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

[PeterDonis]

So two passing ships is considered a set of global inertial frames?

No. The ships are passing each other, so they are both within the confines of any local inertial frame centered on the event at which they pass. That includes both of their rest frames, and a local Lorentz transformation can be used to transform between them. But none of those frames will extend very far in spacetime, so they're not global. (Unless you are assuming that the two spaceships are alone in the universe, no gravitating masses anywhere.)

a frame in a gravitational field and one without are two frames that cannot share such a relationship?

That's right; the gravitational field limits the extent of any local inertial frame.

 

[Buckethead]

That's right; the gravitational field limits the extent of any local inertial frame.

That's incredible. Just incredible. I thought GR was basically a relationship between SR and the Equivalence Principle leading to me thinking that frames involved in gravitational relationships were similar to those involved in SR relationships. I had no idea!

 

[PeterDonis]

I thought GR was basically a relationship between SR and the Equivalence Principle

And that relationship is what restricts inertial frames to be local in the presence of gravity.

leading to me thinking that frames involved in gravitational relationships were similar to those involved in SR relationships

When you say "frames involved in gravitational relationships", what you really mean is "frames" (a better term would be "coordinate charts") that cover a large region, large enough that the strength and/or direction of gravity varies significantly. In such a region there is simply no such thing as a global inertial frame and SR simply no longer applies. This is a very basic fact about GR and is discussed in most textbooks.

 

[PAllen]

There is another observation that can be made on this, based on the Komar mass formula. That is, if you have a composite body, and weigh each element/cell of it locally, sum theses, and then ask how this compares to the mass of the body measured from afar by orbiting test bodies, you will find that the sum of element masses is greater than the mass mass measured from afar. You could choose to describe this e.g. as a cc of water at triple point at standard pressure has less mass in gravitational field (measured from afar) than it does locally, or isolated far from all other bodies.

 

[PeterDonis]

if you have a composite body, and weigh each element/cell of it locally, sum theses, and then ask how this compares to the mass of the body measured from afar by orbiting test bodies, you will find that the sum of element masses is greater than the mass mass measured from afar.

This is true, but I don't think you can interpret it as a claim about the "mass measured from afar" of each individual element of the body.

 

[Klystron]

Thank you for suggesting that. I read "Time Reborn" about a year ago, and again about 2 months ago and probably need to read it a third time. One thing I was disappointed in was how he only fleetingly mentioned relativity of simultaneity in relation to his new theory. It seems they are at odds with each other and I was looking for a good resolution to that in the book. I admire Smolin very much. I think he is a maverick thinker. I also enjoyed his book "The Trouble With Physics" which I also read twice. With regard to Gleick, are you referring to his book "Time Travel"?

With regard to your suggestion: "Energy may be a useful concept to help understand time, rather than mass", could you elaborate a little? How can calling mass energy via E=mc^2 help with explaining the flow of time?

I'm excited about my talk but still have a lot of research to do.

I delayed responding until I could visit my local library. I assumed Smolin from your writing. Your catastrophe ideas reminded me of "Chaos: Making a New Science". While I enjoy Gleick's reporting in the New York Times online, I hadn't read "Chaos" since first publication. Gleick uses equations and naive set theory including Mandelbrot sets to explain his ideas. Not a textbook but colorful.

As others indicate we measure Energy in the form of EM radiation (light) from distant sources; mass not as directly. Distant black hole sounds interesting but consider how Hawking uses the BH model to demonstrate Time in his books without the BH directly affecting Earth.

 

[vanhees71]

Just to be clear, I'm not referring to rest mass. Since time dilates and mass increases for an object that has a velocity relative to an observer, it lead me to wonder if mass also increases (since time dilates) for an object in a gravitational field relative to an observer outside that field.

I've seen a lot of outdated (not necessarily bad but outdated) as well as very bad textbooks on special relativity introducing the erratic idea of a relativistic mass (which Einstein very soon after commiting the sin using this unnecessary concept in his 1905 paper abandoned).

The idea to use such a "concept" in GR, however, I've never seen before. Where does this idea come from? It doesn't matter, I'd rather recommend another book about GR (e.g., to begin the best I know is Landau and Lifshitz vol. II).

 

[pervect]

My position on "relativistic mass" in GR is that relativistic mass is one component of the stress-energy tensor, so it's not useful by itself but it's useful as part of a larger entity. Unfortunately, this observation may not be that helpful to someone who has no idea of what a tensor is. Hopefully it is useful to say that energy, momentum, and stress all contribute to something we call the stress-energy tensor,. The very name "stress energy tensor" seems to scare some readers. For such readers, one can omit using this technical term and simply say that energy, momentum, and stress all contribute to the gravitational field. I think this is understandable, but it seems hard to get people to accept. There are also occasional questions about what "stress" is, as well.

My impression is that people are very attached the idea of F=ma, and looking for some concept of "m" that they can stick into this equation and have it work in special and General relativity. This doesn't actually work, though. Several conceptual evolutions are needed to find the equations that do work. The first evolution involves the use of ordinary differential calculus, to replace the algebraic equation F=ma with the differential equation F=dp/dt. This, with a bit of supporting explanation, suffices for special relativity.

The next evolution involves moving on from ordinary differential equations to partial differential equations. Tensors can be viewed as a way of organizing partial differntial equations, but this obsrevation doesn't help if one isn't familiar with partial differential equations.

My impression though is that even the first step is a hurdle for a lot of readers of PF - they know their algebra, but switching to a calculus based approach loses them. I could be wrong, it's hard to guess what someone does and does not understand from reading their replies to posts, but that's the working conclusion I've come to. Nobody (and I'd have to include myself) really wants to talk about what they don't know. There is also the difficulty that when one doesn't understand something, one doesn't necessarily know what one is missing.

 

[timmdeeg]

My position on "relativistic mass" in GR is that relativistic mass is one component of the stress-energy tensor, so it's not useful by itself but it's useful as part of a larger entity.

Why isn't it suffice to call it just relativistic kinetic energy instead, which depends on the frame of reference as well?

EDIT I think I expressed myself not clearly. I'm not aware of that "relativistic mass" in GR is a component of the stress-energy-tensor. Do we need this term in this context or is it suffice to use the term "kinetic energy"?

 

[vanhees71]

My position on "relativistic mass" in GR is that relativistic mass is one component of the stress-energy tensor, so it's not useful by itself but it's useful as part of a larger entity.

In my opinion that's the problem already in special relativity! I never understood, why I should introduce for a specific component of a four vector (the energy-momentum four-vector) another name. Energy is energy and mass is mass in Newtonian physics, and there's no single good reason for changing the meaning of the quantity "mass" in relativistic physics. Mass is a scalar, energy a time-like component of the four-momentum and thus dependent on the used reference frame.

In general relativity components of tensors (and tensor densities) usually have no specific physical meaning but only the tensors themselves. A lot of misunderstanding and trouble can be avoided when clearly using the correct physical meaning of tensor quantities and not in their components with respect to arbitrary reference frames or bases and co-bases of tangent and co-tangent vector spaces of the space-time manifold.

 

[vanhees71]

Why isn't it suffice to call it just relativistic kinetic energy instead, which depends on the frame of reference as well?

EDIT I think I expressed myself not clearly. I'm not aware of that "relativistic mass" in GR is a component of the stress-energy-tensor. Do we need this term in this context or is it suffice to use the term "kinetic energy"?

If you have free-streaming particles (or "dust") the energy-momentum tensor may be written as
$$T_{\mu \nu} = \mu u_{\mu} u_{\nu},$$
where $u_{\mu}$ is the four-velocity field of the free-streaming "fluid" particles and $\mu$ the scalar mass density, to be measured in the usual way as "mass per spatial volume" in the local rest frame of the fluid cell.

Already for an ideal gas there's not much meaning for "mass density" anymore since here
$$T_{\mu \nu}=(\epsilon+P) u_{\mu} u_{\nu}-P g_{\mu \nu},$$
where $\epsilon$ is the scalar energy density and $P$ the pressure (also a scalar density!).

I've no idea, where you need relativistic masses (or "mass densities") and how you can sensibly define such a quantity in GR.

 

[Ibix]

EDIT I think I expressed myself not clearly. I'm not aware of that "relativistic mass" in GR is a component of the stress-energy-tensor.

Relativistic mass in SR is simply the total energy of a system (give or take a factor of $c^2$). In many reasonable coordinate systems, the tt component of the stress-energy tensor is an energy density, so could be argued to be something like a relativistic mass density. I tend towards vanhees71's dislike for that way of expressing things.

 

[timmdeeg]

@vanhees71 and @Ibix thanks for your answers. I just saw that

The time–time component is the density of relativistic mass, i.e. the energy density divided by the speed of light squared. In the case of a perfect fluid this component is

where $\rho$ is the relativistic mass per unit volume

Perhaps this justifies the usage of the term "relativistic mass" in this special context. How could one call $\rho$, respectively $\gamma{m_0}$ otherwise?

Relativistic mass in SR is simply the total energy of a system (give or take a factor of $c^2$).

Isn't the total energy of a system given by the energy-momentum relation?

 

[Ibix]

Isn't the total energy of a system given by the energy-momentum relation?

Yes. For massive particles it's $\gamma mc^2$, which is $c^2$ times the relativistic mass.

 

[timmdeeg]

Yes. For massive particles it's $\gamma mc^2$, which is $c^2$ times the relativistic mass.

Got it, thanks.

 

[vanhees71]

@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.

 

[timmdeeg]

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 $T^{ii}$.

 

[pervect]

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 $T^{ii}$.

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.

 

[timmdeeg]

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.

 

[PAllen]

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.

 

[pervect]

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.

 

[PAllen]

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.

 

[jartsa]

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.

 

[PeterDonis]

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.

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.

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".

 

[PAllen]

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.

 

[pervect]

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.

 

[PeterDonis]

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

.

 

[PeterDonis]

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).

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.

 

[pervect]

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.

 

[PeterDonis]

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.

 

[nitsuj]

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...