When we study mechanical waves, we essentially have to deal with $dm$ mass increments; then we can "expand" they as:

$$ dm = \rho(x,t)dV.\tag{1}$$

Also, I think that it's pretty direct that the Kinectic energy have a "canonical form":

$$K = \frac{1}{2}dm\Bigg(\frac{\partial u(x,y,z;t)}{\partial t}\Bigg)^2 = \frac{1}{2}\rho(x,y,z;t)\Bigg(\frac{\partial u(x,y,z,t)}{\partial t}\Bigg)^2dV. \tag{2}$$

Furthermore (now considering 1D waves in a rope), the energy carried by the wave propagation is then given by:

$$E_{rope} = \frac{1}{2}\int_{a}^{b}dx\Bigg[\rho \Bigg(\frac{\partial u(x,t)}{\partial t}\Bigg)^2 + T\Bigg(\frac{\partial u(x,t)}{\partial x}\Bigg)^2 \Bigg]. \tag{3}$$

Where $T$ is the tension of the rope.


Since $(3)$ is another way to write the sum of mechanical energy $E= K+U$, I tried to generalize it to every (conservative) force in the system:

$$\mathcal{E} = \frac{1}{2}\int_{a}^{b}dx\Bigg[\rho \Bigg(\frac{\partial u(x,t)}{\partial t}\Bigg)^2 + (\phi_{1}+\phi_{2}+...+\phi_{n}) \Bigg], \tag{4}$$


$$F = - \vec{\nabla} \phi \implies \phi = \int_{\gamma} \langle \vec{F} , d\vec{l}\rangle. \tag{5}$$


My question arises concerning the potential energy. Since we know that Kinectic energy will retain always it's form, potential energy won't. I worked out a example given by $[1]$, of a 1D wave in a column filled with gas media (which isn't too straightforward).

Well, the only force in this system is pressure; $\vec{F} = \vec{p}$. Then,

$$\int_{\gamma} \langle \vec{F} , d\vec{l}\rangle = - \vec{\nabla} \phi \implies \phi = \int_{\gamma} \langle \vec{p} , d\vec{l}\rangle. \tag{6}$$

Also, this problem is an $1$D problem, so the line integral becomes:

$$\phi = \int_{\gamma} \langle \vec{p} , d\vec{l}\rangle= \int_{0}^{L} p_{x}ds = \int_{0}^{L} p_{x}dx. \tag{7}$$

But my reference $[1]$ have given a specific form of $p_{x} = p$:

$$ p = p_{0} - \kappa\frac{\partial u(x,t)}{\partial x}. \tag{8} $$

Well, then it's a matter of do the proper calculations:

$$\phi = \int_{0}^{L} p_{x}dx = \int_{0}^{L} dx \Bigg( p_{0} - \kappa\frac{\partial u(x,t)}{\partial x} \Bigg). \tag{9}$$

So, we can then "re-write" expression $(9)$, since it's a line integral of a scalar field, in the end of the day:

$$\phi = \int_{0}^{L} pds = \int_{0}^{L} p\sqrt{1+\Bigg(\frac{\partial u(x,t)}{\partial x}\Bigg)^2 } . \tag{10}$$


$$\int_{0}^{L} p_{x}\sqrt{1+\Bigg(\frac{\partial u(x,t)}{\partial x}\Bigg)^2 } = \int_{0}^{L} dx \Bigg( p_{0} - \kappa\frac{\partial u(x,t)}{\partial x} \Bigg) \tag{11}.$$

Now, we can Taylor expand the term $\sqrt{1+\Bigg(\frac{\partial u(x,t)}{\partial x}\Bigg)^2 } $ as:

$$\phi \approx \int_{0}^{L} \Bigg( p_{0} - \kappa\frac{\partial u(x,t)}{\partial x} \Bigg) \Bigg(1+\frac{1}{2}\Bigg(\frac{\partial u(x,t)}{\partial x}\Bigg)^2\Bigg)dx \tag{12}.$$

The things become tricky now. If we calculate all the products we reach on the following equation:

$$-\int_{0}^{L}p_{0}ds + \int_{0}^{L}\kappa\frac{\partial u}{\partial x} ds = -\int_{0}^{L}p_{0}dx - \int_{0}^{L} \frac{p_{0}}{2}\Bigg(\frac{\partial u}{\partial t}\Bigg)^2dx - \int_{0}^{L}\kappa \frac{\partial u}{\partial t}dx-\int_{0}^{L}\kappa\frac{1}{2}\frac{\partial u}{\partial x}\Bigg(\frac{\partial u}{\partial x}\Bigg)^2 dx \tag{13}$$


$$-\int_{0}^{L}p_{0}(ds+dx) + \int_{0}^{L}\kappa\frac{\partial u}{\partial x} (ds+dx) =$$

$$= - \int_{0}^{L} \Bigg[ \frac{p_{0}}{2}\Bigg(\frac{\partial u}{\partial t}\Bigg)^2 + \kappa\frac{1}{2}\frac{\partial u}{\partial x}\Bigg(\frac{\partial u}{\partial x}\Bigg)^2\Bigg] dx = $$

$$= - \int_{0}^{L} \Bigg[ \Bigg(\frac{p_{0}}{2} + \kappa\frac{1}{2}\frac{\partial u}{\partial x}\Bigg) \Bigg(\frac{\partial u}{\partial x}\Bigg)^2\Bigg] dx \implies$$

$$ -\int_{0}^{L}p_{0}(ds+dx) + \int_{0}^{L}\kappa\frac{\partial u}{\partial x} (ds+dx) =$$ $$ - \int_{0}^{L} \Bigg[ \Bigg(\frac{p_{0}}{2} + \kappa\frac{1}{2}\frac{\partial u}{\partial x}\Bigg) \Bigg(\frac{\partial u}{\partial x}\Bigg)^2\Bigg] dx \tag{14} $$

Using $(14)$, we can then continue, passing term containing $p_{0}$ to LHS,

$$ -\int_{0}^{L}p_{0}(ds+dx) + \int_{0}^{L}\kappa\frac{\partial u}{\partial x} (ds+dx) +\int_{0}^{L}\frac{p_{0}}{2}\Bigg(\frac{\partial u}{\partial x}\Bigg)^2dx =$$ $$ - \int_{0}^{L} \Bigg[ \Bigg(\kappa\frac{1}{2}\frac{\partial u}{\partial x}\Bigg) \Bigg(\frac{\partial u}{\partial x}\Bigg)^2\Bigg] dx \tag{15} $$

Now, since partial derivative $\partial u/\partial x$ can be consider an ordinary derivative, then it's inverse is $\partial x/\partial u$. We can use this fact to multiply both LHS and RHS by $\partial x/\partial u$ to eliminate $\partial x / \partial u$ :

$$ \frac{\partial x}{\partial u}\Bigg\{-\int_{0}^{L}p_{0}(ds+dx) + \int_{0}^{L}\kappa\frac{\partial u}{\partial x} (ds+dx) +\int_{0}^{L}\frac{p_{0}}{2}\Bigg(\frac{\partial u}{\partial x}\Bigg)^2dx\Bigg\} =$$ $$ \frac{\partial x}{\partial u}\Bigg\{- \int_{0}^{L} \Bigg[ \Bigg(\kappa\frac{1}{2}\frac{\partial u}{\partial x}\Bigg) \Bigg(\frac{\partial u}{\partial x}\Bigg)^2\Bigg] dx\Bigg\} \tag{16} $$

Then, finally, we can write potential energy as:

$$\phi \approx \int_{0}^{L} \kappa\frac{1}{2}\Bigg(\frac{\partial u}{\partial x}\Bigg)^2 dx \tag{17}$$


Using expression $(4)$, $(17)$ and calculations of section III), can I write down the expression for the energy of a wave in a gas column as:

$$\mathcal{E} = \frac{1}{2}\int_{0}^{L}dx\Bigg[\rho_{0} \Bigg(\frac{\partial u(x,t)}{\partial t}\Bigg)^2 + \kappa\Bigg(\frac{\partial u}{\partial x}\Bigg)^2\Bigg]=$$ $$= \frac{\rho_{0}}{2}\int_{0}^{L}dx\Bigg[\Bigg(\frac{\partial u(x,t)}{\partial t}\Bigg)^2 + \frac{\kappa}{\rho_{0}}\Bigg(\frac{\partial u}{\partial x}\Bigg)^2\Bigg] ?\tag{18}$$


Now, $[2]$ says that the velocity of the wave is precisely the coeficient in potential energy (using now the case of the rope):

$$v^{2} = \frac{T}{\rho} \implies$$

$$\implies \frac{\partial ^2 u}{\partial t^2} = \frac{T}{\rho}\frac{\partial ^2 u}{\partial x^2}$$

And my calculations up above returned the same form of the velocity given by $[1]$

$$v^{2} = \frac{\kappa}{\rho_{0}} \implies$$

$$\implies \frac{\partial ^2 u}{\partial t^2} = \frac{\kappa}{\rho_{0}}\frac{\partial ^2 u}{\partial x^2}$$


$[1]$ ALONSO.M; FINN.J.E; University Physics: Fields and Waves. v2. pages 683-685.

$[2]$ KING.C.G; Vibrations and Waves; pages 116-120.

  • $\begingroup$ Where does equation (10) come from? $\endgroup$ Apr 5, 2020 at 3:34

1 Answer 1


Your derivation ends up with a correct result, but I couldn't quite follow your derivation. A few remarks that may help you:

  • The forces acting on a small element of a continuuous 1D medium is $-P(x+dx)S$ on one side and $P(x)S$ on the other sides so the net force is $\frac{\partial P}{dx}S$. Of course you may drop the $S$ term since everyting will be expressed in terms of lineic energy densities (per meter) in the end.

  • The elementary work done by the force may be written as $\delta W=\mathbf{F}.\mathbf{dl}$ but beware that $\mathbf{dl} \neq \mathbf{dx}$: instead $\delta W=\mathbf{F}.\mathbf{du}$, with $\mathbf{du}$ the elementary displacement of your elementary volume$(S)dx$.

  • Finally, you may write $\delta W=\frac{\partial P}{dx}du= \kappa\frac{\partial u^{2}}{dx^{2}} \frac{\partial u}{dx}dx$ and so indeed, $\delta{W}=dU=d \left( \frac{\kappa}{2}(\frac{\partial u}{dx})^{2} \right)$


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