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As we know, most of the mass of ordinary matter comes from the kinetic energy of quarks. This means kinetic energy of quarks contributes to the mass of any object.

However take a look at this question.

Does a moving object curve space-time as its velocity increases?

A moving star does not bend space-time any more than if it were not moving. (One example given was that a moving star won’t collapse into a black hole just because its moving). This seems to be a contradiction. In the first case motion of quarks do contribute to bending (by contributing mass). In the second case, the motion of the star does not. Both make sense. But how do we resolve the contradiction?

Let's take a more macroscopic example. If we had a many asteroids moving in random directions in a region of space, would the motion increase the bending of space-time (say the observer far away from the asteroid field). Imagine that you shrink down the asteroids and increase their number so you have a gas of small particles. This “motion in random directions” translates to internal energy and that contributes to the bending of space time, just as the kinetic energy of quarks contributes to the bending of space-time. And in fact, enough of this can cause a black hole.

The interesting thing is you can’t transform to a frame where none of the asteroids or small particles are in motion. In the case of a single star in motion, you can transform to frame where the star is not moving.

This question seems related: Does relativistic mass exhibit gravitiational effects?

Article by Wilzcek on "mass without mass". I think this article is a good answer to the question. http://www.aip.org/pt/nov99/wilczek.html

Other notes:

In the case of a star, the stress-tensor determines how space-time curves. The stress-tensor is a coordinate independent object. So it doesn’t matter what particular coordinates you use to do your calculations. In other words, whether you are in some frame co-moving with the star or in spaceship moving by the star, calculated metric tensor should be the same.

In the case of a gas, there is no frame in which the particles are all standing still with respect to that frame. You stand still with respect to some particles, but other particles are then moving and so on. The component of the stress-energy tensor change, but the result is the same.

So motion does add to the curvature of space time, but only if you can’t transform it away.

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  • $\begingroup$ You can't fully comprehend these matters on the basis of just the 0,0 element of the stress-enegy tensor, you have to look at the whole mathematical object. $\endgroup$ Commented Sep 6, 2014 at 2:19
  • $\begingroup$ Sure, I agree. But full comprehension is probably not necessary. Some insight is better than nothing. $\endgroup$
    – yalis
    Commented Sep 6, 2014 at 2:38
  • $\begingroup$ The real problem is, that we don't know, if microscopic objects gravitate, at all. No amount of reasoning about gravity at the scale of a star, or the scale of one gram of mass can tell us anything about gravity on the scale of a quark. All we can say about nucleon masses is, that their inertial mass is consistent with the dynamics inside of them. That, however, has absolutely no consequences for the validity of the equivalence principle, and for all we know, it may not even hold at the nuclear level. $\endgroup$
    – CuriousOne
    Commented Sep 6, 2014 at 2:48
  • $\begingroup$ @CuriousOne: This is definitely not the real problem. You could ask a separate question about this. $\endgroup$ Commented Sep 6, 2014 at 4:49
  • $\begingroup$ @FredericBrünner: OK, I play. Please list a couple of experimental limits on the equivalence principle at the level of nuclei. Otherwise, if you can't, please explain to us why we should talk about gravitation at 1e-15m using examples the size of stars? $\endgroup$
    – CuriousOne
    Commented Sep 6, 2014 at 5:08

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I think two concepts are being confused here. The concept of invariant mass, and the concept of relativistic mass. In particle physics the relativistic mass is no longer widely in use as it tends to confuse newcomers.

As we know, most of the mass of ordinary matter comes from the kinetic energy of quarks. This means kinetic energy of quarks contributes to the mass of any object.

This is misrepresented. The mass of the proton comes from the invariant mass of all its constituents, which include gluons and quark antiquark pairs. It is the length of the four vector and it is a Lorenz invariant.

A moving star does not bend space-time any more than if it were not moving

That is because its invariant mass stays invariant.

In the first case motion of quarks do contribute to bending (by contributing mass)

It is not the motion of the quarks that contributes to observable mass, it is the addition of internal four vectors. The conglomerate of four vectors within a proton appear because of the strong interaction between the quarks . No such strong interaction exists in the asteroid examples. Gravity is extremeley weak.

As long as one remains in the Lorenz frame there can be no contribution from motion to gravitational bending. If one goes to beginning of the universe energies where gravity, once it is correctly quantized , is a strong interaction , then one can revisit this. At those energies no asteroids are hypothesized.

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    $\begingroup$ Caveat. It’s important to stay away from the concept of “invariant mass”. This is from SR "it turns out that it is impossible to find an objective general definition for the concept of invariant mass in general relativity. " We must consider the stress-energy tensor. Construct the stress-energy tensor of the given distribution of energy-momentum and everything should follow, regardless of coordinate system. for the star, we can transform to a co-moving frame, so the velocity of the star does not matter. for a gas. You can’t transform to a frame where all the particles are not moving. $\endgroup$
    – yalis
    Commented Sep 6, 2014 at 17:37
  • $\begingroup$ @yalis I agree that with general relativity in the game a more general framework is important. But when one is talking of quarks the scales are such that GR is irrelevant, they exist in the local flat Lorenz space . At energies where unification of all forces makes gravity strong it is a different story. Anyway, as I say in my answer, it is not motion that creates the mass, but the strong interactions modeled as gluons etc. $\endgroup$
    – anna v
    Commented Sep 6, 2014 at 18:42
  • $\begingroup$ @annav Your answer makes it sounds like the kinetic energy of the asteroids doesn't contribute to the bending of space time. It has to otherwise you would be able to get rid of a particles gravitational effect by having it decay into two photons. $\endgroup$
    – Virgo
    Commented Sep 7, 2014 at 4:43
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Actually I am a student of fs.c so I can be wrong. I have been searching the almost same problem for 3 days on internet but what come in my understanding is that bending of space-time depend upon REST MASS not RELATIVISTIC MASS because bending of space-time depend upon energy-momentum as this quantity( energy-momentum) remain same for relative observer (as you have already read ~Does a moving object curve space-time as its velocity increases?) This means energy-momentum is not affected by RELATIVISTIC MASS but depend upon REST MASS which in turn depend UPON ENERGY IN REST FRAME OF REFERENCE As rest frame of reference is that frame of reference in which matter's particle (just like box with particles) have net momentum zero and rest mass depend upon energy. So if u have large box and if u put asteroids in it and then increase their random energy then energy-momentum increase and space-time curve.

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the elephant in the room. one does not "have" kinetic energy, it has to have an energy source, otherwise all matter would have decayed billions of years ago. And we would never have got here to talk about it! So the kinetic energy of quarks have to be supplied with some type of universal field, perhaps among many fields. And this field(s) consequently gives mass.

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  • $\begingroup$ Welcome to PhysicsSE. It would be great if you could make your answer more elaborated and clearer. $\endgroup$ Commented Jul 7, 2017 at 9:17

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