Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\Omega$ be an open subset of $\mathbb{R}^n$. The following

$$\lVert u \rVert_{1, 2}^2=\int_{\Omega} \lvert u(x)\rvert^2\, dx + \int_{\Omega} \lvert \nabla u(x)\rvert^2\, dx$$

defines a norm on $H^1(\Omega)$ space, that is sometimes called energy norm.

I don't feel easy with the physical meaning this name suggests. In particular, I see two non-homogeneous quantities, $\lvert u(x)\rvert^2$ and $\lvert \nabla u(x)\rvert^2$, being summed together. How can this be physically consistent?

Maybe some example could help me here. Thank you.

share|improve this question
2  
If everything in physics would be homogenous, there would be never a characteristic lengthscale (like the radius of an atom) appearing. –  Fabian May 22 '11 at 10:34
6  
In physical applications there may be a unital constant attached to one term or the other which is being ignored in this definition for convenience. –  Qiaochu Yuan May 22 '11 at 10:44
    
@Qiaochu: Your observation is exactly what I was missing. Thank you. @Fabian: Your comment & answer look very interesting but I miss some language to understand it. Can you point me somewhere where I can understand this concept of characteristic lengthscale, please? –  Giuseppe Negro May 23 '11 at 8:10
add comment

2 Answers

up vote 11 down vote accepted

To expand on my comment (see e.g. http://online.itp.ucsb.edu/online/lnotes/balents/node10.html):

The expression which you give can be interpreted as the energy of a $n$-dimensional elastic manifold being elongated in the $n+1$ dimension (e.g. for $n=2$, membrane in three dimension); $u$ is the displacement field.

Let me put back the units $$E[u]= \frac{a}{2}\int_{\Omega} \lvert u(x)\rvert^2\, dx + \frac{b}{2} \int_{\Omega} \lvert \nabla u(x)\rvert^2\, dx.$$ The first term tries to bring the manifold back to equilibrium (with $u=0$), the second term penalizes fast changes in the displacement. The energy is not homogenous and involves a characteristic length scale $$\ell_\text{char} = \sqrt{\frac{b}{a}}.$$ This is the scale over which the manifold returns back to equilibrium (in space) if elongated at some point. With $b=0$, the manifold would return immediately, you elongate it at some point and infinitely close the manifold is back at $u=0$. With $a=0$ the manifold would never return to $u=0$. Only the competition between $a$ and $b$ leads to the physics which we expect for elastic manifold. This competition is intimately related to the fact that there is a characteristic length scale appearing.

It is important that physical laws are not homogenous, in order to have characteristic length scales (like $\ell$ in your example, the Bohr radius for the hydrogen problem, $\sqrt{\hbar/m\omega}$ for the quantum harmonic oscillator, ...). The energy of systems only become scale invariant in the vicinity of second order phase transitions. This is a strong condition on energy functionals to the extend that people classify all possible second order phase transitions.

share|improve this answer
    
Maybe you did not notice my last comment to the main question, so I repeat it here. Unfortunatetly, I don't know this concept of characteristic lengthscale, can you point me somewhere where I can learn more? Thank you. –  Giuseppe Negro May 23 '11 at 17:35
    
@dissonance: only the first person gets notified with the (@). About the characteristic lengthscale: the wikipedia article seems to be rather short and without information. What I can say is even shorter (but hopefully having more information): given the parameters of the systems' energy ($a$ and $b$ above) any combination which has the dimension of length is called characteristic length. Slightly more information, you can get by reading Street-Fighting-Mathematics especially chaper 3. (I assume you are mathematician) –  Fabian May 23 '11 at 17:46
    
Yes, I'm a student of Mathematics. Thank you very much for this pointer. –  Giuseppe Negro May 23 '11 at 17:52
add comment

When u is the velocity field of an viscous, incompressible homogeneous fluid with density set to 1, then the first integral is proportional to the kinetic energy of the fluid flow. The second integral is called enstrophy (see Wikipedia).

Enstrophy is in this situation something different than energy, therefore the term "energy norm" is slightly misleading. For a viscuous flow described by the Navier-Stokes equations, the enstrophy measures how fast the fluid flow dissipates energy due to friction, it is possible to show under some mild assumptions that $$ \frac{d}{dt} \text{energy} = - \text{viscosity} * \text{enstrophy} $$ So, physically speaking, fluid flows with finite energy and finite enstrophy are precisely those whose velocity fields are elements of $H^1$. And physically interesting fluid flows should have finite enstropy so that the process of energy loss due to friction is described by the model.

share|improve this answer
    
So, why would one want to add up an entropy and an enstrophy? (as I read the OP's question this is what he/she asks) –  Fabian May 22 '11 at 12:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.