0
$\begingroup$

If the recent gravitational wave's energy had reached us as visible light, how bright would it have been? Stackexchange complains about the form and brevity of the question so i add something... if it is strong enough to move a heavy mirror albeit by a 10000 th of a proton it should be quite strong and intuitively the light of even a bright star shouldn't have that strength.

$\endgroup$
5
$\begingroup$

The only meaningful way to compare "strength" between light and a gravitational wave is via the energy flux they deliver. (Otherwise, how do you compare a stretch in the geometry of space and time with an electric field?)

In that vein, then, a good representative is the first observation, GW150914, which emitted an energy of about $E=3.0\:M_\odot c^2$ in about $\Delta t=0.1\mathrm{\:s}$ at a distance of some $L=440\:\mathrm{Mpc}$, which comes down to an energy flux of $$ I=\frac{E}{4\pi L^2 \Delta t}\approx 0.23 \:\mathrm{\mu W/cm^2}. $$ This is equivalent to a weak light source, but it's probably visible by naked eye under suitably dark conditions.

That said, it's important to remark that the gravitational wave should not be thought of as "moving a heavy mirror" by any distance, and certainly not as performing work while doing so. Instead, its action is to expand and contract the space between the mirrors (even as they remain stationary as far as they can tell), as explained e.g. in How does gravitational wave compress space time?.

$\endgroup$
  • 1
    $\begingroup$ Yeah, it's at about that order of magnitude, though keep in mind that the full moon only looks as bright as it does because it's generally against a much darker background. Regarding the measurement of the waves, it's like measuring the speed of a car with a Doppler radar: you can argue that the car does perform some work (by stretching out the light) but it's a minuscule part of the total energy it carries. $\endgroup$ – Emilio Pisanty Oct 10 '17 at 12:56
  • $\begingroup$ None that I'm aware of. If you have further questions of a back-and-forth nature, please take them to the site chatroom. (Also: maybe in the dedicated community the notation is standard, but in a general-physics context, the phrase "G-wave" isn't standard.) $\endgroup$ – Emilio Pisanty Oct 10 '17 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.