2
$\begingroup$

I tried looking in vain (at the LHC site and elsewhere) on the net and could not find out if a $7$-TeV proton weighs more than its rest mass.

Can anyone explain and point me towards experiments that have studies the issue. Does a particle traveling at near light-speed weighs more, given that it has a greater gravitational mass?

I understand it has an inertial mass which is 7,450 times greater than its rest mass. Is there a significant increase in weight?


edit after the comments:

to those who are suggesting that weight should not increase, I remind that thermal energy increases the gravitational mass of a body, so, why not Ke?

$\endgroup$
4
  • 2
    $\begingroup$ How are you defining "weigh" and "gravitational mass"? Are you thinking about a scale and gravity? That together with the answers so far should point you in the right direction... $\endgroup$
    – ohneVal
    Feb 10 at 10:57
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/3436/123208 & the links therein. $\endgroup$
    – PM 2Ring
    Feb 10 at 11:02
  • $\begingroup$ @ohneVal, hi :), I am enquiring about concrete experiments $\endgroup$
    – user157860
    Feb 10 at 11:03
  • $\begingroup$ also related: physics.stackexchange.com/q/499510 $\endgroup$
    – Paul T.
    Feb 10 at 15:57

6 Answers 6

4
$\begingroup$

This is a really tricky question. A lot of it comes down to what you define as "weight".

TLDR:

Does a particle traveling at near light-speed weigh more, given that it has a greater gravitational mass?

  • Yes it weighs more, but I don't think that means it has greater "gravitational mass".

I understand it has an inertial mass which is 7,450 times greater than its rest mass. Is there a significant increase in weight?

  • A particle's mass does not change due to its motion. Relative to gravitational scales on Earth, a $7$ TeV proton has not significantly changed its weight. Any change in weight will be experimentally immeasurable by current methods.

inertial and gravitational mass

In classical mechanics inertial mass, $m_I$, is the mass that appears in Newton's second law, $\vec{F}_\mathrm{net} = m_I \vec{a}$, and gravitational mass, $m_G$, is the mass that appears in the gravitational force, $\vec{F}_g = m_G \vec{g}$. The equivalence principle says these two things are equal. That's why things with different masses fall at the same rate in free fall.

In special relativity there is only one mass $m$ the mass of the object. This is sometimes called the rest mass.

mass and motion

There's an outdated concept of relativistic mass, $m_R =\gamma m$, where $\gamma$ is the Lorentz factor of the object. According to this idea, an observer at rest would say the mass of a moving object is greater than its rest mass. This concept turns out to not be useful, because it leads to the wrong form of Newton's second law.

Relativistic momentum is defined $\vec{p} = \gamma m \vec{v} = m_R \vec{v}$. So far so good. But the general way to define Newton's second law is $$\vec{F}_\mathrm{net} = \frac{d\vec{p}}{dt},$$ and this does not lead to $F = m_R a$ as you might hope. The Lorentz factor is a function of the object's speed, so it is not a constant. For a force that is parallel to the object's velocity you would find: $$\vec{F}_\mathrm{net} = \frac{d}{dt}\left(\gamma m \vec{v}\right) = \gamma^3 m \vec{a}. $$

So $\gamma m$ is not the inertial mass of a relativistic object! It gets worse because the scaling in Newton's second law depends on the angle between the force and the velocity. The only useful mass in S.R. is the rest mass.

I'm not even sure inertial mass is well defined in S.R., since there isn't a simple way to write $F = ma$.

mass and gravity

Gravitational mass gets used in two ways.

The first way is about how an object responds to a gravitational field, $\vec{F}_g = m_G \vec{g}$, where $\vec{g}$ is the gravitational field. Near the surface of the Earth $\vec{g} \approx 9.8$ m/s$^2$ down. When you ask if a moving object weighs more, I would say no, because the object responds to an external gravitational field following the equivalence principle.

Even in general relativity, the motion of a very fast object is determined by the geodesic equation, which has the equivalence principle built right in. The concept of relativistic mass is not needed to explain the object's motion.

This example depends on the object being "small" in the sense that it doesn't really affect the background gravitational field. Like Earth is small compared to the Sun, or a baseball is small compared to the Earth. The rest energy ($mc^2$) of the Earth is about $10^{60}$ eV, so a $7$ TeV proton is still small compared to the Earth!

To answer "no", we sort of dodge the question! The proton's motion near the earth is not noticeably changed, mostly because it doesn't weigh noticeably more compared to the Earth.

The second way gravitational mass appears is as a source of gravitational fields. Unlike Newtonian gravity, where only mass matters, in general relativity energy (well, stress-energy) is the source of gravitational fields.

The gravitational field of a point mass is described by the Schwarzschild metric for a black hole. So does the gravity of a black hole change if it moves? One way to state the principle of relativity is that there is no experiment that can distinguish rest from uniform motion. So in the reference frame of the black hole, its gravitational field is identical to it being at rest. This fact his held up in one of the more common ways to define mass in general relativity, the ADM mass. The ADM mass of a black hole moving at constant velocity is the same as one at rest. The ADM mass is the $M$ that appears in the Schwarzschild metric defining its gravitational field.

Even though the $7$ TeV proton is moving very fast, its gravitational mass (by some definition of mass) does not change! This argument relies on the proton being an isolated system, alone in the universe. So maybe we dodged the question again.

Conclusion

Okay, so it sounds a whole lot like the weight of the proton doesn't change, but I said it did at the very top.

The key here is that "weight" applies to a gravitational interaction between two things. Two $7$ TeV protons moving at the same velocity are in a shared reference frame, so their gravitational attraction is identical to two protons at rest. But if one proton is moving relative to the the other, all observers will agree that the total energy of the system is greater than the sum of their masses. The joint gravitational field of both protons will be stronger due to their relative motion. In that sense the moving proton "weighs" more.

In order to observe this extra "weight" we'd need to measure the gravitational attraction between two subatomic particles moving at high speeds relative to each other. And we would need to disentangle the gravitational interaction from all other interactions. That is a very tall order. An experiment that can do that will open up the doors of quantum gravity, and I sure hope somebody figures it out eventually.

$\endgroup$
2
$\begingroup$

As you yourself and others have pointed out, the more energetic a particle, the more strongly it interacts with a gravitational field (i.e. the harder a given gravitational potential 'pulls' on that particle). 'Weight' is simply defined as the gravitational force that acts on a given mass. So, from this definition, it's easy to see that the weight of an energetic particle is larger than that of the same particle at rest.

$\endgroup$
1
$\begingroup$

In principle, one can describe a moving particle as being gravitationally heavier than its rest mass. This is called relativistic mass, but the concept has generally been abandoned by the physics community for not really being useful, see Wikipedia.

A simple demonstration of the concept is the photon. It has zero rest mass, but finite energy, and can thus be affected by gravitational fields. The same idea applies to any particle, the more energetic it is, the more it interacts with gravitational fields.

$\endgroup$
1
  • $\begingroup$ A moving mass is not gravitationally heavier. $\endgroup$ Feb 11 at 15:52
1
$\begingroup$

As an afterthought to your question, you add

I remind that thermal energy increases the gravitational mass of a body

to which I would respond: has that experiment been done? I don’t know that it has. Consider a chunk of copper, with heat capacity $\rm 385\,J\,kg^{-1}\,K^{-1}$. For simplicity, pretend this heat capacity is from room temperature down to absolute zero: a kilogram of copper would have a total thermal energy like $\rm 10^5\,J$. Compare this with its rest energy $mc^2 \approx \rm 10^{17}\,J$. If someone is doing sub-part-per-trillion mass measurements on macroscopic objects over a wide temperature range, I’d love to know about it. I have lots of practical questions.

Measuring the weight of a subatomic particle is challenging, because you have to get it to interact gravitationally, and gravity is the weakest of the (known) forces with which particles interact. You can observe slow/cold particles falling in a gravitational field, including Bose-Einstein condensates released from a laser trap, ultra-cold neutrons which can be stored in an open-topped bucket, and the cesium fountain which is the basis of the atomic clock. But energetic particles tend to just run away before you can weigh them.

If you wanted to weigh a relativistic particle, you’d have to either

  1. measure its gravitational scattering from some heavy mass (but not the Earth, because otherwise CERN would already be correcting for the effect you’re asking about);
  2. run along next to the particle in its rest frame (where it has the rest mass, so that’s no good), or
  3. trap your particle and keep it relativistic.

Probably the best experimental access to (3), precision measurement of a trapped relativistic particle, is mass spectrometry sensitive to the inner electrons bound to heavy nuclei. For example the uranium x-ray $K$-edge, which is roughly the binding energy of the innermost electron shell, is about 115 keV. To the extent that the virial theorem applies, the electron’s kinetic energy should be about half of its binding energy: a 10% correction to the electron mass. The relativistic kinetic energy works out to something like a part-per-million correction to the mass of a hydrogen-like U nucleus bound to a single electron, which in principle seems doable for a charged-ion mass spectrometer. But there are plenty of other poorly-understood details about the interactions of relativistic bound electrons with very heavy nuclei; it would be a real coup to be able to isolate the electron mass correction. Furthermore, mass spectrometers measure inertial mass, not gravitational mass.

There are some folks who would tell you the proton itself is an ensemble of trapped relativistic particles: the “current quark masses” only add up to about 1% of the proton’s mass, so the rest of the proton’s mass “must be” those quarks’ relativistic kinetic energy. That turns out to be a surprisingly subtle statement, because the proton’s interior also includes some “ocean” of quark-antiquark pairs; the fraction of the proton’s mass associated with the “valence quarks” relative to the “sea quarks” changes depending on the momentum you have in mind when you ask the question. Better to leave that example aside.

$\endgroup$
3
  • $\begingroup$ I don't see what quarks moving in a proton have to do with this question. You think their kinetic energy contributes to their weight? $\endgroup$ Feb 11 at 16:00
  • $\begingroup$ That’s a common pop-science explanation, see e.g., but as I wrote here it’s too simplified to be useful. $\endgroup$
    – rob
    Feb 11 at 16:23
  • $\begingroup$ I'm not sure about the relevance of the weight of a proton to this question. Three protons moving in a box relativistiacally weigh more than three at rest. I can't see the relation with three quarks in a proton. $\endgroup$ Feb 11 at 17:18
1
$\begingroup$

It is not true that a boosted black hole has the same ADM mass as a stationary one. A boost changes the lapse at asymptotic spatial infinity. In other words, ADM has an asymptotic energy-momentum tensor that transforms as a 4-vector under asymptotic Poincare transformations (and a boost will also act asymptotically in this way). I recommend Regge and Teitelboim: "Role of surface integrals in the Hamiltonian formulation of general relativity" and Beig and OMucrchadha "The Poincare group as the asymptotic symmetry group of canonical general relativity"

$\endgroup$
-2
$\begingroup$

Let's imagine a space with a uniform gravitational field, say caused by an infinite massive plate. In a frame traveling along with a fast moving proton, you can put a scale beneath the relativistic proton so it will stay at the same height. The scale will point to the same number as when at rest.

A fast proton will cause frame dragging only. A proton that's not moving wrt you will get an acceleration in the direction of the velocity of the the fast proton but not an extra acceleration directly towards it. Two oppositely fast moving protons will pull harder on a stationary proton than two protons at rest.

When two fast moving protons smash head on, a black hole can form. This was used to find out about a "large" extra dimension. If it existed, a tiny hole would form at lower energy. It didn't happen. No large dimension.

You write: "edit after the comments: to those who are suggesting that wight should mot increase, I remind that thermal energy increases the gravitational mass of a body, so, why not Ke?"

The thermal energy contributes to the rest energy of the total container. A box with ultrafast protons will weigh more than with the same protons at rest. It's because the relation with other protons they increase in weight. A scale put beneath a box with two relativistic protons will show more than the sum of the values the scales show for individual protons.
Each on its own won't produce a black hole but two of them can create one, depending on their relative velocity and distance.

So the answer to your question is no. Unless you consider more of them. Than they can make a hole appear.

$\endgroup$
3
  • 2
    $\begingroup$ A scale measures normal force. If you put a scale under a horizontally moving relativistic proton, the proton won’t interact with the scale, and it’ll read zero. $\endgroup$
    – rob
    Feb 10 at 15:13
  • $\begingroup$ @rob The protons move in a uniform gravity field. I will edit. $\endgroup$ Feb 11 at 15:33
  • $\begingroup$ @rob Why it won't interact with the scale? Did you think the proton moves in free space? $\endgroup$ Feb 12 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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