Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

A singularity, by the definition I know, is a point in space with infinite of a property such as density.

Density is Mass/Volume.

Since the volume of a singularity is 0, then the density will thus become infinite because Mass/0 = undefined

However, is it possible to have a singularity with a mass of 0? 0/0 is indeterminate, but would it be possible for a singularity to exist even if its mass and volume were zero?

share|improve this question
    
Not being a physicist I must ask - What would such singularity consist of? Charge? –  Tomáš Zato May 8 '13 at 8:51
add comment

3 Answers

You can take a Reissner-Nordström solution for the charged non-rotating black hole, and put its mass $m=0$. Then it would become a so-called naked singularity. More precisely, singularity is a point where some value ends at infinity, while density of mass being just one option.

For more thorough consideration of Reissner-Nordstrom and Kerr-Newman solutions, refer to Hawking, Ellis: The Large Scale Structure of Space-Time.

Update: The Reissner-Nordström solution in some coordinates, given in Hawking, Ellis, has a form:
$$ds^2=-\bigl(1-\tfrac{2m}{r}+\tfrac{e^2}{r^2}\bigr)dt^2+\bigl(1-\tfrac{2m}{r}+\tfrac{e^2}{r^2}\bigr)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta\,d\phi^2)$$ We can see that it depends on a function $F(r)=1-\tfrac{2m}{r}+\tfrac{e^2}{r^2}$. It is positive on infinity, it tends to $+\infty$ in zero, and it can be entirely positive or not, depending on the sign of $e^2-m^2$. Particularly we can assume $m=0$, and the function will be essentially the same as in case $0<m^2<e^2$. It only would nowhere be increasing - the physical consequences of that are unclear to me. Nothing else would happen to that function and that solution at that point.

We can even consider $m<0$ and still have a correct solution to Einstein-Maxwell equations. Here $m$ has only the sense of the constant of integration. These options are useless if we seek for a metric to describe the empty space outside some spherical massive uncharged body, but if we talk about naked singularities - what do we know about the properties of such singularities? We should examine all possibilities.

share|improve this answer
    
Can you expand on "put its mass m=0"? An extremal Reissner-Nordström black hole doesn't have the mass equal to zero, it has a charge large enough to balance out the non-zero mass. If the mass were zero you'd have a charged massless particle and I'm not sure that's physically possible. –  John Rennie May 8 '13 at 8:51
1  
Formally there is no problem to take a limit $m\to 0$ for the case of $e^2>m^2$, or just immediately use $m=0$ in the formula of Reissner-Nordström metric. Yes this is kind of weird, and we do not know real particles fitting that case, but the formula would still be a valid solution of Einstein-Maxwell equations. I'll expand that in the text of the answer. –  firtree May 8 '13 at 9:21
    
@JohnRennie Done expanding. –  firtree May 8 '13 at 9:53
1  
I don't think this interpretation is quite right. The Reissner-Nordström solution is a valid electrovac solution containing a naked, timelike singularity. It is true, although nontrivial to prove, that the parameter $m$ equals the ADM mass arxiv.org/abs/0708.1958 , so in this case the ADM mass, i.e., the total mass is zero. However, that doesn't mean that there is no mass anywhere. As discussed in the link, the energy density of the electric field acts like a positive mass density, while the charge at the singularity acts like a negative mass density. –  Ben Crowell May 8 '13 at 11:21
    
[The second sentence in the comment above should read "The m=0 Reissner-Nordström solution is..."] –  Ben Crowell May 8 '13 at 11:29
show 2 more comments

This is a great question, although unfortunately it turns out to be very difficult to interpret it in a way that allows a definite answer. The question is ambiguous because of the way mass is defined in relativity. From the way the question is posed, I assume the OP doesn't have a lot of technical background in relativity. However, there is no way to resolve the ambiguities in the question without getting pretty technical.

In relativity, "mass" really means mass-energy. Mass isn't additive. For example, a photon has zero mass, but consider a system consisting of one photon moving to the right and another photon of equal energy moving to the left. This system has a nonzero mass. This follows from the definition of inertial mass in special relativity according to the equation $m^2=E^2-p^2$, in units with $c=1$.

In GR, the source of curvature isn't mass-energy density, it's the stress-energy tensor. Some of the components of the stress-energy tensor correspond to pressure rather than density of mass-energy $\rho$.

The total mass (i.e., mass-energy) of a system in GR is not always a well-defined thing. For an arbitrarily chosen spacetime, there is no way to define the total mass. There are definitions of mass that work (i.e., are conserved and scalar) in special cases, such as an asymptotically flat spacetime. For example, there's the ADM mass.

If we want to define the mass-energy density $\rho$ at a point, we can do that. It's one of the components of the stress-energy tensor. However, there are a couple of limitations here: (1) under a Lorentz boost, a $\rho=0$ can transform into a $\rho\ne0$; (2) a singularity isn't a point in space, it's more like a point removed from space, so we can't define $\rho$ at a singularity.

So for a singular spacetime, we can't define the mass-energy density at the singularity, and in a typical, general case, there is no way to define the total mass, either. You could have a spacetime with a family of observers defined, one at each nonsingular point in spacetime, such that every one of these observers detects $\rho=0$; however, other observers in different states of motion might measure $\rho=0$. This ambiguity only goes away if the whole stress-energy tensor vanishes, i.e., if it's a vacuum solution (not just an electrovac solution like the Reissner-Nordström metric).

It's probably possible to have a singularity such that, in the rest frame of the singularity, $\rho\rightarrow0$ as you approach the singularity, but the pressure blows up to infinity. (This isn't consistent with the equation of state of any known form of matter, and it violates various energy conditions.) However, the statement that $\rho\rightarrow0$ will be false in other frames.

It is definitely possible to take a bunch of massless ingredients such as photons, mix them together (so that the collection as a whole has nonzero mass), and then let them collapse gravitationally into a singularity. But then the ADM mass of the singularity won't be zero.

There are curvature singularities and conical singularities. For any curvature singularity, the energy stored in the gravitational field surrounding the singularity will probably show up as a nonzero ADM mass. A conical singularity might be the best bet for an affirmative answer to the question if you want a zero ADM mass as well as a zero stress-energy tensor everywhere. I don't know for sure whether a spacetime with these properties exists in 3+1 dimensions. I don't think conical singularities can form by gravitational collapse in our 3+1-dimensional universe.

share|improve this answer
    
> There are curvature singularities and conical singularities. < Where to read about them? –  firtree May 8 '13 at 12:36
1  
add comment

I recall a presentation (many years ago at DAMPT) where the presenter claimed that focussing gravity waves could produce a curvature singularity that bore some similarities to a black hole. I've done a quick Google and found this paper, that references two papers by Alekseev:

  • Alekseev, G. A. and Griffiths, J. B. “Gravitational waves with spherical wavefronts,” Classical and Quantum Gravity, 12,, pp.L13-L18 (1995).

  • Alekseev, G. A. and Griffiths J. B., “Exact solutions for gravitational waves with cylindrical, spherical and toroidal wavefronts,” Classical and Quantum Gravity 13, pp. 2191-2209 (1996).

Unfortunately neither of these are on the Arxiv and I can't find any copies that aren't behind paywalls, so I can't be sure these would match your criteria. Still, unless my memory is badly failing me, this does seem a physically reasonable way to create a singularity without any mass being present.

share|improve this answer
    
I think it's well known that colliding gravitational waves can result in a singularity. However, that doesn't mean that there is no mass. A box of photons has mass, even though the photons individually are massless. –  Ben Crowell May 8 '13 at 11:32
    
Well yes, given that the stress-energy tensor treats mass and energy as the same thing. I'm guessing that when the OP says mass they mean it in the non-relativistic use of the word. If $T_{00}$ is zero everywhere I can't think of any physically reasonable way to get a singularity. Is it possible for other elements to be non-zero if $T_{00}$ is zero? –  John Rennie May 8 '13 at 11:38
    
The question of whether $T_{00}=0$ is a completely coordinate-dependent one. One observer could say $T_{00}=0$ at a given point, and a different observer could measure $T_{00}\ne0$. –  Ben Crowell May 8 '13 at 12:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.