'Mass' of an object at different positions in a weak gravitational field

[TobyC]

We had a physics lesson today on nuclear binding energy, where we looked at the mass of two protons and two neutrons independently and compared it with the mass of a helium nuclei, and were told how the discrepancy related to the release of energy during nuclear fusion.

The first question in the textbook then asked us to calculate how much mass a 10kg block gains when it is raised by 2 meters above the ground, using E=mc^2, I guess it was intended as a way of introducing us to the concept of binding energy in a way that was familiar to us.

The first problem I have is that I've read while learning about electromagnetism that the potential energy of a system is not stored with the particles, but is stored in the field, so the block in this example would not gain mass at all, but rather the field would. If the book was correct then as a particle falls its increase in kinetic energy would always be directly countered by a decrease in its potential energy (so that it's mass doesn't change) and then what would stop things exceeding the speed of light?

The second thing that occurred to me is that its mass probably does change, not directly because of its change in energy, but because of the relativistic effects of a gravitational field, in the same way the mass of an object increases as it approaches the speed of light. I know the concept of mass becomes a bit weirder once you get into general relativity and start dealing with non-inertial coordinate systems, but I thought that it if you take the weak field limit where you can treat space-time as having a minkowski metric with a tensor field over the top representing the pertubation, then you could come up with a meaningful definition of mass (and when I'm talking about mass here I'm not referring to gravitation, but inertial mass).

I calculated the three vector momentum for a particle in a weak field by differentiating position with respect to proper time and multiplying by the rest mass, and I got a result which was the product of the particle's velocity and some other complicated function of the field, the rest mass, and the velocity, and I defined this function to be the mass. I thought this should work because if an experiment was conducted somewhere in the field then it would follow the normal Newtonian laws of physics if you interpreted this function to be the mass.

So basically what I came up with is that the mass of an object decreases if you raise its position in a gravitational field, and its dependence on velocity is different to standard special relativity. My expression for the 'mass' of a particle in a gravitational field is this:

$$\frac{m}{\sqrt{1-v^{2}}} - m\phi - m(\phi)(v^{2})$$

Where the units are chosen so that the speed of light equals 1, and m is the rest mass, $$\phi$$ is the gravitational potential, and v is the speed of the particle.

The physical significance of this expression is the mass the particle would appear to have in mechanical processes. For example, if this particle was at a great height and viewed from the ground, when the motion is analysed this is the mass the particle would appear to have.

I tried to check this result online when I derived it, but I couldn't find a reference to it anywhere, so I decided to post here to ask whether I've done anything wrong or overlooked anything, I don't really know a great deal about general relativity. Does the mass of an object really appear to decrease in this way when you raise it in a gravitational field?

The final question I have isn't really related to relativity, but goes back to the original subject of the lesson. If potential energy is stored in the field rather than in the particles, why is a proton more massive on its own than when it is combined in a nucleus?

 

[Mentz114]

Keeping it simple. The potential energy of the raised mass is mgh, and naively converting to mass gives (mgh)/c². ( m = mass, g = grav. acc., h = height)

 

[Bill_K]

The first problem I have is that I've read while learning about electromagnetism that the potential energy of a system is not stored with the particles, but is stored in the field, so the block in this example would not gain mass at all, but rather the field would.

That's correct. Or the system consisting of the Earth plus field plus particle would gain mass.

If the book was correct then as a particle falls its increase in kinetic energy would always be directly countered by a decrease in its potential energy (so that it's mass doesn't change) and then what would stop things exceeding the speed of light?

You should abandon the concept of 'relativistic mass' - it's an old-fashioned idea. And a poor explanation of why objects cannot go faster than light.

The second thing that occurred to me is that its mass probably does change, not directly because of its change in energy, but because of the relativistic effects of a gravitational field

It's the energy that changes. For example a photon trying to climb out of the gravitational well will become red-shifted.

(and when I'm talking about mass here I'm not referring to gravitation, but inertial mass).

The principle of equivalence says they're the same.

If potential energy is stored in the field rather than in the particles, why is a proton more massive on its own than when it is combined in a nucleus?

It's not. Just as in the gravitational case, the total mass of the system is less, not the individual particles.

 

[Jonathan Scott]

In electromagnetism, when you apply forces the rest energy of something typically remains constant but its kinetic energy changes and hence its total energy changes too (which is usually assumed to involve energy transfer to or from a field).

In gravity, it is the total energy which remains constant (at least in a static field). In Newtonian terms, this is equal to the kinetic energy plus potential energy plus rest energy. When you use the weak field approximation in GR, the potential energy effectively becomes the fractional change in rest energy due to the time dilation effect of the gravitational potential.

That means that a static gravitational field effectively changes momentum without changing energy. This is just one of the weird things which makes it notoriously difficult to reconcile with quantum theory.

 

[TobyC]

It's not. Just as in the gravitational case, the total mass of the system is less, not the individual particles.

Are you sure? That's not what we've been told.

 

[TobyC]

You should abandon the concept of 'relativistic mass' - it's an old-fashioned idea. And a poor explanation of why objects cannot go faster than light.

I'm aware that in relativity it makes more sense to talk about a particle's momentum rather than its mass. However, for simple purposes, you can get a definition of mass by dividing the momentum by the velocity, in analogy with Newtonian physics, since particles follow the normal Newtonian laws if you add the condition that they have this mass, and this is the quantity people talk about when they say that the mass of an object increases as you approach the speed of light.

I was trying to do an analogous thing for gravity, for the purposes of answering the physics book question (which I'm pretty sure is wrong anyway), and I really want someone to double check that I've done the calculation right.

 

[Jonathan Scott]

As Mentz114 already mentioned, you can just use mgh/c² to calculate the mass equivalent of the potential energy difference for a height h. The potential energy increases with height, so the effective mass of a higher object is greater than that of a lower object. For heights small enough that g is effectively constant, this holds in General Relativity as well.

 

[TobyC]

As Mentz114 already mentioned, you can just use mgh/c² to calculate the mass equivalent of the potential energy difference for a height h. The potential energy increases with height, so the effective mass of a higher object is greater than that of a lower object. For heights small enough that g is effectively constant, this holds in General Relativity as well.

I think that's what the question was getting at, but I don't understand why this holds?

If we take gravity out of the picture to make things simpler and talk about electrostatics instead, is it true that if you have two oppositely charged particles and move them apart then they become heavier because of the gain in potential energy? Because I thought I'd read that if you are to talk about the position of the potential energy for the purposes of calculating where the mass is then the energy lies in the field, not in the particles.

 

[TobyC]

Keeping it simple. The potential energy of the raised mass is mgh, and naively converting to mass gives (mgh)/c². ( m = mass, g = grav. acc., h = height)

I understand that's what the book's after, I'm just disputing that it is in fact true.

 

[Sam Gralla]

I agree with Toby that the book's claim is probably wrong. At least, I can think of no reason why it would be correct. As others have pointed out, if you view the whole "earth plus mass" system as one body, than the total mass should be less than the sum of the individual masses, due to the "gravitational binding energy". But I see no reason why this energy should get assigned to either object. What book is this? That's pretty sloppy.

You might find it interesting that there is a case in electromagnetism where a body can change rest mass. Consider a magnetic dipole (current loop) in an inhomogeneous magnetic field. We all know that there will be a force on it, causing it to gain kinetic energy. But we also know that magnetic fields can do no work. So where does the kinetic energy come from? It comes from the rest mass! We actually saw this explicitly in http://arxiv.org/abs/0905.2391 (see appendix), where rest mass is defined as an integral over the stuff making up a small but extended body.

 

[PeterDonis]

I understand that's what the book's after, I'm just disputing that it is in fact true.

PMFJI, but I think the answer is true if you define "true" appropriately. ;)

I think the question the book should have asked is:

If you raise a 10-kg block 2 meters above the ground,

(1) How much work do you have to do on the block?

(2) What is the change in [STRIKE]potential[/STRIKE] (edit) total energy of the block-Earth-field system?

The answer everyone's been giving is the answer to (1). You can check that easily by measuring the actual work done when you raise the block. However, I agree that the book's implication, which is that answering (1) also answers the question, "What is the change in potential energy of the *block*?" (as opposed to the (edit) total energy of the (/edit) block-field-Earth system) is, strictly speaking, incorrect. For convenience, we might *talk* about the potential energy of the block, because it's a lot easier for us to move the block than to move the Earth. But strictly speaking, the work done on the block does go into the system as a whole, and can't really be isolated to one part of it.

 

[Jonathan Scott]

I agree with Toby that the book's claim is probably wrong. At least, I can think of no reason why it would be correct. As others have pointed out, if you view the whole "earth plus mass" system as one body, than the total mass should be less than the sum of the individual masses, due to the "gravitational binding energy". But I see no reason why this energy should get assigned to either object. What book is this? That's pretty sloppy.

In GR, it is not possible to assign an observer-independent location to the "gravitational binding energy" of the system as a whole, but from the point of view of someone standing on the earth, the rest energy of a nearby object definitely increases as it is lifted higher because of the change in time dilation, which is approximately given by $(1 - GM/rc^2)$ or $(1 - gh/c^2)$ provided that the height is small enough for g to remain effectively constant. This time dilation was of course confirmed by the Pound-Rebka experiment.

It is also true that from the point of view of a distant observer the sum of the masses making up a system is greater than the effective mass of the system because of gravitational binding energy. However, an observer at the same potential as a given object will see it as having its usual mass.

 

[TobyC]

What book is this? That's pretty sloppy.

It's an AQA A-level textbook.

 

[TobyC]

but from the point of view of someone standing on the earth, the rest energy of a nearby object definitely increases as it is lifted higher because of the change in time dilation, which is approximately given by $(1 - GM/rc^2)$ or $(1 - gh/c^2)$ provided that the height is small enough for g to remain effectively constant.

Right this is what I'm trying to get at! Although I worked out that the mass should decrease as you raise it. I guess I got a sign wrong somewhere. Although doesn't the formula you posted show a decrease in mass too? If on your second formula you increase the height, h, then the mass is going to decrease isn't it?

 

[Jonathan Scott]

Right this is what I'm trying to get at! Although I worked out that the mass should decrease as you raise it. I guess I got a sign wrong somewhere. Although doesn't the formula you posted show a decrease in mass too? If on your second formula you increase the height, h, then the mass is going to decrease isn't it?

I should have been a bit more careful with my notation there as g and h are actually vector quantities and that should have been their dot product: $(1 - \mathbf{g}.\mathbf{h}/c^2)$.

When g is downwards and h is upwards, the dot product is negative, so with a minus sign it comes out positive and the relative mass is slightly greater than 1 compared with the original location.

The fractional time dilation is so tiny compared with 1 that we can use approximations of the form (1+f1)/(1+f2) = (1 + (f1-f2)).

 

[TobyC]

I should have been a bit more careful with my notation there as g and h are actually vector quantities and that should have been their dot product: $(1 - \mathbf{g}.\mathbf{h}/c^2)$.

When g is downwards and h is upwards, the dot product is negative, so with a minus sign it comes out positive and the relative mass is slightly greater than 1 compared with the original location.

The fractional time dilation is so tiny compared with 1 that we can use approximations of the form (1+f1)/(1+f2) = (1 + (f1-f2)).

Hmm... Ok could you post a derivation of that result so I could see where I've gone wrong? Or if you like I'll post how I got to where I got to and you can show me.

 

[Jonathan Scott]

Hmm... Ok could you post a derivation of that result so I could see where I've gone wrong? Or if you like I'll post how I got to where I got to and you can show me.

That works just as in Newtonian physics. The change in potential energy is the work done on an object of mass m moved through displacement $\mathbf{h}$ within a field $\mathbf{g}$. The lifting force is in the opposite direction to the field, $-m\mathbf{g}$, so the work done is $-m\mathbf{g}.\mathbf{h}$. To get the fractional change in energy, which is the same as the fractional time dilation, we divide by the rest energy, $mc^2$, giving $-\mathbf{g}.\mathbf{h}/c^2$, and to convert it to a ratio we add 1.

 

[TobyC]

That works just as in Newtonian physics. The change in potential energy is the work done on an object of mass m moved through displacement $\mathbf{h}$ within a field $\mathbf{g}$. The lifting force is in the opposite direction to the field, $-m\mathbf{g}$, so the work done is $-m\mathbf{g}.\mathbf{h}$. To get the fractional change in energy, which is the same as the fractional time dilation, we divide by the rest energy, $mc^2$, giving $-\mathbf{g}.\mathbf{h}/c^2$, and to convert it to a ratio we add 1.

Right I'm not too sure of why that works or where I've gone wrong, so let me show you what I did.

I found that the momentum of a particle in a weak gravitational field is given by:

$$\frac{mv}{\sqrt{(1 + 2\phi) - (1 - 2\phi)v^{2}}}$$

And then I divided by the speed to get an expression which can be treated as the 'mass':

$$\frac{m}{\sqrt{(1 + 2\phi) - (1 - 2\phi)v^{2}}}$$

And then I found what the mass would be when the particle is stationary:

m$$(1 + 2\phi)^{-(1/2)}$$

And since $$\phi$$ is small then this can be written as:

m - m$$\phi$$

So according to this it would seem that the mass of the particle would decrease as you raised it in the field.

All of this is in units where the speed of light is 1.

 

[Jonathan Scott]

Right I'm not too sure of why that works or where I've gone wrong, so let me show you what I did.

I found that the momentum of a particle in a weak gravitational field is given by:

$$\frac{mv}{\sqrt{(1 + 2\phi) - (1 - 2\phi)v^{2}}}$$

I can't claim familiarity with that expression, but it appears to be the momentum adjusted for the coordinate variation in the speed of light in a specific GR coordinate system. I don't think it's a good place to start from. Where did you get it from?

 

[TobyC]

I can't claim familiarity with that expression, but it appears to be the momentum adjusted for the coordinate variation in the speed of light in a specific GR coordinate system. I don't think it's a good place to start from. Where did you get it from?

Sorry for the delayed reply, I haven't been online for a while.

I got it from the weak field metric:

$$p^{i} = m\frac{dx^{i}}{d\lambda}$$

Where p is the momentum, m is the rest mass, x is the coordinate position, and lambda is the proper time along the particle's path.

So:

$$p^{i} = m\frac{dx^{i}}{dt}\frac{dt}{d\lambda}$$

$$p^{i} = \frac{mv}{\sqrt{-g_{\mu\nu}\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt}}}$$

And since: $$g_{\alpha\beta} = \eta_{\alpha\beta} - 2\phi\delta_{\alpha\beta}$$

then that gives:

$$\frac{mv}{\sqrt{(1 + 2\phi) - (1 - 2\phi)v^{2}}}$$

 

[Jonathan Scott]

I must admit that I'm too rusty in this stuff to be sure about all the plus and minus signs, but I think there's a problem with factors of the coordinate speed of light here.

The complication is that although the standard value of c is always the same, if you want to describe "mass", "momentum" or "energy" in another coordinate system, you have to be careful about factors of the coordinate speed of light.

In fact, even the concept of the "coordinate speed of light" only works in isotropic coordinates, because they have the specific property that the scale factor between local space and coordinate space is the same in all directions.

In isotropic coordinates in a weak field, local clock rates and ruler sizes both vary as (1-Gm/rc²) compared with the coordinate system, at least approximately. This means that the coordinate speed of light varies as the square of this factor.

This means that relative to such a coordinate system, energy values increase with height, but local momentum values (with units the same as Ev/c²) decrease by the same factor and local mass values (with units the same as E/c²) actually decrease by the cube of this factor.

When we look at a mass at a higher potential from the local point of view, it has an increased rest energy, and that is what we are actually interested in. We can also describe the corresponding mass value by dividing by the standard c². However, if we try to describe a momentum or mass value in the external coordinate system, we have to be very careful about factors of the coordinate speed of light, as the one to one equivalence of units only applies in local space, not in the coordinate system.

 

[TobyC]

I must admit that I'm too rusty in this stuff to be sure about all the plus and minus signs, but I think there's a problem with factors of the coordinate speed of light here.

The complication is that although the standard value of c is always the same, if you want to describe "mass", "momentum" or "energy" in another coordinate system, you have to be careful about factors of the coordinate speed of light.

In fact, even the concept of the "coordinate speed of light" only works in isotropic coordinates, because they have the specific property that the scale factor between local space and coordinate space is the same in all directions.

In isotropic coordinates in a weak field, local clock rates and ruler sizes both vary as (1-Gm/rc²) compared with the coordinate system, at least approximately. This means that the coordinate speed of light varies as the square of this factor.

This means that relative to such a coordinate system, energy values increase with height, but local momentum values (with units the same as Ev/c²) decrease by the same factor and local mass values (with units the same as E/c²) actually decrease by the cube of this factor.

When we look at a mass at a higher potential from the local point of view, it has an increased rest energy, and that is what we are actually interested in. We can also describe the corresponding mass value by dividing by the standard c². However, if we try to describe a momentum or mass value in the external coordinate system, we have to be very careful about factors of the coordinate speed of light, as the one to one equivalence of units only applies in local space, not in the coordinate system.

I'm quite confused by that post

I'm just trying to figure out what exactly it is that you're saying. Could we ignore gravity and all of its weird effects for the moment and talk about electrostatics? If you have two stationary charged particles, are you saying that those particles will be more massive the further apart they are from one another?

 

[Jonathan Scott]

I'm quite confused by that post

I'm just trying to figure out what exactly it is that you're saying. Could we ignore gravity and all of its weird effects for the moment and talk about electrostatics? If you have two stationary charged particles, are you saying that those particles will be more massive the further apart they are from one another?

Nothing similar applies for electrostatics. In that case the electromagnetic energy is separate from the original rest energy of the particles, and can be considered to reside in the field.

In gravity, space-time is slightly curved, and on any scale large enough to include gravitational sources it is intrinsically curved in such a way that we cannot locally approximate it as flat, so we have to choose a way to map locations into a flat coordinate system, in a similar way to choosing a map projection when we make a map of the earth.

The shape of space-time around a central dominant mass is described by Schwarzschild's solution to Einstein's Field Equations. Schwarzschild's solution is normally derived using the Schwarzschild coordinate system in which local rulers have a different coordinate size in the radial direction from the tangential direction. This makes it easy to calculate the solution, but for practical astronomical purposes it is easier to work with the "isotropic" coordinate system where the map is the same shape locally as the local space, with the scale factor being the same in all spatial directions.

Provided that the potential is small compared with 1 (as it is everywhere in the solar system), local clocks run slow relative to coordinate time by a factor (1-Gm/rc²) and local rulers appear to be slightly shrunk relative to coordinate distance, by the same factor.

Gravitational effects are primarily due to the clock rate change (time dilation). This has the effect that identical objects at different potentials appear to have different energies, because energy is measured on a scale which is proportional to frequency (and Planck's constant doesn't vary with potential).

If you observe the progress of a light ray relative to the coordinate system, this is affected both by the slowing of the clock rate and by the shrinking of rulers, so the coordinate speed of light is approximately (1-2Gm/rc²) times the local value.

This leads to a complication. The normal definition of "momentum" in Special Relativity is as p = Ev/c² where E is the total energy. This means that if you calculate momentum in a particular coordinate system, you are effectively calculating the value of this expression where all the terms, including c, are as measured in that coordinate system. This means that you can't just ignore factors of c when you try to work out how this varies with position.

I've always been surprised how simple the laws of motion are in this coordinate system (for a weak field):

$$\frac{dp}{dt} = \frac{E}{c^2} \, \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )$$

where $$\mathbf{g}$$ is the equivalent of the Newtonian field, roughly equal to the gradient of the potential.

This says that the rate of change of isotropic coordinate momentum for a particle moving at speed v (in any direction, radial, tangential or in between) is simply (1+v²/c²) times the corresponding Newtonian force.

This seems particularly odd for example for a light beam moving downwards, as it is moving at the coordinate speed of light, which decreases with decreasing radius, so it decelerates as it falls. However, the expression Ec/c² is equal to E/c which increases with decreasing radius, so the momentum increases in the downwards direction, and the rate at which it increases is exactly twice the Newtonian rate, as expected from (1+v²/c²).

This means that if you have an object with the same local energy at different potentials, the energy in isotropic coordinates increases with increasing potential, but if you have an object with the same local momentum at different potentials, the magnitude of the coordinate momentum decreases with increasing potential, and if the object has the same mass, the coordinate mass in mass units (obtained by dividing the coordinate rest energy by the coordinate speed of light) decreases three times as fast with increasing potential.

I'm not completely clear about what coordinate system and other assumptions you were using, but if you calculate momentum in a coordinate system and then use that to calculate energy or mass you need to be aware that the conversion may need to use the coordinate speed of light rather than the standard value. I think this may account for the sign problem in your mass change.

(The usual scheme in GR is to work with masses expressed entirely in energy units, which helps hide such problems but often creates different forms of confusion in the process).

 

[TobyC]

Nothing similar applies for electrostatics. In that case the electromagnetic energy is separate from the original rest energy of the particles, and can be considered to reside in the field.

In gravity, space-time is slightly curved, and on any scale large enough to include gravitational sources it is intrinsically curved in such a way that we cannot locally approximate it as flat, so we have to choose a way to map locations into a flat coordinate system, in a similar way to choosing a map projection when we make a map of the earth.

The shape of space-time around a central dominant mass is described by Schwarzschild's solution to Einstein's Field Equations. Schwarzschild's solution is normally derived using the Schwarzschild coordinate system in which local rulers have a different coordinate size in the radial direction from the tangential direction. This makes it easy to calculate the solution, but for practical astronomical purposes it is easier to work with the "isotropic" coordinate system where the map is the same shape locally as the local space, with the scale factor being the same in all spatial directions.

Provided that the potential is small compared with 1 (as it is everywhere in the solar system), local clocks run slow relative to coordinate time by a factor (1-Gm/rc²) and local rulers appear to be slightly shrunk relative to coordinate distance, by the same factor.

Gravitational effects are primarily due to the clock rate change (time dilation). This has the effect that identical objects at different potentials appear to have different energies, because energy is measured on a scale which is proportional to frequency (and Planck's constant doesn't vary with potential).

If you observe the progress of a light ray relative to the coordinate system, this is affected both by the slowing of the clock rate and by the shrinking of rulers, so the coordinate speed of light is approximately (1-2Gm/rc²) times the local value.

This leads to a complication. The normal definition of "momentum" in Special Relativity is as p = Ev/c² where E is the total energy. This means that if you calculate momentum in a particular coordinate system, you are effectively calculating the value of this expression where all the terms, including c, are as measured in that coordinate system. This means that you can't just ignore factors of c when you try to work out how this varies with position.

I've always been surprised how simple the laws of motion are in this coordinate system (for a weak field):

$$\frac{dp}{dt} = \frac{E}{c^2} \, \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )$$

where $$\mathbf{g}$$ is the equivalent of the Newtonian field, roughly equal to the gradient of the potential.

This says that the rate of change of isotropic coordinate momentum for a particle moving at speed v (in any direction, radial, tangential or in between) is simply (1+v²/c²) times the corresponding Newtonian force.

This seems particularly odd for example for a light beam moving downwards, as it is moving at the coordinate speed of light, which decreases with decreasing radius, so it decelerates as it falls. However, the expression Ec/c² is equal to E/c which increases with decreasing radius, so the momentum increases in the downwards direction, and the rate at which it increases is exactly twice the Newtonian rate, as expected from (1+v²/c²).

This means that if you have an object with the same local energy at different potentials, the energy in isotropic coordinates increases with increasing potential, but if you have an object with the same local momentum at different potentials, the magnitude of the coordinate momentum decreases with increasing potential, and if the object has the same mass, the coordinate mass in mass units (obtained by dividing the coordinate rest energy by the coordinate speed of light) decreases three times as fast with increasing potential.

I'm not completely clear about what coordinate system and other assumptions you were using, but if you calculate momentum in a coordinate system and then use that to calculate energy or mass you need to be aware that the conversion may need to use the coordinate speed of light rather than the standard value. I think this may account for the sign problem in your mass change.

(The usual scheme in GR is to work with masses expressed entirely in energy units, which helps hide such problems but often creates different forms of confusion in the process).

I read on wikipedia that there are varying definitions within GR for 'mass', and maybe this is where the problem is.

I'll try and make it clear exactly what I am trying to calculate here, it may be wrong to call what I have found the mass:

I am working in a coordinate system which has a weak gravitational field (the lab frame on Earth), and I am trying to find the quantity which the acceleration of a stationary particle is inversely proportional to. For example, if two stationary particles with the same charge were exposed to the same electromagnetic field they would initially experience a different acceleration, which would depend on this quantity I am trying to calculate.

Is it definitely wrong that this quantity increases as you raise an object in a gravitational field? Or could I be right in saying it decreases even if I'm using all the wrong terminology?

 

[Jonathan Scott]

I read on wikipedia that there are varying definitions within GR for 'mass', and maybe this is where the problem is.

I'll try and make it clear exactly what I am trying to calculate here, it may be wrong to call what I have found the mass:

I am working in a coordinate system which has a weak gravitational field (the lab frame on Earth), and I am trying to find the quantity which the acceleration of a stationary particle is inversely proportional to. For example, if two stationary particles with the same charge were exposed to the same electromagnetic field they would initially experience a different acceleration, which would depend on this quantity I am trying to calculate.

Is it definitely wrong that this quantity increases as you raise an object in a gravitational field? Or could I be right in saying it decreases even if I'm using all the wrong terminology?

You don't need to worry about the different GR definitions for "mass" as that's very obscure compared with what you're looking at.

For purposes of laws of motion and similar, we are interested in the total energy of a particle, which for Newtonian comparisons can be split into rest energy (energy in the local frame as seen by an observer at the same potential), potential energy (difference from energy in local frame due to time dilation) and kinetic energy.

In GR, we conventionally assume units such that the standard value of c is 1, and then we can talk about mass and energy as essentially being the same thing.

The difference in potential energy between two locations in a static field as measured in an external coordinate system is also the difference between the local rest energy observed by an observer at one location and the apparent local rest energy seen by the same observer looking at the other location. If we use that observer's local value of c, which is the standard value, then the observer also calculates the mass at the other location to have increased by the same proportion. This is the natural way to describe the increase of both energy and mass with higher potential as seen locally.

However, if you calculate the momentum or mass as seen in an external coordinate system, you are effectively calculating a coordinate value of Ev/c² or E/c² using not the standard c but the coordinate speed of light (which is only a meaningful concept when using isotropic coordinates, otherwise that speed varies with direction). If you multiply or divide by factors of the standard c to try to derive the other quantities, your result will be wrong by corresponding factors of (1-2Gm/rc²).

Note that as space and time vary, so do other physical quantities including strengths of electromagnetic fields, with the result that local experiments (such as the acceleration of a charge in a given electrostatic field) are not affected by gravitational potential. A device to measure the mass of a local object will not show a difference at different potentials.

 

[TobyC]

The difference in potential energy between two locations in a static field as measured in an external coordinate system is also the difference between the local rest energy observed by an observer at one location and the apparent local rest energy seen by the same observer looking at the other location.

Why is this though? That's the bit I don't understand.

 

[Jonathan Scott]

Why is this though? That's the bit I don't understand.

If the two points are at distance $r_1$ and $r_2$ from the source, then in terms of coordinate time, the observer clock rate is $(1-GM/r_1 c^2)$ and the rate at the other point is $(1-GM/r_2 c^2)$. From the point of view of any static observer, using any consistent time coordinate, the ratio of these two rates gives the relative rate of time at these two points:

$$\frac{1 - GM/r_2 c^2}{1 - GM/r_1 c^2} \approx 1 - (GM/r_2 c^2 - GM/r_1 c^2)$$

If the difference in r is small enough that the field $\mathbf{g}$ is approximately constant, this can be approximated as $(1 + |g| (r_2 - r_1) / c^2)$.

 

[TobyC]

If the two points are at distance $r_1$ and $r_2$ from the source, then in terms of coordinate time, the observer clock rate is $(1-GM/r_1 c^2)$ and the rate at the other point is $(1-GM/r_2 c^2)$. From the point of view of any static observer, using any consistent time coordinate, the ratio of these two rates gives the relative rate of time at these two points:

$$\frac{1 - GM/r_2 c^2}{1 - GM/r_1 c^2} \approx 1 - (GM/r_2 c^2 - GM/r_1 c^2)$$

If the difference in r is small enough that the field $\mathbf{g}$ is approximately constant, this can be approximated as $(1 + |g| (r_2 - r_1) / c^2)$.

Yeah but why is the mass different?

 

[Jonathan Scott]

Yeah but why is the mass different?

That's because rest energy of identical objects (or the same object after moving it) is always the same when measured locally, but energy units vary in the same way as frequency (by $E = h\nu$), so if clock rates are affected by time dilation, then energy is as well.

 

[TobyC]

That's because rest energy of identical objects (or the same object after moving it) is always the same when measured locally, but energy units vary in the same way as frequency (by $E = h\nu$), so if clock rates are affected by time dilation, then energy is as well.

Maybe the problem is due to the raising and lowering of indices.

I've worked out the time component of the momentum vector for a stationary particle and found it to be:

$$m - m\phi$$

But when you lower the index of this momentum vector (turning it into a one-form) you get the result you've been posting:

$$m + m\phi$$

I think I'm right in saying that it is the components of the one form which are conserved, so when you're talking about 'energy', if you want it to be a conserved quantity it makes more sense to use the bottom formula, but if you want to find the quantity which the acceleration of an object depends on then wouldn't you have to use the top formula? Which shows a mass decrease when you raise an object in the field?

 

[Jonathan Scott]

For a static field, energy varies in exactly the same way as frequency and clock rates in general, and this is described totally by the local time per coordinate time factor in the metric. Energy is very simple, as ratios of frequencies as seen by observers at rest are the same for all observers in any static coordinate system; it's just a matter of counting.

Momentum is more complicated as it involves space and depends much more heavily on the choice of coordinate system. I don't know why you keep trying to work backwards from momentum to mass or energy.

In GR, mass is normally assumed to be directly equivalent to energy, using the standard c as a conversion factor. This works perfectly well in most contexts, but it is not the same as mass expressed consistently within a flat coordinate system being used to map curved space-time.

Mass in coordinate mass units is not very relevant in GR. If you try to describe it in a non-isotropic coordinate system it's different in different directions. It's more convenient to use alternative forms of the force laws in terms of total energy and momentum.

Conservation laws are not relevant here. I don't think raising and lowering indices has any meaningful relevance, but perhaps my brain is getting scrambled by your attempts to work backwards from a complicated result to a simple one.

 

[TobyC]

Momentum is more complicated as it involves space and depends much more heavily on the choice of coordinate system. I don't know why you keep trying to work backwards from momentum to mass or energy.

Energy is the time component of the momentum isn't it?

And if you want the quantity which acceleration depends on won't you have to look at the momentum, since forces are expressed in terms of rate of change of momentum?

 

[Jonathan Scott]

Energy is the time component of the momentum isn't it?

And if you want the quantity which acceleration depends on won't you have to look at the momentum, since forces are expressed in terms of rate of change of momentum?

For any question in which GR plays any part, you need to think about whose point of view you are trying to use.

The original question was about the apparent mass increase due to potential energy, so it is obviously from the point of view of someone near the mass, in terms of their local view. In that case you can just use the Newtonian mgh for the extra energy combined with E = mc². You can also compare local time rates at different points in the Schwarzschild solution to get exactly the same result for the energy difference and hence the effective mass difference in local coordinates (at least for the weak approximation case).

The previous expression you derived for momentum is not relative to local coordinates but rather to some external coordinates, so it is not exactly the same momentum, because of additional factors affecting space and time. It is therefore not surprising that it does not vary in the same way with potential.

 

[TobyC]

For any question in which GR plays any part, you need to think about whose point of view you are trying to use.

The original question was about the apparent mass increase due to potential energy, so it is obviously from the point of view of someone near the mass, in terms of their local view. In that case you can just use the Newtonian mgh for the extra energy combined with E = mc². You can also compare local time rates at different points in the Schwarzschild solution to get exactly the same result for the energy difference and hence the effective mass difference in local coordinates (at least for the weak approximation case).

The previous expression you derived for momentum is not relative to local coordinates but rather to some external coordinates, so it is not exactly the same momentum, because of additional factors affecting space and time. It is therefore not surprising that it does not vary in the same way with potential.

I don't see why the expression I derived is not relative to local coordinates? I defined the gravitational potential to be zero at ground level, so at ground level the metric is just the standard minkowski metric, so the expression should be valid from the point of view of a coordinate system which is located and fixed at ground level, which is the perspective that seems to be implied from the question.

 

[csmyth3025]

I won't pretend to understand the formulas and equations that have been used in this thread. Conceptually, the OP seems to be asking in the mass of an object that is raised to a higher level in a gravitational field decreases. Since this section deals with Special and General Relativity, I'm guessing that the question boils down to whether the equivalent mass/energy of the object decreases as it is raised in a gravitational field.

I would look at this problem from the perspective of a distant observer viewing a large colud of interstellar dust and ice unperturbed by any outside forces. This cloud certainly contains a set amount of mass in the form of dust grains and ice particles. It also contains a set amount of gravitational potential energy. In layman's terms, the constituent particles are far away from each other, but their natural tendency is to "fall" together.

If one were to observe this cloud over the course of several million years, there would be a contraction - very slowly at first, but gradually becoming faster and faster. The gravitational potential energy of the once aimlessly drifting particles is being converted into the kinetic energy of particles moving towards each other.

Let's say, for the sake of simplicity, that this cloud eventually coalesces into a sub-stellar object of 10 Jupiter masses. The particles making up this object are the same particles that comprised the original cloud. To be sure, a lot of energy has been radiated away by the heat of all this stuff crashing in on itself.

After 10 or 20 billion years most of this heat will radiate away and there will be a ball of stuff all mashed together as closely as the various inter-molecular forces will allow.

The question posed by to OP is whether the mass of a piece of this ball will decrease if it is raised some distance from the surface.

We can consider that there has been a lot of energy lost by the various particles making up this ball from where they originally started to their final resting place. To time-reverse this process (even for a small piece of this ball) will require the input of energy.

Does this inputed energy reside in the particle's mass, or in the particle's position relative to the big ball of stuff?

My take on this is that the answer is a qualified "yes" to both questions. Considering that every particle in the universe has some equivalent mass/energy relative to every other particle in the universe, the mass/energy of the piece of the ball that is raised above the surface will increase relative to the ball.

I apologize for this unsophisticated reply - bereft of equations such as it is. I may be totally wrong about all this. If so, please feel free to correct me.

Chris