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Consider two masses $m_1$ and $m_2$ that are connected by a spring. Mass 1 follows the worldline $x_1(\tau)$ while mass 2 follows $x_2(\tau)$. Note that the argument $\tau$ is the proper time in the rest frame of mass 1 and in the rest frame of mass 2 respectively but we do not use a different symbol for both.

I want to calculate how the system reacts to an incident gravitational wave (GW).

Both worldlines fulfill the geodesic equation:

$$m_1\frac{d^2x_1^\mu}{d\tau^2}=f^\mu_1-m_1\Gamma^\mu_{\nu\lambda}(x_1(\tau))\frac{dx_1^\nu}{d\tau}\frac{dx_1^\lambda}{d\tau}$$

$$m_2\frac{d^2x_2^\mu}{d\tau^2}=f^\mu_2-m_2\Gamma^\mu_{\nu\lambda}(x_2(\tau))\frac{dx_2^\nu}{d\tau}\frac{dx_2^\lambda}{d\tau}$$

where $f^\mu_1$ and $f^\mu_2$ are forces that act on the different masses.

How do we calculate the four-forces? Do we need a specific frame for that?

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What is the freely falling system? Weber writes in his paper that it is the center of mass between the two masses. Why should the center of mass be freely falling in the presence of a GW?

In an old-fashioned Newtonian picture, you could say a "freely-falling" system has no forces acting on it besides gravity. As responsibly grown-up relativists, we should say a "freely-falling" system means one following a geodesic in space-time. Since gravitational waves are part of the metric, particles moving with a gravitational wave, without being acted on by an external force, are also following geodesics and so are also freely falling.

For me this looks wrong since in a freely falling system Γ=0 and then the geodesic equations would look different?

As implied by my comment above, $\Gamma$ need not be zero for a freely falling observer.

I didn't follow all of your algebra, but I am sure your last term $\partial_\alpha \Gamma^\mu_{\nu \rho} L^\alpha$ cannot be correct. The reason is that you should have an equation between tensors. The left hand side $\frac{D^2 \xi}{D \tau^2}$ is a tensor, and the first term $\sim R x^2$ is a tensor, but $\partial \Gamma L$ is not a tensor. Therefore, this equation can at best hold in one coordinate system. However there's nothing in your calculation that picks out a special coordinate system that explains why you would get a term that's not a tensor.

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  • $\begingroup$ Thanks for your answer! My understanding of freely falling is that the gravitational effects are eliminated in a local coordinate system in which non-gravitational forces (such as the spring force) can still act on the test mass? And this impies $\Gamma=0$ in this system. This is actually also confirmed in the Weber paper "We use coordinates in which the Christoffel symbols vanish..." $\endgroup$
    – jojo123456
    Aug 5 at 6:09
  • $\begingroup$ I agree the last term looks odd. I will let you know if I find the mistake $\endgroup$
    – jojo123456
    Aug 5 at 6:10
  • $\begingroup$ @jojo123456 There might just be a difference in terminology. For me, a "freely falling system" would be an object or group of objects following geodesics (that you could describe in any coordinate system), while "locally intertial coordinates" or maybe "freely falling coordinates" at a point $x$ would have the metric equal to the flat space metric and the Christoffel symbols vanishing at $x$. We probably agree on the substance. $\endgroup$
    – Andrew
    Aug 5 at 21:55
  • $\begingroup$ I think there must be a more explicitly covariant way to do this calculation, I suspect that will be less error prone or at least make it more transparent what your extra term is. For example, you might be better off starting from the geodesic deviation equation to derive an equation for $D^2 \xi / D\tau^2$. $\endgroup$
    – Andrew
    Aug 5 at 22:00
  • $\begingroup$ This is exactly what Weber is doing in his paper. Have a look. He starts from the geodesic equation (his equation (10)) and then he splits the vector ($\eta^\mu$) which fulfills the geodesic equation into a constant and a varying part $\eta^\mu=r^\mu+\xi^\mu$. This point which I don't understand is, why the double covariant derivative of the constant vector $\frac{\delta ^2r^\mu}{\delta s^2}=0$. One needs this to go from Webers equation (10) to equation (13). I don't understand this!? $\endgroup$
    – jojo123456
    Aug 5 at 22:41

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