Shifting LIGO to the binary. Thought experiment.

[Robittybob1]

Seriously?

 

Are you not aware that the amount of energy carried by a wave is related to the amplitude of the wave.

 

Those two formulas directly relate to the quoted posts

I am not aware that the amount of energy carried by the wave is related to the amplitude.

Are there any other formulas or information on the amplitude of the GW?

[Mordred]

Lets give you a hint. There is little difference between power radiated from an antenna than a GW.

 

At least not in principle one is dipole and the other is quadrupole.

 

The latter also doesn't have charge.

 

Yet many of the formulas of basic physics apply. This includes the terminology. If you didn't recognize that the energy of a wave is carried by the amplitude I would suggest googling basic electromagnetic terminology.

 

Look at the definition and formulas relating frequency, cycle, wavelength, amplitude and polarity.

 

This will help you understand the more complex aspects in GW waves.

 

Keep in mind per individual wave the amplitude is the energy displacement.

 

However a higher frequency via a shorter wavelength will increase the amount of energy per timelength. While the energy/amplitude is per individual wave. Essentially in a given time you receive more waves

Edited March 22, 2016 by Mordred

[Robittybob1]

This site gives me some math to work on. http://www.tapir.caltech.edu/~teviet/Waves/gwave.html "Gravitational waves"

 

And it talks about what you are saying in #27 Mordred Thanks.

 

The GW has an amplitude which is related to the dimensions of the system (first animation).

The amplitude of the LIGO recording is not obvious (to me) how it relates to the amplitude of the GW. They call it "Dimensionless amplitude".

 

 

 

Dimensionless amplitude

The tidal field g' is the physically measurable part of gravitational phenomena: it represents an observable relative acceleration or force between two displaced "test masses". However, when discussing gravitational waves, the most common parameter describing the amplitude is a dimensionless "strain" h = 2∫∫ g' dt ².
What does this quantity mean? Remember that g' is a gravity gradient, so g'd gives the difference in gravity, i.e. the differential acceleration, between two objects separated by a small displacement d. Two time integrals of acceleration give us the instantaneous change in this displacement as a function of time. Thus h is twice the fractional change in displacement between two nearby masses due to the gravitational wave. This change in displacement occurs in the plane transverse to the direction of radiation, and causes a stretch along one axis and a squeeze along the orthogonal axis: this is illustrated below, showing how a ring of freely-floating masses would be disturbed by a passing gravitational wave. The net distortion is twice as much as a stretching or squeezing alone, which is the reason for the factor of 2 in the equations for h.

What is your way of connecting amplitude in the animation to the Dimensionless amplitude?

In the above article they use a similar approach to the problem as I am doing here:

The first term is roughly the size of a black hole of mass M, so the distance r to the system must clearly be much greater. Similarly, v/c is the ratio of the speeds of masses in the system to the speed of light, which must be less than (usually much less than) unity. Thus h approaches unity when one is standing in the immediate vicinity of black holes moving about at lightspeed, and is less for any other circumstance.

 

So it is a valid approach to move the LIGO to the same spot as they talk about "in the immediate vicinity of black holes".

.

[imatfaal]

Rob - please please do a little background reading before trying to jump in the deep end. It is getting embarrassing that you are posing novel though experiments without any knowledge of the basics.

 

A strain in physics is often a dimensionless quantity which measures the relative change in shape or size due to an external influence. The fact that is is relative ie [change in length] divided by [original length] means that it has no units eg 1 nanometre / 1 metre = 10e-9 ; as there are no units to this number this is referred to as dimensionless. Most school-children come across strain in first year physics in terms of Young's Modulus - but it is roughly the same thing when dealing with the change in size of a light path both absent gravitational radiation and during gravitational radiation

[Robittybob1]

Rob - please please do a little background reading before trying to jump in the deep end. It is getting embarrassing that you are posing novel though experiments without any knowledge of the basics.

 

A strain in physics is often a dimensionless quantity which measures the relative change in shape or size due to an external influence. The fact that is is relative ie [change in length] divided by [original length] means that it has no units eg 1 nanometre / 1 metre = 10e-9 ; as there are no units to this number this is referred to as dimensionless. Most school-children come across strain in first year physics in terms of Young's Modulus - but it is roughly the same thing when dealing with the change in size of a light path both absent gravitational radiation and during gravitational radiation

I had already understood that but what is the connection between the amplitude of the strain and the amplitude of the G-Wave produced by the binary? I have no problem with the frequency bit, so the wavelength will be speed of light divided by frequency, but as the binary masses infall their separation is getting less so the amplitude, if based on dimensions of the binary, should be getting smaller yet the chirp amplitude is getting larger. I'll read up on this a bit more, as you say.

[Strange]

I had already understood that but what is the connection between the amplitude of the strain and the amplitude of the G-Wave produced by the binary?

 

The strain is the measured amplitude of the gravitational waves.

 

I have no problem with the frequency bit, so the wavelength will be speed of light divided by frequency, but as the binary masses infall their separation is getting less so the amplitude, if based on dimensions of the binary, should be getting smaller yet the chirp amplitude is getting larger.

 

The amplitude is related to the energy being radiated, which is related to the distance, masses, orbital speeds, and angular momentum of the pair.

[Mordred]

Or in the case of strain the amount of displacement.

[Robittybob1]

Lets give you a hint. There is little difference between power radiated from an antenna than a GW.

 

At least not in principle one is dipole and the other is quadrupole.

 

The latter also doesn't have charge.

 

Yet many of the formulas of basic physics apply. This includes the terminology. If you didn't recognize that the energy of a wave is carried by the amplitude I would suggest googling basic electromagnetic terminology.

 

Look at the definition and formulas relating frequency, cycle, wavelength, amplitude and polarity.

 

This will help you understand the more complex aspects in GW waves.

 

Keep in mind per individual wave the amplitude is the energy displacement.

 

However a higher frequency via a shorter wavelength will increase the amount of energy per timelength. While the energy/amplitude is per individual wave. Essentially in a given time you receive more waves

 

 

The strain is the measured amplitude of the gravitational waves.

 

 

The amplitude is related to the energy being radiated, which is related to the distance, masses, orbital speeds, and angular momentum of the pair.

Well I have been thinking, and not yet done any reading. The strain (dimensionless) is the measured amplitude of the GWs. That still makes me query the concept sorry.

The amplitude of the strain "is related to the energy being radiated, which is related to the distance, masses, orbital speeds, and angular momentum of the pair". That seems to make sense.

Or in the case of strain the amount of displacement.

Displacement of the Mirrors/masses in the LIGO, is that what you mean? Otherwise displacement of what? I'm about to look up Google scholar to see if someone has discussed this.

[Mordred]

Coordinate displacement of each particle.

 

Take a rigid rod, apply force at one end but not the other you produce strain.

 

Were dealing with coordinate change of spacetime each particle being essentially attached to a specific coordinate.

Edited March 23, 2016 by Mordred

[Robittybob1]

Where do the rates of "fall off" come from? What does purely transverse mean?
Quote http://www.tapir.caltech.edu/~teviet/Waves/gwave.htm

By analogy with electromagnetic dipole radiation, we can say the following things about gravitational waves:
Whereas static fields have both radial and transverse components, the radiative fields are purely transverse.
Whereas static fields fall off as 1/r³, the radiative fields fall off only as 1/r, and soon completely dominate over the static fields.

 

 

Does gravity ever fall off at [latex]\frac{1}{r^3}[/latex] ?

.

Edited March 23, 2016 by Robittybob1

[Strange]

Does gravity ever fall off at [latex]\frac{1}{r^3}[/latex] ?

 

Tidal forces do.

[imatfaal]

Coordinate displacement of each particle.

 

Take a rigid rod, apply force at one end but not the other you produce strain.

 

Were dealing with coordinate change of spacetime each particle being essentially attached to a specific coordinate.

 

I would have to disagree with your definition of strain here Mordred. Strain in these terms is the relative change of a measured quantity of length due to an external influence - your definition "force at one end but not the other you produce strain" is closer to the notion of material stress (ie the internal force caused by an external strain). A rigid rod - assuming you meant ideally rigid - suffers no strain as it does not change dimension.

 

In LIGO the strain is the change of the measured distance divided by the expected distance (ie dimensionless). The results are given as strain x10e-21 which seems about right for a displacement of ~10e-18m over distance of ~10e3m

[Mordred]

Yeah I agree it was a poor example of amplitude displacement. I didn't even consider "ideally rigid"

 

But stress and strain are two different formulas.

 

[latex]stress=\frac{f}{a_{cross,section}}[/latex]

 

[latex]strain=\frac{\Delta L}{L}[/latex]

 

My example was still poorly worded though lol

Edited March 23, 2016 by Mordred

[Robittybob1]

 

I would have to disagree with your definition of strain here Mordred. Strain in these terms is the relative change of a measured quantity of length due to an external influence - your definition "force at one end but not the other you produce strain" is closer to the notion of material stress (ie the internal force caused by an external strain). A rigid rod - assuming you meant ideally rigid - suffers no strain as it does not change dimension.

 

In LIGO the strain is the change of the measured distance divided by the expected distance (ie dimensionless). The results are given as strain x10e-21 which seems about right for a displacement of ~10e-18m over distance of ~10e3m

In the thought experiment we would be using interference of lasers light to measure the difference in length of the two arms at right angles to each other. We would expect so much greater movement closer to the binary so we will have to count the whole wavelength changes as well as the fractional ones.

[Robittybob1]

 

You could use Newtonian gravity to answer that (incorrectly). In which case, they would be in a stable orbit like any other pair of masses.

 

But, of course, they would not be creating gravitational waves, which is why the orbit is stable.

And LIGO would detect nothing because there would be no gravitational waves.

A dumbbell shaped mass spinning end for end will also create gravitational waves, you could keep the masses the same distance apart, so if that was spinning in space without air resistance, I presume it would slow down its rotation as it radiated gravitational energy. Would LIGO placed at the point of rotation be able to detect this sort of situation even if the forces were calculated using Newton's gravitational force equation? I think it would for the test masses would be pulled outward and then relaxed in a quadrupolar like fashion.

So when arm1 was inline the test masses will be recording maximum pressure on the ends, whereas in arm2 the test masses would not be pushing on the ends for they are orthogonal to the radius in this phase This pressure would result in a separation of the masses and this can be recorded as a positive shift in the interference pattern inside the interferometer.

 

https://en.wikipedia.org/wiki/Gravitational_wave

 

 

In general terms, gravitational waves are radiated by objects whose motion involves acceleration, provided that the motion is not perfectly spherically symmetric (like an expanding or contracting sphere) or rotationally symmetric (like a spinning disk or sphere). A simple example of this principle is a spinning dumbbell. If the dumbbell spins around its axis of symmetry, it will not radiate gravitational waves; if it tumbles end over end, as in the case of two planets orbiting each other, it will radiate gravitational waves. The heavier the dumbbell, and the faster it tumbles, the greater is the gravitational radiation it will give off. In an extreme case, such as when the two weights of the dumbbell are massive stars like neutron stars or black holes, orbiting each other quickly, then significant amounts of gravitational radiation would be given off.

 

So when arm2 was inline the test masses will be recording maximum pressure on the ends, whereas in arm1 the test masses would not be pushing on the ends for they are orthogonal to the radius in this phase This pressure would result in a separation of the masses and this can be recorded as a negative shift in the interference pattern inside the interferometer.

 

This pattern seems to be producing 2 positive peaks per orbit and 2 negative peaks per orbit, which is similar to gravitational waves of two waves per orbit, 2 positive crests and 2 negative crests.

Edited March 25, 2016 by Robittybob1