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Why is the intensity proportional to the square of the amplitude What
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Why is the intensity proportional to the square of the amplitude?
What is the concept behind this? Are we squaring the amplitude to get an area or something analogous to an area?
After the concept is explained, how does this relate to probably amplitudes (wave functions) in quantum physics?
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Expert:
Fred
replied 7 years ago.
Hello,
When we say that intensity is proportional to the square of the amplitude, well we are talking about RMS (rootmean square) amplitude. It is defined as the square root of the mean over time of the vertical distance (squared also) from rest state of the graph.
In other words, it is easier to use RMS than ''peak'' amplitude to calculate an intensity, because in real life, graphs do not always show a clean sinusoidal waveform. In these cases, it gets much easier to use RMS.
As an example, if you take sound waves, well these are waves that travel in the air and create a movement that is usually described as dB (decibels). This movement or displacement of the air is based on a logarithmic scale which make the intensity of the wave in the air proportional to the square of the amplitude.
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Customer:
replied 7 years ago.
So the intensity is like the "area" underneath the curve formed by the peak amplitude of the wave?
How does this relate to the squaring of a wave function being equal to the probability of finding a particle at a certain position at a certain time?
Expert:
Fred
replied 7 years ago.
Well, RMS amplitude is not exactly the area under the graph wave, but it is like taking each point amplitude, squaring it and take the average if all points. So, if you take perfect sinusoidal wave, well, your result will be ''0'' since it goes up and down symetrical. TaXXXXX XXXXXke that, you can kind of say that it representes the area under the curve/wave.
Well, this is a postulate. I will try to explain it.
To determine where a particle is in a defined ''box'', you have to know it's ''state''. If we take normal modes, such as the classical waveform, then we can say that the probability (P) or finding a particle is :
P(x) = (A(x))^2 A : complex amplitude
Now, in quantum mechanic, the complex amplitude is also called the wave function. If you replace it you get :
P(x) = (Wave function)^2
Hope it was clear enough, but it is not always easy to summarize these postulates.
Waiting for some feedback,
Fred
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Customer:
replied 7 years ago.
Well, RMS amplitude is not exactly the area under the graph wave, but it is like taking each point amplitude, squaring it and take the average if all points. So, if you take perfect sinusoidal wave, well, your result will be ''0'' since it goes up and down symetrical. TaXXXXX XXXXXke that, you can kind of say that it representes the area under the curve/wave.
For the paragraph above...at what point do you take the square root?
Can you define "box" and "state" a little differently?
Expert:
Fred
replied 7 years ago.
Okay, the exact way to do it is to square the amplitude at the defined point, and then calculate the average of all the squared amplitudes.
You do not use square root, the final answer is the average of all squared amplitudes. RMS is very different from spiked amplitude, it has to be specified, but let's be honest, in real life, symetric and perfect sinusodial waveform do not exist (or almost). Therefor, almost only RMS is used to calculate intensity.
Box : it was just to define a coordinate system, or to replace the at a certain moment and place. Just use it as "environment"; or "space".
State : this is how the particle is initially placed in the "defined space" or "box". Or in generic terms, it's the initial state of the object (not taken as chemical state, but geometric (position and time) state).
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Customer:
replied 7 years ago.
Okay, last reply from me...let's go back to the beginning sort of
I get that you average the squared amplitudes...By why square the amplitude..what is the concept, the point of squaring the amplitude to find intensity?
Since the definition of intensity is energy per unit area and amplitude is the amount of energy at a point, then amplitude squared is the energy within that area?
In relationship to and electron,in the physical sense, [wavefunction] gives the amplitude of the wave associated with the electron. Is this amplitude a type of radius from the nucleus? And does [wavefunction]2 determines the probability of finding the moving electron in a given region or it gives the probability density. what is the given region?
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Fred
replied 7 years ago.
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