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Aharanov-Bohm question

  1. Aug 6, 2010 #1
    Every resource I can find that discusses the Aharonov-Bohm effect states that the phase shift in SI units is given by
    [tex]\Delta\varphi=\frac{e}{\hbar}\oint{A\cdot dx}
    but in Ahranov and Bohm's original 1959 paper (and their subsequent 1961 paper) it is given as
    [tex]\Delta\varphi=\frac{e}{c\hbar}\oint{A\cdot dx}
    I don't understand where the factor of c comes from. Does anyone have any ideas?
    Last edited: Aug 6, 2010
  2. jcsd
  3. Aug 6, 2010 #2


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    Probably different systems of units.
  4. Aug 6, 2010 #3
    That's what I thought, but it's usually SI units in which c appears explicitly, but the SI one is without the c.
    Last edited: Aug 6, 2010
  5. Aug 6, 2010 #4
    another thing...without c in the equation the units match. That is, the right hand side of the equation has no units, as does the left. With c in the equation, the change in flux strangely acquires the units of s/m.
  6. Aug 6, 2010 #5
    As far as I can tell, it turns out that the original paper is in Heaviside-Lorentz units. Still trying to figure out exactly why this causes a factor of c to appear though.
  7. Aug 6, 2010 #6


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    Sure AB use the rationalized Gaussian units, also known as Heaviside-Lorentz units. While the SI units are good for practical purposes in engineering, the Heaviside-Lorentz units are the best in view of the general structure of electromagnetism as given by Maxwell's equations (and in QED too of course).

    The basic additional unit of electromagnetism compared to mechanics is electric charge, and this is defined by the Coulomb force of point particles at rest. This force is given by the Coulomb potential

    [tex]V(\vec{x}_1,\vec{x}_2)=\frac{q_1 q_2}{4 \pi |\vec{x}_1-\vec{x}_2|}.[/tex]

    The electric and magnetic fields [tex]\vec{E}[/tex] and [tex]\vec{B}[/tex] have the same units as is natural from a relativistic point of view. The Lorentz force on a particle in an external electromagnetic field [tex](\vec{E},\vec{B})[/tex] thus reads

    [tex]\vec{F}=q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right ).[/tex]

    Finally the (microscopic) Maxwell equations read

    [tex]\vec{\nabla} \times \vec{E} + \frac{1}{c} \partial_t \vec{E}=0,[/tex]
    [tex]\nabla \cdot \vec{B}=0[/tex]
    [tex]\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}[/tex]
    [tex]\nabla \cdot \vec{E}=\rho[/tex]

    This shows that the Heaviside-Lorentz units combine the merits of the SI (being "rational", i.e. there are no irrational factors of [tex]4 \pi[/tex] in the basic equations, but in the Coulomb law where it belongs due to the spherical symmetry of a point charge) and Gaussian units (reflecting the physical structure of electromagnetism better than the SI; the six components of the electromagnetic field, [tex](\vec{E},\vec{B})[/tex] have all the same units, and there are no artificial constants like [tex]\epsilon_0[/tex] and [tex]\mu_0[/tex] in vacuo).
  8. Aug 6, 2010 #7
    Hey, thanks for that response. It's very helpful. I'm still having trouble understanding how the units work out though. Even though the Heaviside-Lorentz units are different to SI, they must still make sense from a physical perspective. Speed must have units of distance over time; Potential must have units of energy per unit charge; Magnetic flux must have units of field-area, etc. My point being that, as I mentioned earlier, when I work out the units of the expression given in Aharonov and Bohm, the phase difference ends up being in units of time over distance instead of being unit-less.

    The only thing I can think that I'm missing is that hbar is no longer in units of energy-time, or something like that. But considering the constants and variables still have the same physical meanings, I don't see how something like that can change.
  9. Aug 6, 2010 #8
    I think that answers my question. It still trips me out that the quantities can be defined so differently in different unit systems.
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