Exploring Curved Space and Gravity: Understanding the Geodesic Line in Spacetime

In summary, the conversation discusses the concept of geodesic lines in spacetime and how they are not necessarily straight lines in space. It also explores the curvature of spacetime and how it affects the trajectory of objects in Earth's gravity. The idea of "curved time" is introduced, and the concept of proper time as an extremum is discussed. The conversation ends with a suggestion to use the calculus of variations to further explore these concepts.
  • #1
Ratzinger
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The GR equ. tells us that a test particle will follow a geodesic line in spacetime, which is not a geodesic line in space. Usually space is flat, but that does not imply the geodesic line of a test particle in space is a straight line (right?). Since if I throw a ball the gravity of Earth makes the ball go a parabolic line.

So why go balls curved trajectories in the presence of Earth gravity? What does GR say about little balls in Earth's gravity? Can we speak here about space or spacetime curvature that causes its parabolic trajectory?

I know my thinking went somewhere utterly wrong here. Could someone clearify?
Thanks
 
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  • #2
Ratzinger said:
The GR equ. tells us that a test particle will follow a geodesic line in spacetime, which is not a geodesic line in space. Usually space is flat, but that does not imply the geodesic line of a test particle in space is a straight line (right?).
In flat spacetime, a geodesic is always a straight worldline. In flat space but not flat spacetime (like a spatially flat expanding universe), this isn't true, although I'm not sure whether the particle's path through space would still always be straight in this case.
Ratzinger said:
So why go balls curved trajectories in the presence of Earth gravity, but neither does Earth bend spacetime, nor space (appreciably)?
The Earth definitely curves spacetime appreciably, if you tried to model it as flat you wouldn't get those curved trajectories. A geodesic path always maximizes the proper time, ie the time elapsed on a clock that takes that path, so one way of thinking about the path of a thrown object is that there's a tension between the clock "wanting" to spend as much time as possible at greater heights because it ticks slower when it's closer to the ground due to gravitational time dilation, but not "wanting" to move upwards too quickly because then velocity-based time dilation will slow it down. The parabolic path is the one that ideally balances these opposing tensions to maximize the number of ticks of the clock that travels that path (I saw this explanation in one of Feynman's books).
 
  • #3
Ratzinger said:
The GR equ. tells us that a test particle will follow a geodesic line in spacetime, which is not a geodesic line in space. Usually space is flat, but that does not imply the geodesic line of a test particle in space is a straight line (right?).
First, the ball does not follow anything in 'space', if by space you mean a space-like hypersurface of simultaneity (defined in the observers frame of reference). Anything on that hyper-surface is fixed in time and does not move, you have taken a snap shot of it.
Since if I throw a ball the gravity of Earth makes the ball go a parabolic line.
So why go balls curved trajectories in the presence of Earth gravity? What does GR say about little balls in Earth's gravity? Can we speak here about space or spacetime curvature that causes its parabolic trajectory?
I know my thinking went somewhere utterly wrong here. Could someone clearify?
Thanks
Secondly, you are seeing the ball follow a parabola in 4D space-time, the problem is your time dimension is not to scale.

Suppose you throw a ball so that it lands 100 metres away, after rising to 2.5 metres altitude before falling to the ground ~ 1.5 seconds later. To visualise the trajectory with the time dimension to scale, the time duration would be equal to the distance to the Moon (1.5 light seconds away).

So, visualise your parabola stretched out 100 metres in one direction, 2.5 metres in another and ~ 384,000,000 metres in the other, it is pretty straight! The Earth's gravitational field only deviates from flat space-time by a factor of around 7x10-10 ([itex]\frac{GM}{rc^2}[/itex]).

I hope this has helped.

Garth
 
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  • #4
Garth said:
First, the ball does not follow anything in 'space', if by space you mean a space-like hypersurface of simultaneity (defined in the observers frame of reference). Anything on that hyper-surface is fixed in time and does not move, you have taken a snap shot of it.
But you can also do a "foliation" of spacetime into a series of spacelike hypersurfaces, and then see how an object's position in space changes at different instants. Of course there is no unique way to divide spacetime into a series of instants, that depends on your choice of coordinate system.
 
  • #5
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  • #6
Ratzinger said:
The GR equ. tells us that a test particle will follow a geodesic line in spacetime, which is not a geodesic line in space. Usually space is flat, but that does not imply the geodesic line of a test particle in space is a straight line (right?). Since if I throw a ball the gravity of Earth makes the ball go a parabolic line.
So why go balls curved trajectories in the presence of Earth gravity? What does GR say about little balls in Earth's gravity? Can we speak here about space or spacetime curvature that causes its parabolic trajectory?
I know my thinking went somewhere utterly wrong here. Could someone clearify?
Thanks

Basically it is "curved time", the fact that the rate at which clocks ticks depends on the altitude of the clock, that causes objects to follow a curved trajectory.

Formally, one can do the analysis with "only" the calculus of variations.

We say that the path an object takes makes the proper time an extremum. Suppose we have a particle following a trajectory r(t) in (for example) the Schwarzschild metric. Then we can write (using geometric units, in which c=1 for convenience)

[tex]
d\tau^2 = g_{00} dt^2 - g_{11} dr^2
[/tex]
[tex]
\tau = \int \sqrt{ g_{00} - g_{11} \left( \frac{dr}{dt} \right)^2} dt
[/tex]

where g_00 and g_11 are functions of r. For weak fields and slow velocities, we can ignore the effects of g_11 entirely, and assume that it is unity. Thus we are taking into account only g_00, which represents the fact that clocks tick at different rates at different altitudes. We then have a basic problem in the calculus of variations

http://mathworld.wolfram.com/CalculusofVariations.html

i.e. we are extremizing the intergal

[tex]
\int L(r,\dot{r},t) dt \hspace{.5 in} \dot{r} = \frac{dr}{dt}
[/tex]

with
[tex]
L(r,\dot{r},t) = \sqrt{g_{00}(r) - \dot{r}^2}
[/tex]

which imples that the falling body obeys the Euler-Lagrange equations

[tex]
\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) = \frac{\partial L}{\partial r}
[/tex]

You can solve this in detail if you like (for the specific examle you'll need to look up how g_00 depends on r by looking up the Schwarzschild metric). T

he only points I want to make are very general ones, that the "force" term, [itex]\partial L/\partial r[/itex], results from the fact that g_00 is a function of r, and is in fact proportional to [itex]\partial g_{00}/\partial r[/itex], and that the momentum term is very close to the SR form for momentum, with some small corrections for the metric coefficient g_00.

You can also include the effects of g_11 if you're really ambitious, and show why they are small for low velocities/weak fields.
 
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  • #7
Thanks, gentlemen. You are all great people.

And did I mention recently how much I like this site?
 
  • #8
Ratzinger said:
The GR equ. tells us that a test particle will follow a geodesic line in spacetime, which is not a geodesic line in space. Usually space is flat, but that does not imply the geodesic line of a test particle in space is a straight line (right?). Since if I throw a ball the gravity of Earth makes the ball go a parabolic line.
So why go balls curved trajectories in the presence of Earth gravity? What does GR say about little balls in Earth's gravity? Can we speak here about space or spacetime curvature that causes its parabolic trajectory?
I know my thinking went somewhere utterly wrong here. Could someone clearify?
Thanks
You don't have to have a curved spacetime to have a ball follow a parabolic trajectory in space. In fact a ball will follow a curved spatial trajectory in a uniform gravitational field and a uniform g-field has zero spacetime curvature.

By the way, particles follow geodesics of [bextremal[/b aging, not maximal.

Pete
 
  • #9
pmb_phy said:
You don't have to have a curved spacetime to have a ball follow a parabolic trajectory in space. In fact a ball will follow a curved spatial trajectory in a uniform gravitational field and a uniform g-field has zero spacetime curvature.
Is this is basically the same as noting that in SR an object will follow a curved trajectory in an accelerated coordinate system, and then also noting that the equivalence principle of GR means you can explain the observations of any accelerated observer in terms of a g-field?
pmb_phy said:
By the way, particles follow geodesics of [bextremal[/b aging, not maximal.
What are examples of spacetimes where a geodesic corresponds to minimal aging as opposed to maximal? Would they come up in any realistic physical situations?
 
  • #10
JesseM said:
Is this is basically the same as noting that in SR an object will follow a curved trajectory in an accelerated coordinate system, and then also noting that the equivalence principle of GR means you can explain the observations of any accelerated observer in terms of a g-field?[/qupte]That is exactly right. The answer to the question posed had nothing to do with curved spacetime.
What are examples of spacetimes where a geodesic corresponds to minimal aging as opposed to maximal? Would they come up in any realistic physical situations?
There is an example in MTW. I'll see if I can find it. But think of the photon sphere around a black hole. Let there be a light emitter/detector at rest on that sphere. When the light is emitted let there be a particle of non-zero proper mass released from the emitter/detector such that it comes right back there at the same time the photon is detected which circumvented the photon shere. The worldline of the photon has zero proper time. The worldline of the particle had a propertime greater than zero. Yet each of the particles started and ended at the same event.

Pete
 
  • #11
I should probably note yet again, that "curvature", especially in scare quotes, does not necessarily mean a non-zero Riemann curvature tensor.

Speaking losely, it can apply to any situation where the metric coefficients vary with position.

For an example of this usage, see for instance MTW pg 187, chapter 7, section 3. The title of this chapter is "Gravitational redshift implies space-time is curved".

Reading this section, it is clear that the "curvature" being described in this chapter does necessarily mean a non-zero curvature tensor. MTW, and many other authors, use the term "curvature" to describe any situation where the metric coefficients vary with position.

In fact gravitational red-shift does occur in situations where the Riemann curvature tensor is zero, such as the Rindler metric of an accelerating observer (also sometimes called a uniform gravitational field).

One can either believe that several well-recognized texts are "wrong" and go to great efforts to "correct" them, or one can believe that "curved space-time" does not always mean "a non-zero curvature tensor". I have chosen the later course, pmb has chosen the previous course.
 
  • #12
pmb_phy said:
There is an example in MTW. I'll see if I can find it. But think of the photon sphere around a black hole. Let there be a light emitter/detector at rest on that sphere. When the light is emitted let there be a particle of non-zero proper mass released from the emitter/detector such that it comes right back there at the same time the photon is detected which circumvented the photon shere. The worldline of the photon has zero proper time. The worldline of the particle had a propertime greater than zero. Yet each of the particles started and ended at the same event.
Pete
But we already know that particles that start at the same position with different initial speeds and directions will follow different geodesics--just think of two particles orbiting the sun whose orbits intersect in two distinct locations, if the orbits were arranged right they could meet at each crossing and yet (correct me if I'm wrong) they could have aged different amounts between the two events of their meeting, but I don't think this would contradict the idea that each particle's path between the two events was the one that maximized its aging, given each particle's speed and direction at the first crossing. So why does the photon example show that the slower-than-light particle's proper time was minimized? Would it be true in this case that if you had another particle with the same initial speed and direction as the first one, but which had a non-gravitational force acting on it so it deviated from the first particle's path before reuniting with it later, that the particle deviating from the geodesic path would have actually aged more rather than less as long as they were both inside the photon sphere? Would the twin paradox work in reverse inside the photon sphere, in other words?
 
  • #13
JesseM said:
But we already know that particles that start at the same position with different initial speeds and directions will follow different geodesics...
The principle of extremal aging refers to two events. I.e. the proper time as measured on a clock which passes through both events.
So why does the photon example show that the slower-than-light particle's proper time was minimized?
The proper time along the geodesic of any photon is zero and no positive number has a value less than zero. I could tell you to look in the glossary of Taylor and Wheeler's Exploring Black Holes if you don't believe me but note that I'm the one who wrote that glossary. :biggrin:
Would it be true in this case that if you had another particle with the same initial speed and direction as the first one, but which had a non-gravitational force acting on it so it deviated from the first particle's path before reuniting with it later, that the particle deviating from the geodesic path would have actually aged more rather than less as long as they were both inside the photon sphere? Would the twin paradox work in reverse inside the photon sphere, in other words?
I havne't calculated it bu that would seem so. Note that the principle of extremal aging refers to geodesics not arbitrary worldlines.

Pete
 
  • #14
JesseM said:
What are examples of spacetimes where a geodesic corresponds to minimal aging as opposed to maximal? Would they come up in any realistic physical situations?

I'm not going to give specific examples for spacetime, but I can give a specific example for space, and I can talk a little about what happens in general.

First, consider the analgous situation in 2-dimensional spaces (not spacetimes). For points p and q in flat R^2, a geodesic (i.e., a straight line) is a local minimum (in terms of length) for curves that run from p to q. For point p and q in curved 2-dimensional spaces (i.e., positive definite Reimannian manifolds), it is not always the case that a geodesic that joins p and q is a local minimum in terms of length.

As an example, consider S^2, the 2-dimensional surface of a 3-dimensional ball, and, for concreteness, take the surface to be the surface of the Earth. Take p to be the north pole and q to be Greenwich England.

The shortest route from the north pole to anywhere is along the appropriate line of longitude, and the shortest route from the north pole to Greewich is south along the 0 line of longitudeprime (prime meridian). This route is a geodesic and a local minimum for length. Call it the short geodesic.

There is, however, another geodesic route that starts at the north pole and ends at Greenwich. From the north pole, go south along the 180 line of longitude to the south pole pole and then north along the 0 line of longitude from the south pole to Greenwich. This route is also a geodesic, but clearly is not a local minimum for length. Call this the long geodesic. Taken together, the short and long geodesics comprise the unique great circle that runs through both the north pole and Greenwich.

How can the diffence between the short geodesic and the long geodesic be characterized? First consider the long geodesic. Any geodesic that starts the north pole, and that is infintesimally close to the long geodesic crosses the long geodesic before it gets to Greenwich, i.e., at the south pole. (Note: these "close" geodesic start at he north pole, but do not go through Greewich). The south pole is what is called a conjugate point.

Now consider the short geodesic. Any geodesic that starts at the north pole, and that is infinitesimally close to the short geodesic does *not* cross the short geodesics between the north pole and greenwich. The short geodesic does not have any conjugate points.

This gives the required characterization: a geodesic between any point p and q of a 2-dimensional space is a local minimum for length if and only if the geodesic has no conjugate points.

Something similar is true for spacetime: a timelike geodesic between events p and q is a local maximum for proper time if and only if there are no conjugate points (to p or q) between p and q. See Wald for a proof. This is very much related to the singularity theorems of Penrose and Hawking, as the focusing nature of gravity (assuming an appropriate energy condition) often causes conjugate points to occur.

Regards,
George
 
  • #15
Ratzinger said:
if I throw a ball the gravity of Earth makes the ball go a parabolic line.
What does GR say about little balls in Earth's gravity?
Can we speak here about space or spacetime curvature that causes its parabolic trajectory?
GR says almost nothing here.
You shouldn’t expect so much from GR & space-time they say very little about parabolic curves of balls, or Galileo’s cannonballs, on earth. GR affects compared to the predominantly Newtonian effects are just too small.

First - in the ideal case the paths are NOT parabolic – they are elliptical as they attempt to establish an elliptical orbit around the center of the earth. Of course the surface of the Earth gets in the way but before it hits it is basically on an orbital path.

Second – We never get to see the “ideal” because AIR gets in the way. Due to the air resistance the horizontal speed is slowly but only partly reduced all the way to infinity.
Maximum speed in the vertical is also limited by terminal velocity. Since the horizontal speed is never 100% eliminated the curve only approaches a vertical line. Thus it’s parabolic.

Guys, you don’t have to create the most complicated answers, just a little insight to a proper concept is so much simpler.
I’m sure Occam would agree.
 
  • #16
pmb_phy said:
The principle of extremal aging refers to two events. I.e. the proper time as measured on a clock which passes through both events.
Yes, but doesn't it refer to taking two paths between the two events, starting with the same initial speed and direction from the first event, one path being a geodesic and one being a non-geodesic? I don't see why it would be relevant that, starting with different initial speed and directions from the first event, you could have two different paths which were both geodesics and which would reunite at a different event with different proper time elapsed. Again, this would seem to be possible for two different orbits around the sun that cross at two points--in this case you might have an observer A who ages less than an observer B between their two meetings, but that wouldn't prove that observer A's path did not maximize his aging given his initial speed and direction, would it?
pmb_phy said:
The proper time along the geodesic of any photon is zero and no positive number has a value less than zero.
But I don't see how this is relevant to the issue of extremal aging--if you could address my example of the two orbits that intersect twice it would help me out.
pmb_phy said:
Note that the principle of extremal aging refers to geodesics not arbitrary worldlines.
Again, I thought it referred to a variational principle where you show that any deviations from the geodesic path would give a non-extremal aging (similar to showing that variations from the path that an object takes in classical mechanics would give a non-extremal value for the action integral).
 
  • #17
RandallB said:
GR says almost nothing here.
You shouldn’t expect so much from GR & space-time they say very little about parabolic curves of balls, or Galileo’s cannonballs, on earth. GR affects compared to the predominantly Newtonian effects are just too small.
GR reduces to Newtonian mechanics in the limit of small spacetime curvature and small velocities, just like SR reduces to Newtonian mechanics in the limit of small relative velocities. So when dealing with a mass like the earth, the geodesic path GR predicts for a cannonball should be almost identical to the parabolic path predicted by Newtonian mechanics.
 
  • #18
JesseM said:
Yes, but doesn't it refer to taking two paths between the two events, starting with the same initial speed and direction from the first event, one path being a geodesic and one being a non-geodesic?
No.
I don't see why it would be relevant that, starting with different initial speed and directions from the first event, you could have two different paths which were both geodesics and which would reunite at a different event with different proper time elapsed.
That's the idea.

But I don't see how this is relevant to the issue of extremal aging--if you could address my example of the two orbits that intersect twice it would help me out. Again, I thought it referred to a variational principle where you show that any deviations from the geodesic path would give a non-extremal aging (similar to showing that variations from the path that an object takes in classical mechanics would give a non-extremal value for the action integral).
The variation from the path is infinitesimal and I believe it yields a maximum aging when the events are close together. The idea is that extremal aging tells you that amoung all the worldlines which pass through two events, those which are geodesics are those for which the proper time is extremal.

Your example of crossing orbits is another example of two geodesics which may have different proper times associated with them and each geodescic is one for which the proper time is an extremal. This means that if you vary either path just a tad then the proper time would be either slightly higher or slightly lower than the one on the geodesic. If you chose alpha as an index to paramaterize worldlines which are close to the geodesic then plotter the proper time vs the index alpha then you'd see that those values of alpha for which the curve has a zero derivative corresponds to a geodesic. The curve doesn't have to have a maximum there. It could have a minimum of a point of inflection.

Pete
 
  • #19
JesseM said:
Yes, but doesn't it refer to taking two paths between the two events, starting with the same initial speed and direction from the first event, one path being a geodesic and one being a non-geodesic?
pmb_phy said:
No.
But doesn't your description below that you take a geodesic path and then "vary it just a tad" match what I was saying above? Isn't the idea that if you vary it slightly from the true geodesic path, you will always get a smaller proper time? (assuming you're dealing with a spacetime where extremal = maximal) In contrast, if you take two different initial velocities then you can have two different valid geodesics, but the fact that the second geodesic may have more proper time between the event of the particles departing and the event of their reuniting than the first geodesic does not show that the first geodesic was not a maximal one, correct? So in order for the comparison of two paths to be at all relevant to showing that one path has maximal proper time, it seems critical that both paths start out parallel (same initial velocity) and then one path is a non-geodesic one.
JesseM said:
I don't see why it would be relevant that, starting with different initial speed and directions from the first event, you could have two different paths which were both geodesics and which would reunite at a different event with different proper time elapsed.
pmb_phy said:
That's the idea.
Why? Do you agree that if you have two geodesic paths A and B that intersect at two events, then the fact that path A may have greater proper time in no way disproves the idea that path B was a maximal one? If so, then in what way is comparing two geodesic paths relevant to deciding whether one of them is maximal or not?
pmb_phy said:
The variation from the path is infinitesimal and I believe it yields a maximum aging when the events are close together.
But even if you want the variation to be infinitesimal, by having the particles start out with infinitesimally different initial velocities, it seems you could have two infinitesimally different paths which were each valid geodesics, and yet the fact that path A had infinitesimally greater proper time would not show that path B was not a maximal one. Am I wrong about that? If not, then again, it seems the relevant comparison is between two paths which start out with the same initial velocity, but then one departs slightly from the geodesic path, and in this case you are guaranteed that the non-geodesic path will have a slightly smaller proper time (again, assuming the geodesic path was one that maximized the proper time rather than minimized it, I'm not objecting to the idea that there are spacetimes where a geodesic is minimal rather than maximal).
pmb_phy said:
The idea is that extremal aging tells you that amoung all the worldlines which pass through two events, those which are geodesics are those for which the proper time is extremal.
But not "extremal when compared to other geodesic paths between the same two events", just "extremal when compared to non-geodesic paths that start out parallel to the geodesic path", it seems to me.
pmb_phy said:
Your example of crossing orbits is another example of two geodesics which may have different proper times associated with them and each geodescic is one for which the proper time is an extremal. This means that if you vary either path just a tad then the proper time would be either slightly higher or slightly lower than the one on the geodesic.
But see my example of two infinitesimally close geodesics--since they are both geodesics, it can't be true that they are each extremal when compared with each other, which again is why I always understood it as critical that you only compare nearby paths which start out with exactly the same initial velocity, and then you will see that given that initial velocity, there is a unique extremal path that starts out tangent to that velocity. Comparing paths with different initial velocities would seem to be apples and oranges, it would be of no use in showing that a given path is "extremal".
 
  • #20
JesseM said:
But even if you want the variation to be infinitesimal, by having the particles start out with infinitesimally different initial velocities, it seems you could have two infinitesimally different paths which were each valid geodesics, and yet the fact that path A had infinitesimally greater proper time would not show that path B was not a maximal one. Am I wrong about that?
It occurs to me that my thinking could be going wrong here, perhaps there are no spacetimes where given a geodesic between two events, it would be possible to find another arbitrarily close geodesic between the same events...this wouldn't be true in flat spacetime, for example. I was thinking about gravitational lensing and how geodesic paths starting at different angles behind the lens will be focused at a single point in front of the lens, but I don't know if objects traveling these paths will arrive at the same time as well, so that you really have a collection of arbitrarily-close geodesics between the same two points in spacetime. I was also thinking about great circles on a sphere, but perhaps geodesics in spacetime are fundamentally different from geodesics in space in this way. If indeed there are no other geodesics among the paths arbitrarily close to a given geodesic, then I suppose the geodesic could be the extremal one out of all of these paths, even if the other paths did not start with the same initial velocity. Do you know the answer to this question about whether a geodesic path between two events can have arbitrarily close geodesic "neighbors" between the same two events?
 
  • #21
JesseM said:
GR reduces to Newtonian mechanics in the limit of small spacetime curvature and small velocities, just like SR reduces to Newtonian mechanics in the limit of small relative velocities. So when dealing with a mass like the earth, the geodesic path GR predicts for a cannonball should be almost identical to the parabolic path predicted by Newtonian mechanics.
Well isn’t that the point – why split hairs about the small insignificant details of GR at speeds where the effects disappear in the real issue here. After all, the original post #1 question was “Can we speak here about space or space-time curvature that causes its parabolic trajectory?”

And the answer to that has to be NO, because it doesn’t. Neither GR nor Newtonian predicts your “parabolic paths” at these low speeds!
They predict elliptic paths. Only the special case exact minimum escape velocity produces a parabolic trajectory. The parabolic trajectories under question are due to Air Resistance not Newton or GR.
 
  • #22
RandallB said:
Well isn’t that the point – why split hairs about the small insignificant details of GR at speeds where the effects disappear in the real issue here.
Reading the original post, my guess is the poster wanted a conceptual answer to how GR would explain such paths, in terms of curved spacetime or something else. Even though GR makes almost identical quantitative predictions as Newtonian physics in certain situations, the conceptual explanation for these predictions is still different, with GR explaining the paths in terms of geodesics in curved spacetime and Newtonian gravity explaining them in terms of a gravitational force.
RandallB said:
And the answer to that has to be NO, because it doesn’t. Neither GR nor Newtonian predicts your “parabolic paths” at these low speeds!
They predict elliptic paths. Only the special case exact minimum escape velocity produces a parabolic trajectory. The parabolic trajectories under question are due to Air Resistance not Newton or GR.
Newtonian gravity predicts parabolic paths in a constant G-field, and I'm guessing GR does too. Of course the G-field produced by a planet decreases with distance from the surface, but for a situation like tossing a ball a few feet in the air the difference in the strength of gravity between the surface and the point of the ball's maximum height is minimal, so constant gravity works fine as an approximation. If you're going to complain about "splitting hairs" in noting the difference between GR and Newtonian gravity in low-mass/low-velocity situations, then noting the difference between a constant G-field and a decreasing G-field in a situation like one involving a tossed ball should also be viewed as splitting hairs.
 
  • #23
JesseM said:
I was also thinking about great circles on a sphere ... Do you know the answer to this question about whether a geodesic path between two events can have arbitrarily close geodesic "neighbors" between the same two events?

These are the things that I wrote about in my previous post in this thread.

If 2 events are conjugate, then a geodesic through the events can have arbitrarily close geodesic neighbours between the same two events.

Consider the north and south poles on the Earth. The geodesics are lines of longitude, and 2 lines of longitude through the poles can be arbitarily "close".

Regards,
George
 
  • #24
JesseM said:
Reading the original post, my guess is the poster wanted a conceptual answer to how GR would explain such paths, in terms of curved spacetime or something else.

...Newtonian gravity predicts parabolic paths in a constant G-field, and I'm guessing GR does too.
... then noting the difference between a constant G-field and a decreasing G-field in a situation like one involving a tossed ball should also be viewed as splitting hairs.
But I didn’t note a difference between a constant G-field and a decreasing G-field.

I agree that is the question being asked. And conceptually GR and Newton both do not predict parabolic trajectories here. Conceptually the declining horizontal speed due to air resistance does. I don’t see where a G-field that doesn’t increase during the fall helps.

The only conceptual approximation I see that could create a parabolic without the air resistance would be to assume a flat surface to Earth with an infinite deep fall.

I’m not sure what a constant G-field would do – assuming adequate height and speed, maybe something like a elliptic orbit with a
large precession or maybe a spiral.
 
  • #25
George Jones said:
These are the things that I wrote about in my previous post in this thread.
If 2 events are conjugate, then a geodesic through the events can have arbitrarily close geodesic neighbours between the same two events.
Consider the north and south poles on the Earth. The geodesics are lines of longitude, and 2 lines of longitude through the poles can be arbitarily "close".
Regards,
George
On a sphere, the geodesics between two opposite points like the north and south pole would all have equal length--is this also true of the proper time of different geodesics between conjugate events in spacetime? I guess this would also be a way of insuring that all nearby paths to a geodesic will have shorter or equal proper time (assuming a spacetime where extremal = maximal), even if some nearby paths are also geodesics themselves.
 
  • #26
RandallB said:
But I didn’t note a difference between a constant G-field and a decreasing G-field.
Right, but you noted a difference between a path that's a consequence of constant-G (parabolic) and a path that's a consequence of decreasing-G (elliptical).
RandallB said:
I agree that is the question being asked. And conceptually GR and Newton both do not predict parabolic trajectories here. Conceptually the declining horizontal speed due to air resistance does. I don’t see where a G-field that doesn’t increase during the fall helps.
The only conceptual approximation I see that could create a parabolic without the air resistance would be to assume a flat surface to Earth with an infinite deep fall.
I’m not sure what a constant G-field would do – assuming adequate height and speed, maybe something like a elliptic orbit with a
large precession or maybe a spiral.
It's got nothing to do with air resistance, in fact I'm assuming there are no forces besides gravity acting on the object. In a constant G-field the acceleration in the vertical direction should be constant (because the gravitational force on a mass m will always be -mg), so you have:

[tex]a(t) = -g[/tex]

Integrate that with respect to t to find vertical velocity as a function of time, and you get:

[tex]v(t) = -gt + v_0[/tex]

Integrate again with respect to t to find vertical position as a function of time:

[tex]y(t) = (-g/2) t^2 + v_0 t + y_0[/tex]

([tex]v_0[/tex] and [tex]y_0[/tex] represent the velocity and height at t=0)

If you graph position vs. time, this gives you a parabola. And if the object had an initial horizontal velocity as well as a vertical velocity, then the horizontal velocity would be constant since the gravitational force only acts in the vertical direction, so the object would describe a parabolic path through space as well.
 
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  • #27
JesseM said:
a path that's a consequence of constant-G (parabolic)
a path that's a consequence of decreasing-G (elliptical)
It's got nothing to do with air resistance
Where do these assumptions come from??
Exactly how is the curve of the minimum escape velocity not parabolic? – it’s certainly not the result of a constant-G.

As to your derivation of a parabolic you did that assuming a “flat surface to Earth with an infinite deep fall” as I said would work in post #24.

ONE LAST TIME – the only time Newton or GR predicts a parabolic for a spherical world is at the minimum escape speed.

I don’t know if you’re having trouble admitting in writing that you got something wrong – or you need to hear it from someone else.
So I’ll let someone else step in, to help you work it out.
Good Luck.
 
  • #28
RandallB said:
Where do these assumptions come from??
Exactly how is the curve of the minimum escape velocity not parabolic? – it’s certainly not the result of a constant-G.
I didn't say that Newtonian gravity with decreasing G will never predict a parabolic path, I was just responding to your claim that 'Neither GR nor Newtonian predicts your “parabolic paths” at these low speeds!'
RandallB said:
As to your derivation of a parabolic you did that assuming a “flat surface to Earth with an infinite deep fall” as I said would work in post #24.
It isn't obvious to me what you mean by "infinite deep fall"--even if the Earth was a flat 2D surface that extended to infinity in all directions, the gravitational pull of this surface would still decrease with distance. But OK, do you agree that if we assume constant G, then we get parabolic trajectories? And do you agree that constant G is a perfectly reasonable approximation for situations like tossing a ball in the air? If so, then I'll repeat (with a slight modification) my earlier argument: "If you're going to complain about "splitting hairs" in noting the difference between GR and Newtonian gravity in low-mass/low-velocity situations, then noting the difference between the predictions of a constant G-field and the predictions of a decreasing G-field in a situation like one involving a tossed ball should also be viewed as splitting hairs."
RandallB said:
ONE LAST TIME – the only time Newton or GR predicts a parabolic for a spherical world is at the minimum escape speed.
Only we weren't talking specifically about the gravity from a "spherical world", you just introduced that restriction in this post. We were simply talking about what is predicted when you toss a ball up, and if you argue that using Newtonian mechanics is a reasonable approximation (even though we know these laws are always going to make slightly incorrect predictions), then I'm arguing it's also a reasonable approximation to assume a constant G-field (and indeed, this approximation is made all the time in physics textbooks). Also, as noted earlier, constant G-fields do crop up in GR when you want to analyze things from the perspective of an accelerating observer (as seen by inertial observers) in flat spacetime, in that case it is not even an approximation.
 
  • #29
JesseM said:
Only we weren't talking specifically about the gravity from a "spherical world", you just introduced that restriction in this post. We were simply talking about what is predicted when you toss a ball up, .
I give up -- tossing a ball on Earth assumes we are part of the flat Earth society ?
What is all this talk about geodesic’s and GR then?

“It isn't obvious to me what you mean by "infinite deep fall"—“ just how deep a hole do you allow the ball to fall into confirm your parabolic – needs to be infinite to approach the perpendicular line to your infinite surface doesn’t it?
Nothing tricky there. And what difference does constant G or an increasing G as it falls mean they will both speeds approach infinity thus approach a perpendicular to the surface thus both give a parabolic. Of course that means much faster than “c” that with flat Earth you really can’t apply GR very well.

Good grief tossing a ball around is not a GR space-time event. If you take air resistance out of it you will completely change the shape of the curve, what’s the justification for making that approximation it’s not going to relate to reality.
And when you do take the air out it’s elliptic. "."
Then go ahead - Transform the the spherical elliptic to your flat Earth - and it will look more hyperbolic than parabolic.
Think it through do the work don't make these false assumptions.
 
  • #30
JesseM said:
On a sphere, the geodesics between two opposite points like the north and south pole would all have equal length

Here's an interesting question that I think is related. On the sphere, we have a one parameter family of geodesics that go from the north-south pole, all of which have the same length. (The sole parameter is the starting angle).

Now, suppose we apply a diffeomorphism to the sphere to distort it into an ellipsoid. We imagine that the north pole is at x=y=0, z=1, and we apply the diffeomrphism y' = ky to the y-axis to "stretch" it.

Now we have (I think!) at least two geodesics from the north pole to the south pole that have different lengths (one in the 'x' direction, one in the stretched 'y' direction).

The question is whether we still have a one-parameter family of geodesics that go from the north pole to the south. Imagining such a smooth family of geodesics of varying lengths would seem to violate the "stationary" property of the length of a geodesics, however.
 
  • #31
RandallB said:
I give up -- tossing a ball on Earth assumes we are part of the flat Earth society ?
That would be like me saying "I give up -- tossing a ball on Earth assumes spacetime is not curved? What is this, the anti-relativity society?" in response to your claim that Newtonian mechanics was OK for dealing with the example of a tossed ball. Of course it is perfectly valid to use Newtonian mechanics in this situation, and the reason is that Newtonian mechanics is a valid approximation in this case, even if we know it isn't strictly speaking correct. Similarly, constant G is a valid approximation in the case of a tossed ball on earth, even though we know the G-field is actually decreasing very slightly as you move above the surface. If you disagree with the use of approximations, then your argument about Newtonian mechanics is hypocritical. Also, like I said, the approximation of a constant G-field is used in most classical mechanics textbooks (I can provide examples if you don't believe me)--do you want to accuse all these textbook authors of being flat-earthers?
RandallB said:
What is all this talk about geodesic’s and GR then?
But we weren't talking exclusively about geodesics and GR, we were also talking about Newtonian mechanics, a subject which you introduced. Do you deny that Newtonian mechanics is only valid as an approximation, that it does not capture the precise truth?
RandallB said:
“It isn't obvious to me what you mean by "infinite deep fall"—“ just how deep a hole do you allow the ball to fall into confirm your parabolic – needs to be infinite to approach the perpendicular line to your infinite surface doesn’t it?
Huh? I am only talking about the case of throwing a ball a few feet above a flat surface using the assumption that the G-field is constant above the surface--where would any infinite quantities enter into this? If you're saying that a "parabolic trajectory" needs to be infinite because a parabola is an infinite curve, that's not what physicists mean when they use the term "parabolic trajectory", all that's important is that the trajectory looks like a section of a parabola.
RandallB said:
Nothing tricky there. And what difference does constant G or an increasing G as it falls mean they will both speeds approach infinity thus approach a perpendicular to the surface thus both give a parabolic.
Again, I'm just talking about tossing the ball upwards from the surface, watching it travel up a finite distance on a parabolic trajectory, then fall back down.
RandallB said:
Good grief tossing a ball around is not a GR space-time event.
Sure it is, GR should give the most accurate possible analysis of this situation. You don't need GR to analyze it, but then we're once again back on the issue of approximations, and constant G-field is a perfectly good approximation for analyzing this situation. (but again, I think the poster wanted a conceptual understanding of how GR would explain the trajectory, which means that neither the Newtonian gravity approximation nor the constant G-field approximation would be useful in answering his question).
RandallB said:
If you take air resistance out of it you will completely change the shape of the curve
"Completely"? For an ordinary rubber ball the trajectory will still look parabolic, air resistance increases as a function of velocity and at the speeds the ball is moving it makes very little difference. For example, look at this classroom demonstration in which a parabola is projected onto a screen and then a tennis ball is tossed so that its trajectory follows the curve of the parabola.
RandallB said:
what’s the justification for making that approximation it’s not going to relate to reality.
Of course it is, this is routinely demonstrated in science classes. Here's a page from CERN's high school teaching materials section explaining how to do a similar demonstration:

A thrown ball follows a parabolic path

It would be pretty embarrassing if the experiment CERN recommends for high school physics teachers totally failed to work, no?
RandallB said:
And when you do take the air out it’s elliptic. "."
Yes, if the ball could fall right through the earth, it would make an ellipse, but the tiny section of this ellipse that's above the surface will look almost exactly like the parabola that you'd predict if you just assumed the G-field was constant. Have a look at the diagrams on http://www.cpo.com/Weblabs/projecta.htm from the Cambridge Physics Outlet to see how this works.
RandallB said:
Think it through do the work don't make these false assumptions.
The assumption of a constant G-field is no more "false" than the assumption of a Newtonian force law, both are perfectly valid approximations in the case of a ball tossed a few feet in the air on the surface of the earth. Again, the constant G-field assumption is made in classical mechanics textbooks everywhere, I can give you examples if you like.
 
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  • #32
but again, I think the poster wanted a conceptual understanding of how GR would explain the trajectory
Yes, that was my intention. Also, I found out by now that every GR text has a few pages that go under the headline the Newtonian limit which just adresses my original problem quite well.
 
  • #33
JesseM - Sorry for being silent for so long but do to this darn back injury I can't sit at the computer for that long and I'm in the midst of re-writing my paper on the concept of mass in relativity.

Is there a question you have for me at this point? I'm unable to sit and read the entire discission to see if this is the case. I'd be glad to respond.

Pete
 
  • #34
JesseM said:
(but again, I think the poster wanted a conceptual understanding of how GR would explain the trajectory,
It would be pretty embarrassing if the experiment CERN recommends for high school physics teachers totally failed to work, no? Yes, if the ball could fall right through the earth, it would make an ellipse, but the tiny section of this ellipse that's above the surface will look almost exactly like the parabola
Well at least you admit the real path is an ellipse
The original question in post one was for clarification on if and how Newton’s or GR etc caused parabolic curves at low speeds tossing a ball on earth.

The simple answer is - they don’t – air resistance does change the ellipse into a parabolic (or if we go back to the sphere a spiral to the center). We don’t get to experience them being tossed about in a vacuum where they would not move in a parabolic.

What is the point it distorting the truth with a bunch of hand waving and assumptions just to shoehorn in a GR explanation where it doesn’t belong. It does little good to cut out the top of an ellipse and call it a parabola. Cut out a smaller section in a demonstration and it looks “almost exactly like” a section of a circle – so let's call it that? Just because it’s approximately right doesn’t make it right.

When the truth of GR is miss-applied like this it’s no wonder some winding up supposing there is something wrong with GR, when they do figure things out correctly. When the real problem was they got a misapplication or GR in the first place – not some flaw in GR.

You do more damage to the understanding of GR when you do this, than you do good.

And no it wouldn’t surprise me to see an explanation intended to describe what a parabola looks like; to be misapplied in a short hand explanation of how a real parabola is formed and what one really is; leaving out the part that it is just “almost exactly like” one.
 
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  • #35
RandallB said:
GR says almost nothing here.
You shouldn’t expect so much from GR & space-time they say very little about parabolic curves of balls, or Galileo’s cannonballs, on earth.
That is very untrue. GR describes all gravitational phenomena especially trajectories that particles take in a gravitational field.
GR affects compared to the predominantly Newtonian effects are just too small.
It is the deviations from the predictions of Newton's laws due to GR effects that are too small to measure by normal means. But GR quite adequately describes the motion of a ball following a geodesic, the spatial trajectory being nearly parabolic.
First - in the ideal case the paths are NOT parabolic – they are elliptical as they attempt to establish an elliptical orbit around the center of the earth.
He was speakikng about the approximation made when the trajectory stays near the surface of the earth. One can't simply forget the approximation one is speaking about when answering a question. And since you want to be exact, the ball doesn't follow an ellipse either. It follows a spatial trajectory which will depend on the Sun, Moon, Mars etc. But you choose to ignore small efects like this just as the person speaking about the ball following a parabolic section in the approximation when the Earth's field is approximated by a uniform g-field. However the questioner didn't even say that he was on the surface of the earth. Who knows! Maybe he's inside the Earth in a spherical cavity dug out to do experiments. In such a cavity the field is uniform and it does follow a geodesic and the spacetime corresponding to the interior of the cavity is flat.
Guys, you don’t have to create the most complicated answers, just a little insight to a proper concept is so much simpler.
I’m sure Occam would agree.
The question is on GR and is a perfectly good question.

Pete
 
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