Why do objects follow curved trajectories under Earth's gravity?

Ratzinger
Messages
291
Reaction score
0
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
 
Last edited:
Physics news on Phys.org
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).
 
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 (\frac{GM}{rc^2}).

I hope this has helped.

Garth
 
Last edited:
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.
 
Last edited by a moderator:
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)

<br /> d\tau^2 = g_{00} dt^2 - g_{11} dr^2 <br />
<br /> \tau = \int \sqrt{ g_{00} - g_{11} \left( \frac{dr}{dt} \right)^2} dt<br />

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

<br /> \int L(r,\dot{r},t) dt \hspace{.5 in} \dot{r} = \frac{dr}{dt}<br />

with
<br /> L(r,\dot{r},t) = \sqrt{g_{00}(r) - \dot{r}^2}<br />

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

<br /> \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) = \frac{\partial L}{\partial r}<br />

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, \partial L/\partial r, results from the fact that g_00 is a function of r, and is in fact proportional to \partial g_{00}/\partial r, 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.
 
Last edited:
Thanks, gentlemen. You are all great people.

And did I mention recently how much I like this site?
 
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
 
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:

a(t) = -g

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

v(t) = -gt + v_0

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

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

(v_0 and y_0 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.
 
Last edited:
  • #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.
 
Last edited by a moderator:
  • #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.
 
Last edited:
  • #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
 
  • #36
The simple answer is the one Jesse gave. Various non-idealities such as "mascons" make the actual orbital path of a body non-ellilptical as well as non-parabolic, even ignoring air resistance. (Mascons, aka mass concentrations, are due to the fact that the Earth is not an ideal homogeneous sphere, but contains areas of higher and lower density, which slightly but measurably affect its gravitational field. They are present on both the Earth and moon, though mascons were first observed, IIRC, when trying to calculate landing orbits for the apollo program.)

The OP in this thread (Ratzinger) already knew that objects fall in parabolic paths in the low-velocity, low height approximation, the only one who appears to be really confused about this point is RandallB.

Unfortunately, I don't know of any way to un-confuse poor Randall (I and various other people have tried without success).

Jesse is definitely not "distorting the truth with handwaving" as RandallB claims, it is just RandallB "not getting it".
 
Last edited:
  • #37
pervect said:
The simple answer is the one Jesse gave. Various non-idealities such as "mascons" make the actual orbital path of a body non-ellilptical as well as non-parabolic, even ignoring air resistance.
(Ratzinger) already knew that objects fall in parabolic paths in the low-velocity, low height approximation, the only one who appears to be really confused about this point is RandallB.
it is just RandallB "not getting it".
NOT you too.
Your solution is to add yet another epicycle ("mascons") onto the epicycles already loaded up on this issue.
Of course the origin question understood that parabolic are observed! The question was how do you explain the parabolic curves – I assume Ratzinger was smart enough to know they should only occur with an escape velocity – hence the confusion as to how they come to be parabolic again at these low speeds.
With enough epicycles piled on here I assume you’ll eventually be able to convince yourself to go back to Ptolemy’s solution!

Why duck the real solution when is hiding right out in plain sight. Go back to basic physics 101 with fiction and air resistance.

Or do some reading on the Newton vs. Hook debates; Newton was long embarrassed by Hooke besting him in applying air resistance to the idea of imagining things falling to the center of the earth.
In this case you don’t even need to throw the ball just drop it leave gravity in place and assume you have a clear path to the center of the Earth solve for without air resistance and then with stationary air (not rotating with the point dropping the ball) resistance. If you cannot draw a picture for yourself find the one by Hooke, he solved this one over three hundred years ago!

This is to simple and obvious to be impressed by someone willing to shoot the messenger as “not getting it”.

Newton and GR work just fine at these levels to predict NON-parabolic elliptical paths. If you really don’t get that I’ll let keep it.
 
Last edited:
  • #38
RandallB said:
Well at least you admit the real path is an ellipse
I think the real path is only an ellipse if you use Newtonian gravity, which is just as much of an approximation as the constant G-field is. We know that GR predicts Mercury's orbit is not a perfect elipse--the perihelion changes slightly with each orbit--so I would assume that this effect does not just vanish abruptly at some point, presumably if you calculated the exact path of a tossed ball which was able to travel straight through the earth, its perihelion would precess slightly too. Of course the difference from the Newtonian prediction and the GR prediction might be only a nanometer with each orbit or something, but the difference between the constant-G prediction of a ball's path between the time it is tossed and the time it hits the ground again would probably also differ from the decreasing-G prediction by only a microscopic amount (do you admit, by the way, that your earlier statement 'If you take air resistance out of it you will completely change the shape of the curve' was incorrect?)
RandallB said:
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 original poster affirmed my guess that his question was about the conceptual question of how GR would explain curved paths in Earth gravity, therefore any answer that does not involve GR would be unhelpful.
RandallB said:
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).
Would air resistance make it parabolic, even in theory? Not if the ball was tossed high enough that it would reach terminal velocity...what equation for air resistance as a function of velocity are you assuming here? It's possible that the idealized equation for air resistance at low velocities would transform it into an exact parabola, but I haven't seen this calculation before; in any case, any simple equation for air resistance as a function of velocity is probably just going to be an approximation as well.
RandallB said:
We don’t get to experience them being tossed about in a vacuum where they would not move in a parabolic.
Again, if you toss a ball at ordinary velocities air resistance has virtually no effect on the path. That classroom demonstration where you toss a ball and it moves along a parabola projected on the wall would work just fine in a vacuum--do you disagree?
RandallB said:
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.
Except there would be no mathematical justification for saying it looked like a circle, that would just be a matter of innacurate eyeballing. On the other hand, you can show show that in the limit as ratio between the height the ball is tossed above the surface and the radius of the planet approaches zero (ie the ball is tossed only a small height compared to the planet's radius), the difference between the parabolic path predicted by the constant-G assumption and the small section of the ellipse predicted by the decreasing-G assumption should approach zero. This is exactly analogous to how the difference between GR's prediction and Newtonian gravity's prediction approaches zero in the limit as mass and velocities approach zero; all good approximations should be justifiable in terms of such limits.
 
  • #39
JesseM said:
I think the real path is only an ellipse if you use Newtonian gravity ... (do you admit, by the way, that your earlier statement 'If you take air resistance out of it you will completely change the shape of the curve' was incorrect?) The original poster affirmed my guess that his question was about the conceptual question of how GR would explain curved paths in Earth gravity, therefore any answer that does not involve GR would be unhelpful.
Wrong he confirmed the question was about how GR explained “the curve” that curve in question being parabolic! And as I said before IT DOES NOT.
Would air resistance make it parabolic, even in theory? Not if the ball was tossed high enough that it would reach terminal velocity...what equation for air resistance as a function of velocity are you assuming here? Again, if you toss a ball at ordinary velocities air resistance has virtually no effect on the path. That classroom demonstration where you toss a ball and it moves along a parabola projected on the wall would work just fine in a vacuum--do you disagree?
What on Earth are you talking about (small pun) how high you toss or drop a ball for a non rotating Earth has nothing to do with any kind of curve! It’s the horizontal component of initial speed that will give a curve and the horizontal air resistance that gives the parabolic. Staying in just the vertical is only going to give a straight line. And sure you need me to say it twice? IN A VACUUM IT’S ELLIPTIC not parabolic!
Except there would be no mathematical justification for saying it looked like a circle, that would just be a matter of innacurate eyeballing.
Exactly, just as there is no justification for inaccurately eyeballing a ellipse and calling it a parabolic curve!
I don’t know why you and pervect have not figured out you’ve conceptually blown this one badly – question is when you do figure out the concept correctly, are you willing to admit it and explain it here.
I can recommend as I did before find some old book that includes the diagram by Robert Hooke, who helped Newton get it right before he wrote the Principia. Not that he was any nicer about it than you guys, he was really rather abusive in embarrassing Newton in front of the rest of the Royal Society for some time over it.
It shouldn’t be hard to find, it’s famous. He shows three curves in it:
-One marked A,B,C,D,A for the circle an object follows that is supported by the structure of earth.
-A second marked A,F,G,H,A for the elliptical path that would be followed when all of Earth's structure is removed but gravity remained.
-And a third marked A,I,K,L,M,N,O,P,C for when only air and it’s resistance was replaced for the entire structure of earth.
It’s this third one that can be reduced to parabola if you redo the coordinates for a flat Earth - that will not happen in the non-air case.

So (even with constant g) don’t expect me to buy your argument that GR or Newton will produce a parabolic except in the case that they both agree on, escape v.

Just like I don’t believe pervect’s claim he can apply a nonlinear spring to mercury so that the Newtonian will duplicate the precession of GR – when he hasn’t even tried to produce the function of such a non-liner spring in the other thread.

So if you can match the diagram by Hooke and explain him wrong in his explanation of air resistance I’d like to see the detail - you’d be the first since Dec of 1679! – you’ll be famous.
 
Last edited:
  • #40
RandallB said:
Wrong he confirmed the question was about how GR explained “the curve” that curve in question being parabolic! And as I said before IT DOES NOT.
Not exactly, but approximately. Just like GR doesn't say curves are exactly elliptical, although they are approximately so. If he had asked how GR explains elliptical paths, would you just thunder that GR "DOES NOT" predict such paths, or would you understand that he was speaking approximately?
RandallB said:
What on Earth are you talking about (small pun) how high you toss or drop a ball for a non rotating Earth has nothing to do with any kind of curve!
Do you understand the concept of approximations based on limits in physics? If so, do you disagree that in the limit as the height the ball is tossed becomes very small compared to the radius of the planet, the path of the ball (in the absence of air resistance and other complicating factors) will approach a perfect match to the parabolic curve predicted by the constant-G assumption? Please answer this question yes or no.
RandallB said:
It’s the horizontal component of initial speed that will give a curve and the horizontal air resistance that gives the parabolic.
Again, what mathematical calculation are you using to justify this? What specific equation are you assuming for air resistance, and can you show rigourously that the combined effects of gravity and air resistance produce a parabola? I have never seen such a claim before, it's possible it's correct but if you don't have some math to back it up it seems unfounded.

Secondly, you didn't answer my earlier question--do you agree that even in the absence of air resistance, the path of a tossed ball (with both horizontal and vertical velocity) will be very very close to the perfect parabola predicted by the constant-G assumption? Do you agree, for example, that the classroom demonstrations I linked to earlier would work just fine in a vacuum, and that the precise elliptical path predicted by Newtonian gravity would not depart from the parabolic path predicted by the constant-G assumption by more than a tiny amount? Again, please answer this question yes or no, it's not meant to be rhetorical.
RandallB said:
Staying in just the vertical is only going to give a straight line.
The path will just be a straight line through space, but if you graph position vs. time it will be very close to a parabola again. And if you're just talking about the path through space alone, then the path won't look like an ellipse either when the ball's velocity has no horizontal component.
RandallB said:
And sure you need me to say it twice? IN A VACUUM IT’S ELLIPTIC not parabolic!
Not exactly, according to GR--again, think of the precession of the perihelion of Mercury's orbit, showing that the orbit is not a perfect ellipse. But the ellipse is a valid approximation because in the Newtonian limit the orbit predicted by GR becomes arbitrarily close to an ellipse. Similarly, the parabola is a valid approximation because in the limit as the height of the ball gets arbitrarily small compared to the radius of the planet, the elliptical path predicted by Newtonian gravity becomes arbitrarily close to the parabolic path predicted by the constant-G assumption. Do you disagree with either of these statements about limits? Please answer yes or no.
RandallB said:
Exactly, just as there is no justification for inaccurately eyeballing a ellipse and calling it a parabolic curve!
If you can show that in a certain limit the difference between the parabolic curve and the elliptical curve approaches zero, then it is reasonable to use the parabolic curve in situations where you are close to that limit. Do you think there is ever a rigorous justification for using approximations in physics? For example, do you think it is ever justified to use Newtonian mechanics when we know that these predictions differ slightly from those of GR (for example, GR will say that a planet's orbit is not precisely elliptical), and that GR's predictions are the more accurate ones? If you do think that the Newtonian approximation is all right, then you're being hypocritical in your attack on the constant-G approximation; if you don't agree that limits can be used to rigorously justify the use of approximations in any case, then you are disagreeing with pretty much the whole physics community on this one.
RandallB said:
I don’t know why you and pervect have not figured out you’ve conceptually blown this one badly
..and pmb_phy, and every textbook on classical mechanics have "blown it" apparently, according to you. The use of approximations justified by mathematical limits is routine in physics, and the parabolic path approximation is a perfectly typical example of this.
RandallB said:
I can recommend as I did before find some old book that includes the diagram by Robert Hooke, who helped Newton get it right before he wrote the Principia. Not that he was any nicer about it than you guys, he was really rather abusive in embarrassing Newton in front of the rest of the Royal Society for some time over it.
It shouldn’t be hard to find, it’s famous. He shows three curves in it:
-One marked A,B,C,D,A for the circle an object follows that is supported by the structure of earth.
-A second marked A,F,G,H,A for the elliptical path that would be followed when all of Earth's structure is removed but gravity remained.
-And a third marked A,I,K,L,M,N,O,P,C for when only air and it’s resistance was replaced for the entire structure of earth.
OK, so presumably this is the basis for your claim that air resistance produces a parabola. If you're not sure what assumption Hooke made about the equation for air resistance as a function of velocity (presumably he used an equation which doesn't take into account the phenomenon of terminal velocity, for example), can you at least tell me the name of the book so I can look this up myself?
RandallB said:
It’s this third one that can be reduced to parabola if you redo the coordinates for a flat Earth - that will not happen in the non-air case.
So (even with constant g) don’t expect me to buy your argument that GR or Newton will produce a parabolic except in the case that they both agree on, escape v.
I don't say that Newton or GR produce a parabolic exactly, just like GR doesn't produce elliptical orbits exactly. Once again, are you completely unfamiliar with the concept of approximations based on limits?
RandallB said:
So if you can match the diagram by Hooke and explain him wrong in his explanation of air resistance I’d like to see the detail - you’d be the first since Dec of 1679! – you’ll be famous.
Sigh. Where have I ever said anything about Hooke being wrong? I simply said that I had never seen anyone claim that air resistance would produce an exact parabola, and asked you for more detail. I did say that this claim would obviously be wrong if you used a totally accurate equation for air resistance that takes into account terminal velocity, but presumably Hooke was using an approximation of some sort, if indeed your memory of what he said is accurate.

In any case, my claim was never about what the exact Newtonian prediction would be, it was always about an approximation based on what Newtonian mechanics predicts in the limit as the height that the ball is tossed becomes very small compared to the radius of the planet. It is you who is disagreeing with the entire physics community in claiming that the whole concept of limits based on approximations is unjustified, or that it is no more rigorous than just eyeballing a path and saying it looks like a certain curve.
 
  • #41
For God's sake. Everyone knows that under decent, reasonable approximations the trajectory for an ordinary sized ball tossed with an ordinary human-sized toss in the Earth's gravitational field will result in a parabolic path.

Now, Newton's theory explains this via a constant force, and force being the first time-derivative of momentum. The OP's questions was how does GR explain this, with all it's machinery of spacetime and geodesics.

The OP did not ask, "what are the differences between the two approaches," since he/she (as well as everyone else) knows these differences will not be appreciable, and since we have already made very reasonable assumptions in claiming that Newton's theory predicts parabolic trajectories, we can assume these differences are negligible.

The OP wanted to know how we could arrive at the answer of parabola straight from GR machinery.

EDIT: so my point was, stop confusing the issue! Incidentally, I think someone answered the OP question way back in page 1...
 
  • #42
The OP did not ask, "what are the differences between the two approaches," since he/she...
I'm male, even though I think posting under a female name would give more replies. So if you like, keep thinking I'm a twenty years old brazilian girl gotting interested in GR at the Newtonian limit and struggling with the conceptual understanding.

I think someone answered the OP question way back in page 1...
I think post no. 2 cleared me up already. Also found this link http://astro.isi.edu/notes/gr.pdf . It's called GR for the faint of heart, seven pages long, page five handles my original question.
 
Last edited by a moderator:
  • #43
Ratzinger said:
I'm male...

I just always thought it could (not necessarily that it would) be quite offensive if I assumed someone was a certain sex. It's certainly true that those with names that imply they are female get more attention on this forum.
 
  • #44
JesseM said:
Please answer this question yes or no.
How many times do I need to say NO, are you not reading?
Based on you’re quick reply you’re certainly not thinking.

GR reduces to the Newtonian in our real world yet you the micro almost impossible to detect the flaw in Mercury’s orbit as justification for making irrational approximations to convert ecliptics into parabolas - I don’t believe I’m seeing from someone of science.
I was giving you Hooke as a means to help you see your problem with just a little common sense. But you don’t seem to have the time to look or think.

You need help finding a book ask a librarian,
As to common sense get face to face with a good sensible instructor maybe you can work it out.

This is too simple and basic, to fuss over here any more.
 
  • #45
RandallB said:
How many times do I need to say NO, are you not reading?
OK, you were responding to this:
JesseM said:
Do you understand the concept of approximations based on limits in physics? If so, do you disagree that in the limit as the height the ball is tossed becomes very small compared to the radius of the planet, the path of the ball (in the absence of air resistance and other complicating factors) will approach a perfect match to the parabolic curve predicted by the constant-G assumption?
So are you saying "no" you don't understand the concept of approximations based on limits, or don't think such approximations have any place in physics? Or are you saying "no" you don't agree that in the limit as the height of the ball becomes very small compared to the radius of the planet, its path approaches that of a parabola? If the latter, would you disagree that in the limit as the height of the ball becomes small compared to the planet's radius, the difference in the gravitational force between the point where the ball is tossed and the maximum height it reaches will approach zero? Would you disagree that in the idealized case where the difference in the gravitational force between different heights is zero (ie constant-G), the ball's path will be a perfect parabola in a vacuum? If you disagree with either of these statements, I'd be quite happy to prove them; if you agree with both but don't see how they imply that the path of a tossed ball will approach a parabola in the limit as the difference between the ground and the maximum height becomes arbitrarily small compared to the radius of the planet, then you really need to think about it a little more, it's pretty trivial.

If you want to see some outside confirmation, look at the section titled "low energy trajectories" of the wikipedia article on Orbital Equations:
If the central body is the Earth, and the energy is only slightly larger than the potential energy at the surface of the Earth, than the orbit is elliptic with eccentricity close to 1 and one end of the ellipse just beyond the center of the Earth, and the other end just above the surface. Only a small part of the ellipse is applicable.

...

The part of the ellipse above the surface can be approximated by a part of a parabola, which is obtained in a model where gravity is assumed constant. This should be distinguished from the parabolic orbit in the sense of astrodynamics, where the velocity is the escape velocity. See also trajectory.
RandallB said:
GR reduces to the Newtonian in our real world
Not exactly, no--only in the limit. In any real example, there will be a slight difference between GR's prediction and the Newtonian prediction. Do you disagree?
RandallB said:
yet you the micro almost impossible to detect the flaw in Mercury’s orbit as justification for making irrational approximations to convert ecliptics into parabolas - I don’t believe I’m seeing from someone of science.
The parabolic path predicted by the constant-G assumption will also differ from the elliptical path predicted by Newtonian gravity by only a "micro almost impossible to detect" amount for a situation like a ball tossed a few feet in the air in a vacuum, and in the limit as the height the ball is tossed becomes arbitrarily small compared to the radius of the planet, the difference between the between the two predictions also becomes arbitrarily small (just for the section of the path between the ball leaving the ground and hitting it again, not if you extrapolate the path through the planet). Again, this is a very obvious fact that any physicist could confirm for you, and everyone else who has chimed in on this thread has agreed it is trivially true as well; if you disagree with this, you need to do some reviewing of basic Newtonian mechanics, and possibly of how limits work in calculus.

If you like, we could also work out some numerical examples involving a ball tossed a few meters up, to show how the difference between the parabolic-path prediction and the elliptical-path prediction would be miniscule. If I calculated the equations for both paths in the case of a ball tossed a meter up, and looked at 10 different times while the ball was in the air and showed that the height at each time predicted by both equations differed by only a microscopic amount, would this lead you to reconsider your position?
RandallB said:
I was giving you Hooke as a means to help you see your problem with just a little common sense. But you don’t seem to have the time to look or think.
You need help finding a book ask a librarian,
Hard to do if you don't tell me what the title of the book is. In any case, you obviously don't remember the actual derivation since you never respond to my request to provide Hooke's equation for air resistance, and if you "think" a bit you will see that if Hooke took into account the phenomenon of terminal velocity there's no way a falling object's path could stay parabolic for long, since an object moving at terminal velocity will have a constant speed. So if Hooke did provide a proof of what you say he did, he must have used an equation for air resistance that becomes totally inaccurate at high velocities. Anyway, I have my doubts that your memory of what Hooke proved is accurate in the first place, since you have been very vague on the details, including where you read it.

In any case, the question of whether an object's path could be made precisely parabolic by assuming some approximate equation for air resistance is irrelevant to what I was arguing, namely that the path of an object thrown at a small height is approximately parabolic, because the actual Newtonian path gets arbitrarily close to a parabola in the limit as the height becomes small.
RandallB said:
This is too simple and basic, to fuss over here any more.
It should be, but your refusal to listen to anyone else or answer questions in detail has made this discussion go on forever. If you continue to deny the obvious while refusing to answer any of my questions or take me up on the offer to look at an actual numerical example, I think I'll just report your post under the "wrong claims" rule, since any admin with an understanding of classical mechanics will surely agree it's wrong to deny that the path of a tossed ball becomes arbitrarily close to a parabola in the limit as the height the ball is tossed becomes very small compared to the radius of the planet.
 
Last edited:
  • #46
JesseM said:
. .
O good grief you can’t find your way in a library? You need to use more than wikipedia. Try biographies like “Isaac Newton and His Times” by Gail Christianson or others that include details on the Newton - Hooke letters of 1679. The diagram there comes from others copies of his letters, just as many other complete biographies use it as well, just find a good one. “provide Hooke's equation for air resistance” why would you think Hooke would use formulas to define his “Elleptueid” curves for these cases, this was years before Principia made them ellipses and decades before calculus was published.

As to you PROOF - I’m waiting

Pick and altitude for your peek vertical height and define the trajectory by the speed of the object perpendicular to the radius to that point.

Just to be sure we are on the same page – free falling objects like this are described by Newtonian and GR alike as conic sections (circular orbit, elliptic orbit, parabolic trajectory, hyperbolic trajectory).
With the key speed being Vc that of a circular orbit, at this speed there is no change in altitude (circular orbit)
Above Vc our described point is the perigee of an elliptic orbit.
Except where the speed is greater than (√2)Vc where it’s not an orbit at all but an escape hyperbolic trajectory.
And the special case of the speed being exactly (√2)Vc for an escape on a parabolic trajectory.
Describing this parabolic path in some detail; if you send objects in opposite directions from this point the limit of both these trajectories are parallel lines with the mid point line between them being the extended line of the radius to our starting point and perigee.

Now for speed less than Vc we have elliptic orbits again this time with our point being the apogee or peak altitude.
If you disagree with any of the above and cannot clear up you confusion in wikipedia let us know.


NOW for your proof; somehow at lower speeds the hyperbolic returns?
Please be complete in describing this parabolic – do use two objects going in opposite directions.
1) define the limit lines of each
2)Are these limits parallel to each other?
3)What reference line(s) can we compare them to as perpendicular or parallel too?
4)And at what speed below which do these slower speeds change from the expected elliptic orbits into your parabolic trajectory, defining some a factor of Vc will be fine.

This I got to see.
 
Last edited:
  • #47
RandallB said:
O good grief you can’t find your way in a library?
Of course I can, but again, a library is not much use to me until you tell me what book I should be looking for, which you haven't up till now.
RandallB said:
You need to use more than wikipedia. Try biographies like “Isaac Newton and His Times” by Gail Christianson or others that include details on the Newton - Hooke letters of 1679. The diagram there comes from others copies of his letters, just as many other complete biographies use it as well, just find a good one. “provide Hooke's equation for air resistance” why would you think Hooke would use formulas to define his “Elleptueid” curves for these cases, this was years before Principia made them ellipses and decades before calculus was published.
I'll look up the book--but if Hooke didn't calculate how air resistance would affect the path of a moving object by using a formula giving air resistance as a function of velocity, then would the assumptions he made about how air resistance affects the paths of moving objects be considered remotely accurate today?
RandallB said:
As to you PROOF - I’m waiting
Well, the thing I offered to prove was that as the ratio between the height of the tossed ball and the radius of the Earth approaches zero, the difference in gravitational force between the point the ball was tossed and its maximum height will approach zero. Do you disagree with this? If so I can provide a proof, it's pretty simple to demonstrate (provided that in your limit, the radius of the planet does not also approach zero).

I also offered to look at a numerical example where we compared the predicted path of Newtonian gravity with the predicted path of uniform gravity, so we could check how small the difference is. Do you want me to do this? If so, please first answer my earlier question:
If I calculated the equations for both paths in the case of a ball tossed a meter up, and looked at 10 different times while the ball was in the air and showed that the height at each time predicted by both equations differed by only a microscopic amount, would this lead you to reconsider your position?

Finally, for a more general proof about when the constant-G approximation works, I came across this paper (summarized http://www.lhup.edu/~dsimanek/scenario/secrets.htm):

http://arxiv.org/abs/physics/0310049

The paper does show that you need a few extra conditions than the one I mentioned to make the limit work, though. For one, not only does the ratio between the vertical height and the planet's radius have to approach zero, but so does the ratio between the horizontal distance the ball travels and the Earth's radius; this is a fairly obvious one, but I forgot about it. They also show that it is necessary to assume the maximum curvature of the ball's path is much greater than the curvature of the Earth's surface, a condition I wasn't aware of. But if all these assumptions hold, the paper shows that the difference between the parabolic path predicted by the constant-gravity assumption and the elliptical path predicted by Newtonian gravity will approach zero in the limit.
RandallB said:
Pick and altitude for your peek vertical height and define the trajectory by the speed of the object perpendicular to the radius to that point.
Just to be sure we are on the same page – free falling objects like this are described by Newtonian and GR alike as conic sections (circular orbit, elliptic orbit, parabolic trajectory, hyperbolic trajectory).
I agree that this is true in Newtonian mechanics, I don't think it's always true in GR, because the perihelion of a non-circular orbit will precess--see the animated diagram at Perihelion Advance of Mercury.
RandallB said:
With the key speed being Vc that of a circular orbit, at this speed there is no change in altitude (circular orbit)
Above Vc our described point is the perigee of an elliptic orbit.
Except where the speed is greater than (?2)Vc where it’s not an orbit at all but an escape hyperbolic trajectory.
And the special case of the speed being exactly (?2)Vc for an escape on a parabolic trajectory.
Describing this parabolic path in some detail; if you send objects in opposite directions from this point the limit of both these trajectories are parallel lines with the mid point line between them being the extended line of the radius to our starting point and perigee.
Now for speed less than Vc we have elliptic orbits again this time with our point being the apogee or peak altitude.
If you disagree with any of the above and cannot clear up you confusion in wikipedia let us know.
Your description of the exact paths predicted by Newtonian gravity is fine.
RandallB said:
NOW for your proof; somehow at lower speeds the hyperbolic returns?
For the millionth time, do you have no understanding of the difference between exact results and approximations based on limits? At lower speeds, the exact trajectory predicted by Newtonian gravity will always be an ellipse. However, for a ball tossed a small height above the Earth's surface, the difference between the exact piece of an ellipse predicted by Newtonian gravity and the parabola predicted by the uniform-G assumption will approach zero in the limit described in the paper I linked to above (the limit where the the height the ball is tossed becomes arbitrarily small compared to the radius of the earth, as does the horizontal distance it travels, but the maximum curvature of the path does not become arbitrarily close to the curvature of the earth). Thus the parabolic path is a perfectly good approximation even though it is never exactly correct for any finite height, just like Newtonian orbits are perfectly good approximations for GR even though they are never exactly correct for any gravitational source with finite mass/energy.
RandallB said:
Please be complete in describing this parabolic – do use two objects going in opposite directions.
How can one use two objects going in opposite directions for a tossed ball? Do you want one ball tossed upward and another falling downward through the earth? That won't work, because the limit only works for the small section of the path that's above the earth. Do you want two falling objects moving in opposite horizontal directions from the peak of the path of a single ball, or what?
RandallB said:
1) define the limit lines of each
What exactly is a "limit line"? Googling this phrase, and searching for it on the mathworld site, it doesn't look like it's a standard mathematical term. Are you thinking of something like the asymptotes of a hyperbola, or the tangent line to a curve at a particular point?
RandallB said:
4)And at what speed below which do these slower speeds change from the expected elliptic orbits into your parabolic trajectory, defining some a factor of Vc will be fine.
Try thinking about what I mean when I keep talking (over and over and over) about the idea of approximations based on limits, and the fact that the path will never be exactly parabolic for any finite V below the escape velocity, it is just that the difference between the exact elliptical path and the parabolic approximation approaches zero in the limit as the height and horizontal distance become small compared to the radius of the earth, with the extra caveat mentioned in the paper above that the maximum curvature of the path does not become arbitrarily close to the curvature of the Earth in the limit.
 
Last edited:
  • #48
JesseM said:
for a more general proof about when the constant-G approximation works, I came across this paper

http://arxiv.org/abs/physics/0310049

The paper does show that you need a few extra conditions ...
...I don't think it's always true in GR,
Are we starting to waffle here ?
Are you about to say you didn’t intend assumptions to actually be accurate?
For the millionth time, do you have no understanding of the difference between exact results and approximations based on limits? the parabolic path is a perfectly good approximation even though it is never exactly correct for any finite height, just like Newtonian orbits are perfectly good approximations for GR
I question your counting as much as I disagree with the usefulness these approximations limits or otherwise.

Just look at the web DOC you suggested – see figure 1
The diagram on the left cuts out a section of an ellipse that looks much more like a hyperbolic than a parabolic. But then the less than faithful redrawing in a “flat” frame on the right might look parabolic, so what if you can distort the view to make look like anything someone wants. You aren’t going to let that pass as real science when they already plainly agree the true path is an ellipse are you?
The point of the paper is complaining that most texts do a bad job of defining hyperbolic curves at low speeds! Are they talking about CREN??
And then inexplicably to me they proceed to interpret a parabola from the apogee of an ellipse! Parabolas don’t have an apogee just a perigee! Why don’t they use the IDENTICAL in shape “parabola” cut from the other side of the ellipse they display at perigee! Then describe a “parabolic” path displaying some kind of negative gravity as the ball is rising in altitude! Their concluding proviso that it shouldn’t apply for speeds much less than escape? They really mean speeds much less than circular – this is just a pure diversion from the reality. I consider it no more than a magician’s need to justify a misdirection even if it has fooled themselves as well.

I’ve already covered the point that a segment of an ellipse can look like part of a circle now here we have what looks like a hyperbola till they distorted it to seem like a parabola. SO WHAT, who cares that you can contort the measuring frame to make something like else. Don’t you want to deal with reality?
Or is a flat map of the world a good enough approximation to convince you that Greenland is larger than the USA and Iceland is nearly as large as Texas!
The graininess of observation (approximation) here required to fail to see elliptical paths is huge by miles more than the graininess of the mercury observations – so stop making that comparison it’s just foolish.

How can one use two objects going in opposite directions for a tossed ball? Do you want one ball tossed upward and another falling downward through the earth? ...
Do you want two falling objects moving in opposite horizontal directions from the peak of the path of a single ball, or what?
Of course what’s hard about that and why do you continue to define a tossed ball by it’s vertical direction or speed?? That is totally meaningless here. When defining your toss, I don’t care what angle – the only thing we need is its horizontal speed at apogee or perigee when vertical speed reaches zero. So just pick a horizontal speed to start with and pitch it flat to the north from the equator. To see the whole curve just toss a ball north and a second ball south same speeds what’s the problem, you can’t be worried about air resistance.
What exactly is a "limit line"?
A parabolic trajectory is on a path to reach “apogee” 180 on the opposite side of the Earth (or main body being orbited) that apogee on a straight line from our stating perigee point though the center of orbit as a major axis to whatever distance needed to reach the apogee point. Just like the elliptic path that as that orbit moves away from the major axis it cannot start to move back towards it until the tangent of the path is parallel to that major axis line at some distance from it. The point at which it does become parallel defines the mid point of the ellipse and the crossing point for the minor-axis making a perpendicular line across the major-axis to reach the point where the ball tossed the other way has also reached a parallel tangent in its path. The issue with a parabolic is it can never move toward an apogee or even back towards what would be the major axis as its tangent never reaches this line except as a limit at infinity. What are you using to define the limit of this parabolic, if it’s an asymptote what do you use as references to it? Does it ever become perpendicular to the Earth's surface? – how and when?
.
Approximations are only useful if they accurately describe true results. And I see no useful propose here for these misrepresentations.
Using approximations to convince people that GR actually predicts parabolic paths at speeds much below escape velocity or circular velocity should be considered malpractice – I don’t care if it’s you or if CERN actually did it – it’s just wrong and grossly misleading of the truth in reality.

As I said before, but not a million times yet, GR only predicts a parabola at escape velocity not at lower speeds. Galileo’s parabolas require air resistance.
 
  • #49
JesseM said:
for a more general proof about when the constant-G approximation works, I came across this paper

http://arxiv.org/abs/physics/0310049

The paper does show that you need a few extra conditions ...
...I don't think it's always true in GR,
RandallB said:
Are we starting to waffle here ?
Are you about to say you didn’t intend assumptions to actually be accurate?
Not waffling, just admitting that I was wrong about the correct definition of the limit--it's not just the limit as the height becomes small compared to the radius of the planet, but also the limit as the horizontal distance becomes small compared to the radius, and the maximum curvature of the curve does not become small compared to the curvature of the planet. In my defense, I had mainly been thinking about the simplest case where the ball is tossed vertically and it's the graph of position vs. time that you want to approach a parabola--in this case, the only limit you need to worry about is the height being small compared to the radius of the planet.

As for my second statement that "I don't think it's always true in GR" (referring to planets traveling in ellipses), if you think this is "waffling" you haven't been paying attention to my earlier posts, I've said all along that elliptical orbits would only be valid in GR as an approximation in the Newtonian limit. For example, in post #38 I said:
I think the real path is only an ellipse if you use Newtonian gravity, which is just as much of an approximation as the constant G-field is. We know that GR predicts Mercury's orbit is not a perfect elipse--the perihelion changes slightly with each orbit--so I would assume that this effect does not just vanish abruptly at some point, presumably if you calculated the exact path of a tossed ball which was able to travel straight through the earth, its perihelion would precess slightly too. Of course the difference from the Newtonian prediction and the GR prediction might be only a nanometer with each orbit or something, but the difference between the constant-G prediction of a ball's path between the time it is tossed and the time it hits the ground again would probably also differ from the decreasing-G prediction by only a microscopic amount (do you admit, by the way, that your earlier statement 'If you take air resistance out of it you will completely change the shape of the curve' was incorrect?)
Then in post #40 I emphasized this point many times:
JesseM said:
Just like GR doesn't say curves are exactly elliptical[/color], although they are approximately so. If he had asked how GR explains elliptical paths, would you just thunder that GR "DOES NOT" predict such paths, or would you understand that he was speaking approximately?

...

RandallB said:
Originally Posted by RandallB
And sure you need me to say it twice? IN A VACUUM IT’S ELLIPTIC not parabolic!
Not exactly, according to GR--again, think of the precession of the perihelion of Mercury's orbit, showing that the orbit is not a perfect ellipse. But the ellipse is a valid approximation because in the Newtonian limit the orbit predicted by GR becomes arbitrarily close to an ellipse.

...

For example, do you think it is ever justified to use Newtonian mechanics when we know that these predictions differ slightly from those of GR (for example, GR will say that a planet's orbit is not precisely elliptical[/color]), and that GR's predictions are the more accurate ones?

...

I don't say that Newton or GR produce a parabolic exactly, just like GR doesn't produce elliptical orbits exactly. Once again, are you completely unfamiliar with the concept of approximations based on limits?
And in post #45 I repeated this:
JesseM said:
RandallB said:
GR reduces to the Newtonian in our real world
Not exactly, no--only in the limit. In any real example, there will be a slight difference between GR's prediction and the Newtonian prediction. Do you disagree?
OK, back to your post:
JesseM said:
For the millionth time, do you have no understanding of the difference between exact results and approximations based on limits? the parabolic path is a perfectly good approximation even though it is never exactly correct for any finite height, just like Newtonian orbits are perfectly good approximations for GR
RandallB said:
I question your counting as much as I disagree with the usefulness these approximations limits or otherwise.
Whether it is "useful" or not is not the issue here, the issue is just whether the notion of an approximation becoming arbitrarily close to accurate in a certain limit is a mathematically rigorous one, and it is. You're free to accept this but still not use these approximations because "anything less than perfect accuracy is EEEVIL" or whatever, although I'll note again that if this is your attitude it's hypocritical of you to say Newtonian mechanics is appropriate to use in any situation (even, say, planetary orbits), because we know it always differs slightly from the predictions of GR, even the difference between Newtonian predictions and GR predictions becomes arbitrarily small in certain limits. Again, I'm pretty sure that GR would say the orbits of planets are never exactly elliptical, so that'd be an example of what I'm talking about.
RandallB said:
Just look at the web DOC you suggested – see figure 1
The diagram on the left cuts out a section of an ellipse that looks much more like a hyperbolic than a parabolic. But then the less than faithful redrawing in a “flat” frame on the right might look parabolic, so what if you can distort the view to make look like anything someone wants.
In the limit as the width and height of the region you're considering becomes arbitrarily small compared to the radius of the planet, the polar coordinates become arbitrarily close to cartesian coordinates--for example, the angle between different radial lines in the polar coordinate drawing approaches zero in the region, and the curvature of the constant-radius lines approaches zero in the region. Do you disagree?
RandallB said:
You aren’t going to let that pass as real science when they already plainly agree the true path is an ellipse are you?
For the last time Randall, I agree the true path in Newtonian mechanics is an ellipse too, my argument (and theirs) is just that in a certain well-defined limit, the true elliptical path becomes arbitrarily close to the approximate parabolic path. Similarly, in GR the true path of an orbit is not an ellipse, but in a certain well-defined limit, the true non-elliptical path predicted by GR becomes arbitrarily close to the elliptical path predicted by Newtonian mechanics.
RandallB said:
The point of the paper is complaining that most texts do a bad job of defining hyperbolic curves at low speeds! Are they talking about CREN??
The CERN page did not even attempt to define what the appropriate limit is where the parabolic approximation becomes valid, it just took for granted that tossing a ball on Earth is a situation where it's fine to use this approximation, and the paper would validate this assumption, even if the authors may feel it's important to make the nature of the limit explicit.
RandallB said:
And then inexplicably to me they proceed to interpret a parabola from the apogee of an ellipse! Parabolas don’t have an apogee just a perigee!
So what? The point is that the apogee of the ellipse becomes arbitrarily close to the perigee of the approximate parabola in this limit, and they demonstrate this mathematically. There's no rule that says it must be the perigee of an ellipse which approaches the perigee of a parabola in the limit. If you disagree with their proof, please point out which step you think is incorrect.
RandallB said:
Why don’t they use the IDENTICAL in shape “parabola” cut from the other side of the ellipse they display at perigee!
Why should they? The perigee would be somewhere deep beneath the surface of the earth, and would have no relevance to the question of whether the portion of the path immediately above the surface becomes arbitrarily close to a parabola in some limit. In any case, if mathematics proves that the apogee of an ellipse becomes arbitrarily close to the perigee of a parabola in some well-defined limit, it doesn't make any sense to "dispute" this proof by spluttering "but one is an apogee and the other is a perigee!"
RandallB said:
Then describe a “parabolic” path displaying some kind of negative gravity as the ball is rising in altitude!
Er, what? If the initial velocity of the ball at the surface is in the upward direction, then of course it will rise (this is just as true of the elliptical path as it is of the parabolic approximation), positive gravity just says that everything must accelerate towards the Earth (as the ball is doing, since its upward velocity is decreasing as it goes up), not that every object's velocity must point towards the Earth at all moments. "Negative gravity" would be if it was accelerating upwards, ie if its velocity away from the Earth was increasing as it travelled.
RandallB said:
Their concluding proviso that it shouldn’t apply for speeds much less than escape?
That's not an additional proviso, they just say that this proviso is equivalent to the condition on the radius of curvature of the object's path vs. the Earth that they mentioned earlier, namely \alpha \epsilon &lt;&lt; 1.
RandallB said:
They really mean speeds much less than circular – this is just a pure diversion from the reality. I consider it no more than a magician’s need to justify a misdirection even if it has fooled themselves as well.
Why is this a "misdirection"? As they say earlier in the paper, "The condition \alpha \epsilon &lt;&lt; 1, then, is simply the condition that the motion is very slow compared to typical “orbital” (as opposed to “trajectory”) motions." Would you disagree that a ball tossed by human hands near the surface of the Earth will have a velocity very small compared to that of a ball orbiting the earth?
RandallB said:
I’ve already covered the point that a segment of an ellipse can look like part of a circle now here we have what looks like a hyperbola till they distorted it to seem like a parabola. SO WHAT, who cares that you can contort the measuring frame to make something like else. Don’t you want to deal with reality?
I know of no well-defined limit where the elliptical path predicted by Newtonian gravity can be shown to become arbitrarily close to a circular path--do you? This certainly would not be true in the case of the limit they discuss in their paper, where the height and horizontal path length are very small compared to the radius of the planet, and the maximum curvature of the path is much greater than the curvature of the planet.
RandallB said:
The graininess of observation (approximation) here required to fail to see elliptical paths is huge by miles more than the graininess of the mercury observations – so stop making that comparison it’s just foolish.
No it isn't. Again, I'd be happy to look at an actual numerical example where we consider a ball tossed with a small initial velocity, and find the difference between the exact elliptical path and the approximate parabolic path at a bunch of different points along the trajectory. I am confident we'd find that the difference is microscopic--so again, if you take me up on this offer, then please tell me in advance whether you'd reconsider your position if it turns out I'm correct that the difference would be microscopic in such an example. I would certainly reconsider my position on the usefulness of this approximation if it turned out the difference was non-microscopic (I would not reconsider the position that as the height and horizontal length of the path becomes arbitrarily small compared to the radius of the planet, the parabolic approximation becomes arbitrarily close to perfect, but if it turned out the difference was non-microscopic in the case of a ball tossed a few meters than I'd conclude that this height was not close enough to 'arbitrarily small' compared to the radius of the Earth for the approximation to be useful for typical examples).
RandallB said:
Of course what’s hard about that and why do you continue to define a tossed ball by it’s vertical direction or speed?? That is totally meaningless here. When defining your toss, I don’t care what angle – the only thing we need is its horizontal speed at apogee or perigee when vertical speed reaches zero. So just pick a horizontal speed to start with and pitch it flat to the north from the equator. To see the whole curve just toss a ball north and a second ball south same speeds what’s the problem, you can’t be worried about air resistance.
OK, I see what you're saying--you want two balls tossed in opposite directions from their maximum height, with their vertical velocity set to be zero at the moment they're tossed. Yes, then the two balls together will give the same curve as a ball tossed upwards and horizontally from the surface with a certain velocity--but if the two situations give the same path, what difference does it make? Perhaps you're just suggesting that we start from the apogee because it's easier to calculate the equation of the elliptical path with this information?
JesseM said:
What exactly is a "limit line"?
RandallB said:
A parabolic trajectory is on a path to reach “apogee” 180 on the opposite side of the Earth (or main body being orbited) that apogee on a straight line from our stating perigee point though the center of orbit as a major axis to whatever distance needed to reach the apogee point. Just like the elliptic path that as that orbit moves away from the major axis it cannot start to move back towards it until the tangent of the path is parallel to that major axis line at some distance from it. The point at which it does become parallel defines the mid point of the ellipse and the crossing point for the minor-axis making a perpendicular line across the major-axis to reach the point where the ball tossed the other way has also reached a parallel tangent in its path. The issue with a parabolic is it can never move toward an apogee or even back towards what would be the major axis as its tangent never reaches this line except as a limit at infinity. What are you using to define the limit of this parabolic, if it’s an asymptote what do you use as references to it? Does it ever become perpendicular to the Earth's surface? – how and when?
You're talking about what would happen to the ball after it falls right through the surface of the earth, but in case I haven't made this clear, the parabolic approximation is only supposed to become arbitrarily close to the true elliptical path for the small subsection of the path above the surface of the earth, I completely agree that the parabolic path will begin to wildly diverge from the elliptical path if you extrapolate it through the crust and down past the center of the earth! In this case, as you say, the elliptical path will eventually turn around and start approaching the major axis again, finally hitting it at the perigee, while the parabolic path will never do so. The approximation was never meant to hold for such an extrapolated path, just for the section of the path that is actually physically meaningful for a ball tossed near the surface (since a real ball won't be able to travel right through the surface of the Earth like a neutrino).
RandallB said:
As I said before, but not a million times yet, GR only predicts a parabola at escape velocity not at lower speeds. Galileo’s parabolas require air resistance.
And as I've said, the fact that the path of a tossed ball looks like a parabola has nothing to do with air resistance. Again, if we were to actually calculate the difference between an elliptical path and a parabolic path for a ball tossed above the surface (not continuing the paths past the surface) at ordinary velocities, I am confident that the difference will be microscopic. Are you willing to actually look at such an explicit numerical calculation and reconsider your position if it turns out I am right?
 
Last edited:
  • #50
JesseM said:
And as I've said, the fact that the path of a tossed ball looks like a parabola has nothing to do with air resistance. Again, if we were to actually calculate the difference between an elliptical path and a parabolic path for a ball tossed above the surface (not continuing the paths past the surface) at ordinary velocities, I am confident that the difference will be microscopic.
And as I've said well within the microscopic range of a circle and a hyperbola as well
Are you willing to actually look at such an explicit numerical calculation and reconsider your position if it turns out I am right?
Sure if you can work out the numbers that translate these curve shapes to a flat surface I'll look at them. Why not include what you expect the curve shape with air resistance to be as well, if it’s not a parabola it be worth knowing what it matches up best with, ellipse at perigee, circle, ellipse at apogee, hyperbola.
 
Back
Top