Can gravity exist without mass?

[MeJennifer]

I don't think we're going to agree, so I'll wait and see if anyone else has any views.

These discussions about whether acceleration in flat spacetime is gravity started already in 2003 on this forum. It is all arguments about terminology and not physics. A waste of time if you ask me.

 

[Mentz114]

These discussions about whether acceleration in flat spacetime is gravity started already in 2003 on this forum.

That is not what we're talking about.

A waste of time if you ask me.

It's not compulsory to take part.

 

[pmb_phy]

A point is not a sphere ! You cannot call a 'point mass' a sphere.

I didn't say that a point mass was a sphere. In fact it never entered my mind. I said that the distribution of mass is spherically symetric. That means that if you rotate the coordinates about the point of symmetry about any axis passing through that point and if the mass distribution does not change then the distribution has a a spherical symmetry to it. The reason the term "spherical" is ued is because of the coordinate system that this is usually expressed in, not because the object is spherical. You can, in fact, have a spherical object with a non-spherical mass distribution. Any mass density that has the form $\rho = \rho(r)$ has a spherically symmetry to it.

The constant that appears in the integration is associated with the mass by ansatz to make a 'connection with the physical world'.

That is correct. In mathematical language its called boundary conditions. In the present case the constant of integration has the physical interpretation of the mass of the body generating the gravitational field. This constant may very well be zero as you are assuming here. Once that constant is set to zero the spacetime becomes flat because the problem becomes empty spacetime which then has zero spacetime curvature.

I don't think we're going to agree, so I'll wait and see if anyone else has any views.

People agree or disagree when it comes to things such as interpretation of some physical phenomena (like the physical interpretation of the wave function in quantum mechanics) or how a term should be defined etc. The result of a calculation is not subject to such opinions. It has a very definite answer to it. In the beginning you started with the assumption that the spacetime is devoid of matter (i.e. the stress-energy-momentum tensor vanishes at all events in spacetime). That means that there is no matter at the coordinate system. So either there is a mass point there or there is no mass point. No other possibility exists. The derivation of the Schwarzschild metric requires boundary conditions which you didn't state in your original derivation. That made it incomplete. The actuall derivation of the Schwarzschild metric starts with the bondary conditions explicitly stated somewhere along the line. A Schwarzschild solution which has no physical connection to the real world cannot be said to actually exist and therefore cannot be said to be a spacetime devoid of matter. If we impose a physical interpretation, i.e. make this a real physical problem to solve then there is indeed a point mass at the origin. In fact the very paper you referenced stated as such.

And no. We don't have to agree at all. Wouldn't the world be quite boring if everyone agreed on everything?

It is all arguments about terminology and not physics.

Some of the most important concepts in math and physics are about terminology. That doesn't make the physics any less important. Let us not forget that Einstein himself made some major changes in physics by getting deeper into the definitions of certain things like "time" and "simultaneous" and "length" etc. Ignoring the simple things means that we're not paying attention to the basic foundations of our lovely science.

Best wishes

Pete

 

[Mentz114]

Hi Pete,

This constant may very well be zero as you are assuming here.

I am not assuming this. Where do I say this ?

I stand by my statement the Schwarzshild metric is not the exterior space-time of a spherically symmetric mass distribution. Matter has an equation of state. What is the equation of state of a 'point mass' ?


MeJennifer,
I'm sorry I snapped at you. Frustration at being misquoted.

 

[pmb_phy]

Howdy Mentz114

I am not assuming this. Where do I say this ?

My mistake. I should have said that you implied it rather than stated it. Recall your first post in this thread
https://www.physicsforums.com/showpost.php?p=1691532&postcount=12

The Schwarzschild space-time has no matter, but lots of gravity. So do all vacuum solutions of EFE. Is this not gravity without matter ?

In https://www.physicsforums.com/showpost.php?p=1693337&postcount=24
you also stated the condition Setting T_00 = 0. If you see the energy density equal to zero everywhere then the mass-density is zero everywhere as is the resulting total mass. However I neglected to notice something when you stated that. The zero vacuum is not defined by T₀₀ = 0 but by $$T_{\mu\nu}$$ = 0 (as well as having a zero cosmological constant).

Therefore, if you hold that the mass is zero then it implies that the constant is zero since they are equal.

I stand by my statement the Schwarzshild metric is not the exterior space-time of a spherically symmetric mass distribution.

I disagree of course. I can't ever comprehend why you'd make such a statement. Did you read this in a GR text somewhere? Do you know of a GR text which states this? If eitehr is true then I'd love to read it. Can you pleae give me a reference to some place in the physics literature where I can find such an assertion? Thanks.

What is the equation of state of a 'point mass'?

The equation of state of a body is not required in order to find the gravitational field. All that is required is the stress-energy-momentum (SEM) tensor. There is nothing in Einstein's field equations which describes the equation of state.

Regarding the concept of a spherically symmetric mass distribution: A mass distribution function $$\rho$$ is said to have spherical symmetry if and only if the $$\rho$$ is a function of r only, i.e. \rho = \rho(r). Since the delta function $$\delta(r)$$ is just such a function then it is a spherically symmetric mass distribution, by definition. If you know of a text or research article which states otherwise please post the reference. I'd love to read it. Thanks.

Best wishes

Pete

 

[pmb_phy]

These discussions about whether acceleration in flat spacetime is gravity started already in 2003 on this forum. It is all arguments about terminology and not physics. A waste of time if you ask me.

Hi MeJennifer

Let me give you an example of why this is not a waste of time. If some one believes that where there is a gravitational field there is spacetime curvature then its very tempting to take that to mean that this is true in all possible instances of a gravitational field. One physicist who did this assumed that the spacetime of a uniform gravitational field had a non-zero spacetime curvature. Therefore when he stated what he thought was a definition of a uniform gravitational field (which he stated as dz/d$\tau$ = constant) he arrived at a curved spacetime. Had he understood the true meaning of what a uniform gravitational field was then he would have been surprised when he got a curved spacetime and he'd therefore modify his assumption of what a uniform gravitational field was. In the case I'm thinking of he not only got the answer wrong but both the editor and the referees missed this error and as such the paper was published. A very bad paper too. All resulting because he didn't understand the relationship between gravity and spacetime curvature and their definitions. This led the author to make other wrong conclusions on a very basic postulate of relativity. The article is

Nonequivalence of a uniformly accelerating reference frame and a frame at rest in a uniform gravitational field , Edward A. Desloge, Am. J. Phys. 57, 1121 (1989). The abstract reads

A general expression is obtained for the space-time interval between neighboring events in a one-dimensional space in which it is possible to set up a rigid reference frame. Particular expressions are then obtained for the interval for the special cases of a rigid frame at rest in a uniform gravitational field and a rigid frame uniformly accelerating in field-free space. The two expressions are not equivalent and are used to show why, how, and to what extent observations made in a rigid enclosure at rest in a gravitational field are not equivalent to observations made in a rigid enclosure that is uniformly accelerating in field-free space. Two facts of particular interest that are demonstrated in the course of the analysis are the following: (i) Two spatially separated particles that are simultaneously released from rest and allowed to fall freely in a uniform gravitational field will not remain at rest with respect to one another. (ii) Uniformly accelerating reference frames and inertial frames are the only possible one-dimensional rigid frames in flat space-time.

Pete

 

[pmb_phy]

By the way Mentz114, the SEM tensor is not zero everywhere. It is only zero in that region of spacetime where there is a vacuum. That means that it is zero outside the matter distribution. Therefore what you've been talking about is the exterior solution. That means that it applies to the regions outside the Sun, Planet etc. It does not apply to regions inside the gravitating body and it sure doesn't apply to the origin for a point particle.

Pete

 

[Mentz114]

Hi Pete,
you are misunderstanding still. I am talking about a spherically symmetric mass not a spherically symmetric space-time. A ball of matter, not a point. The Scwarzschild metric is not the space-time of a ball of matter. If it were, we would have a non-zero T_00. All I've shown ( unless my maths is faulty) is that the Schwarzschild metric cannot be the metric of any space containing matter ( which we knew in any case).

In https://www.physicsforums.com/showpos...7&postcount=24
you also stated the condition Setting T_00 = 0. If you see the energy density equal to zero everywhere then the mass-density is zero everywhere as is the resulting total mass. However I neglected to notice something when you stated that. The zero vacuum is not defined by T00 = 0 but by = 0 (as well as having a zero cosmological constant

Therefore, if you hold that the mass is zero then it implies that the constant is zero since they are equal. ).

I don't understand these remarks at all. All I've done is some simple maths which you don't seem to grasp.

The equation of state of a body is not required in order to find the gravitational field. All that is required is the stress-energy-momentum (SEM) tensor. There is nothing in Einstein's field equations which describes the equation of state.

I disagree. You can't write the EMT ( or SEM as you prefer to call it) without the equation of state. See the FLRW solution, where the equation of state is crucial in evaluating the metric.

Please don't bang on about your delta function. $$\rho\delta(r)$$ is not a mass distribution because it is infinitisimal. A ball of matter is described by a radial step function with a non-zero R_0, not a delta function.

I'm very busy at work so I'm giving up on this. You fail to understand what I mean by a spherical mass distribution and you deny simple maths - what can I do ?

Best wishes to you too,

M

 

[pmb_phy]

I neglected to say that the Black holes that I was referring to are of the kind that didn't form by gravitational collapse. I was thinking of the kind that have existed from the beginning of the universe, i.e. microblack holes. Collapsed stars are not singularities, although there is no way to determine that from observation.

Pete

 

[Mentz114]

Hi Pete,
I've been thinking about this and looking at other solutions of the EFE and I have to say I'm not so sure of my ground anymore. With the Tolman solution for instance, it is assumed that the field outside a collapsing star is Schwarzschild and remains so throughout the collapse. Maybe it really is.

M

 

[pmb_phy]

Hi Pete,
I've been thinking about this and looking at other solutions of the EFE and I have to say I'm not so sure of my ground anymore. With the Tolman solution for instance, it is assumed that the field outside a collapsing star is Schwarzschild and remains so throughout the collapse. Maybe it really is.

M

Hi Mentz

No worries my friend. I too have been rethinking my comments above. For now I'll have to retract my assertions regarding the stress-energy-momentum tensor of a black hole through all of spacetime. I spoke of it because I recall reading a paper on some sort of averaging process over diract functions. However I don't understand the particulars so its more or less high gugu physics to me. If I don't understand something fully then I'd rather not make assertions about it as I did here. I therefore retract my comments regarding this topic.

I also didn't understand that you were unaware that the field outside a collapsing star is Schwarzschild and remains as such throughout the collapse.

Pete

 

[NYSportsguy]

Vacuum Energy

Notice I said 'in some theories'. I believe that in some theories of quantum physics and possibly in string theory it might be possible to detect virtual particles forming and annhilating from the vacuum. Some explanations of the Casimir effect suggest vacuum energy (as opposed to EM) might be detectable. Hawking radiation from black hole evaporation if detected would in some sense indicate the existence of the spacetime vacuum. All of this is speculation at this time I believe but allows the possibility in principle.

Vacuum energy is still a type of "energy" nonetheless and as long as there is energy somewhere there will always be gravity as a result of it. According to Einstein, gravity is a RESULT of matter or energy. If neither were around there would be no gravity because in essence, space-time would not be "curved" or "indented" in any way.

Hope this helps.

 

[Nickelodeon]

Vacuum energy is still a type of "energy" nonetheless and as long as there is energy somewhere there will always be gravity as a result of it.

We might need another term here. By default, gravity is an effect caused by the presence of matter. A similar effect to gravity could be caused by energy with no associated matter present.

 

[pmb_phy]

We might need another term here. By default, gravity is an effect caused by the presence of matter. A similar effect to gravity could be caused by energy with no associated matter present.

You can find more on this at http://en.wikipedia.org/wiki/Einstein_field_equations. See section on the cosmological constant.

Pete

 

[NYSportsguy]

Well we know E= MC^2. Thus in theory energy = mass. They are interchangeable.

So if mass can cause gravity, so can energy. This is a proven fact.

Vacuum energy causes gravity.

Now if there was no energy or no mass in our universe, then gravity would not be present. The concept of gravity would still be alive however...but none would be present.

 

[Freezeezy]

Pete and Mentz, I have to say that I love all of your comments. They are truly rational and clearly thought out.

I must say that to stay in accordance with SR and GR, Gravity does not exist without matter. Mass distorts space-time and thus gravity can be evaluated... Of course... you can look at the lack of mass meaning that gravity is just null... not void.

So choose what you want...

I say that gravity exists (Zero gravity is still gravity... like zero is still a number... which took the Romans years to figure out) when no mass is present.

 

[campal]

someone might have posted this already, sorry if they have. but doesn't a black hole have gravity and no mass, just a rip in space time?

 

[pmb_phy]

Pete and Mentz, I have to say that I love all of your comments. They are truly rational and clearly thought out.

Thanks Freezeezy. That is kind of you to say. Its nice to know that all the posting I do here is not in vain.

someone might have posted this already, sorry if they have. but doesn't a black hole have gravity and no mass, just a rip in space time?

No. A black hole definitely has mass. It isn't necessarily a rip in spacetime either. A star collapsing might collapse into a black hole and yet leave no singularity.

Pete

 

[NYSportsguy]

The reason a black hole is a black hole to begin with is because it has TREMENDOUS mass and density...lol.

Another thing, I truly think "black holes" are conclusive evidence that the universe is flat in shape and not positively or negatively curved as some cosmologists and astrophysicists hypothesize.

 

[George Jones]

TNo. A black hole definitely has mass. It isn't necessarily a rip in spacetime either. A star collapsing might collapse into a black hole and yet leave no singularity.

Why do you say this? Can you give a referenece?

Another thing, I truly think "black holes" are conclusive evidence that the universe is flat in shape and not positively or negatively curved as some cosmologists and astrophysicists hypothesize.

Why do you say this? Can you give a referenece?

 

[pmb_phy]

Why do you say this?

As matter falls into a black hole it slows down and, as measured by an outside observer, never gets to or past the event horizon.

Can you give a referenece?

I'll look for one. Perhaps I'm wrong.

Pete

 

[George Jones]

As matter falls into a black hole it slows down and, as measured by an outside observer, never gets to or past the event horizon.

It seems that you mean the purely classical case of stellar collapse. I thought you meant either the semi-classical case of Hawking radiation, something completely non-classical like loop quantum gravity or string theory.

In the classical case, the existence of an infinite redshift surface does not preclude the existence of spacetime singularities. In fact, in the purely classical case of stellar collapse, once the collapse proceeds far enough (as it does for sufficiently massive stars), I know of no way to evade the Penrose-Hawking singularity theorems. A person falling along with the collapsing surface of the star will rapidly find himself inside an event horizon, and will almost as rapidly himself up close and personal with a singularity.

If Hawking radiation is considered, the mainstream view is that a singularity still forms. A small minority of physicists think that Hawking radiation will carry away enough mass/energy to prevent the formation of an event horizon and singularity; see https://www.physicsforums.com/showthread.php?p=1534827#post1534827".

Quantum theories of gravity are not sufficient developed for the emergence of a consensus view on status of spacetime singularities inside event horizons.

 

[pmb_phy]

It seems that you mean the purely classical case of stellar collapse.

The term classical includes relativistic. If you mean Newtonian then no, the slowing down of matter and stopping outside the event horizon is not a Newtonian phenomena.

In the classical case, the existence of an infinite redshift surface does not preclude the existence of spacetime singularities.

True.

In fact, in the purely classical case of stellar collapse, once the collapse proceeds far enough (as it does for sufficiently massive stars), I know of no way to evade the Penrose-Hawking singularity theorems.

Sorry but I never heard of them. But this is an observer dependent phenomena. Whether something crosses the event horizon depends on who is doing the observing.

A person falling along with the collapsing surface of the star will rapidly find himself inside an event horizon, and will almost as rapidly himself up close and personal with a singularity.

True. But the subject is in regards to observers outside the event horizon. Not about observers who are falling through it.

Pete

 

[MeJennifer]

Note that black holes can only completely form in an open universe.

 

[Haelfix]

Thats a bit dubious MeJennifer. I know why you would say that, but I think its a bit of a mathematical technicality more so than a real physical effect. In physics, we tend to cut infinities off as a general rule.

A big star collapses, it pops out a black hole. Which is why we likely had black holes 10 billion years ago, even though we weren't sure whether or not we were open, closed or critical. Indeed, they may have evaporated already, depending on how you model quantum effects.

 

[George Jones]

The term classical includes relativistic.

Yes, usually, and that is how I used the term

Sorry but I never heard of them. But this is an observer dependent phenomena. Whether something crosses the event horizon depends on who is doing the observing.

The Penrose-Hawking singularity theorems are infamous in general relativity. Misner, Thorne, and Wheeler gives a brief treatment, Wald's text gives a detailed treatment, and they are at least mentioned in many introductory GR texts.

True. But the subject is in regards to observers outside the event horizon. Not about observers who are falling through it.

No, the subject is singularities inside black holes.

In post #48, you said

A star collapsing might collapse into a black hole and yet leave no singularity.

In classical general relativity, this just isn't true.

 

[pmb_phy]

The Penrose-Hawking singularity theorems are infamous in general relativity. Misner, Thorne, and Wheeler gives a brief treatment, Wald's text gives a detailed treatment, and they are at least mentioned in many introductory GR texts.

Do you have a precise reference?

Pete

 

[George Jones]

Do you have a precise reference?

I won't have the same spacetime coordinates as my relativity books until Monday, and I can give you a bunch of references then.

The only reference that I have at home is The Edge of Infinity: Beyond the Black Hole, by Paul Davies, which has a very good popular-level treatment of singularities. I really like this book, and it gives a tremendous (and quite accurate) explanation of Penrose's first singularity theorem. I even recommend it as a complement to mathematical references for physics types encountering for the first time the technical details of singularity theorems.

 

[pmb_phy]

I won't have the same spacetime coordinates as my relativity books until Monday, and I can give you a bunch of references then.

The only reference that I have at home is The Edge of Infinity: Beyond the Black Hole, by Paul Davies, which has a very good popular-level treatment of singularities. I really like this book, and it gives a tremendous (and quite accurate) explanation of Penrose's first singularity theorem. I even recommend it as a complement to mathematical references for physics types encountering for the first time the technical details of singularity theorems.

Great! Thanks George! A short time back I got into a discussion about black hole singularities and I became convinced that it wasn't possible to state that a black hole had a singularity since there would be no way to observe them. I also read something in Kop Thorne's book "Black Holes & Time Warps" regarding frozen stars. I guess I gave up too soon and for no good reason. Thanks for catching me on this. Greatly appreciated.

Pete

 

[quantumcore]

well empty space does not mean that there is zero energy , and as relativity says energy is equivalent to mass , and mass create gravity . hence if space -time is there then i think gravity is also there.