How to compute the energy of scalar wave equation

[xortdsc]

Hi,

considering the scalar wave equation
$$
{ \partial^2 u \over \partial t^2 } = c^2 \nabla^2 u
$$
(where ∇^2 is the (spatial) Laplacian and where c is a fixed constant)

how can I derive the potential and kinetic energy for a given state u and u' ?

Thanks and cheers

 

[vasilis]

Hi,

First of all notice that this PDE is linear - that is, the unknown function u(x,t) appears in powers of one only. Therefore this a free (non-interacting, V(x)=0) scalar theory which can be easily solved (diagonalized) via a Fourier Transformation. An easy way to see that there is no interaction, and therefore the Potential energy is zero, is to notice that the solutions of this equation are plane waves and any linear superposition of these is also a plane wave which means that the two waves pass through each other without any mixing occurring.

A quick way to derive the KE is to guess the Lagrangian which yields the corresponding PDE. If you have studied special relativity before, it might help because you are looking at a relativistic wave equation which is Lorentz invariant. Notice that

\begin{equation}
0=(\partial^{2}_{t}-∇^{2})u(t,x) = η ^{\mu\nu} \partial_{\mu}\partial_{\nu} u(t,x)
\end{equation}

which is clearly invariant under SO(1,3) transformations.
Thus if you try to write down a Lorentz invariant action for your free scalar field u(t,x) you will realize that the Lorentz symmetry is so restrictive which "uniquely" determines the form of the action to be (modulo total derivatives):

\begin{align}
S&=\int dt \mathcal{L}\\
\mathcal{L}&=\int d^{3}x\frac{1}{2}η ^{\mu\nu} \partial_{\mu}u\partial_{\nu} u
\end{align}

Since there is no potential then the "conventional" KE is equal to the Lagrangian.

To be more precise by definition the velocity is defined as the time derivative of u. But since this is a relativistic theory it is convenient to combine space and time derivatives and call the whole thing the velocity of u. However when one computes the Hamiltonian of the system H = KE + PE then time acquires a special role in this description and one may interpret the contribution coming from (∇u)^{2} as a PE - in particular, as the energy cost of "shearing" in space.

An alternative and more laborious way to obtain the KE is to integrate the PDE (it may help to first multiply by u and then integrate by parts) and find the conserved energy.

 

[vasilis]

In case you haven't studied SR:

\begin{equation}
\mathcal{L}=\int d^{3} x\frac{1}{2}\Big( (\dot u)^2 -(∇u)^{2})
\end{equation}

 

[dauto]

that question cannot be answered without some knowledge of the physical meaning of the variable u.

 

[xortdsc]

Hi,

First of all notice that this PDE is linear - that is, the unknown function u(x,t) appears in powers of one only. Therefore this a free (non-interacting, V(x)=0) scalar theory which can be easily solved (diagonalized) via a Fourier Transformation. An easy way to see that there is no interaction, and therefore the Potential energy is zero, is to notice that the solutions of this equation are plane waves and any linear superposition of these is also a plane wave which means that the two waves pass through each other without any mixing occurring.

A quick way to derive the KE is to guess the Lagrangian which yields the corresponding PDE. If you have studied special relativity before, it might help because you are looking at a relativistic wave equation which is Lorentz invariant. Notice that

\begin{equation}
0=(\partial^{2}_{t}-∇^{2})u(t,x) = η ^{\mu\nu} \partial_{\mu}\partial_{\nu} u(t,x)
\end{equation}

which is clearly invariant under SO(1,3) transformations.
Thus if you try to write down a Lorentz invariant action for your free scalar field u(t,x) you will realize that the Lorentz symmetry is so restrictive which "uniquely" determines the form of the action to be (modulo total derivatives):

\begin{align}
S&=\int dt \mathcal{L}\\
\mathcal{L}&=\int d^{3}x\frac{1}{2}η ^{\mu\nu} \partial_{\mu}u\partial_{\nu} u
\end{align}

Since there is no potential then the "conventional" KE is equal to the Lagrangian.

To be more precise by definition the velocity is defined as the time derivative of u. But since this is a relativistic theory it is convenient to combine space and time derivatives and call the whole thing the velocity of u. However when one computes the Hamiltonian of the system H = KE + PE then time acquires a special role in this description and one may interpret the contribution coming from (∇u)^{2} as a PE - in particular, as the energy cost of "shearing" in space.

An alternative and more laborious way to obtain the KE is to integrate the PDE (it may help to first multiply by u and then integrate by parts) and find the conserved energy.

In case you haven't studied SR:

\begin{equation}
\mathcal{L}=\int d^{3} x\frac{1}{2}\Big( (\dot u)^2 -(∇u)^{2})
\end{equation}

Thank you very much for the extensive explanation.
Since I havn't studied either mathematics nor SR this sounds utterly complicated, but I'll try to work myself into it. :)

 

[xortdsc]

that question cannot be answered without some knowledge of the physical meaning of the variable u.

Is that so ? I would have thought that it should be determined by the differential alone as it should only reflect the amount of work necessary to bring it into a given state. Of course the zero-point of the energy function is arbitrary, but only relative differences between different states are of importance.

 

[AlephZero]

Is that so ? I would have thought that it should be determined by the differential alone as it should only reflect the amount of work necessary to bring it into a given state.

I'm with dauto here. The problem is that your wave equation doesn't say anything about what "it" is that you are "bringing into a given state".

Take a specific example like continuum mechanics. The wave speed c only depends on the ratio of the elastic modulus and the density of the continuum. For a given amplitude of motion u, the PE and KE depend on the absolute values of modulus and density.

This question has nothing to do with special relativity, zero point energy, etc IMO.

I suppose you could invent some normalized measure of energy density (e.g take the mass density as 1 in an arbitrary unit system) but as an engineer I'm not very interested in that sort of arm waving mathematics

 

[xortdsc]

I'm with dauto here. The problem is that your wave equation doesn't say anything about what "it" is that you are "bringing into a given state".

"It" is u and u' and how to "bring it into a given state" should be exactly defined by the supplied PDE, right ?

Take a specific example like continuum mechanics. The wave speed c only depends on the ratio of the elastic modulus and the density of the continuum. For a given amplitude of motion u, the PE and KE depend on the absolute values of modulus and density.

This question has nothing to do with special relativity, zero point energy, etc IMO.

I suppose you could invent some normalized measure of energy density (e.g take the mass density as 1 in an arbitrary unit system) but as an engineer I'm not very interested in that sort of arm waving mathematics

Well, don't think of this system as being a real physical system. Let's consider it as an abstract, made-up system. The reason I want an energy function even when it lacks an actual physical correspondance is for analysis purposes. This way I can see if certain structures of u and u' yield local minima in the energy function and therefore are stable structures. Of course there are no local minima in linear equations, but my actual equation is slightly more complicated and non-linear and has local minima. I just wanted to understand the process of finding such energy function starting with a simpler linear PDE.

 

[Jano L.]

AlephZero and dauto are right when we talk about energy in its physics meaning, i.e. when there is connection to other energies, work, etc. Then the wave equation by itself is not sufficient to state that there is energy at all.

However, if you seek energy in the sense of a quantity calculated by some integral of the function and its derivatives that is conserved in time or changes in time according to some surface integral of those fields, then such quantity may be found from the equation itself.

Whether this is possible depends on the kind of equation as well. For wave equation, it is possible to derive

$$
\frac{d}{dt}\int_V \frac{1}{2}(\partial_t u)^2 + \frac{1}{2}c^2(\Delta u)^2 \,dV = \oint_S -\partial_t u \nabla u \cdot d\mathbf S
$$

so the integral on the left-hand side has some formal resemblance to energy in continuum mechanics. But here it is just math.

This kind of quantity may not exist for other equations, for example the heat conduction equation
$$
\partial_t u = k\Delta u
$$
does not have such integral.

 

[dauto]

AlephZero and dauto are right when we talk about energy in its physics meaning, i.e. when there is connection to other energies, work, etc. Then the wave equation by itself is not sufficient to state that there is energy at all.

However, if you seek energy in the sense of a quantity calculated by some integral of the function and its derivatives that is conserved in time or changes in time according to some surface integral of those fields, then such quantity may be found from the equation itself.

Whether this is possible depends on the kind of equation as well. For wave equation, it is possible to derive

$$
\frac{d}{dt}\int_V \frac{1}{2}(\partial_t u)^2 + \frac{1}{2}c^2(\Delta u)^2 \,dV = \oint_S -\partial_t u \nabla u \cdot d\mathbf S
$$

so the integral on the left-hand side has some formal resemblance to energy in continuum mechanics. But here it is just math.

This kind of quantity may not exist for other equations, for example the heat conduction equation
$$
\partial_t u = k\Delta u
$$
does not have such integral.

Yes, it's possible to find conserved quantities. And those quantities might be related to energy, but they might mean something else. Impossible to tell their meaning without some knowledge of the meaning of u. For instance, take a sound wave. If u is the displacement of the air, than ∂u/∂t is the speed, and (∂u/∂t)² is the kinetic energy. On the other hand, for the same sound wave, if u is the pressure displacement, than ∂u/∂t means something else.

 

[AlephZero]

There is nothing wrong with converting your differential equation into a weak form involving integrals, and then find some terms in the equation that look analogous to "energy", and use that to help you solve the problem.

But IMO energy is something you can measure in joules, not an arbitrary mathematical expression.

 

[xortdsc]

Okay, then possibly the mistake was simply to call it "energy" when I was really talking about some abstract conserved scalar to measure the "tension" of the system derived from the PDE.

Thank you for the insight.

 

[the_wolfman]

When studying differential equations it is common to define an "energy" regardless of the physical application.
In the context of the original post its clear what is meant by "potential" and "kinetic" energy.

You don't have to like it. Grumble all you want. But is it a common convention.