Gravitational Attraction & Conservation of Energy

[yuiop]

Sorry yuiop but I believe your earlier analysis was closer to the mark. It was an initial confusion over using I = md² (d the bar length) that had me making the mistake of finding an f^-3 rather than f^-2 dependence on inertial mass for radial bar orientation, assuming that angular velocity goes as f^-1 in that instance, but evidently that cannot be applied here. [EDIT: yes it is applicable - angular velocity in this bar orientation is independent of f!]

If I express I = md² as I = mr² and insert that expression into L = Iω you get L = mr²ω = mr²(v/r) = mrv which I believe is the form you were using, but they are all essentially interchangeable. However, the L = mrv form allows us play around with the instantaneous velocity at a given position which I will come to.

Anyway the instantaneous definition L = rxp = rx(mv) is always valid in this kind of situation - one could suddenly stop the flywheel at any angular orientation and that definition must apply to the angular momentum transferred from flywheel to environment.

OK, just to clear up a few things. Angular momentum is never converted to linear momentum. I learned that the hard way. Angular momentum and linear momentum are independently conserved. If the rotating bar with 1kg masses on the end of it collides with two stationary masses of 1kg each, the unattached masses move of with the same velocity as the instantaneous velocities of the masses on the bar and the bar stops rotating of course. The linearly moving masses still have an effective angular momentum around an imaginary axis (and ever increasing radius) and there is no change in angular momentum. The total linear momentum before the collision was zero and after the collision it is still zero because the masses are heading in opposite directions. We cannot break down the angular momentum into sub components, but we can split the linear momentum. If the collision occurs when the bar is horizontal, the velocity is vertical and the mass scales as 1/f^3 and the velocity scales as f^2 for any vertically moving mass whether it is part of a flywheel or an independent linearly moving object. This inertial factor of 1/f^3 for the inertial mass is consistently used for instantaneous angular momentum, linear momentum, collisions, linear kinetic energy and angular kinetic energy and gravitational mass. The inertial scaling factor of 1/f is used for acceleration or collision of masses in the horizontal direction and again this is universally applied for horizontal motion a gravitational field.

So it remains 'true' that by standard SC's, sensibly imposing conservation of L, inertial mass goes as f^-1 for radial axis orientation, but as f^-2 for transverse axis orientation (any bar angle). And that is in actuality nonsense. It must be invariant wrt to rotation axis orientation.

Yes it is nonsense, but for a different reason. Imagine two flywheels, one vertical and one horizontal that are meshed together at their edges by gear teeth. At the point where they mesh at the lower edge of the vertical flywheel the mass at the bottom edge is moving horizontally in parallel with the mass on the edge of the horizontal flywheel and at the same speed so the must be subject to the same gamma factor which if 1/f for low v. Vertically moving masses scale by an inertial factor of 1/f^3 consistently. When you apply these scaling factors, angular momentum for a flywheel is conserved at any altitude at any orientation (using L = Iω) and angular (or linear) kinetic energy scales as f relative to the rest mass for any orientation. I agree that it does not seem entirely satisfactory that the angular momentum is different when the bar is vertical than when it is horizontal, but it all works out in the larger scheme of things. If you do some research you will find references to inertial mass scaling as gamma and gamma cubed for horizontal and vertical mass (or parallel and transverse mass in SR) in the literature. We do not usually call it relativistic mass these days, so you can think of it as the directional inertia scale factor of a given rest mass.

Note that in an earlier post, I calculated the vertical inertial scaling factor of 1/f^3 independently by dividing the force of gravity by the acceleration of gravity. Note that I also independently calculated the horizontal inertial scaling factor of 1/f for a linear projectile, so at least there is some consistency, as those calculation agree with the conclusions obtained by analysing the flywheel scenario you presented.

P.S> If you are still bothered about the instantaneous angular momentum being different at dfifferetn locations on the vertical flywheel consider the case of a rolling flywheel in SR. It the centre of the flywheel is moving horizontally at v, then the bottom of the flywheel is stationary, the top of the flywheel is moving at 2*v and the two edges left and right of the centre are moving at something like sqrt(2v^2) so it should be obvious that the instantaneous angular momentum in the SR case is not constant at all points on the rim either, so we do not need to blame SC's for the mess, as it is an intrinsic feature of relativity even when gravity is not involved.

 

[yuiop]

To more directly address the question in the OP:

Imagine you measure inertial mass (using inertial ballance) of an object far from a gravitating mass which is at rest relative to the object. Then you release the object and let it inertially fall towards the gravitating mass with no atmosphere. On the surface of the mass you stop the object so that not even an atom from it was lost during the fall. You measure its inertial mass again (with the inertial ballance again). Will it be the same as it was in space?

the answer is that yes, the locally measured mass using an inertial balance (measuring osculation periods of a mass on the end of a flexible rod) will be exactly the same low down in a gravitational well as it is faraway in space. The same is true for any other local measurement of mass that you can perform. There is no loss of mass with loss of potential energy as was suggested earlier in the thread. Schwarzschild coordinate calculations of the behaviour of a test mass by a distant observer suggest the horizontal inertia of the mass increases by the gravitational gamma factor and the vertical inertia increases by a factor of gamma cubed. No measurements reveal an apparent loss of mass. If mass was reduced by a factor of gamma after being brought to a stop and dispersing the kinetic energy, then the force of gravity on the object would be GM/r^2 and this has serious consequences, as it implies it would be possible to hold station at the event horizon of a BH and we would have to rethink our entire concept of what a BH is. Then again, loss of mass also suggests that any object arriving at the event horizon would be all kinetic energy and no rest mass, which is equally problematic.

In short, the answer to the title question of this thread "Re: Do objects higher in the gravitational field have higher inertial mass than lower" is no.

 

[Jonathan Scott]

I haven't had time to catch up with all the side-tracks on this thread, but I have a few comments.

In a static isotropic metric (even if the time and space factors differ from GR) the rate of change of coordinate momentum always points exactly towards the gravitational source, so the field cannot affect the coordinate angular momentum of a test mass as a whole or of its components.

I thought we all agreed inertial mass measured locally is unaffected by gravity.

The fact that apparent accelerations are modified by gamma factors in various coordinate systems is irrelevant, as the acceleration of a test particle in a gravitational field does not depend on its mass, and depends on the choice of coordinate system.

Conceptually, one could measure the inertial mass of a test particle at one potential from another by use of a light string or rod, for example by hanging a mass in a tower and measuring the mass of the suspended object by varying the force and determining the resulting acceleration (assuming the gravitational force to be constant). In practice, this is far too small to measure, but theory indicates that the inertial mass measured in this way would be lower at a lower potential. The Pound-Rebka experiment does the equivalent of this with photons and demonstrates that photons created with a locally standard energy at a lower potential have lower energy when observed at a higher potential.

 

[ZirkMan]

I thought we all agreed inertial mass measured locally is unaffected by gravity.

The fact that apparent accelerations are modified by gamma factors in various coordinate systems is irrelevant, as the acceleration of a test particle in a gravitational field does not depend on its mass, and depends on the choice of coordinate system.

Conceptually, one could measure the inertial mass of a test particle at one potential from another by use of a light string or rod, for example by hanging a mass in a tower and measuring the mass of the suspended object by varying the force and determining the resulting acceleration (assuming the gravitational force to be constant). In practice, this is far too small to measure, but theory indicates that the inertial mass measured in this way would be lower at a lower potential. The Pound-Rebka experiment does the equivalent of this with photons and demonstrates that photons created with a locally standard energy at a lower potential have lower energy when observed at a higher potential.

Jonathan, just to make sure I understand. Is it correct to sum up your answer to the question like this:

A. if we measure the inertial mass in space and note the result number and then we let fall the experiment (the test object + the inertial balance) as described and repeat it in the lower gravitational level and compare the results the numbers will be the same. (this is what I understand to be comparisons of local measurements).

B. if we measure the inertial mass in space and note the result number and then we let fall only the test object as described and repeat the experiment in such a way that we lower the measuring rods of the inertial balance from higher gravitational potential and then we make the measurements the results numbers will differ. The inertial mass lower will be smaller compared to the measurement in space (this is what I understand to be comparisons of relative non-local measurements).

Correct?

 

[Dale]

In special relativity Minkowski coordinates, angular momentum is a rank-2 tensor$$L^{ab} = x^aP^b - x^bP^a$$(the wedge product). The three components (L²³, L³¹, L¹²) form the 3-vector angular momentum. In curved spacetime general coordinates, I believe it is undefined (the expression above won't work because x^a isn't a 4-vector (i.e. in the tangent space), it's the coordinates of an event on the curved manifold).

Hi DrGreg, thanks, that is interesting. Do you know if there is a further generalization of that expression to curved spacetimes? If not, then I guess you would simply need to use the stress-energy tensor and not make a simplification of talking about angular momentum of an extended object.

 

[Passionflower]

...the acceleration of a test particle in a gravitational field does not depend on its mass

That is because a test particle is defined as a particle that has negligible mass.

What you are saying is a tautology not science.

 

[Jonathan Scott]

Jonathan, just to make sure I understand. Is it correct to sum up your answer to the question like this:

A. if we measure the inertial mass in space and note the result number and then we let fall the experiment (the test object + the inertial balance) as described and repeat it in the lower gravitational level and compare the results the numbers will be the same. (this is what I understand to be comparisons of local measurements).

B. if we measure the inertial mass in space and note the result number and then we let fall only the test object as described and repeat the experiment in such a way that we lower the measuring rods of the inertial balance from higher gravitational potential and then we make the measurements the results numbers will differ. The inertial mass lower will be smaller compared to the measurement in space (this is what I understand to be comparisons of relative non-local measurements).

Correct?

Yes, I think that's right (assuming that by "in space" you are just referring to the original location as opposed to the lower location).

 

[Q-reeus]

In curved spacetime general coordinates, I believe it is undefined (the expression above won't work because xa isn't a 4-vector (i.e. in the tangent space), it's the coordinates of an event on the curved manifold).

Getting down to specifics, does undefined equate to generally nonconserved? Was under the impression that L is a strictly conserved quantity even in GR - fire a spinning test mass at say a notional BH and final state will have the BH acquiring precisely the initial spin+orbital L inherent in the test mass (neglecting any tiny GW's given off). If that were not so, it should not be hard to come up with a cyclic perpetual motion scenario.

 

[Q-reeus]

OK, just to clear up a few things. Angular momentum is never converted to linear momentum. I learned that the hard way. Angular momentum and linear
momentum are independently conserved.

Of course, but there was no suggestion on my part otherwise. What is important is that one cannot simply accept an 'average' L that otherwise varies with instantaneous bar angular orientation - we impose that L is invariant for any such instantaneous angle. Otherwise a cyclic process could generate an unlimited amount of angular momentum.

If the collision occurs when the bar is horizontal, the velocity is vertical and the mass scales as 1/f^3 and the velocity scales as f^2 for any vertically moving mass whether it is part of a flywheel or an independent linearly moving object.

Velocity scaling as f² is correct, but inertial mass scaling other than as 1/f² leads to L not being invariant wrt potential. Remember - moment arm is unaffected by potential in this orientation (btw I/we should have used I = 1/2md² not md² for moment of inertia of bar-masses, but that's unimportant re principle here).

This inertial factor of 1/f^3 for the inertial mass is consistently used for instantaneous angular momentum, linear momentum, collisions, linear kinetic energy and angular kinetic energy and gravitational mass.

A mixed bag - even within this specifically radial motion orientation case. Apart from my last comment re invariant L where f^-2 scaling is needed, it is true that mass scaling as f^-3 gives for this orientation the correct coordinate scaling as f for KE (whether angular or linear KE is immaterial), when defined as simply 1/2mv² (v<<c) for coordinate or local measure. A serious problem; a single scaling factor aught to apply. It most certainly does not work for gravitational mass. As per #20, apply collapsing masses scenario there to a radial path, and we find that assuming active mass m_i = passive mass m_p, it scales as f. But *only* for this radial path case - using SC's. Of course one could rejig things to have an f^-3 or f^-2 m_a, m_p scaling, - but only by making the 'universal' constant G scale as f⁸ or f⁶ respectively!

The inertial scaling factor of 1/f is used for acceleration or collision of masses in the horizontal direction and again this is universally applied for horizontal motion a gravitational field.

This gets back to my input in #22 (I mistakenly confused your gamma with my f there); consistency is simply lacking overall if SC's are applied. We found the Newtonian force F = m₁m₂G/r² scales as f. There are various options for this particular transverse path case - make G scale as f, or m₁, m₂ scale each as f^1/2, etc. Either way, no way does gravitational mass scaling as 1/f make any sense. Yet that scaling does work here for inertial mass yielding correctly invariant L and KE scaling as f. An ad hoc hodge podge.

...Vertically moving masses scale by an inertial factor of 1/f^3 consistently.

Do not agree. I went back and reading #21 can see how you got that result. That acceleration goes as f³ is correct, but wrong to then apply m = F/a by using the *local* definition for F. Coordinate F value scales as f, and that's the one to use. Which then gives inertial mass scaling as f^-2 (radial direction).

When you apply these scaling factors, angular momentum for a flywheel is conserved at any altitude at any orientation (using L = Iω) and angular (or linear) kinetic energy scales as f relative to the rest mass for any orientation. I agree that it does not seem entirely satisfactory that the angular momentum is different when the bar is vertical than when it is horizontal, but it all works out in the larger scheme of things...

No - see my earlier comment re cyclic process that would allow.

Note that in an earlier post, I calculated the vertical inertial scaling factor of 1/f^3 independently by dividing the force of gravity by the acceleration of gravity. Note that I also independently calculated the horizontal inertial scaling factor of 1/f for a linear projectile, so at least there is some consistency, as those calculation agree with the conclusions obtained by analysing the flywheel scenario you presented.

Not consistently, for reasons given before. Transverse 1/f scaling yes, radial 1/f³ scaling no - 1/f² 'works' in the SC setting.

All in all this is screaming loudly to me that Schwarzschild metric (not just SC's) is where the problem lies. A properly isometric metric (for a static spherical mass) imo cures all these difficulties and more - but then we do not have GR.

 

[Q-reeus]

In a static isotropic metric (even if the time and space factors differ from GR) the rate of change of coordinate momentum always points exactly towards the gravitational source, so the field cannot affect the coordinate angular momentum of a test mass as a whole or of its components.

We agree on that.

I thought we all agreed inertial mass measured locally is unaffected by gravity.

I sure do. We are on about what coordinate values work out.

The fact that apparent accelerations are modified by gamma factors in various coordinate systems is irrelevant, as the acceleration of a test particle in a gravitational field does not depend on its mass, and depends on the choice of coordinate system.

Maybe you should have a crack then at getting consistent results for the variously spin axis oriented twin-mass flywheel case we have been looking at.

Conceptually, one could measure the inertial mass of a test particle at one potential from another by use of a light string or rod, for example by hanging a mass in a tower and measuring the mass of the suspended object by varying the force and determining the resulting acceleration (assuming the gravitational force to be constant). In practice, this is far too small to measure, but theory indicates that the inertial mass measured in this way would be lower at a lower potential. The Pound-Rebka experiment does the equivalent of this with photons and demonstrates that photons created with a locally standard energy at a lower potential have lower energy when observed at a higher potential.

Lower energy does not equate to lower inertial mass here - the opposite is so. To repeat, consider seeing whether you can make your position work consistently for flywheel case - general orientation!

 

[yuiop]

I thought we all agreed inertial mass measured locally is unaffected by gravity.

Yes, I think we are all agreed on that with the stress on the local part. It is coordinate or distant measures that are under dispute here. Also, I think it is fair to state that some of the earlier posts hinted at locally measured mass reducing or at least did not make it clear that when they suggested the mass reduces they were talking about coordinate measurements and even then they are wrong because coordinate weight increases.

The fact that apparent accelerations are modified by gamma factors in various coordinate systems is irrelevant, as the acceleration of a test particle in a gravitational field does not depend on its mass, and depends on the choice of coordinate system.

True, but we should have a consistent way of defining force in a given coordinate system and a consistent way of defining rest mass that is coordinate independent. That is what we are looking for.

Conceptually, one could measure the inertial mass of a test particle at one potential from another by use of a light string or rod, for example by hanging a mass in a tower and measuring the mass of the suspended object by varying the force and determining the resulting acceleration (assuming the gravitational force to be constant). In practice, this is far too small to measure, but theory indicates that the inertial mass measured in this way would be lower at a lower potential.

If we have the test mass attached to a tether as you describe the force higher up would not be the force measured lower down reduced by the redshift factor. I can confidently state this, because if it was true I would be able to lower a test mass on the end of a tether to an event horizon and hold it stationary there using only a finite force.

The Pound-Rebka experiment does the equivalent of this with photons and demonstrates that photons created with a locally standard energy at a lower potential have lower energy when observed at a higher potential.

All that demonstrates is that a system lower down has a lower coordinate energy than when it higher up. You are using the fact that all energy contributes to the gravitational effect and the mass energy equivalence, to jump to the conclusion that mass lower down weighs less. While it might seem a reasonable superficial conclusion it is not correct when looked at in detail. When a small mass (m) is lowered towards a large mass (M) the combined gravitational effect of (M+m) is lower than when they were further apart (if the binding energy is radiated away) so the combined mass of the total system is less, but the lost potential energy was not initially localised in m but is stored in the gravitational field. Gravity itself has an effective mass and contributes to the total gravity of the system in a weird and non linear way.

The fact is that if we lower a battery (or other energy store) the stored energy appears to be reduced when viewed from higher up, whether we use the energy to power a light source, spin up a flywheel or launch a projectile, but if we try to launch the battery itself (with its reduced energy content) it behaves as if has more inertial mass or if we drop the battery it behaves as if it has more mass, which can appear paradoxical because we usually assume less energy = less mass = less weight = less inertia but it does not work like that in this case. I am of course talking about distant coordinate measurements because locally the battery and its energy content does not appear to have changed at all.

Lets try the converse. Let say we raise a small mass m up a tower away from a much larger mass M. This requires we put more energy into the system to raise the small mass. The total energy and mass of the system (M+m) has increased from the point of view of a distant observer. This increase in mass of the total system is not present in M or m, or though we can note that the coordinate stored energy of m appears to have increased, there is no way we can measure an increase in the mass of m locally or from a distance. The additional energy must therefore be elsewhere in the system. One candidate for the location of the energy is in the tower which is now under increased stress supporting the battery at the top compared to when it was at the bottom.

 

[DrGreg]

Hi DrGreg, thanks, that is interesting. Do you know if there is a further generalization of that expression to curved spacetimes? If not, then I guess you would simply need to use the stress-energy tensor and not make a simplification of talking about angular momentum of an extended object.

Sorry, I don't know much more about this topic than what I wrote.

In GR, you can still have local conservation laws (in the form of differential equations) but for non-local conservation there has to be an appropriate symmetry present. In the case of angular momentum, that means rotational symmetry. (Yuiop alluded to this in another recent thread The weight of flywheel, post #9.)

 

[Jonathan Scott]

If we have the test mass attached to a tether as you describe the force higher up would not be the force measured lower down reduced by the redshift factor. I can confidently state this, because if it was true I would be able to lower a test mass on the end of a tether to an event horizon and hold it stationary there using only a finite force.

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.

 

[Jonathan Scott]

You are using the fact that all energy contributes to the gravitational effect and the mass energy equivalence, to jump to the conclusion that mass lower down weighs less. While it might seem a reasonable superficial conclusion it is not correct when looked at in detail. When a small mass (m) is lowered towards a large mass (M) the combined gravitational effect of (M+m) is lower than when they were further apart (if the binding energy is radiated away) so the combined mass of the total system is less, but the lost potential energy was not initially localised in m but is stored in the gravitational field. Gravity itself has an effective mass and contributes to the total gravity of the system in a weird and non linear way.

The fact is that if we lower a battery (or other energy store) the stored energy appears to be reduced when viewed from higher up, whether we use the energy to power a light source, spin up a flywheel or launch a projectile, but if we try to launch the battery itself (with its reduced energy content) it behaves as if has more inertial mass or if we drop the battery it behaves as if it has more mass, which can appear paradoxical because we usually assume less energy = less mass = less weight = less inertia but it does not work like that in this case. I am of course talking about distant coordinate measurements because locally the battery and its energy content does not appear to have changed at all.

Lets try the converse. Let say we raise a small mass m up a tower away from a much larger mass M. This requires we put more energy into the system to raise the small mass. The total energy and mass of the system (M+m) has increased from the point of view of a distant observer. This increase in mass of the total system is not present in M or m, or though we can note that the coordinate stored energy of m appears to have increased, there is no way we can measure an increase in the mass of m locally or from a distance. The additional energy must therefore be elsewhere in the system. One candidate for the location of the energy is in the tower which is now under increased stress supporting the battery at the top compared to when it was at the bottom.

This contains multiple errors.

The inertial mass of a total system (including any effective energy in associated fields, described by pseudotensors in GR) as seen from an external static coordinate system is decreased when its internal potential energy is decreased, provided of course that the relevant energy was removed from the system rather than being stored internally somewhere.

In the case where the objects being moved are of negligible mass compared with the gravitational source, a semi-Newtonian model of potential and kinetic energy provides a consistent model. If you want to also consider the source mass, you then also find you need to assume an effective energy density in the field as well, which complicates matters.

A tower supporting something does not store an amount of energy which is anywhere near to the potential energy. The stored energy is roughly equal to half of the weight times the distance by which the tower contracted when the weight was placed on it. If the object is placed on an elastic tower and released without damping, it will oscillate up and down by twice that distance, and the kinetic energy of the oscillation then has to be removed to stop it bouncing. The ratio of the stored energy to the total potential energy is therefore similar to the ratio of that displacement (with the factor of a half if at rest) to the height of the tower, and can be made smaller by the use of stiffer materials.

 

[sonmage]

From yuiop: "the lost potential energy was not initially localized in m but is stored in the gravitational field" This is the key question that underlies Zirkman's original question: In water falling on a turbine where does the kinetic energy come from ? Does it come from the Earth's gravitational field or does it come from the water and any fields it might have? The traditional view of GR is that the kinetic energy comes from the gravitational field. But I am not sure whether that is part of the essential core of GR or is an explanatory lore that has been added on afterwards. In any case I would like to give the arguments that the kinetic energy comes out of the total energy that was in the water before it fell. The first argument is that the total energy of an object in free fall appears to be constant. If the Earth in its elliptical orbit picked up its kineticc energy from the sun's gravitational field, the Earth's mass would be greater when it was closer to the sun. This could be detected by a change in the orbit of the moon or orbiting space ship. The second reason for concluding that the kinetic energy comes from the water is that the fractional reduction in clock rate is exactly equal to the fraction of kinetic energy over total energy. Thus, dT/T = KE/mc^2 = GMdr/r^2c^2 , where T is clock rate, KE is kinetic energy, M is the mass of the Earth and r is the distance to the center of the earth. Since clock rate is a property of matter and the change in clock rate of accelerated particles also correlates with change in mass (albeit negatively) these correlations are strong evidence that kinetic energy is part of the total energy of the matter particles.

 

[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.