How Do Geometric Shapes Influence the Path of Gravity?

[Dale]

So, my question is, why does time care whether we go "Earthwards" or "upwards"?

Let's consider a somewhat related question that may be easier for you to visualize where we are considering curved coordinates on a flat space. Specifically, let's consider geodesics in a plane (straight lines) from the perspective of polar coordinates. Now, in terms of polar coordinates all geodesics get deflected "outwards". Why do polar coordinates care whether a geodesic goes outwards or inwards? Do you have an intuitive feel for that?

 

[Hoku]

I'm optimistic right now because it seems like the geometric understanding, as well as the "force" reconcilation, that I'm looking for is close. Then again, I could be painfully wrong.

I've been having trouble finding time to finish this post. I should be able to get back to it later today.

 

[Hoku]

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.

Specifically, let's consider geodesics in a plane (straight lines) from the perspective of polar coordinates. Now, in terms of polar coordinates all geodesics get deflected "outwards". Why do polar coordinates care whether a geodesic goes outwards or inwards?

I think the only thing time "cares" about is that it isn't repeated. Otherwise, I don't think either coordinate cares what happens on the "canvas". Time requires movement "upwards" on the graph, but I don't think it cares about direction in space.

I've been up and down with this all day (no graph pun intended). Sometimes optimistic that progress is happening (thanks to the torus diagrams) and sometimes worried that I'm still missing it. Here's how I've organized the rest of this post:
First, I've re-establish the primary problem. Second, I've brought in answers that have been provided. Third, I've paraphrased and elaborated on these answers in case there's a flaw in my interpretation. Fourth and finally, I've described exactly why these answers confuse me. For the first time, I think I can do this effectively and that's where my greatest hope lies.

1) Primary problem:
If gravity is not a force, then how are we stuck to the Earth?

2) Answer:
Because we are following a geodesic into it. We are never at rest because we are always moving in the "time" direction, which curves down (Earthwards).

3) Answer Paraphrased:
A geodesic is a path in spacetime. Because space and time are inseparable, following a geodesic in time necessitates movement in space in some way. This is why we "stick" to the Earth. We're trying to follow the geodesic spatially, but electromagnetism prevents our movement, which leaves us pressed into the earth.

4) Why this doesn't make sense:
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.

I also wanted to address posts #28 and #30, but I think I've made a big enough mess.

 

[yuiop]

This is fascinating; I never thought I could get or needed a deeper understanding of buoyancy!

@Kev: I truly wish you had been there years ago with that explantion for me. It would have saved me time and a headache! Maybe you can help me with one another type of diagram...

I'm poring over Gravitation (MTW) right now, and I'm trying to learn how to interpret Kruskal Diagrams (Kruskal-Szekeres vs. Schwarzschild). Given time I can do it, while referring to MTW, but to blunt... it's heavy! Can you recommend and article or book that might help with this?

Hi Dragger,
I think it would be great to discuss the KS metric with you. I should warn you that my personal interpretation is not the conventional one, but in keeping with rules of this forum I will try and keep to the conventional interpretation. I think you should also start a new thread to avoid hijacking this thread and specify the aspects of the KS metric that you find puzzling.

I know that if you go in a circle, you are not "going back", however, you DO go from 0m, 1m, 2m, 3m and then BACK to 2m, 1m, 0m. So you ARE passing through the "same" places in that sense. If I jump up to reach a branch, I end up, essentially back where I started; not in time, but certainly in space. I guess another way to approach this confusion is to ask, "why must movement in time restrict us to one direction in space (mass-wards)?"

Hi Hoku,
Have another look at the link you posted earlier http://www.adamtoons.de/physics/gravitation.swf

Click play and watch the motion of the particle on the rectangular time and space chart on the left. The particle rises to its maximum height and falls back down again. Now imagine playing the animation backwards (reversing time) and it is easy to see that the particle still rises and falls back even when time is going backwards! So it is not the direction of time that determines this rising and falling path. It is the way the time dimension curves with respect to the space dimensions (together with the initial location and velocity of the particle) that determines the path followed by the particle. This curving of the time and space dimension is illustrated by the bulbous cylindrical surface in the 3D diagram on the right. Here time is the circular cross section of the distorted cylinder and wraps around, as you can see by watching the particle cycle around. If the animation was longer, the particle would oscilate to and fro indefinitely (assuming a tunnel through the massive body and no friction). This is the natural state of the particle with NO forces acting on it and it is following a geodesic. If the tunnel is blocked the particle is no longer following a geodesic and it feels a force acting on it (the same force you feel acting on your feet when you are standing) and so it no longer has inertial motion, which is defined as the motion of a particle with no forces acting on it.

If you mentally cut off the the right third of the distorted cylinder in the adamtoons animation, you essentially get the curved spacetime funnel shape in the link I gave earlier here http://www.rpi.edu/dept/phys/Dept2/Courses/ASTR2050/CurvedSpacetimeAJP.pdf"

4) Why this doesn't make sense:
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.

You should not get too hung up on the effect of the stars and the Sun in this situation. You will still be pressed to the Earth even if the Sun and stars are absent and if the Earth is not spinning. There is no requirement to have an "outlet" for movement in space, but as long as you are not exactly following a geodesic, you will feel a force acting upon you. Put another way, if you require that a particle must have inertial motion (and feel no forces acting on it), then the particle must follow its geodesic though spacetime and any movement in the space diirections that is not exactly the same as the spatial components of the geodesic path, will not meet that requirement and so the spin of the Earth abouts its own axis and the orbit of the Earth around the Sun will not do the trick.

 

[Frame Dragger]

@kev: Understood, I'll do that this weekend or monday. Thanks!

@Hoku: Remember that Earth, along with everything including our local cluster, is hurtling "outwards" from the presumed BB point of origin, at a fairly impressive rate relative to that presumed origin point.

Even if we were not, moving in time-like geodesics is what we do. We could be alive, utterly still in a vacuum, and we wouldn't REALLY be utterly still. To be alive, even on the (biological) microscropic level, requires constant motion and exchanges of molecules and charges, etc. All of that follows a time-like geodesic too. To meet our definition of LIFE, you must follow a time-like geodesic (for starters only) to participate in thermodynamic processes' which underly life (we eat, we radiate heat, and create material waste with less chemical energy than what "went in").

Kev is right in forgetting celestial bodies, but you can really just think of a single cell in a medium (agar) being alive. If we can call it alive, it's experiencing the passage of time, does not exactly follow a space-like geodesic, and will experience a ficticious force called gravity, even if it is only from the medium, and its own mass. Reduce to absuridity as needed!

Anyway, remember the parable of the apple, but now apply it to curved time. To "get" from one moment to the next requires that we follow time-like geodesics, or we're not getting anywhere at all.

 

[Dale]

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?

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?

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.

 

[yuiop]

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 through 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.

 

[Hoku]

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.

 

[Frame Dragger]

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?

 

[Hoku]

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!

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).

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.

[...] 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.

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.

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?

 

[Frame Dragger]

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. 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.

 

[Dale]

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.

 

[Hoku]

I need to pick this apart.

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?

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"?

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.

 

[Hoku]

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."

 

[yuiop]

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.

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.

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.

 

[Frame Dragger]

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"! 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!

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!

 

[Dale]

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.

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).

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).

 

[Hoku]

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.

 

[Hoku]

If we create a tunnel right through 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.

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.

 

[Frame Dragger]

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).

 

[Dale]

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).

 

[Hoku]

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.

 

[Dale]

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.

 

[Frame Dragger]

So... was I correct?... if not, please do correct me, I'm sure that I have enough misconceptions already!

 

[Hoku]

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-dimensional so I'd think cutting through the "inside" of the sphere should be an option for an accelerated observer.

 

[Frame Dragger]

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-dimensional 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?

 

[Dale]

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-dimensional 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?

 

[Hoku]

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?

 

[Hoku]

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...

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.

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.

 

[yuiop]

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.

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.

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.

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-dimensional 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.