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?
Best Answer
This is a really tricky question. A lot of it comes down to what you define as "weight".
TLDR:
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There's an outdated concept of relativistic mass, $m_R =\gamma m$, where $\gamma$ is the Lorentz factor of the object.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.