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Re: [Phys-L] gravitational waves



Why is the word "intensity, given by me to the quantity g? Because intensities of waves (radio, sound,light, etc.) are usually expressed in (W/m^2), while the unit of g is (m/s^2).

Ludwik
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On Apr 10, 2016, at 10:59 AM, Ludwik Kowalski wrote:

P.S. The term "intensity," given by me to the g(x,t) quantity, is probably not appropriate. Would the name "strength" be OK? If not then what should I use?

Ludwik
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On Apr 10, 2016, at 10:34 AM, Ludwik Kowalski wrote:

Thank you for the comment, Bernard. Let me be quantitative.

1) I am treating the source disk (the source of the field at x=0) as it were a point particle of mass M. The detector is at the distance x from the source. Therefore the field intensity, g, at any given distance from the source‚ is given by Newton's law:

g=G*M/x^2, where G is the universal gravitational constant.

2) The distance x is time dependent, due to some unspecified process. That is why g is also time dependent.

3) Te frequency of longitudinal gravitational waves (in my gedanken setup) is the same as the frequency at which the x is changing.

4) What is wrong with this classical reasoning?

Ludwik Kowalski



On Apr 9, 2016, at 6:33 PM, Bernard Cleyet wrote:


On 2016, Apr 08, , at 23:39, John Denker <jsd@av8n.com> wrote:

On 04/08/2016 08:35 PM, Ludwik Kowalski wrote:

The distance x, (between the two disk-like pistons in my model),
changes because the the source disk is oscillating. The gravitational
wave intensity, the amplitude of g(t), at any given x, is inversely
proportional to the x^2,

Suppose the source disk oscillates through a distance k.
I claim the amplitude of g-field oscillations scales like
k/x^3 (not 1/x^2). You can obtain that result via dimensional
analysis, or via a Taylor series : k (d/dx) (1/x^2).


This reminds me of the counter intuitive energy loss of an oscillator with quadratic dissipation for each cycle is proportional to the cube of the initial amplitude for each cycle.

This assumes a rather high Q and a short period oscillator.

bc
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