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Given a manifold $\mathcal{M}$ with coordinates $\psi : \mathcal{M} \rightarrow \mathbb{R^4}$ , $\Psi(p)= (r,\theta ,\phi,t)$ for $ p \in \mathcal{M}$

Suppose we have the active transformation $F : \mathcal{M} \rightarrow \mathcal{M}$ in coordinates given by

$$\Psi \circ F\circ \Psi^{-1}(r,\theta ,\phi,t)=(f(r),\theta ,\phi,t)$$

where is $f$ a function in $\mathbb{R}$

Given the Schwarzschild metric $$ \Psi \circ g\circ \Psi^{-1}(r,\theta ,\phi,t) =-\left(1-\frac{1}{ r}\right)dt^2+\left(1-\frac{1}{ r}\right)^{-1}dr^2+r^2\left(d \theta^2 +\sin^2 \theta d \phi^2\right) \;. $$ we can induce another metric $h$ by

$$\Psi \circ h\circ \Psi^{-1}(r,\theta ,\phi,t)=\Psi \circ F^*g\circ \Psi^{-1}(r,\theta ,\phi,t)=-\left(1-\frac{1}{ f(r)}\right)dt^2+\left(1-\frac{1}{ f(r)}\right)^{-1}f'(r)^2dr^2+f(r)^2\left(d \theta^2 +\sin^2 \theta d \phi^2\right) \;.$$

where $f'(r)=\frac{df(r)}{dr }$

Then for a curve $\gamma(\tau )=(2,0,0,\tau) \in \mathcal{M} $ we would have

then $h(\gamma'(\tau), \gamma'(\tau))=1-\frac{1}{ f(2)}$ so we have
$$\int_{[0, 1]} h(\gamma'(\tau), \gamma'(\tau))d\tau = -\int_{[0, 1]} \left(1-\frac{1}{ f(2)}\right)d\tau $$

where $\gamma'(\tau)=\frac{d\gamma}{d\tau }$

and $$\int_{[0, 1]} g(\gamma'(\tau), \gamma'(\tau))d\tau = \frac{1}{ 2} $$

Since they both satisfy the Enstein equation and they do not give the same line element, how to know which solution should we chose for a given problem.

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  • $\begingroup$ I'm not sure I understand your question. The two solutions are equivalent, so you can use either one. Are you worried because the integrals give different values? I don't know why you expect them to, and anyway the Schwarzschild coordinates don't work at $r=1$. $\endgroup$ – Javier Oct 15 '19 at 15:38
  • $\begingroup$ If the two solutions are equivalent the integrals should equal because the length of a curve associated with a particle has the meaning of time $\endgroup$ – amilton moreira Oct 15 '19 at 16:06
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Assuming that $f$ is well behaved, so that the transformation is a diffeomorphism, then this is the same spacetime described in different coordinates.

Since they both satisfy the Enstein equation and they do not give the same line element, how to know which solution should we chose for a given problem.

They don't give the same line element because they're expressed in different coordinates. Similarly, the line element for the Euclidean plane in Cartesian coordinates is different from the line element expressed in polar coordinates. Which coordinate system to use is a matter of convenience.

These would not normally be considered distinct solutions to the field equations. They're the same solution. The choice of coordinates for a given problem could be based on what coordinates make the metric look simpler, or which ones respect the symmetry of the spacetime.

If the map isn't a diffeomorphism, then things can get more nontrivial. For example, the external Schwarzschild spacetime can be extended through the coordinate singularity at the horizon, and can also be extended to the maximal extension, which includes an Einstein-Rosen bridge.

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    $\begingroup$ They are expressed in same coordinate. $F$ is an active transformation $\endgroup$ – amilton moreira Oct 15 '19 at 15:07
  • $\begingroup$ I will edit my question $\endgroup$ – amilton moreira Oct 15 '19 at 15:16
  • $\begingroup$ @amiltonmoreira: The distinction you're making is a physically vacuous one. It doesn't matter whether you regard this as an active transformation or a passive one. $\endgroup$ – user4552 Oct 15 '19 at 19:16

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