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Setup

Consider the metric for a spherically symmetric and isotropic spacetime: $$ds^2 = -B(r)c^2dt^2 + A(r)dr^2 + r^2d\Omega^2.\tag{$\Delta$}$$ With this metric we can find (with the geodesic equation) the following equation's, that characterize the movement of a particle/photon $$\begin{align*} \theta &= \pi/2,\\ r^2 \frac{d\phi}{d\lambda} &= const. \equiv l,\\ B(r)\frac{d t}{d\lambda} &= const. \equiv F,\\ A(r) \left(\frac{d r}{d\lambda}\right)^2 + \frac{l^2}{r^2}-\frac{F^2}{B(r)}&= const. \equiv -\varepsilon, \end{align*}$$ where the value of $\varepsilon$ is either $0$ (photon) or $c^2$ (particle). If we specialize to the Schwarzschild case we find $A=1/B$ and $B=1-r_s/r$, $r_s\equiv GM/c^2$. In this case we can bring everything together to $$\frac{1}{2}\left(\frac{d r}{d\lambda}\right)^2 + V_{\rm eff}(r) = const.\quad V_{\rm eff}(r)\equiv -\frac{a\varepsilon}{r}+\frac{l^2}{2r^2}-\frac{al^2}{r^3}.$$

Question

I have trouble understanding the relationship between $t$ and $\lambda$. As far as I understand eq. $(\Delta)$ is from the perspective of an external observer, meaning that an external observer chooses a coordinate system $(t,r,\theta,\phi)$ and describes the motion of the object in question as the evolution from some initial spacetime point $(t_0, r_0, \theta_0, \phi_0)$ parametrized by $\lambda$. Let us now for example say that we are dealing with a massive particle, then we can choose $\lambda$ to be the proper time of the particle $\tau$ (I think at least we can do this). We then have $dr/d\tau$, which I would read as the radial velocity of the particle as function of the time measured in the reference frame of the particle. But what if I would like to know the radial velocity in my reference frame, aka $dr/dt$? To calculate that I would need $d\tau/dt$, which can't be determined in general (or can it? I think it should be possible if we assume that the particle moves only in radially.) So, what I would like to know is what exactly the connection between the two quatities $t$ and $\lambda$ is.

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  • $\begingroup$ rs=2GM/c², not GM/c² $\endgroup$ – Gendergaga Jan 18 '20 at 23:22
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For convenience, I'll be using units $c=1$ and only consider a massive particle.

As mentioned by Ben, coordinates are largely arbitray. However, there may exist a timelike coordinate, for example $x^0 = t$ with $$g_{00} < 0$$

Note that I said may exist, as it is not a given that one of your coordinates will be time-like (eg 1+1 Minkowski space can be spanned by two null vectors instead of a time-like and a space-like one).

If there is a time-like coordinate $x^0$, it will define a family of observers with 4-velocities $$ \mathbf{u} = \frac1{\sqrt{-g_{00}}}\partial_0 $$

From the perspective of this family of observers, a particle with 4-velocity $\mathbf{\dot x} = d\mathbf x/d\tau$ will have a relative velocity (as expressed by its Lorentz factor) of $$ \gamma = -g(\mathbf{\dot x},\mathbf u) $$ If $g$ has no cross-terms, this is $$ \gamma = -g_{00}\, {\dot x}^0\, u^0 $$ or in your case $$ \gamma = B\cdot \frac {dt}{d\tau} \cdot \frac1{\sqrt B} = \frac F{\sqrt B} $$ where we have used your equations of motion to substitute $$ \frac{dt}{d\tau} = \frac FB $$ This last equation determines the relationship between coordinate time and eigentime of the particle. However, this coordinate time is 'unphysical', ie does not correspond to the eigentime of any observer as $g_{00} \not= -1$ (cf the lapse function of the ADM formalism).

Finally, one should also note that a family of observers does not necessarily integrate to a family of worldlines at fixed spatial coordinates: If $g$ has cross-terms, $\mathbf u$ would induce spatial motion (cf the shift vector of the ADM formalism).

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As far as I understand eq. (Δ) is from the perspective of an external observer, meaning that an external observer chooses a coordinate system

No, that isn't how it works. Coordinate systems do not relate to observers, and GR doesn't have global frames of reference. See How do frames of reference work in general relativity, and are they described by coordinate systems?

But what if I would like to know the radial velocity in my reference frame, aka dr/dt?

No, you don't have a global reference frame, and dr/dt isn't interpreted that way. Coordinates like $r$ and $t$ are just labels. A derivative like dr/dt isn't really a velocity. It has no special physical significance.

So, what I would like to know is what exactly the connection between the two quatities t and λ is.

There is no special connection, and t does not have the role of a time coordinate in some global frame of reference. There is no such thing as a time coordinate in general relativity. Basically t is just a label. In the Schwarzschild coordinates, it just happens to have some properties that make it similar to a time coordinate.

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  • $\begingroup$ First of all, thank you for the answer! I'm having a hard time wrapping my head around the statement "dr/dt isn't really a velocity". Why exactly should this be true? And if, how comes that it is treated sometimes as such. This question came up in my head when I was trying to solve exercise 5.5 in Carroll's book and found this solution. Eq. (8) and (9) there refer explicitly to $dr/dt$ and $dr/d\tau$ as being velocities. What makes these expressions different from my example above? $\endgroup$ – Sito Jan 18 '20 at 20:58
  • $\begingroup$ but you can have a family of observers/vierbein/section of the frame bundle/local trivialization of the tangent bundle/... $\endgroup$ – Christoph Jan 18 '20 at 21:34
  • $\begingroup$ @Christoph I'm sorry, but I can't follow you.. Could you maybe explain in a bit more detail what you mean.. $\endgroup$ – Sito Jan 18 '20 at 22:09
  • $\begingroup$ In your example dr/dt is the shapirodelayed velocity, and dr/dτ is the coordinate celerity. If you want the local velocity relative to a fiducional observer who is at rest to the black hole see the last 3 lines at yukterez.net/f/einstein.equations/files/2.html $\endgroup$ – Gendergaga Jan 18 '20 at 23:27

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