Particle Mass: Newsletter Challenge (September 2013)

E=MC² and quantum physics mean that some of those masses has been turned to energy and vice versa. Do you mean in one point in time or taken over certain time frame?

These particles being relativistic in nature may have comparable or greater mass than the rest mass of the individual particles, if however the mass is expressed via Dirac equations and these are assumed to be accurate, and further that the particles are expressed as part of a system, then it could be possible that these expressions would be equivalent....

No, because the system mass doesn't include the contribution mass of theoretical 'dark matter' which is not taken into account as part of the particle mass.

Wrong? Probably, and likely on more than one level.

_{Jack - Not a particle expert by any stretch of the imagination.}

Vectorially, maybe, but not by absolute values. [???]

From the last sentence I take it that we deal with something relativistic! For me this leads to the conclusion that the particle has not even a defined mass, ergo the mass of the system itself is undefined.

I remember we had once a question about closed system and temperature and height. Here we only have a closed system. Do we miss parameters as to if the closed system is in equilibrium?

My answer is "No"

Am I close?

Relativistic is simply a loose definition for something that has enough velocity to be impacted by the Theory of Relativity.

That point varies depending on the degree of precision you want to work to, but does not mean that the object must be traveling at C. It could be any fraction of C depending on what your target precision is.

From the Webster Dictionary, "moving at a velocity such that there is a significant change in properties (as mass) in accordance with the theory of relativity <a relativistic electron>"

The key word is "significant" and that depends on what you define as significant for the problem space you are working in.

Your final answer (to my understanding) is correct, but I think not exactly for the reasons you stated.

Symantics.

Considering the individual particles as being steady relative to our frame -hence considering their rest mass- then all moving particles have a total mass greater than the sum of their rest masses. ( Relativistic mass: m=m_o/√(1-v²/c²), where m_o is the rest mass. )

However, it depends of what the questioner means by saying "individual particles". Does this mean that we consider every particle as "being alone" (i.e. like all the other particles have vanished)???. If so, then this means that each "individual particle" has the same velocity as in the case of participating in the group. (I.e. we could consider it as a kind of a "snapshot" of the group.) In this case, of course, the total mass of all particles are equal to the sum of their masses.

A step further: Each particle has an ultra-tiny gravitational field (due to its mass). This gravitational energy is "translating" to mass (m=E/c²) which contributes to the total mass of the particles.

An interesting theoretical question but there are a few ambiguities here I wish to address first before I toss my answer into the pool. I presume that the meaning of this system being a closed system is that all forces operating onto this system are part of this system. No outside influences should be considered in the answer, regardless of there actually being other influences in effect. All of the particles of concern have a rest mass. The momentum of zero rest mass particles produced by the system due to whatever relativistic effect should be ignored since only the mass and not the momentum is the question.

I believe that the mass of this system then should be considered as the sum of the masses of these particles. However, as the problem is stated this is a moot point for two reasons. First, it is given that the particles are moving freely. This implies that the particles do not even interact with each other. This is a contradiction of the other given datum that this is a closed system. The premise is faulty. Second, the mass of any system is only relevant if something outside of the system interacts with the system. A closed system does not interact outside of itself.

That isn't the only reason this is a poor challenge question.

Another reason is that challenge questions should not depend on esoteric knowledge, heavy computation, etc. Instead, they should be the sort of problem that might seem hard at first, but yields to the right clever insight. This problem is basically an either you know it or you don't proposition.

A challenge question FOR AN ENGINEER should contain some calculations. After all this is the main feature of our trade. We do not design by "clever insight". So, a little thinking together with a little math goes the way of engineering.

This is thinking outside the box.

As a closed system, the mass is the mass of the sum of particles. As a closed system, the relative velocities are relative ONLY to the other particles in the system. The system is homogeneous, reletivistically, so, the mass is the sum of masses of particles.

I was tempted to answer in similar manner but the question is a about a group of particles, each moving randomly. If they are each moving randomly, they do not share an inertial reference frame!

Generally, all the though experiments I have seen use a "jar", which is the inertial frame of reference.

However, you could also pick any one particle and use it as a reference point. It should work out the same.

The whole point of the challenge question is that the whole system's invariant mass and the relativistic mass are only equal when all particles are at rest (no kinetic motion). Otherwise the relativistic mass is always higher.

How is this for an example? Assume the three quarks inside a proton are moving at relativistic speeds, 99.9999% of the speed of light (c). But the proton, which is our system, is at rest. What we measure as the "rest" mass of the system includes the relativistic mass of the quarks, or "the total system energy".

Hadrons are examples where the mass of the system is about 85 times that of the sum of the masses of the individual particles. Protons and neutrons are classified as hadrons, and consist of three quarks. The rest mass of the three quarks in a proton is only about 11 (MeV/c*c), while the mass of a proton is about 938 (MeV/c*c). So is matter 98.8% relativistic mass?!

Great first post. You raise a good point.

Welcome to CR4!

I am not sure about your figures.

If I remember correctly, the mass of the three quarks represents something like 10-20% of the total proton rest mass.

The remainder of the mass or energy is the sea of virtual particles that briefly pop into and out of existence inside that proton.

The figures for the mass of the three quarks and the total mass of a proton are from http://en.wikipedia.org/wiki/Quark Look for the section titled "Mass".

You raise an important point, that when dealing with particles the size of quarks, particles pop into existence, and disappear just as quickly. Needless to say, there could be quite a bit going on inside a proton that could contribute or subtract from the mass of the proton "system".

Also, in my reading on sub-atomic physics I have read that a positron (an anti-mater electron) can be thought of as an electron moving backwards in time. Can anyone else confirm this?

Yes, and that was what the Feynman diagram I posted shows (at least at the input side).

The downward traveling -e is the positron.

Consider two similar systems with the same number of the same type of particles. (Lets say, 2 'identical' systems composed entirely of neutrons.) In System A the neutrons are moving slowly; in System B they are moving with relativistic speeds.

Clearly the system with the fast-moving neutrons (B) will have more momentum and thus the total system energy (which is equivalent to mass) is greater than for the other system (A).

So no, the mass (the invariant mass, given by the total system energy divided by c²) will not just be the sum of the 'rest masses' of the individual particles.

I guess it also depends on how one defines mass. So does relativistic neutrons create a greater gravitational attraction than non-relativistic neutrons. I haven't the faintest idea how one could possibly measure this possible change in gravitational attraction.

I have a faint idea, it is called the Einstein's Field Equations.

Those 16 or so equations will allow you to calculate the effect, but I have never tried doing that as it is beyond my mathematical abilities.

The kinetic energy of relativistic mass does indeed contribute to gravitational mass according to General Relativity. If your math skills are strong you might give it a whirl.

As for defining mass, you can define mass in kilograms (SI units) or in energy units (i.e., eV or eV/c²) since General Relativity's mass-energy equivalence equation applies. I think the latter may make things more clear for this challenge question.

Very nice, but; the system is a closed one, How do you contain the neutrons inside?

Amclaussen.

Engineer by profession, scientist by vocation. And Murphy was overly optimistic.

A closed system need not have boundaries.

The universe is considered a closed system, for example.

You can create a virtual universe to any specification you want for any problem set you design.

The definition of a closed system is simply a system that has no external influences.

If you were to take any arbitrary point in this closed system, and calculate the masses of each particle--taking into account the relativistic effects of individual speeds in relationship to the point of reference--the sum of masses has to be equal and a constant (regardless of which point of reference you use). It is meaningless to try to look at each particle individually. The sum of the rest masses of each particle will not be the sum of the masses of the system in motion because there is energy input to set them in motion. This energy input is seen in the form of a mass change of each moving particle. This answer ignores the uncertainty that Heisenberg mentions.

--John M.

IMHO, If "relativistic speed" refers to light speed, then you are asking if we can find the mass of light particles. Since physics class experiments with reflectors and absorbers in a vacuum suggest that light can generate kinetic energy, the "particles" exhibit properties of mass. Therefore it should be possible to determine their mass, if you can define the quantity n.

DonaldSmith goes smaller, but, going larger, the complete universe would qualify as the group of n particles in the question, n would be very large though.

I have recently watched a video of Lawrence Krauss talking about this subject.

Very interesting stuff!

He was talking about finding that mass was imparted by the "Higgs Field" or dark matter.

Makes me wish that I had paid more attention in science class. (I was bored with memorizing lists of facts.)

I'm having trouble pasting links so, if you search you tube for "lawrence Krauss Greatest story ever told" you will see it.

http://www.youtube.com/watch?v=gw7v6EzJDNM

if the particles are in a closed box, so the speed is turned to kinetic energy every time a paticle hits the box, than the total mass of the particles should be

sum ( mi*(1+sqrt(1+v/c)²)):

In an open environment the total mass of the particles should be sum mi!

what is the measured mass of the particle if i measure the mass of this particle near light velocity and the measurement is travelling the same speed like the particle?

A GROUP OF N(NEGATIVE) PARTICLES IN CLOSED SYSTEM (COMPLETE CIRCUIT) WHICH HAS MASS OF GROUP EQUAL TO MASS OF SUM OF PARTICLE MASS (MASS OF AN ELECTRON IS CONSIDERED ZERO AND ZERO MULTIPLIED BY ANY NUMBER EQUALS ZERO) WILL EXHIBIT RELATIVISTIC SPEED

PS: NOT SHOUTING JUST LIKE LOOK OF ALL CAPS.

Do you also like being offensively stupid? Your answer is wrong both in form and content.

At least you didn't shout this time. Thank you.

You knew full well that your all-caps post would be considered offensive. That's why you put in that ridiculous caveat in the first place. Now you have just compounded the offense.

Welcome to CR4!

I would suggest not using all caps. Your message will be better received and by a wider audience, too.

The actual rest mass of an electron is 9.11×10⁻³¹kg. So, they are not really massless.

If you want to learn more on the subject you can try doing searches on invariant mass and relativistic mass.

The bottom line for this puzzle is the system mass is the total system energy. Remember Einstein's E=mc² or mass/energy equivalence.

Well at the risk of creeping too far out on a limb, weight and mass are two different things....If we are measuring weight, then we must include the anti-matter, which might should be anti-gravity in nature, this then possibly rendering the particles weightless when measured as a system....

I am confused. Where did weight come into this? Did I slip up and use the word weight somewhere instead of mass? If so, I meant mass, I am sure.

My understanding of the definition of weight was the amount of "downward" force that an object exerts due to the force of gravity.

So, weight changes depending where you stand on Earth or on another planet, for example. However, rest mass is constant.

Yes you are correct, I think somebody said something about weight and I just added it to the conversation....It was carelessly posted....my apology tendered...

Wasn't offended. Just though I might have misspoke.

weight is the force of a gravitational environment to a mass!

Okay, I'll that that GA for my answer now. Thanks.