Gravitational Attraction & Conservation of Energy

[ZirkMan]

I would like to better understand the nature of gravitational attraction and the law of conservation of energy.

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?

 

[K^2]

Inertial mass will be slightly lower on the ground. Basically, potential energy does contribute to the mass of an object, same as every other form of energy.

 

[Jonathan Scott]

I would like to better understand the nature of gravitational attraction and the law of conservation of energy.

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?

A local measurement will give the same result anywhere (ignoring such complications as energy due to stress of compression when held at rest in a gravitational field).

However, relative to a distant observer, the total energy of the object decreased at the point when the kinetic energy from the fall was removed to bring the object to rest. This means that the effective inertial mass of the object and its strength as a gravitational source have both decreased slightly as seen from a distance.

For a small object near a large gravitational source, one can get a useful model of how it works by assuming that the fractional change in energy is exactly determined by the time dilation effect, which causes all local clocks and frequencies to be slower closer to the gravitational source, where the fractional change in clock rate is approximately -GM/rc² at distance r from source mass M.

However, this model isn't entirely satisfactory, because by the time dilation model, the time dilation effect on the source mass caused by the change in position of the test mass would result in the source mass being changed in energy by exactly the same amount. There are various other ways of trying to account for the location and flow of gravitational energy in GR, but it's a complex subject.

 

[ZirkMan]

A local measurement will give the same result anywhere (ignoring such complications as energy due to stress of compression when held at rest in a gravitational field).
However, relative to a distant observer, the total energy of the object decreased at the point when the kinetic energy from the fall was removed to bring the object to rest. This means that the effective inertial mass of the object and its strength as a gravitational source have both decreased slightly as seen from a distance.

Thank you. If I may, I would like to understand the difference between the local and the distant measurements. Why is there a difference in this case? Should not both the local and the distant observer see the same value on the inertial balance?

For a small object near a large gravitational source, one can get a useful model of how it works by assuming that the fractional change in energy is exactly determined by the time dilation effect, which causes all local clocks and frequencies to be slower closer to the gravitational source, where the fractional change in clock rate is approximately -GM/rc² at distance r from source mass M.

I suppose there is also the gravitational length contraction involved too. To get a complete picture of what is observed in the described situation, can we say that in the case of the gravitationaly falling object there is a mechanism that transforms its inertial mass to kinetic energy and at the same time slows down its time and contracts its length for a distant observer?

 

[sweet springs]

Hello.

Will it be the same as it was in space?

Yes, it is the same. Let us think of inertia measurements of the still matter in original gravitation system and in the local inertia system i.e. the elevator just starts falling. They give the same value.

Regards.

 

[Jonathan Scott]

Thank you. If I may, I would like to understand the difference between the local and the distant measurements. Why is there a difference in this case? Should not both the local and the distant observer see the same value on the inertial balance?

In the distant case, the effective change in mass is relative to equipment at the location of the distant observer. One could for example in theory measure this difference using a rod connecting the object to an inertial balance at the distant observer's location. In practice, the difference in mass is too small to measure directly, but this effect has been confirmed indirectly in the Pound-Rebka experiment which uses resonance effects to measure the fractional energy difference between photons emitted in an apparatus at one gravitational potential and received at a higher potential and those created at the higher potential.

I suppose there is also the gravitational length contraction involved too. To get a complete picture of what is observed in the described situation, can we say that in the case of the gravitationaly falling object there is a mechanism that transforms its inertial mass to kinetic energy and at the same time slows down its time and contracts its length for a distant observer?

The total energy of a falling object (or photon) as seen by any observer does not change while it is falling (at least in a static field). In Newtonian terms, potential energy converts to kinetic, but the total remains constant. It is the act of stopping the falling object which changes its total energy by taking away its kinetic energy.

An observer on the ground would say that the object originally had extra inertial mass because of its higher potential but after falling and being stopped it has the normal inertial mass. The distant observer would say that it originally had the normal inertial mass, but now has less. The difference in opinion is entirely due to their different clock rates.

 

[sonmage]

For a small object near a large gravitational source, one can get a useful model of how it works by assuming that the fractional change in energy is exactly determined by the time dilation effect, which causes all local clocks and frequencies to be slower closer to the gravitational source, where the fractional change in clock rate is approximately -GM/rc2 at distance r from source mass M.

The problem with the time dilation model is that with motion mass increases as clock rate decreases, exactly the opposite of what is observed with gravity.

 

[Ich]

The problem with the time dilation model is that with motion mass increases as clock rate decreases, exactly the opposite of what is observed with gravity.

Why, no, it works fine. As Jonathan Scott explained, the total energy is conserved in this model. You have rest mass and kinetic energy. The sum is conserved, the kinetic energy increases, rest mass gets less, so the ratio relativistic mass (=total energy) / rest mass increases, exactly as is observed.

 

[Jonathan Scott]

The problem with the time dilation model is that with motion mass increases as clock rate decreases, exactly the opposite of what is observed with gravity.

The effect of motion (kinetic energy) simply compensates for changes in gravitational (potential) energy, so it makes sense that it is opposite.

When an observer watches a falling object, its total energy remains constant. However, its effective inertial mass decreases (as its potential energy decreases) and its kinetic energy increases by the same amount. The kinetic energy increase is indeed the same factor as the decrease in the time rate of the falling object as seen by the fixed observer due to velocity-related time dilation.

If the falling object is stopped, the kinetic energy is removed, so what is left is the decreased inertial mass.

 

[sonmage]

Perhaps I misunderstood the model. I believe the time dilation model deals entirely with a falling object. In that case it makes sense that the clock rate determines the energy transfer from potential to kinetic. I was thinking in a broader sense of a single reality that includes particles accelerated to speeds near to c. In that situation mass increases with a reduction in clock rate. In this case it is impossible for the clock rate to determine energy change because energy and clock rate are negatively correlated.

 

[yuiop]

K^2, Ich and Jonathan seem to agree that the rest mass or inertial mass is less when a dropped object is brought to rest lower down. I would like to clarify or at least quantify this issue with some actual equations and try to come to an understanding of where and how gravitational potential energy is stored.

The Newtonian expectation is that the weight of a test object is:

$$\frac{GMm_0}{r^2}$$

where $m_0$ is the rest mass of the test object. One GR expression for weight (or proper force on an object at rest with respect to a gravitational field) that I have seen quoted is:

$$\frac{GMm_0}{r^2} \frac{1}{\sqrt{1-2GM/r}}$$

in units where c=1. Now if we drop an object and then bring it rest in such a way that its falling kinetic energy is dispersed, then if its rest mass reduces by a factor of $\sqrt{(1-2GM/r})$ then this suggests that weight of a stationary object reduces to the Newtonian expectation. Is that what we would actually measure using a set of bathroom scales? In some ways that seems to make sense as that conclusion seems to match the notion of "surface gravity" in discussions of Hawking radiation. It also seems to sit nicely the fact that the falling velocity of an object dropped from infinity is exactly in agreement with the Newtonian expectation $v = \sqrt{2GM/r}$ even in GR, when measured by a local observer.

Now there is that issue of where the potential energy is stored. In SR the total energy is given by:

$$E = \sqrt{m_0^2 + \frac{m_0^2 v^2}{(1-v^2)} }$$

Now in GR, does the total energy include an additional term for the gravitational potential energy and is this additional energy stored somewhere "out there" in the gravitational field, or is the potential energy already embodied in the rest mass?

What I do know is that for a falling body there is an expression in GR for total energy:

$$E = m_0 \sqrt{ (dr/dt) ^2 + (1-2GM/r)}$$

that is conserved for a free falling object. This is obvious when an object is dropped from infinity with velocity $dr/dt = \sqrt{2GM/r}$ because when this is substituted into the above expression the equation becomes $E = m_0$. Now the $m_0\sqrt{1-2GM/r}$ term is called the potential energy of the object and it seems to suggest that the potential energy of an object is in fact its inertial mass. Is that correct?

 

[Jonathan Scott]

Perhaps I misunderstood the model. I believe the time dilation model deals entirely with a falling object. In that case it makes sense that the clock rate determines the energy transfer from potential to kinetic. I was thinking in a broader sense of a single reality that includes particles accelerated to speeds near to c. In that situation mass increases with a reduction in clock rate. In this case it is impossible for the clock rate to determine energy change because energy and clock rate are negatively correlated.

The clock rate has an effect on the observer, not on the test particle, which has constant energy when on a free fall path (which doesn't necessarily mean literally falling downwards, but could be an orbit or just a fly-past).

This applies even for test particles at relativistic speeds in a static field. The total energy is constant, which can be seen as a change in the kinetic energy being balanced by a change to the potential energy part of the rest mass.

It even applies in a way for photons, where the total energy is still constant, but it can't be split into kinetic and potential because the photon does not have rest mass.

 

[Ich]

Hi yuiop,

I couldn't quite identify what exactly you'd like to have clarified. So let me just sum up:
You have E=m+E_kin. E is conserved in free fall.
The outside observer measures E as the inertial and gravitative mass of the object.
If the object is being stopped, E decreases to m. E_kin is set free and could be sent to the observer. So E_kin is accounted for (the observer got it), and m is accounted for (down there, as measured by scales or gravimeters).
The binding energy is just E_kin, one is converted into the other.

And, of course, the basic tenet of relativity still stands: measured locally, the rest mass m never changes.

 

[yuiop]

Hi yuiop,

I couldn't quite identify what exactly you'd like to have clarified. So let me just sum up:
You have E=m+E_kin. E is conserved in free fall.
The outside observer measures E as the inertial and gravitative mass of the object.
If the object is being stopped, E decreases to m. E_kin is set free and could be sent to the observer. So E_kin is accounted for (the observer got it), and m is accounted for (down there, as measured by scales or gravimeters).

So would the scales read

$$\frac{GMm_0}{r^2}$$

or

$$\frac{GMm_0}{r^2} \frac{1}{\sqrt{1-2GM/r}}$$ ?

How is this decreased E actually measured and who measures it?

 

[Ich]

They'd read the decreased value. Either by measuring the force on the rope that fixes the mass, or by trying to accelerate the system (M+m0), or by measuring the gravitation of the system. For all purposes, for the outside world the mass of the system (M+m0) is less than the sum of M and m0, and the difference can be accounted for. It is the energy that has left the system when m0 lost its kinetic energy.

 

[Naty1]

There are various other ways of trying to account for the location and flow of gravitational energy in GR, but it's a complex subject.

I think Scott and Ich got about as far as we can.

On one hand GR doesn't allow us to associate a local energy density with the gravitational field. On the other hand, taking the gravitational potential energy of a particle to be zero at infinite distance from the center of attraction, where gravity is zero, and decreasing as it approaches another body, it takes positive work to separate the bodies. So if I separate two objects, the total mass [energy] of them will increase...the issue is where the energy resides and we don't have a clear answer to that. The location of the additional energy is not clear in General Relativity.

 

[sonmage]

I think we all agree on the following:
(1) An object in free fall retains constant energy even if is moving at relativistic speeds
(2) When the object is stopped it loses its kinetic energy
(3) When a particle is accelerated to relativistic speeds its inertial mass increases and its kinetic energy is derived from the accelerator's EM fields
(4) The rest mass (E = mc^2) remains the same when measured with local units.
(5) Clock rate is reduced by motion and a reduction of altitude
Some uncertainty or disagreement seems to remain regarding ZirkMan's original question which is: Where does the kinetic energy of the falling object come from? Does it come from the gravitational field or does it come from the potential energy which was part of the object before it fell?
In my opinion the reason for the differences is a difference in scientific philosophy. Some physicists believe that a reality that cannot be observed or measured is meaningless. They can believe that clock rate really changes with motion and gravitational position because we have very accurate clocks and these changes can be measured. But our measurement of mass is not so good and we believe them only when they are large enough to measure such as in nuclear reactions or accelerated particles. I am more of a realist and believe that reality can be deduced logically even when it is too small to measure. So in my opinion, Zirkman, an object at high altitude has more mass than at low altitude. But the difference in mass between an object in space and one on the surface of the Earth is 1 part in 1.4 billion. The difference in clock rate is exactly the same but we can measure it so we believe it.

 

[Naty1]

I think we all agree on the following:
(1) An object in free fall retains constant energy even if is moving at relativistic speeds
(2) When the object is stopped it loses its kinetic energy
(3) When a particle is accelerated to relativistic speeds its inertial mass increases and its kinetic energy is derived from the accelerator's EM fields
(4) The rest mass (E = mc^2) remains the same when measured with local units.
(5) Clock rate is reduced by motion and a reduction of altitude

2,3,4,5 are either incorrect or too loosely worded to get much agreement.

I'm not going to try to correct all those...but let's look at [5] as an example. Here is a contra statement which illustrates why #5 is stated incorrectly: Proper time does not vary with speed or altitude. I think your wording is related to the example in this discussion; if so, it's helpful to say so.

Also, KE in general has nothing to do with 'accelerators EM fields'...whatever that means.

 

[ZirkMan]

So in my opinion, Zirkman, an object at high altitude has more mass than at low altitude. But the difference in mass between an object in space and one on the surface of the Earth is 1 part in 1.4 billion. The difference in clock rate is exactly the same but we can measure it so we believe it.

We have the same kind of open question.
If the mass of the fallen object is less than it was in space after it was stopped at the surface (and its KE was removed) then theoretically after an inertial fall to a very massive mass the object can loose a substantial percentage of its mass. Yet, according to the principle of relativity, locally it should not matter and the object should behave without any structural change (ignoring the stress energy of being stopped in a strong gravitational field). If this is the case the question is indeed in what form the kinetic energy was stored in space? I'm fine with the higher clock rate answer. Only I'm not sure if this is the complete picture and if for example length contraction is not involved too? Space and time are one, right?

 

[Q-reeus]

A spinning flywheel, spin axis oriented radially wrt a stationary spherical mass, is lowered into the gravitational well where frequency redshift factor f = (1-2GMr^-1c^-2)^1/2 applies. Interpreted using standard Schwarzschild coordinates, a distant observer notes that rotational speed drops as f, but flywheel radius is unchanged. If we accept conservation of angular momentum holds, there is no choice - contrary to the position of previous postings, inertial mass increases by factor f^-1, not decreases.

Perform the same lowering operation but with spin axis orthogonal to radial direction, and to put it mildly the situation becomes somewhat problematic. Suppose the flywheel is now just two concentrated masses rotating around the midpoint of a joining bar. At the instantaneous angular position where bar is radially oriented, to first order the peripheral velocity of both masses is in coordinate measure reduced by factor f as for the first case. This is required because a local observer must have that bar length, angular velocity, and thus rim speeds are independent of spin axis orientation or instantaneous bar angular orientation. However in coordinate measure the bar length - the flywheel moment arm, is in this orientation reduced by factor f, implying inertial mass should be greater by a factor f^-2 to preserve instantaneous angular momentum!

When the instantaneous bar orientation is horizontal, we have in coordinate measure that bar length is unchanged, but mass rim speeds are reduced by factor f², implying now inertial mass is increased here also by a factor f^-2 so as to preserve instantaneous angular momentum. What is the reason for this sad and sorry state of affairs? My own view, not shared by others, is that Schwarzschild geometry is just wrong. Presumably there is an 'official fix' which preserves both angular momentum and the Kosher status of SM, and it would be interesting to know what that 'fix' is.

 

[yuiop]

Lets say we have a heavy charged battery connected to a laser device and if we run the battery to depletion the energy emitted by the laser is E. This represent the stored energy in the battery. After recharging the battery we drop it into a gravitational well on the end of a rope coiled around a free wheeling massless spool. Eventually we engage the spool to a generator and bring the battery device to rest and recover energy from the generator. Now we turn the laser light on and shine it back up towards the top and the total light energy collected back up there is E*√(1-GM/(rc²)) and the missing energy is equivalent to the energy recovered by the generator when bringing the battery to rest and represents the kinetic energy of the falling battery or equivalently the lost potential energy of the lowered battery. So far so good and all energy is accounted for and conserved. Additionally we note that loss in potential energy is equivalent to the loss in stored chemical energy of the battery.

Now we carry out a similar experiment but this time we use the battery to power a small linear accelerator with a test mass projectile. When the device is activated lower down, the distant observer notes that the terminal velocity of the projectile is slower than when the same experiment was carried out higher up. Since we are fairly sure the stored energy in the battery lower down is reduced by the gravitational gamma factor it is reasonable to assume that the kinetic energy of the fired projectile is also reduced by the same factor, as measured by a distant observer. (A local observer does not see any change.) The Newtonian equation for kinetic energy is:

$$KE = \frac{1}{2}mv^2$$

and is fairly accurate for v<<c even in a strong gravitational field. For the distant observer the velocity decreases by the gravitational gamma factor, but this term is squared and to ensure the overall energy is consistent with gamma reduction in stored energy we require that the inertial mass increases by the gamma factor with reduced altitude so that:

$$KE = \frac{1}{2}m(\gamma) \frac{v^2}{\gamma^2} = \frac{1}{2} \frac{mv}{\gamma}$$

where $\gamma$ represents $1/\sqrt{1-2GM/(rc^2)}$.

If we use the full SR kinetic energy equation valid for all v<c:

$$KE = \sqrt{\left(m_0^2c^4 + \frac{m_0^2 v^2 c^2}{(1-v^2/c^2)} \right) } - m_0 c^2$$

we reach the same conclusion for measurements by the distant observer, that if the KE and terminal velocity are reduced by the gamma factor, then the inertial mass increases by the gamma factor:

If we are interested in the gravitational mass of our lowered object, one way to do this is to weigh the test mass and then measure its acceleration when released and obtain its mass from the relation mass = force/acceleration. If the force measured on local scales is $f = \gamma*GMm_0/r^2$ and the local acceleration is $a=\gamma*GM/r^2$ then the locally measured mass (m) is equal to $m_0$ where $m_0$ is the rest mass measured at infinity. This suggests that loss in potential energy does not manifest itself as a loss in gravitational mass as measured locally. For the distant observer the force is still $f = \gamma*GMm_0/r^2$ and the acceleration is $a=\gamma^{(-2)}*GM/r^2$, so as far as the distant observer is concerned the gravitational mass is $m = m_0 \gamma ^{3}$. No observer actually measures a reduction in weight or gravitational mass with reduced altitude.

So how does this loss in potential energy manifest itself? It appears that when an object is lowered, that any stored energy (in all it forms including gravitational, chemical, nuclear, pressure etc) is reduced by the gravitational gamma factor as seen by a distant observer, but is not apparent to a local observer. This loss is equivalent to the energy that can be gained by the act of lowering the object. As others have already mentioned, the combined gravitational mass of the large massive body (M) and the small test body (m) is reduced when the an object falls and comes to a stop, if the kinetic energy is radiated away as heat after impact. This loss in energy represents the binding energy and also represents the energy required to separate the objects again.

 

[Q-reeus]

...If we are interested in the gravitational mass of our lowered object, one way to do this is to weigh the test mass and then measure its acceleration when released and obtain its mass from the relation mass = force/acceleration. If the force measured on local scales is $f = \gamma*GMm_0/r^2$ and the local acceleration is $a=\gamma*GM/r^2$ then the locally measured mass (m) is equal to $m_0$ where $m_0$ is the rest mass measured at infinity. This suggests that loss in potential energy does not manifest itself as a loss in gravitational mass as measured locally. For the distant observer the force is still $f = \gamma*GMm_0/r^2$ and the acceleration is $a=\gamma^{(-2)}*GM/r^2$, so as far as the distant observer is concerned the gravitational mass is $m = m_0 \gamma ^{3}$. No observer actually measures a reduction in weight or gravitational mass with reduced altitude...

But then there is this: Two masses lying horizontally are allowed to gravitationally close on each other. Heat energy released in this inelastic collision and radiated 'to infinity' is reduced by your gamma factor, relative to the same situation performed far from the central mass. Since the horizontal distance is not affected by potential, we must conclude the gravitational attraction is reduced by gamma. How should one split this gamma factor? It cannot be attached as gamma to both active and passive mass. Regardless, now try the same kind of thing but with the path oriented radially (in a neutral bouyancy flotation tank if you like). You will end up with a quagmire. Which is why I am no fan of Schwarzschild metric!

 

[yuiop]

If we accept conservation of angular momentum holds, there is no choice - contrary to the position of previous postings, inertial mass increases by factor f^-1, not decreases.

This part agrees with my conclusion in the first part of #21 that the horizontal inertia of mass increases by f^-1, where f=√(1-2GM/r).

Perform the same lowering operation but with spin axis orthogonal to radial direction, and to put it mildly the situation becomes somewhat problematic. Suppose the flywheel is now just two concentrated masses rotating around the midpoint of a joining bar. At the instantaneous angular position where bar is radially oriented, to first order the peripheral velocity of both masses is in coordinate measure reduced by factor f as for the first case. This is required because a local observer must have that bar length, angular velocity, and thus rim speeds are independent of spin axis orientation or instantaneous bar angular orientation. However in coordinate measure the bar length - the flywheel moment arm, is in this orientation reduced by factor f, implying inertial mass should be greater by a factor f^-3 to preserve instantaneous angular momentum!

I think that should be greater by a factor of f^-2 if angular momentum is conserved, because the bar length decreases by a factor of f and the velocity decreases also by a factor of f.

When the instantaneous bar orientation is horizontal, we have in coordinate measure that bar length is unchanged, but mass rim speeds are reduced by factor f², implying now inertial mass is increased by a factor f^-2 so as to preserve instantaneous angular momentum.

This part seems OK also if angular momentum is conserved. The most disturbing part is that when the bar is vertical and the tip mass is moving horizontally the mass appears to be increased by a factor of f^-2 while the tip mass of the flywheel with the vertical spin axis has a horizontally moving mass that has only increased by a factor of f^-1. This is an inconsistency that forces us to conclude that angular momentum might not be conserved in Schwarzschild coordinates.

One consolation is that in relativity in general, momentum is not independently conserved. If we look at the kinetic energy of the tip masses and assume that the KE is the same in all orientations (being a scalar) we can conclude that horizontally moving moving mass has an inertia that scales proportional to f^-1 and that vertically moving mass has an inertia that scales proportional to f^-3 in Schwarzschild coordinates. This is consistent with my analysis in the latter part of #21. This suggests that angular momentum around a vertical axis is conserved, but around a horizontal axis it increases by a factor of f^-1 in Schwarzschild coordinates, whether the bar is horizontal or vertical.

EDIT: Actually that last part is wrong. Around the horizontal axis, angular momentum is increased by f^-1 when the bar is horizontal and decreased by f when the bar is vertical and on average the angular momentum is conserved for a complete rotation. This is better viewed in terms of angular momentum (L) expressed as L = Iω where I is the moment of inertia and ω is the angular velocity measured by timing the interval to complete one revolution at any point on the rim. For angular velocity to be conserved, the moment of inertia of the vertical flywheel has to increase by a factor of f^-1, because the angular velocity decreases by f. The angular momentum is now conserved in SC, when expressed in terms of I and ω, whether the flywheel is orientated vertically or horizontally. When we try to analyse the angular momentum for individual point masses and instantaneous velocities, things get a bit messy, but the same is true if we analyse a moving flywheel in SR when its rotation axis is not parallel to the linear motion.

 

[isaac200]

The inertial mass will be the same anywhere because it is only weight that varies with gravitational energy.

 

[Q-reeus]

I think that should be greater by a factor of f-2 if angular momentum is conserved, because the bar length decreases by a factor of f and the velocity decreases also by a factor of f.

Yes - thanks I have edited that bit in #20.

This suggests that angular momentum around a vertical axis is conserved, but around a horizontal axis it increases by a factor of f-1 in Schwarzschild coordinates, whether the bar is horizontal or vertical.

Well if the standard SC's are valid consistent results should obtain if sticking with that coordinate system throughout. We are not seeing it I would suggest! There is no way angular momentum can actually have such an orientation dependence. One can use the isometric form of SC's but while spin axis and bar angle orientation dependence for angular momentum appears fixed my hunch is at the expense of angular momentum no longer being preserved with changing gravitational potential (gamma/redshift factor). Not if one insists rotational KE obeys gamma factor, which it must.

 

[yuiop]

Yes - thanks I have edited that bit in #20.

Well if the standard SC's are valid consistent results should obtain if sticking with that coordinate system throughout. We are not seeing it I would suggest! There is no way angular momentum can actually have such an orientation dependence. One can use the isometric form of SC's but while spin axis and bar angle orientation dependence for angular momentum appears fixed my hunch is at the expense of angular momentum no longer being preserved with changing gravitational potential (gamma/redshift factor). Not if one insists rotational KE obeys gamma factor, which it must.

Hi Q-Reeus, yes, I think a detailed analysis of angular momentum and angular kinetic energy in isotropic coordinates would be interesting and possibly enlightening. I made an error in the last part of my last post, so I have edited it and added further analysis that goes some way to overcoming the problem with angular momentum appearing to be orientation dependent in SC's.

 

[Passionflower]

The inertial mass will be the same anywhere because it is only weight that varies with gravitational energy.

Nope, two identical objects, same atomic composition, temperature etc, at different positions will have a different mass.

However one could ague that because they are positioned differently in a gravitational field they cannot be identical.

 

[Q-reeus]

Hi Q-Reeus, yes, I think a detailed analysis of angular momentum and angular kinetic energy in isotropic coordinates would be interesting and possibly enlightening. I made an error in the last part of my last post, so I have edited it and added further analysis that goes some way to overcoming the problem with angular momentum appearing to be orientation dependent in SC's.

Here is that bit reproduced:

EDIT: Actually that last part is wrong. Around the horizontal axis, angular momentum is increased by f-1 when the bar is horizontal and decreased by f when the bar is vertical and on average the angular momentum is conserved for a complete rotation. This is better viewed in terms of angular momentum (L) expressed as L = Iω where I is the moment of inertia and ω is the angular velocity measured by timing the interval to complete one revolution at any point on the rim. For angular velocity to be conserved, the moment of inertia of the vertical flywheel has to increase by a factor of f-1, because the angular velocity decreases by f. The angular momentum is now conserved in SC, when expressed in terms of I and ω, whether the flywheel is orientated vertically or horizontally. When we try to analyse the angular momentum for individual point masses and instantaneous velocities, things get a bit messy, but the same is true if we analyse a moving flywheel in SR when its rotation axis is not parallel to the linear motion.

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!]

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. 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.

As per comments in #20, this situation was surely looked at way back in the earliest days of GR, so I'm wondering what the standard answer is. :zzz:
[hoo boy - finally edited it all right I think!]

 

[Dale]

If we accept conservation of angular momentum holds

Angular momentum is not globally conserved in curved spacetime. That is, in fact, the operating principle of Gravity Probe B.

 

[Q-reeus]

Angular momentum is not globally conserved in curved spacetime. That is, in fact, the operating principle of Gravity Probe B.

And you believe that applies to our flywheel scenario? What then is your finding as to the correct expression for L as a function of gravitational potential, and additionally do you find any dependence on rotation axis orientation? And how does that then relate back to a specific relation for potential dependence of inertial mass?
[Take again the simpler case of a flywheel with rotation axis radially oriented. Rotational KE = 1/2 Iω² = 1/2Lω (v<<c). Unless L is invariant wrt gravitational potential here, we have that KE is *not* reduced by the usually agreed factor f = (1-2GM/(rc²))^1/2, because f is for sure the factor by which ω is reduced! So unless you have some new physics to share, L is not a function of gravitational potential in given scenario - curved spacetime not withstanding.]

 

[sonmage]

So there you have it Zirkman, you can get any answer you want as long as it is impossible to do the experiment because the difference in mass, if any is too small to measure. One thing we can be sure of: all tests have shown that inertial and gravitational mass are equivalent and that mass and energy are equivalent. If a particle gains energy it gains mass. If it loses energy it loses mass. I'll hold that view until somebody proves that the atomic bomb didn't work

 

[Q-reeus]

...One thing we can be sure of: all tests have shown that inertial and gravitational mass are equivalent and that mass and energy are equivalent...

Only as locally measured. Or you can get consistency with your position applied to coordinate measure of inertial mass in the flywheel scenario? Be my guest!

 

[Dale]

And you believe that applies to our flywheel scenario? What then is your finding as to the correct expression for L as a function of gravitational potential, and additionally do you find any dependence on rotation axis orientation?

I don't know that there is a unique correct expression. I suspect the non uniqueness of parallel transport doesn't allow one to be defined, but it isn't something I have studied in depth.

 

[Dale]

Actually, thinking about it more I am not even sure that angular momentum can be represented as a four-vector. I cannot think of what the timelike component would be. I need to do some digging.

 

[DrGreg]

Actually, thinking about it more I am not even sure that angular momentum can be represented as a four-vector. I cannot think of what the timelike component would be. I need to do some digging.

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).