How do accelerometers measure acceleration without using relative motion?

[PAllen]

So you imply your method counts as an accelerometer, so long as the spring is infinitesimal in length, in a strong enough gravitational gradient all you would need is 1mm to measure a noticable pull on the spring... As your sensitivity to the measurement increases it becomes smaller and small that is needed.

Correct. The more extreme the curvature (GR equivalent of Newtonian gradient), the smaller the accelerometer must be to measure proper acceleration.

I am neither a scientist or a mathematician so I apologize that my calculus terminology is lacking...

In a free-falling frame my acceleration I never change velocity, since A change of no velocity per proper time equals no acceleration, If I fire a rocket to escape(the not pull of) gravity, and use a frame that accelerates thus, then I still measure no change in velocity per proper time...so what is an example of proper acceleration please?

Now the object that is 50m from the surface does change its velocity(in my free-falling coordinate system 100m from surface) per proper time. So does an object 150m from surface, seems to slowly accelerate away? I am probably misusing proper acceleration here, but can you help clear that up?

Here you are being limited by your math back ground, and I'm not sure I can help. I'll try.

A world line is a path through space and time. It represents, for example, the complete history of you, sitting down, running, flying in a plane or a rocket. Even if use coordinates where your world line is always of the form (t,0,0,0) (that is, you are always at the origin of your coordinate system, the only change in your coordinates being your wristwatch time), this does not make either your 4-velocity or 4-acceleration zero. The reason is that the derivative I referred to is really the covariant derivative (this involves a term called the connection as well as ordinary derivatives). One qualitative statement I can make is that at any point on your world line, there is a unique geodesic tangent to it. This geodesic would be the free fall path of an object you let go of at that point. To the extent that object moves away from you, you have proper acceleration.

 

[Zula110100100]

DH: With respect to my free-falling frame, at no point in this am I saying there is one correct coordinate system.

(that is, you are always at the origin of your coordinate system, the only change in your coordinates being your wristwatch time), this does not make either your 4-velocity or 4-acceleration zero

Right, so my 4-velocity would be a constant (ct, 0, 0, 0)?

 

[PAllen]

DH: With respect to my free-falling frame, at no point in this am I saying there is one correct coordinate system.



Right, so my 4-velocity would be a constant (ct, 0, 0, 0)?

Yes, you could construct such coordinates that your 4-velocity would be always be (1,0,0,0) in units of c=1. However, your proper acceleration vector would not, therefore, be zero unless an object you let go of stayed with you.

 

[Zula110100100]

What are the dr and dt in the swartzchild equation relative to, the center of the body M?