# Determining the energy radiated by gravitational waves in a simple system

• rcatalang
rcatalang
Homework Statement
We have two massive particles of mass ##m_1## and ##m_2## that approach each other from infinity up to a certain point, ##x_f##. One has to calculate the energy radiated through gravitational waves.
Relevant Equations
Equation of motion: ##\ddot{x}=\frac{-GM}{x^2}##.

Energy lost (disregarding constants): ##E\propto \int_{t=0}^{t_f} \frac{(\dot{x})^2}{x^4}##
I have tried simplifying the integral (turning into an integral in terms of position variables) using the equation of motion. It's easy to show:

$$\frac{1}{x^4}=\frac{(\ddot{x})^2}{(GM)^2}$$

And therefore one can write:

$$\frac{1}{(GM)^2}\int_{t=0}^{t_f} (\dot{x}\ddot{x})^2 dt = \frac{1}{(GM)^2}\int_{\infty}^{x_f} \dot{x}(\ddot{x})^2 dx$$

But I have no idea about how to continue further with this integral. Any ideas on how to solve this integral?

Let me show you how to work this out. For brevity I'll just assume both masses are equal, ##m_1 = m_2 = m##, but you can easily re-scale the following results for the case of non-equal masses.

The idea is to assume that, since we're in linearized theory, the dynamics given by the Newtonian equations of motion are roughly correct. So consider the two masses to be at ##\pm x(t)## (and starting at ##\pm x_0##, which we'll later take to be infinity). The separation at time ##t## is just ##2x(t)##, and applying Newton III for either particle gives:$$m\ddot{x} = - \frac{Gm^2}{(2x)^2}$$You can multiply through by ##\dot{x}## and integrate both sides, which gets you something similar to an energy equation:$$\frac{1}{2} \frac{d}{dt} (\dot{x}^2) = \frac{d}{dt} \left( \frac{Gm}{4x} \right)$$Using the initial conditions gives you:$$\dot{x}^2 = \frac{Gm}{2} \left( \frac{1}{x} - \frac{1}{x_0} \right)$$Now we return to general relativity briefly. The GW power is proportional to ##\dddot{Q}_{ij} \dddot{Q}_{ij}##, where ##Q_{ij}## is the trace-free part of the quadrupole tensor ##I_{ij}##, defined by:$$I_{ij} = \int \rho(\mathbf{x}) x_i x_j$$where ##\rho(\mathbf{x}) = m\delta(x(t)) + m\delta(-x(t))## in this case. You can see that this only has one non-zero component, ##I_{xx}##,$$I_{xx} = 2mx(t)^2$$So we'll need to find an expression for ##\dddot{I}_{ij}##. Churning out all three derivatives gives you:$$\dddot{I}_{xx} = 12m \dot{x} \ddot{x} + 4m x \ \dddot{x}$$That may look slightly unwieldy, but we do actually have everything we need. Recall that ##\ddot{x} = -GM/(4x^2)##, which you can differentiate again to get ##\dddot{x}##. If you plug everything in, you get\begin{align*}
\dddot{I}_{xx} = - \frac{GM \dot{x}}{x^2} = -\frac{GM}{x^2} \sqrt{\frac{GM}{2}} \sqrt{\frac{1}{x} - \frac{1}{x_0}}
\end{align*}Since this is a homework problem, you have do some work as well. What is the trace free part ##\dddot{Q}_{ij} = \dddot{I}_{ij} - \frac{1}{3} \mathrm{Trace}(\dddot{I}) \delta_{ij}##? Note that ##\dddot{Q}_{ij}## also has ##yy## and ##zz## components. Therefore what is the gravitational power ##P(x)## as a function of ##x##?

Once you have the power ##P(x)##, you can easily integrate$$dE = P(x) dt = P(x) \frac{1}{\dot{x}} dx$$since we know ##\dot{x}## as a function of ##x##...

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