view accepts the "initial conditions" without a need to explain how they came about. The force view goes beyond this by saying that the force is what established the initial contidions in the first place.
 
  • #53
Hoku said:
This is exactly the diverging point between the force view and the geometric view. The geometric view accepts the "initial conditions" without a need to explain how they came about. The force view goes beyond this by saying that the force is what established the initial contidions in the first place.
This is not correct, the force view requires initial conditions also. In fact, the initial conditions are the same in both cases, a position and a velocity.
 
  • #54
So... was I correct?... if not, please do correct me, I'm sure that I have enough misconceptions already!
 
  • #55
DaleSpam said:
E.g. consider geodesics between two points on a sphere, there are generally two geodesics which connect the two points, both are great arcs but one is much longer than the other. There are non-geodesic paths which are shorter than the larger of the two geodesics (but no non-geodesic path is shorter than the short geodesic).
Things are different on a sphere because you don't have the option to "cut through" the middle "ball" section. You are restricted to traveling along the surface. But spacetime IS 3-dimentional so I'd think cutting through the "inside" of the sphere should be an option for an accelerated observer.
 
  • #56
Hoku said:
Things are different on a sphere because you don't have the option to "cut through" the middle "ball" section. You are restricted to traveling along the surface. But spacetime IS 3-dimentional so I'd think cutting through the "inside" of the sphere should be an option for an accelerated observer.

Ah ah ah... spacetime is 4-dimensional. 3+1. That might be your confusion on the issue?
 
  • #57
Hoku said:
Things are different on a sphere because you don't have the option to "cut through" the middle "ball" section. You are restricted to traveling along the surface. But spacetime IS 3-dimentional so I'd think cutting through the "inside" of the sphere should be an option for an accelerated observer.
As Frame Dragger said, spacetime is 4 dimensional (3+1), however the point of the sphere example is not the dimensionality but just the familiarity. You already understand geometry on the curved 2D surface of a sphere, so many of those concepts translate directly to geometry in a curved 4D spacetime.

Do you understand the idea that geodesics are local minima and there can globally exist non-geodesic paths which are shorter than a geodesic path?
 
  • #58
DaleSpam said:
This is not correct, the force view requires initial conditions also. In fact, the initial conditions are the same in both cases, a position and a velocity.
I understand what you are saying here and I do agree. Whether you use Newtonian equations or Relativity equations you are using the same "initial conditions" to make predictions. To better understand where I'm coming from, I'll call, again, on the quote from Wikipedia on "gravitation":
[general relativity] equates free fall with inertial motion, and describes free-falling inertial objects as being accelerated relative to non-inertial observers on the ground. In Newtonian physics, however, no such acceleration can occur unless at least one of the objects is being operated on by a force.
A free-falling, inertial object has position and velocity. The force view says that "force" is what gives it that whereas the geometric view says there is no force involved. Here's another analogy:

A gear serve the purpose of moving things. It's the gears relative position and velocity with the other gears, etc. that is useful for making predictions. However, a moving gear that is not engaged will have no meaningful position or velocity. It is pure force without application. Once it is APPLIED we can "define" the "initial conditions" that are meaningful for making predictions. These defined "initial conditions" are the same for both the force view and the geometric view. The only difference is that the "force" view suggests that force is inherent to the initial conditions and the geometric view simply accepts the initial conditions as primary. Does that make more sense?
 
  • #59
Frame Dragger said:
Ah ah ah... spacetime is 4-dimensional. 3+1. That might be your confusion on the issue?
I don't think this is a problem. I'll call upon something kev said and then elaborate my understanding of it...
kev said:
As Dalespam has pointed out, we normally consider the time dimension when considering geodesics but it is OK to talk about paths in space as long as it is clear that that is what we are doing (i.e. considering only the spatial components of the geodesic).
The "time" portion of spacetime is a dimension of "change". So I believe we can discuss spacetime in 3-dimensions as long as we understand that the "time" component unfolds within it. The idea of taking a "shortcut" through a 3-dimensional "ball" is still compatible in 4-dimensional spacetime.
DaleSpam said:
Do you understand the idea that geodesics are local minima and there can globally exist non-geodesic paths which are shorter than a geodesic path?
I'm pretty sure I do. I think my whirlpool analogy and "ball shortcut" idea demonstrate this. Do you agree? I feel like I understand it.
 
  • #60
Hoku said:
So it is agreed that falling up and down through the center of the Earth is a "straight" geodesic. I will use this in my argument that gravity is a "force". What is happening at the crucial moment when an object stops it's momentum and returns to the center?
It is a straight path in 3D space, but it is a curved geodesic in 3+1D space that includes time, as illustrated in the nice animation in the link you gave earlier.

I think a lot of the confusion here revolves around the definition of force. One definition given in Newtonian mechanics is f = ma, which is mass times change in velocity per unit time or mass times change in location per unit time squared. Now one difficulty with this definition is considering the force acting on an object resting on a table for example. It is going nowhere, so its change in location per unit time is zero so the force acting on it must be zero, yet we know there is a force acting on it even when it is stationary on the table. Einstein resolved this difficulty by defining acceleration as that which is measured by an accelerometer. An accelerometer shows an acceleration when resting on the surface of a table, so it clearly indicates there is a force acting on it even when it is going nowhere. When the accelerometer is dropped, it indicates no acceleration and this indicates that no force is acting on it when it is falling.
Hoku said:
It is "change in the motion of a particle". It is a change in velocity. These things are the result of a force.
As above, if the definition of force involves "change in the motion of a particle" then the object resting on the table has no force acting on it, which you know is not true. The relativistic definition of acceleration and force is almost a complete reversal of the Newtonian concept, but a more logical definition when you consider the example of the table.
Hoku said:
I'm going to use another diving analogy. Divers have a pressure gauge to determine how many atmospheres of pressure they are under. When at the surface, the pressure gauge says "0". When at 66 feet the gauge says, "2". Does this mean there is no pressure at the surface? No. There is a pressure of "1". "1" is the absolute pressure at the surface and "3" is the absolute pressure at 66 feet, even though the gauge says "2". But the pressure gauge isn't concerned with pressure at the surface so it doesn't factor it in. I believe this is exactly the same with the force of gravity.
Yes, but we can construct a pressure gauge calibrated for zero using a vacuum and detect the non zero pressure at the surface of the sea. Now if we calibrate an accelerometer for zero in flat space far from any gravitational forces, then we still can not detect any acceleration when the accelerometer is falling in a gravitational field. In other words the zero acceleration indicated by an accelerometer in free fall is a true zero or absolute zero and not a gauge zero. There is no way to construct an accelerometer so that it can detect the acceleration of free falling.

An observer sealed in a box that is dropped down a shaft going through the centre of a massive body and left to oscillate up and down, can not detect anything different from the experience of another observer in a similar sealed box inside a peacefully orbiting spacestation.
Hoku said:
Things are different on a sphere because you don't have the option to "cut through" the middle "ball" section. You are restricted to traveling along the surface. But spacetime IS 3-dimentional so I'd think cutting through the "inside" of the sphere should be an option for an accelerated observer.
The surface of a ball analogy used in relativity is usually meant to simulate a 2D surface embedded in a higher dimension that the observers on the surface do not have access to. Taking a short cut through the ball is equivalent to taking a shortcut though the 5th or 6th dimension to get to your friend's house. Those extra dimension are simply not available to us (with current technology :P)

As Dalespam has indicated, there can be a number of geodesics that connect two events in spacetime, but the path that has longest proper time (as indicated by a clock carried from one event to the other) involves no acceleration as measured by an onboard accelerometer (i.e is inertial).

[EDIT] I thought I better add that there can be more than one "shortest" path connecting one point to another as is obvious by considering the multiple shortest routes from the North Pole to the South Pole.
 
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  • #61
"There is no way to construct an accelerometer so that it can detect the acceleration of free falling."

Hmmm, purely in the realm of "in a far future", if one could reference sources of gravitational waves, and define a coordinate system accordingly... you could probably detect acceleration in a gravitational field. It would still be artificial, based on a kind of "fixed stars" replacing gravitational radiation with stars.

Still... I wonder if that would work? We can measure other pseudoforces after all...
 
  • #62
Hoku said:
The only difference is that the "force" view suggests that force is inherent to the initial conditions and the geometric view simply accepts the initial conditions as primary. Does that make more sense?
Again, this is not correct. Both views are the same in this respect. In both cases the physics of a situation can be divided into the laws and the initial conditions. In both cases the initial conditions consist of the masses and their initial positions and velocities. In one case the law describes the forces and in the other case the law describes the geometry, but the laws and the initial conditions do not have any different relationship to each other in either approach. The two views are completely equivalent with the only difference being that the geometric approach is more accurate in certain circumstances.
 
  • #63
Frame Dragger said:
We can measure other pseudoforces after all...
No, you can't. In all cases what you measure is the acceleration due to the sum of the real forces only.
 
  • #64
DaleSpam said:
No, you can't. In all cases what you measure is the acceleration due to the sum of the real forces only.

Ah, well then, so much for that!
 
  • #65
Hoku said:
I will use this in my argument that gravity is a "force". What is happening at the crutial moment when an object stops it's momentum and returns to the center? It is "change in the motion of a particle". It is a change in velocity. These things are the result of a force.
Sorry Hoku, I didn't respond to this earlier. However, I would like to point out that my baseball pitch description works also for this. All you have to do is to change the spatial distance from 60 feet to 0 feet.

DaleSpam said:
Now, let's think about the global definition of a geodesic and how this applies to throwing a baseball given this information about the spacetime interval. First, just applying the formula you can see that a good baseball pitch goes about 60 feet in the space direction, but about 400 million feet in the time direction. That is the reason why the small curvature in the time direction becomes so important. Now, we are claiming that the spacetime path from the release of the pitch to the contact of the ball with the bat is a geodesic (neglecting air resistance) meaning that it is the shortest path through spacetime connecting the release and the contact, or equivalently that it maximizes the amount of time that a clock attached to the baseball would record. So, from what we know of time dilation, we can make a "rule of thumb" or two: the ball will tend to go as slow as possible on average since that will minimize the velocity time dilation, similarly the ball will tend to go as high as possible on average since that will minimize the gravitational time dilation. However, there is a trade-off between these two, the higher it goes the faster it needs to go on average in order to get up and back in time, and the slower it goes the lower it needs to travel on average in order to reduce the distance. The best trade-off is the parabolic path, this path spends most of its time going as high and slow as it can without "overdoing" it and requiring overcompensation on some other part of the path. This is the path which minimizes the spacetime "distance" by maximizing the time measured on a clock between the release and the contact. It is in this sense that it is a geodesic "straight line" in a curved spacetime.
So with 0 feet in the spatial direction we still identify a geodesic in the same way, the path which maximizes the time recorded on a clock, so again we want to go high and slow. This path has the "crutial moment" you describe above, but there is nothing special about this moment. It is still going straight through a curved spacetime, following the path which minimizes the spacetime interval squared. That point is just one point on the minimal path.
 
  • #66
Sometimes it's harder for me to get to this thread that I'd like but it's not because I'm not chomping at the bit to do so. I've been dedicated to this, heart and soul, since it began in the "3+1" thread. Also, please know that I won't just "abandon" the thread without wrapping it up. I know I didn't need to say this, but it makes me feel better that I did.

What I want to do right now is give a brief description, in laymans terms, of curved spacetime as I've come to understand it from our discussions. I also want to pursue the discussion of what a "force" is, as kev was describing it in post #60, but that will wait. Ok, here I go:

Spacetime is viewed as a single "fabric". But it's important to keep in mind that a "fabric" is made of two groups of threads; one group that goes crossways and a second group perpendicular to this that goes up and down. With spacetime, one "group" is space and the other "group" is time. This is why spacetime is separated into "3"+"1". So, space and time are different from one another, however, their threads are so entwined that if you do something to one group, it necessarily affects the other. Furthermore, without both threads, there is nothing meaningful to either. Consequently, they can't be separated without essentially "disintegrating" the entire universe. Given this, let's look at "curved" spacetime.

There is no way to avoid curvature of spacetime. When two objects are near enough, curvature is the only way that spacetime can be experienced. It is impossible to "simulate" flat spacetime in curved spacetime. However, this doesn't mean that both space and time must be curved simultaneously. It only means that the net value between them will always be curved. Let's look at "straight" space and "straight" time separately.

It's not difficult to imagine straight space, just draw a straight line from one point to another. Straight time isn't that much more difficult to understand but, first, we need to understand how we know when time is "straight". It's quite simple, really. Straight time is measured by constancy. If a clock maintains it's speed and doesn't slow down or speed up, then it's moving in a "straight line". The interesting thing about this, is that time is only "straight" when it's measured in a circle that has the essential point of gravity at its center. From this, we can understand how curved spacetime is unavoidable. If we go straight in space, we can no longer following the "circle" that keeps time straight. If we follow the circle that keeps time straight, we can no longer move straight in space. So, one way or another, spacetime will always be curved.

We are what is called "timelike", which means that we are naturally inclined to follow a path that keeps time straight. (This doesn't really make sense, though, because our bodies "want" to go straight in space to the center of the Earth, which is not straight in time, as I defined it. *heavy sigh*)
 
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  • #67
Hoku said:
Spacetime is viewed as a single "fabric". But it's important to keep in mind that a "fabric" is made of two groups of threads; one group that goes crossways and a second group perpendicular to this that goes up and down. With spacetime, one "group" is space and the other "group" is time. This is why spacetime is separated into "3"+"1". So, space and time are different from one another, however, their threads are so entwined that if you do something to one group, it necessarily affects the other. Furthermore, without both threads, there is nothing meaningful to either. Consequently, they can't be separated without essentially "disintegrating" the entire universe. Given this, let's look at "curved" spacetime.

There isn't a unique 3+1 separation of spacetime. At every point in spacetime, there are many threads that could be considered timelike, not just one.
Hoku said:
There is no way to avoid curvature of spacetime. When two objects are near enough, curvature is the only way that spacetime can be experienced. It is impossible to "simulate" flat spacetime in curved spacetime. However, this doesn't mean that both space and time must be curved simultaneously. It only means that the net value between them will always be curved. Let's look at "straight" space and "straight" time separately.

Space and time must be curved simultaneously. The curvature is that of spacetime. You will hear people say "time is curved", but that is as an approximation within one particular 3+1 split of spacetime.

Hoku said:
It's not difficult to imagine straight space, just draw a straight line from one point to another. Straight time isn't that much more difficult to understand but, first, we need to understand how we know when time is "straight". It's quite simple, really. Straight time is measured by constancy. If a clock maintains it's speed and doesn't slow down or speed up, then it's moving in a "straight line". The interesting thing about this, is that time is only "straight" when it's measured in a circle that has the essential point of gravity at its center. From this, we can understand how curved spacetime is unavoidable. If we go straight in space, we can no longer following the "circle" that keeps time straight. If we follow the circle that keeps time straight, we can no longer move straight in space. So, one way or another, spacetime will always be curved.

A clock never slows down or speeds up. You should distinguish between proper time and coordinate time.

Hoku said:
We are what is called "timelike", which means that we are naturally inclined to follow a path that keeps time straight. (This doesn't really make sense, though, because our bodies "want" to go straight in space to the center of the Earth, which is not straight in time, as I defined it. *heavy sigh*)

Timelike means that if a light beam intersects us at any point in spacetime, we will always measure it to be traveling faster than us.
 
  • #68
atyy said:
There isn't a unique 3+1 separation of spacetime. At every point in spacetime, there are many threads that could be considered timelike, not just one.
I wonder if you're not understanding my analogy. I know that there are many different "time thread" options based on velocity, that's part of what makes it a fabric. But your use of the word "timelike" here is also confusing, based on your definition of it at the end of your post. At any rate, time and space are differentiated, if only from the -+++ formula, so I think this does make a unique separation.
atyy said:
Space and time must be curved simultaneously. The curvature is that of spacetime.
I understand that this is true in inertial objects, but I think that you can "straighten" one or the other of them with acceleration.
atyy said:
A clock never slows down or speeds up. You should distinguish between proper time and coordinate time.
I agree that you can only say "slowed down" or "sped up" when comparing proper time to coordinate time. I'm just trying to say that if there is time dialation, time is not "straight".
atyy said:
Timelike means that if a light beam intersects us at any point in spacetime, we will always measure it to be traveling faster than us.
I guess I meant to say that we follow timelike geodesics.
 
  • #69
atyy said:
Timelike means that if a light beam intersects us at any point in spacetime, we will always measure it to be traveling faster than us.
Are you saying that the only thing defining a timelike geodesic is it's restriction to velocities below the speed of light? That's not complicated at all. Is it really that simple?
 
  • #70
kev said:
As above, if the definition of force involves "change in the motion of a particle" then the object resting on the table has no force acting on it, which you know is not true.
I think there IS change in motion of a particle resting on a table. The change is that it now isn't moving. It's inertial velocity is being held in check. The force to do this must be constant, which means that a particle resting on a table is undergoing a constant "change in motion" just to be "resting".
kev said:
The relativistic definition of acceleration and force is almost a complete reversal of the Newtonian concept[...]
This seems like another classic "measurement problem" that we encounter in science.
kev said:
Now if we calibrate an accelerometer for zero in flat space far from any gravitational forces, then we still can not detect any acceleration when the accelerometer is falling in a gravitational field. In other words the zero acceleration indicated by an accelerometer in free fall is a true zero or absolute zero and not a gauge zero.
Here is that measurement problem, again. I wonder if the problem is that the an inertial particle, whether in flat space or curved, always has curved space around it. All particles curve space, right? They all produce "gravity". If all inertial movement is the result of different "gravities" interacting, then the particle is really just along for the ride. All the "work" is done in the spacetime surrounding the particle. As long as the gravities are free to be natural, the particle is "protected" by it's own "gravitational shield". Does this make sense?
 

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