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I'm trying to conceptually understand what the amplitude term in the wave equation for a gravitational term represents, which is depicted as $A = A_0\cos(\omega t-kx)$ where $A_0$ is the amplitude term and the wave equation itself is $$\left(\frac{\partial^2 A}{\partial^2 x}+\frac{\partial^2 A}{\partial^2 y}+\frac{\partial^2 A}{\partial^2 z}\right) = \frac{1}{v^2}\frac{\partial^2A}{\partial^2t}$$

For example, if two 100 solar mass blackholes are rotating around one another the gravitational wave they would produce would have a greater magnitude or intensity than a gravitational wave produced by two 10 solar mass neutron stars.

As such, would the intensity (the magnitude) of the gravitational wave that is being produced by each system be encapsulated in the amplitude term of the wave equation?

Or does the wave equation not deal with the intensity (the magnitude) of a gravitational wave and that the amplitude is something totally different and not a representation of intensity or magnitude?

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  • $\begingroup$ Have you learned what a spacetime “metric” is? $\endgroup$
    – Ghoster
    Commented Nov 9, 2023 at 4:44
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    $\begingroup$ The amplitude of a gravitational wave is the maximum deviation of the metric components from their background (i.e., usually flat-spacetime) values. $\endgroup$
    – Ghoster
    Commented Nov 9, 2023 at 7:43
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    $\begingroup$ Notice that what you have written is the homogeneous wave equation (but, please, use a partial time derivative) with no source term. So, it is quite difficult to understand what's the effect of the source mass on the amplitude of the wave. $\endgroup$ Commented Nov 9, 2023 at 12:02
  • $\begingroup$ @AstoundingJB, in terms of the wave equation, what would a source term look like as an example and would it be encapsulated in the amplitude term? $\endgroup$
    – Tivity
    Commented Dec 9, 2023 at 0:34
  • $\begingroup$ @Tivity : Refer for example to the Einstein's quadrupole formula. On the left-hand side, you have the (tensor) field, and on the right the source term, proportional to the second time-derivative of the quadrupole moment. $\endgroup$ Commented Dec 11, 2023 at 13:58

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To get started, let's think about a simpler wave first. For a wave on string, the amplitude tells you the displacement from the rest position of the string, $\Delta y$. If a string is strung horizontally parallel to the $x$-axis at height $y_0$ and has a vertical sine-wave propagating along it, the vertical position of each part of the string would be: $$ y(t,x) = y_0 + \Delta y(t,x),$$ where $\Delta y$ is the wave which acts as a perturbation to the resting string position. $$\Delta y(t,x) = A_0 \sin(\omega t - kx)$$

A gravitational wave is a perturbation a background spacetime. Spacetime is described by a metric, $g_{\mu\nu}$, that effectively tells us how to measure distances (and time intervals). In the simplest formulation the background is flat, empty space described by the Minkowski metric $\eta_{\mu\nu}$. The total spacetime metric would be the sum of the background spacetime and the wave perturbation: $$g_{\mu\nu}(t,\vec{x}) = \eta_{\mu\nu} + h_{\mu\nu}(t,\vec{x}),$$ where $h_{\mu\nu}$ is the gravitational wave. For a sinusoidal gravitational wave, $$h_{\mu\nu}(t,\vec{x}) = A_{\mu\nu} \sin(\omega t - \vec{k}\cdot\vec{x})$$ The wave is a function of time, $t$, and the full 3D location, $\vec{x}$. The $\mu$ and $\nu$ are indices denoting the tensor components, e.g. $(t,x,y,z)$. Since it is 4D and there are two indices there are 16 components in the wave (in the usual choice of coordinate system GR predicts only four of them are non-zero, and in any coordinate system only two are independent). The amplitude of the wave tells us how to measure distances in the spacetime. The wave will stretch and squeeze space time, so as the wave passes the distance between freely floating particles will change. A larger amplitude wave will cause a bigger change in the distance.

Any shaped wave can satisfy the wave equation. To find out which systems make which waves, you need to connect the wave equation to a source term. For a wave a on a string, the source term might describe a driving force which moves one end of the string in a particular pattern. For gravitational waves the source term describes the motion and masses of particles that make up the system.

For a binary system:

  • the orbital frequency is proportional to the gravitational wave frequency
  • faster speed, i.e. higher orbital frequency, produces higher amplitude waves (for two systems with the same mass)
  • higher mass produces higher amplitude (for two systems with the same frequency)

It can get more complicated than this if we include more details about the motion, like the spins of the individual objects or the ellipticity of the orbits.

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  • $\begingroup$ Given that the amplitude of a gravitational wave is determined by the orbital frequency and mass of the object, and considering that typical rotational systems involve two objects with the same orbital frequency but different masses. Then wouldn't each individual object produce its own gravitational wave with its own distinct amplitude? Hence, to accurately represent the gravitational wave generated in a binary system, wouldn't it be necessary to add together two wave equations depicting the superimposition of each individual wave each with its own amplitude? $\endgroup$ Commented Nov 9, 2023 at 19:14
  • $\begingroup$ Because GR is a non-linear theory, adding two solutions of the field equations does not produce a new solution. The correct thing to do is consider the whole system all together. The wave equation describes the propagation of waves, and in GR comes from making a linear approximation of the full field equations. To a very good approximation, two propagating low amplitude waves superimpose just fine. But the creation of GWs by a BH binary happens in a decidedly non-linear regime, and you cannot superimpose two individual BH solutions to figure it out. $\endgroup$
    – Paul T.
    Commented Nov 10, 2023 at 14:42
  • $\begingroup$ Is there an equation the computes the value of $A$ in $A\sin(\omega t - kx)$. In other words, How is $A$ derived? $\endgroup$ Commented Nov 10, 2023 at 21:28
  • $\begingroup$ You should search for existing answers, and if you cannot find anything ask that as a new question. $\endgroup$
    – Paul T.
    Commented Nov 10, 2023 at 22:36
  • $\begingroup$ @PaulT. Would the source term be encapsulated within the coefficient $A$ in $A\sin(\omega t - kx)$? $\endgroup$
    – Tivity
    Commented Dec 9, 2023 at 0:43

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