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I Are Evanescent Gravitational Waves Measurable

  1. May 20, 2017 #1

    Paul Colby

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    Hi,

    (all discussions here are in the extreme weak field approximation about Minkowski space)

    For the last couple of years I've been looking into the production and reception of radio frequency gravitational waves. It's kind of a retirement project the main goal of which is to get a better understanding of GW through futile calculations. I've gotten to the point where one of the approaches I've been looking at might actually may lead to a feasible measurement. Feasible here means likely beyond my resources but well within the existing technology.

    As singularly hopeless as this may sound, it may be possible to produce and subsequently detect evanescent gravitational waves. Evanescent waves are non-propagating near fields. They occur in all wave phenomena in one guise or another. For an isolated time harmonic GW source these fields die off as ##(kr)^{-5}## whereas the radiation fields die off as ##(kr)^{-1}##.

    The particular example I've been considering uses garden variety quartz crystal resonators in the HF radio band. A typical 4MHz fundamental mode crystal is a quartz disk 8mm in diameter and about 0.42mm thick. There are two circular electrodes plated about 4mm in diameter plated on the front and back faces of the disk. When a sinusoidal voltage is applied a volume shear mode is excited in the plane of the disk along the "x" crystal axis (and no I don't know which way this axis points or even if it's direction is controlled by the manufacturer. Generation of gravitational waves isn't in the spec sheet).

    Back to evanescent GW. The experiment would be to place two 4MHz crystals near one another. Near in this case is can-to-can which puts the crystal center to center distance at 3.45mm. A quick calculation yields,

    ##(kr)^{-5} = (\frac{\lambda}{2\pi r})^5 = 5\times 10^{17}##​

    The complete back of the envelop is,

    ##16 \pi G c^{-4} = 4.15\times 10^{-43} \frac{s^2}{kg\;m}##

    ##T_{x y} = 3.35\times 10^{6} \times Q## Pa

    This is for a drive current of 0.5 A (exceeds the spec but these things are cheep). Typical Q values run 50,000 to 100,000 so ##T_{x y}## is also a "big" number. Also I'd like to mention that ##T_{x y}## depends on the mechanical stress induced in the material and is not really a direct function of the crystal mass. The metric strain is the integral of the green function times the ##T_{x y}## so a factor on the crystal volume appears,

    ##V = 2\pi R^2 = 2.11\times 10^{-8} m^3##

    The metric strain is approximately,

    ## h_{x y} = G V T k (kr)^{-5} \approx \times 10^{-22}## [edit: I've removed Q which is in T. Q is included in T]

    Working a similar estimate for the received voltage gives about 50 pico volts. This prompted me to ask people who might know good approaches to measuring at these levels. I posted this yesterday on the Electrical Engineering forum,

    Signal measurement ~50 pico volts
     
    Last edited: May 20, 2017
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  3. May 20, 2017 #2

    mfb

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    Aren't these near fields just the regular Newtonian gravity? Regular variations of the mass distances have been tested at various frequencies, with "0" as the easiest measurement (Cavendish).
     
  4. May 20, 2017 #3

    Paul Colby

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    No. There are no waves in Newtonian gravity evanescent or otherwise. The gravitational field of a 8mm x 0.42mm disk is quite small. The Newtonian tidal forces associated with such a disk even smaller yet. The density of quartz is 2.65 that of water if I recall correctly. That's a net mass of the disk is ##5.6\times 10^{-5}## kg.
     
  5. May 20, 2017 #4

    mfb

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    4 MHz gives a wavelength of 75 meters, and no object moves at relativistic speeds. The Newtonian term (1/r potential) is by far the dominant part of the gravitational interaction. It doesn't matter if you call it evanescent wave or Newtonian gravity. It is there, and it is orders of magnitude stronger than everything else.
     
  6. May 20, 2017 #5

    Paul Colby

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    Could be the case. I'm not certain if in electromagnetism referring to evanescent waves as just time dependent Van der Waals forces would be very acceptable or an accurate depiction of the physics. That said, I think my shear values may be a bit unrealistic so I'll have to check my numbers.

    [edit: Okay why does ##k## show up in the result and how exactly does it appear in the Newtonian calculation?]
     
    Last edited: May 20, 2017
  7. May 20, 2017 #6

    Paul Colby

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    Okay, I'm going argue no, these are not time dependent tidal forces of Newtonian origin. According to the wiki

    https://en.wikipedia.org/wiki/Tidal_force

    The leading term goes like ##r^{-3}## and of course the speed of light doesn't appear in a Newtonian calculation so there is no opportunity for the wavelength to appear. Now there are ##(kr)^{-3}## terms along with all the others up to ##(kr)^{-5}## in the full calculation so Newtonian tidal terms are likely present but not dominant at the distances of interest to me.
     
  8. May 20, 2017 #7

    mfb

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    I don't follow your calculations (do you have a step by step version with all the formulas before plugging in numbers?), but if you get terms that are larger than Newtonian gravity and you don't propose having a black hole handy they are wrong.

    One possible cross check: What happens with your formulas if you let the frequency go to 0? Does the signal strength diverge?
     
  9. May 20, 2017 #8

    Paul Colby

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    Hum, no bias there.. What is that statement based on?

    The calculation is based on the radiation integral,

    ##h_{\mu\nu}(t,r) = -4Gc^{-4}e^{i\omega t}\int T_{\mu\nu}(r') \frac{e^{-ik|r - r'|}}{4\pi|r-r'|} d^3r'## ​

    which is in the Lorentz Gauge ##\partial^\mu h_{\mu\nu} = 0## since ##\partial^\mu T_{\mu\nu} = 0##. This result is gauged to the transverse traceless coordinates. This is done with a 4'th order differential projector acting on the space components of ##h##. That's a straight forward derivation of gauge functions ##\xi_\mu## which are harmonic (i.e. obey the scalar wave equation and have an ##e^{i\omega t}## time dependance). It's kind of long so I'll refer you to the standard text on the subject (and then ask you to tell me which one because I haven't found this anywhere for general fields as above).

    There are mutually inconsistent assumptions at ##\omega = 0## and the TT-gauge projection operator I used is clearly undefined in this limit. I agree 0 would be a more comfortable (that is, the only physically acceptable) limit. Great suggestion, thanks.
     
  10. May 20, 2017 #9

    pervect

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    I wrote a rather long post in which I estimated that the total power radiated in gravitational waves from a vibrating quartz crystal would be on the order of 3*10^-41 watts or so. This is based on an internal power flow of 10^6 watts in the crystal (I'm not positive I calculated this correctly, but the number is so low that I'd need to be off by factors of billions for it to matter.)

    I won't go into more details than this unless there is some interest, because it struck me, on re-reading the thread as being off-topic. What's more to the point is this. If we assume that the far field (M/r) gives a radiated power of such a small amount, how can we possibly hope to detect terms due to the near field on the order of (M/r)^5? It would seem that (M/r)^5 should be much smaller than (M/r).
     
  11. May 20, 2017 #10

    Paul Colby

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    If ##M/r## is the order parameter then clearly not. This is tantamount to assuming ##answer = 0## in everyones mind which I question. I'm explicitly dropping the constant terms and considering only terms arising from time harmonic components. As I point out, the stress terms developed in the material don't depend on the mass. The mass appears in the sound velocity in the quartz so it's in there but not in the scale of the stress. Anyway, taking limits like ##k\rightarrow 0## have to be handled correctly. It's not clear I've done that. I need to look at the TT gauge transformation in this light. So until this is understood I don't have much more to say.
     
  12. May 20, 2017 #11

    mfb

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    You can use the Post-Newtonian expansion, for example. For weak fields, expand in h. Alternatively, expand in terms of speed. Without black holes and without fast motion, the first order dominates.
    If the signal strength diverges for ##\omega \to 0##, then something is wrong. Otherwise drive your crystal with 1 µHz (doesn't even have to be resonant) and observe the bending of spacetime with the naked eye.

    I think the problem is your r-5 scaling assumption. The approximation you do might be valid for the immediate vicinity of the crystal - it is not valid for the distance of 1 wavelength. Instead of scaling down to smaller length scales, we could use r-5 scaling to find the deviation at 75 meters.

    Using your integral:
    $$h_{\mu\nu}(t,r) = -4Gc^{-4}e^{i\omega t}\int T_{\mu\nu}(r') \frac{e^{-ik|r - r'|}}{4\pi|r-r'|} d^3r'$$
    and ##T_{xy} = 3.35 MPa \cdot Q## with ##Q=100,000## we can make a quick direct estimate. Assume that the denominator is always s=4mm, assume that the numerator is always about 1 (a very good assumption for crystal sizes much smaller than 75 m), and then just plug everything in.

    $$|h_{xy}| = -4 G c^{-4} T_{xy} \frac{V}{4\pi s} \approx 6 \cdot 10^{-39}$$
    WolframAlpha

    That is 17 orders of magnitude smaller than your estimate in the first post, supporting my guess that the r-5 scaling was applied incorrectly.

    Edit: We can also see that this contribution scales with 1/r and the strain does not depend on the frequency. While strain from tension is not directly Newtonian (not a mass as source), it is still looks very similar to Newton.
    A radiative part only comes from the r'-dependent complex phase in the numerator, it is much smaller than the non-radiative term, and even that one is completely negligible.
     
    Last edited: May 20, 2017
  13. May 20, 2017 #12

    Paul Colby

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    What I've done may be wrong but what you are doing I think is worse. A gauge must be chosen in which the linear strain equations,

    ##\rho \ddot{u}_n = \partial^mT_{n m}## (*)​

    where ##n## and ##m## are summed over the spatial coordinates hold as written, or, in a gauge where these equations are modified to be consistent. This is not a small deal as the linear strain,

    ##S_{n m} = \partial_n u_m + \partial_m u_n##​

    is itself very close in form to a gauge transformation so these choices matter. Certainly the problem may be solved in any gauge but must be solved correctly in the chosen gauge. I work the TT-gauge in order that, in the absence of interatomic forces, Equations (*) or

    ##\ddot{u}_n=0##​

    holds identically. Recall I'm making the assumption,

    ##T_{\mu\nu}(r,t) = T_{\mu \nu}^\text{static}(r) + e^{i\omega t} T_{\mu \nu}(r)##​

    where I ignore entirely ##T^\text{static}##. To compute the TT gauge harmonic functions, ##\xi_\mu(r,t)## must be found so that,

    ##h^\text{TT}_{t\mu} = h_{t\mu} - \partial_t \xi_\mu - \partial_\mu\xi_t = 0.##​

    Frequent use of ##\partial^t \xi_\mu = ik\xi_\mu## and the like for ##\partial^t h_{t \mu} = ik h_{t \mu}## are used. Clearly, all the spatial components must vanish as ##k\rightarrow 0##. I'll slog through my derivation and try to either see how to take the limit or where I've blundered.
     
    Last edited: May 20, 2017
  14. May 21, 2017 #13

    Paul Colby

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    It's pretty clear what the issue with ##k\rightarrow 0## is. The back of the envelop calculation throws the volume integral under the buss for expedience and references only the dominate radial dependence of the TT-gauge green function. Liberal use of relations like,

    ##h_{t,t} = \frac{1}{k^2} \partial^n\partial^m h_{nm}##​

    on the time harmonic components have to be used with care in the 0 frequency limit. The harmonic stress energy tensor component is also constrained by a conservation relation. Basically the bottom line is the full radiation integral with a consistent sources must be done. In the 0 frequency limit all harmonic components must vanish since the ##\omega = 0## components are already in the discarded static terms. A non-vanishing term here would clearly be a double counting.

    On the one hand my not jumping to a tried and true post Newtonian formalism is justified by the extreme near field calculation I'm attempting. On the other the result must be consistent with known results. The green function I'm working with is valid to all orders in ##r## as it is a closed form expression. I've directly checked it satisfies the required scalar wave equations along with all the gauge constraints of the TT gauge. Somewhere in there is a correct answer and sure it's likely tiny.
     
  15. May 22, 2017 #14

    Vanadium 50

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    Is your mistake. Radiation is, by definition, a far field effect. This is inarguable, since it is the definition. Using a near-field approximation to estimate a far-field effect can only produce nonsense.
     
  16. May 22, 2017 #15

    Paul Colby

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    Perhaps you aren't familiar with evanescent waves? It's a common term/phenomena in many fields but I haven't seen it used in gravitation. As I've said the radiation terms are well beyond experimental reach. Here I'm raising the question about the non-radiating ones. Isn't this a fair question?
     
  17. May 22, 2017 #16

    Paul Colby

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  18. May 22, 2017 #17

    Paul Colby

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    This is a really good observation. A similar issue arrises with the ##k\rightarrow 0## limit in EM, at least with the way I did it. Consider the E-field produced by 1 amp current being driven into a 100pF capacitor at 4 MHz. Now keep the current constant by arbitrarily raising the applied voltage as the frequency is reduced. In the ##k\rightarrow 0## limit this voltage diverges as does the accumulated charge on the plates. This is the EM analog of the calculation that confused me.

    Also it's clear from the above example that I need to provide a better answer than I gave in post #3. Part of the field is clearly just a time dependent Newtonian one. Part is provided by GR. The question is what percentage is this GR part versus the "Cavendish" part. A number can and should be put on this value.
     
    Last edited: May 22, 2017
  19. May 30, 2017 #18

    Paul Colby

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    The motivation to use the TT-gauge is to preserve the Newtonian form of continuum mechanics. The time harmonic radiation integral,

    ##\bar{h}_{\mu\nu}(r) = -16\pi G \int T_{\mu\nu}(r')\frac{e^{-ik|r-r'|}}{4\pi|r-r'|}d^3r'##​

    yields a time harmonic metric strain which is not transverse or traceless. The ##T_{\mu\nu}## in the above contains just the time harmonic part of the stress energy. I've worked through 3 methods to obtain the TT-gauge strain from the one given above. These are 1) solve for the gauge transforms 2) use the gauge invariance of the curvature which leads to,

    ##h^{TT}_{\mu\nu}(r) = \frac{2}{k^2}R_{\mu 4\nu 4}(r)##​

    which by far is the most elegant. Or, 3) my way, guess and check, arguably the least defendable. All these lead to the same 4th order differential operator. All lead to the same ##(kr)^{-5}## near field dependence.

    For completeness the differential operator, ##\Delta_{nmpq}## is given where lower case latin letters are as usual restricted to 1,2,3. I use two definitions to help reduce writing,

    ##\Delta^2_{nm} = \epsilon_{nmp}\partial^p##​

    ##\Delta^1_{nm} = \partial_n\partial_m+k^2\delta_{nm}##​

    where ##\epsilon_{nmp}## is the fully antisymmetric symbol on 3 dimensions. The full operator is,

    ##\Delta_{nmpq} = \frac{1}{2k^4}(\Delta^1_{np}\Delta^1_{mq}+k^2\Delta^2_{np}\Delta^2_{mq})##​

    The TT-gauge result is computed,

    ##h^{TT}_{nm}(r) = \Delta_{nmpq} (\bar{h}_{pq}(r)).##​

    Since for space coordinates ##\eta_{nm} = \delta_{nm}## I've played fast and loose and written everything as subscripts. Doing this with the time components can and will lead to errors.

    The above completely defines where I'm getting my green function radial dependencies from. The next post will hopefully address the question in the second half of post #17.
     
    Last edited: May 30, 2017
  20. May 30, 2017 #19

    Paul Colby

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    The initial motivation for this thread was a realization that the gravitational field equations appear to say something quite counter to what I had always assumed as the case. Part of the realization comes from learning the field equation in more detail and part stems from my learning some continuum mechanics, something I always considered too icky to look at.

    The approach to the question first raised in #2 is to construct a mechanical example which 1) produces a time harmonic metric strain 2) produces identically 0 time harmonic strain or force in the Newtonian theory. The case initially given fails to do this because while the material density remains fixed the crystal boundary changes with time leading to a truly negligible time harmonic near field force.

    Consider instead a toroidal disk of inner radius, ##R_1##, and outer radius, ##R_2##, fashioned from an isotropic material like steel of thickness, ##\tau##. Let ##u(r)e^{i\omega t}## be the displacement from the reference configuration for a harmonic vibrational mode of angular frequency, ##\omega##. The mode of interest is written in cylindrical coordinates, ##{r,\phi,z}##, as

    ##u_{\phi}(r) = U(r)\hat{\phi}##​

    where ##U(r)## is the sum of two cylindrical Bessel functions such that the boundary conditions are met at ##R_1## and ##R_2## the inner and out disk radius, respectively. This mode is found by solving the Lame equation,

    ##\mu\nabla^2 u +(\mu+\lambda)\nabla(\nabla\cdot u) = 0##​

    using the separation of variables. Since the material motion is always tangent to the disk boundary, the boundary of the disk remains constant with time. Also, since ##\nabla\cdot u=0##, the density of the material is constant as is the custom with shear modes. As a result there is identically 0 time dependance of the Newtonian gravitational field in and about the disk. However the gravitational field equations appear to yield a non-zero answer. This mode yields a time dependent stress component,

    ##T_{\phi r} = \frac{du_\phi}{dr}##​

    which leads to a non-vanishing metric strain with the same ##(rk)^{-5}## dependence.
     
  21. May 30, 2017 #20

    mfb

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    I don't think physicsforums is the right place for developing the calculations. If you think it can be measured, why don't you publish it?
     
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