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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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