I have read explanations of this but haven't really understood. Given a spacetime $(M,g)$ I have read that if I represent the metric in some coordinates $(x,y,z,t)$ as $g(x,y,z,t)$ and then in another coordinate system as $g'(x',y',z',t'),$ that $g'(x,y,z,t)$ (now using the old coordinates) will also solve the Einstein equations. Now $g$ and $g'$ are two different metrics on the manifold and so should predict different physics, but somehow they are the same? If I measure the distance between two points on the manifold I should get different answers if the metrics are different, shouldn't I? This is the part I don't really understand. Is it something like $(M,g')$ isn't a solution, but only $(M',g')$ where $M'$ is diffeomorphic to $M$?

To give an explicit example, let's look at the Schwarzschild metric: \begin{equation} ds^2=-\left(1-\frac{2GM}{c^2 r}\right)c^2dt^2+\left(1-\frac{2GM}{c^2 r}\right)^{-1}dr^2+r^2\left(d \theta^2 +\sin^2 \theta d \phi^2\right) \;. \end{equation} Apparently (according to http://faculty.luther.edu/~macdonal/HoleArgument.pdf) \begin{equation} ds^2=-\left(1-\frac{2GM}{c^2 f(r)}\right)c^2dt^2+\left(1-\frac{2GM}{c^2 f(r)}\right)^{-1}f'(r)^2dr^2+f(r)^2\left(d \theta^2 +\sin^2 \theta d \phi^2\right) \; \end{equation} is also a solution for any diffeomorphism $f$ though distances aren't the same. Of course when deriving the Schwarzschild metric it seems (according to Carroll's book) $(r,\theta,\phi,t)$ are just symbols that you only interpret after you find the metric, which is confusing. Note I didn't do a change of coordinates here, I changed the actual metric. Are they really both solutions on the same manifold?

  • $\begingroup$ Why do you think changing the coordinate system changes the metric? It changes only its coordinate expression. $\endgroup$ – ACuriousMind Jul 18 '14 at 18:26
  • $\begingroup$ If you move to a different inertial coordinate system, you should expect to measure distances differently. The metric is transformed as a function of the velocity vector. This is in keeping with the concepts of length contraction and time dilation. $\endgroup$ – KidElephant Jul 18 '14 at 18:32
  • $\begingroup$ @ACuriousMind I didn't just change the coordinate system, I changed the metric also, ie. if $p$ is a point on the manifold, then we have $g'(p)\ne g(p)$. $\endgroup$ – JLA Jul 18 '14 at 18:36

Changing the coordinates of a metric does not change the underlying physics, nor the solution. A metric $g$ describing a manifold $M$, if under diffeomorphisms is transformed to a new metric $g'$ it will still describe the original $M$.

  • $\begingroup$ But if $p$ is a point on the manifold, then we have $g'(p)\ne g(p)$. $\endgroup$ – JLA Jul 18 '14 at 18:35

When you change coordinates, you just choose a different atlas for $\mathcal{M}$. This does not change the metric $g$, as it is defined independent of coordinates.

Now, your Schwarzschild example has little to do with this, since there you are talking about a diffeomorphism $t : \mathcal{M} \rightarrow \mathcal{M'}$. Yet, if $\mathcal{M'}$ is not already endowed with a metric, your natural choice for a metric $g'$ on it is

$$ g'(\mathrm{d}t(X),\mathrm{d}t(Y)) := g(X,Y) $$

which means requiring that $t$ is an isometry.

So, when you choose $t$ to be an automorphism of $\mathcal{M}$, and only act upon the coordinate $r$ as a smooth $f(r)$, the induced metric on the target manifold (which is $\mathcal{M}$ itself) is exactly the second metric you wrote down if the source manifold is endowed with the usual Schwarzschild metric. As these two are isometric, they describe the same physics. In particular, it implies that $g'$ is also a solution to the Einstein equations. Thus, both $(\mathcal{M},g)$ and $(\mathcal{M},g')$ are allowed spacetimes.

So, I just skimmed the paper, and, as it turns they really just do a coordinate change: They change the radial coordinate as $f(r') = r$. So, you've the metric given w.r.t. $(t,r,\phi,\theta)$ by your first equation, and you've the metric given w.r.t. $(t,r',\phi,\theta)$ by your second equation. They're the same, since, for any point $p$, the coordinates are related as $f(r') = r$. They are not two different metrics, just different coordinate expression.

  • $\begingroup$ So take two points $p,p'$ on $M$. Using $g, g'$ you measure two different distances between them, but since they are isometric there is no observable difference in the physics? $\endgroup$ – JLA Jul 18 '14 at 19:02
  • $\begingroup$ To get the exact same physics, you must use $p,p'$ with $g$ and $t(p),t(p')$ with $g'$. $(\mathcal{M},g)$ and $(\mathcal{M},g')$ are two different things, you can't plug in points on the one into the metric of the other. $\endgroup$ – ACuriousMind Jul 18 '14 at 19:07
  • $\begingroup$ I don't understand that, why do you call them the same thing if they are two different things? You seem to have two different metrics on the same manifold, hence should be able to use the same points. $\endgroup$ – JLA Jul 18 '14 at 19:13
  • 1
    $\begingroup$ But they only mention the coordinate transformation thing later, on page two they seem to be literally considering different metrics on the same manifold. They even write $G(r)$ and $G'(r)$, and since they are using the same letter I can only assume that $r$ represents the same point on the manifold, resulting in two different metrics. Which part of this isn't right? $\endgroup$ – JLA Jul 18 '14 at 19:55
  • 1
    $\begingroup$ On page 2: "Define a coordinate change $f(r') = r$.". They then proceed to consider $g$ and $g'$ as two different metrics on $\mathcal{M}$, as by my isometry argument, but they don't say that plugging $p$ into $g$ and $g'$ considered that way gives the same results: Case in point, on page 3, top: "The solutions $g$ and $g'$ are physically distinguishable". Their writing isn't the best, that much I will say. $\endgroup$ – ACuriousMind Jul 18 '14 at 20:01

Now g and g′ are two different metrics on the manifold and so should predict different physics, but somehow they are the same?

The field equations require four coordinate conditions for a unique solution.

For the Schwarzschild line element, a coordinate condition is that the surface area of each sphere is $4\pi r^2$.

For the line element of the 2nd solution you provide, a different coordinate condition is used; the surface area of each sphere is $4\pi f^2(r)$ so this line element belongs to a different spacetime manifold but one that is a static solution for a vacuum spherically symmetric spacetime.

From the linked paper:

General relativity considers $G$ and $G'$ to live on different spacetime manifolds, say $M$ and $N$. To distinguish them, rename their $r$ coordinates to $r_M$ and $r_N$.

Map an event $E_N \in N$ at $r_N$ to the event $E_M \in M$ with $r_M = f (r_N)$, with the other coordinates unchanged. Then the transformation maps $G'$ to $G$.

  • $\begingroup$ Well I'm glad you see what I mean. So then are the two spacetimes distinguishable? I mean, can we tell which one we "live" in? I think part of my confusion arises from the fact that the coordinates don't seem to be defined before finding for the metric. $\endgroup$ – JLA Jul 18 '14 at 22:42
  • $\begingroup$ @JLA, of the 10 independent metric components, 6 describe the geometry in a coordinate independent way leaving 4 coordinate degrees of freedom so one can use these coordinate degrees of freedom to choose coordinates in which the line element is in its simplest possible form. Solving the field equations in these coordinates determines the functional form of the metric components in this coordinate system. One can perform a coordinate transformation from $r$ to $r'$ such that the 2nd line element is in the same form as the first showing that the coordinate independent geometry is identical. $\endgroup$ – Alfred Centauri Jul 18 '14 at 23:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.