B Does the mass of an object increase in a gravitational field?

  • #51
timmdeeg said:
@vanhees71 and @Ibix thanks for your answers. I just saw thatPerhaps this justifies the usage of the term "relativistic mass" in this special context. How could one call ##\rho##, respectively ##\gamma{m_0}## otherwise?

Isn't the total energy of a system given by the energy-momentum relation?
Energy is energy and mass is mass. The same holds for the respective densities. Of course, ##T^{00}##, is a tensor-field component, i.e., a frame-dependent quantity and thus bears no particular physical meaning in general relativity. Only invariant objects, i.e., tensors or tensor fields have a clear physical meaning and interpretation.
 
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  • #52
vanhees71 said:
Energy is energy and mass is mass. The same holds for the respective densities. Of course, ##T^{00}##, is a tensor-field component, i.e., a frame-dependent quantity and thus bears no particular physical meaning in general relativity.
How do we deal with cases where work is done on an object.

I would expect that the work done to stretch a spring increases the energy of the spring and thus should be represented by ##T^{00}##. On the other hand the energy added to the spring due to stretching is shear stress and then should be represented by ##T^{ik}##. Could you elaborate a bit on this. Whereby I'm not sure if this is possible in simple terms.

Another example is work done to compress or expand a gas such (adiabatic) that its internal energy changes. Is that subject to ##T^{00}##? On the other hand pressure is represented by ##T^{ii}##. o_O
 
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  • #53
timmdeeg said:
How do we deal with cases where work is done on an object.

I would expect that the work done to stretch a spring increases the energy of the spring and thus should be represented by ##T^{00}##. On the other hand the energy added to the spring due to stretching is shear stress and then should be represented by ##T^{ik}##. Could you elaborate a bit on this. Whereby I'm not sure if this is possible in simple terms.

Another example is work done to compress or expand a gas such (adiabatic) that its internal energy changes. Is that subject to ##T^{00}##? On the other hand pressure is represented by ##T^{ii}##. o_O

Take the case with one spatial dimension. I'll assume a flat space-time, a Minkowskii metric, to make things slightly easier.

Let the coordinates be (t,x). We will introduce both numerical and alphabetic subsripts, so that ##x^0=t## and ##x^1=x##.

Then ##T^{00} = T^{tt}## is the energy density, ##T^{01}=T^{10}= T^{tx}=T^{xt}## is the momentum density, and ##T^{11}=T^{xx}## is the stress. In the one-dimensional case, we can interpret a negative value of ##T^{xx}## as tension , and a positive value as compression. There aren't any other components to worry about. Our one-dimensional object (we can think of it as a bar if we take the limit of zero cross section) is either under tension or compression, there's no other form of stress with only one spatial dimension.

Because it's flat space-time, we can use ordinary partial derivatives for the continuity equation. We can write in tensor notation the expression for the continuityu equation that represents the conservation of energy-momentum as ##\partial_a T^{ab}=0##.

Using numerical notation and the Einstein convention we can expand this tensor equation as:

$$\partial_0 \, T^{00} + \partial_1\, T^{10} = 0 \quad \partial_0 \, T^{01} + \partial_1 \, T^{11} = 0$$

or using symbolic notation as

$$\partial_t \, T^{tt} + \partial_x \, T^{xt} = 0 \quad \partial_t \, T^{tx} + \partial_x \, T^{xx} = 0$$

Here ##\partial_t## represents ##\frac{\partial}{\partial t}## and ##\partial_x## represents ##\frac{\partial}{\partial x}##

In a non-flat space-time, we'd need to write ##\nabla_a \,T^{ab}=0## instead, so we'd have to replace the partial derivatives with covariant derivatves.

These partial differential equations are basically the replacement for F=ma. As I implied in previous remarks, when we go to the continuum limit, we need to replace ordinary differential equations with partial differential equations.
 
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  • #54
pervect said:
Take the case with one spatial dimension. I'll assume a flat space-time, a Minkowskii metric, to make things slightly easier.

Let the coordinates be (t,x). We will introduce both numerical and alphabetic subsripts, so that ##x^0=t## and ##x^1=x##.

Then ##T^{00} = T^{tt}## is the energy density, ##T^{01}=T^{10}= T^{tx}=T^{xt}## is the momentum density, and ##T^{11}=T^{xx}## is the stress. In the one-dimensional case, we can interpret a negative value of ##T^{xx}## as tension , and a positive value as compression. There aren't any other components to worry about. Our one-dimensional object (we can think of it as a bar if we take the limit of zero cross section) is either under tension or compression, there's no other form of stress with only one spatial dimension.
Very helpful, thanks.
 
  • #55
Going back to the question posed in the OP "From a frame of reference outside of a gravitational field, does the mass of an object near a gravitational field increase? ", I would like to reiterate my position that for the most reasonable way to make this question meaningful, the answer is that it is smaller by the gravitational red shift factor, rather than larger or the same.

I define measuring the mass of something from afar by examining orbits of test bodies. This seems both more sensible and feasible than other notions. Then, given an object defined by some number of atoms at some temperature and pressure measured locally, starting far away from some large body, you lower it slowly via pulley to the surface of the body, and detach and roll in the pulley. You find:

1) The locally measured mass is the same as it was locally measured when the object was far away (what happened to potential energy? it was consumed doing work on the pulley system as the object was lowered; it could have been usefully extracted to generate current and charge a battery for example).

2) The increase in mass of original body plus object, measured from afar by test bodies, is the locally measured object mass divided by the gravitational red shift factor.

All statements above are good approximations for non-extreme cases.
 
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  • #56
PAllen said:
Going back to the question posed in the OP "From a frame of reference outside of a gravitational field, does the mass of an object near a gravitational field increase? ", I would like to reiterate my position that for the most reasonable way to make this question meaningful, the answer is that it is smaller by the gravitational red shift factor, rather than larger or the same.

I define measuring the mass of something from afar by examining orbits of test bodies. This seems both more sensible and feasible than other notions. Then, given an object defined by some number of atoms at some temperature and pressure measured locally, starting far away from some large body, you lower it slowly via pulley to the surface of the body, and detach and roll in the pulley. You find:

1) The locally measured mass is the same as it was locally measured when the object was far away (what happened to potential energy? it was consumed doing work on the pulley system as the object was lowered; it could have been usefully extracted to generate current and charge a battery for example).

2) The increase in mass of original body plus object, measured from afar by test bodies, is the locally measured object mass divided by the gravitational red shift factor.

All statements above are good approximations for non-extreme cases.

My attitude towards this is in some sense very similar, and in other senses, very different.

If we have a two body system, I would agree that the total mass of the two-body system, bound by gravity, is less than the sum of the masses of the two bodies if they are not bound by gravity. In the weak field case, we can even attribute this mass difference to the Newtonian binding energy. It doesn't really quite work in the strong field, there isn't really a good strong-field notion for binding energy. There is, however a good strong-field notion of the mass of a system of bodies.

Where I differ is in the whole idea that we should assign masses to bodies. I would argue that it turns out to be much less misleading to associate mass with a system than with a body. Then we don't expect the mass of a system of bodies to be the sum of the masses of the bodies themselves. In fact, to even define the mass of a body by the method Pallen proposes (which I think is a reasonable method), we need to isolate the bodies from each other, then measure the mass of each isolated body.

When the bodies are isolated, we can use the method Pallen describes to measure their individual masses. Then we add these masses together, to get the total mass. We compare this total computed sum to the observed mass of the bound system, and find that the mass of the bound system is lower than this sum of the isolated masses.

In this view, gravity is a nonlinear interaction. The non-linear interaction makes the mass of a system of bodies bound by gravity change as we move the bodies relative to each other. The closer they get, the lower the system mass.
 
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  • #57
pervect said:
My attitude towards this is in some sense very similar, and in other senses, very different.

If we have a two body system, I would agree that the total mass of the two-body system, bound by gravity, is less than the sum of the masses of the two bodies if they are not bound by gravity. In the weak field case, we can even attribute this mass difference to the Newtonian binding energy. It doesn't really quite work in the strong field, there isn't really a good strong-field notion for binding energy. There is, however a good strong-field notion of the mass of a system of bodies.

Where I differ is in the whole idea that we should assign masses to bodies. I would argue that it turns out to be much less misleading to associate mass with a system than with a body. Then we don't expect the mass of a system of bodies to be the sum of the masses of the bodies themselves. In fact, to even define the mass of a body by the method Pallen proposes (which I think is a reasonable method), we need to isolate the bodies from each other, then measure the mass of each isolated body.

When the bodies are isolated, we can use the method Pallen describes to measure their individual masses. Then we add these masses together, to get the total mass. We compare this total computed sum to the observed mass of the system, and find that the mass of the system is lower than this sum.

In this view, gravity is a nonlinear interaction. The non-linear interaction makes the mass of a system of bodies bound by gravity change as we move the bodies further apart from each other.
Note that my comment at the end of my post: "All statements above are good approximations for non-extreme cases" is meant, more explicitly, to specify:

1) mass object <<< mass body (e.g. planet)
2) mass distant observer and apparatus <<< mass body
3) Einstein linearized gravity is accurate to required precision for the problem at hand.

Note, this applies to e.g. lowering an asteroid to the surface of uranus.

The point I mean to stress is that contrary to any ill conceived notion of relativistic mass in SR, this (IMO, well motivated way of measuring mass from a distance) says, mass of an object with some local definition, on the surface of a massive body, is less (measured from afar) than the mass of an equivalent object far away from a massive body (measured either locally or from afar).

All of this is captured in the Komar mass integral, which effectively discounts locally measured mass of an element by the gravitational redshift factor.
 
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  • #58
pervect said:
Where I differ is in the whole idea that we should assign masses to bodies. I would argue that it turns out to be much less misleading to associate mass with a system than with a body.

If we want to know the path of a test mass orbiting a system, we need to know how mass is distributed in the system. I'm talking about a system that is not spherically symmetric.

We need to know whether or not an iron atom at the center of a planet has the same gravitational mass as an iron atom at the top of a mountain on that planet.
 
  • #59
pervect said:
there isn't really a good strong-field notion for binding energy. There is, however a good strong-field notion of the mass of a system of bodies.

I don't think the question of whether the fields are weak or strong is relevant for either of these.

The key factor for mass is whether or not the system is stationary or asymptotically flat (or both). Those two cases are the cases where there is a well-defined mass for the system (the Komar mass for the stationary case and the ADM mass or Bondi mass for the asymptotically flat case). This is indeed independent of whether the fields are weak or strong; but for systems that are neither stationary nor asymptotically flat, there is, AFAIK, no well-defined notion of mass.

The requirement for there to be a well-defined binding energy is that the system can be viewed as a composite of subsystems, each of which has a known mass in isolation: then the binding energy is just the difference between the mass of the total system and the sum of the masses of the subsystems. But this works even if the fields are strong.

pervect said:
In this view, gravity is a nonlinear interaction. The non-linear interaction makes the mass of a system of bodies bound by gravity change as we move the bodies further apart from each other.

You don't need nonlinearity for this. It happens in the weak field approximation, in which nonlinearities are negligible. To move the bodies further apart from each other, you have to add energy to the system: that energy shows up as the increased mass.

PAllen said:
All of this is captured in the Komar mass integral

Yes, this is the one that intuitively "adds up" all of the local pieces of the system (and taking into account, as you say, the redshift factor). But, as noted above, it only works for stationary systems. For a non-stationary system, there is no well-defined notion of "gravitational redshift factor".
 
  • #60
PeterDonis said:
Yes, this is the one that intuitively "adds up" all of the local pieces of the system (and taking into account, as you say, the redshift factor). But, as noted above, it only works for stationary systems. For a non-stationary system, there is no well-defined notion of "gravitational redshift factor".
This is where validity of weak field approximations do come into play (in particular, the regime of validity of linearity). To the extent they are good for something like the solar system, you can continue to treat e.g. Neptune as stationary as well as its equilibrium state after an object has been added to it (by the method I described), even though the solar system as a whole is clearly not stationary.

My key point is simply to address, to first order accuracy, the OP question. That is, given that for motion in SR an object's total energy (which some call its relativistic mass) increases in proportion to its observed time dilation (all per the frame in which the motion is evident). So they ask is there something similar for an object (possibly a radioactive body, so also a clock) experiencing gravitational time dilation on a planet's surface? My answer remains, to first order, the mass measured at a distance will be less by the dilation factor, contrary to the OP expectation.
 
  • #61
I've been thinking my remarks over, and the use of the term "isolated system" is important to my argument. I believe, though, that my thinking is that an isolated system basically IS just one that has an asymptotically flat background metric, at a minimum. Probably it'd be good to add some constraints on gravitational radiation, so that a system that was strongly radiation gravitational radiation wouldn't be isolated because the gravitational radiation was escaping the system.

So it's an issue of semantics, and I'm using the term "isolated system" because I think it's more layman-friendly than "asymptotically flat". But the question arises, are the two notions really the same?

Possibly my thinking is wrong - it's not something I read in a textbook. But currently, I cannot think of any counterexamples. Perhaps someone else can, if so it would be very interesting. I suppose at this point I am proposing that we can think of the idea that the terms "asymptotically flat" and "isolated system" are the same as a conjecture, and try to disprove the idea by finding a counterexample.

Failure to find a counterexample won't necessarily prove anything, but it makes it plausible. And finding a counterexample would be interesting. Part of the issue is semantics - I'm not sure there is a formal definition for "isolated system", in fact, that's sort of what we're trying to figure out.

Birkhoff's theorem is the starting point of my thinking. It says that any spherically symmetric solution of Einstein's field equations in a vacuum must be static and asymptotically flat. Now if we could argue that an isolated system, viewed from a long distance, should be spherical symmetrical, we'd be done.

I don't think this quite works though. Clearly, gravitational radiation won't be spherically symmetrical. But we are already adding some constraints on gravitational radiation in considering the system to be isolated.

Additionally, we can note that the presence of gravitational radiation won't necessarily spoil asymptotic flatness, as long as it dies out fast enough with increasing distance. And I'd expect that to happen.

Anyway, none of this is going to replace a serious study of ADM mass, Bondi mass, and Komar mass as they are currently defined in General relativity. But it might make the discussion more accessible, IF we can accept that an "isolated system" has asymptotic flatness.
 
  • #62
PAllen said:
for motion in SR an object's total energy (which some call its relativistic mass) increases in proportion to its observed time dilation (all per the frame in which the motion is evident). So they ask is there something similar for an object (possibly a radioactive body, so also a clock) experiencing gravitational time dilation on a planet's surface? My answer remains, to first order, the mass measured at a distance will be less by the dilation factor, contrary to the OP expectation.

To me this is a superficial comparison anyway. The analogy between "time dilation due to motion" and "time dilation due to gravity" has so many problems that my preference is to just reject it altogether

.
 
  • #63
pervect said:
my thinking is that an isolated system basically IS just one that has an asymptotically flat background metric, at a minimum. Probably it'd be good to add some constraints on gravitational radiation, so that a system that was strongly radiation gravitational radiation wouldn't be isolated because the gravitational radiation was escaping the system.

Not just gravitational radiation: any kind of radiation. I think a good working definition of "isolated system" is a system that is asymptotically flat, and whose ADM mass equals its Bondi mass (the latter condition is what excludes any kind of radiation escaping).

pervect said:
the presence of gravitational radiation won't necessarily spoil asymptotic flatness, as long as it dies out fast enough with increasing distance

That's correct (and it goes for any kind of radiation, not just gravitational). The presence of radiation just makes the Bondi mass less than the ADM mass.
 
  • #64
On further thought, I think I may have a counterexample to my own idea. A closed universe is arguably isolated, I think, but it's not necessarily asymptotically flat. So the issue of the existence or nonexistence of the mass of a closed universe rather spoils my idea. The generally accepted idea (or at least the idea as expressed in MTW's gravitation) is that there isn't a meaningful notion of the mass of a closed universe. So - back to the idea board.
 
  • #65
pervect said:
A closed universe is arguably isolated

I would say not, because to me, "isolated" implies that there could be an observer "outside" the system somewhere who can measure its properties. That's not true of a closed universe. But the term "isolated" does not have an exact technical meaning, so this is really a matter of personal preference.
 
  • #66
Nugatory said:
No. Under the conditions you have specified, the reception events will be separated by less than one day of proper time along our worldline (assuming that a "day" is defined to be the number of cesium-clock seconds that will be counted by a clock on Earth during one rotation).
liked that post because "day" was used in quotes and defined as...a day..on Earth...measured in defined seconds; cesium "cycles".

that level of detail "inherits" more ambiguity...:woot:
 
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