Can gravity exist without mass?

In summary, the conversation explores the concept of whether or not gravity exists in the absence of matter and energy. One participant argues that if something cannot be measured or detected, it is not a subject for science, while another suggests that in some theories, it may be possible to detect spacetime without reliance on matter and energy. The conversation also delves into the definition of gravity and how it relates to the existence of matter and spacetime. Ultimately, there is no clear answer to the initial question, and the discussion highlights the complexities and nuances involved in understanding fundamental concepts in physics.
  • #1
JinChang
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This is sort of a chicken-egg conundrum or asking if a tree truly falls if no one's there to confirm it, so there may not be an answer, but here goes...

Does gravity exist if the universe is devoid of all matter, including the dark-matter? I suppose it might given that we are now talking about the dark-energy, but even that would mean no gravity can be detected for all intents and purposes if the distribution of dark-energy were uniform, wouldn't it?

So, does gravity exist only if there are means to detect it (i.e. via something like matter)? Also, without getting too philosophical, does "reality" consist only of things that can be measured (sure seems like it)? If so, since the very act of measurement requires having a reference (call it ground state), is everything we can measure all relative to some ground state/value? Finally, assuming that the logic holds thus far, can this reference or the ground-state vary (universally) and still produce the same measurement as before?

I guess what my questions are leading to is whether or not there can be a single Universal Constant that varies, but is tied to other Constants in such a way that relative values being measured (with the Universal Constant) are consistent so that the universe as we know it doesn't break down.

I'm probably reiterating things in layman's terms that have already been addressed technically elsewhere, so I apologize if this post is a repeat or too trivial. I also make this post because I find it fascinating that nature inevitably give rise to many chicken-or-the-egg scenarios where existence of one depends on the other without leaving little or no clue as to how either one initially got started.
 
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  • #2
I think it's pretty safe to say that if you can't measure something, even in principle, then it isn't a subject for physics (or more generally, science). I personally believe this should be considered an axiom.

As an example, any talk of a deity is not a subject for science since there is no way to prove the deity exists or not.

The more modern way to say this is any postulate must be falsifiable in principle before it can be considered a theory. This is essential for science to work.

As to your particular question regarding the existence of gravity in the absense of all matter and energy (I added energy since energy affects spacetime as well) I feel the answer should be 'no', as it would be impossible, even in principle, to detect it.

If you rephased the question slightly to 'does spacetime still exist in the absense of all matter and energy' the answer might be different since in some theories there are ways to detect spacetime in principle.
 
  • #3
paw,

Thanks for the response.

Your last statement is interesting since it's somewhat related to my original question. What are some ways of detecting space-time without reliance on existence of matter and energy? My initial thought would suggest that it's impossible.

Also, since we know that Matter and Gravity are both real, why shouldn't Gravity exist with Matter out of the picture? Or, is Gravity dependent on both Matter and Space-Time being present to be considered "real"?
 
  • #4
JinChang said:
Your last statement is interesting since it's somewhat related to my original question. What are some ways of detecting space-time without reliance on existence of matter and energy? My initial thought would suggest that it's impossible.

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.

JinChang said:
Also, since we know that Matter and Gravity are both real, why shouldn't Gravity exist with Matter out of the picture? Or, is Gravity dependent on both Matter and Space-Time being present to be considered "real"?

I'm suggesting that without matter or energy there'd be no way in principle to detect gravity therefore it would not be a matter for physics. This isn't the same thing as saying spacetime wouldn't exist. Simply that it would be a matter of philosophy.
 
  • #5
It depends on what one means by "gravity". Most textbooks use the equivalence principle to say that an observer inside an accelerating elevator in flat space-time experiences "gravity". This is an example of a sort of "gravity" that requires no mass.

However, this definition of "gravity" isn't really very deep, though it is commonly used. The question tends to devolve into a lot of semantic argument about the meaning of the words being used.
 
  • #6
pervect,

I think you nailed the source of confusion for layperson like myself when trying to understand the topics covered on these threads. Nearly all aspects of physics seem to depend on the semantics being used, which leads to different people having different understanding (however slight) of the topic. You can just imagine what that does to person like me when trying to keep up (sigh...).

To put it in other terms, it's like trying to understand ying-and-yang where the ying's existence depends on the yang (^^). One must accept both in order to make sense of it.
 
  • #7
The electromagnetic field gravitates. I'm not sure if electromagnetic energy can be emitted from anything but charged massive particles though. So, if there is no mass in your model, I'm not sure it's possible to consider the electromagnetic field's gravitational field.

Either way, both mass and light are expected to cause gravitation, in the real universe.
 
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  • #8
pervect said:
It depends on what one means by "gravity". Most textbooks use the equivalence principle to say that an observer inside an accelerating elevator in flat space-time experiences "gravity". This is an example of a sort of "gravity" that requires no mass.

I have to disagree with you and say what you have outlined in this instance as "gravity" does require mass.
The observers mass.
 
  • #9
paw said:
I think it's pretty safe to say that if you can't measure something, even in principle, then it isn't a subject for physics (or more generally, science). I personally believe this should be considered an axiom.
Not neccesarily. Sometimes the knowledge doesn't exist to make predictions for certain things. It may happen that some day the knowledge will be gained. For example: consider the multiple universe theory of quantum mechanics. We don't have the means for detecting such universes as of yet. For this reason the theory isn't taken too seriously. However there may come a day that humans will be able to construct wormholes. If it is possible to travel from one universe to another then we may yet have access to some of those universes. We'd then be able to study some of them to see if the multiple universe theory is in agreement with observations. It seems me that one construct a wormhole from one universe to the other. If such wormholes can be construced and they are stable enough for such a use then it seems to me that such a wormhole can connect different universes.
As an example, any talk of a deity is not a subject for science since there is no way to prove the deity exists or not.
First of all science itself does not have the means to ever prove any theory is true. However that doesn't mean that evidence can't be found which is consistent with observations supposedly made by ancient people. For example; consider the story of Moses. The Bible tells us that Pharaoh got upset with Moses which led to the demand that the Hebrews make their bricks with half the straw they usually did. Later Pharaoh demanded that they make bricks with no straw at all. If this is true then one would expect to find bricks which were made in different ways at the same construction area. E.g. one with normal bricks, one with less straw than normal and one with no straw at all. Such areas of construction has been discovered. What they found was normal bricks, bricks with less straw and bricks with no straw. Instead of straw what they found was other material such as twigs, small branches etc. Therefore what they found is consistent with what is told in the Bible. That is exactly how science works. Yoiu can read more about such discoveries in the journal Biblical Archeology. In fact when one studies the universe itself then one is often dumbfounded by what they find. E.g. if one of several constants of nature were slightly different than the actual values then life could not exist in the universe. Dr. Owen Gingerich, an astrophysicist at the Harvard Smithsonian Center of Astrophysics, wrote a book on this topic. I read it and found it a very interesting read.
The more modern way to say this is any postulate must be falsifiable in principle before it can be considered a theory. This is essential for science to work.
No such requirement is found in science that I'm aware of. I.e. it is not part of the scientific method.
If you rephased the question slightly to 'does spacetime still exist in the absense of all matter and energy' the answer might be different since in some theories there are ways to detect spacetime in principle.
I don't see how. If nothing exists then there is nobody around to run an experiment to test it and there would be no equipment to do the detecting. Observer and the observed cannot be separated ... at least I can't imagine how it would be done and am fairly certain that Observer and the observed cannot be seperated is a well known concept to most, if not all, physicists.

Pete
 
  • #10
JinChang said:
Does gravity exist if the universe is devoid of all matter, including the dark-matter?
I know that gravity, according to general relativity, can't exist in teh absence of mass since mass is referred to as the source of gravity. The term "matter" is one that is not well defined in physics but Eintein used the term to refer to that which is presence at points in spacetime where the stress-energy-momentum tensor is non-zero.

Pete
 
  • #11
paw said:
'does spacetime still exist in the absense of all matter and energy'
The simplest empty spacetime in general relativity is the Minkowski spacetime.

With some coordinate transformations we can make this spacetime locally identical with the Einstein homogeneous static universe (the solution with the famous cosmological constant).
 
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  • #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 ?
 
  • #13
Mentz114 said:
The Schwarzschild space-time has no matter, but lots of gravity. So do all vacuum solutions of EFE. Is this not gravity without matter ?
First off a Schwarzschild spacetime does have matter present. The Earth has a Schwarzschild spacetime outside the Earth. The mass of such a spacetime is determined by the orbits of test particles around the center of the gravitating body.

And no, that is not gravity without matter. I believe that the question referred to whether the presence of a gravitational field is generated by the presence of mass. There is no assumption here about the gravitational field at point A being generated by the mass at point A. Gravitational field at A is generated by mass at other locations. Thus a Shwarzschild spacetime is generated by the presence of the mass which generates that spacetime.

Pete
 
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  • #14
Mentz114 said:
The Schwarzschild space-time has no matter, but lots of gravity. So do all vacuum solutions of EFE. Is this not gravity without matter ?
The Schwarzschild solution represents the spherically symmetric empty spacetime outside a spherically symmetric massive body. Obviously if that body has zero mass the spacetime outside is a Minkowski spacetime.

pmb_phy said:
The Earth has a Schwarzschild spacetime outside the Earth.
That is incorrect, the Earth spins and thus a Schwarzschild solution cannot be used. Even if the Earth would not spin we could only use the Schwarzschild solution if there were no other bodies around. In general relativity we cannot simply add the Schwarzschild solutions of multiple bodies in a linear fashion. The only exception I can think of is the Schwarzschild lattice closed universe as described by Lindquist and Wheeler (rest his soul).
 
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  • #15
Pete,
Thus a Shwarzschild spacetime is generated by the presence of the mass which generates that spacetime.
Why isn't the mass represented by a non-zero EMT ? It's easy to show that no distribution of matter could generate a Schwarzschild space-time**.

The properties of the Schwarzschild space-time arise purely from its symmetry.

However, it can be joined onto an interior Sch. solution.

** I can show this.
 
  • #16
MeJennifer said:
That is incorrect, the Earth spins and thus a Schwarzschild solution cannot be used.
Thanks. That is correct. I was neglecting the Earth's rotation as an approximation. With bodies such as the Earth one must make aproximations since the Earth isn't really a perfect sphere and is rotating.
Even if the Earth would not spin we could only use the Schwarzschild solution if there were no other bodies around.
We can use the Schwarzschild solution as an excellant approximation by considering objects such as the moon to be test particles, i.e. objects which don't significantly alter the Schwarzschild solution.
Wheeler (rest his soul).
Wheeler died? I was unaware of that and very sad to hear of it. Yes - rest his soul indeed.

Pete
 
  • #17
Mentz114 said:
Pete,
Why isn't the mass represented by a non-zero EMT ?
What does "EMT" mean?
It's easy to show that no distribution of matter could generate a Schwarzschild space-time**.
That is quite wrong. Any spherically symmetric distribution of matter will generate such a field, a black hold being one of them and black holes do have mass.
The properties of the Schwarzschild space-time arise purely from its symmetry.
It arises due to a spherically symmetric distribution of matter.
However, it can be joined onto an interior Sch. solution.

** I can show this.
Okay. Please do so. I'm certain that you're using a different definition of mass than, say, MTW does. MTW quite literally defines "mass" of a gravitating body according to the orbits of test particles. Since such orbits are identical to mass distributions for a spherically symmetric spacetimes like the Schwarzschild spacetime it therefore has mass by definition. I anxiously await your proof to the contrary. Since your using a different definition than MTW please prove that their definition is wrong and yours is right. After all its not like you're saying that you dislike their definition. You're implying that it is quite erroneous.

Pete
 
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  • #18
pmb_phy said:
Wheeler died? I was unaware of that and very sad to hear of it. Yes - rest his soul indeed.
He died last Sunday.
 
  • #19
pmb_phy said:
Not neccesarily. Sometimes the knowledge doesn't exist to make predictions for certain things...

And until that day arrives any postulate must remain a postulate. I'm not saying we can't speculate but we need to recognize that we ARE speculating.

pmb_phy said:
First of all science itself does not have the means to ever prove any theory is true. However that doesn't mean that evidence can't be found which is consistent with observations supposedly made by ancient people. For example; consider the story of Moses...

I'm sorry but I will not argue religion. Religion is not science. My comment was neutral and for illustrative purposes only. I should have chosen a different illustration.

pmb_phy said:
No such requirement is found in science that I'm aware of. I.e. it is not part of the scientific method.

I'm surprised you'd say that. I'd say it's the most fundamental part of the scientific method. If a postulate does not allow falsification it will forever remain a postulate. A theory is a postulate which does allow falsification. A good theory is one which has survived falsification.

pmb_phy said:
I don't see how. If nothing exists then there is nobody around to run an experiment to test it and there would be no equipment to do the detecting. Observer and the observed cannot be separated ... at least I can't imagine how it would be done and am fairly certain that Observer and the observed cannot be seperated is a well known concept to most, if not all, physicists.

Now you are just nit picking. I made it quite clear that I was speculating what might be possible in other circumstances.
 
  • #20
paw said:
I'm sorry but I will not argue religion.
It was you that raised the subject, not I.
Religion is not science.
Huh? Nobody said religion was science. That doesn't mean that one can't address the other. E.g. historians use scientific knowledge to gain information about the past. Archaeologists are just such scientists.
My comment was neutral and for illustrative purposes only.
I see. Well we're not mind readers here.

Pete
 
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  • #21
pmb_phy said:
First off a Schwarzschild spacetime does have matter present.

Grrr. Pete should know better than to say things like this.
 
  • #22
Pete,

EMT is 'energy-momentum tensor'. It is the [tex]T^{\mu\nu}[/tex] on the rhs of the EFE.
That is quite wrong. Any spherically symmetric distribution of matter will generate such a field, a black hole being one of them and black holes do have mass.
Not according to the equations.
I'm certain that you're using a different definition of mass than, say, MTW does. MTW quite literally defines "mass" of a gravitating body according to the orbits of test particles. Since such orbits are identical to mass distributions for a spherically symmetric spacetimes like the Schwarzschild spacetime it therefore has mass by definition. I anxiously await your proof to the contrary. Since you're using a different definition than MTW please prove that their definition is wrong and yours is right. After all its not like you're saying that you dislike their definition. You're implying that it is quite erroneous.
I am using a different definition of mass from MTW. Their definition is operational and inferred from test particles, as you say. As I said, it is easy to show that no distribution of mass can give the Shwarzschild solution. I'm not sure this thread is the place for me to do it, so I'll PM or email you something.

M
 
  • #23
Exterior Schwarzschild is a vacuum solution... no matter (i.e. T_ab is zero) in this region.
http://books.google.com/books?q=schwarzschild+solution+vacuum

(It's not often appreciated that if you cut out a portion of a spacetime, the result is still a spacetime with the same metric and curvature of the original in the remaining region.)
 
  • #24
Starting with a line element that shows spherical symmetry,

[tex]ds^2 = -c^2B(r)dt^2 + B(r)^{-1}dr^2 + r^2d\theta^2 + r^2sin^2(\theta)d\phi^2[/tex]

the 00 component of the Einstein tensor
[tex]E_{\mu\nu} = R_{\mu\nu} - \frac{R}{2}g_{\mu\nu}[/tex]

is,

[tex]E_{00} = B(r)r^{-2}-B(r)^{2}r^{-2}-B(r)\partial_{r}B(r)r^{-1}[/tex]

The 00 component of the field equations is thus

[tex]B(r)r^{-2}-B(r)^{2}r^{-2}-B(r)\partial_{r}B(r)r^{-1} = \kappa T_{00}[/tex] ----------(1)

Setting T_00 = 0, gives a differential equation

[tex]\partial_{r}B(r) = r^{-1}-B(r)r^{-1}[/tex]

which has the solution,

[tex]B(r) = 1 - \frac{A}{r}[/tex] where A is a constant of integration. This is the Schwarzschild solution, and is only valid if T_00=0.

If T_00 is non-zero, then equation (1) would read,

[tex]\partial_{r}B(r) = r^{-1} - B(r)r^{-1} - \frac{r\kappa T_{00}}{B(r)}[/tex] ---------------- (2)

and the above solution is no longer valid, for any non-zero value of T_00.

This is using a hammer to crack a nut, but I originally did this calculation to see if equation 2 has any solutions - so far I haven't found any.

These calculations have not been checked by anyone else, so there could be errors. Do not take on trust.
 
  • #25
Gravity without matter

I am wholly confused! SR and GR dictate E=mc2. Matter inclusively requires mass. G is a function relative to both rest mass and relativistic acceleration. E in and of itself exhibits a gravitational effect. Where there is energy there is gravity in the absence of matter.
 
  • #26
Mentz114 said:
Starting with a line element that shows spherical symmetry,

[tex]ds^2 = -c^2B(r)dt^2 + B(r)^{-1}dr^2 + r^2d\theta^2 + r^2sin^2(\theta)d\phi^2[/tex]

the 00 component of the Einstein tensor
[tex]E_{\mu\nu} = R_{\mu\nu} - \frac{R}{2}g_{\mu\nu}[/tex]

is,

[tex]E_{00} = B(r)r^{-2}-B(r)^{2}r^{-2}-B(r)\partial_{r}B(r)r^{-1}[/tex]

The 00 component of the field equations is thus

[tex]B(r)r^{-2}-B(r)^{2}r^{-2}-B(r)\partial_{r}B(r)r^{-1} = \kappa T_{00}[/tex] ----------(1)

Setting T_00 = 0, gives a differential equation

[tex]\partial_{r}B(r) = r^{-1}-B(r)r^{-1}[/tex]

which has the solution,

[tex]B(r) = 1 - \frac{A}{r}[/tex] where A is a constant of integration. This is the Schwarzschild solution, and is only valid if T_00=0.

If T_00 is non-zero, then equation (1) would read,

[tex]\partial_{r}B(r) = r^{-1} - B(r)r^{-1} - \frac{r\kappa T_{00}}{B(r)}[/tex] ---------------- (2)

and the above solution is no longer valid, for any non-zero value of T_00.

This is using a hammer to crack a nut, but I originally did this calculation to see if equation 2 has any solutions - so far I haven't found any.

These calculations have not been checked by anyone else, so there could be errors. Do not take on trust.
I'll get back to you on this. In the meantime I'd like you to coinsider the fact that your initial assumptions also hold for flat spacetime (Minkoswki spacetime) and, at best, you've showed that there is not unique solution to Einstein's equation. There is more to it than this but I'm not seeing it at the moment. Let me study this and get back to you.

Until then please evaluate the constant of integration that you failed to evaluate. Until you evaluate that constant the solution is incomplete. When you evaluate this consant you're going to find that its the mass of the gravitatating body! :smile:

Note that it appears to me that you're deriving the relation for the spacetime of a non-rotating black hole. Such an object has a singularity at the origin. The mass distribution is non-zero, i.e. there is a delta function around the origin which accounts for the mass distribution. The stress-energy-momentum tensor does not vanish at the origin. You've not only ignored that but you've chosen a defintion which is not the one use in GR. Therefore you've only shown a different result which results from your own definition and have therefore merely expressed your opinion without finding an error in what I've stated above.

Pete
 
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  • #27
mojocujo said:
I am wholly confused! SR and GR dictate E=mc2. Matter inclusively requires mass. G is a function relative to both rest mass and relativistic acceleration. E in and of itself exhibits a gravitational effect. Where there is energy there is gravity in the absence of matter.
Not according to Einstein. As I mentioned above Einstein defined the term "matter" as that which is present where the stress-energy-momentum tensor does not vanish. Therefore where there is energy there is matter, by definition. In any case you're ignoring the mass-energy relation which says that where there is energy there is mass. In fact MTW refer to T00 as the mass density as well as the mass-energy density.

Pete
 
  • #28
Pete,

would you agree that T_00 for a spherical mass (pressureless) is given by a density times a radial step function ( c=1 in this expression),

[tex]T_{00} = \Theta( r - R_0)\kappa\rho[/tex]

Clearly, this will not allow the Schwarzschild metric as a solution. As you have said, it must be the solution for a black hole, which is not a spherical mass distribution.

I suppose one could say that the step function is zero outside R_0, so setting T_00 to zero is justifiable. But that loses the character of the mass distribution.

Until then please evaluate the constant of integration that you failed to evaluate.
The only way is to go to the weak field limit and show that A=2Gm/c^2 gives Newton's gravity. There is nothing in the derivation itself to suggest this.

If you have not already read it, I recommend Schwarzschild's original paper (in English) which is reprinted in the arXiv under physics/9905030.
 
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  • #29
Mentz114 said:
Pete,

would you agree that T_00 for a spherical mass (pressureless) is given by a density times a radial step function ( c=1 in this expression),

[tex]T_{00} = \Theta( r - R_0)\kappa\rho[/tex]

Clearly, this will not allow the Schwarzschild metric as a solution.
No. I would not agree. The delta function in this case is [itex]\delta(r) \rho[/itex].

Note: For some reason the rho in the expression for the mass distribution is not appearing. Why is that?
As you have said, it must be the solution for a black hole, which is not a spherical mass distribution.
You're kidding me right? The Schwarzschild metric is the solution for a non-rotaing uncharged black hole. As such the mass distribution in such a case is [itex]\delta(r)\rho[/itex] which is a spherically symmetric mass distribution.
The only way is to go to the weak field limit and show that A=2Gm/c^2 gives Newton's gravity. There is nothing in the derivation itself to suggest this.
It is implicit in the starting assumptions. Otherwise the constant of integration is zero and you're left with flat spacetime.
If you have not already read it, I recommend Schwarzschild's original paper (in English) which is reprinted in the arXiv under physics/9905030.
I read it several years ago. In any case you should consider the subject of that paper and its derivation. Just look at the title On the Gravitational Field of a Mass Point according to Einstein’s Theory. Notice that the author also speaks of the constant of integration that I've been talking about. Notice what he says about it The latter contains only the constant [itex]\alpha[/itex] that depends on the value of the mass at the origin.

If memory serves then the mass comes out of the metric when one uses Einstein's equation at the origin which will then show that there must be a delta function at r = 0 if the mass density is non-zero. The only way for it to be zero is when the constant of integration is zero which is a degenerate case of a Schwarzschild solution in which the curvature of the spacetime is zero everywhere and the spacetime is flat.

I'm still waiting for you to evaluate the constant of integration.

Pete
 
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  • #30
Pete,

No. I would not agree. The delta function in this case is .
Not much point in carrying on then. Clearly the words 'spherical mass distribution' mean something different to you.

You're kidding me right? The Schwarzschild metric is the solution for a non-rotaing uncharged black hole.
I'm not denying that.

As such the mass distribution in such a case is which is a spherically symmetric mass distribution.
A point is not a sphere ! You cannot call a 'point mass' a sphere.

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

I don't think we're going to agree, so I'll wait and see if anyone else has any views.
 
  • #31
Mentz114 said:
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. :smile:
 
  • #32
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.
 
  • #33
Mentz114 said:
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 [itex]\rho = \rho(r)[/itex] 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 boundry 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? :smile:
MeJennifer said:
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
 
  • #34
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.
 
  • #35
Howdy Mentz114
Mentz114 said:
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 [tex]T_{\mu\nu}[/tex] = 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 [tex]\rho[/tex] is said to have spherical symmetry if and only if the [tex]\rho[/tex] is a function of r only, i.e. \rho = \rho(r). Since the delta function [tex]\delta(r)[/tex] 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
 
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