Trivial sign problem Please set me straight

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In summary, the conversation is about calculating the capacitance of a set of spherical shells with radii b > a. The electric field vector is determined by placing a charge +Q on the inner shell and -Q on the outer shell, causing the electric field to point outward. The potential is found by integrating against the electric field, which results in a minus sign. However, some people neglect the minus sign to get the correct answer for capacitance, while others take the absolute value. It is better to integrate against the field to ensure an increase in potential.
  • #1
tshafer
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I'm calculating the capacitance of a set of spherical shells of radii b > a. To do this, I place a charge +Q on the inner shell and -Q on the outer shell so that I get the electric field vector pointing outward
[tex]
\vec E = \frac{Q}{4\pi\varepsilon_0} \frac{\hat r}{r^2}.
[/tex]

Finding the potential should be trivial... I do the integral
[tex]
V = -\int_a^b \vec E \cdot d\vec r = -\frac{Q}{4\pi\varepsilon_0} \int_a^b \frac{dr}{r^2} = -\frac{Q}{4\pi\varepsilon_0}\left( \frac{1}{a} - \frac{1}{b} \right).
[/tex]

This is off by a minus sign and I cannot understand why... reversing the limits of integrations, of course, kills the sign but introduces a sign flip in the dot product as E then opposes dr. This should be trivial... help? :)
 
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  • #2
tshafer said:
I'm calculating the capacitance of a set of spherical shells of radii b > a. To do this, I place a charge +Q on the inner shell and -Q on the outer shell so that I get the electric field vector pointing outward
[tex]
\vec E = \frac{Q}{4\pi\varepsilon_0} \frac{\hat r}{r^2}.
[/tex]

Finding the potential should be trivial... I do the integral
[tex]
V = -\int_a^b \vec E \cdot d\vec r = -\frac{Q}{4\pi\varepsilon_0} \int_a^b \frac{dr}{r^2} = -\frac{Q}{4\pi\varepsilon_0}\left( \frac{1}{a} - \frac{1}{b} \right).
[/tex]

This is off by a minus sign and I cannot understand why... reversing the limits of integrations, of course, kills the sign but introduces a sign flip in the dot product as E then opposes dr. This should be trivial... help? :)

Are you sure there is a sign error? If you have +Q at a and -Q at b, which way does the Electric field point?


EDIT -- If you are moving in the direction of the E-field, your potential is decreasing. Still seems like the signs are correct to me.
 
  • #3
It would have to point from =Q to -Q (outward radial), no? That's why I'm confused... I could take the absolute value to get the correct value for the capacitance but still... maybe it's confusion in calculating capacitance? As best practice should one always do the integration against the field (which would ensure an increase in V) ?

Thanks!

EDIT: In studying for finals I am comparing to Jackson solutions... most people glibly neglect the minus sign out front to get the right answer, whilst some other souls take the absolute value. It seems better to me to simply integrate against the field?
 
  • #4
tshafer said:
It would have to point from =Q to -Q (outward radial), no? That's why I'm confused... I could take the absolute value to get the correct value for the capacitance but still... maybe it's confusion in calculating capacitance? As best practice should one always do the integration against the field (which would ensure an increase in V) ?

Thanks!

EDIT: In studying for finals I am comparing to Jackson solutions... most people glibly neglect the minus sign out front to get the right answer, whilst some other souls take the absolute value. It seems better to me to simply integrate against the field?

Yes, I'm inclined to integrate against the field as well. After all, C = Q/V, and to get a +Q, you need to integrate from the -Q to the +Q location. Going that direction also gives you a +V, hence a +C. I suppose you can to the other direction by symmetry, and get a -Q/-V to get the same +C.
 
  • #5
Great, thanks! It's things like this that I really should have nailed down by now... but better to get it now than right before the written exam, I suppose.
 

Related to Trivial sign problem Please set me straight

1. What is the "Trivial Sign Problem"?

The Trivial Sign Problem is a problem in theoretical physics that arises when trying to apply Monte Carlo methods to systems with a complex action, such as quantum chromodynamics. In these systems, the integral over the action can have both positive and negative contributions, resulting in a cancellation of terms and making the calculation difficult.

2. How does the Trivial Sign Problem affect scientific research?

The Trivial Sign Problem can significantly hinder scientific research in fields such as quantum chromodynamics and condensed matter physics. It limits the applications of Monte Carlo methods, which are widely used in these fields for simulating complex systems and predicting their behavior.

3. What are some proposed solutions to the Trivial Sign Problem?

Several proposed solutions to the Trivial Sign Problem include reweighting techniques, complexification of the action, and approximations of the partition function. However, these solutions come with their own limitations and further research is needed to fully solve the problem.

4. How does the Trivial Sign Problem relate to the sign problem in statistical mechanics?

The Trivial Sign Problem is a specific case of the sign problem in statistical mechanics, where the partition function has both positive and negative contributions. However, the sign problem in statistical mechanics is more general and can also occur in other systems, such as fermionic systems with chemical potential.

5. Is there any progress being made towards solving the Trivial Sign Problem?

Yes, there is ongoing research and progress being made towards solving the Trivial Sign Problem. Some recent developments include the use of machine learning techniques and new algorithms to improve the efficiency of Monte Carlo simulations. However, it is still a challenging problem and more work is needed to fully overcome it.

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