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When calculating the energy difference between the normal and the superconducting state in a superconductor at zero magnetic field, the result is as follows:

enter image description here

Now I'm quite confident of this result, as it is the same as my textbook tells me it is. The problem is that when I do a dimensional analysis, I find

enter image description here

The dimension on the left is energy, the dimension on the right is vacuum permeability times H-field squared. The units of the first are found on Wikipedia, $N/A^2$, as are the units of the second, $A/m$ (Ctrl+F for ``The H-field is measured in'').

As you can see, the lengths don't cancel out, we have $[L]^2$ on the left and $[L]^{-1}$ on the right. What am I doing wrong? Is the first equation not in SI?


Edit: In case it might be of any help, the derivation of the above equation is (using the Meissner effect in the final step, $M=-H$) as follows:

enter image description here

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Remember that your expression gives the energy difference per unit volume. So you need an additional factor of $\left[L^{-3}\right]$ on the left hand side.

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    $\begingroup$ That was exactly what I found a minute ago, you beat me to it :p. $\endgroup$
    – Betohaku
    Jul 13, 2013 at 15:52
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    $\begingroup$ @Betohaku Great! I find it's always more rewarding to get to that Aha-moment on my own. So I haven't robbed you of that :) $\endgroup$
    – Wouter
    Jul 13, 2013 at 15:55
  • $\begingroup$ Just a bit frustrating that the authors use the term Gibbs energy instead of Gibbs energy density.. $\endgroup$
    – Betohaku
    Jul 13, 2013 at 15:58
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    $\begingroup$ Yeah, it can sometimes be a cause for confusion when authors make things like that implicit. Especially when you're first learning about something, it's important to stop at every equation you come across and think: what exactly am I looking at here? Doing some quick dimensional analysis as you have done for this equation is a useful thing (and a good habit) in that respect, especially for relatively simple equations. For more complicated equations dimensional analysis can take some time, though you should still do it when in doubt. $\endgroup$
    – Wouter
    Jul 13, 2013 at 16:07

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