4 ms·
If R is the shape of space, is it fair to say that gravity waves from a black holes merger are large waves in R, like large waves in a lake we could surf on; wh
by ry454 6y ago
If R is the shape of space, is it fair to say that gravity waves from a black holes merger are large waves in R, like large waves in a lake we could surf on; while there are also tiny ripples that we describe with F? In other words, F is really R with large waves subtracted.
- tobinfricke 6y ago> In other words, F is really R with large waves subtracted. No, they are completely independent. Here's another analogous example. Suppose you have a two-dimensional rubber sheet. You put a coordinate system on it, so any position on the rubber sheet can be identified by coordinates (x, y). You deform the rubber sheet so it has a curvy shape - anything you want. You can have a tensor R(x,y) that tells you the curvature of the rubber sheet as a function of position (x,y). And a scalar T(x,y) that tells you the temperature of the rubber sheet at (x,y). These are completely independent quantities. Different fields.
- ry454 6y agoT is indeed independent from R, but how do we know that F and R are also independent? Would theory fall apart if we try to model F as microscopic modulations of R? I'm trying to look at R and F from the signal processing point of view: if we apply Fourier transform to both, we'd see that R is described by low frequency band (planetary scale distortions), while F would be ideally described by a high frequency band.
- TeMPOraL 6y agoI wonder about something different - if R and F are independent, and R also tells you the shape of space, what would happen if EM and gravity waves crossed through each other? My guess is the EM wave would be modulated somehow, within the intersection volume, as if crossing into a different matter medium, while gravity wave would be unaffected.
- pa7x1 6y agoBoth effects can occur; GW <-> EMW. You can search for Gertsenshtein Effect, there is also this old paper by Zel'dovich: http://jetp.ac.ru/cgi-bin/dn/e_038_04_0652.pdf http://jetp.ac.ru/cgi-bin/dn/e_038_04_0652.pdf
- tobinfricke 6y agoThat's correct. The "shape of space" can act as a lens for light. Even in static situations this can be quite interesting: https://en.wikipedia.org/wiki/Einstein_ring#:~:text=An%20Einstein%20ring%2C%20also%20known,to%20come%20from%20different%20places https://en.wikipedia.org/wiki/Einstein_ring#:~:text=An%20Ein.... It's possible that you could observe a kind of "shimmer" in the position of background stars as a gravitational wave passes between you and them. In fact there is an attempt to detect gravitational waves this way through pulsar timing: https://en.wikipedia.org/wiki/Pulsar_timing_array https://en.wikipedia.org/wiki/Pulsar_timing_array
- Aerroon 6y ago>You can have a tensor R(x,y) that tells you the curvature of the rubber sheet as a function of position (x,y). And a scalar T(x,y) that tells you the temperature of the rubber sheet at (x,y). These are completely independent quantities. Different fields. I get that it's an analogy, but I don't think they would be completely independent quantities here. There is likely going to be some correlation with temperature and curvature of the rubber sheet. It would likely pale in comparison to almost everything else though, but on the topic of discussion that might be a large enough correlation. Also, don't high energy photons curve spacetime?
- tobinfricke 6y ago> There is likely going to be some correlation with temperature and curvature of the rubber sheet Yeah, that's a failure of my analogy. Deforming a rubber sheet does create heat, but deforming spacetime does not create light.
- namanyayg 6y agoBut R(x) is required to determine x -- whereas F(x) is not. Thus, F and R are fundamentally different.