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My cursory studies of general relativity had introduced me to the concept of how gravity is associated with the warping of spacetime and that the presence of matter in space warps space and this warping is observed as gravity. And this left me with the belief that you have to have matter to have mass and thus gravity. And only when you have matter do you have mass.

But now I'm reading Kip Thorne's The Science of Interstellar, and in Thorne's description of black holes he says".. matter as we know it gets stretched and squeezed out of existence." and "all that is left is warped space and time"

So according to Thorne (and others) black holes contain no matter, yet they have mass and exert gravitational forces. So my initial belief must therefore be wrong.

How can mass exist without matter?

Black holes are created from collapsing stars that once contained very large quantities of matter. Through the process of collapse and formation of the black hole, mass is conserved, but matter is not? Where did the matter go?

docscience
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7 Answers7

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Treating only

How can mass exist without matter?

and not the context, it is important to understand the usual relativistic definition of mass as the norm of the energy-momentum four-vector $$ \left( m c^2 \right)^2 = \left|\mathbf{p}\right|^2 = E^2 - \vec{p}^2 c^2\,.$$

In this context any system of photons not all pointing in the same direction has mass without having any "matter" in a conventional understanding.

dmckee --- ex-moderator kitten
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  • I understand that photons have a rest mass, but can you please elaborate regarding moving photons and direction? I see in this equation momentum, $p$ with an arrow over it as a (four?) vector in the right most term as the only indication of directionality, but it's squared, so is this then a dot product which loses sense of direction? And what in the equation says you do or do not have matter? Just the fact you've selected photons? – docscience Jan 17 '15 at 23:01
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    Individual photons do not have a rest mass (or even a rest frame). Notationally, the bold face represents a four-vector and the arrow a three-vector. Alas, if you are not familiar with the energy-momentum four vector and its relationship to mass this answer is probably not very helpful. – dmckee --- ex-moderator kitten Jan 17 '15 at 23:21
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    That said, photons have four-momenta $(E,E\hat{p}/c)$ (meaning that their mass is zero), but two or more taken together (i.e. considered as a system) can have non-zero mass. The condition is that their momenta are not all in the same direction (that is $\hat{p}_1 \ne \hat{p}_2$). The easiest way to see this is by choosing back-to-back photons of the same energy. Their combined four-vector is $(2E,\vec{0})$ so that the mass of the system is $2E/c^2$. – dmckee --- ex-moderator kitten Jan 17 '15 at 23:27
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"..matter as we know it gets stretched and squeezed out of existence." and "all that is left is warped space and time"

As you can't observe anything inside Event Horizon of a Black Hole, it can be seen as the matter goes out of existence as it can't be located in our Spacetime. That's it.

Earth is a Spoon
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I think Kip Thorne is actually talking about something about which there is no wide scientific consensus.

When a star collapses into a black hole, what remains is a complete vacuum with a single singular point (a singular hoop in the case of a spinning star). The singular point or hoop is a point where the equations break down - from a strict methodological viewpoint it should not be considered a part of the space-time. But it is this very point where all the matter was before the final stages of the collapse and where some physicists would say "all the matter was squeezed into".

Mathematical physicists call these space-times "vacuum space-times" because they are solutions to vacuum Einstein equations and the warping around such singular points is supplied by a set of boundary conditions. I.e., the space-time ends at the singular point inside the black hole and you only impose the warping of the space-time in an infinitely close neighborhood and at infinity. Just by imposing these conditions you recover the complete structure of the black hole.

You are faced with three possibilities of handling this squeeze-out of matter. First, you can say that all the matter is squeezed into the singular point. Second, you say that the point is actually a singular hole in the space-time pinched in there by a collapse of matter and that the matter was, indeed, squeezed out of existence. And third, you say that the laws are actually slightly different, so that this singular squeeze-out never happens in reality and is only an artefact of extending a set of effective laws beyond their range of applicability.

The first two possibilities actually bear the same physical implications for the surrounding warped space-time, so physicists do not really make a distinction between them. However, the most widely accepted possibility is the third one, and the theory which should prevent the singular squeeze-out is supposed to be quantum gravity.

Void
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  • In his new book, Thorne is writing for the everday person but categorizes his discussions as either Truth, Educated Guesses or Speculation. For this particular subject he has categorized it as truth. My family is originally from Missouri - so to me its almost all speculation. To me facts are something you are willing to stake your life (not necessarily your career) on :) – docscience Jan 17 '15 at 19:02
  • Well, I haven't read the book so I am just speculating but I think the assertion of truthfulness Kip Thorne is presenting is slightly more flexible. It could be that relativity is not the correct theory of gravity, the observations we have are somehow misinterpreted etc. But the scenario of a collapse has been studied so thoroughly by mathematicians and physicists that it is completely safe to say that if relativity applies at all scales, this "squeeze-out" of matter must happen. The thing is that most physicists believe relativity does have a scale where different laws start to apply. – Void Jan 18 '15 at 11:03
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Mass is not a conserved quantity, as special relativity has taught us. What is conserved is energy, and that is how we get energy out of nuclear reactors. General relativity goes further by postulating that space time itself creates the format of matter and energy,

At its core are Einstein's equations, which describe the relation between the geometry of a four-dimensional, pseudo-Riemannian manifold representing spacetime, and the energy–momentum contained in that spacetime. Phenomena that in classical mechanics are ascribed to the action of the force of gravity (such as free-fall, orbital motion, and spacecraft trajectories), correspond to inertial motion within a curved geometry of spacetime in general relativity; there is no gravitational force deflecting objects from their natural, straight paths. Instead, gravity corresponds to changes in the properties of space and time, which in turn changes the straightest-possible paths that objects will naturally follow. The curvature is, in turn, caused by the energy–momentum of matter. Paraphrasing the relativist John Archibald Wheeler, spacetime tells matter how to move; matter tells spacetime how to curve.

So it is not a simple matter. ;)

anna v
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  • How mass is not conserved? – Paul Feb 07 '15 at 15:10
  • @Paul Example The mass of the deuteron is smaller than the sum of the masses of a proton and a neutron, of which it is composed. http://hyperphysics.phy-astr.gsu.edu/hbase/particles/deuteron.html – anna v Feb 07 '15 at 15:19
  • But energy will be conserved in a system and energy is mass by $E=mc^2$? – Paul Feb 08 '15 at 02:39
  • Mass of a system is always conserved because energy and momentum is conserved and (Esystem)2= (psystemc)2+ (msystemc2)2 , I dont know where I am wrong. http://profmattstrassler.com/articles-and-posts/particle-physics-basics/mass-energy-matter-etc/mass-and-energy/ – Paul Feb 08 '15 at 03:36
  • @Paul you are ignoring that (msystemc2)2 is variable unless it is an elementary particle. m of the system is a dot product in four dimensions. In a similar way that a dot product in three dimensions depends on the angles of the two vectors and even though the length of the vectors does not change the dot product is variable. – anna v Feb 08 '15 at 04:02
  • In the E=mc**2 form m is variable as it it the relativistic mass, changes with velocity . It is not the same as the mass in the formula of your second comment, and the relativistic formula is not used by current physics studies because of the confusions that arise. – anna v Feb 08 '15 at 04:06
  • @Paul when the system changes, as in a decay for example, msys changes. Conservation implies checking over changes. The energy is not lost, but the dot product changes because of changes in momenta in the new conficuration – anna v Feb 08 '15 at 04:09
  • Take a ''isolated'' box with matter and antimatter inside it,after sometime there will be photons(due to Annihilation) it but mass ,energy and momentum of the box will remain same for a given frame of reference.http://physics.stackexchange.com/questions/137316/photons-inside-a-box we always talk about conservation for a particular frame of reference. – Paul Feb 08 '15 at 06:25
  • The wiki article is confusing me really http://en.m.wikipedia.org/wiki/Conservation_of_mass – Paul Feb 08 '15 at 07:10
  • imo it is written in a very confusing way by somebody who wants to stick to the classical conservation of mass concept. A closed system has a given invariant mass and it will not change unless there is external interaction. At its rest frame the invariant mass is equall to the energy of the system. that is all. – anna v Feb 08 '15 at 07:27
  • A closed system has a given invariant mass and it will not change unless there is external interaction. THEN MASS IS CONSERVED,THATS WHAT WE MEAN BY CONSERVATION.?? – Paul Feb 08 '15 at 07:36
  • @Paul It is a total mass. It is not the sum of the individual masses in the system that is conserved. In classical physics we have conservation of mass. If we add a kilo of salt in ten kilos of water we will have 11 kilos of mass. This is what conservation of mass means classically. It does not hold with special relativity conditions. – anna v Feb 08 '15 at 07:59
  • Isn't there a quantum mechanical experiment in which two particles with mass collide and as a result of the collision (two or three) photons are produced? – jjack Apr 03 '15 at 19:16
  • @jjack yes. example electron positron annihilation http://en.wikipedia.org/wiki/Electron%E2%80%93positron_annihilation – anna v Apr 04 '15 at 03:22
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The common experiences attributed to gravity are simply the effects of a curved spacetime. Spacetime curvature can persist and propagate even in regions far from any mass or matter.

What mass (or more generally, energy density, momentum density, and stress) does is change how curvature at one time is related to the curvature before and nearby to be different than it otherwise would be.

An analogy would be electromagnatism. Electromagnetic fields can exist, persist, and propagate far from any charges, but charges allow the relationships between fields at different points in spacetime to be different than would otherwise be allowed. In fact, based on the relationships between the fields we can find out what net charges are around.

Similarly for curvature, give the metric, we can compute the Einstein tensor based on the changes in the metric (in space and time). From that we can find out whether there is net energy, momentum or stress in a region.

OK. So that's how mass and matter is related to spacetime (they can cause energy density, momentum density, and stress). But if you are far from any such sources, you don't know what (if anything) is causing that curvature, because that curvature is simply caused by earlier and nearby curvature.

Going back to our analogy with electromagnetism, it's like if you saw an electromagnetic wave zip by. It probably was caused by some charges in the past. But the equations themselves can't tell you that, if it has just been zipping around the universe forever, then it will continue until some charge or current somewhere makes it propagate different.

So when you notice a curved spacetime you can compute the metric and then the Einstein tensor and see whether it is zero, or whether it is electrovacuum or whether it is something else not corresponding to mass or matter. But that doesn't tell you whether there is mass or matter elsewhere else in the past that is responsible in some way for what you see in the here and now. There might be, there might not be. If what you see is an electromagnetic wave, then the analogy is strong in this example, and it cold have been caused by some stress, momentum, or energy in the past, or maybe it's always been zipping around. And looking at the curvature out here in the now doesn't and won't tell us that.

An extreme case of ignorance would be a stationary, static, spherically symmetric spacetime in a shell like region of vacuum. We can't tell from the solution in the shell whether it is from a black hole or from some matter. In the case of a black hole, we don't know what is on the inside, only the curvature on the outside. And something left on the outside in the vacuum must be the kind of curvature that propagates itself through the vacuum. Which is the kind that doesn't need mass or matter (to propagate or persist, though maybe it interacted with mass or matter in the past or elsewhere)

There was a research program called "mass without mass" by Wheeler that looked at solutions that from far away and long periods of time looked like there was some mass or matter in there, but didn't have any.

Timaeus
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mass is a part of more broad quantity stress energy tensor.and stress energy tensor $Tμν$ can be derived from varying the action with respect to metric(which defines spacetime) . $$T_{\mu\nu}:= \frac{-2}{\sqrt{-g}}\frac{\delta (\sqrt{-g} \mathcal{L}_\mathrm{M})}{\delta g^{\mu\nu}} = -2 \frac{\delta \mathcal{L}_\mathrm{M}}{\delta g^{\mu\nu}} + g_{\mu\nu} \mathcal{L}_\mathrm{M}.$$ so the main entity is metric.so it is correct to say that matter as we know it gets stretched and squeezed out of existence." and "all that is left is warped space and time.

in many other places also we derive $T_{\mu \nu}$ by varying the action.more you can read anywhere.

Hare Krishna
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    A metric does not 'define' a manifold; not all properties follow from it, and there isn't even a unique or default choice for a manifold. – JamalS Jan 17 '15 at 18:47
  • which properties can not follow from metric.the more general definition of stress energy tensor is the variation of action with respect to metric. – Hare Krishna Jan 18 '15 at 05:40
  • I never said the formula for the stress-energy you gave is wrong, it's more or less fine, ignoring some subtleties. I object to the incorrect statement that a metric defines a manifold, or spacetime. Also, your statement about matter and 'all that is left is warped space and time' is completely nonsensical. Your last statement is also a repetition of what you said in the first two lines. – JamalS Jan 18 '15 at 10:27
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W+- and Z weak bosons have big masses and are no matter.

Juan
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