- #1
H Smith 94
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I am currently studying a course on waves, which has a real ambiguity in the lecture notes. Essentially, I don't know how the professor got from equation \ref{eq:surf_x-y} to equations \ref{eq:vel_u} and \ref{eq:vel_w}. I have tried to work backwards to find a method but am not sure of its validity.
Please note that this is not a homework exercise, so full answers are appreciated.
Given formulas
The step is from the surface height of a 1-dimensional wave, given by \begin{equation}\label{eq:surf_x-y}\eta(x,t) = A \cos(kx - \omega t)\end{equation} to a velocity field $$\mathbf{v}(x,z,t) = (u(x,z,t),w(x,z,t)),$$ where \begin{equation}\label{eq:vel_u}u(x,z,t) = A \omega \frac{\cosh(k(H+z))}{\sinh(k H)}\, \cos(kx - \omega t)\end{equation} and \begin{equation}\label{eq:vel_w}w(x,z,t) = A \omega \frac{\sinh(k(H+z))}{\sinh(k H)}\, \sin(kx - \omega t).\end{equation}
My workings
Given this as true, one finds a position function \begin{equation}\label{eq:pos_r}\phi(x,z,t) = \frac{A \omega}{k} \frac{\cosh(k(H+z))}{\sinh(k H)}\, \sin(kx - \omega t)\end{equation} from the integration $$\phi(x,z,t) = \int \mathbf{v}(x,z,t) \cdot\mathrm{d}\mathbf{r},$$ from which follows the assumption that we have multiplied equation \ref{eq:surf_x-y} by a ##z##-dependent factor $$f(z) = \frac{\omega}{k}\,\frac{\cosh(k(H+z))}{\sinh(k H)}.$$
'Fudged method'
Now, the only way I can see of getting from \ref{eq:pos_phi} to ##\phi(x,z,t)=\eta(x,t)\,f(z)## is using the following method, which finds that:
\begin{equation}\label{eq:vel_vec}\mathbf{v}(x,z,t) = \frac{1}{k}\,\frac{\partial}{\partial t}\,\nabla \phi(x,z,t) = \left(\begin{array}{c}\frac{\partial}{\partial t}\frac{\partial \phi}{\partial x} \\ \frac{\partial}{\partial t}\frac{\partial \phi}{\partial z}\end{array}\right).\end{equation}
Attempt at physical justification
I'm trying to justify this in physical terms by using the relation for phase speed ##c##, which states that ##c = \frac{\omega}{k}##, meaning we can infer from equation \ref{eq:vel_vec} that if $$\mathbf{v}(x,z,t) = \frac{\mathbf{\omega}}{k}$$ and so \begin{equation}\mathbf{\omega} = \frac{\partial}{\partial t}\,\nabla \phi(x,z,t).\end{equation}
I am taking this to mean we have a scalar function ##\phi(x,z,t)##, whose gradient gives the position ##\mathbf{r}(x,y,t)=(r_x(t),r_y(t))## of particles in the wave, meaning its rate of change would give us the velocity in the respective directions. If this is true, what is the physical meaning of ##\phi##?
Please note that this is not a homework exercise, so full answers are appreciated.
Given formulas
The step is from the surface height of a 1-dimensional wave, given by \begin{equation}\label{eq:surf_x-y}\eta(x,t) = A \cos(kx - \omega t)\end{equation} to a velocity field $$\mathbf{v}(x,z,t) = (u(x,z,t),w(x,z,t)),$$ where \begin{equation}\label{eq:vel_u}u(x,z,t) = A \omega \frac{\cosh(k(H+z))}{\sinh(k H)}\, \cos(kx - \omega t)\end{equation} and \begin{equation}\label{eq:vel_w}w(x,z,t) = A \omega \frac{\sinh(k(H+z))}{\sinh(k H)}\, \sin(kx - \omega t).\end{equation}
My workings
Given this as true, one finds a position function \begin{equation}\label{eq:pos_r}\phi(x,z,t) = \frac{A \omega}{k} \frac{\cosh(k(H+z))}{\sinh(k H)}\, \sin(kx - \omega t)\end{equation} from the integration $$\phi(x,z,t) = \int \mathbf{v}(x,z,t) \cdot\mathrm{d}\mathbf{r},$$ from which follows the assumption that we have multiplied equation \ref{eq:surf_x-y} by a ##z##-dependent factor $$f(z) = \frac{\omega}{k}\,\frac{\cosh(k(H+z))}{\sinh(k H)}.$$
'Fudged method'
Now, the only way I can see of getting from \ref{eq:pos_phi} to ##\phi(x,z,t)=\eta(x,t)\,f(z)## is using the following method, which finds that:
\begin{equation}\label{eq:vel_vec}\mathbf{v}(x,z,t) = \frac{1}{k}\,\frac{\partial}{\partial t}\,\nabla \phi(x,z,t) = \left(\begin{array}{c}\frac{\partial}{\partial t}\frac{\partial \phi}{\partial x} \\ \frac{\partial}{\partial t}\frac{\partial \phi}{\partial z}\end{array}\right).\end{equation}
Attempt at physical justification
I'm trying to justify this in physical terms by using the relation for phase speed ##c##, which states that ##c = \frac{\omega}{k}##, meaning we can infer from equation \ref{eq:vel_vec} that if $$\mathbf{v}(x,z,t) = \frac{\mathbf{\omega}}{k}$$ and so \begin{equation}\mathbf{\omega} = \frac{\partial}{\partial t}\,\nabla \phi(x,z,t).\end{equation}
I am taking this to mean we have a scalar function ##\phi(x,z,t)##, whose gradient gives the position ##\mathbf{r}(x,y,t)=(r_x(t),r_y(t))## of particles in the wave, meaning its rate of change would give us the velocity in the respective directions. If this is true, what is the physical meaning of ##\phi##?