0
$\begingroup$

If I understand it correctly, curvature in a region of spacetime is due to concentration of energy and momentum in that region. If I further understand things (I don't) a gravitational wave is a transient, radiating region of curvature in spacetime.

Ignoring for the moment the absurdly small curvature caused by charging even the largest capacitor, I believe it is correct to say that a charged capacitor, which locally stores energy, curves spacetime.

Now presume that we have a device that can detect such small curvature, a distance d away from our capacitor. If we charge and discharge that capacitor, will the detector detect transient changes in the gravitational field? Is that the same as saying it is detecting a gravitational wave?

$\endgroup$
2
$\begingroup$

You are right in principle, but as you also mention, the resulting curvature would be undetectably tiny. This is because spacetime can be thought of as being extremely "stiff", meaning it takes truly gigantic energies to make it flex or bend enough to be detected by even the most phenomenally sensitive device (like LIGO).

$\endgroup$

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.