This is probably a simple question, but what does the subscript $0$ mean in the following expression?

$$V=\frac{1}{2}\left(\frac{d^2 V}{{dq_i}{dq_j}}\right)_0$$


The subscript zero here probably just means either

  1. that one should put $q=0$ after differentiating twice,

  2. or that the Hessian $\left. \partial_i\partial_j V \right|_{q=q_{(0)}}$ is evaluated at point $q_{(0)}$ that is a stationary point $\left. \partial_i V \right|_{q=q_{(0)}}=0$.

One would have to see the precise context to be sure.

  • $\begingroup$ I would guess (2), but yes context would help. I'm betting it's the second derivative of a potential evaluated at the equilibrium point. $\endgroup$ – kηives Sep 27 '12 at 3:55

Usually that means "initial" or if you're in Europe it can mean either "nought" or initial. All three mean the same thing though. An example would be $v_0$ which would be initial velocity.


The zero subscript does not HAVE to mean initial. As I mentioned in a comment, we can take an example from electrostatics where epsilon nought ($ε_0$) is the permittivity of free space, not any initial value.

  • $\begingroup$ Even in Europe it means initial. In fact, we are more likely than Americans to use $_0$ rather than $_{in}$ or $_{initial}$, etc. $\endgroup$ – Peteris Sep 26 '12 at 4:07
  • $\begingroup$ Strange. I thought it had something to do with equilibrium position. I should mention that I found this expression in the context of the chapter on small oscillations. $\endgroup$ – Joebevo Sep 26 '12 at 4:10
  • $\begingroup$ @Peteris I'll make the change; I didn't know that. Thanks. $\endgroup$ – Stumbleine75 Sep 27 '12 at 2:56
  • 2
    $\begingroup$ @ThroatOfWinter57 Don't you mean nought, not not? Pronunciation aside, I'm pretty sure the word refers to the zero-ness of the subscript, rather than any logical negation. This is a common misunderstanding though: see english.SE. $\endgroup$ – user10851 Oct 27 '12 at 8:25
  • $\begingroup$ Oops, I guess. It's changed now though. Also, and I just now realized this, but in Physics we have Epsilon nought which obviously does not mean initial permittivity of free space! I'll make a futile edit. $\endgroup$ – Stumbleine75 Feb 26 '13 at 2:39

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