Propagation of gravitational waves

In summary: The solution is as follows.First, I expressed ##c## as follows:##c=f(H, \rho_{a}, \rho_{w}, t, a, \lambda)##, where ##H## is the depth of water, ##t## is time.I used the ##Buckingham-\pi## Theorem.Now, since ##\frac{\rho_{a}}{\rho_{w}} \approx 0.001## the density of air is usually ignored. So now we have the following dimensions in the variables of ##f_1##:##[\rho_{a}^*] = ML^{-4}####[\rho_{w
  • #1
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Stumbled upon this problem lately. Maybe someone could help me clarify some subtleties I do not see?

1.
Consider the propagation speed ##c## of periodic surface of gravity waves with wavelength ##\lambda## and amplitude ##a## in water of depth ##H##. Let ##\rho_{a}## and ##\rho_{w}## be the density of air and water. Ignoring viscosity and surface tension, use diminutional analysis to find an expression for the propagation speed ##c##. Since ##\frac{\rho_{a}}{\rho_{w}} \approx 0.001## the density of air is usually ignored. How does your expression simplify in this case? What about in the small amplitude limit ##a \rightarrow 0##?

Homework Equations

: [/B]none

The Attempt at a Solution

:[/B]

1. I expressed ##c## as follows:

##c=f(H, \rho_{a}, \rho_{w}, t, a, \lambda)##, where ##H## is the depth of water, ##t## is time.

I used the ##Buckingham-\pi## Theorem.

2. I non-dimensionalized ##\pi_1## as follows: ##\pi_1 = \frac{\lambda}{a}##.

3. Now, ##c^* = f_1(\rho_{a}^*, \rho_{w}^*, t, \frac{\lambda}{a})## - I divided by ##a## which has units of length. So now we have the following dimensions in the variables of ##f_1##:

##[\rho_{a}^*] = ML^{-4}##
##[\rho_{w}^*] = ML^{-4}##
##[t] = T##
And ##\frac{\lambda}{a}## is dimensionless.
Where M is in kilograms, L is in meters, T is in seconds.

So, ##\pi_2 = c^{*\alpha}\rho_a^{*\beta}\rho_w^{*\gamma} t^{\delta}## (is this correct?)

4. Now, as per the ##Buckingham-\pi## Theorem, we have

##[\pi_2]=T^{-\alpha+\delta} M^{\beta + \gamma} L^{-4\beta - 4\gamma}##

So,

##-\alpha + \delta = 0##
##\beta + \gamma = 0##
##-4\beta - 4\gamma = 0##

Thus ##\alpha = \delta##, ##\beta = -\gamma##. And now we have:

##\pi_2 = (ct)^{\alpha}(\frac{\rho_{a}}{\rho_{w}})^{\beta}##.

So we could express the dimensionless equation as

##F(\frac{\lambda}{a}) = ct(\frac{\rho_{a}}{\rho_{w}})^{\beta}##. (I omitted the *'s).

I think there is or are some errors in my deduction because if we use the same method that I described but eliminate the density of air or the amplitude of waves, we get ##{\pi_2=1}##.

Thank you for your attention!
 
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  • #2
Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
  • #3
Have you thought about any other physical parameters for your analysis? For instance, does the Earth's gravity play a part in your analysis? (You can answer this by considering how gravity waves in fluid propagate if you were on a planet with little gravity etc.)

In Buckingham Pi analysis, you can always over fit the dimensionful quantities, then by interrelations between them exclude them in your refinement.
 
  • #4
I have now solved this problem. I had actually misinterpreted the problem, as I thought it was about relativistic waves rather than water waves :)
 
  • #5


Hello,

Thank you for sharing your thoughts on the propagation of gravitational waves. I am happy to help clarify some of the subtleties you have encountered.

Firstly, it is important to note that gravitational waves are different from gravity waves. Gravitational waves are ripples in the fabric of spacetime, while gravity waves are surface waves on a fluid.

In the context of gravity waves, your attempt at using dimensional analysis to find an expression for the propagation speed is a good approach. However, there are a few points that need to be clarified.

1. The Buckingham-π Theorem states that if there are n variables in a problem and k fundamental dimensions, then there will be n-k independent dimensionless variables. In this case, you have correctly identified the four fundamental dimensions (M, L, T, and ML^-4). However, you have only identified three variables (H, ρa, ρw) and one constant (t). Therefore, there should be three independent dimensionless variables, not two as you have assumed.

2. In step 2, you have correctly identified the first dimensionless variable, π1 = λ/a. However, in step 3, you have made a mistake by assuming that c* is dimensionless. This is not correct. The propagation speed c has dimensions of LT^-1, and therefore c* = c/a also has dimensions of LT^-1.

3. In step 3, you have also incorrectly assumed that ρa* and ρw* are dimensionless. In fact, they have dimensions of ML^-4. Therefore, the correct expression for π2 should be π2 = c*α ρa*β ρw*γ tδ.

4. In step 4, you have made another mistake by assuming that the only dimensionless variable in the problem is λ/a. In fact, there are two more independent dimensionless variables that can be formed from H, ρa, ρw, and t. These are H/a and ρa/ρw. Therefore, the correct expression for π2 should be π2 = c*α H*β ρa*γ ρw*δ t*ε, where α, β, γ, δ, and ε are exponents to be determined.

5. Using the Buckingham-π Theorem, we can now equate the dimensions of π1 and π2 to determine the ex
 

1. What are gravitational waves?

Gravitational waves are ripples in the fabric of spacetime, caused by the acceleration of massive objects. They were predicted by Albert Einstein's theory of general relativity and have been recently detected by scientists.

2. How are gravitational waves propagated?

Gravitational waves are propagated through space at the speed of light. They are created when massive objects, such as black holes or neutron stars, accelerate and create distortions in the fabric of spacetime. These distortions then travel outwards, carrying energy with them.

3. How are gravitational waves detected?

Gravitational waves are detected by using incredibly sensitive instruments called interferometers. These instruments use lasers to measure tiny changes in the length of their arms caused by passing gravitational waves. When a gravitational wave passes through, it causes a tiny amount of stretching and squeezing in the fabric of spacetime, which can be measured by the interferometer.

4. What are the potential applications of gravitational wave detection?

Gravitational wave detection has the potential to provide us with a new way of observing and studying the universe. By studying the properties of gravitational waves, scientists can gain a better understanding of the behavior of massive objects, such as black holes and neutron stars. This can also lead to advancements in fields such as astrophysics, cosmology, and even technology.

5. Can gravitational waves be used for communication?

No, gravitational waves cannot be used for communication. Unlike electromagnetic waves, which can be manipulated and modulated for communication purposes, gravitational waves are not affected by electric or magnetic fields and cannot be controlled in the same way. Additionally, gravitational waves are incredibly weak by the time they reach Earth, making them unsuitable for long-distance communication.

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