Long planet and Galileo thought experiment

[yuiop]

A long time ago, Galileo stated the vertical motion of a falling body is not influenced by its horizontal motion. Is that still true in moderns terms, taking relativity into account?

For example, if a particle is fired horizontally at 0.8c, will it hit the ground at the same time as a particle dropped vertically? (Assume a very long flat planet)

 

[YellowTaxi]

A long time ago, Galileo stated the vertical motion of a falling body is not influenced by its horizontal motion. Is that still true in moderns terms, taking relativity into account?

For example, if a particle is fired horizontally at 0.8c, will it hit the ground at the same time as a particle dropped vertically? (Assume a very long flat planet)

Seems odd. He was the first to predict that falling bodies are influenced by rotation of the planet Earth. And that an object dropped from the top of tower of Piza Tower would fall to the East (slightly). Can only assume he was thinking (connected with this realisation) in idealised terms where there's no rotation of the planet your standing on.

 

[yuiop]

Seems odd. He was the first to predict that falling bodies are influenced by rotation of the planet Earth. And that an object dropped from the top of tower of Piza Tower would fall to the East (slightly). Can only assume he was thinking (connected with this realisation) in idealised terms where there's no rotation of the planet your standing on.

I think Galileo gave this version of relativity. Let's say a small cannon ball is dropped from the top of the mast of a stationary ship and it lands at the base of the mast in 2 seconds. Galileo thought the ball would also land at the base of the mast in 2 seconds if the ship was moving, so without reference to some landmarks they could not tell if they were moving. He then suggested if there was an observer on the shore (stationary), he would also see the ball take 2 seconds to drop, but he would see the ball follow a parabolic path if the ship was moving, while the people on board the ship always see it fall straight down. That, I think, is why he concluded horizontal motion does not alter the time it takes an object to fall vertically. Now is that still true given what we know now about relativity?

 

[JesseM]

I think Galileo gave this version of relativity. Let's say a small cannon ball is dropped from the top of the mast of a stationary ship and it lands at the base of the mast in 2 seconds. Galileo thought the ball would also land at the base of the mast in 2 seconds if the ship was moving, so without reference to some landmarks they could not tell if they were moving. He then suggested if there was an observer on the shore (stationary), he would also see the ball take 2 seconds to drop, but he would see the ball follow a parabolic path if the ship was moving, while the people on board the ship always see it fall straight down. That, I think, is why he concluded horizontal motion does not alter the time it takes an object to fall vertically. Now is that still true given what we know now about relativity?

If the region of space and time you're looking at is large enough so tidal forces are insignificant (as would be true if you made the size of the planet as large as possible so it looked very close to an infinite flat plane in the region you're considering), then sure. After all, the equivalence principle says that a freefalling observer moving downwards along with the two objects would see them behave just as the would in SR, and in SR if you have one object at rest and another moving horizontally, the vertical distance between them won't change.

 

[yuiop]

If the region of space and time you're looking at is large enough so tidal forces are insignificant (as would be true if you made the size of the planet as large as possible so it looked very close to an infinite flat plane in the region you're considering), then sure. After all, the equivalence principle says that a freefalling observer moving downwards along with the two objects would see them behave just as the would in SR, and in SR if you have one object at rest and another moving horizontally, the vertical distance between them won't change.

Well, here is why I think Galileos proposal does not hold.

Say we have 2 towers of the same height, separated by a distance of 0.8 light seconds. Now let's say it takes a ball one second to fall straight down from the top of a tower to its base. When the ball (A) is fired horizontally from tower A at 0.8c it should arrive at the base of tower B one second. Now if a ball is released simultaneously from tower B the moment ball A is fired from the top of tower A, then both balls should arrive simultaneously at the base of tower B.

Now if we have an observer that is moving at 0.8c in the same direction as ball A then ball A appears to be falling vertically to him. Now in his reference frame, the ball is released earlier from tower B due to the relativity of simultaneity. So if ball B has a head start over ball A, then both balls can not arrive simultaneously at the base of tower B if both balls are falling at the same rate. Clearly, the falling rate IS influenced by the horizontal motion, disproving Galileo's proposal.

In fact, I think it works out that ball B lands before ball A is even fired according to the observer moving relative to the surface.

 

[JesseM]

Well, here is why I think Galileos proposal does not hold.

Say we have 2 towers of the same height, separated by a distance of 0.8 light seconds. Now let's say it takes a ball one second to fall straight down from the top of a tower to its base. When the ball (A) is fired horizontally from tower A at 0.8c it should arrive at the base of tower B one second. Now if a ball is released simultaneously from tower B the moment ball A is fired from the top of tower A, then both balls should arrive simultaneously at the base of tower B.

Now if we have an observer that is moving at 0.8c in the same direction as ball A then ball A appears to be falling vertically to him. Now in his reference frame, the ball is released earlier from tower B due to the relativity of simultaneity. So if ball B has a head start over ball A, then both balls can not arrive simultaneously at the base of tower B if both balls are falling at the same rate. Clearly, the falling rate IS influenced by the horizontal motion, disproving Galileo's proposal.

In fact, I think it works out that ball B lands before ball A is even fired according to the observer moving relative to the surface.

You have to remember that according to the equivalence principle, only a local freefalling frame can be treated as equivalent to an inertial frame in SR. So if tidal forces are negligible in the region of spacetime containing both towers and the balls dropping from top to bottom, then we can treat this as being equivalent to a situation in flat spacetime where we have the ground accelerating in the vertical direction while the two balls move inertially, and where in the frame where they are released simultaneously, they both have the same initial vertical velocity relative to the ground at that moment. To make things simple one could stick to frames where they are both at rest in the vertical axis while the ground accelerates up to meet them.

So, viewed in this way, in your scenario we're starting out by looking at an observer at rest relative to B, who observes A to be moving towards B at 0.8c and reaching B's position at the exact moment that the ground accelerates up and hits both of them. Now, it's true that if we switch to A's rest frame, in this frame B and A are released at different moments (B released earlier than A) and yet A and B both hit each other and the ground simultaneously. However, trying to sketch a 3D spacetime diagram of this situation, I think it's also true that the ground is not level in this frame--the vertical position of the two towers is not equal at the moment A is released. The math to show this would be a little tough though, you'd have to use the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html to deal with the motion of the bases of the two towers, assuming that in the frame of B they both start at 0 velocity at the moment A and B are released simultaneously and then accelerate towards them, then do a Lorentz transform into A's frame to figure out where the bases of the towers are at the same moment (in this frame) that A is released.

 

[yuiop]

Hi JesseM,

Your adaptation of the problem is very helpful and takes some concerns about different time rates between the top and bottom of the towers due to gravitational time dilation, out of the equation.

If I may, I would like to add a third ball (C) to the experiment to show there is still an anisotropic acceleration. Ball C is released from the top of tower A at the same time A is fired. Release of ball C and firing of A is simultaneous in both frames. The arrival of ball A and ball B at the base of tower B (by your argument) is also simultaneous. However, because the towers are spatially separated, ball C does not hit the ground simulateously with balls A and B. The problem is, balls A and C were released simultaneously, but did not hit the ground simultaneously, according the observer comoving with ball A.

As I see it, the observer comoving with ball A sees the towers as the same height initially, but sees tower B accelerating upwards before tower A and the ground curving upwards from tower A to tower B.

 

[JesseM]

If I may, I would like to add a third ball (C) to the experiment to show there is still an anisotropic acceleration. Ball C is released from the top of tower A at the same time A is fired. Release of ball C and firing of A is simultaneous in both frames. The arrival of ball A and ball B at the base of tower B (by your argument) is also simultaneous. However, because the towers are spatially separated, ball C does not hit the ground simulateously with balls A and B.

Well, it hits simultaneously in the rest frame of ball C and ball B. In the rest frame of ball A, it's true that it doesn't hit simultaneously, but again I think this is because the ground is not level in this frame--if you calculated the vertical position of the base of each tower at the moment C was dropped and A was fired in A's rest frame, I think you'd find they're different at this moment.

 

[yuiop]

Well, it hits simultaneously in the rest frame of ball C and ball B. In the rest frame of ball A, it's true that it doesn't hit simultaneously, but again I think this is because the ground is not level in this frame--if you calculated the vertical position of the base of each tower at the moment C was dropped and A was fired in A's rest frame, I think you'd find they're different at this moment.

I agree, when C is is dropped the base of tower B has already moved upward. To Anne (the observer comoving with ball A) ball B had a head start and is already part way down tower B when Ball C is dropped. To anne, ball B must be falling slower than ball A because C had a head start and yet does not arrive before ball A. Anne also thinks ball C is falling slower than ball A because balls A and C started out at the same time but ball A hits the ground first.

After a few quick calculations, in Anne's frame, ball A takes 0.6 seconds to fall to the ground (compared to 1 second in the ground frame) and balls B and C take 1.6666 seconds to fall (while taking 1 second in the ground frame.)

It seems that Galileo's proposal is right in the rest frame of the massive body (the planet) but does not hold when the observer is moving relative to the massive body.

For example, if Anne is in a closed lab, she would note that it takes one second for a particle to fall in her lab when it is at rest with the planet, but takes only 0.6 seconds when her lab is moving relative to the planet. (She would detect her velocity relative to the massive planet without looking out the window). An observer on the ground sees that her clock is slow due to her relative motion and by his clock the ball takes one second to fall in her lab.

On the other hand, if an apple takes one second to fall a given height on Earth, then to an observer far out in space moving at 0.8c relative to the Earth, the Earth and apple are moving relative to him and it takes 1.6666 seconds for the apple to fall by his measurements, as would be expected by taking time dilation into account.

 

[JesseM]

I agree, when C is is dropped the base of tower B has already moved upward. To Anne (the observer comoving with ball A) ball B had a head start and is already part way down tower B when Ball C is dropped. To anne, ball B must be falling slower than ball A because C had a head start and yet does not arrive before ball A. Anne also thinks ball C is falling slower than ball A because balls A and C started out at the same time but ball A hits the ground first.

Why do you say Anne thinks C is falling slower? After all, if Anne is at rest relative to A, doesn't she always say the vertical position of C is identical to the vertical position of A at all times? It's the ground that's moving up to meet them here. What's not clear to me is whether Anne sees the bases of the towers at different heights at any given time but still both accelerating vertically at the same rate (and moving sideways at constant velocity in Anne's frame), like a sloped surface accelerating upwards uniformly, or if Anne not only sees them at different heights but also accelerating up at different rates. We'd have to do the calculations to figure this out--I suppose these could be made easier by making the towers fairly short in height, say 100 meters, so that you could assume in B and C's frame that the bases both have height y(t) = t^2 * (4.9 meters/second^2) as in Newtonian physics (where y=0 is the position of the two bases at the moment the balls are let go at the top of the towers in B and C's frame), without needing to use the relativistic formula for distance as a function of time since the time is too short for relativistic effects to matter significantly.

 

[yuiop]

Why do you say Anne thinks C is falling slower? After all, if Anne is at rest relative to A, doesn't she always say the vertical position of C is identical to the vertical position of A at all times?

If Anne sees the base of tower B as being at a different different height to the base of tower A, (when she is at tower A) due to the spatial separation of B from A and the relativity of simultaeinity, then when she is closer to tower B she will see the base of tower A as being at a different height to an imaginary tower comoving with her for the same reasons.

 

[JesseM]

If Anne sees the base of tower B as being at a different different height to the base of tower A, (when she is at tower A) due to the spatial separation of B from A and the relativity of simultaeinity, then when she is closer to tower B she will see the base of tower A as being at a different height to an imaginary tower comoving with her for the same reasons.

Sure, but like I said, it's not clear to me that this rules out the possibility that she observes both towers accelerating upwards at the same rate in her frame (with tower A being at a lower height than tower B at all times, as if the two towers are on a sloping hillside), I still think we'd need to do the math to check.

 

[yuiop]

Sure, but like I said, it's not clear to me that this rules out the possibility that she observes both towers accelerating upwards at the same rate in her frame (with tower A being at a lower height than tower B at all times, as if the two towers are on a sloping hillside), I still think we'd need to do the math to check.

Anne probably does see both towers accelerating upwards at the same rate which she could also view as both balls B and C (the ones falling straight down their respective towers) as falling at the same rate, but that rate is slower than the falling rate of the ball she is co-moving with (ball A). For example if ball B starts moving downward (or tower B starts moving upward) a full half second before ball A is fired, then it takes ball B 1.5 seconds to drop the same distance as ball A drops in one second (because they both finish at the same time and place. Ball C starts dropping at the same time as A is fired but lands half a second later than ball A. So both balls B and C fall at the same rate (both towers accelerating upward at the same rate) and they both take 1.5 seconds to fall the height of a tower, while ball A only takes 1 second to fall the height of a tower, so ball A is falling faster from Anne's point of view, while all 3 balls fall at exactly the same rate (all start at the same time and all finish at the same time) to an observer that is stationary with respect to the towers. See the asymmetry?

 

[JesseM]

Anne probably does see both towers accelerating upwards at the same rate which she could also view as both balls B and C (the ones falling straight down their respective towers) as falling at the same rate, but that rate is slower than the falling rate of the ball she is co-moving with (ball A). For example if ball B starts moving downward (or tower B starts moving upward) a full half second before ball A is fired, then it takes ball B 1.5 seconds to drop the same distance as ball A drops in one second (because they both finish at the same time and place.

That doesn't make sense to me--if A and B have the same y-coordinate (vertical height) at all times in B's rest frame, then they must have the same y-coordinate at all times in A's rest frame which is also Anne's rest frame, no? If we inertially extend the paths of A and B backwards so that instead of being dropped from the tops of the towers they simply flew by them inertially, then it's true that in Anne's frame B passes the top of tower B before A passes the top of tower A, but if the ground is accelerating up evenly as I suggested, then it'd just be because the top of tower A is at a lower height than the top of tower B, due to the slope of the "hillside" in this frame. In Anne's frame, it is still true that when B was passing by the top of tower B, then whatever the y-coordinate of the top of tower B, ball A was also at that y-coordinate at the same moment in Anne's frame.

Ball C starts dropping at the same time as A is fired but lands half a second later than ball A. So both balls B and C fall at the same rate (both towers accelerating upward at the same rate) and they both take 1.5 seconds to fall the height of a tower, while ball A only takes 1 second to fall the height of a tower, so ball A is falling faster from Anne's point of view, while all 3 balls fall at exactly the same rate (all start at the same time and all finish at the same time) to an observer that is stationary with respect to the towers. See the asymmetry?

No, if it is true that the entire ground is accelerating up at a constant rate in Anne's frame (which still might not be true, I don't know), then all three balls always occupy the same y-coordinate, the only reason C hits the ground at a later time in Anne's frame is because it hits the ground at a lower y-coordinate due to the slope of the ground in Anne's frame. You could set up a similar situation in ordinary Newtonian physics, dropping three balls from the same height above sea level at a single moment, but dropping them onto a sloping hill so that two of them hit the ground before the other one.

 

[Jorrie]

That, I think, is why he [Galileo] concluded horizontal motion does not alter the time it takes an object to fall vertically. Now is that still true given what we know now about relativity?

From a GR perspective, Galileo was wrong of course, especially if he could launch the ball horizontally at 0.8c, as you suggested in your OP. Galilean and relativistic analyzes would be giving different results.

Pervect has shown https://www.physicsforums.com/showpost.php?p=1046874&postcount=17" some time ago that both radial and transverse velocities influence accelerations, which influence the time it will take for the ball to drop a given height.

$$\frac{d^2 r}{d t^2} = \frac {3 m{{\it v_r}}^{2}}{ \left( r-2\,m \right) r} + \left( r-2\,m \right) \left( {{\it v_{phi}}}^{2}-{\frac {m}{{r}^{3}}} \right)$$

v_r = dr/dt and v_phi = dphi/dt

I've roughly calculated the radial acceleration in relation to the usual 1g for a particle traveling precisely horizontally near Earth's surface at 0.8c and got ~ 2.2g. A particle approaching c horizontally will experience a radial acceleration of ~3g.

 

[JesseM]

I've roughly calculated the radial acceleration in relation to the usual 1g for a particle traveling precisely horizontally near Earth's surface at 0.8c and got ~ 2.2g. A particle approaching c horizontally will experience a radial acceleration of ~3g.

But will it actually seem to accelerate towards Earth 3 times faster than a particle dropping straight down, as seen by someone on the ground? In other words, can you see particles dropping at different rates in a single frame?

 

[yuiop]

From a GR perspective, Galileo was wrong of course, especially if he could launch the ball horizontally at 0.8c, as you suggested in your OP. Galilean and relativistic analyzes would be giving different results.

Pervect has shown https://www.physicsforums.com/showpost.php?p=1046874&postcount=17" some time ago that both radial and transverse velocities influence accelerations, which influence the time it will take for the ball to drop a given height.


I've roughly calculated the radial acceleration in relation to the usual 1g for a particle traveling precisely horizontally near Earth's surface at 0.8c and got ~ 2.2g. A particle approaching c horizontally will experience a radial acceleration of ~3g.

Pervect is rarely wrong, but he might be on this occasion because his equation violates the equivalence principle.

Case 1:
An elevator (lab) in space far from any gravitational sources and not experiencing any acceleration. Some particles are fired across the lab at say 0.5 c. they hit a spot on the opposite side of the lab and the spot is marked. A laser is alligned so that the beem hits the same spot and the particles are parallel to the laser beem for the entire journey.

Case 2:

The same elevator is dropped in a gravitational field. The particles, the photons and the elevator all drop at the same rate, and the particles and photons all hit the same spot on the wall as when the elevator was not accelerating. If the particles and photons did not fall at the same speed they hit different spots on the wall. The free falling elevator would not be the same as the elevator far out in space that is not accelerating.

This is the very foundation of GR. It was Einstein's "happiest thought"! Do the GR equations really violate that?

[EDIT] I am off course talking about the particles and photons moving horizontally with respect to the gravitational "field".



[EDIT][EDIT] I have spotted the error in my own argument (before anyone else points it out). My assumption that Pervect might sometimes be wrong like other mortals is flawed :P

Actually, the particles and photons may follow different paths because they start out at different times if they are to arrive at the far end at the same time. I'll have to give this some more thought ;)

 

[yuiop]

That doesn't make sense to me--if A and B have the same y-coordinate (vertical height) at all times in B's rest frame, then they must have the same y-coordinate at all times in A's rest frame which is also Anne's rest frame, no?

No. because what is simultaneous in B's rest frame (y coordinates) is not simultaneous in A's rest frame.

If we inertially extend the paths of A and B backwards so that instead of being dropped from the tops of the towers they simply flew by them inertially, then it's true that in Anne's frame B passes the top of tower B before A passes the top of tower A, but if the ground is accelerating up evenly as I suggested, then it'd just be because the top of tower A is at a lower height than the top of tower B, due to the slope of the "hillside" in this frame. In Anne's frame, it is still true that when B was passing by the top of tower B, then whatever the y-coordinate of the top of tower B, ball A was also at that y-coordinate at the same moment in Anne's frame.

The ground is not accelerating upwards evenly as you suggest in Annes frame. See below.

No, if it is true that the entire ground is accelerating up at a constant rate in Anne's frame (which still might not be true, I don't know),

In Anne's frame, tower B accelerates upwards for an initial period before tower A starts to accelerate upwards. In Anne's frame the ground must appear to bend, to accommodate tower B accelerating upward, while tower A (The top and base and the ground near tower A) remains stationary, due to the (lack of) simultaneity in Anne's frame. Anne sees the ground between tower A and tower B getting progressively more bent/curved.

then all three balls always occupy the same y-coordinate, the only reason C hits the ground at a later time in Anne's frame is because it hits the ground at a lower y-coordinate due to the slope of the ground in Anne's frame. You could set up a similar situation in ordinary Newtonian physics, dropping three balls from the same height above sea level at a single moment, but dropping them onto a sloping hill so that two of them hit the ground before the other one.

There is no reason why the heights of the towers should appear different to Anne. The lengths of the towers is transverse to her horizontal motion so there is no length contraction in that direction. There is a small amount of length contraction of the towers due the upward velocity of the towers relative to Anne, but this will be equal for both towers. This small vertical length contraction of the towers does not stop balls A and B arriving simultaneously at the base of tower B or ball C arrving late at the base of tower A.

 

[JesseM]

No. because what is simultaneous in B's rest frame (y coordinates) is not simultaneous in A's rest frame.

I wasn't talking about simultaneity though--A, B and C occupy the same y-coordinate at all times in the frame of B and C, and this is also true in A's frame. Imagine two measuring-rods parallel to the x-axis, moving alongside each other at the same y-coordinate--wouldn't you agree that they are alongside each other and at the same y-coordinate in the rest frame of each rod? If so, just imagine that B and C are mounted on one rod, while A is mounted on the other.

In Anne's frame, tower B accelerates upwards for an initial period before tower A starts to accelerate upwards.

But if we want to deal with inertial frames where the laws of SR hold, we should not imagine either ball being suddenly "dropped", but rather moving inertially at all times, even before they pass the tops of the towers (you could also just imagine taking the perspective of an inertial observer who has been moving inertially since the beginning of time, and then suddenly sees the ball come to rest in her frame at the moment it is let go). So in the inertial frame of B and C, they have been moving inertially for an infinite time prior to passing the tops of the towers, and the towers have been accelerating up to meet them that whole time. In the inertial frame of A, this is also true, although in this frame B and C pass the tops of the towers at different times due to the "slope", and the towers are accelerating upwards and moving sideways at constant velocity. The only question is whether both towers are accelerating up at the same rate in A's frame, or at different (perhaps changing) rates.

In Anne's frame the ground must appear to bend, to accommodate tower B accelerating upward, while tower A (The top and base and the ground near tower A) remains stationary, due to the (lack of) simultaneity in Anne's frame. Anne sees the ground between tower A and tower B getting progressively more bent/curved.

If we are dealing with inertial frames, then both tower A and tower B have been accelerating up for all time before balls A and C pass the top of tower A and ball B passes the top of tower B (at different times in Anne's frame).

There is no reason why the heights of the towers should appear different to Anne.

Sure there is. In Anne's frame A and B have the same vertical height at all times, but B passes the top of tower B before A passes the top of tower A, so the only way this can be true is if the top of tower A is lower than the top of tower B at the moment B passes the top of tower B. I'm sure if you actually keep track of the coordinates of the tops of the towers in the B/C frame and then do the Lorentz transformation on these coordinates to find how they behave in the A frame, the Lorentz transformation would confirm this. Again, the only thing I'm not sure about is whether both towers are accelerating up at the same constant rate in the A frame, as they are in the B/C frame.

The lengths of the towers is transverse to her horizontal motion so there is no length contraction in that direction.

I didn't say there was. I'm talking about the ground which appears level in the B/C frame appearing sloped in the A frame, so both the top and the bottom of tower A are at a lower height on the y-axis than the top and the bottom of tower B in the A frame. As an analogy, if in one frame we have two ships moving upward at constant speed along the y-axis and zero velocity along the x-axis in this frame, with the ships moving side by side so they're both at the same height on the y-axis at any given moment in this frame, then if we transfer into another frame moving along the x-axis of the first, in this frame the two ships will not be at the same height on the y-axis at any given moment, because of the relativity of simultaneity.

There is a small amount of length contraction of the towers due the upward velocity of the towers relative to Anne, but this will be equal for both towers. This small vertical length contraction of the towers does not stop balls A and B arriving simultaneously at the base of tower B or ball C arrving late at the base of tower A.

Again, I'm not talking about length contraction, I'm talking about the base of tower A being lower than the base of tower B in Anne's frame. This allows A, B, and C to maintain the same height at all times in Anne's frame, and yet C to take longer to reach the base of tower A than B and C take to reach the base of tower B.

 

[yuiop]

I wasn't talking about simultaneity though--A, B and C occupy the same y-coordinate at all times in the frame of B and C, and this is also true in A's frame. Imagine two measuring-rods parallel to the x-axis, moving alongside each other at the same y-coordinate--wouldn't you agree that they are alongside each other and at the same y-coordinate in the rest frame of each rod? If so, just imagine that B and C are mounted on one rod, while A is mounted on the other.

If the two measuring rods are accelerating upwards tranverse to Anne's motion along the x-axis axis and if the measuring rods are spatially separated along the x-axis then they wil not have the same y coordinates at all times. We need 3 measuring rods. One for each ball. There is no reason why ball B should "share" a measurement rod with ball C anymore than ball B should share a measurement rod with ball A.

But if we want to deal with inertial frames where the laws of SR hold, we should not imagine either ball being suddenly "dropped", but rather moving inertially at all times, even before they pass the tops of the towers (you could also just imagine taking the perspective of an inertial observer who has been moving inertially since the beginning of time, and then suddenly sees the ball come to rest in her frame at the moment it is let go). So in the inertial frame of B and C, they have been moving inertially for an infinite time prior to passing the tops of the towers, and the towers have been accelerating up to meet them that whole time.

That is not true. To an observer on the ground stationary with respect to the towers, the ball are fixed to the tops of the towers before they are dropped. To the moving observer that is also true. She can not see the towers moving up to meet the balls before they are dropped.

 

[Jorrie]

But will it actually seem to accelerate towards Earth 3 times faster than a particle dropping straight down, as seen by someone on the ground? In other words, can you see particles dropping at different rates in a single frame?

Yep, I believe it is so. Such a particle will be in a hyperbolic orbit relative to Earth and we know that the angular bending is twice what a Newtonian corpuscle calculation would have yielded (from a single distant (Schwarzschild) frame's point of view). Bring in the full relativistic effects and distill out the effective instantaneous acceleration, and the acceleration is ~3 times Newton's for v_r = 0 and v_phi => c.

I've reworked Pervect's equation into a pseudo-Newtonian acceleration and got the following:

$$a = -\frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}-\frac{3v_r^2}{c^2(1-2GM/(rc^2))}+\frac{2 r^2 v_{phi}^2}{c^2}\right)$$

I did crosscheck the result numerically with his.

(Edit: I originally had a typo in the last term in the bracket and a c^2 missing elsewhere)

 

[JesseM]

If the two measuring rods are accelerating upwards tranverse to Anne's motion along the x-axis axis and if the measuring rods are spatially separated along the x-axis then they wil not have the same y coordinates at all times. We need 3 measuring rods. One for each ball. There is no reason why ball B should "share" a measurement rod with ball C anymore than ball B should share a measurement rod with ball A.

But we're not talking about accelerating measuring rods--the balls are moving inertially in the freefall frame where the laws of SR work, remember? It is the ground that's accelerating, not the balls.

But if we want to deal with inertial frames where the laws of SR hold, we should not imagine either ball being suddenly "dropped", but rather moving inertially at all times, even before they pass the tops of the towers (you could also just imagine taking the perspective of an inertial observer who has been moving inertially since the beginning of time, and then suddenly sees the ball come to rest in her frame at the moment it is let go).

That is not true. To an observer on the ground stationary with respect to the towers, the ball are fixed to the tops of the towers before they are dropped. To the moving observer that is also true. She can not see the towers moving up to meet the balls before they are dropped.

Again, an observer "on the ground stationary with respect to the towers" is not moving inertially, she is accelerating. If you want to deal with the problem using the laws of SR, you must use a freefall frame where the ground is accelerating while the balls are moving inertially after being dropped. If you don't like the idea of the balls moving inertially for all time and just passing the tops of the towers (though I hope you see that this cannot make any difference in terms of what happens after they have passed the tops of the towers), then see the part I put in bold above. Again, if you want to use the laws of SR, you have to pick the frame of an idealized inertial observer who has been "freefalling" (from the perspective of someone on the ground) for all time.

 

[JesseM]

Yep, I believe it is so. Such a particle will be in a hyperbolic orbit relative to Earth and we know that the angular bending is twice what a Newtonian corpuscle calculation would have yielded (from a single distant frame's point of view).

But if you're talking about orbits then you're talking about a region large enough that you can't use the laws of SR since tidal forces become significant--that's not what we're talking about in this problem. We want to deal with a small enough region relative to the size of the planet that the Earth appears "flat" in this region (that was the point of the 'long planet' of the title), and the laws of SR will work fine for a freefalling observer in this region.

 

[Jorrie]

But if you're talking about orbits then you're talking about a region large enough that you can't use the laws of SR since tidal forces become significant--that's not what we're talking about in this problem.

When dealing with a point particle, even a moving one, why should tidal forces be an issue? Even if the gravitational field appears uniform, the instantaneous gravitational acceleration of the particle is influenced by d_phi/dt, according to pervect's analysis. The transverse velocity term is not influenced by g_00, it seems.

We want to deal with a small enough region relative to the size of the planet that the Earth appears "flat" in this region (that was the point of the 'long planet' of the title), and the laws of SR will work fine for a freefalling observer in this region.

Can you use SR when you have a static observer looking at a particle in free fall in a gravity field? By the equivalence principle alone, you may get away with it, but if there is any spatial curvature...

-J

 

[yuiop]

Beautiful drawing

Attached is a beautifully crafting illustration of the equivalence principle (if a do say so, myself :p )

The blue balls are falling vertically while the green blob represents a photon moving left to right and and the black blob is a particle moving at 0.5 c left to right. The drawing clearly shows that particles all fall at exactly the same rate regauardless of horizontal velocity or mass. This is always true for an observer at rest with gravitational field. While the falling rate of the falling particle is independent of the horizontal velocity of the particle relative to the massive body, it is not independent of the velocity of the observer relative to the massive body, and to such an observer the falling rate is dependent on the horizontal velocity of the particles as discussed earlier in the thought experiment involving Anne and the towers.

This may be the velocity dependence that comes out of Pervect's equation. However, if Pervect's equation predicts a particles falling rate is affected by its horizontal velocity when the observer is at rest with the massive body, then it disagrees with the equivalence principle.

 

[Jorrie]

The drawing clearly shows that particles all fall at exactly the same rate regauardless of horizontal velocity or mass. This is always true for an observer at rest with gravitational field.

Despite your very neat drawings, I have difficulty understanding what you mean by an observer at rest in the gravitational field. The views that you show correspond to an observer that falls with the particles, with the lab accelerating past, or do I understand it wrongly? Such an observer will surely observe the particles to hit the floor at the same time.

To an observer riding on the 'floor' of the accelerating lab, the particles with horizontal movement will display conic section paths. The biggest problem is that the clock readings and definitions of simultaneity of the two observers will differ and I do not think that the floor observer will agree that the particles all hit the floor at the same time.

... if Pervect's equation predicts a particles falling rate is affected by its horizontal velocity when the observer is at rest with the massive body, then it disagrees with the equivalence principle.

I'm not sure if the simultaneity issue that I mentioned has anything to do with pervect's solution, but what I'm sure of is that his equations work in Schwarzschild coordinates, whether the particles are in orbit or not. If your accelerating lab represents a small chunk of Schwarzschild coordinates, then pervect's equation apply. If it doesn't, then your scene is not be quite GR compliant, I think!

Maybe the issue is that 'a uniform gravitational field' and 'a long, flat Earth', are not quite Schwarzschild coordinate compliant?

-J

 

[yuiop]

Despite your very neat drawings, I have difficulty understanding what you mean by an observer at rest in the gravitational field. The views that you show correspond to an observer that falls with the particles, with the lab accelerating past, or do I understand it wrongly? Such an observer will surely observe the particles to hit the floor at the same time.

Yes, free falling vertically (but with no horizontal motion) for the lower diagram titled "accelerating rocket".

To an observer riding on the 'floor' of the accelerating lab, the particles with horizontal movement will display conic section paths. The biggest problem is that the clock readings and definitions of simultaneity of the two observers will differ and I do not think that the floor observer will agree that the particles all hit the floor at the same time.

Agree with conic paths. Agree the two observers will measure different intervals for the particles to fall. Do not agree that the observer in the lab will measure any difference in the rate particles fall on the left hand side compared to the particles on the right hand side of the rocket or for a particle moving from left to right.

I'm not sure if the simultaneity issue that I mentioned has anything to do with pervect's solution, but what I'm sure of is that his equations work in Schwarzschild coordinates, whether the particles are in orbit or not. If your accelerating lab represents a small chunk of Schwarzschild coordinates, then pervect's equation apply. If it doesn't, then your scene is not be quite GR compliant, I think!

Maybe the issue is that 'a uniform gravitational field' and 'a long, flat Earth', are not quite Schwarzschild coordinate compliant?

-J

Well, I guess the model being discussed here is for an infinitessimal section of space that approximates a flat gravitational field. Sure there are differences when considering a spherical massive body on a larger scale.


However, the question was "Is the vertical falling rate of a particle independent of its horizontal velocity in a flat gravitational field?" and someone suggested that Pervect's equation says the answer is no. I suggest that the answer is yes and that Pervect's equation does not apply to a flat field, which is what you are hinting at ( I think)

By flat , I mean horizontally uniform. There is of course still a vertical gravitational gradient in this context.


To the question "Will an observer moving horizontally relative to a massive body see particles with different horizontal velocities fall at different rates?" I think the answer is yes. The diagram attached to this post illustrates the point as discussed in my previous posts.


One amusing thought is how we would measure the rate that a photon falls at. If we built a LIGO type device (Say 1000 times longer) how would we make sure the device was flat? We couldn't use the usual method of using a laser to determine a straight line, because that would defeat the purpose. Spirit level maybe?

 

[yuiop]

From a GR perspective, Galileo was wrong of course, especially if he could launch the ball horizontally at 0.8c, as you suggested in your OP. Galilean and relativistic analyzes would be giving different results.

Pervect has shown https://www.physicsforums.com/showpost.php?p=1046874&postcount=17" some time ago that both radial and transverse velocities influence accelerations, which influence the time it will take for the ball to drop a given height.

$$\frac{d^2 r}{d t^2} = \frac {3 m{{\it v_r}}^{2}}{ \left( r-2\,m \right) r} + \left( r-2\,m \right) \left( {{\it v_{phi}}}^{2}-{\frac {m}{{r}^{3}}} \right)$$


v_r = dr/dt and v_phi = dphi/dt

I've roughly calculated the radial acceleration in relation to the usual 1g for a particle traveling precisely horizontally near Earth's surface at 0.8c and got ~ 2.2g. A particle approaching c horizontally will experience a radial acceleration of ~3g.

Assuming Pervect's equation is derived from the Scharzchild metric in the curved spacetime of a spherical massive body, then my guess is that the radial accelerations you mention are upwards rather than downwards. Even in Newtonian physics a satellite in orbit has zero downwards acceleration due to its horizontal velocity because of "centrifugal force". In that sense it is obvious that a particle with extreme horizontal velocities will fall slower in spherical spacetime. A particle moving horizontally faster than its orbital speed will presumably experience increasing dr/dt rather than falling.

The exception is in the extreme case near a black hole where increasing angular velocity makes a particle fall even faster when on the wrong side of the effective potential peak. I read that somewhere. Not sure if it is true.

 

[Jorrie]

Assuming Pervect's equation is derived from the Scharzchild metric in the curved spacetime of a spherical massive body, then my guess is that the radial accelerations you mention are upwards rather than downwards.

Even in Newtonian physics a satellite in orbit has zero downwards acceleration due to its horizontal velocity because of "centrifugal force". In that sense it is obvious that a particle with extreme horizontal velocities will fall slower in spherical spacetime.

We are comparing Galilean/Newtonian with relativistic gravitational accelerations. If you work it through, you will find that Newtonian gravity makes the particle with added velocity at periapsis deviate faster (upwards) from the circular orbital path than with relativistic gravity. The Newtonian equivalent of pervect's equation is simply:

$$\frac{d^2 r}{d t^2} = r v_{phi}^{2}-{\frac {m}{{r}^{2}}$$

which is the simple sum of the positive centrifugal acceleration and the negative gravitational acceleration.

[Edit]: If you subtract the Newton radial acceleration from the relativistic one for v_r=0, we get:

$$\Delta \frac{d^2 r}{d t^2} = -2mv_\phi^2 + {\frac {2m^2}{{r}^{3}}$$

This clearly shows a negative (towards the mass) influence of $v_\phi$ on the radial acceleration (compared to Newton's). [/Edit]

When you reduce the (originally circular) orbital velocity, the relativistic orbit deviates faster downwards from the circle than the Newtonian orbit. From both cases I conclude that the relativistic particle with initially only horizontal velocity will fall faster in a normal gravitational field than what Newton would have predicted. This is part of the reason for the precession of relativistic orbits.

The exception is in the extreme case near a black hole where increasing angular velocity makes a particle fall even faster when on the wrong side of the effective potential peak. I read that somewhere. Not sure if it is true.

That's right. As I have it, with Schwarzschild radial parameter $r < 6GM/c^2$, circular orbits suffering a slight loss in angular velocity will spiral into the hole. From $r < 4GM/c^2$, adding angular velocity will make them spiral in too.

-J

 

[JesseM]

When dealing with a point particle, even a moving one, why should tidal forces be an issue?

We're talking about comparing the rate at which two or more spatially separated point particles fall, so tidal forces are definitely an issue.

Even if the gravitational field appears uniform, the instantaneous gravitational acceleration of the particle is influenced by d_phi/dt, according to pervect's analysis. The transverse velocity term is not influenced by g_00, it seems.

I'd like to hear pervect's opinion on that--it seems to me that if we restrict our experiments to a local region, than GR is supposed to reduce to SR in that region, so no effects related to the curvature of spacetime should be relevant.

Can you use SR when you have a static observer looking at a particle in free fall in a gravity field? By the equivalence principle alone, you may get away with it, but if there is any spatial curvature...

The static observer would be equivalent to an accelerating observer in SR. And again, GR is always supposed to reduce to SR if you zoom in on an arbitrarily small region of space and look at it for an arbitrary small period of time; in this region it should always be possible to find a locally inertial coordinate system where the surfaces of simultaneity have no spatial curvature, regardless of what the larger spacetime surrounding this region looks like.

 

[Jorrie]

And again, GR is always supposed to reduce to SR if you zoom in on an arbitrarily small region of space and look at it for an arbitrary small period of time; in this region it should always be possible to find a locally inertial coordinate system where the surfaces of simultaneity have no spatial curvature, regardless of what the larger spacetime surrounding this region looks like.

I can't argue against "zoom(ing) in on an arbitrarily small region of space and look(ing) at it for an arbitrary small period of time".

It would be interesting to know what would happen to Galileo's cannon balls if they were dropped near the event horizon of an arbitrarily large black hole, so that the region that we measure becomes arbitrarily small in comparison. I have a hunch that the horizontally moving ball will drop faster, as observed by a static observer ('hovering' his spaceship there), because I think pervect's equations and Schwarzschild geometry says that.

I suppose one must specify the test somewhat better and then compute the accelerations, but that's a bit much for a Friday evening in my valley...

-J

 

[Jorrie]

It would be interesting to know what would happen to Galileo's cannon balls if they were dropped near the event horizon of an arbitrarily large black hole, so that the region that we measure becomes arbitrarily small in comparison. I have a hunch that the horizontally moving ball will drop faster, as observed by a static observer ('hovering' his spaceship there), because I think pervect's equations and Schwarzschild geometry says that.

Let's use Kip Thorne's "Gargantua" from his "Black holes and time warps", sporting 15 trillion solar masses and set up the lab at 1.0001 Rs (like Kip), where the g-force is still 10g, but the tidal forces are negligible. The horizon circumference is 29 light years, so our test arena size should also be negligible by comparison. Since Gargantua is spinning very slowly, the spacetime around it is almost perfectly Schwarzschild and pervect's equations should be valid in Schwarzschild coordinates, i.e. by an observer at 'infinity'.

With the event horizon practically flat underneath us and the gravitational field practically homogeneous, set up Cartesian coordinates with the x-axis normal to the radial and the y-axis parallel to it. We can now determine the instantaneous Cartesian acceleration when a ball is dropped without and with horizontal velocity (must be far below the orbital velocity, which is v_o= 0.707c in this case).

In the Newton case, the acceleration will simply be

Eq. 1
$$\frac{d^2 r}{d t^2} = -\frac{m}{r^2}$$

irrespective of horizontal (or vertical) velocity.

In the relativistic case, with dr/dt = 0, pervect's radial acceleration reduces to:

Eq. 2
$$\frac{d^2 r}{dt^2} = \left( r-2\,m \right) \left(v_{\phi}^{2}-{\frac {m}{{r}^{3}}} \right)$$

To convert this to a Cartesian coordinate acceleration, we firstly have to add the negative centripetal acceleration $rv_\phi^2$ (because it is inherent in spherical coordinates):

Eq. 3
$$\frac{d^2 r}{dt^2} = \left( r-2\,m \right) \left(v_{\phi}^{2}-{\frac {m}{{r}^{3}}} \right) - r^2v_\phi^2 = -\frac {m}{r^2}\,-\, 2 mv_\phi^2\,+\, \frac {2m^2}{r^3}$$

with a final, most useful form:

Eq. 4
$$\frac{d^2 r}{dt^2} = -\frac {m}{r^2}\left(1 - \frac {2m}{r} + 2r^2v_\phi^2\right)$$

Strictly, one should also replace r with y and $rv_{\phi}$ with $v_x=dx/dt$, the horizontal velocity, but that this is not necessary to see the effect that we are after.

It is clear that the negative gravitational acceleration has a larger magnitude in the presence of horizontal velocity in Schwarzschild coordinates. The final question: is this also true in the local coordinates of the static experimenter? My take (without proof*) is that Schwarzschild coordinate radial acceleration converts to the static local observer's coordinates by a factor $(1-2m/r)^{-1.5}$, i.e.:

Eq. 5
$$\frac{d^2 r}{d\tau^2} = -\frac {m}{r^2(1-2m/r)^{1.5}}\left(1 - \frac {2m}{r} + 2r^2v_\phi^2\right)$$

making no difference to the conclusions that the ball with horizontal velocity has a higher instantaneous vertical acceleration than a ball without.

Note* I think this acceleration conversion factor is common in the literature, but I could not quickly find a reference. Anyone spotting a blunder?

-J

 

[JesseM]

I can't argue against "zoom(ing) in on an arbitrarily small region of space and look(ing) at it for an arbitrary small period of time".

It would be interesting to know what would happen to Galileo's cannon balls if they were dropped near the event horizon of an arbitrarily large black hole, so that the region that we measure becomes arbitrarily small in comparison. I have a hunch that the horizontally moving ball will drop faster, as observed by a static observer ('hovering' his spaceship there), because I think pervect's equations and Schwarzschild geometry says that.

But that would seem to be a clear violation of the equivalence principle--don't you agree the EP says that if you pick a sufficiently small region of a larger curved spacetime so that the curvature is negligible in that region, than the laws of physics for a freefalling observer in this region should reduce to those of an inertial observer in SR? And don't you agree this would apply to picking a small region that includes the event horizon?

 

[Jorrie]

But that would seem to be a clear violation of the equivalence principle--don't you agree the EP says that if you pick a sufficiently small region of a larger curved spacetime so that the curvature is negligible in that region, than the laws of physics for a freefalling observer in this region should reduce to those of an inertial observer in SR? And don't you agree this would apply to picking a small region that includes the event horizon?

I may be missing something, but the issue is not free-falling observers here; it is free-falling objects being measured by observers static in a gravitational field. I've posted my rationale in my previous post, at about the same time that you posted. If I'm wrong there (which is very possible), I would like to learn where and why.

-J

 

[JesseM]

I may be missing something, but the issue is not free-falling observers here; it is free-falling objects being measured by observers static in a gravitational field.

Yes, but according to the equivalence principle, if we're talking about free-falling objects as seen by "observers static in a gravitational field" in a small region of spacetime where curvature becomes negligible, this should be equivalent to inertial objects as seen by observers accelerating upward at a constant rate in SR.