Gravitational Attraction & Conservation of Energy

[PAllen]

I think the answer is no. I can only assume you misread the question and did not notice that the tether is 1km long in both instances.

I think I am right. The difference is the 'potential' change between 1 million and 1 million + 1 km (from SC radius) is infinitesimal; while 1 km and 2 km from SC radius have very large 'gravitational potential difference'. The former amounts to measurement in flat spacetime; the latter not even close. I didn't emphasize this as I thought the purpose of the formulation was obvious.

 

[yuiop]

At 1 million+1 km out, do we observe a larger stretch on the spring than at 2 km out?

General physical arguments suggest the answer is yes, the spring would have larger stretch at the greater distance. ...

I think the answer is no. ...

I think I am right. The difference is the 'potential' change between 1 million and 1 million + 1 km (from SC radius) is infinitesimal; while 1 km and 2 km from SC radius have very large 'gravitational potential difference'. The former amounts to measurement in flat spacetime; the latter not even close. I didn't emphasize this as I thought the purpose of the formulation was obvious.

Let's do the calculations. Let's say that that the gravitational force (F) acting on the one kilogram mass stretches the spring by one picometres at one million km outside the EH (The additional 1km of the tether is not important at this radius). The redshift factor at this radius is unity to the accuracy we require here.

The radius of a one Solar mass black hole is approximately 3km. The redshift factor at 1 km outside the event horizon is:

\frac{1}{\sqrt{1-\frac{3km}{(3+1)km}}} = 2

The gravitational force acting on the mass 2km outside the event horizon is (GM/r^2)*2 = F * (10^6)^2/(4)^2*2 = F * 625,000,000,000 Newtons.

The stretch of the spring measured locally at altitude 4km is 625,000,000,000 picometres (0.625 metres).

Now we raise the spring balance by 1km and attach it to the mass by a 1km tether and find that the spring stretches by the same amount (if the tether is considered to have negligable mass and if we ignore errors due to the stretching of the spring itself).

Note the claim I am making here. The locally measured stretch of a spring attached by a massless tether to a mass at a given fixed radius from a black hole, is independent of the length of the tether.

Summary:

At 1,000,001 Kms out, the spring stretches by 1 picometre.
At 2 Kms out, the spring stretches by 625,000,000,000 picometres.

Now we can answer the question "At 1 million+1 km out, do we observe a larger stretch on the spring than at 2 km out?" and the answer is no.

The stretch of the spring when the mass is further away from the gravitational source is reduced by the Newtonian 1/r^2 relationship and additional reduced by the GR redshift factor.

 

[PAllen]

Let's do the calculations. Let's say that that the gravitational force (F) acting on the one kilogram mass stretches the spring by one picometres at one million km outside the EH (The additional 1km of the tether is not important at this radius). The redshift factor at this radius is unity to the accuracy we require here.

The radius of a one Solar mass black hole is approximately 3km. The redshift factor at 1 km outside the event horizon is:

\frac{1}{\sqrt{1-\frac{3km}{(3+1)km}}} = 2

The gravitational force acting on the mass 2km outside the event horizon is (GM/r^2)*2 = F * (10^6)^2/(4)^2*2 = F * 625,000,000,000 Newtons.

The stretch of the spring measured locally at altitude 4km is 625,000,000,000 picometres (0.625 metres).

Now we raise the spring balance by 1km and attach it to the mass by a 1km tether and find that the spring stretches by the same amount (if the tether is considered to have negligable mass and if we ignore errors due to the stretching of the spring itself).

Note the claim I am making here. The locally measured stretch of a spring attached by a massless tether to a mass at a given fixed radius from a black hole, is independent of the length of the tether.

Summary:

At 1,000,001 Kms out, the spring stretches by 1 picometre.
At 2 Kms out, the spring stretches by 625,000,000,000 picometres.

Now we can answer the question "At 1 million+1 km out, do we observe a larger stretch on the spring than at 2 km out?" and the answer is no.

The stretch of the spring when the mass is further away from the gravitational source is reduced by the Newtonian 1/r^2 relationship and additional reduced by the GR redshift factor.

I don't see that your calculation applies to my scenario. It seems I was not sufficient clear in my explanation. I am measuring inertial mass at a distance (1 km), not passive gravitational mass. I'll try again:

The test mass and observer 1 km away are initially in free fall in such a way that the observer initially, instantaneously sees zero speed on the tether. Then, a combination of forces to the end of the spring attached to the tether and supporting forces on the test body are applied such that the acceleration of the end of the tether next to the observer is exactly 9.81 m/sec^2 (in the direction away from the central mass). Under these conditions, the spring will have some stretch that is dependent on the inertial mass of the test body as observed by the observer at 1km further away from central mass.

[edit: More clarification on division of forces: at some starting moment, the observer is in free fall with zero radial speed; the test body has force on it adjusted such that essentially zero force on the spring will keep the tether stationary relative to the observer. Now, additional force is applied to the spring such that the tether end has 9.81 m/sec^2 outward radial acceleration measured by the observer.]

 

[yuiop]

[edit: More clarification on division of forces: at some starting moment, the observer is in free fall with zero radial speed; the test body has force on it adjusted such that essentially zero force on the spring will keep the tether stationary relative to the observer. Now, additional force is applied to the spring such that the tether end has 9.81 m/sec^2 outward radial acceleration measured by the observer.]

OK let's say we have an a very rigid rod 1km long that connects your hand to the weight lower down with no spring. Now let's say the the gravitational gamma factor ratio over that distance is 2. When you accelerate the top end of the rod by 9.81 m/s^2 the lower end of the rod (as measured by an observer local to the weight at the bottom) will accelerate by 8*9.8 m/s^2 and so it will feel as if you are accelerating a much larger mass or in other words it feels like the mass has more inertia. It will require gamma^3 more force to accelerate the top end of the rod by 9.8m/s^2 than it will take to accelerate the same mass by the same amount if it was in your hand at the top position. If we put a spring at the top end of the rod with the weight back at the bottom, it will stretch gamma^3 more than when we attach the weight directly to the bottom of the spring and dispense with the rod. I am of course assuming a massless rod.