I am currently studying Optics, fifth edition, by Hecht. In chapter 2.9 Spherical Waves, the author says the following:

$$\dfrac{\partial^2}{\partial{r}^2}(r \psi) = \dfrac{1}{v^2} \dfrac{\partial^2}{\partial{t}^2} (r \psi) \tag{2.71}$$ Notice that this expression is now just the one-dimensional differential wave equation, Eq. (2.11), where the space variable is $r$ and the wavefunction is the product $(r \psi)$. The solution of Eq. (2.71) is then simply $$r \psi(r, t) = f(r - vt)$$ or $$\psi(r, t) = \dfrac{f(r - vt)}{r} \tag{2.72}$$ This represents a spherical wave progressing radially outward from the origin, at a constant speed $v$, and having an arbitrary functional form $f$. Another solution is given by $$\psi(r, t) = \dfrac{g(r + vt)}{r}$$ and in this case the wave is converging toward the origin. The fact that this expression blows up at $r = 0$ is of little practical concern. A special case of the general solution $$\psi(r, t) = C_1\dfrac{f(r - vt)}{r} + C_2 \dfrac{g(r + vt)}{r} \tag{2.73}$$ is the harmonic spherical wave $$\psi(r, t) = \left( \dfrac{\mathcal{A}}{r} \right) \cos k(r \mp vt) \tag{2.74}$$ or $$\psi(r, t) = \left( \dfrac{\mathcal{A}}{r} \right) e^{ik(r \mp vt)} \tag{2.75}$$ wherein the constant $\mathcal{A}$ is called the source strength. At any fixed value of time, this represents a cluster of concentric spheres filling all space. Each wavefront, or surface of constant phase, is given by $$kr = \text{constant}$$ Notice that the amplitude of any spherical wave is a function of $r$, ware the term $r^{-1}$ serves as an attenuation factor. Unlike the plane wave, a spherical wave decreases in amplitude, thereby changing its profile, as it expands and moves out from the origin. Figure 2.27 illustrates this graphically by showing a "multiple exposure" of a spherical pulse at four different times. The pulse has the same extent in space at any point along any radius $r$; that is, the width of the pulse along the $r$-axis is a constant. enter image description here

I don't understand this part:

The pulse has the same extent in space at any point along any radius $r$; that is, the width of the pulse along the $r$-axis is a constant.

I don't understand what is meant by "the pulse has the same extent in space at any point along any radius $r$". Is the author claiming that the pulse at $r = t_1$ has the same width as the pulse at $r = t_4$? That doesn't look to be true to me.

I would greatly appreciate it if people would please take the time to explain this.


2 Answers 2


What the quote is saying is that the amplitude of the pulse decreases, but the $r$ coordinate interval that it occupies stays the same. The picture might not be too clear but while the height of the pulse goes down, its width is indeed the same at all times.

You can see this by taking a function $f(x)$ that is only nonzero for $a < x < b$. If we use it to construct a spherical wave, we would get $\psi(r,t) = f(r-vt)/r$. This is only nonzero when the argument of $f$ is between $a$ and $b$; that is, when

$$a + vt < r < b + vt,$$

and the difference between the maximum and minimum values of $r$ is always $b-a$, independently of $t$.


"the same extent" means (See the textbook 2.9 Spherical Waves and Figure 2.28:

the resulting wavefronts of a spherical wave are concentric spheres that increase in diameter as they expand out into the surrounding space.

i.e., the wavefront surface of a constant phase $\mathbf k \cdot \mathbf r$ (=kr in this case) is a sphere and the amplitude of $\psi(\mathbf r)=\psi (r,\theta,\phi) = \psi (r)$ is independent of $\theta,\phi$. Eq. $(2.68)$.

"the width of the pulse along the r-axis is a constant":

The pulse width is one $\lambda = \frac {2 \pi}{k}$ which takes one time period $\mathbf T (= \frac {1}{f}=\frac{\lambda}{v} $) to travel from t1 position to t3 position and then propagating from t2 to t4.


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