In summary, @BvU asks you to calculate the potential energy of a sphere with a uniformly positive charge distribution. You can do this by using the relation U=k(3/5)((Q^2)/R). This will give you a mass equivalent.
  • #1
Adams2020
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3
Homework Statement
The masses of protons and neutrons are different. Suppose a proton is a sphere with a uniformly positive charge distribution. Can the mass difference between protons and neutrons be due to the electrical potential energy of the protons? Justify your answer with a simple calculation.
Relevant Equations
I'm not sure
I do not really know the relationship between potential energy and mass difference.
Isn't the difference in mass of protons and neutrons due to their quarks? (the neutron is made of two down quarks and an up quark and the proton of two up quarks and a down quark.)
Please help.
 
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  • #2
Adams2020 said:
I do not really know the relationship between potential energy and mass difference.
When in doubt, use ##E=mc^2\quad## 😁
Suppose a proton is a sphere with a uniformly positive charge distribution.
How much energy does that give you ?
 
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  • #3
Adams2020 said:
Isn't the difference in mass of protons and neutrons due to their quarks? (the neutron is made of two down quarks and an up quark and the proton of two up quarks and a down quark.)
Up and down quarks have masses of a few ##MeV##; whereas, neutrons and protons have a mass of about ##1GeV##.
 
  • #4
BvU said:
How much energy does that give you ?
i don't know
 
  • #5
Adams2020 said:
i don't know
As per the rules, you'll have to put in some effort.
 
  • #6
DrClaude said:
As per the rules, you'll have to put in some effort.
I asked for guidance, not a complete answer.
If someone can not guide, no problem.
 
  • #7
Adams2020 said:
I asked for guidance, not a complete answer.
If someone can not guide, no problem.
But you are getting guidance. @BvU asked you to calculate the potential energy of a sphere with a uniformly positive charge distribution. Can you do that?
 
  • #8
DrClaude said:
But you are getting guidance. @BvU asked you to calculate the potential energy of a sphere with a uniformly positive charge distribution. Can you do that?
Regarding the potential energy of a charged sphere, I know the following relation:
U = k(3/5)((Q^2)/R)
But I do not know how to use this to solve.
 
  • #9
What’s your background in physics?
 
  • #10
vela said:
What’s your background in physics?
Im a physics student and this exercise was given to us by a professor of particle physics.
 
  • #11
So you've already taken upper-division quantum mechanics and electromagnetism?
 
  • #12
vela said:
So you've already taken upper-division quantum mechanics and electromagnetism?
yes
 
  • #13
OK. The reason I'm asking is because your initiative and level of engagement seems unusually low for the typical student at your level of education. It just feels like something is off here. I thought perhaps you were taking this class before you were properly prepared for it, but you seemed to have taken the usual prerequisites.

I'll refer you again to BvU's reply above for the next step.
 
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  • #14
BvU said:
How much energy does that give you ?
Regarding the potential energy of a charged sphere, I know the following relation:
U = k(3/5)((Q^2)/R)
But I do not know how to use this to solve.
 
  • #15
Adams2020 said:
how to use this to solve
You simply fill in the values and the dimensions to calculate a result: the amount of energy associated with the charge configuration. And, via ##E= mc^2##, a mass equivalent.

From the way the problem statement is formulated, you can already suspect that this isn't going to explain the mass difference :smile: (which, of course, you have already looked up :rolleyes: ? )
 
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  • #16
If I'm not missing something, the neutron is heavier than the proton, so even without doing any calculations we should know that the positive electrostatic self energy of the model of the proton as a uniformly charged sphere actually does the opposite of explain the mass difference!
 
  • #17
BvU said:
You simply fill in the values and the dimensions to calculate a result: the amount of energy associated with the charge configuration. And, via ##E= mc^2##, a mass equivalent.

From the way the problem statement is formulated, you can already suspect that this isn't going to explain the mass difference :smile: (which, of course, you have already looked up :rolleyes: ? )
Suppose I get a number for potential energy this way. How can this number explain the difference in mass to me?🤔
 
  • #18
etotheipi said:
If I'm not missing something, the neutron is heavier than the proton, so even without doing any calculations we should know that the positive electrostatic self energy of the model of the proton as a uniformly charged sphere actually does the opposite of explain the mass difference!
So we can not explain the mass difference in this way?:frown:
 
  • #19
Well you'd need to ask someone who knows a bit of particle physics to get a proper answer, but I think a simple explanation is that the quark compositions are different, and u has a mass of 2.3MeV whilst d has a mass of 4.8MeV. In any case, most of the mass of a proton or neutron (over 900MeV) is due to the strong interaction (see here).

(N.B. that we take zero electric potential energy to be at infinite separation [where there is zero electric interaction], so bound states have reduced mass whilst configurations with high electric potential energy have increased mass. On the other hand, the zero potential energy of strong interaction between quarks and gluons is taken at zero separation [e.g. like a spring], so you end up with positive potential energy associated with this bound state, and the strong force actually increases the mass here).
 
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  • #20
Adams2020 said:
Justify your answer with a simple calculation.
The exercicse wants you to do a calculation. What did you actually calculate so far ?
 
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  • #21
BvU said:
The exercicse wants you to do a calculation. What did you actually calculate so far ?
Nothing. I do not know exactly what to calculate. All I know is a relationship to potential energy. Suppose I got it, what does it have to say and how does it explain the mass difference?
 
  • #22
See #14. Do not reply with "suppose I get a number", but do the calculation
 
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  • #23
BvU said:
See #14. Do not reply with "suppose I get a number", but do the calculation
Using the relation sent here and the following data, I obtained a value for the electric potential energy of the proton, which is:
1.6 * 10^ (-13) J

R= 8.7 * 10 ^ (-16) m
k= 9 * 10 ^ (19) Nm^2/c^2
Q= 1.6 * 10 (-19) c

What does this number say?
 
  • #24
If this number is somewhat equal to the mass difference between proton and neutron (with the correct sign and in the same units), you can answer the question
Can the mass difference between protons and neutrons be due to the electrical potential energy of the protons?
affirmativeley :smile:Side nite:

Adams2020 said:
U = k(3/5)((Q^2)/R
Others find ##1\over 2## ; where does your ##3\over 5## come from ?
 
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  • #25
BvU said:
Others find ##1\over 2## ; where does your ##3\over 5## come from ?
Feynman agrees with the factor 3/5 for a uniform sphere of charge.
 
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  • #26
So you have to calculate the energy yourself carefully. In physics an argument that the one or other authority made a statement, doesn't count ;-).

Hint: The energy density of an electrostatic field is
$$u=\frac{\epsilon_0}{2} \vec{E}^2.$$
You can simplify the task to evaluate the total energy by using
$$\vec{E}=-\vec{\nabla} \Phi,$$
plugging this into the integral and then use Gauss's Law
$$\vec{\nabla} \cdot \vec{E}=-\frac{1}{\epsilon_0} \rho.$$
 
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  • #27
And if in doubt, you can always try doing it explicitly, $$
E_r(r) = \begin{cases}
\frac{kQr}{R^3} & r \leq R \\
\frac{kQ}{r^2} & r> R
\end{cases}$$to find$$U = \frac{\varepsilon_0}{2} \int_{\mathbb{R}^3} E^2 dV = \frac{\varepsilon_0}{2} \int_0^{4\pi} \int_0^{\infty} E^2 r^2 dr d\Omega = 2\pi \varepsilon_0 \int_0^\infty E^2 r^2 dr$$
 
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  • #28
BvU said:
If this number is somewhat equal to the mass difference between proton and neutron (with the correct sign and in the same units), you can answer the question
After equalizing the units, the mass difference is almost the same as this number. Thanks for your guidance
 
  • #29
BvU said:
If this number is somewhat equal to the mass difference between proton and neutron (with the correct sign and in the same units), you can answer the question...
affirmativeley :smile:

The sign is incorrect, so this certainly cannot account for the mass difference. The protons are actually less heavy than the neutrons, not the other way around. Did I miss something?
 
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  • #30
etotheipi said:
And if in doubt, you can always try doing it explicitly, $$
E_r(r) = \begin{cases}
\frac{kQr}{R^3} & r \leq R \\
\frac{kQ}{r^2} & r> R
\end{cases}$$to find$$U = \frac{\varepsilon_0}{2} \int_{\mathbb{R}^3} E^2 dV = \frac{\varepsilon_0}{2} \int_0^{4\pi} \int_0^{\infty} E^2 r^2 dr d\Omega = 2\pi \varepsilon_0 \int_0^\infty E^2 r^2 dr$$
Do I have to use both electric fields? That is, I have to use both the electric field inside the sphere and outside the sphere ?
 
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  • #31
Adams2020 said:
Do I have to use both electric fields? That is, I have to use both the electric field inside the sphere and outside the sphere ?

Yeah, you would need to split the integral into$$U = 2\pi \varepsilon_0 \int_0^\infty E^2 r^2 dr = 2\pi \varepsilon_0 \int_0^R \frac{k^2 Q^2 r^4}{R^6} dr + 2\pi \varepsilon_0 \int_R^\infty \frac{k^2 Q^2}{r^2} dr$$
 
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  • #32
etotheipi said:
Yeah, you would need to split the integral into$$U = 2\pi \varepsilon_0 \int_0^\infty E^2 r^2 dr = 2\pi \varepsilon_0 \int_0^R \frac{k^2 Q^2 r^4}{R^6} dr + 2\pi \varepsilon_0 \int_R^\infty \frac{k^2 Q^2}{r^2} dr$$
The results of this way should be the same as the previous way, right?
 
  • #33
I've not seen andy previous way in this thread, only guesses about factors, which haven't been before either ;-)).
 
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  • #34
Adams2020 said:
The results of this way should be the same as the previous way, right?
Yes, the result is exactly the same. I got the previous formula.
Thank you and those who guided me in this exercise.
 
  • #35
etotheipi said:
The sign is incorrect, so this certainly cannot account for the mass difference. The protons are actually less heavy than the neutrons, not the other way around. Did I miss something?
You did not. But it's not your exercise !
 
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<h2>What is mass difference due to electrical potential energy?</h2><p>Mass difference due to electrical potential energy refers to the change in mass of a system when it gains or loses electrical potential energy. This change in mass is a result of the conversion of electrical potential energy into mass, as described by Einstein's famous equation E=mc^2.</p><h2>How does electrical potential energy affect mass?</h2><p>Electrical potential energy affects mass by converting into mass, according to Einstein's equation E=mc^2. This means that when a system gains or loses electrical potential energy, there will be a corresponding change in its mass.</p><h2>What is the relationship between electrical potential energy and mass?</h2><p>The relationship between electrical potential energy and mass is described by Einstein's equation E=mc^2, which states that a change in energy (E) will result in a change in mass (m). In other words, as electrical potential energy is gained or lost, there will be a corresponding change in mass.</p><h2>Can mass be created or destroyed through electrical potential energy?</h2><p>No, mass cannot be created or destroyed through electrical potential energy. Instead, electrical potential energy is converted into mass, meaning that the total mass of a system will remain constant.</p><h2>How is mass difference due to electrical potential energy measured?</h2><p>Mass difference due to electrical potential energy can be measured using a variety of methods, such as mass spectrometry or nuclear reactions. These techniques allow for the precise measurement of changes in mass due to changes in electrical potential energy.</p>

What is mass difference due to electrical potential energy?

Mass difference due to electrical potential energy refers to the change in mass of a system when it gains or loses electrical potential energy. This change in mass is a result of the conversion of electrical potential energy into mass, as described by Einstein's famous equation E=mc^2.

How does electrical potential energy affect mass?

Electrical potential energy affects mass by converting into mass, according to Einstein's equation E=mc^2. This means that when a system gains or loses electrical potential energy, there will be a corresponding change in its mass.

What is the relationship between electrical potential energy and mass?

The relationship between electrical potential energy and mass is described by Einstein's equation E=mc^2, which states that a change in energy (E) will result in a change in mass (m). In other words, as electrical potential energy is gained or lost, there will be a corresponding change in mass.

Can mass be created or destroyed through electrical potential energy?

No, mass cannot be created or destroyed through electrical potential energy. Instead, electrical potential energy is converted into mass, meaning that the total mass of a system will remain constant.

How is mass difference due to electrical potential energy measured?

Mass difference due to electrical potential energy can be measured using a variety of methods, such as mass spectrometry or nuclear reactions. These techniques allow for the precise measurement of changes in mass due to changes in electrical potential energy.

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