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I've always been a little uneasy with the notion of direction of wave propagation, for some reason. I guess it's always been defined 'intuitively' and I want to know the limits of the concept. To this end, what can we say about the direction of propagation of the wave

$$X = \sin(\frac{1}{x-t})$$

in the limit as $|x-t|\to 0$?

My reasoning is as follows. Pick an arbitrary value of $(x,t)$ and write $X_f=X(x,t)$. We want to find the coordinates in $(x,t)$ space of $X_f$ at time $t+\textrm{d}t$ when $\textrm{d}t>0$. Let's call these coordinates $(x',t')=(x',t+\textrm{d}t)$.

We hence require $\sin(\frac{1}{x'-(t+\textrm{d}t)})=\sin(\frac{1}{x-t})$ which gives $x-t=k_n(x'-t-\textrm{d}t)$ where $k = 1+n\pi$ some $n\in\mathbb{Z}$.

Writing $\textrm{d}x=x'-x$ we arrive at $\textrm{d}x=(k-1)(t-x)+k\mathbb{d}t$. Normally we would impose a continuity condition that $\textrm{d}x=\textrm{O}(\textrm{d}t)$ to conclude that $k=1$ and the waves move to the right.

But what happens if $|t-x|=\textrm{O}(\textrm{d}t)$? Then surely we can pick any $k$ we want to get the waves moving to right or left! Does the whole concept just break down here, and if so why? Is there any subtlety I have missed? And could such an example crop up physically?

Many thanks, and apologies if I have made any errors!

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3 Answers

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An important aspect of finding a proper solution to the wave equation is that it should work for any waveform, not just one specific choice. So you can't really specify $X(x,t) = \sin[1/(x-t)]$ and make an argument based on that. You have to use a general function $f(x-t)$. If this has the value $F_0$ at some point $(x_0, t_0)$ (and for this purpose "undefined" counts as a valid value), then other points $(x, t)$ at which the function has the same value will be those for which $x - t = x_0 - t_0$, and in general, only those. This defines a line in the $(x,t)$ plane, and the wave propagates along that line.

For certain specific functions $f$, you might be able to find other points not on this line for which the function has the same value - in other words, you might be able to find points $(x_2, t_2)$ where $f(x_2, t_2) = F_0$ but $x_2 - t_2 \neq x_0 - t_0$. Those other points form other lines in the $(x,t)$ plane, along which $x_2 - t_2$ has some value other than $x_0 - t_0$. These different lines don't cross each other, and the wave can't "jump" from one line to another, so the existence of those other lines doesn't really matter. It doesn't change the fact that the wave propagates along the lines where $x_2 - t_2$ stays fixed.

For example, consider your function $X(x,t) = \sin[1/(x-t)]$. Pick a point $(x_0,t_0)$, set $F_0 = \sin[1/(x_0 - t_0)]$, and the wave propagates along the line $x' - t' = x_0 - t_0$. But there are other points where $F_0 = X(x',t')$, some of which are given by

$$F_0 = \sin\biggl(\frac{1}{x_0 - t_0} + 2\pi n\biggr) = \sin\biggl(\frac{1}{x' - t'}\biggr)$$

for $n \in\mathbb{Z}$. If you set the arguments of the two sine functions equal, you find

$$x_0 - t_0 + 2\pi n(x_0 - t_0)(x' - t') = x' - t'$$

which is a little different from the expression you had. The different values of $n$ label the different lines in the $(x,t)$ plane that different points on the wave propagate along.

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I like this answer - it directly addresses what I was getting at. There is a hidden subtlety and you hit on it perfectly! Thanks! –  Edward Hughes Aug 19 '12 at 17:59
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I guess it's always been defined 'intuitively'

But it is not -- it follows from the wave equation. The lines $x\pm ct = const$ are two characteristics of the wave equation, which lead you to the d'Alembert's formula.

...propagation of the wave $$X = \sin(\frac{1}{x-t})$$

If you "blindly" substitute $f(x)=2\sin(1/x),\, g(x)=0,\, c = 1$ into d'Alembert's formula then you'll get this solution: $$u(x,t) = \sin(\frac{1}{x+t})+\sin(\frac{1}{x-t})$$ which are left- and right-propagating waves respectively.

The only thing that you might be worried about is the behaviour of the function at $x\pm t = 0$. But it is not the problem of the whole framework, but of that particular initial conditions.

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Your wave signal is ill defined: you can't mix x and t (x-v*t would be better), and it is unphysical to use a signal with a singularity since the energy needed would go to infinity.

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While I doubt that the OP is intentional using "natural" units, time and space have the same units when $c = 1$ as is preferred in particle physics and relativistics. –  dmckee Aug 16 '12 at 17:42
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I was in fact intentionally using natural units. My question was motivated by various examples in the Zwiebach book on String Theory. –  Edward Hughes Aug 16 '12 at 21:26
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