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Old Jun 3rd 2018, 05:03 AM   #1
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Relationship between amplitude and distance ?

What is the relationship between amplitude and distance (expressed in terms of proportionality) for a wave travelling from a point source outward:
i) as a 3D spherical wavefront
ii) as a 2D circular wavefront
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Old Jun 4th 2018, 03:31 AM   #2
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For 3D waves, an inverse square-law with distance is probably the most appropriate. Something like this:

$\displaystyle A(r) \propto \frac{1}{r^2}$.

Therefore,

$\displaystyle A(r) = \frac{A_0}{r^2}$

For 2D, the intensity is spread out over a circumference rather than a spherical shell, so the relationship probably becomes

$\displaystyle A(r) \propto \frac{1}{r}$.

$\displaystyle A(r) = \frac{A_0}{r}$



EDIT: this is wrong... see the next post

Last edited by benit13; Jun 4th 2018 at 06:22 AM. Reason: Error
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Old Jun 4th 2018, 03:42 AM   #3
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Have you got a reference for that?
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Old Jun 4th 2018, 05:51 AM   #4
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It is a purely geometrical issue.
the "same" amplitude is spread out over a larger region.
for a 2D wave it is the relationship between radius and circumference (of a circle)
for a 3D wave it is the relationship between radius and surface area (of a sphere).

You only asked for the proportionality, thus the "PI" portion of the equation is not required.
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Old Jun 4th 2018, 06:21 AM   #5
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Originally Posted by justme View Post
Have you got a reference for that?
Oooops... I made a mistake... it's the intensity that varies with the inverse square law, not the amplitude. The amplitude will vary as the square root of intensity, so

$\displaystyle A(r) \propto \frac{1}{r}$ for a spherical wavefront and

$\displaystyle A(r) \propto \frac{1}{r^{1/2}}$ for a circular wavefront.

Note: it's a routine problem in 1st year undergraduate texts to calculate, for example, that the extra-terrestrial radiation at the Earth's surface is ~1400 W/m2. E.g.

Solar luminosity = $\displaystyle 3.828 \times 10^{26}$ W
Earth-Sun distance = $\displaystyle 1.5 \times 10^{11}$ m

Solar luminosity is spread out over a sphere with radius equal to the Earth-Sun distance, so

$\displaystyle L_{Earth} = \frac{L_{\odot}}{A} = \frac{L_{\odot}}{4\pi r^2} = \frac{3.828 \times 10^{26}}{4\pi \times (1.5 \times 10^{11})^2} = 1354 $W/m2

Most textbooks on waves explain the behaviour.
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Old Jun 4th 2018, 08:48 AM   #6
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Originally Posted by justme View Post
What is the relationship between amplitude and distance (expressed in terms of proportionality) for a wave travelling from a point source outward:
i) as a 3D spherical wavefront
ii) as a 2D circular wavefront
Why don't you ask the same question for the E=hf photon.
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Old Jun 4th 2018, 08:54 AM   #7
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Originally Posted by Farsight View Post
Why don't you ask the same question for the E=hf photon.
Perhaps the OP is investigating something else?
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Old Sep 21st 2018, 03:52 AM   #8
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The energy spreads out in a spherical shell, the energy density decreases as the square of the distance from the source. That's your amplitude decrease. However, the frequency and wavelength stay the same as long as it keeps traveling at the speed of light.
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