6 ms·

Dumb question, is the idea that gravitational waves are not very common from natural sources, so there's less...interference? I didn't get why this method of co

Dumb question, is the idea that gravitational waves are not very common from natural sources, so there's less...interference? I didn't get why this method of communication is somehow special (beyond being theoretically interesting)?

tobinfricke 6y ago
It's special because it's a different field. Radio/light are oscillations of the electromagnetic field, and sound is an displacement in a physical medium. This is an entirely different medium: the metric tensor of spacetime itself. And the universe is, by and large, completely transparent to gravitational waves. So these signals would not be blocked by intervening matter. A similar thing could be said about the prospect of sending messages using neutrinos. But it is only just now barely possible to detect the loudest gravitational wave signals with extreme difficulty and machines costing on the order of a billion dollars. It will remain impossible for humans to generate gravitational wave signals of a detectable amplitude for the foreseeable future. Source: PhD in gravitational wave detectors.
- ry454 6y ago
  A lame question from someone who barely remembers college-level physics: isn't that "electromagnetic field" just waves in the spacetime metric-tensor?
  - tobinfricke 6y ago
    The electromagnetic field is a function F(x), where F is the field tensor (which groups together the electric field E and the magnetic field B) and x is the position in space. Maxwell's equations tell you how F (or E and B) evolve in time, and give you wave-like solutions. In a continuous medium like air, water, or a solid, you could describe a "displacement field" d(x) that tells you how far the piece of material that used to be at position x is displaced from its equilibrium position. This is governed by the equations of physical acoustics, which also have wave-like solutions: sound. The Riemmann curvature tensor R is a function R(x) of spacetime coordinates x, which tells you the curvature of spacetime at location x. Einstein's equations tell you how R evolves, and it also turns out to have wave-like solutions: gravitational waves. In radio, it's F (or E and B) that's waving. In gravitational waves, it's R. What's confusing is that R also tells you "the shape of space," and you need to use it to figure out "where" the coordinate x corresponds to.
    - 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 ago
        T 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 ago
        I 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.
        6y ago
- csunbird 6y ago
  If we were able to catch neutrinos effectively, could we use neutrinos for the same purpose?
  - lou1306 6y ago
    We already had a working point-to-point neutrino beam [0]. But I'm not a physicist and I don't really know if you could modulate it in any meaningful way. [0] https://en.wikipedia.org/wiki/CERN_Neutrinos_to_Gran_Sasso https://en.wikipedia.org/wiki/CERN_Neutrinos_to_Gran_Sasso
    - formerly_proven 6y ago
      Neutrinos have energy (as particles), so they should have a wavelength (as a wave), so frequency modulation works. And you can measure intensity, so AM works, too, and you can build all other modulation schemes from these two.
deleted 6y ago
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tobinfricke 6y ago

It's special because it's a different field. Radio/light are oscillations of the electromagnetic field, and sound is an displacement in a physical medium. This is an entirely different medium: the metric tensor of spacetime itself. And the universe is, by and large, completely transparent to gravitational waves. So these signals would not be blocked by intervening matter. A similar thing could be said about the prospect of sending messages using neutrinos. But it is only just now barely possible to detect the loudest gravitational wave signals with extreme difficulty and machines costing on the order of a billion dollars. It will remain impossible for humans to generate gravitational wave signals of a detectable amplitude for the foreseeable future. Source: PhD in gravitational wave detectors.

ry454 6y ago
A lame question from someone who barely remembers college-level physics: isn't that "electromagnetic field" just waves in the spacetime metric-tensor?
- tobinfricke 6y ago
  The electromagnetic field is a function F(x), where F is the field tensor (which groups together the electric field E and the magnetic field B) and x is the position in space. Maxwell's equations tell you how F (or E and B) evolve in time, and give you wave-like solutions. In a continuous medium like air, water, or a solid, you could describe a "displacement field" d(x) that tells you how far the piece of material that used to be at position x is displaced from its equilibrium position. This is governed by the equations of physical acoustics, which also have wave-like solutions: sound. The Riemmann curvature tensor R is a function R(x) of spacetime coordinates x, which tells you the curvature of spacetime at location x. Einstein's equations tell you how R evolves, and it also turns out to have wave-like solutions: gravitational waves. In radio, it's F (or E and B) that's waving. In gravitational waves, it's R. What's confusing is that R also tells you "the shape of space," and you need to use it to figure out "where" the coordinate x corresponds to.
  - 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 ago
        T 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 ago
        I 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.
        6y ago
csunbird 6y ago
If we were able to catch neutrinos effectively, could we use neutrinos for the same purpose?
- lou1306 6y ago
  We already had a working point-to-point neutrino beam [0]. But I'm not a physicist and I don't really know if you could modulate it in any meaningful way. [0] https://en.wikipedia.org/wiki/CERN_Neutrinos_to_Gran_Sasso https://en.wikipedia.org/wiki/CERN_Neutrinos_to_Gran_Sasso
  - formerly_proven 6y ago
    Neutrinos have energy (as particles), so they should have a wavelength (as a wave), so frequency modulation works. And you can measure intensity, so AM works, too, and you can build all other modulation schemes from these two.

deleted 6y ago

[deleted]