# Reconciling quarks and the uncertainty principle

1. Jun 30, 2013

### NLB

As a result of some precise experimental data, we now know that the mass of the quark is not naively 1/3 the mass of the proton. The most recent estimates for the mass of the quark is:

Masses of the current quarks:
= 2 - 8 MeV/c2 = 1.0 - 1.6 GeV/c2 = 168 - 192 GeV/c2
= 5 - 15 MeV/c2 = 0.1 - 0.3 GeV/c2 = 4.1 - 4.5 GeV/c2

reference: http://en.wikipedia.org/wiki/Current_quark_mass

Also see:

Researchers reported in Physical Review Letters, April 2010, the mass of the 3 lightest quarks.
The team finds that an up quark weighs 2.01 +/- 0.14 megaelectron-volts, whereas a down
quark weighs 4.79 +/- 0.16 MeV. That’s 0.214% and 0.510% of the mass of the proton,
respectively.

reference: http://news.sciencemag.org/sciencenow/2010/04/mass-of-the-common-quark-finally.html [Broken]

Even though there is a bit of disagreement on the range of the quark mass, let’s be generous and assume the heavier mass is right, at 8 MeV for the up quark, and 15 MeV for the down.

We also know the quarks can not go faster than the speed of light. Let's assume that their velocity is between 0 and close to the speed of light.

And we know the radius of the nucleon is 0.841 fm, the accurate measurement made as of January 2013.

How do these experimentally-observed facts relate to the Heisenberg uncertainty principle?

Converting MeV into MKS mass, the up quark is 1.43E-29 kg. If it s going close to the speed of light, its momentum is 3E8*1.43E-26=4.28E-21, in MKS units. This its uncertainty in the momentum. Its position is within the nucleon, which is a diameter of 1.68E-15. This is its uncertainty in position.

This gives:
(Delta x) (Delta p)= (4.28E-21)*(1.68E-15)=7.2E-36

This is way below h-bar, which is 1.05E-34 in MKS units. A violation of the uncertainty principle.

How is this reconciled?

Last edited by a moderator: May 6, 2017
2. Jun 30, 2013

### Dreak

Heisenberg is about measuring momentum/position at the same time.
You can measure it's position 'exactly', or you can measure it's momentum 'exactly', but you can't measure them both at the same time, it's there where you have Heisenberg.

Or did I misunderstood you?

3. Jun 30, 2013

### Staff: Mentor

p=mv is a nonrelativisic formula. It is not true for relativistic motion, where you need $p=\gamma m v$ with $\displaystyle \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. p is not limited, even if v is.

Apart from that, don't forget the gluons and sea quarks in the proton. They contribute to the proton mass, too.

4. Jul 1, 2013

### NLB

Hello Dreak, thank you for your comment. In other words, you are saying that the particle can exist with an delta-x and a delta-p below the limit of the uncertainty principle, but as long as we do not directly measure x and p simultaneously, then it does not violate uncertainty principle? Is that consistent with the Coppenhagen Interpretation?

I have too often heard other interpretations of the uncertainty principle--specifically that the particle can not exist if we know that it's delta-x and delta-p are always constrained to be less than the uncertainty principle limit, regardless of whether or not someone is actually measuring it.

Thank you again for helping to better understand this.

NLB

5. Jul 1, 2013

### NLB

Hello mfb, Thank you for your reply. Yes, relativistic effects would kick in, but I was just being generous. If we say the maximum speed of the particle is say 1/10 the speed of light, or even 1/100 the speed of light, so that relativistic effects do not come into play, then the product of (delta-x)*(delta-p) is even worse. Saying it is close to the speed of light, but only say .9c, and using gamma as you recommended, we still have the same problem, but somewhat mitigated. I was simply not wanting to complicate the question with relativistic effects. Once again thank you.

6. Jul 1, 2013

### Staff: Mentor

Actually, the uncertainty principle is not about measurements. It is purely a statement about the particle itself.

Just check v=.99c, v=.999c and v=.9999c, and see what you get. There is nothing mitigated, you just look in the wrong parameter range.
Quarks in a proton are ultra-relativistic. There is no way to understand the proton mass without relativistic effects.

7. Jul 1, 2013

### Bill_K

Better yet, instead of trying to guess the velocity, take the momentum value you get from the uncertainty relation and calculate what velocity would be necessary to produce that momentum.

8. Jul 1, 2013

### Dreak

No, but you can give it this interpretation in this case, can't you?

"In quantum mechanics, the uncertainty principle is any of a variety of mathematical inequalities asserting a fundamental limit to the precision with which certain pairs of physical properties of a particle known as complementary variables, such as position x and momentum p, can be known simultaneously."

9. Jul 1, 2013

### Bill_K

There's really nothing deep going on here. The relationship between momentum and the deBroglie wavelength is p = h/λ. If you confine a particle in a box of side Δx, the wavefunction for the ground state fits one half wavelength in the box, so p = h/2Δx and p Δx = h/2.

10. Jul 2, 2013

### tom.stoer

The HUP does not say anything about the momentum, but about the uncertainty of the momentum

11. Jul 2, 2013

### NLB

12. Jul 2, 2013

### Bill_K

I answered this question, but you ignored it. Take the Δp you get from the uncertainty relation and use it to calculate the velocity.

13. Jul 2, 2013

### The_Duck

I'm not sure what you mean by "effective mass," but the quoted masses are in fact rest masses. Perhaps this is the source of your confusion?

Mass does not appear in the uncertainty principle. Only momentum and position. Mass is only important in that you need to know it to compute momentum.

No, this is the key mistake you have been making. You seem to be under the impression that the maximum momentum of a particle with mass $m$ is $mc$. This is wrong, at least if $m$ is the rest mass, and the $m$ we are talking about is in fact the rest mass. The momentum of a particle with rest mass $m$ and velocity $v$ is actually

$p = mv / \sqrt{1 - v^2 / c^2}$.

For example, if $v = 0.999c$, then using the above formula we get $p \approx 22 mc$.

Your intuition is misleading you here: there is actually a huge difference between 0.999c and 0.9999c. Try plugging these speeds into the formula for $p$ above. I get

$v = 0.999c \Rightarrow p \approx 22 m c$

$v = 0.9999c \Rightarrow p \approx 71 m c$

This is quite a large uncertainty! It is much larger than the momentum uncertainty if you know $v$ is between 0 and 0.9c, say.

Last edited: Jul 2, 2013
14. Jul 2, 2013

### Staff: Mentor

"Mass" is always "rest mass"*. Those 8 MeV/c^2 are m0 (written as m).

*except in ancient text books and bad TV documentations.

15. Jul 3, 2013

### NLB

Bill_K: "I answered this question, but you ignored it. Take the Δp you get from the uncertainty relation and use it to calculate the velocity."

Hello Bill, No I directly answered your question. Sorry if you did not realize that. I will give a more detailed reply this time, so there is no misunderstanding about what I am saying. This is attached.

This will also answer some of the other questions as well.

#### Attached Files:

• ###### HUP.docx
File size:
115.3 KB
Views:
105
16. Jul 3, 2013

### NLB

If 8 MeV/c^2 is the rest mass of the quark, and it goes .9999c, then this is an effective mass of 566 MeV/c^2, for the mass of a single quark. There are three quarks in a proton. This puts the proton mass at 1700 MeV/c^2, or in MKS, the mass of the proton would be 3.03E-27. But we know the rest mass of the proton is: 1.66E-27 in MKS. If the uncertainty in the momentum of the quarks is between 21 and 71 mc, but still confined within the proton, we would see the rest mass of the proton fluctuating wildly. We do not. Relativity is not the solution.

17. Jul 3, 2013

### The_Duck

Note that physicsts today avoid talking about "effective mass," because it can lead to a lot of linguistic confusion. We would just say that the *total energy* of the quark is 566 MeV. 8 MeV of that comes from the rest mass, and the remaining 558 MeV is kinetic energy.

So probably the quarks are not moving quite that fast. We also have to remember that the potential energy of the quarks will also contribute to the rest mass of the proton.

Keep in mind that I didn't claim that this was the actual range of momenta of quarks inside protons. I was merely pointing out that the uncertainty in momentum can be much larger than you seemed to have assumed.

No. The kinetic energy of the quarks is only part of the rest mass of the proton. There is also their potential energy from the strong interaction. They are constantly converting kinetic and potential energy back and forth, in such a way that the total energy is fixed (energy is conserved, after all) but the fluctuations (uncertainty) in the kinetic energy and momentum are quite large.

It is. This situation is often given as an example/exercise at the beginning of a course in relativistic quantum mechanics. It is usually posed like something this:

"The proton is made of quarks and is about $10^{-15}$ meters across. Use the uncertainty principle to estimate the typical kinetic energy of quarks inside the proton and compare the answer you get to the proton's rest mass."

The desired answer is something like:

The uncertainty principle gives $\Delta p \sim \hbar / \Delta x$. Plugging in $\Delta x = 10^{-15} m$ we get $\Delta p \sim 200 MeV/c$. Since this is very large compared to the rest mass of the quarks, we have, using the relativistic formula for total energy, $E = \sqrt{p^2c^2 + m^2c^4} \approx \sqrt{p^2 c^2} = pc$. So the total energies of the quarks (which are essentially all kinetic energy) are of order 200 MeV. 3*200 = 600 MeV which is similar to the mass of the proton, which makes sense: apparently some significant fraction of the proton's rest mass is from the kinetic energy of the quarks.

Edit: since I gave the answer to that exercise, here is a similar one:

In the Standard Model the electron is a point particle, but it might actually be a composite particle of finite size. Experiments give an upper bound of something like $10^{-20} m$ on the radius of the electron. Suppose the electron is composed of 2 "preons" orbiting each other at a radius of $10^{-20} m$. Use the uncertainty principle to estimate the preons' kinetic energy and compare it to the electron's rest mass. Conclude that if the electron is composite, the preons' (negative) potential energy must somehow be finely tuned to cancel their (positive) kinetic energy to give the observed electron mass of 511 keV/c^2.

Last edited: Jul 3, 2013
18. Jul 3, 2013

### Bill_K

Thanks, Duck. This is exactly the calculation that I've been trying to get him to do.