How does Gravitational Mass work?

[geordief]

I have just come across this terminology and wonder how it  works (also eager to "blood" my use of the term   )

So,if we have a massive body far removed from other such bodies,I understand  that it will  curve the surrounding spacetime(or create a gravitational field)

So my question is "Does this gravitational mass dissipate as a result of having caused this effect"?

And ,perhaps relatedly does the Gravitational field created by this massive body  expand  at the speed of c from the massive body in a similar way to how an em wave does (as an expanding   sphere)?

If it does ,does it carry away the mass of the body so that it dissipates eventually? 

[MigL]

Mordred or Markus may be able to better explain the technicalities, but basically the term is used to differentiate between the property 'gravitational mass' which results in gravitational effects, and the property 'inertial mass' which is responsible for resistance to inertia change. However, the equivalence principle of GR states that the  ( local ) gravitational force due to a gravitating body is equivalent to the perceived force due to acceleration ( non inertial FoR ).

The field does not need to 'expand' as it is already there, however changes in the field propagate at c just like all other information.
Nor does the property of mass dissipate with 'use' or time, as it is due to the configuration of the system.

The force ( due to gravitational mass )does fall off as 1/r^(n-1) where n is the number of dimensions expanding into ( so 1/r^2 ).

[Markus Hanke]

5 hours ago, geordief said:

I have just come across this terminology and wonder how it  works (also eager to "blood" my use of the term   )

My advice to you would be to stay clear of this type of terminology, most especially in GR, as it will only add more confusion, and does nothing to enhance clarity of understanding. In Newtonian gravity, mass is a straightforward concept, but in GR the question “how much mass is in a given region of spacetime?” becomes very non-trivial and tricky to answer. There are quite a number of concepts of “mass” in GR, and they are all different, and most apply only to specific circumstances. The reason why this is so is because GR is a non-linear theory, which physically means that the gravitational field is self-interacting; energy-momentum is a source of gravity, but so is the gravitational field itself. This means that you cannot really isolate and localise the source that “makes” a given gravitational field; in most circumstances, almost all of it will come from the central body, but there is also some contribution by the field itself. This makes defining concepts such as “gravitational mass” in a precise manner very difficult. It can be done under some circumstances, but you will then need to consider the entire region of spacetime, not just the central body.

5 hours ago, geordief said:

So my question is "Does this gravitational mass dissipate as a result of having caused this effect"?

Again, the answer to this depends on what definition of “gravitational mass” you use, but will be no in most cases. A counterexample would be a scenario where the gravitational system has a quadrupole (or higher multipole) moment, in which case energy-momentum will be dissipated via gravitational radiation. This changes the gravitational field in the immediate vicinity of the system.

5 hours ago, geordief said:

And ,perhaps relatedly does the Gravitational field created by this massive body  expand  at the speed of c from the massive body in a similar way to how an em wave does (as an expanding   sphere)?

You cannot just create a massive body out of nowhere, so the question is not physically meaningful. However, if it were possible, then the answer would be yes, the change in the gravitational field would expand outwards at the speed of light. Note that this would be pretty bad news, because a very sudden and large change in gravity would result in an expanding shock front that contains a narrow region of extreme tidal forces. This would potentially be a very destructive phenomenon. 

5 hours ago, geordief said:

If it does ,does it carry away the mass of the body so that it dissipates eventually?

No it wouldn’t.

[md65536]

See https://en.wikipedia.org/wiki/Gravitational_wave

Quote

Gravitational waves are disturbances in the curvature of spacetime, generated by accelerated masses, that propagate as waves outward from their source at the speed of light. [...] Gravitational waves transport energy as gravitational radiation,

Read the section "Sources", as that explains some cases where gravitational waves are emitted (or "in most cases" aren't).

In particular, in the case of black hole mergers, a lot of mass can be lost as gravitational waves. This comes from the extreme acceleration of the masses rotating around each other.

Quote

On 11 February 2016, the LIGO collaboration announced the first observation of gravitational waves, from a signal detected at 09:50:45 GMT on 14 September 2015 of two black holes with masses of 29 and 36 solar masses merging about 1.3 billion light-years away. During the final fraction of a second of the merger, it released more than 50 times the power of all the stars in the observable universe combined. [...] The mass of the new merged black hole was 62 solar masses. Energy equivalent to three solar masses was emitted as gravitational waves.

That's about 5% of the mass, radiated away. But once merged, the mass stops accelerating and losing mass to gravitational waves, as the animations and charts show.

[geordief]

Oops ,missed the first replies

3 hours ago, Markus Hanke said:

My advice to you would be to stay clear of this type of terminology, most especially in GR, as it will only add more confusion, and does nothing to enhance clarity of understanding. In Newtonian gravity, mass is a straightforward concept, but in GR the question “how much mass is in a given region of spacetime?” becomes very non-trivial and tricky to answer.

So gravitational mass is a property of the entire system? Is it not possible to consider a massive body so far removed from other bodies  that it is only its own field that is in play*?

And can we stipulate that this body is homogenous and non rotating? Does that change anything? (Do I have to consider  the forces inside the atom as sources for the field?)

 

*Maybe isolated fields are also physically meaningless?

Edited December 9, 2019 by geordief

[swansont]

You may be overthinking this. In Newtonian terms, gravitational mass is what you find in F=GMm/r^2

Inertial mass is what you find in F=ma

We find that they are the same thing. If you put the gravitational force equation into an inertial equation, you can cancel the masses.

The gravitational field you get from this will be the same as from GR, in most cases. (e..g. for the earth and the solar system, the relativistic effects are very small). There is no "dissipation" of the mass — energy is conserved. It sounds like you might be presenting a scenario where mass suddenly appears, but that's unphysical. Mass doesn't do that.

[geordief]

24 minutes ago, swansont said:

You may be overthinking this. In Newtonian terms, gravitational mass is what you find in F=GMm/r^2

Inertial mass is what you find in F=ma

We find that they are the same thing. If you put the gravitational force equation into an inertial equation, you can cancel the masses.

The gravitational field you get from this will be the same as from GR, in most cases. (e..g. for the earth and the solar system, the relativistic effects are very small). There is no "dissipation" of the mass — energy is conserved. It sounds like you might be presenting a scenario where mass suddenly appears, but that's unphysical. Mass doesn't do that.

I hope I wasn't doing that.

If we take a small body with mass in a pre-existing * gravitational  field, this body will (I am supposing) contribute its own field in its infinitesimal degree   to it.

 

Does this infinitesimally small contribution continue without  drawing on some energy reserves inside the body itself (ie does  the body use up its sources of mass or does it replenish them from the prexisting "external" (?) gravitational field?

 

Do different sources of gravitational fields "top up" or "drain" each other in this way?

 

*Ideally I would have this pre-existing field to be as small as possible for purposes of clarity. 

Edited December 9, 2019 by geordief

[swansont]

1 hour ago, geordief said:

I hope I wasn't doing that.

If we take a small body with mass in a pre-existing * gravitational  field, this body will (I am supposing) contribute its own field in its infinitesimal degree   to it.

 

Does this infinitesimally small contribution continue without  drawing on some energy reserves inside the body itself (ie does  the body use up its sources of mass or does it replenish them from the prexisting "external" (?) gravitational field?

A mass can't exist without its gravitational field already existing. AFAIK, you can't separate the two as if they are separate effects.

 

1 hour ago, geordief said:

Do different sources of gravitational fields "top up" or "drain" each other in this way?

Gravity obeys superposition. Fields add, and they are vectors.

There is a point in the earth-sun system where the gravitational force is zero on a point mass, because the forces are equal and opposite. Same for the earth and moon. (It's not the halfway point, though, as incorrectly claimed in the movie Space Cowboys)

 

 

[studiot]

56 minutes ago, swansont said:

Gravity obeys superposition. Fields add, and they are vectors.

There is a point in the earth-sun system where the gravitational force is zero on a point mass, because the forces are equal and opposite. Same for the earth and moon. (It's not the halfway point, though, as incorrectly claimed in the movie Space Cowboys)

Swansont has a superb 'feel' for how things interact.

the neutral point is called a lagrangian point and whilst I'm sure you can google this and get lots of pretty pictures you get a feel by looking at some calculations as here

https://iwant2study.org/lookangejss/02_newtonianmechanics_7gravity/YJC2016 H2 Phy Topic 7 G field - Tutorial (Teacher).pdf

[Mordred]

2 hours ago, swansont said:

 

Gravity obeys superposition. Fields add, and they are vectors.

This is extremely important, in a real sense one can think of spacetime curvature that we often call gravity as the sum of all fields via the stress momentum tensor. Or if you prefer the Ricci curvature tensor 

[geordief]

4 minutes ago, Mordred said:

This is extremely important, in a real sense one can think of spacetime curvature that we often call gravity as the sum of all fields via the stress momentum tensor. Or if you prefer the Ricci curvature tensor 

Which are those fields?

How many of them are there?

[swansont]

33 minutes ago, geordief said:

Which are those fields?

How many of them are there?

We're talking about gravitational fields.  Other kinds of fields would not contribute. 

[Markus Hanke]

8 hours ago, geordief said:

So gravitational mass is a property of the entire system?

Yes, mass as it appears in - for example - the Schwarzschild metric is a global property of the entire spacetime.

8 hours ago, geordief said:

Is it not possible to consider a massive body so far removed from other bodies  that it is only its own field that is in play*?

Of course. But the mass will still be a global property of that region of spacetime.

8 hours ago, geordief said:

And can we stipulate that this body is homogenous and non rotating? Does that change anything?

Sure, that’s an important piece of information. While the internal makeup of the body is irrelevant so long as it remains spherically symmetric, the presence of angular momentum would change the geometry of the surrounding spacetime.

8 hours ago, geordief said:

Do I have to consider  the forces inside the atom as sources for the field?

They are already taken into account in the total mass of an atom of a given element.

4 hours ago, swansont said:

Gravity obeys superposition. Fields add, and they are vectors.

Just to add to this, purely to avoid possible confusion - this is the Newtonian picture. In GR, the gravitational fields are not vectors but tensor fields, and they don’t linearly add, meaning that the sum of two solutions (i.e. metrics) to the field equations is not usually itself a valid solution.
However, as was quite correctly pointed out, in the weak field regime (such as our solar system), the Newtonian picture is usually adequate, since the nonlinear effects are quite small. 

1 hour ago, Mordred said:

This is extremely important, in a real sense one can think of spacetime curvature that we often call gravity as the sum of all fields via the stress momentum tensor. Or if you prefer the Ricci curvature tensor 

Apologies, but I don’t quite understand this comment...? In vacuum outside a central mass, both \(R_{\mu \nu}=0\) and \(T_{\mu \nu}=0\). What do you mean by sum of all fields?

Edited December 9, 2019 by Markus Hanke

[Mordred]

In essence all forms of energy/mass density contribute. All energy/mass density can be represented by a field.

For example a photon field, Higgs field etc, etc all contribute.

[Markus Hanke]

11 hours ago, Mordred said:

In essence all forms of energy/mass density contribute. All energy/mass density can be represented by a field.

For example a photon field, Higgs field etc, etc all contribute.

Thanks, I get you now - you meant that the energy-momentum tensor is the sum of all relevant contributions: \(T_{\mu \nu}:=T_{EM}+T_{matter}+...\)
I would like to add to this with the reminder that the Einsteins equations are a local constraint on the metric - so the above is evidently correct in the interior of energy-momentum distributions. But in a vacuum region outside such a distribution, the energy-momentum tensor field identically vanishes - instead, the role of the (now distant) sources is then played by the boundary conditions which one has to impose to obtain a solution to the field equations.

What this essentially means is that the Einstein equations alone do not uniquely determine the geometry of spacetime - they only place a local constraint on it. One has to supply extra constraints (i.e. boundary conditions) to obtain a unique geometry. I think this is a really important thing to realise.

[studiot]

17 hours ago, Markus Hanke said:

Just to add to this, purely to avoid possible confusion - this is the Newtonian picture. In GR, the gravitational fields are not vectors but tensor fields, and they don’t linearly add, meaning that the sum of two solutions (i.e. metrics) to the field equations is not usually itself a valid solution.

 

4 hours ago, Markus Hanke said:

What this essentially means is that the Einstein equations alone do not uniquely determine the geometry of spacetime - they only place a local constraint on it. One has to supply extra constraints (i.e. boundary conditions) to obtain a unique geometry. I think this is a really important thing to realise.

 

16 hours ago, Mordred said:

In essence all forms of energy/mass density contribute. All energy/mass density can be represented by a field.

For example a photon field, Higgs field etc, etc all contribute.

 

One consequence of moving from a distributed system of point masses to a Field where configuration of the participants plays a part is the need to move also to a mass density function.
where contributions from the participants is summed. The more so for nonlinear equations.
In these circumstances you cannot simply say that if you move a 'mass' at point A to another point B the total mass will remain the same or that the effect on the rest of the system of this mass will be the same at point B as it is at point A, because of the reconfiguration of the system.

This is similar to the move to a charge/current density function as opposed to simply charge/current in electrodynamic theory, where J is taking over from I as the basic electric 'dimension'.

And yes, as always, boundary conditions are all important.

Edited December 10, 2019 by studiot

[geordief]

25 minutes ago, studiot said:

In these circumstances you cannot simply say that if you move a 'mass' at point A to another point B the total mass will remain the same

(if I am allowed to "snipe"), would such a reconfiguration of a system  change the overall value for  the system's mass and does it follow that ,for the largest system  of which we have knowledge, that its mass  changes all the time ? 

Would that mean that ,around the time of BB that value was smaller,even vanishingly small? (or larger even?)

Edited December 10, 2019 by geordief

[studiot]

8 minutes ago, geordief said:

Would that mean that ,around the time of BB that value was smaller,even vanishingly small? (or larger even?)

As you know I am not a cosmologist and I cna't answer this question, even if they think they can, though in my lifetime I have seen their answer change many times.

10 minutes ago, geordief said:

(if I am allowed to "snipe"), would such a reconfiguration of a system  change the overall value for  the system's mass and does it follow that ,for the largest system  of which we have knowledge, that its mass  changes all the time ? 

It might and then again it might not. It depends upon the new and old configurations.

Don't forget I am simply pointing out for instance that in Newtonion Mechanics there are an infinite number of configurations of a given system with some property the same, eg  moment of inertia.

If configuration is important in mass/energy (eg potential energy) then this is also true of such systems.

[geordief]

4 minutes ago, studiot said:

.

If configuration is important in mass/energy (eg potential energy) then this is also true of such systems.

As we run the clock backwards does that rigorously show that the configurations get simpler? (and so the value for system mass based on that parameter must get less)

.

Studiot quote: "As you know I am not a cosmologist and I cna't answer this question, even if they think they can, though in my lifetime I have seen their answer change many times" 

 

(If ,as you seem to  say the consensus for the value of system mass around the BB   has changed over the last few decades that seems to indicate it is not a simple question.)

[studiot]

16 minutes ago, geordief said:

As we run the clock backwards does that rigorously show that the configurations get simpler? (and so the value for system mass based on that parameter must get less)

.

Studiot quote: "As you know I am not a cosmologist and I cna't answer this question, even if they think they can, though in my lifetime I have seen their answer change many times" 

 

(If ,as you seem to  say the consensus for the value of system mass around the BB   has changed over the last few decades that seems to indicate it is not a simple question.)

 

If there was a BB was a controvertial subject a few decades ago.

#and there were ( perhaps still are) plenty of other hypotheses to go round.

[Mordred]

1 hour ago, geordief said:

(if I am allowed to "snipe"), would such a reconfiguration of a system  change the overall value for  the system's mass and does it follow that ,for the largest system  of which we have knowledge, that its mass  changes all the time.

Would that mean that ,around the time of BB that value was smaller,even vanishingly small? (or larger even?)

We have to be careful here. At the time of the BB all particle species would be in thermal equilibrium and indistinguishable from one another.

 The extreme high temperature would result in all particle species to be indistinct from photons. So you wouldn't have a rest or invariant mass however you would have an extremely high inertial or variant mass.

 This is prior to electroweak symmetry breaking so even leptons wouldn't have invariant mass until the Higgs field decouples.

Secondly the volume being so miniscule for our observable universe portion at time of BB would make curvature meaningless. In essence one can  describe the instant of the BB via its temperature and density. 

[geordief]

12 minutes ago, Mordred said:

We have to be careful here. At the time of the BB all particle species would be in thermal equilibrium and indistinguishable from one another.

 The extreme high temperature would result in all particle species to be indistinct from photons. So you wouldn't have a rest or invariant mass however you would have an extremely high inertial or variant mass.

 This is prior to electroweak symmetry breaking so even leptons wouldn't have invariant mass until the Higgs field decouples.

Secondly the volume being so miniscule for our observable universe portion at time of BB would make curvature meaningless. In essence one can *  describe the instant of the BB via its temperature and density. 

Does being indistinguishable mean uncountable? Would there be less than at other times or just the same number in a smaller confine?

I don't think I was trying to describe any "instant" ,just the  period immediately following  after that estimated  "instant"**.Does your description apply to that "epoch"?

*Do you mean "can only"?

**ie T+10^-43 seconds  and after.

Edited December 10, 2019 by geordief

[Mordred]

Indistinguishable means you cannot identify one particle type from another. One way to understand this is think of a Bose Einstein condensate but at high temperature.

Edited December 10, 2019 by Mordred

[Mordred]

Here is a short video on Bose Condensates what essentially happens when particles become indistinguishable is that there Compton or DeBoglie wavelengths become identical.

https://www.google.com/url?sa=t&source=web&rct=j&url=https://m.youtube.com/watch%3Fv%3Du8wNSVxYZGI&ved=2ahUKEwjH_cyO3qvmAhWGHDQIHb4mBscQwqsBMAN6BAgHEA8&usg=AOvVaw0B9EXy8zTbUrcJ75t_Bq0C

Lol side don't pay too much attention to the particle animations the graphics are a bit off lol.

Edited December 10, 2019 by Mordred

[geordief]

On 11/30/2019 at 4:07 PM, studiot said:

We identify this property as the resistance of a body to impressed forces even in the total absence of the gravitational force as already noted.
It is called the inertial mass.

We do not know why when we take the numerical value of the inertial mass that value works correctly in equations for the gravitational effects.

From Einstein's Popular Exposition

https://www.bartleby.com/173/19.html

"According to Newton’s law of motion, we have
(Force) = (inertial mass) × (acceleration),
where the “inertial mass” is a characteristic constant of the accelerated body. If now gravitation is the cause of the acceleration, we then have
(Force) = (gravitational mass) × (intensity of the gravitational field),
where the “gravitational mass” is likewise a characteristic constant for the body. From these two relations follows:

(acceleration)=[(gravitational mass)/(inertial mass)] x(intensity of the gravitational field)
 
  If now, as we find from experience, the acceleration is to be independent of the nature and the condition of the body and always the same for a given gravitational field, then the ratio of the gravitational to the inertial mass must likewise be the same for all bodies. By a suitable choice of units we can thus make this ratio equal to unity. We then have the following law: The gravitational mass of a body is equal to its inertial mass."

 

Can I take it from what you said that Einstein was not really being mathematically rigorous? Is he perhaps  "jumping ahead" and just presenting a plausible mathematical argument for  the lay public ?

I found it hard to follow his equation and if you say that this relationship is more experimental than logical ,I will be relieved.....unless this is indeed rigorous I will have to re examine it.