Long planet and Galileo thought experiment

[Jorrie]

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.

But, does "curvature becomes negligible" mean there is no curvature? I guess one must go infinitesimally small to have no curvature near a black hole. Whether that makes Galileo's cannon balls fall differently in a finite lab when they have horizontal velocity is not clear, i.e., can one use Schwarzschild coordinates inside such a very small finite lab?

On a more practical level, one may ask: does particles moving at ~c in linear accelerators here on Earth suffer 3g downward acceleration? I'm not sure if is even measurable, but to me it seems that pervect's equations under discussion suggest that.

-J

 

[JesseM]

But, does "curvature becomes negligible" mean there is no curvature? I guess one must go infinitesimally small to have no curvature near a black hole.

Well, one could say that to the extent that the predictions of SR in the region would be slightly in error, one could make the amount of error be as small as desired by making the region of spacetime sufficiently small. One could also achieve the same effect just by keeping the size of the region constant while making the black hole bigger and bigger. The curvature doesn't go to infinity near the event horizon, so for any finite region of spacetime that contained the horizon, by making the black hole large enough one could make the error small--one part in a trillion, say.

Whether that makes Galileo's cannon balls fall differently in a finite lab when they have horizontal velocity is not clear, i.e., can one use Schwarzschild coordinates inside such a very small finite lab?

I'm sure you could, although Schwarzschild coordinates in the small region of spacetime might look very different from more "natural" coordinates usually chosen for accelerating observers in SR, like Rindler coordinates. But it should at least be true that constant height in the lab (where height is along the direction of the pull of gravity) would correspond to constant Schwarzschild radius.

On a more practical level, one may ask: does particles moving at ~c in linear accelerators here on Earth suffer 3g downward acceleration? I'm not sure if is even measurable, but to me it seems that pervect's equations under discussion suggest that.

By the equivalence principle I think a photon that starts out at the same height as a falling object (in the coordinate system of the observer at a fixed gravitational radius), and is initially moving horizontally in the object's own freefalling frame, would stay at the same height as the falling objects at later times, at least if the observer at fixed radius is using a definition of simultaneity which matches that of the freefalling observer in the horizontal direction.

 

[yuiop]

Relativistic submarine paradox

Note: I have edited this post to correct my mistaken assumption that that if a particle falls a given distance in gamma less time in a moving frame then it requires gamma squared less gravitational force for that to happen. Upon reflection, in relativity force is $$F' = (dp/dt)' = (dmdx/dt^2)' = (dm y)(dx)/(dt^2 y^2) = F y^{-1}$$. In other words gamma less force results in gamma squared less acceleration which gives the required gamma less falling interval.



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Hi,

To try and shed more light on the subject I have referred to a paper by Matsas http://arxiv.org/PS_cache/gr-qc/pdf/0305/0305106v1.pdf where he "solves the submarine paradox. Another simplified version of the paper is here. http://arxiv.org/PS_cache/arxiv/pdf/0708/0708.2488v1.pdf

The analysis done by Matsas is relevant here because in his own words:

" ... we will look for a background with planar symmetry. This is necessary in order to avoid the appearance of centrifugal effects which are not part of the submarine paradox. This is accomplished by the Rindler spacetime."

After a lengthy calculation he concludes that that the forces acting on the submarine as measured by an observer co-moving with the submarine experiences a total force of:

$$F' = -mg \gamma (\gamma - 1/ \gamma)$$

where $$-mg \y^2$$ is the force accelerating the sub downwards and mg is the buoyancy force acting upwards on the sub. mg is also the magnitude of the forces that are in equilibrium when the submarine is stationary with respect to the seabed.

He then does a Lorentz transformation and concludes that the total forces acting on the moving submarine as measured by an observer stationary with water and the sea bed as:

$$F= -mg (\gamma - 1/ \gamma)$$

We see in this last equation that both the gravitational force acting on the moving submarine and the bouyancy force is greater by gamma compared to when it is not moving.

Now we adapt the experiment by removing the water and buoyancy forces.

Lets say it takes one second for the non moving sub to fall vertically to the seabed when released, as timed by an observer on the seabed. Now if the sub is moving horizontally at 0.8c the sub takes 1*(0.6) = 0.6 seconds to fall to the seabed as timed by that same observer.

In Earth strength gravity the correction required to allow for the difference in clock rates due to height is of the order of parts per billion. This is insignificant compared to the reduction of 40% in falling time interval due to the horizontal velocity. Clearly we should not get too hung up on whether the observer is free falling or not.

Now the time it takes to fall from the point of view of an observer co-moving with the submarine is the time measured by the seabed observer reduced by gamma. This requires the downward acceleration in the moving submarine frame to be greater by gamma than in the seabed frame.

Analysis of the buoyancy force
===============================================================

When an object with no horizontal motion falls the time period for the object to fall is increased by gamma as viewed by an observer moving horizontally relative to the object. This suggests the force of gravity is reduced by gamma squared in this situation.

We can apply this observation to the buoyancy calculation. One description of buoyancy is that the upward force is proportional to the weight of water displaced. When the sub is moving wrt the water it is length contracted so it displaces gamma less water so there is gamma less upward force from the pov of the observer on the seabed. (mg/y)

To an observer on the sub there is gamma more mass of water displaced due to length contraction of the water relative to the length of the submarine. To this observer the water is co-moving with sea bed so it experiences a gravitational force that is reduced by gamma. Theses two effects cancel out and the weight of the displaced water is simply mg. (in agreement with Matsas)

Matsas used a different method. He reasoned that when the sub is moving relative to the seabed observer, it is length contracted relative to the water so there are gamma less columns of water supporting the sub. To an observer on the sub there gamma more columns of water underneath the sub than when the sub was at rest with water. Each of these columns exert an upward force that is reduced by gamma due to the relative motion of the water so the net effect is that the upward buoyancy force is the same as when the sub was stationary wrt the water.

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So along the way we have concluded:

When the massive body is moving relative to the observer (and the test particle is not) the effective force of gravity is F' = F y^2

When the particle is moving wrt the observer (and the massive body is not) then the effective force of gravity on the particle is F' = F y

When the test particle and the massive body are both moving with the same velocity relative to the observer then the effective force of gravity on the particle is F' = F / y

The last observation is a requirement of Special Relativity.

An equation that wraps all these apparently contradictory observations into one unified result is:

$$F' = \frac{GMm}{R^2} \frac{(1-Vv/c^2)^2}{(1-V^2/c^2)\sqrt{1-v^2/c^2}}$$

where V is the velocity of the massive body (M) wrt to the observer and v is the velocity of the test particle (m) with respect to the observer.

It is assumed here that M is much greater than m so that we do have to be concerned with acceleration of M towards m.

The principle involved to formulate that equation is that the force of gravity between M and m increases by gamma squared as a function of the relative velocity (V-v) of M and m to each other (using the relativistic subtraction formula) and at the same time, the force of gravity acting on m is reduced by a factor of gamma as a function of the velocity (v) of the test particle (m) relative to the observer.

The equation is a bit easier to visualise when expressed as


$$F' = \frac{GMm}{R^2} \frac{\sqrt{1-v^2}}{(1-(V-v)^2)}$$

where it is understood that (V-v) is the relativistic subtraction of v from V.


=============================================================

The problem I still have is that this experiment that is designed to be as close as possible to a gravitational field simulating an upwardly accelerating rocket shows there are significant differences, and the Equivalence Principle breaks down under even these supposedly ideal conditions.

 

[yuiop]

...

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

Hi Jorrie,

In my last post (now edited to correct a major error ) Matsas concludes in his paper that a horizontally moving object will experience $\sqrt{1-v^2/c^2}$ greater gravitational "force" than an object falling purely vertically. This presumably translates into a greater downward acceleration of (gamma squared) for the horizontally moving object.


If we assume the observer is basically at the same altitude as the falling particle and illiminate the gravitational gamma factor (weak field aproximation?) then I think your equation simplifies to

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


Is that broadly in agreement with the conclusion reached by Matsas when we allow for the little details?

Edit: I'm guessing the last formula is the binomial expansion aproximation of the Matsas result for v << c.

 

[yuiop]

If we assume the observer is basically at the same altitude as the falling particle and illiminate the gravitational gamma factor (weak field aproximation?) then I think your equation simplifies to

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


Is that broadly in agreement with the conclusion reached by Matsas when we allow for the little details?

Edit: I'm guessing the last formula is the binomial expansion aproximation of the Matsas result for v << c.

OK, I've checked it out. Substitute linear velocity for angular velocity $v/r = v_\phi$ and the equation becomes the binomial aproximation of:


$$\frac{d^2 r}{d tau^2} = -\frac {GM\gamma^4}{R^2}$$

and allow for the time dilation due to the horizontal motion wrt the ground observer

$$\frac{d^2 r}{dt^2} = -\frac {GM\gamma^2}{R^2}$$

Matsas calculates the force of gravity acting on an object with horizontal motion to be greater by gamma. The question is, does that agree with an acceleration greater by gamma squared? A lot depends on what we assume the mass to be.

It still bothers me that the Equivalence Principle is defeated so easily even in flat spacetime :(

Taking the results from my earlier post on Matsas:

1) When the massive body is moving relative to the observer (and the test particle is not) the effective force of gravity is F' = F y^2

2) When the particle is moving wrt the observer (and the massive body is not) then the effective force of gravity on the particle is F' = F y

3) When the test particle and the massive body are both moving with the same velocity relative to the observer then the effective force of gravity on the particle is F' = F / y

then the acceleration in the same 3 cases is

1) a' = F y^2 / m = a y^2

2) a' = F y / (m y) = a

3) a' = (F/y)/(m y) = a /y^2

in which case the equivalence principle can be saved

...but I am not sure gravitational "force" and relatavistic mass can be handled in that way

Is there anyone that can do the relevant calculations using four-force and four-acceleration and invariant mass to get a more formal conclusion?

 

[Jorrie]

In my last post (now edited to correct a major error ) Matsas concludes in his paper that a horizontally moving object will experience $\sqrt{1-v^2/c^2}$ greater gravitational "force" than an object falling purely vertically.

I did not spot that in the Matsas paper. I also presume you meant a $1/\sqrt{1-v^2/c^2}$ factor? I also did not notice where they talk about radially free falling vs. free falling with horizontal movement, just static with buoyancy and moving horizontally with buoyancy. Can you perhaps guide me to the place in the text?

If we assume the observer is basically at the same altitude as the falling particle and illiminate the gravitational gamma factor (weak field aproximation?) then I think your equation simplifies to

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

Is that broadly in agreement with the conclusion reached by Matsas when we allow for the little details?

No, I think they are two different things, the one free falling and the other with buoyancy. But yes, the equation that you gave is my interpretation for the weak field, high speed limit, like high speed particles on Earth's surface.

I did not fully study all your of latest posts, but interestingly, I do get around 0.67 seconds, using my interpretation, for the example that you worked out for a particle with horizontal speed 0.8c. Remember that a particle at 0.8c travels ~ halfway to the moon in 0.67 seconds, so it is no longer an "infinitesimally small" lab test. So maybe the equivalence principle is not supposed to hold?

-J

 

[my_wan]

You said the balls were released simultaneously. Simultaneous relative to who? That will be the frame that sees the balls land at the same time so long as that observer didn't move significantly during the time the balls fell. Of course another frame of reference will not agree that they landed at the same time. That frame of reference didn't see the balls dropped at the same time either.

Yes Galilean Relativity is still valid in the limit of weak gravity and normal relative speed.

 

[yuiop]

I did not spot that in the Matsas paper. I also presume you meant a $1/\sqrt{1-v^2/c^2}$ factor? I also did not notice where they talk about radially free falling vs. free falling with horizontal movement, just static with buoyancy and moving horizontally with buoyancy. Can you perhaps guide me to the place in the text?

Fair point. The Matsas paper is the closest thing I can find to a treatment of gravity in flat spacetime. (I am using mejennifer's definition of a cylinder as a flat in GR), but it only gives static gravitational force balanced by static buoyancy force. I am not sure how we calculate acceleration from a measured gravitational force in SR. The last part of post #41 is my attempt to convert static gravitational force to acceleration. For example if we had a one kilogram mass as measured by scales on Earth and that same mass weighed 2 kilograms using the same scales on another massive body, then what would we expect the acceleration of the mass to be if we dropped it on the other massive body?


(Yes, I meant $1/\sqrt{1-v^2/c^2}$ )

No, I think they are two different things, the one free falling and the other with buoyancy. But yes, the equation that you gave is my interpretation for the weak field, high speed limit, like high speed particles on Earth's surface.

I did not fully study all your of latest posts, but interestingly, I do get around 0.67 seconds, using my interpretation, for the example that you worked out for a particle with horizontal speed 0.8c. Remember that a particle at 0.8c travels ~ halfway to the moon in 0.67 seconds, so it is no longer an "infinitesimally small" lab test. So maybe the equivalence principle is not supposed to hold?

-J

I only used one second as round number. It could be one microsecond if we wished. In flat space I do not think there is any real requirement to have an infinitessimal small region (horizontally anyway). In one second an object falls about 5 meters in Earth gravity. The height dr should be minimal so that the radius R from the centre of the gravitational body is considered essentailly constant from the top to the bottom of the experiment.


I am also curious if a horizontal photon will take less time to fall 5 meters than the particle moving at 0.8c ?

 

[yuiop]

You said the balls were released simultaneously. Simultaneous relative to who?

Simultaneous as measured by the observer that has no horizontal motion relative to the larger gravitational body.

That will be the frame that sees the balls land at the same time so long as that observer didn't move significantly during the time the balls fell. Of course another frame of reference will not agree that they landed at the same time. That frame of reference didn't see the balls dropped at the same time either.

Pretty much agree with the above. The question is, will a particle moving horizontally fall at the same rate as a particle with no horizontal motion in ideal flat space? Consideration of a similar experiment in an acclerating rocket creating an artificial "gravity" suggests it should, in line with the Equivalence Principle.

Yes Galilean Relativity is still valid in the limit of weak gravity and normal relative speed.

We are wondering what happens with horizontal motion that is a significant fraction of the speed of light eg 0.8c

 

[my_wan]

Simultaneous as measured by the observer that has no horizontal motion relative to the larger gravitational body.

Yes and that will be the frame in which the balls land at the same time.

Pretty much agree with the above. The question is, will a particle moving horizontally fall at the same rate as a particle with no horizontal motion in ideal flat space? Consideration of a similar experiment in an acclerating rocket creating an artificial "gravity" suggests it should, in line with the Equivalence Principle.

Yes. Even though from the perspective of the above observer time slows down for the balls. An observer riding the balls would say their clock didn't slow down but the distance to the target ground was shorter so it still matches.

One of the mistakes to avoid is if you define a frame of reference then you can't think of an observer in another frame of reference as seeing what happened in the frame that's not their own.

We are wondering what happens with horizontal motion that is a significant fraction of the speed of light eg 0.8c

Yes it makes it weirder to think about but as long as you don't mix up frames of reference then, in flat space-time, the space in one frame will dilate in proportion to the time in the other frame (and visa versa) such that the end result is the same. Add Gravity and curved spaces and the distortions don't completely balance out anymore.

 

[yuiop]

...

Yes it makes it weirder to think about but as long as you don't mix up frames of reference then, in flat space-time, the space in one frame will dilate in proportion to the time in the other frame (and visa versa) such that the end result is the same. Add Gravity and curved spaces and the distortions don't completely balance out anymore.

So if a ball takes one second to fall (aproximately 5 meters) vertically to the surface of a massive body that is flat and almost infinitely long (as measured by an observer with no horizontal motion relative to the massive body) how long would it take for a ball with a horizontal velocity of 0.8 c to fall as measured by that same observer?

Assume the radius of the massive body is similar to that of the Earth with gravitational acceleration of aproximately 1g as on Earth. (Also assume the horizontal motion is parallel to the long axis of the long massive body)

 

[Jorrie]

For example if we had a one kilogram mass as measured by scales on Earth and that same mass weighed 2 kilograms using the same scales on another massive body, then what would we expect the acceleration of the mass to be if we dropped it on the other massive body?

For a static local observer, the 2 times weight should translate to twice the acceleration (it's twice the force, after all).

In flat space I do not think there is any real requirement to have an infinitesimal small region (horizontally anyway).

Hmm... This may be problematic in the presence of a non-negligible gravitational field, where anything more than an infinitesimal small region is not flat spacetime anymore.

I am also curious if a horizontal photon will take less time to fall 5 meters than the particle moving at 0.8c ?

Depending on how you define the '5 meters fall', the 0.8c particle will usually take less time, because it will curve more towards the massive body than the photon. With fast horizontal particles, one cannot measure the acceleration by timing the fall over a particular distance. I think you have to measure the curvature of the particle's trajectory and its speed and then determine the acceleration analytically in your chosen frame of reference (preferably Schwarzschild coordinates, which can be easily converted to other coordinate definitions).

-J

 

[yuiop]

For a static local observer, the 2 times weight should translate to twice the acceleration (it's twice the force, after all).

Ok, so in our original experiment, Anne weighs the mass of the ball to be 1 kg and the acceleration to be 1g when she is at rest with the tower. When the ball is moving horizontally relative to the tower there is a static force of 1kg*y before the ball is released as measured by an observer at rest with the tower, according to Matsas. Now to Anne, in her horizontally moving lab the weight is suspended from a spring to make a crude weighing scale. To her the static force is increased by y^2 s so the ball weighs y^2 more as far as she is concerned. The spring is visibly stetched by a factor of y^2. The tower observers can see the spring scale but they know that transverse force in a moving frame is less by a factor of gamma than the proper force measured in the co-moving frame so that agrees with their opinion that the force is Fy. When Anne releases the ball she measures the acceleration to be greater by a factor by a factor of y^2 compared to the original case and this equates to a falling time that is reduced by 1/y by her measurement compared to when she was at rest with the tower and dropped it vertically. So the falling time is 0.6 seconds by her timing. If the falling time by Anne's measurement is 0.6 seconds then the falling time measured by Bob at rest with tower must be one second by Lorentz transformation. Therefore the horizontally moving ball falls at the same rate as the purely vertically falling ball.

Relativity requires that the acceleration measured by Anne is greater than the acceleration measured by Bob to be greater by y^2. If Bob measures the acceleration of the horizontally moving ball to be greater by y (compared to a purely vertically falling ball) then that requires Anne to measure the acceleration to be y^3 greater than purely vertically falling ball. Somehow that does not seem to agree with Anne measuring the static force acting on the ball to be only greater by y^2.

Hmm... This may be problematic in the presence of a non-negligible gravitational field, where anything more than an infinitesimal small region is not flat spacetime anymore.

If we time the interval it takes a ball to fall vertically 5 meters on the Earth using one clock at the top and another at the bottom, the correction to take general relativity into account is of the the order of parts ber billion. Can we call that negligible for now?

 

[yuiop]

New gyroscope experiment.

2 gyroscopes experiment:

One is spinning while the other is not.
Both have there main spin axis aligned vertically.
Both are dropped vertically in a vacuum cylinder.

General relativity requires that the spinning gyroscope (that has allmost all its constituent elements moving horizontally) should fall faster than the non spinning gyroscope.

This experiment was actually carried out in this paper. http://arxiv.org/PS_cache/gr-qc/pdf/0111/0111069v1.pdf

They got a null result. Both gyroscopes fell at the same rate. So either General Relativity is wrong, or the interpretation of general relativity (that an object with horizontal motion falls faster than an object without) is wrong... or the experiment was carried out incorrectly.

 

[my_wan]

So if a ball takes one second to fall (aproximately 5 meters) vertically to the surface of a massive body that is flat and almost infinitely long (as measured by an observer with no horizontal motion relative to the massive body) how long would it take for a ball with a horizontal velocity of 0.8 c to fall as measured by that same observer?

Essentially the same one second. When we talk about time slowing down for the ball it is only the balls as compared to the observer. It doesn't change anything about the time that the observer actually measures themself. In the balls frame the time wasn't slow, the distance between where it was dropped and where it landed was shorter. So even though in the balls frame it landed sooner the equivalence principle required it to because the distance was shorter. In the balls frame the flat ground appears curved upward so the landing spot appears to be a higher elevation than straight down from the tower. The result is that, considering only SR, the equivalence principle works for either observer, even though both observer disagree on distance and time.

Assume the radius of the massive body is similar to that of the Earth with gravitational acceleration of aproximately 1g as on Earth. (Also assume the horizontal motion is parallel to the long axis of the long massive body)

Not real sure but I assume I understood the intentions of those conditions.

 

[my_wan]

For a static local observer, the 2 times weight should translate to twice the acceleration (it's twice the force, after all).

No. If you dropped two balls side by side instead of one does that mean that they will accelerate faster? Why not, they are twice the weight of one ball? This is what the principle of equivalence says: the gravitational acceleration is in no way dependent on the weight.

 

[Jorrie]

No. If you dropped two balls side by side instead of one does that mean that they will accelerate faster? Why not, they are twice the weight of one ball? This is what the principle of equivalence says: the gravitational acceleration is in no way dependent on the weight.

If you check #43 and #47 again, you'll notice that this not the experiment wer'e talking about. It's dropping a 1kg mass on Earth versus dropping it on a planet with twice the surface gravity.

-J

 

[Jorrie]

General relativity requires that the spinning gyroscope (that has allmost all its constituent elements moving horizontally) should fall faster than the non spinning gyroscope.

This experiment was actually carried out in this paper. http://arxiv.org/PS_cache/gr-qc/pdf/0111/0111069v1.pdf

They got a null result. Both gyroscopes fell at the same rate. So either General Relativity is wrong, or the interpretation of general relativity (that an object with horizontal motion falls faster than an object without) is wrong... or the experiment was carried out incorrectly.

Hmm... this is as close to a 'infinitesimal region' of space we can get. If the spacetime curvature is negligible, the differences should also be negligible. I do not think GR predicts that there should then be a difference.

I'm ready to concede to JesseM that the effects of GR in the case of horizontal movement only show up over extended regions of space, where spatial curvature cannot be ignored (especially in the horizontal direction). If you make the size of the gravitating body approach infinity, there will be no effect even over large regions of space. Remember that with high linear horizontal speed, the horizontal movement is many orders larger than the vertical movement - that's where the gyroscope case is different.

Pervect's equations and my interpretation are probably correct for relatively large areas in comparison with the massive body, like orbits or portions of orbits. The equations give the correct relativistic orbits, but should be used with care on small scales.

I think the buoyancy problem is something quite different. The Rickard M. Jonsson paper (http://arxiv.org/PS_cache/arxiv/pdf/0708/0708.2488v1.pdf) that you referenced clearly shows that even the 'submarine effect' reverses when you're are outside the photon radius (3GM/c^2) in a 'real spherical ocean' - meaning there is spatial curvature.

-J

 

[yuiop]

Hi all,

Not got much time. Thanks for the comments.

I was wondering if there is a simple Rindler spacetime solution to all this? I think it is supposed to handle these sort of conditions.

 

[my_wan]

If you check #43 and #47 again, you'll notice that this not the experiment wer'e talking about. It's dropping a 1kg mass on Earth versus dropping it on a planet with twice the surface gravity.

-J

Oh, failed to notice. Your response was quiet reasonable then.

 

[Jorrie]

Gargantua violates the EP?

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). Corrective note: the v_o = 0.707c is an error; there are no orbits at r = 1.0001Rs. One can use any local velocity < c.
...

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

Since nobody raised an issue on this equation (based on pervect's derived radial acceleration), I did put in some numbers and was quite surprised! Firstly, I rewrote it in local Cartesian (x,y) coordinates (as defined in the above quote):

Eq. 6
$$\frac{d^2 y}{d\tau^2} = -\frac {m}{y^2(1-2m/y)^{1.5}}\left(1 - \frac {2m}{y} + 2(1-2m/y)v_x^2\right)$$

The last term is multiplied by (1-2m/y) to convert the local horizontal velocity (v_x) back to the Schwarzschild tangential velocity $rv_\phi$ required by the quoted equation 5.

This acceleration was numerically integrated over one second, starting at y=1.0001Rs, first for a static particle (v_x = 0) being dropped and then for Kev's particle with v_x = 0.8c (horizontal) velocity, over the same drop in the static x,y frame. I found that the first particle dropped a vertically height of 51 meters in the one second, at a practically constant acceleration of -101.717 m/s^2 (~10g, as Kip used). It obviously reached a speed (v_y) of -101.717 m/s in the one second, not relativistic at all.

The particle moving at v_x = 0.8c dropped the 51 meters in 0.66 seconds, also at a practically constant acceleration of -231.916 m/s^2, reaching a vertical speed (v_y) of -153.06 m/s, still not relativistic. The significance of this is that one can ignore the effect of v_y on dtau and on the vertical acceleration. This particle traveled a horizontal distance of some 158,400 km (0.53 ls), insignificant relative to the 29 ly circumference black hole. We are still in an insignificantly small region relative to the hole (practically infinitesimal).

So, if these calculations are correct, Galileo was wrong - a horizontally moving particle does fall faster than the particle dropped from rest. Before everyone jumps in and flame the calculations, consider this: the test is performed at r=1.0001Rs, far inside the "photon radius", where no stable orbits are possible and any increase in tangential velocity (kinetic energy) drags the object faster into the hole. Qualitatively, there is no problem here - horizontal velocity makes things fall faster in that region and it becomes more severe the closer to the event horizon, where even light's trajectory becomes purely vertical.

Quantitatively, doing the experiment at the photon radius r = 3m with v_x = 1 indeed gives a circular orbit with a gravitational acceleration of -2.349 m/s^2. At r=4m, using v_x = 0.707, the orbit is circular with a gravitational acceleration of -0.82 m/s^2, corresponding to the local centripetal acceleration for a circular orbit at that radius ($ca = v_x^2/(r-2)$. I also found that the increase factor in the acceleration did not change if I increased the mass of black hole without limit. Sobering…

This result is probably controversial and may be wrong, but I'll be very interested to get comments on the validity of the equations for the scenario described, rather than just that it violates the equivalence principle.

-J

 

[Jorrie]

Gargantua correction

Quantitatively, doing the experiment at the photon radius r = 3m with v_x = 1 indeed gives a circular orbit with a gravitational acceleration of -2.349 m/s^2. At r=4m, using v_x = 0.707, the orbit is circular with a gravitational acceleration of -0.82 m/s^2, corresponding to the local centripetal acceleration for a circular orbit at that radius ($ca = v_x^2/(r-2)$.

There obviously cannot be circular orbits in the local Cartesian frame, just curved paths that represent a tiny portion of the Schwarzschild orbits at those radii. It's better to say that a local Cartesian centripetal acceleration of $ca = -v_x^2/(r-2m)$ relative to the center of Gargantua would have produced the partial orbits as quoted above.

(The previous centripetal equation has a typo, but I cannot edit the original any longer, so I had to post again - my apologies)

-J

 

[DukeL]

Equivalence principle violated or not?

I'm a newly registered but long time reader of PF and has followed this conversation with interest, but it seems to have ground to a halt now. This made me register to try and get closure on the matter. Recapping some of the discussion between JesseM, Jorrie and others:

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?

In a reply to Kev on the spinning gyroscopes that was dropped, Jorrie then apparently accepted JesseM's view:

I'm ready to concede to JesseM that the effects of GR in the case of horizontal movement only show up over extended regions of space, where spatial curvature cannot be ignored (especially in the horizontal direction). If you make the size of the gravitating body approach infinity, there will be no effect even over large regions of space. Remember that with high linear horizontal speed, the horizontal movement is many orders larger than the vertical movement - that's where the gyroscope case is different.

However, later Jorrie responded with a detailed calculation, based on, but not quite equivalent to the 'pervect' equation that he quoted and apparently changed his mind:

So, if these calculations are correct, Galileo was wrong - a horizontally moving particle does fall faster than the particle dropped from rest. Before everyone jumps in and flame the calculations, consider this: the test is performed at r=1.0001Rs, far inside the "photon radius", where no stable orbits are possible and any increase in tangential velocity (kinetic energy) drags the object faster into the hole. Qualitatively, there is no problem here - horizontal velocity makes things fall faster in that region and it becomes more severe the closer to the event horizon, where even light's trajectory becomes purely vertical.

This left me totally confused. Is the equivalence principle violated in infinitesimal regions of spacetime near the horizon of a very large black hole, or not?

 

[yuiop]

Hi DukeL,

During the course of this thread I have come to the conclusion that the equivalence principle is a "one way relationship". While an observer in a closed artificially accelerated lab could not be certain that he is not experiencing acceleration due to the proximity of a massive gravitational body, the reverse is not always true. An observer in a large lab that is stationary with respect to a large gravitational body can detect differences from the artificial acceleration due to tidal influences. Those differences are minimised by considering an infinesimal region. However, it is questionable whether it is possible to find a sufficiently small region within the massive gravitational curvature found near the event horizon of a black hole if there are other limitations such as the Planck length (if that is a real physical limitation of nature). Such a violation of the EP would probably not be a fatal wound for GR if we consider the "one way" relationship I mentioned before. It is clear that on large scales the EP is violated in GR in the reverse direction. I am as interested as you, to know if the EP is violated in an infinitesimal region near the event horizon ;)

 

[Jorrie]

I'm a newly registered but long time reader of PF and has followed this conversation with interest, but it seems to have ground to a halt now. This made me register to try and get closure on the matter. Recapping some of the discussion between JesseM, Jorrie and others: ...

This left me totally confused. Is the equivalence principle violated in infinitesimal regions of spacetime near the horizon of a very large black hole, or not?

Hi DukeL, welcome as a registered member of PF now!

I think the answer lurks in the difficulty of defining an 'infinitesimal frame' in which to perform the 'Gargantua experiment' that I described (as Kev has also pointed to for the event horizon). My statement: "So, if these calculations are correct, Galileo was wrong ..." did not say the equivalence principle is wrong, but rather that any experiment he could have performed on Earth with relativistic speed particles would not have been in confined to an 'infinitesimal frame'.

One has to have a reasonably long horizontal separation in order to have any measurable drop and especially a measurable difference in drop time. Then spatial curvature bedevils the test - i.e., it is no longer a pure SR problem and the equivalence principle only applies approximately. Only in the OP's fictional "very long, flat Earth" (essentially infinitely large) can the EP be applied directly in this type of test.

Anyway, since I stand by my result, that's the only conclusion that I could come to.

-J

 

[DukeL]

One has to have a reasonably long horizontal separation in order to have any measurable drop and especially a measurable difference in drop time. Then spatial curvature bedevils the test - i.e., it is no longer a pure SR problem and the equivalence principle only applies approximately. Only in the OP's fictional "very long, flat Earth" (essentially infinitely large) can the EP be applied directly in this type of test.

In your calculating post above, you said that when you increased the size of the black hole without limit, the acceleration with a 0.8c horizontal velocity remained about 2.3 times the 'straight drop' acceleration. This does not correlate with your "essentially infinitely large" statement above, not so?

I think part of the problem is that your calculation in a Cartesian frame does not work in the Schwarzschild spacetime around a practical black hole of any size.

 

[DukeL]

During the course of this thread I have come to the conclusion that the equivalence principle is a "one way relationship". While an observer in a closed artificially accelerated lab could not be certain that he is not experiencing acceleration due to the proximity of a massive gravitational body, the reverse is not always true.

Hi kev, I guess if a lab is really small and the massive body is really big, neither observer would be able to say for sure which is which. If the hole is arbitrarily big, doesn't the tidal forces at the horizon become arbitrarily small in a small lab?

 

[Jorrie]

In your calculating post above, you said that when you increased the size of the black hole without limit, the acceleration with a 0.8c horizontal velocity remained about 2.3 times the 'straight drop' acceleration. This does not correlate with your "essentially infinitely large" statement above, not so?

Hi DukeL.

Yep, but I think the problem is that one cannot practically "increase the size of the black hole without limit" in physical calculations (they blow up). I suppose one can do it in the formulas, which will then give you a uniform gravitational field and the equivalence principle applies.

I think part of the problem is that your calculation in a Cartesian frame does not work in the Schwarzschild spacetime around a practical black hole of any size.

Agreed, but I have checked this roughly and the error introduced is about 4% in the time difference, relatively small compared to the 33% time difference between the two different 'drops'.

I'm pretty sure (but haven't checked) that if you do a full relativistic orbit calculation at r = 1.0001Rs, with and without $d\phi/d\tau = 0.8c/r$ angular starting velocity, you will roughly get the same drop time difference (say for a decrease in r of one meter).

-J

 

[DukeL]

I'm pretty sure (but haven't checked) that if you do a full relativistic orbit calculation at r = 1.0001Rs, with and without $d\phi/d\tau = 0.8c/r$ angular starting velocity, you will roughly get the same drop time difference (say for a decrease in r of one meter).

Jorrie, I've tried to get my head around this by taking pervect's original derivation, simplified for zero radial velocity, i.e.

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

Looking at it, the angular velocity $v_{\phi}$ of the object can only have any significant influence if $v_{\phi}^2$ it is in the same order of mag. as $m/r^3$, because they are subtracted.

Using your 15 trillion suns black hole at radial range r=1.0001Rs and $v_{\phi}= 0.8/(r-Rs)$, I get the following values:

$v_{\phi}^2\approx 10^{-38}$ and $m/r^3 \approx 10^{-34} \gg v_{\phi}^2$ (in geometric units meter$^{-2}$).

If I haven't made any calculation errors, this seems to rule out any significant influence of the angular velocity on the radial acceleration in your example, casting doubt on your calculations.

 

[yuiop]

In the example of the accelerating rocket experiencing constant proper born rigid acceleration as plotted on a Minkowski diagram, the acceleration experienced by the observers is proportional to 1/R. Normally gravitaitional acceleration is proportional to 1/R^2. I have recently discovered that the gravity perpendicular to a an infinitely long cylnder attenuates proportional to 1/R. This is an interesting coincidence and adds to the possibility of the spacetime of the cylindrical gravitational body being an exact equivalent of the accelerating rocket. It would be interesting if someone could figure out the gravitational potential of the cylinder and what implications that has to adapting the Schwarzschild metric from the spherical to the cylindrical case.

 

[Jorrie]

I have recently discovered that the gravity perpendicular to a an infinitely long cylinder attenuates proportional to 1/R. This is an interesting coincidence and adds to the possibility of the spacetime of the cylindrical gravitational body being an exact equivalent of the accelerating rocket.

I guess this is correct for the spatial dimension orthogonal to the acceleration vector and parallel to the cylinder's long axis, but not in the other orthogonal dimension (orthogonal also to the cylinder's axis), where the spatial curvature remains. In the rocket's case, there is no curvature in either of the two orthogonal directions.

-J

 

[yuiop]

I guess this is correct for the spatial dimension orthogonal to the acceleration vector and parallel to the cylinder's long axis, but not in the other orthogonal dimension (orthogonal also to the cylinder's axis), where the spatial curvature remains. In the rocket's case, there is no curvature in either of the two orthogonal directions.

-J

I see your point and this presents the equivalence principle with a problem because there appears to be no (even hypothetical) gravitational body that can duplicate all the measurements made inside the cabin of the accelerating spaceship.

The gravity orthogonal to an infinite flat slab does not diminish at any distance from the slab so we can not use that hypothetical gravitational body either.

How do we save the equivalence principle?

 

[Jorrie]

The gravity orthogonal to an infinite flat slab does not diminish at any distance from the slab so we can not use that hypothetical gravitational body either.

How do we save the equivalence principle?

I guess a very large spherical body does approximate the uniform gravitational field that corresponds to the equivalence principle as far as tidal effects are concerned. I suppose even the inverse square law would be very difficult to observe in a small lab.

I do not think there is a body representing a pure uniform gravitational field - it does not even help to make it spherically 'infinite', because then the gravitational field will be the same everywhere, like for the 'infinite' square slab - actually, you can't be outside an infinite 3D thing, so there might be no gravitational effects, not so?

-J

 

[Jorrie]

If I haven't made any calculation errors, this seems to rule out any significant influence of the angular velocity on the radial acceleration in your example, casting doubt on your calculations.

Hi DukeL.

It does not look like you made any computational errors, but I'm still battling to convert what you wrote (for the Schw. coordinate frame) to the local frame. Will come back to you later...

-J

 

[Jorrie]

Hi again DukeL.

It does not look like you made any computational errors, but I'm still battling to convert what you wrote (for the Schw. coordinate frame) to the local frame. Will come back to you later...

It appears that is not necessary to worry about the conversion between Schwarzschild and local coordinates in the solution, after all. The issue lies in the centripetal acceleration inherent in polar coordinates, e.g., one cannot say that an object moving with $d\phi/dt > 0$, but with $dr/dt=0$ has zero acceleration. One can quite legally replace the polar coordinates with rectangular x,y,z coordinates and see that, as I indicated here:

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): (Note should have read $-rv_\phi^2$)

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

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.

Eq. 4 should hence be written as:

$$\frac{d^2 y}{dt^2} = -\frac {m}{y^2}\left(1 - \frac {2m}{y} + 2v_x^2\right)$$

I found the following acceleration values for the case that you calculated for the distant observer's frame:

$$\frac{d^2 r}{dt^2}= -1.13\times 10^{-21}$$

$$-rv_\phi^2 = -1.45\times 10^{-21}$$

$$\frac{d^2 y}{dt^2} = -2.58\times 10^{-21}$$

All in geometric units of meter$^{-1}$, indicating the source and magnitude of the increase in acceleration in the presence of angular velocity. Does this make any sense?

-J