# Accelerometer on a dropped object

expandotron
Why does the accelerometer tell me that the dropped object does not accelerate to the ground but that the ground accelerates to the dropped object?

Mentor
Why does the accelerometer tell me that the dropped object does not accelerate to the ground but that the ground accelerates to the dropped object?
The reason “why” depends on the theoretical formulation you use. In the Newton Cartan formulation of classical physics and in the standard formulation of general relativity it is because in an invariant sense it is the ground that is accelerating upwards. In those formulations gravity is not a real force locally. The only real force is the pressure pushing up on the ground, and the dropped object is not experiencing any force at all. Thus the ground accelerates up and hits the object.

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• vanhees71, Richard R Richard and atyy
Keeping to standard Newtonian language, an accelerometer does not measure inertial acceleration. Instead it measures something called the specific force. This is because uniform gravity and acceleration cannot be distinguished (in some sense) in Newtonian physics, which is a principle of equivalence. These lecture notes by Jaime Peraire give a more detailed explanation: https://dspace.mit.edu/bitstream/handle/1721.1/60691/16-07-fall-2004/contents/lecture-notes/d14.pdf

As @Dale mentioned, in general relativity, there is a different concept called the proper acceleration. An accelerometer measures the proper acceleration.

So the accelerometer does not measure inertial acceleration, but instead measures specific force (standard Newtonian formulation) or the proper acceleration (general relativity). The Newton-Cartan formulation of mechanics that @Dale mentioned may provide a way to see how the two different concepts of specific force and proper acceleration are related (but I don't know Newton-Cartan well enough off the top of my head).

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• vanhees71 and Dale
expandotron
Thank you Dale and atyy.

Why does the accelerometer tell me that the dropped object does not accelerate to the ground but that the ground accelerates to the dropped object?
Accelerometers measure acceleration relative to free fall.

Gravity is proportional to mass, so a uniform gravitational field accelerates all parts of a accelerometer equally, and thus acceleration by a uniform gravitational field cannot be detected based on the movement / forces between those parts.

• vanhees71, Dale and atyy
alan123hk
Why does the accelerometer tell me that the dropped object does not accelerate to the ground but that the ground accelerates to the dropped object?

Just like when we stand on the ground, we can feel gravity. When sitting in a car, we can also feel the force when the car accelerates and decelerates. When we dive from a high place, in a free fall state, we will not feel any other force except air resistance. What the accelerometer detects is just the same as ours.

So maybe we can understand how the accelerometer works by asking ourselves.

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• vanhees71, Dale and atyy
expandotron
The pressure pushing up on the ground?

• vanhees71
Mentor
The pressure pushing up on the ground?
Yes. Consider a rock on the ground. The ground is pushing up on the rock (the normal force) so it accelerates up.

The dirt under the rock is pushed up by the bedrock under the dirt, and pushed down by the rock on top, but the upward force from the bedrock is much larger than the downward force from the surface rock, so the dirt accelerates upward too.

• vanhees71
expandotron
Matter continuously expands outward at an accelerating rate?

• Dale
Mentor
Matter continuously expands outward at an accelerating rate?
As @A.T. said, if the surface of the Earth were accelerating in flat spacetime then it would indeed be expanding. However, spacetime is curved, so although the surface of the Earth is continuously accelerating outward it is not expanding.

Homework Helper
Gold Member
Adhering for the moment to classical physics, it helps to think of the accelerometer physically as a spring with one end attached to the accel's body while the other is attached to a mass (called the "proof mass"). The accel's output is the extent of the spring's stretch or compression.
So when the body is laid flat on a table there is no spring stretch and the output is zero.

If I now accelerate the body horizontally, the spring will stretch or compress and the output is dv/dt.

Hold the body vertically; the spring is stretched by the force of gravity on the proof mass. The spring stretches and the output stretch corrresponds to 1g. There is no vertcal acceleration of the accel but the output reads 1g thanks to the pull of gravity on the proof mass which is held in place by the spring while the accel's body is held by the hand (i.e. push by the Earth).