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Why is g = 9.8 and from where/what triggers the upward acceleration?
Дата публикации: 28-09-2026 08:58:51
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- Level: High School
- Thread starter Thread starter Abi1007
- Start date Start date Sep 25, 2026
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The below image shows how the local geometry in the above animation above fits into the global picture. The two green apples still hanging on the branches have both proper acceleration pointing outwards, without moving apart.
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The below image shows how the local geometry in the above animation above fits into the global picture. The two green apples still hanging on the branches have both proper acceleration pointing outwards, without moving apart.
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Proper acceleration is not relative. We have an upward proper acceleration. Full stop.
Relative to a local freely-falling frame we also have an upward coordinate acceleration.
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Roberto Pavani
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Fair enough, I missed a comma.
“We have an upward proper acceleration, with ‘upward’ defined relative to a local freely-falling frame.”
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The proper acceleration is equal to the (momentary) coordinate-acceleration with reference to a momentarily co-moving inertial coordinate-system.
The diagram shows two instantaneous co-moving inertial reference-frames ##S## and ##S'##.
A worldline of constant spacetime-distance from the origin of a Minkowski diagram is the following hyperbola:
$$x(t)^2 - c^2t^2 =...
Details:
That can be done to understand Rindler-coordinates.
The diagram shows two instantaneous co-moving inertial reference-frames ##S## and ##S'##.
A worldline of constant spacetime-distance from the origin of a Minkowski diagram is the following hyperbola:
$$x(t)^2 - c^2t^2 =...
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When you say 'the ground doesn't have a geodesic because it experiences proper acceleration,' you are playing a semantics game. Spacetime curvature around a static mass is stationary. The Earth’s surface is not expanding, it is not moving, and its coordinate acceleration through space is exactly zero. And as for the zero sense on contact , when I say the ground absorbs your geodesic, it means that the moment you make solid contact, your independent free-fall trajectory is arrested. You become a static, unified part of the Earth’s bound molecular matrix. Your worldline is now physically locked to the Earth's worldline.This makes zero sense. Nothing changes about gravity itself by the contact.This is just wrong. The ground doesn't have a geodesic worldline, because just like the things resting on the ground, it experiences proper acceleration upwards.
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You can. The following experiment proved, that the surface of Earth has a proper upwards acceleration of ##9.81 \ m/s^2)##.
https://en.wikipedia.org/wiki/Pound–Rebka_experiment
Reason:
Yes, but then it would not be an acceleratometer.
You can also construct an accelerometer with a lamp and a light sensor, mounted at constant distance.
A proper acceleration leads to a Doppler frequency-shift, because after the finite time, a light pulse needs from the lamp to the sensor, the sensor is already moving faster than the lamp, when it sent out the light pluse.
If such an accelerometer is at rest relative to the surface of Earth, then the measured...
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I think you are confusing coordinate artifacts for physical motion. Coordinates can be changed on a whim, so coordinate dependent things are not “actual physical”, including coordinate acceleration.
The proper acceleration is the actual physical acceleration. It is measured by actual physical accelerometers, requires actual physical forces, and can have actual physical effects.
The coordinate acceleration is only math and can be changed arbitrarily with no physical effects. Coordinates cannot be detected, only assigned.
It certainly does not. Remove the ground and the measurement drops to zero despite the gravitational pull remaining. Remove gravity and push the accelerometer with a real force and the reading is non-zero despite having zero gravity.
Accelerometers do not detect the gravitational force, nor other fictitious forces. They only detect real physical forces.
Sure you can. That is what spacetime curvature does.
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It is movement on spacetime that reffers to the Time field , the geocoordinates remain the sameI think you are confusing coordinate artifacts for physical motion. Coordinates can be changed on a whim, so coordinate dependent things are not “actual physical”, including coordinate acceleration.The proper acceleration is the actual physical acceleration. It is measured by actual physical accelerometers, requires actual physical forces, and can have actual physical effects.
The coordinate acceleration is only math and can be changed arbitrarily with no physical effects. Coordinates cannot be detected, only assigned.
It certainly does not. Remove the ground and the measurement drops to zero despite the gravitational pull remaining. Remove gravity and push the accelerometer with a real force and the reading is non-zero despite having zero gravity.
Accelerometers do not detect the gravitational force, nor other fictitious forces. They only detect real physical forces.
Sure you can. That is what spacetime curvature does.
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To repeat what everyone else has said, it is the proper acceleration that is the direct observable. This is the lesson you are supposed to have taken from Einstein's lift thought experiment. If you are in a closed box (such as a lift) there is no local way to distinguish between the cases that the lift is at rest on the surface of a planet or in a rocket in deep space under thrust. In both cases you can measure identical weight from a given mass. Nor is there any way to distinguish between free falling above an (airless) planet and floating inertially in deep space. And in any of these cases you may consider yourself to be at rest or undergoing coordinate acceleration.
The only physically meaningful quantity here is the proper acceleration.
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