4
$\begingroup$

I was thinking about the way you measure gravitational waves. What you do is you measure how they affect a bunch of particles. Affecting something means transferring some mass/energy to it, according to my layman knowledge.

So let's do a thought experiment. You have a massive spaceship, without any thrust. It flies through almost completely homogeneous universe, gravity wise.

Without gravitational waves, this is easy. Gravity pulls in all directions, so it cancels and you fly forever.

With gravitational waves. Hmm, imagine that on every planet there's a device measuring gravitational waves. Of course, everything is a device measuring gravitational waves, but actual devices help imagination here. So your movement affects all those devices by transferring tiny amounts of energy to them. That energy sure doesn't pop out of thin air - or vacuum - so you're going to miss it.

My conclusion is that for every tiny amount of energy, your relative velocity to the measuring device should decrease.

Is that correct or not and why?

Note that I appreciate there's certainly more to it - eg. decrease in your velocity should happen with light-speed delay. Hard to wrap my mind around that.

$\endgroup$
  • $\begingroup$ Question: do you understand the mathematics behind General Relativity? If you do, everything that there is to say is expressed in terms of the "gravitational wave" solutions of the linearized Einstein's equations. If you don't, I am sorry to say that understanding complicated physical phenomena in terms of analogies is sometimes a bad idea. Math is much more efficient as it is compact, precise and unambiguous. Please don't take this personally, it is what I would've said to anybody. $\endgroup$ – Prof. Legolasov Dec 5 '16 at 14:32
  • 2
    $\begingroup$ @SolenodonParadoxus I don't take this personally, but maybe then you're not the right person to answer questions by laymen tagged [thought-experiment]. Besides, according to what you said this question would never exist if I did understand the mathematics behind it, so it's pointless to wonder that I don't. $\endgroup$ – Tomáš Zato Dec 5 '16 at 14:37
  • 1
    $\begingroup$ Well, its not like I voted to close (I didn't even downvote your question). Maybe somebody else will answer it. What I'm saying is: there is a better way of understanding these phenomena, which is: studying General Relativity. Analogies can only take you this far... Anyways, good luck with your question. $\endgroup$ – Prof. Legolasov Dec 5 '16 at 14:39
  • $\begingroup$ @SolenodonParadoxus I know that what you say is true and I really did not take any offense. But it will take me a while before touching any complex physics math again. $\endgroup$ – Tomáš Zato Dec 5 '16 at 14:44
2
$\begingroup$

I think I understand your question, but just to be sure: you're saying that the gravitational waves produced by this spaceship will lead to it slowing down. The first problem is that a normal spaceship with a constant velocity, or even acceleration won't produce gravitational waves (GW) because GW require an accelerating quadrupole moment.

You're correct that an object producing GW ends up transferring energy to the detectors... this was actually one of the deciding arguments that GW were a real, observable phenomenon. GW carry energy, which is extracted from the system producing it. In the case of a binary, the energy comes from the orbit, which causes the binary to tighten and eventually coalesce. This isn't quite analogous to 'drag', per se, because 'drag' specifically refers to dissipative interaction with the background medium. When an electron accelerates, it produces EM waves, causing it to lose energy... but I don't think we would via that as 'drag' in any way.

But, I think that in a spacetime with gravitational waves propagating in the $+\hat{z}$ direction, and a spaceship traveling in the $-\hat{z}$ direction... the spaceship would experience an (outrageously, likely-never-detectable-by-even-future-technology) deceleration force from non-linear coupling with the GW. That sounds more like a drag force.

$\endgroup$
  • $\begingroup$ So if a heavy object passes near you, you can't detect it by it's gravitational waves as long as it takes straight path (except it's in in curved space)? $\endgroup$ – Tomáš Zato Dec 5 '16 at 15:17
  • $\begingroup$ @TomášZato it's even a little harder than that! A spherically symmetric ship, even if it's accelerating (i.e. curved path, or changing speed) still won't radiate. You also need an asymmetric mass distribution... but I guess even an oblong/cylindrical spaceship would be fine... I think. $\endgroup$ – DilithiumMatrix Dec 5 '16 at 16:34
-3
$\begingroup$

Do gravitational waves create 'drag' in space?

No.

I was thinking about the way you measure gravitational waves. What you do is you measure how they affect a bunch of particles. Affecting something means transferring some mass/energy to it, according to my layman knowledge.

It isn't quite true. If you were sitting in a spaceship and a gravitational wave passed through, it would make you and your clocks go slower for a little while. You might notice that some distant pulsar speeded up a little. But soon the gravitational would have moved on, without losing any energy. In theory you could detect a massive dark object passing by behind you via the same method.

So let's do a thought experiment. You have a massive spaceship, without any thrust. It flies through almost completely homogeneous universe, gravity wise. Without gravitational waves, this is easy. Gravity pulls in all directions, so it cancels and you fly forever.

No problem.

With gravitational waves. Hmm, imagine that on every planet there's a device measuring gravitational waves. Of course, everything is a device measuring gravitational waves, but actual devices help imagination here. So your movement affects all those devices by transferring tiny amounts of energy to them.

It doesn't actually transfer energy to them. You could say some of the gravitational wave energy is device energy as the gravitational wave passes through the device. But the gravitational wave soon passes through taking that energy away.

That energy sure doesn't pop out of thin air - or vacuum - so you're going to miss it. My conclusion is that for every tiny amount of energy, your relative velocity to the measuring device should decrease. Is that correct or not and why?

It isn't correct I'm afraid. As the gravitational wave passes through, the velocity of the things inside clocks etc is reduced. Those clocks run slower so they measure other things like distant pulsars to be going faster. You run slower too, along with everything else. We call it time dilation.

Note that I appreciate there's certainly more to it - eg. decrease in your velocity should happen with light-speed delay. Hard to wrap my mind around that.

What's easy to wrap your head round is that the speed of light reduces as the gravitational wave passes through. See the second paragraph here where Einstein says the speed of light is spatially variable in a gravitational field. For a gravitational wave as opposed to a field, the speed of light is temporally variable. But because everything slows down, you can't measure any local difference. This Baez article is worth a read: Is The Speed of Light Everywhere the Same?

$\endgroup$

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.