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On page 5 of Bondi et. al. (1962) (https://doi.org/10.1098/rspa.1962.0161), a suitable method for solving the gravitational wave equation is demonstrated for the case of the scalar wave equation. Starting with

$$\partial^2 Q/\partial t^2 = \nabla^2Q, \tag{1}\label{1}$$

Bondi et. al. “separate the part $Q_n$ proportional to the surface harmonic $S_n$ and introduce a null variable $u$ by the relation $u=t-r$.” Thus, they claim that $\eqref{1}$ change to

$$2\left(\frac{\partial^2Q_n}{\partial r\partial u} + \frac 1r\frac{\partial Q_n}{\partial u}\right) = \frac{\partial^2 Q_n}{\partial r^2} + \frac 2r\frac{\partial Q_n}{\partial r} - \frac{n(n+1)}{r^2}Q_n.\tag{2}\label{2}$$

The RHS of $\eqref{2}$ is just the radial Laplacian of $Q_n$, which is assumed to be proportional to $S_n$ and thus independent of the angle coordinates $\theta$ and $\phi$.

My question is how the LHS of $\eqref{2}$ can match the LHS of $\eqref{1}$. I have that $\partial_t =\partial_u+\partial_r$ from writing the explicit Jacobian and confirming with a different algebraic approach shown in the addendum. Thus, $$\partial_t^2 Q_n = \left(\partial_u + \partial_r\right)^2 Q_n = \left(\partial_u^2 +2\partial_r\partial_u + \partial_r^2\right)Q_n,\tag{3}\label{3}$$ so it would suffice to show that $(\partial_u^2 + \partial_r^2)Q_n = 2r^{-1}\partial_u Q_n$. I tried rewriting $\partial_u^2$ back in terms of $\partial_t$ (with $\partial_u = \partial_t -\partial_r$) and eliminating $\partial_t^2$ with the RHS of $\eqref{2}$. However, on the RHS of $\eqref{2}$, the differential operator with coefficient $2r^{-1}$ is not $\partial_u$ but $\partial_r = \partial_t - \partial_u$ like on the LHS, and, moreover, I can’t see how I the rightmost term of the RHS of $\eqref{2}$ would be canceled.

Is this simply a matter of changing variables or am I missing additional assumptions?

Addendum:

Writing the Jacobian to transform the differentials,

$$\begin{pmatrix} \partial_u \\ \partial_r\end{pmatrix} = \begin{pmatrix} t_u & r_u \\ t_r & r_r \end{pmatrix} \begin{pmatrix} \partial_t \\ \partial_r \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \partial_t \\ \partial_r \end{pmatrix},$$

it follows that $\partial_u = \partial_t - \partial_r$ and $\partial_t = \partial_u + \partial_r$. (Alternatively, we can calculate $\partial_u = \partial_u|_r + \partial_u|_t = \partial_t|_r - \partial_r|_t = \partial_t|_r - \partial_r|_t + \left( \partial_t|_u -\partial_r|_u\right) = \partial_t-\partial_r$ because $\partial_t|_u = \partial_t|_r + \partial_r|_t = \partial_r|_u$.)

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  • $\begingroup$ Please don't use images in place of text and mathematical equations. $\endgroup$
    – Frobenius
    Aug 4, 2022 at 20:46
  • $\begingroup$ @Frobenius I’ve changed the image to text and mathematical equations. $\endgroup$
    – Rodrigo
    Aug 5, 2022 at 9:40
  • $\begingroup$ Well done Rodrigo !!! $\endgroup$
    – Frobenius
    Aug 5, 2022 at 11:35

2 Answers 2

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You are not transforming the differentials correctly. $\partial/\partial t|_r=\partial/\partial u|_r$ and $\partial/\partial r|_t=\partial/\partial r|_u-\partial/\partial u|_r$.

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  • $\begingroup$ I showed how I transformed the differentials. Could you elaborate on where my mistake would have been? $\endgroup$
    – Rodrigo
    Aug 4, 2022 at 22:29
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    $\begingroup$ How is $\partial r/\partial u)_r=-1$? You need to show what is being held constant in the partial differential $\endgroup$
    – CWPP
    Aug 4, 2022 at 23:49
  • $\begingroup$ Because $r = r(t,u) = t - u$, it follows that $\partial r/\partial u = -1$. At least, this is how I usually see it in the mathematical literature to which I’m used, where I rarely see the explicit notation of variables that are held constant. Could you show how what you obtain with your method differs? $\endgroup$
    – Rodrigo
    Aug 4, 2022 at 23:59
  • $\begingroup$ The functions in the Jacobian are functions of the variables $r$ and $u$ . The suffix notation I am using shows the variable being held constant. It's confusing in this case that one of the new variables has the same name as one of the old ones $\endgroup$
    – CWPP
    Aug 5, 2022 at 4:33
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    $\begingroup$ It might be clearer to define the transformation as $s=r$ and $u=t-r$. Then the Jacobian is $\partial(t,r)/\partial(u,s)$. $\endgroup$
    – CWPP
    Aug 5, 2022 at 4:58
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The impeccable answer of @CWPP identifies the problem and leads to the straightforward correct answer; I'm only writing this as a footnote with an explicit expansion of his point to avoid the logical minefield that the OP has walked into.

(1) has devolved to functions independent of the angular variables, so, then, $$ \left ( \partial_t^2 - \partial_r^2 -{2\over r} \partial_r + {n(n+1)\over r^2}\right ) Q_n =0 . \tag{1'} $$

Variables are changed to $(r,t)\mapsto (s=r,u=t-r)$, hence $$ \partial_r=\partial_rs~ \partial_s+ \partial_r u ~ \partial_u= \partial_s -\partial_u, \\ \partial_t= \partial_t s ~ \partial_s + \partial_t u~ \partial_u= \partial_u , $$ whence $$0=\left ( \partial_u^2 - (\partial_s-\partial_u)^2 -{2\over s} (\partial_s -\partial_u) + {n(n+1)\over s^2}\right ) Q_n \\ = \left ( 2 \partial_u \partial_s + {2\over s}\partial_u - \partial_s ^2 -{2\over s} \partial_s + {n(n+1)\over s^2}\right ) Q_n \tag{2} . $$ At the very end, r supplants s.

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