# LHC particle mass increase

1. Aug 26, 2009

### Force1

Is it proper to say that when the LHC accelerates particles, those particles experience a mass increase? What is the Special Relativity explanation of the mass increase, i.e. do the particles have increased mass when observed in the rest frame of the detector or do the particles have the same mass and greater kinetic energy in the frame of the detector?

2. Aug 26, 2009

### Staff: Mentor

It depends on which definition of "mass" you use.

This is a correct statement for the "relativistic mass" that most pop-science books and some introductory physics textbooks talk about, but relatively few physicists use.

This is a correct statement for the "invariant mass" (usually called "rest mass" in pop-science books and some intro physics textbooks) that most physicists mean when they say simply "mass."

3. Aug 26, 2009

### Force1

I should have said relativistic mass then and physicists would say what, apparent mass?
OK, so can we say that the relativistic or apparent mass (if that term is correct) of the particles in the detector is increased by the kinetic energy added by acceleration in separate frames, different from the frame of the detector, and when we view the particles in the frame of the detector the difference between rest mass and apparent mass is the kinetic energy?

4. Aug 26, 2009

### humanino

The reason people eventually gave up on the concept of "relativistic mass" is precisely because it is not "apparent" ! There is nothing more than $E^2=p^2c^2+m^2c^4$ and Lorentz transformation of $(E,p)[/tex] leaving [itex]m$ fixed. The "relativistic mass" is the result of multiplying the mass by the Lorentz dilatation factor $\gamma=\frac{1}{\sqrt{1-\beta^2}}$ "as if" the mass were a time-component, but in fact it is not, the energy is a time-component and it happens to be mass only when the mass is at rest. The "relativistic mass" is the equivalent mass an object at rest would have corresponding to the true total energy of the moving mass.

In fact, wikipedia reproduces an interesting historical comment by Einstein. I did not read the article in details.

5. Aug 26, 2009

### ZikZak

Most physicists, I think, would say "energy." There is nothing massy about "relativistic mass."

6. Aug 26, 2009

### Force1

So for physicists at the LHC, the mass of the particles they view in the detector are total energy divided by the speed of light squared. The term relativistic is a point of contention in that relativistic mass is not a fundamental concept of the theory of GR spacetime, although mistakenly used that way by some. “There is disagreement over whether the concept is pedagogically useful”, but most or many physicists have stopped or rarely use the term “relativistic” in circumstances like with the LHC? Is that correct?

7. Aug 26, 2009

### humanino

Certainly not, unless they did not evolve for 50 years. The mass is a constant, similar to the electric charge, caracteristic of a particle properties under space-time symmetry (Loretnz group representation). Please read https://www.worldscientific.com/phy_etextbook/6833/6833_02.pdf [Broken] on the question, which you could have found for instance from wikipedia.

Last edited by a moderator: May 4, 2017
8. Aug 26, 2009

### Force1

From the PDF:

E sub o=mc^2 (1)

E=mc^2 (2)

E^2-p^2c^2=m^2c^4 (5)

P=v E/c^2 (6)

“Thus we obtain in the nonrelativistic limit the well known Newtonian equations for momentum and kinetic energy. This means that m in equation 5 is the ordinary Newtonian mass. Hence, if I were to use m sub o instead of m, the relativistic and nonrelativistic notation would not match.”

“If the notation m sub o and the term ‘rest mass’ are bad, why then are the notation E sub o and the term ‘rest energy’ good? The answer is, because mass is a relativistic invariant and is the same in different reference systems, while energy is the fourth component of a four-vector (E,p) and is different in different reference systems. The index 0 in E sub o indicates the rest system of the body.”

“Let us look again at the equations 5 and 6, and consider them in the case when m=0, the extreme ‘anti-Newtonian’ case. We see that in this case the velocity of the body is equal to that of light: v=c in any reference system. There is no rest frame for such bodies. They have no rest energy; their total energy is purely kinetic."

So from that PDF, “equations 5 and 6 describe the kinematics of a free body for all velocities from 0 to c, and equation 1 follows from them directly. Every physicist who knows special relativity will agree on this.”

“On the other hand, every physicist and many nonphysicists are familiar with ‘the famous Einstein formula E=mc^2.’ But it is evident that equations 1 and 2, E sub o=mc^2 and E=mc^2, are absolutely different. According to equation 1, m is constant and the photon is massless. According to equation 2, m depends on energy (on velocity) and the photon has mass m=E/c^2.”

So (5) and (6) give us equation 1 which is one of Einstein’s great discoveries, energy is equal to rest mass times the speed of light when velocity and momentum are 0, i.e. Newtonian and nonrelativistic. But as Einstein said in a letter to Lincoln Barnett, 19 June 1948, “It is not good to introduce the concept of mass M=m/(1-v^2/c^2)^1/2 of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ‘rest mass’ m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion.”

So can you say how the physicist at the LHC would describe the mass/energy of the accelerated particles? Is it simply that they would refer to the energy of the particles and not m, not M, but total energy?

Last edited by a moderator: May 4, 2017
9. Aug 26, 2009

### A.T.

Re: Velocity translating into mass?

Who said energy creates protons?
In this case it doesn't, it just increases mass as they are both equivalent. But in general you can create matter+antimatter just from energy.

10. Aug 26, 2009

### humanino

Again, in the modern point of view there is one mass and it is constant. It is a caracteristic of the particle, it defines what we call a particle (from a representation of the Lorentz group). The total energy of the particle includes kinetic energy, and this energy can be arbitrarily high (at least without taking into account gravity, otherwise one may hit a stability limit and form a black-hole). When we accelerate a particle, we simply increase its kinetic energy.

11. Aug 26, 2009

### Force1

To test my understanding let me say it; The physicists at the LHC would say that a particle from the representation of the Lorentz group has one mass and that is the rest mass as in E sub o = rest mass times the speed of light squared, and the total energy of said particle would be the energy of the rest mass plus the kinetic energy added by accelerating said particle which could be extremely high at relativistic speeds. Is that correct?

If not I want to keep trying :). But if that is correct, then you mentioned "without taking into consideration gravity". How would gravity change things and can you mention something about the stability limit relative to a black hole forming :surprised:.

12. Aug 26, 2009

### humanino

I think so, yes.
That's a qualitative guess that I made mostly to illustrate that the "energy can increase without limit" is a "for all purpose" statement concerning protons at the LHC. Beyond experimental constraints, more serious care must be taken. A reasonable guess is that from the atom scale to the LHC scale is an increase which is (much) less than the necessary increase to go from LHC to conventional gravity scale scenario. However, it could also be, in more exotic scenarios, that the gravity scale occurs sooner. Since we do not know the correct quantum theory for gravity, there is room for improvisation. It is in any case fairly certain that if you keep increasing the particle's energy, gravity will come into the game at some point. The formation of a black-hole is a dramatic example. It can not occur merely from kinematics BTW. We would need our putative very very very energetic particle to interact with something around.

13. Aug 26, 2009

### Force1

Does this qualitative guess relate to the conceivable presence of an amount of energy in a particle that reaches some as yet undefined energy limit beyond which there could be gravitational events that we don't yet understand?

14. Aug 26, 2009

### humanino

Almost. I was trying to convey that this energy is not really intrinsic to the particle : it is in the velocity difference between the particle and us (or the lab).