# Is it possible to have a singularity with zero mass?

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?

• Not being a physicist I must ask - What would such singularity consist of? Charge? – Tomáš Zato May 8 '13 at 8:51

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.

• 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
• 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
• 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

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.

• > There are curvature singularities and conical singularities. < Where to read about them? – firtree May 8 '13 at 12:36
• – Ben Crowell May 8 '13 at 13:19

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.

• 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

BLACK HOLE FORMATION At the final nanosecond in the formation of a black hole, essentially all of the matter confined within the envelope of a developing Schwarzschild sphere (whatever its size) can no longer be transformed into particle kinetic-energy. Special relativity does NOT allow an inertial system to exceed the speed of light; and particles colliding at speeds approaching that of light cannot absorb a further increase in the momentum produced by gravitational forces. At this point, gravitational energy is transformed directly into radiation (entropy) rather than particle momentum.

This critical event, representing a change-in-state from matter to radiation, is preceded by an exponential increase in the momentum of particles confined within the volume of a contracting star (or any object). As particle velocities approach the speed of light, and as distance and time between particle collisions approach zero, the energy-density of a collapsing object will reach a limit where the transformation of gravitational force can no longer be defined in terms of particle collisions.  As the distance between colliding particles gets smaller, quantum mechanical factors require that the uncertainty in particle momentum correspondingly gets larger. At a critical point in this combination of events, the collision-distance between particles has decreased to a nanometer range that corresponds to the frequency of particle collisions; and particle interactions can be expressed in terms of a series of discrete quantum-mechanical wave-functions (distance and time between particle collisions) that approach the restricting value of 'h' (the Planck constant), producing a potential, quantum-mechanical catastrophe.

In order to preserve thermodynamic continuity, the thermodynamics of the system must change; consequently, particle matter is transformed into radiant energy by means of quantum mechanical processes. Gravitational energy is now expressed as a function of the total radiation-energy distributed over the surface of the ensuing Schwarzschild sphere, and any additional energy impacting the Black-hole produces an expansion of the Schwarzschild radius, while maintaining a constant energy-density, and a constant boundary acceleration, corresponding to a constant (Unruh) temperature.

A clue to the transformation of kinetic energy to radiation is seen in the function: e^hf/KT (from the Planck "key" to the ultraviolet paradox). In this function, particle kinetic-energy, "KT", increases due to an increase in particle velocity and effective particle temperature; the frequency of particle collisions, represented by "hf", increases with particle density due to increased gravitational confinement within a developing Schwarzschild object (Black Hole). But temperature and frequency do not rise to infinity, as one might expect from the Planck relationship.

The formation of a Schwarzschild boundary is coincidental with a maximum particle acceleration and a maximum temperature. This critical event represents the maximum energy-density (rather than the maximum energy) permitted by nature. Temperatures cannot rise beyond this critical point. Instead, these variables now become Black-hole constants and are conserved in all Black-Holes, regardless of their size and total energy. Subsequent to the formation of a Black-hole, temperature, acceleration, gravity and energy-density remain constant at their maximum values...even as more energy is added and the Schwarzschild envelope grows correspondingly larger.