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Consider the following line element:

$$ds^{2} = K(x,y,z,t)(-dt^2+dx^2)+M(x,y,z,t)dxdt+dy^2+dz^2$$

Then the lagragian method give to us the lagrangian from line element:

$$\mathcal{L}^2 = K(x,y,z,t)(-\dot{t}^2+\dot{x}^2)+M(x,y,z,t)\dot{x}\dot{t}+\dot{y}^2+\dot{z}^2 \tag{1}$$

where, for example, $\dot{x} := \frac{dx}{d\lambda}$ and $\lambda$ is an affine parameter.

Given now one of the four Lagrange equations:

$$\frac{\partial \mathcal{L}^2}{\partial t}-\frac{d}{d\lambda}\Big( \frac{\partial \mathcal{L}^2}{\partial \dot{t}} \Big) = 0$$

Consider just a part of this expression:

$$-\frac{d}{d\lambda}\Big( \frac{\partial \mathcal{L}^2}{\partial \dot{t}} \Big)$$

then, $$-\frac{d}{d\lambda}\Big( \frac{\partial}{\partial \dot{t}} [K(x,y,z,t)(-\dot{t}^2+\dot{x}^2)+M(x,y,z,t)\dot{x}\dot{t}+\dot{y}^2+\dot{z}^2] \Big) = -\frac{d}{d\lambda}\Big( -2K\dot{t}+M\dot{x}\Big) = 2K\ddot{t}-M\ddot{x} \implies $$

$$ 2K\ddot{t}-M\ddot{x} = 0 \tag{2}$$

My question is: both $(1)$ and $(2)$ are valid? I mean, did I calculate right?

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    $\begingroup$ You seem to have overlooked implicit $\lambda$-dependence. $\endgroup$ – Qmechanic Dec 30 '18 at 18:18
  • $\begingroup$ But the expression (1) is ok? $\endgroup$ – M.N.Raia Dec 30 '18 at 18:23
  • $\begingroup$ No I don’t think so.. I’ll answer when I’m on my computer $\endgroup$ – InertialObserver Dec 30 '18 at 19:52
  • $\begingroup$ If you want to calculate the geodesics, your ansatz is worng. you have to use the metric and calculate the geodesics with Christoffel symboles. $ G=\left[ \begin {array}{cccc} -K \left( t,x,y,z \right) &1/2\,M \left( t,x,y,z \right) &0&0\\ 1/2\,M \left( t,x,y,z \right) &K \left( t,x,y,z \right) &0&0\\ 0&0&1&0 \\0&0&0&1\end {array} \right] $ $\endgroup$ – Eli Dec 31 '18 at 10:55
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    $\begingroup$ @Eli no, he is using the geodesic Lagrangian method. If done properly, it works. See my answer below. $\endgroup$ – magma Dec 31 '18 at 18:13
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The Lagrangian is $$ L= \frac{1}{2} g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$$

and the Euler/Lagrange eqs. are $$\frac{d}{d\lambda}\frac{\partial L}{\partial(dx^\mu/d\lambda)} = \frac{\partial L}{\partial x^\mu} $$

Now the part in the $t$ component that you considered

$$-\frac{d}{d\lambda}\Big( \frac{\partial}{\partial \dot{t}} [K(x,y,z,t)(-\dot{t}^2+\dot{x}^2)+M(x,y,z,t)\dot{x}\dot{t}+\dot{y}^2+\dot{z}^2] \Big) = -\frac{d}{d\lambda}\Big( -2K\dot{t}+M\dot{x}\Big) = 2K\ddot{t}-M\ddot{x} +2\frac{dK}{d\lambda} \dot{t}-\frac{dM}{d\lambda}\dot{x}$$ Where the total derivatives $\frac{dK}{d\lambda}$ and $\frac{dM}{d\lambda}$ are calculated as $\frac{\partial K}{\partial x^\alpha}\frac{d x^\alpha}{d\lambda}$ and $\frac{\partial M}{\partial x^\alpha}\frac{d x^\alpha}{d\lambda}$ , respectively.

So , to answer directly your questions:

(1) is almost right (2) was definitely incomplete (and therefore wrong), since you forgot the indirect dependence of K and M on $\lambda$

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