# Mass dilation determination

1. Aug 30, 2008

### cos

In his book 'Einstein's Universe' Nigel Calder wrote that relativistic mass increases in the region of 40,000 times a particle's rest mass have been determined.

Is this the maximum GF attained in particle acceleration experiments?

How is such a figure arrived at?

I read somewhere that it is determined on the basis of the amount of energy that has to be applied laterally to the particle in order to make it change direction. Is this correct?

Bill

2. Aug 30, 2008

### HallsofIvy

Staff Emeritus
Basically yes. It is measured by having the moving particle slam into one that is motionless relative to the observer and observing the motion after the collision. Essentially it is a measurement of momentum.

3. Aug 30, 2008

### JesseM

In this thread bernhard.rothenstein described (or quoted, I'm not sure) a method of measuring relativistic mass using a magnetic field which exerts a force perpendicular to the direction of motion, not sure if this is actually used experimentally. It is true as a theoretical matter that the relation between a force f applied perpendicular to a particle's motion and the particle's coordinate acceleration a would be f = m*gamma*a, where m is the rest mass, and of course relativistic mass is defined as m*gamma. On the other hand, for a force parallel to a particle's direction of motion the relation is f=m*(gamma^3)*a, which is one of the reasons that the concept of "relativistic mass" may be seen as misleading to students.

4. Sep 1, 2008

### bernhard.rothenstein

Thanks for quoting me. The thread was
When we speak about a physical quantity we should define it and it is compulsory to propose a procedure of measuring it.
But how does one measure relativistic mass? The answer lies in the fact that relativistic mass (as well as proper mass) is never really measured directly, nor is energy measured directly. As Jammer1 wrote, “As in the last analysis all measurements in physics are kinematic in nature.” So, one does not measure mass nor energy. One calculates them from measured kinematic quantities. Suppose you know the strength of a uniform magnetic field B. Launch a charged particle, of magnitude charge q, into the field such that the velocity is perpendicular to the field lines. The charge will move in a circle of radius r. B is known while r and v are measurable. Then use the cyclotron relation p=qBr to find p. Then m=p/v. Multiply by c2 to get the relativistic energy of the particle E=mc2. We have indicated a measurement procedure for a physical quantity on which physicists wish to ban.2
References
1 Max Jammer, Concept of Mass in Classical and Modern Physics, (Harper and Row, 1964, Dover 1997)
2 Lev Okun, “The concept of mass,” Phys.Today, 42, 31-36 (1989)

Do you aggree with the lines above?

The procedure presented above does not work in the case of a motion parallel to the magnetic field, because in that case no unioform motion could take place and so we are not able to find out a kinematic relationship between E and B.

5. Sep 1, 2008

### atyy

Of course not! I believe you should correct the last sentence to "which some physicists wish to ban" :tongue2:

6. Sep 1, 2008

### bernhard.rothenstein

The leading in baning is Lev Okun I quote at the end of my thread.

7. Sep 1, 2008

Staff Emeritus
Relativistic mass is energy. (That's one of the reason most physicists don't like the term - having two names for the same thing is confusing) So $$\gamma = E/m$$. It's as simple as that: measure the energy and divide by the (known) mass.

Electrons at particle accelerators have been seen with energies in excess of 300 GeV. That gives one a $$\gamma$$ of about 600,000.

Neutrinos have been produced with energies above 50 GeV. We don't know exactly how much a neutrino weighs, but it's almost certainly less than 0.3 eV (which would give a $$\gamma$$ of about 150 billion, and probably less than a tenth that, for a $$\gamma$$ in excess of $$10^{12}$$.

Ultrahigh energy cosmic rays have energies of $$10^{21}$$ eV, and assuming these are initiated by protons would give a $$\gamma$$ around $$10^{12}$$ as well.

8. Sep 1, 2008

### ZikZak

This seems to be a completely pointless statement. The reason "relativistic mass" is a Bad Idea has nothing to do with it being undefinable, or being definable, not being measurable, but with its usage in low-level classes leading to, at best, a merely superficial semi-Newtonian understanding of SR, and at worst, a complete misunderstanding of it.

Yes, your proposed experiment measures the total energy of the particle, what some like to call the "relativistic mass." However, when students are taught this concept, this is not the experiment that they have in mind. Relativistic mass is typically used in high school as a heuristic kludge to explain why particles cannot be accelerated to c, i.e. that their inertia increases, rather than the correct reason which is simply that velocities don't add linearly. So ask a student who has been told this nonsense what you get when you take F/a for that particle he's trying to get up to c and the answer will ALWAYS be the incorrect $m= \gamma m_0.$ Ask what the pilot of a constant-force rocket feels as he approaches c and the answer will ALWAYS be the incorrect "decreasing feeling of acceleration."

9. Sep 1, 2008

### bernhard.rothenstein

As I have mentioned in a previous thread the proposed measuring procedure has nothing to do with acceleration.
You reduce the problem to a semantic one. I could mention that it is a procedure to measure the total energy (energy=relativistic massxc2.
Many threads posted by learners concerning "photon has no mass" are due the fact that they do not know the concept of relativistic mass". I consider that it would be better to say "the photon has no rest mass."

10. Sep 1, 2008

### JesseM

I agree that in principle the method you describe should work, since for a charge moving in a circle in a uniform field the magnetic force is always perpendicular to its motion (it's a centripetal force), and for a force perpendicular to an object's motion we have the relation force = (relativistic mass)*acceleration (though as I said this wouldn't work for a force parallel to the object's motion). I don't know how well the method would work as a practical method for measuring relativistic mass or energy, measuring the radius of a particle's circular motion might be difficult, especially with quantum uncertainty taken into account.

11. Sep 1, 2008

### ZikZak

There's no real reason why it shouldn't be practical. It is, in fact, the basic idea behind the bubble chamber, which you can build in your garage.

12. Sep 1, 2008

### JesseM

But the bubble chamber is designed to detect trails left by particles created in collisions, not circles made by particles moving in a magnetic field, right? Obviously a bubble chamber could detect such circles if they were large enough, but what I'm wondering is whether it would have enough spatial resolution to detect changes in the size of the circle due to relativistic effects, which might be fairly small, and also whether the particles would really move in stable circles or if small disruptions might send them careening off or moving in some complicated helical pattern.

13. Sep 1, 2008

### Staff: Mentor

When I was a grad student, I worked on a neutrino experiment that used a bubble chamber. We used the BC to detect the interactions themselves, of course, but we also used track curvature to measure the momenta of the outgoing charged particles, which formed the basis for much of the analysis.

You don't need to see an entire circle, or even a large fraction of it, in order to measure the locations of several points and calculate the radius of curvature with useful precision.

Many of these particles were highly relativistic, by the way, typically pions ($m_0 c^2$ = 140 MeV) or muons ($m_0 c^2$ = 106 MeV) with energies up to 50 GeV or more.

14. Sep 1, 2008

Staff Emeritus
You don't need to see the full circle to measure the radius of curvature - an arc will do. This was the standard way to measure the momentum of a particle in the days of bubble chamber experiments.

Momentum is the key quantity - the radius of curvature is inversely proportional to the momentum. This is related to the energy (a term that is more universally used and understood than "relativistic mass") by the expression $$p = E \beta /c = mc \beta \gamma$$. So JesseM's question can be expressed as whether or not one can measure the curvature accurately enough to see that $$\gamma \gt 1$$

15. Sep 1, 2008

### JesseM

OK, that makes sense. I was just asking about whether this method would make practical sense since I don't know a lot about bubble chamber experiments or how energy is measured in particle collisions.

16. Sep 1, 2008

Staff Emeritus
That should be gamma > 1 (obviously).

17. Sep 1, 2008

### Staff: Mentor

I might as well add that that the bubble chamber I was working with was pretty big. It was the 15-foot bubble chamber at Fermilab, which used a magnetic field of a few tesla, I think (have to look in my dissertation at my office tomorrow to be sure). You need plenty of room to measure the curvature of a 50-GeV muon, even in that strong a magnetic field.

18. Oct 11, 2008

### vibhuav

Maybe I am not reading this right. Are you saying that the mass does not really increase? I have in front of me, the book "Gravity from the ground up" by Bernard Schutz, where he gives an example of two automobiles of 1000kg rest mass colliding head on at 100km/hr (pg 191). The calculated increase in mass is 8 picograms (4pg in each car), and he says explicitly: "...the rest mass of the wreck is larger than the sum of the two original rest mass by 8pg. This 8pg of mass takes the form mainly of extra chemical energy in the deformed structures of the cars. It is a real mass: it would show up in a precision weighing of the wreck, and it would contribute to the inertia if the rescue vehicles try to push the mess off the road."

If my understanding is correct, I wanted to ask a follow-up question: in what form of matter does the new mass exist? hydrogen? carbon? What is the "extra chemical energy" he is talking about?

19. Oct 12, 2008

### Jonathan Scott

I would assume he means that the new mass would consist of additional electromagnetic energy in chemical bonds which have been stretched or rearranged. I think that much of the additional mass would actually be in the form of additional kinetic energy of atoms jiggling around, otherwise known as heat, which would rapidly dissipate to the surroundings.

In general, rest mass is NOT exactly conserved. Energy and momentum are conserved, and rest mass is the magnitude of the energy-momentum four-vector. If you add two four-vectors together, the magnitude of the sum is nearly equal to the sum of the magnitudes if the directions are similar, such as when both relate to non-relativistic speeds and hence to Newtonian mechanics, but these two expressions diverge when the velocities become less similar.

20. Oct 12, 2008

### Staff: Mentor

There is no such thing as "the mass" in relativity, except in so far as people agree on it, and there is no universal agreement.

In non-relativistic physics, we can associate with any object a single quantity called "mass" which can be used in a variety of different kinds of calculations, for example in calculating the acceleration of an object that has a certain force applied to it (via F = ma), or in calculating the gravitational force that one object exerts on another (via Newton's law of gravitation).

In the relativistic regime, these classical calculations do not give correct results, using the classical definition of "mass". Furthermore, there is no way to redefine the classical mass in a way that we can still use all the classical calculations with the new definition of mass to give correct results in all cases.

So, we can do one of the following:

1. We can use the "classical mass" for all calculations, but change the equations used in those calculations.

2. We can keep all the classical equations but define different kinds of "mass", with different values for any given object (depending on the object's velocity, for example) for each calculation.

3. We can use a mixture of 1 and 2: define different kinds of "mass" for some calculations, while changing the equations for other kinds of calculations.

Today, physicists who deal with relativistic objects usually use option 1. If you ask a particle physicist, "what is the mass of a proton that has been accelerated by the Large Hadron Collider?", he will almost certainly answer, "938.3 MeV/c^2", regardless of its speed. I say "almost certainly" because if he knows you're a non-physicist, he might feel the need to answer in terms of "relativistic mass" as described below. But if he's talking to another physicist, he will certainly say "938.3 MeV/c^2".

Historically, people have used option 3, and so we have different kinds of "mass" which are appropriate for different calculations. Most commonly, we have the "rest mass" a.k.a. "invariant mass" which requires equations that are different from the classical ones for most all calculations, and the "relativistic mass" which we can use in some classical equations, to give correct results.

Many or most popular-level treatments of relativity use "relativistic mass". Among university introductory-level textbooks, it's a mixed bag: some use both "relativistic mass" and "invariant mass" and some use only "invariant mass."

On this forum, and elsewhere, we have people with a wide variety of backgrounds, so you cannot count on them to agree on what they mean by "mass," without further description. So when people like you ask about "mass" in relativity, you get a variety of answers, and people start arguing about the "proper" definition of "mass", and pretty soon the original question is forgotten.

So... instead of asking of what happens to "the mass" in relativity, you had better ask about specific phenomena or experiments. Are you interested in how the effects of gravity change, or how the effects of a force on an object's acceleration change, or how an object's energy is different, or how an object's momentum is different, or what?