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B Toy GW detector model

  1. Aug 24, 2016 #1

    Paul Colby

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    A simple harmonic oscillator model consisting of two masses connected with a massless spring is provided in

    [edit] I hate cut and past on the mac. Always seems broken
    Slide 30
    http://webs.um.es/bussons/GW_lecture_KG.pdf

    The left hand side of the equation of motion is simply the classical one while the driving term on the right is the contribution of a weak GW on an otherwise flat background. My question is how is the driving term computed? Normally I would start with a relativistic Lagrangian and expand the metrics therein. Not having this I would assume that the Hook law force is ##-ks## where ##s## is the (proper) distance between the masses. When illuminated by a GW this distance becomes ##s = \xi+L_oh(t)##, ##L_o## being the equilibrium mass separation. Clearly this doesn't yield the answer given which I assume is correct.
     
    Last edited: Aug 24, 2016
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  3. Aug 24, 2016 #2

    PeterDonis

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    It's the second time derivative of the deviation from equilibrium of the proper distance between the test masses, due to the GW. In other words, it's the "force" (per unit mass) exerted by the GW on the test masses.

    (Btw, it's worth noting that this toy model doesn't apply as it stands to interferometric detectors, i.e., to LIGO, VIRGO, LISA, etc. In those detectors, the test masses are not connected by springs and their motion, at least as far as detecting GWs goes, is intended to be free fall.)
     
  4. Aug 24, 2016 #3

    Paul Colby

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    Thanks. I'm really missing something basic here. In the weak limit a particle at rest in the lab stays at rest. This is because the ##\Gamma^\mu_{\alpha,\beta}## all involve the particles coordinate velocity and the metric deviation has only space components. So, to first order,

    ##\frac{d^2x^i}{dt^2} = 0##

    A GW comes by and each mass will experience a force due to the connecting spring no longer being it's equilibrium length. Each mass feels the force of the other through a compressed or extended spring. This adds a term to the spring force, not an inertial one like the derivative term suggests. What am I missing??

    Your comment about the interferometers is spot on with my understanding. The mirror are suspended inertially so the stay at rest in the lab coordinates. It's the distance between them that is modulated causing the light to fringe shift.
     
  5. Aug 24, 2016 #4

    PeterDonis

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    I don't understand what you mean here. The ##\Gamma## coefficients involve derivatives of the metric, which means derivatives of the metric perturbation ##h##. They have nothing to do with coordinate velocity.

    If you are referring to the terms involving ##\Gamma## in the geodesic equation, those do not involve coordinate velocity; they involve the derivatives of coordinates with respect to the particle's proper time, i.e., components of the particle's 4-velocity. But in this toy model, when a GW passes the test masses do not move on geodesics; see below.

    It is the spring force. And, as you say, this force is felt by each mass; i.e., each mass is no longer moving on a geodesic.

    I don't understand what you mean here either. The response of each test mass to the spring force will be determined by the inertia of the test mass and the spring constant. What's the issue?
     
  6. Aug 24, 2016 #5

    Paul Colby

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    Could be because I'm in error. But, here is the logic. A point particle not acted upon by a force follows the equation of a geodesic.

    ##\frac{d^2x^\mu}{ds^2}+\Gamma^\mu_{\alpha,\beta}\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds} = 0##

    In our coordinates and for our wave in the TT gauge we may take ##s = t##. The 4-velocity of a point at rest in our coordinates is ##\frac{dx^\mu}{ds} = (1,0,0,0)##
    When I put all these into the Gamma term I get zero coordinate acceleration for all time. With no spring the particles follow geodesics. With a spring they are pushed by the spring. How does this spring force acquirer a second order time derivative. The only candidate is the ##\Gamma^x_{t,t}## and that's zero (I believe)?
     
  7. Aug 24, 2016 #6

    Paul Colby

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    Yep I checked ##\Gamma^x_{t,t}=0## for a weak GW in the TT gauge
     
  8. Aug 24, 2016 #7

    PeterDonis

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    Yes, as long as no GW is passing.

    I'm still not sure what the issue is. When the spring force is acting (because a GW is passing), the particles don't follow geodesics. So the geodesic equation does not describe their motion. So the fact that ##\Gamma^x{}_{tt} = 0## doesn't mean the particles must stay at constant coordinates for all time. And they don't.
     
  9. Aug 24, 2016 #8

    Paul Colby

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    Well, the geodesic equation is an ODE. At ##t = 0## let's say the GW arrives and the initial condition is ##\frac{dx}{dt}=0## for the velocity to begin to change the acceleration must be non-zero. The acceleration is a quadratic function of the coordinate velocity for a geodesic in this example. There is no way for an isolated mass to acquire a coordinate velocity (in these coordinates). Now, all this changed when there is a spring connecting the two masses and we seem to agree on the general mechanism. What I find confusing is how this (coordinate) force could become proportional to the second derivative of the metric strain as shown on the chart.

    Now, the geodesic equation I quoted above actually appears in non-geodesic motion as,

    ##\frac{dx^{\mu}}{ds^{2}}+\Gamma{^\mu}_{\alpha\beta}\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=F^\mu##
    where ##F^\mu## is the applied 4-force at least in the references I've read over the years

    For free inertial masses I believe this is the case in this example. If not I have a major gap in my understanding that I need to chase down. If I look at charts 10 and 11 in the talk it even indicates the inertial point particle locations when illuminated by a GW are ##\pm x_o## which is independent of time according to my understanding.

    Anyway thanks for the help. I've been confused on this issue for a number of years so a few more probably won't hurt :frown:
     
  10. Aug 24, 2016 #9

    PeterDonis

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    More precisely, a test object that starts out at rest in these coordinates, and always moves on a geodesic (i.e., always feels no force), will never acquire any coordinate velocity in these coordinates; it will stay at the same spatial coordinates forever. So any change in its proper distance from any other similar test object, also at rest in these coordinates and moving on a geodesic, must be due to a change in the metric itself, i.e., to a GW passing (assuming that's the only kind of spacetime curvature we're considering).

    Ah, I see; you're wondering how the second derivative of the coordinate ##x^\mu## gets "translated" into the second derivative of the metric strain. See below.

    The first term here has a typo, it should be ##d^2 x^\mu / ds^2##. With that correction, yes, this equation is the "force equation" for non-geodesic motion, given a choice of coordinates so that the ##\Gamma##s are well-defined. (The only caveat is that the "force" here is actually force per unit mass, i.e., acceleration.) And at the start, when the GW is just starting to pass by and there is no strain, the term in ##\Gamma## is zero (because the only nonzero 4-velocity component is the ##t## component and there are no nonzero ##\Gamma##s with lower indexes ##tt##). So the "force equation" is just

    $$
    \frac{dx^{\mu}}{ds^{2}} = F^\mu
    $$

    What is ##dx^\mu / ds^2##? Let's suppose that we put one test mass at ##x^i = 0## for ##i = 1, 2, 3##, and a second test mass at ##x^1 = L##, ##x^2 = x^3 = 0##; i.e., we align the two masses and the spring along the ##x## axis. The proper distance between the two masses is then ##D = \sqrt{g_{xx}} L##, which before the GW passes is just ##L##, since ##g_{xx} = 1## if no GW is present.

    Now, if the spring were not there, so that the test masses were both moving geodesically, then the proper distance between them while the GW is passing would be ##D = \sqrt{1 + h_{xx}} L##. For small ##h## we can write this as ##D = \left( 1 + \frac{1}{2} h_{xx} \right) L##. The second derivative of this is then ##d^2 D / ds^2 = \frac{1}{2} L d^2 h_{xx} / ds^2##. If we set ##h_{xx} = h_+ e^{i \omega t}##, then we get the expression on the RHS of the equation you referred to in your OP.

    Now consider what happens if we put the spring in. The expression for ##d^2 D / ds^2## above defines what geodesic motion would be; i.e., it defines the motion that the test masses would have if there were no spring. So with the spring present, that same expression defines the "driving force" that the spring is resisting. For example, if the spring had an infinite spring constant, so that it maintained the two test masses at exactly the same proper distance ##L## no matter what, then the equation above would give the coordinate acceleration of the masses, in the coordinates we're using, where geodesically moving test masses would have zero coordinate acceleration.

    This is all heuristic, but to delve much deeper into it would require more math than is appropriate for a "B" level thread.
     
  11. Aug 24, 2016 #10

    Paul Colby

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    Hey thanks. My motivation in assignment of this thread to "B" level is my belief that I might be missing something conceptual. Your physical description is completely in line with what I understand even if the mathematics is lacking[1]. The proper response for me is to start a new thread with a derivation of the equation of motion based on the understanding I do have. This has to lead to an answer which is at variance with what's in the charts. With all the steps laid out it should be an easy matter for people to point to the step that's wrong and set me straight.

    [1]The conceptual question I have with what you said in #9 is why are you taking the second derivative of ##D## with time to proper time to obtain the force due to the spring? This isn't what one does to compute Hook's law which is the source of the force in the first place. So, this is a conceptual question not a mathematical one. Doing as you suggest implies the force of the spring on the mass depends on the rate of change of the rate of change of the proper length of the spring. This makes no sense to me.
     
  12. Aug 24, 2016 #11

    PeterDonis

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    I didn't. I did that to obtain the driving force, the force that the spring is resisting. The force that the spring itself exerts is ##- k \delta D##, where ##\delta D## is the change in proper distance from the equilibrium value. In the equation of motion in the PDF you linked to in the OP, that is what gives rise to the ##\omega_0^2 \xi## term on the LHS. The term I took the second derivative of ##D## to obtain is the driving force on the RHS of that equation.

    In other words, conceptually, the change in the metric due to the GW passing acts just the same as if we had grabbed the test masses and pulled them apart or pushed them together. If we modeled that kind of driving force, we would do it, conceptually, using ##F = m a##, i.e., the mass of the test mass times its acceleration--i.e., the second time derivative of the displacement. The point is that the GW works, conceptually, the same way--the change in the metric, from the point of view of the spring, is "grabbing" the test masses and pulling them apart or pushing them together, and the spring responds with its own force to that change. So the driving force (per unit mass) due to the GW is just the "acceleration" due to the change in the proper distance caused by the change in the metric, i.e, the second time derivative.
     
    Last edited: Aug 24, 2016
  13. Aug 24, 2016 #12

    Paul Colby

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    Okay, I found a paper written by the guy who gave the talk. The term on the right is the tidal force which is proportional to the second time derivative and is clearly the force you're speaking of. So, I clearly don't understand the principles, thanks. I need to look more at the geodesic deviation equations to really understand. I recall reading gobs about it in MTW but never really forgave them for the whole "Bongs of a Bell" mantra.
     
  14. Aug 24, 2016 #13

    PeterDonis

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    Yes. Calling it a "force" is somewhat confusing, though, because, as I discussed in previous posts, an object that actually fully responds to the "force" will feel no force at all--this would be a test mass that moves on a geodesic, and stays at the same coordinate position, and whose proper distance from another similar test mass changes exactly as the metric changes. Whereas an object that does not fully respond to the "force"--such as the test masses connected by the spring, which are constrained by the spring--will actually feel a force because whatever is constraining it, the spring in this case, is making it move on a non-geodesic worldline.
     
  15. Aug 25, 2016 #14

    Paul Colby

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    But this does leave me with a conundrum. In the limit of no spring there is a huge driving force but no work done on the masses. That's spooky. My goal is to be able to do a continuum dynamics calculation for a complex structure. If it's this difficult to see for two point masses I have little hope of getting through such a calculation. The answer is bound to be simple like add curvature tensor times the mass density somewhere....you wouldn't happen to just know the answer would you...
     
  16. Aug 25, 2016 #15

    PeterDonis

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    This is why the term "driving force" can be confusing. It's not really a force in the GR sense, because objects moving solely under its influence (i.e., no spring) feel no force, are weightless, and move on geodesics. But it can appear similarly to an ordinary driving force in an equation of motion, with an appropriate choice of coordinates.

    No, it isn't. It's complicated. AFAIK there is no general closed form solution known for a continuous distribution of matter with a GW passing through it; it has to be solved numerically.
     
  17. Aug 25, 2016 #16

    Paul Colby

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    Well, to be fair having thought on your comments (which have been very helpful, thanks) it is just like gravity strangely enough. Good old garden variety gravitation is an acceleration "field" which in Newtonian terms applied a force per unit volume of ##\rho \hat{g}## where ##\rho## is the mass density.

    Alway been my intension to use a FEM code like FreeFEM which allows one to define the equations being solved. It's starting to look like I need to add something like ##\rho\ddot{h}_{nm}## as a (stress?) field to the continuum equations of linear elasticity equations.
     
  18. Aug 25, 2016 #17

    PeterDonis

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    Just like Newtonian gravity, yes. Which is weird since we're dealing with a phenomenon that is not even present in Newtonian gravity.

    That would seem like a reasonable starting point, yes. Unfortunately I'm not familiar enough with the literature in this area to know if this case has already been studied, or what model was used if it has.
     
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