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For example, the differential operator Laplacian is $$\nabla^2 = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}.$$

My questions are:

  1. Is it scale-invariant?

  2. what is scale-invariant?

UPDATE

The 2nd-order derivative operator $\frac{\partial^2}{\partial x^2}$,$\frac{\partial^2}{\partial y^2}$, they just compute the limit like this $$\frac{\partial^2}{\partial x^2} = lim_{\Delta x \rightarrow 0}\frac{\Delta(\frac{\partial}{\partial x})}{\Delta x}$$, right? So I think, no matter at which scale I measure it, $\Delta x$ will always be the same, since $\Delta x\approx 0$, right? If so, $\nabla^2$ should be scale-invariant?

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Scale invariance (roughly) means invariance of a quantity independent of the scale used to measure it (e.g. meter or cm scale used for measuring length). The given operator has dimension of mass^2 and hence is not scale invariant. In general dimensionful quantities can never be scale invariant (e.g. lenth. A 1 meter long stick is 100 cm long on cm scale). There are some dimensionless quantities too whose definition itself depends upon choice of some scale and hence which are not scale invariant. One example is charge of an electron. It depends upon from how close you look at it. –  user10001 Sep 16 '13 at 14:36
    
@user10001, thanks for the explanation of scale-invariance. From your comment, I assume scale-invariance should belong to ratio-like quantities, right? –  loganecolss Sep 16 '13 at 14:41
    
@user10001, Get back to my question, since the $\nabla^2$ operator has the 2nd-order derivative, so it has dimension of $mass^2$? why? –  loganecolss Sep 16 '13 at 14:43
    
Yes, dimensionless quantities whose definition doesn't involve choice of any scale are scale invariant; e.g. (length of stick A)/ (length of stick B). { reply to comment2: Mass=inverse length in natural units ($\hbar=c=1$)} –  user10001 Sep 16 '13 at 14:48
    
@user10001, why do we need to make a differential operator dimensionless? What's the significant difference between applying the dimensional and dimensionless $\nabla^2$ on the same function? And you might need to add your comment as an answer, ;-) –  loganecolss Sep 16 '13 at 15:15

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