Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Harri on 03/07/2021 19:45:33
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Just watched TV and saw a bowling ball and feather dropped from a height and of course the ball lands first. The same experiment is repeated but in a vacuum chamber. The ball and feather land at the same time! Shouldn't the feather and ball just hang suspended side by side in a vacuum until a force acts on them to either move them of pull them to the ground?
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Gravity is acting on them, pulling them to the ground.
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Shouldn't the feather and ball just hang suspended side by side in a vacuum until a force acts on them to either move them of pull them to the ground?
But gravity does act on them, not the air. The absence of air just eliminates the friction, that friction being the only thing that retards the progress of the feather more than it does the ball.
Expressed in the terms of GR, there is no force of gravity, but the ground accelerates upward to the stationary ball and feather, and there's no way it can get to one sooner than the other if they're both stationary. The acceleration of the ground can be measured with any accelerometer.
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So what am I actually seeing? The force of gravity acting on the ball and feather pulling them down or the ground accelerating up to the stationary ball and feather?
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There is no falling, only attraction. Accelleration.
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There is no falling, only attraction. Accelleration.
That's what falling is: acceleration due to gravitational attraction.
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But gravity does act on them, not the air. The absence of air just eliminates the friction, that friction being the only thing that retards the progress of the feather more than it does the ball.
Expressed in the terms of GR, there is no force of gravity, but the ground accelerates upward to the stationary ball and feather, and there's no way it can get to one sooner than the other if they're both stationary. The acceleration of the ground can be measured with any accelerometer.
Er, no. The model sort of works in the vacuum case, but if there is air in the tube, why does the ground accelerate faster towards the ball than to the feather? and if our chum Evan does the same experiment at the same time in Australia, why doesn't the earth split open as the ground rushes simultaneously towards both balls?
An important strength of GR is that it gives the same result as newtonian mechanics for mesoscopic objects and low speeds.
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The model sort of works in The model sort of works in the vacuum case, but if there is air in the tube, why does the ground accelerate faster towards the ball than to the feather?
Because the ground accelerates the air up with it, and that wind pushes the feather away much in the same way it doesn't put the ball.
Consider a centrifuge in deep space (away from gravity). You have two rooms spinning around a central axis, one with air, the other in a vacuum. You drop the feather and rock in each room, and in the vacuum case the two hit the floor at the same time, but in the air case, the rock hits first because the floor is accelerating the air which in turn exerts an accelerating force on both objects towards the axis (up). The feather, massing much less, accelerates away from the floor far more than does the more massive rock, per Newton's f=ma. No GR involved.
and if our chum Evan does the same experiment at the same time in Australia, why doesn't the earth split open as the ground rushes simultaneously towards both balls?
You're confusing coordinate acceleration with proper acceleration here.
Same example as above. Both rooms have a proper acceleration towards each other, but their separation remains fixed, so using the rotating frame of reference as we do in labs here on Earth, the coordinate acceleration of the floor is zero and it is the feather and rock that accelerates (not quite straight) to the floor.
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Hi. I think I've got to support Halc on this one.
Er, no. The model sort of works in the vacuum case, but if there is air in the tube, why does the ground accelerate faster towards the ball than to the feather?
Good question. It's because both the ground and the air is rushing toward those objects. The air is getting some support from the ground (or from the bottom of the tube). The air can deform and move past the objects and at that point, the objects are experiencing a force (friction). The feather experiences a greater force (per unit mass). Neither the feather or the ball remain stationary but the acceleration is greater for the feather.
and if our chum Evan does the same experiment at the same time in Australia, why doesn't the earth split open as the ground rushes simultaneously towards both balls?
Another good question. The gravitational field is not uniform as you cross planet earth (it completely reverses direction). Your local inertial frame is different to Evan's as far as GR is concerned. Locally, Evan's ground is rushing upward and so is your own. If you consider Evan's ground in your own local inertial frame then the whole planet is rushing toward your objects in the tube.
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Because the ground accelerates the air up with it,
So why does it accelerate nitrogen molecules upwards, but not feather and ball molecules?
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If you consider Evan's ground in your own local inertial frame then the whole planet is rushing toward your objects in the tube.
and therefore away from the objects in Evan's tube, which is pointing in the opposite direction to mine.
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The acceleration of the ground can be measured with any accelerometer.
So I'm sitting in my airplane, which has an accelerometer, and you drop a bowling ball or a feather on the runway, and the plane takes off, or at least registers positive g. Saves a lot of fuel.
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So why does it accelerate nitrogen molecules upwards, but not feather and ball molecules?
Relative to a local inertial frame, all of those accelerate (proper) upward. If you put an accelerometer on any of them, it shows acceleration upward. There's no force pushing in the down direction.
So I'm sitting in my airplane, which has an accelerometer
It does? I mean, you can put one in there, but I was unaware of it being equipped with a device that measures proper acceleration. I can assume such a device. Heck, cell phones have them.
and you drop a bowling ball or a feather on the runway, and the plane takes off, or at least registers positive g. Saves a lot of fuel.
Not following you. The plane parked should register positive g. Taking off, it should register some component forward as well as the force from the engines are greater than the wind pressure accelerating it backwards. No idea what you're trying to illustrate with that and with the objects you left back there on the runway.
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I think it's obvious from my questions and responses that I read popular science books and I am in no way a scientist. I live in, and I am a product of my logical world where apparently illogical things occur. Illogical to my logical upbringing that is. The writer attempts to explain the nature of these illogical things without the math and depth of scientific fact and sometimes it just doesn't translate into understanding.
Could I ask, what if we return to the vacuum chamber where a bowling ball and feather are suspended from the ceiling. We then create a vacuum by withdrawing the air in the chamber. This time, midway between the objects and the floor we pass a light beam across their path. On release of the ball and feather what would reach the beam first, the ground or the objects? Or both simultaneously?
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It does? I mean, you can put one in there, but I was unaware of it being equipped with a device that measures proper acceleration. I can assume such a device. Heck, cell phones have them.
And you'd be mad to undertake aerobatics without one.
Not following you.
Nor I you, skipper.
Expressed in the terms of GR, there is no force of gravity, but the ground accelerates upward to the stationary ball and feather, and there's no way it can get to one sooner than the other if they're both stationary. The acceleration of the ground can be measured with any accelerometer.
Now the mutual acceleration between the bowling ball and the ground must be g m/s^2 because that's what we can measure by plotting separation vs time. So if the ball stays stationary and the ground suddenly starts hurtling upwards at g, the accelerometer in my plane will move from +1g to +2g because it is sitting on the surface of a planet that is now moving (it certainly works on the deck of a carrier that's plunging up and down). Let's drop the ball (or accelerate the planet) from 10 meters. Now the planet hits the ball and stops moving, but the plane has some upward speed (196 m/s!) so it lifts off at several times its normal climb rate, without troubling the engine at all!
Of course there is a boring newtonian explanation, but that is so 17th century.
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This time, midway between the objects and the floor we pass a light beam across their path. On release of the ball and feather what would reach the beam first, the ground or the objects? Or both simultaneously?
Whilst I'm teasing the experts, I'll let you into a newtonian secret. The planet and the ball and the feather all accelerate towards their mutual center of mass*, and would reach it at the same time if there were nothing in the way. But fortunately for you terrestrial beings, the planet is so much more massive than any artefact that the barycenter is pretty close to the middle of the planet and the time difference between the ball and the feather hitting the ground whether released separately or together is negligible.
*the big question is why?
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Could I ask, what if we return to the vacuum chamber where a bowling ball and feather are suspended from the ceiling. We then create a vacuum by withdrawing the air in the chamber. This time, midway between the objects and the floor we pass a light beam across their path. On release of the ball and feather what would reach the beam first, the ground or the objects? Or both simultaneously?
Depends on the frame in which the light beam is stationary. If it's mounted to something fixed relative to the ground, obviously the falling objects will pass through it. But if it is stationary in the local inertial frame of the falling objects (i.e. falling with them), the the beam will intersect the accelerating ground and the stationary objects will never reach it.
Now the mutual acceleration between the bowling ball and the ground must be g m/s^2 because that's what we can measure by plotting separation vs time. So if the ball stays stationary and the ground suddenly starts hurtling upwards at g, the accelerometer in my plane will move from +1g to +2g because it is sitting on the surface of a planet that is now moving
Sorry. Your accelerometer on the plane should read zero if there's no force (the ground) pushing upward on it, just as it would in deep space. So place it on the 1g accelerating ground and it should read 1g do to that one force acting on its wheels. You know this, but apparently you consider yourself 'teasing me'.
Let's drop the ball (or accelerate the planet) from 10 meters. Now the planet hits the ball and stops moving
As you said, you're teasing. You know this is nonsense.
but the plane has some upward speed (196 m/s!) so it lifts off at several times its normal climb rate, without troubling the engine at all!
In the accelerating frame of the ground on which the plane is parked, it gains no speed at all. But yes, in the local inertial frame of a ball dropped in a vacuum from at least 2 km up, the plane very much acquires a speed of 196 m/sec after 20 seconds, and yes, all without help from the engine (useless in the vacuum I put it in). I imagine the fast moving plane will damage itself when it collides with the sufficiently large (up until the moment of impact) stationary ball. At that moment, the planet surface continues to accelerate and the ball is no longer stationary in that frame.
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*the big question is why?
Too big 8)
This time, midway between the objects and the floor we pass a light beam across their path. On release of the ball and feather what would reach the beam first, the ground or the objects? Or both simultaneously?
Think. What have you fastened the light beam equipment to?
Edit: whoops, clashed with @Halc reply
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Your accelerometer on the plane should read zero if there's no force (the ground) pushing upward on it, just as it would in deep space.
Wrong. It must read 1g on the runway because the force acting on the wing strut is 1 g x the mass of the wing and the force on the undercarriage is 1 g x the mass of the entire aircraft. When flying straight and level the lift provided by the wings equals the total mass of the aircraft x 1 g by definition of "straight and level". At the top of a gliding loop it will read 0 because the entire structure is in free fall.
Attached photo is the nearest one I have to straight and level showing a g meter. The artificial horizon (top right) and turn indicator (center bottom) show pretty close to S&L, and the g meter (bottom left) is close to 1. Interesting aircraft, incidentally, a Victa Airtourer, manufactured by a company that made lawnmowers.
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Could I ask, what if we return to the vacuum chamber where a bowling ball and feather are suspended from the ceiling. We then create a vacuum by withdrawing the air in the chamber. This time, midway between the objects and the floor we pass a light beam across their path. On release of the ball and feather what would reach the beam first, the ground or the objects? Or both simultaneously?
Okay, we will assume that the source of the light is fixed relative to the ground. The light is turned on. The light, after leaving the source, will curve downward due to gravity. However since light travels so fast, it would take extremely accurate measurements to even notice that it hits the far wall a bit lower.
But what happens to the beam if we let it continue on past the chamber wall? If we assume that the ground extends as a flat plane, and that gravity acts perpendicular to that plane, the light will, at some distant point, hit the ground. The time it would take to do this would be equal to the time it would take for the ball and feather to fall to the height of the source (at least to close approximation and if we assume that gravity strength does not differ over height). But, the beam will have dropped a tint bit by the time it reaches the middle of the chamber, so they will not quite have reached the light beam yet.
So, under these conditions, the light will hit the ground just slightly ahead of the objects reaching the beam.
If we take into account the fact that gravity drops off with height above the surface, this means that the objects fall at a slightly less rate of acceleration than the light beam. If the chamber is tall enough, this can lead to a significant difference.
However, we also assumed that the ground was an flat plane that gravity acted perpendicular to.
The ground however, is the surface of a sphere, and gravity acts toward the center of that sphere.
The amount that gravity can curve the light beam is much, much less than the curvature of the Earth. So the light beam ends up getting further and further from the Earth's surface as it travels, and ends up heading out into space, never hitting the ground*.
So the question relies a great deal on your base assumptions.
* This doesn't just apply to light. Any object that is traveling fast enough, even if it starts off moving parallel to the ground, will end up never falling to the ground. Look up Newton's cannon for further explanation.
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And an apology! v = √2as = 14 m/s or so, not 196, after 10m at 1g! Forgot the square root....Even so, that's one heck of a rate of climb - 2600 ft/min beats a Spitfire!
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Let's drop the ball (or accelerate the planet) from 10 meters. Now the planet hits the ball and stops moving
As you said, you're teasing. You know this is nonsense.
Ipsi dixit. And both Newton and Einstein agree.
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Hi all,
So if we use Harri's example of a light beam passing through the experiment and also factor in Colin's point
Think. What have you fastened the light beam equipment to?
lets say you use sunlight as your light source, passing through a pinhole into the room/lab, streaming across the mid point would that not discriminate the points under discussion ?
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lets say you use sunlight as your light source, passing through a pinhole into the room/lab, streaming across the mid point would that not discriminate the points under discussion ?
The pinhole is fixed to a wall which is fixed to the floor and so they move together, see explanation by @Janus
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Hi all
lets say you use sunlight as your light source, passing through a pinhole into the room/lab, streaming across the mid point would that not discriminate the points under discussion ?
The pinhole is fixed to a wall which is fixed to the floor and so they move together, see explanation by @Janus
Janus started with:
(Okay, we will assume that the source of the light is fixed relative to the ground)
Therefore the angle of the light stream will be altered if the earth/lab was moving significantly. Rather than the other way round.
Given the walls will have thickness. It would be like trying to line up the light source through 2 pinholes.
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Given the walls will have thickness. It would be like trying to line up the light source through 2 pinholes.
is that a problem?
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Hi all,
So if we use Harri's example of a light beam passing through the experiment and also factor in Colin's point
Think. What have you fastened the light beam equipment to?
lets say you use sunlight as your light source, passing through a pinhole into the room/lab, streaming across the mid point would that not discriminate the points under discussion ?
Then there would be aberration and the effect gravity had on the light before it entered the pin hole. But all that would do is alter the initial angle that the light exits the hole relative to the chamber. It would be the same as a fixed light source that wasn't aimed at a point directly across from it.
This can be simply accounted for by arranging to perform your experiment when the combination of the two above effects leads to light entering the pin hole parallel to the floor of the chamber.
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Hi All
So if I was following the logic correctly
Given the walls will have thickness. It would be like trying to line up the light source through 2 pinholes.
is that a problem?
Then Harri was trying to get a handle on, was what was actually accelerating the bowling ball and feather or the Earth, given Halcs and Alans counter arguments with GR and Newtons laws of motion.
But first I think we need to do away with the dropping of light in the gravity field as being a separate point, and isn't really the point under consideration, as the light beam suggested can be treated as straight enough to discern whats accelerating.
Given we could very easily set the ball and feather to drop say 10 metres and the points observed of the stream light passing across the vacuum the deviation from straight/horizontal would be negligible, I.E across the length of the football pitch at Wembley 105 metres it would drop 6.017 x10^-13 metres.
So if you used the center mark on the vacuum tube as suggested by Harri as the equivalent as your traveler in a set of boning rods you would expect the line of sunlight to accelerate down the tube at 9.81 m^-2 if you postulate only the earth is accelerating.
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Hi.
you would expect the line of sunlight to accelerate down the tube at 9.81 m^-2 if you postulate only the earth is accelerating.
It does. However one photon (one bit of light) is travelling with horizontal velocity ~ c. It's only in the tube for a tiny fraction of a second. The Earth and the floor of the tube hasn't had much time to move up toward that photon.
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Hi all,
Hi.
you would expect the line of sunlight to accelerate down the tube at 9.81 m^-2 if you postulate only the earth is accelerating.
It does. However one photon (one bit of light) is travelling with horizontal velocity ~ c. It's only in the tube for a tiny fraction of a second. The Earth and the floor of the tube hasn't had much time to move up toward that photon.
Ok just to clarify are you therefore stating, the beam of sunlight will reach the bottom of the tube before the ball and the feather, which is dropped from the top of the tube ?
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Ok just to clarify are you therefore stating, the beam of sunlight will reach the bottom of the tube before the ball and the feather, which is dropped from the top of the tube ?
Janus gave a good explanation earlier. Reply #20 I think. If you followed the first photons that were put into the tube and allowed them to continue their path out beyond the other side of the tube, then they hit the ground before the feather and the ball would hit the ground. Janus sets out all the assumptions (e.g. the surface of the earth remains flat, rather than being the curved surface we know it is).
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Hi all
So ES as I stated earlier and put a value to the drop due to gravity, of light across the length of the pitch at wembley, or if you prefer the drop of a photon for a meter of horizontal travel at earths surface gives approx 5.458 x10^-17 m
But this is not whats under consideration, if the Earth is really accelerating upwards as you are postulating, the beam of sunlight where it intersects the 10 m tube at the midway point I.E
At 5 m then in your scenario the beam will intersect the bottom of the tube at point 0
1.0096 seconds after the ball/feather is released
and ball/feather reaching that point 0.418 seconds later
Also If what you and others are postulating here was correct you would be able to increase the sunset time at lets say; latitude London.
By approximately 25 seconds just by dropping an object from 10m above sea level.
???
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Also If what you and others are postulating here was correct you would be able to increase the sunset time at lets say; latitude London.
By approximately 25 seconds just by dropping an object from 10m above sea level.
???
I don’t understand how you reach that conclusion
Yes indeed as you say “???”
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Hi all,
Morning Colin
So
Also If what you and others are postulating here was correct you would be able to increase the sunset time at lets say; latitude London.
By approximately 25 seconds just by dropping an object from 10m above sea level.
???
I don’t understand how you reach that conclusion
Yes indeed as you say “???”
If it’s postulated that the surface of the earth accelerates up to meet the ball and feather then if you did this at a sea shore or a large body of water from a height of 10 m just as the sun dropped below the horizon, and the earth really did lift, then it would bring part of the sun back into view.
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Hi.
@gem
The surface of the earth is always accelerating upward at 1g. It does this whether you drop the ball and the feather or not.
Whatever effect is happening to the suns rays due to this acceleration of the earth's surface, it doesn't happen more (or less) just by dropping objects from 10m above the earth's surface.
Nice picture of a sunset by the way, thanks for including that.
Best wishes to you.
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Hi all,
Morning Colin
So
Also If what you and others are postulating here was correct you would be able to increase the sunset time at lets say; latitude London.
By approximately 25 seconds just by dropping an object from 10m above sea level.
???
I don’t understand how you reach that conclusion
Yes indeed as you say “???”
If it’s postulated that the surface of the earth accelerates up to meet the ball and feather then if you did this at a sea shore or a large body of water from a height of 10 m just as the sun dropped below the horizon, and the earth really did lift, then it would bring part of the sun back into view.
The Sun is 149.6 million km away. A change of 10 meters in your viewing position would only result in an angular shift of 0.00001378 seconds of arc. That's magnitudes less than even the angular resolution of the Hubble telescope.
As a comparison, a 1 meter high wave on the horizon would subtend 68 seconds of arc from your viewpoint.
In other words, waves on the ocean would make the edge of the Sun appear and disappear by a far greater amount than the Earth shifting 10 m.
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The surface of the earth is always accelerating upward at 1g.
Both in the UK and in Australia. It's amazing that we can still talk to one another at an ever-increasing distance. Thanks to relativity, our relative speed cannot exceed c, but the fact that you can still send parcels by ship
(about 10-8c) in time for Christmas is a positive mark for the post office.
Or maybe the earth really is flat.
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Hi all
So Janus
Hi all,
Morning Colin
So
Also If what you and others are postulating here was correct you would be able to increase the sunset time at lets say; latitude London.
By approximately 25 seconds just by dropping an object from 10m above sea level.
???
I don’t understand how you reach that conclusion
Yes indeed as you say “???”
If it’s postulated that the surface of the earth accelerates up to meet the ball and feather then if you did this at a sea shore or a large body of water from a height of 10 m just as the sun dropped below the horizon, and the earth really did lift, then it would bring part of the sun back into view.
The Sun is 149.6 million km away. A change of 10 meters in your viewing position would only result in an angular shift of 0.00001378 seconds of arc. That's magnitudes less than even the angular resolution of the Hubble telescope.
As a comparison, a 1 meter high wave on the horizon would subtend 68 seconds of arc from your viewpoint.
In other words, waves on the ocean would make the edge of the Sun appear and disappear by a far greater amount than the Earth shifting 10 m.
So Janus, your disputing the method of measuring earths diameter by difference in time of sunset by height is not detectable ?
http://astronomy.nmsu.edu/geas/lectures/lecture10/slide05.html
There’s plenty of links on line that would suggest it is
And ES
Hi.
@gem
The surface of the earth is always accelerating upward at 1g. It does this whether you drop the ball and the feather or not.
Whatever effect is happening to the suns rays due to this acceleration of the earth's surface, it doesn't happen more (or less) just by dropping objects from 10m above the earth's surface.
Nice picture of a sunset by the way, thanks for including that.
Best wishes to you.
If the surface of the earth is always accelerating up at 1 g
Then what’s it’s velocity ?
If you state it accelerates upwards to the objects it must have a change in velocity, if it’s not detected then your in the realms of flat earth argument,
Given the method I am describing to measure the diameter of the earth is a well known counter argument to the flat earth believers.
https://www.omnicalculator.com/physics/flat-vs-round-earth
You want to have a sniff at what you’re selling.
👃💩
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If the surface of the earth is always accelerating up at 1 g
Then what’s it’s velocity ?
Very Large with respect to a local inertial frame established at the surface of the earth 10 years ago (in cosmological co-ordinate time).
Medium w.r.t. an inertial frame established at the surface of the earth yesterday.
0 w.r.t. a local inertial frame established at the surface of the earth just now.
If you state it accelerates upwards to the objects it must have a change in velocity, if it’s not detected then your in the realms of flat earth argument,
It does have a change of velocity (see above). Actually, we are in the realms of a flat earth argument. It was stated many replies back that the equivalence principle applies Locally and not globally where there are tidal forces or a non-uniform gravitational field. Further back than that, Janus described how the photons would fall by assuming the surface of the earth remained flat indefinitely but also described the more realistic situation where the surface is curved.
Locally, the surface of the earth is accelerating upward at 1g with respect to an inertial frame as defined in GR (this is a frame in free-fall).
I haven't followed the link you provided, sorry. I've not heard of omnicalculator before and I'm a bit cautious of pointing my web browser at it.
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Whilst folk are throwing the equivalence principle at each other, it might be worthwhile stating it for the benefit of newcomers to this bear pit. Wikipedia is succinct:
In the theory of general relativity, the equivalence principle is the equivalence of gravitational and inertial mass, and Albert Einstein's observation that the gravitational "force" as experienced locally while standing on a massive body (such as the Earth) is the same as the pseudo-force experienced by an observer in a non-inertial (accelerated) frame of reference.
Which is very dull because it doesn't invoke a flat earth hurtling at ever-increasing speed towards whatever you choose to drop, or a spherical planet tearing itself to pieces every time a Kiwi hits a six or converts a try.
Back in the benighted Sixties, one campaign slogan of the NAACP (I'm not allowed to say what the initials stand for because that would nowadays be considered offensive to its own members) was "equivalent is not the same". Well worth remembering in this context.
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So Janus, your disputing the method of measuring earths diameter by difference in time of sunset by height is not detectable ?
http://astronomy.nmsu.edu/geas/lectures/lecture10/slide05.htm
That is a completely different measurement than the one you suggested. Determining the diameter of the Earth from sunset time at ground level vs. head level, as shown in the link depends on the Earth's rotation, Not on a displacement of the Earth in position. It also depends on the distance of the horizon as viewed from two different heights (ground and head). At typical head height, the horizon is ~ 5 km away, (vs no distance at ground level) this results in a viewing angle difference of ~70 arc sec. This measurement is totally independent of the Sun's distance.
However, any change in viewing angle caused by the Earth being displaced by 10 m does depend on the distance to the Sun. And would result in the extremely small value I gave in my Earlier post.
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Then Harri was trying to get a handle on, was what was actually accelerating the bowling ball and feather or the Earth, given Halcs and Alans counter arguments with GR and Newtons laws of motion.
But first I think we need to do away with the dropping of light in the gravity field as being a separate point, and isn't really the point under consideration, as the light beam suggested can be treated as straight enough to discern whats accelerating.
The light beam is a red herring, you are right to ignore it. Any light source - including a pinhole - is still connected to the same frame as the ground. If the light falls, it won’t tell you whether the surface is accelerating upwards or not. I suggest you reread the posts by @Janus
However, we do need to drag you and @Harri out of this gravitational well you’ve fallen into:
If it’s postulated that the surface of the earth accelerates up to meet the ball and feather
The problem is that you have to consider things from a relativity point of view and ask “relative to what”.
In SR you are used to understanding that you can’t tell whether you are moving, only that you are moving relative to something else. Einstein gave a comparison between gravity and acceleration, saying you can’t tell the difference.
Just for clarification, as Alan says, Einstein never said they were the same, only equivalent. He also assumed a uniform gravitational field (which is approximately true in a small volume like an elevator) so no tidal effects or variations of g with height.
So, you can’t tell the difference between gravity and acceleration, but how do you know if you are accelerating? You feel the force. If you are in a car which is accelerating the seat pushes your body forward and you feel the force on your back. If you don’t feel the force you must be at rest or moving at constant speed.
There is an instrument that will tell you if you are accelerating, an accelerometer. Imagine a ballbearing in a horizontal tube, if the car you are in accelerates the ball, due to inertia, will try to stay where it is as the tube moves forward and if there is a sensor in the tube you can measure the force between ball and tube f=ma, hence measure a, acceleration. However, tip the tube vertically and it will register an acceleration of 1g, the same as if the car had been accelerating at 1g; but let the tube freefall and it will register 0g, it is not accelerating.
The acceleration we measure in this way is called proper acceleration and is what is used in GR, and freefall is the GR equivalent of an inertial frame in SR - no acceleration.
The problem is that you are used to measuring what we call coordinate acceleration, that is the rate of change of speed. Sitting in your car if the speedo shows increasing speed you are accelerating; but wait a minute speed is relative in SR so how do we know we are moving? Take 2 spaceships side by side in deep space, are they at rest, moving at same speed, or accelerating? The only one we can measure is acceleration using our trusty accelerometer, so this is what we use in GR.
However, coming back down to earth, does the earth move for @gem?
Well, if you are measuring the ball and feather relative to the surface of the earth (coordinate measurement) then they are accelerating towards you at the same 9.8m/s2 - just remember to wear your vacuum suit or we get vacuum packed gem!
So if your reference frame (coordinates) is the vacuum chamber, fixed to the earth, with a light source or pinhole fixed to the side of the chamber then not only the ball and feather, but also the light beam are falling (relative to those coordinates - the frame of the earth’s surface).
So, the surface of the earth is not moving upwards relative to the earth’s centre, but it is accelerating if we use proper acceleration as our measurement.
So next time the earth moves for you, don’t assume it’s the ground coming up to meet you ;)
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Hi all
Yes thanks Colin I agree with what you’re saying, However why you believe I think the earth and the tube are rising up is beyond me,
Hi.
you would expect the line of sunlight to accelerate down the tube at 9.81 m^-2 if you postulate only the earth is accelerating.
It does. However one photon (one bit of light) is travelling with horizontal velocity ~ c. It's only in the tube for a tiny fraction of a second. The Earth and the floor of the tube hasn't had much time to move up toward that photon.
It is others whom stated it.
Which I was highlighting was ridiculous and misleading.
Compared to your statement Colin:
(So, the surface of the earth is not moving upwards relative to the earth’s centre)
Yes quite agree 🧐💋
As you say Alan covered this,some times people get mixed up what equivalence actually is, and state things which are misleading.
And Janus you seem to be arguing the lifting of the local area
(💋which isn’t happening💋)
is undetectable via a local change in height.
The point I was making if you look back to the boning rods/line of sight is we are not in an elevator with no windows and if parts or all of the earth moved upwards as ridiculously suggested by others, we would be able to a detect that movement by the same principle of the boning rods/line of sight.
I suppose what you’re definition of local is comes into it, but if the ground came up by 10m every where in your line of sight from the centre of the earth, it would be detectable,.
If it came up across the whole earth 🌍 it would be detectable
Because the day would become longer.
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Compared to your statement Colin:
(So, the surface of the earth is not moving upwards relative to the earth’s centre)
Yes quite agree 🧐💋
You still have to be very clear what you are actually talking about.
My statement is a coordinate statement relative to earth centre based on velocity (and acceleration) as a 3-vector, GR deals with 4-velocity which is a 4-vector, not 3. In that system the surface of the earth can be considered to be moving up towards to ball and feather even though there is no spacial movement of the surface. Note, this is a short very incomplete answer, but you get the drift.
Don’t let it drag you down, looked at properly it can be quite uplifting ;)
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HI all,
Compared to your statement Colin:
(So, the surface of the earth is not moving upwards relative to the earth’s centre)
Yes quite agree 🧐💋
You still have to be very clear what you are actually talking about.
My statement is a coordinate statement relative to earth centre based on velocity (and acceleration) as a 3-vector, GR deals with 4-velocity which is a 4-vector, not 3. In that system the surface of the earth can be considered to be moving up towards to ball and feather even though there is no spacial movement of the surface. Note, this is a short very incomplete answer, but you get the drift.
Don’t let it drag you down, looked at properly it can be quite uplifting ;)
MMMM yes I must say all this proper acceleration is slowing me down. but its more uplifting than you know.
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:)