Jump to content

Recommended Posts

Posted

Wouldn't that be the phase variation? [latex]e^{-i \omega t}[/latex] is used to generate a sinusoidal evolution in time - an addition to the exponent [latex]e^{-i (\omega t +k)}[/latex]phase shifts the wave whilst leaving the frequency/wavelength untouched (they depend on t and omega).

 

Have you tried reading the original Papadopoulos and Esposito paper referred to ?

Posted

Wouldn't that be the phase variation? [latex]e^{-i \omega t}[/latex] is used to generate a sinusoidal evolution in time - an addition to the exponent [latex]e^{-i (\omega t +k)}[/latex]phase shifts the wave whilst leaving the frequency/wavelength untouched (they depend on t and omega).

 

Have you tried reading the original Papadopoulos and Esposito paper referred to ?

 

This is the original paper:

 

http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1981ApJ...248..783P&data_type=PDF_HIGH&whole_paper=YES&type=PRINTER&filetype=.pdf

Posted

As I thought - alpha is just the phase. It is the common usage of wave equation in complex form

 

if W is a plane wave moving in space x and evolving in time t with

A as peak amplitude

k as the wave number

little omega as the angular frequency

and

alpha as the initial phase

you get

 

[latex]W(x,t) = A \left(Cos(kx-\omega t + \alpha)\right) [/latex]

 

You can rewrite this in complex form as

 

[latex]Z(x,t) = A e^{-i(kx-\omega t +\alpha)}[/latex]

 

As the wave moves throught space and time you get a sinusoidal variation - but you need to know what the initial phase angle was; and you can note that if you set x (distance) and t (time) to zero you get just alpha ie the initial phase angle. Remember that adding a phase angle seems to push the wave form back to the left on a plot

  • 2 weeks later...
Posted (edited)

 

As I thought - alpha is just the phase. It is the common usage of wave equation in complex form

 

if W is a plane wave moving in space x and evolving in time t with

A as peak amplitude

k as the wave number

little omega as the angular frequency

and

alpha as the initial phase

you get

 

[latex]W(x,t) = A \left(Cos(kx-\omega t + \alpha)\right) [/latex]

 

You can rewrite this in complex form as

 

[latex]Z(x,t) = A e^{-i(kx-\omega t +\alpha)}[/latex]

 

As the wave moves throught space and time you get a sinusoidal variation - but you need to know what the initial phase angle was; and you can note that if you set x (distance) and t (time) to zero you get just alpha ie the initial phase angle. Remember that adding a phase angle seems to push the wave form back to the left on a plot

Thank You. But for calculating alpha= cos inverse w(0)/A (As you pointed out) I would require amplitude in distance. Now since I am trying to calculate in hypothetical situation How can I calculate amplitude if I know mass and speed of object generating it ?

For the original equation 2b I had calculated alpha in Decibels since it was dimensionless.

Edited by pennieloot
Posted

I think you need to spend a little time learning some basics before you run with gravitational waves - unfortunately, I don't have the academic knowledge to help you here. I cannot begin to understand why you would use the decibel (which is basically a log ratio between measured value and standard value) - the phase angle is just that an angle. Perhaps play around with simple trigonometric equations to ensure you are clear with sinusoidal waves first

 

y=Cos(x) y=Sin(x) y=Sin(x+pi/2) y=Cos(x+k)

Posted

I think you need to spend a little time learning some basics before you run with gravitational waves - unfortunately, I don't have the academic knowledge to help you here. I cannot begin to understand why you would use the decibel (which is basically a log ratio between measured value and standard value) - the phase angle is just that an angle. Perhaps play around with simple trigonometric equations to ensure you are clear with sinusoidal waves first

 

y=Cos(x) y=Sin(x) y=Sin(x+pi/2) y=Cos(x+k)

 

But if you read equation 2b alpha has to be dimensionless.

Posted

 

But if you read equation 2b alpha has to be dimensionless.

 

What dimension does an angle in radians have? Exactly. The decibel is far from the only dimensionless form of measure.

Posted

 

What dimension does an angle in radians have? Exactly. The decibel is far from the only dimensionless form of measure.

Sorry it was a typo. I meant amplitude not alpha. What would you suggest as the most appropriate dimensionless form of measure for amplitude?

Posted

Sorry it was a typo. I meant amplitude not alpha. What would you suggest as the most appropriate dimensionless form of measure for amplitude?

 

Almost anything that can vary with the variation being sinusoidal - the most obvious examples are Volts per metre aka Newtons per Coulomb and Teslas aka Newtons per metre per Ampere.

 

edit - to be clear the above is not with regard to the equation in your paper but wave equations in general and the two most obvious EMR

edit 2 . In your equation, the amplitude is in Teslas or Newtons per meter per Ampere as the field is the magnetic field

Posted (edited)

 

Almost anything that can vary with the variation being sinusoidal - the most obvious examples are Volts per metre aka Newtons per Coulomb and Teslas aka Newtons per metre per Ampere.

 

edit - to be clear the above is not with regard to the equation in your paper but wave equations in general and the two most obvious EMR

edit 2 . In your equation, the amplitude is in Teslas or Newtons per meter per Ampere as the field is the magnetic field

 

 

Oh ok. I actually thought that amplitude was for gravity wave. So for calculating arbitrary constant as told by you , what do I take distance as(for amplitude) ? Or rather how do I calculate distance?

Edited by pennieloot
Posted

 

 

Oh ok. I actually thought that amplitude was for gravity wave. So for calculating arbitrary constant as told by you , what do I take distance as ? Or rather how do I calculate distance?

 

Sorry - I just glanced at the paper and didn't read in as much depth as I should; frankly I saw it was a magnetic field and stopped there assuming that it was a varying B.

 

a is the amplitude of the gravitational wave and is thus dimensionless and is the strain between two test objects I believe

 

Again without reference to your papers and being basics of gravitational waves. We look at g' - this is the relative acceleration of two test objects separated by a small distance. What we want for the amplitude is h the strain (just like the materials science strain) which is the change in displacement divided by the distance between the objects. If you integrate acceleration (remember g' is an acceleration) twice with respect to time you get a displacement

 

[latex]

h=2 * \iint g' \cdot dt^2 = 2 * \frac{change\ in\ displacement}{displacement}

[/latex]

  • 3 weeks later...
Posted

 

Sorry - I just glanced at the paper and didn't read in as much depth as I should; frankly I saw it was a magnetic field and stopped there assuming that it was a varying B.

 

a is the amplitude of the gravitational wave and is thus dimensionless and is the strain between two test objects I believe

 

Again without reference to your papers and being basics of gravitational waves. We look at g' - this is the relative acceleration of two test objects separated by a small distance. What we want for the amplitude is h the strain (just like the materials science strain) which is the change in displacement divided by the distance between the objects. If you integrate acceleration (remember g' is an acceleration) twice with respect to time you get a displacement

 

[latex]

h=2 * \iint g' \cdot dt^2 = 2 * \frac{change\ in\ displacement}{displacement}

[/latex]

 

Thanks for the explanation that clears alot.

 

Also after equation 2b it states that resonance occurs at twice the larmor frequency and since exact 2:1 resonance's probability is almost zero, would there be a net increase or accelerating resonance even if the wave frequency was more than twice (lets say thrice or even 4-5 times) of larmor frequency ?

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.