Gravitational Attraction & Conservation of Energy

[Jonathan Scott]

If you're not already confused, here's another point to consider ...

In the Newtonian view and in the semi-Newtonian model with GR, the fact that a free-falling object has constant energy means that NO energy is being transferred, only momentum. Any change in the kinetic energy is balanced by a change in the potential energy, which is effectively part of the rest energy.

This is very different from electromagnetic forces, where the rest energy remains fixed and the kinetic energy adds to it.

 

[sonmage]

I like what you say Jonathan, but my perspective is slightly different. In my view the fundamental particles have a very precise energy rerquirement so that every electron is exactly like every other electron. But this precise requirement changes with changes in motion and gravitational field. When the electron is moving fast relative to the universe it requires more energy. When the gravitational field gets stronger it requires less energy. When protons and neutrons are packed in a nucleus they require less energy. So when the excess energy is released we call that nuclear or kinetic energy. When extra energy is required at high altitudes we call that potential energy.

 

[yuiop]

You seem to have missed an important point here. I'm not talking about the weight of the mass, which depends on the gravitational field, and will therefore generally be greater closer to the source. I'm talking about the inertial mass, which can for example in theory be determined by the response to a change in the force in the tether.

OK let's attach our test mass to an inertial balance higher up by a lightweight, but long and fairly rigid rod. The longer the period of the oscillation of the inertial balance, the greater the inertial mass. Let's say local to the test mass, the test mass is observed to oscilate with a period of T, then higher up local to the inertial balance, the apparatus is observed to oscilate with an increased period of T*f where f is the gravitational gamma red shift factor of 1/√(1-2GM/rc^2). This means using your method, the measured inertial mass of the test weight higher up is greater by the red shift factor, than when measured locally lower down and not reduced as you have been suggesting.

 

[Jonathan Scott]

OK let's attach our test mass to an inertial balance higher up by a lightweight, but long and fairly rigid rod. The longer the period of the oscillation of the inertial balance, the greater the inertial mass. Let's say local to the test mass, the test mass is observed to oscilate with a period of T, then higher up local to the inertial balance, the apparatus is observed to oscilate with an increased period of T*f where f is the gravitational gamma red shift factor of 1/√(1-GM/r^2). This means using your method, the measured inertial mass of the test weight higher up is greater by the red shift factor, than when measured locally lower down and not reduced as you have been suggesting.

Your suggestion does not take into account the fact that as the inertial balance is at a higher potential, the clocks and forces involved in the mechanism relate to that potential, not to the lower potential.

Regardless of the details of the mechanism, if you have a balance which is calibrated with a standard mass at the higher potential, then in theory it would register a lower value (so for a conventional inertial balance, it would oscillate faster than for the standard mass, compared with a local clock) when connected to an identical object at a lower potential.

The difference in the oscillation rate as seen from different potentials is not relevant. It is the difference between the oscillation rate in the two different situations (with the same balance connected to local mass or lower mass), as seen from either location, which is important.

Your red-shift factor time dilation factor is a bit mixed up. I'd use (1-GM/rc^2) as the relative clock rate at distance r from mass M. If you want to be more accurate in Schwarzschild coordinates, you can use √(1-2GM/rc^2).

 

[yuiop]

Your suggestion does not take into account the fact that as the inertial balance is at a higher potential, the clocks and forces involved in the mechanism relate to that potential, not to the lower potential.

It was you that suggested that we use a measurement of forces higher up via a tether to a mass lower down to show an effective reduction in inertial mass. When I demonstrated that the resulting measurement shows that the inertial mass measured by this this method actually is greater rather than reduced you respond that it does not relate to forces and clocks lower down. Of course it does not. If we wanted to relate to clocks and forces at the same potential as the test mass, we would simply make a local measurement and dispense with the tether. When we do that, we so no change in the inertial mass at any height.

Regardless of the details of the mechanism, if you have a balance which is calibrated with a standard mass at the higher potential, then in theory it would register a lower value (so for a conventional inertial balance, it would oscillate faster than for the standard mass, compared with a local clock) when connected to an identical object at a lower potential. The difference in the oscillation rate as seen from different potentials is not relevant. It is the difference between the oscillation rate in the two different situations (with the same balance connected to local mass or lower mass), as seen from either location, which is important.

Now we are getting to the crux of the matter. I would suggest that the oscillation frequency higher up would be identical to the oscillation period measured lower down using local clocks. The frequency of the inertial balance is a primitive form of clock will appear to "tick" at the same rate locally, anywhere in the field.

Your red-shift factor time dilation factor is a bit mixed up. I'd use (1-GM/rc^2) as the relative clock rate at distance r from mass M. If you want to be more accurate in Schwarzschild coordinates, you can use √(1-2GM/rc^2).

Yes you are right. That was a typo. I have now corrected it. Thanks. It does not however change the material argument about a lowered mass have the same inertia measured locally as it did higher up and a greater inertia when measured from higher up at a distance. No one measures a given mass to have less inertia when it is lowered in a gravitational field.

 

[PAllen]

It does not however change the material argument about a lowered mass have the same inertia measured locally as it did higher up and a greater inertia when measured from higher up at a distance. No one measures a given mass to have less inertia when it is lowered in a gravitational field.

This is the crux of the discussion. I have never calculated, in a way I am satisfied with, nor read through a calculation I am satisfied with, of the following question (therefore I propose no answer):

Assuming GR, plus some notion of atoms:

Imagine 10^10 atoms of palladium at 0 degrees c measured locally to the palladium. Imagine pulling on it with a 'massless' tether from 1 km further away from the only large mass in the universe. Imagine the 'pulling' produces a local (to far end of tether) coordinate acceleration relative to free fall of 9.81 m/sec^2. We measure the stretch on a spring as this is done. We perform this experiment at 1 million km from a solar mass SC event horizon , and at 1 km beyond a solar mass event horizon (that is, palladium 1 km outside horizon, spring 2 km outside). Question:

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

 

[PAllen]

This is the crux of the discussion. I have never calculated, in a way I am satisfied with, nor read through a calculation I am satisfied with, of the following question (therefore I propose no answer):

Assuming GR, plus some notion of atoms:

Imagine 10^10 atoms of palladium at 0 degrees c measured locally to the palladium. Imagine pulling on it with a 'massless' tether from 1 km further away from the only large mass in the universe. Imagine the 'pulling' produces a local (to far end of tether) coordinate acceleration relative to free fall of 9.81 m/sec^2. We measure the stretch on a spring as this is done. We perform this experiment at 1 million km from a solar mass SC event horizon , and at 1 km beyond a solar mass event horizon (that is, palladium 1 km outside horizon, spring 2 km outside). Question:

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. But what I haven't been able to do is demonstrate this in a convincing way using the SC metric.

 

[Jonathan Scott]

... It does not however change the material argument about a lowered mass have the same inertia measured locally as it did higher up and a greater inertia when measured from higher up at a distance. No one measures a given mass to have less inertia when it is lowered in a gravitational field.

This area is certainly tricky, but Kenneth Nordtvedt asserts that the effective inertial mass of a system is decreased by its binding energy, and so is the gravitational mass, and the consistent relationship between the two was verified by a null "Nordtvedt effect" result in the Apollo-era Lunar Laser Ranging experiments.

According to Nordtvedt, the effective decrease in inertial mass can be alternatively viewed as being due to "self-acceleration" linear frame-dragging within the object. For more details, see his article "Some considerations on the varieties of frame dragging", which can be found in "Nonlinear Gravitodynamics" in Google Books.

 

[yuiop]

Imagine 10^10 atoms of palladium at 0 degrees c measured locally to the palladium. Imagine pulling on it with a 'massless' tether from 1 km further away from the only large mass in the universe. Imagine the 'pulling' produces a local (to far end of tether) coordinate acceleration relative to free fall of 9.81 m/sec^2. We measure the stretch on a spring as this is done. We perform this experiment at 1 million km from a solar mass SC event horizon , and at 1 km beyond a solar mass event horizon (that is, palladium 1 km outside horizon, spring 2 km outside). Question:

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 can only assume you misread the question and did not notice that the tether is 1km long in both instances.

 

[Q-reeus]

Having got no feedback from #43-45 was going to let this one die gracefully, but best to set the record straight as to my current position. Now realize conservation of L only holds for the total system = flywheel+central mass, not flywheel on it's own. That accords with the Machian principle of action-reaction: if central mass M acts to alter the inertia of flywheel masses, it must work both ways and therefore one expects by analogy with EM induction that say lowering a flywheel down a potential well will induce a frame-dragging change in L of the central mass generating that potential. Thus flywheel L must decrease, as unlike EM induction, the sign is such that induced L of the central mass has the same sign of the flywheel L.

And of course coordinate velocties v need to be expressed as normalized wrt to coordinate c for fully consistent expressions for KE, L etc., although at low speeds the simpler expressions involving just v work ok. Overall I maintain the above still presents real trouble re consistency for standard Schwarzschild coordinate evaluation, but somewhat differently than I had got earlier. Enough for now.

 

[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.