every equation i found in internet for gravitational wave radiation is for binary system which is need 2 mass in order to find strain (h).

my question is: if we have one mass which is oscillate , how about this one for calculating strain(h) ?

for example: for a single mass 1.000.000 kg & frequency 100 times per second

if there is formula or reference link , let us know.

  • $\begingroup$ en.wikipedia.org/wiki/Quadrupole_formula has the relevant equations. But the configuration of your oscillator is unclear. Is it just moving back & forth on a line? How far does it travel on each oscillation? The calculations for binary systems assume that the centre of mass doesn't move, but that's not the case for your oscillator. $\endgroup$
    – PM 2Ring
    Sep 24, 2023 at 12:50
  • $\begingroup$ Define what you mean by a single source and how its mass distribution oscillates. $\endgroup$
    – ProfRob
    Sep 24, 2023 at 14:00
  • $\begingroup$ @PM2Ring 1-what is 'r' in that formula? 2-i can put kg mass in 'I' in the formula? 3-frequency is not important? 4-this formula give us strain? $\endgroup$ Sep 24, 2023 at 14:59
  • 2
    $\begingroup$ The formula is given in en.wikipedia.org/wiki/Quadrupole_formula . If you don't understand that formula then I suggest you reformulate your question to ask about that. $\endgroup$
    – ProfRob
    Sep 24, 2023 at 15:41
  • 2
    $\begingroup$ Note that you have to conserve momentum, so if you're pushing a mass back and forth, you also have to account for the motion of yourself pushing it. This is basically why gravitational waves are sourced by time-varying mass quadrupoles. Time-varying mass dipoles are not even allowed. $\endgroup$
    – Sten
    Sep 24, 2023 at 16:11

1 Answer 1


Suppose you have a mass $m$ moving up and down along the z-axis according to:

$$ z(t) = A \cos\omega t $$

The traceless part of the quadrople moment tensor then has components

\begin{align} I_{xx}^T &= -\frac{1}{3} m A^2 \cos^2\omega t,\\ I_{yy}^T &= -\frac{1}{3} m A^2 \cos^2\omega t,\\ I_{zz}^T &= \frac{2}{3} m A^2 \cos^2\omega t,\\ \end{align}

with all other components zero.

This can be plugged into quadrupole formula:

$$ \bar{h}_{ij}(t,r) = \frac{2 G}{c^4 r} \ddot{I}_{ij}(t-r/c)$$

to find the spatial part of the trace reversed perturbation of the metric, $\bar{h}_{ij}$.

The average power emitted in gravitational waves is $$\frac{16G}{15c^5} A^4m^2\omega^6.$$

Note that the quadrupole formula is an approximation only valid in the non-relativistic approximation, i.e. $A\omega\ll c.$

Suppose I am waving my hand up and down, and want to know the gravitational waves produced. For sake of simplicity we put $m = 1 kg$, $A=1 m$, and $\omega = 1Hz$, the emitted power is then about $1.5\times10^{-52} W$. That is very small. To see how small, lets suppose that gravity is in fact quantized. The energy of a single quantum (i.e. graviton) should then be $\hbar\omega\approx 1\times10^{-34} J$. So, I'd expect to emitted less then one graviton per Hubble time. So while you might sometimes see a gravitational wave physicist on stage claim that wave their hands about would generate gravitational waves, this might not quite be true due to quantum effects!

  • $\begingroup$ thanks a zillion !!! really good answer. can you calculate strain (h) also so that i can see how formula works for that? for your aboved example $\endgroup$ Sep 25, 2023 at 16:39
  • $\begingroup$ also what is A ? you meant in 1 square meter deposit??? $\endgroup$ Sep 25, 2023 at 16:42
  • $\begingroup$ A is the amplitude of the oscillation. $\endgroup$
    – TimRias
    Sep 25, 2023 at 18:17
  • $\begingroup$ can also caculate strain (h) for your example ? thanks $\endgroup$ Sep 25, 2023 at 18:22
  • 1
    $\begingroup$ I can, but so can you. Everything you need is here. $\endgroup$
    – TimRias
    Sep 25, 2023 at 18:25

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