How Does the Schwarzschild Metric Transform in Kruskal-Szekeres Coordinates?

[yuiop]

I am trying to understand a derivation posted by Pervect a long time ago (I suspect Pevect is no longer active) that involves differentiation and I was hoping someone here could fill in some of the steps to make it clearer.

The original posts by Pervect are here:

https://www.physicsforums.com/newreply.php?do=newreply&p=1019121

https://www.physicsforums.com/showpost.php?p=1020560&postcount=5


I have taken the liberty of posting a cut down version of Pervects posts here. I have added equation numbers and comment. The comments in blue are just notes and the comments in magenta are the places where I am stuck and need some more explanation.

Let us start with the initial Schwarzschild metric:

$$(Eq1.1) \hskip 1 cm ds^2 = -(1-2M/r) dt^2 + 1/(1-2M/r) dr^2 + r^2 d\Omega^2$$

Now make the following variable substitutions. (These are from Wald, not that it particularly matters BTW).

(Eq1.2) (r/2M - 1) exp(r/2M) = X^2 - T^2
(Eq1.3) (t/2M) = ln [(X+T)/(X-T)]

It is useful to know at this point that the last two equations are easily obtained from the Kruskal coordinates which are defined as:

$$X=e^{(r/2M)}*\sqrt{(r/2M-1)}*cosh(t/4M)$$

$$T=e^{(r/(2M))}*\sqrt{(r/(2M)-1)}*sinh(t/4M)$$

REf: http://en.wikipedia.org/wiki/Kruskal-Szekeres_coordinates

Note that these are of the form

(Eq1.4) (X+T)(X-T) = f(r)
(Eq1.5) (X+T)/(X-T) = g(t)

hence we can solve them for X+T = sqrt(f*g) and X-T = sqrt(f/g)

So from Eq 1.2 and Eq1.3

$$X+T= \sqrt{(r/2M - 1)} e^{(r/4M)}e^{(t/4M)}$$

$$X-T= \sqrt{(r/2M - 1)} e^{(r/4M)}e^{(-t/4M)}$$

We can then write:

$$(Eq1.6) \hskip 1 cm dr = 4\, \left( 2\,X{\it dX}-2\,T{\it dT} \right) {M}^{2} \left( {e^{{ \frac {r}{2M}}}} \right) ^{-1}{r}^{-1}$$

$$(Eq1.7) \hskip 1 cm dt = 2\,M \left( 2\,X{\it dT}-2\,T{\it dX} \right) \left( {\frac {r}{ 2M}}-1 \right) ^{-1} \left( {e^{{\frac {r}{2M}}}} \right) ^{-1}$$

OK, this is one part I need help with. How does Pervect obtain dr and dt in Eq1.6 and Eq1.7?
Is he using ordinary or partial differential equations?
Is his result correct?
I am not too good with differentials so do not worry about over explaining ;)

Now we can re-write the Schwarzschild metric, with r(X,T) implicictly defined by

(Eq1.8) (r/2M - 1) exp(r/2M) = X^2 - T^2

as

$$(Eq1.9) \hskip 1 cm \frac{32 M^3 e^{r/2M}}{r}(-dT^2 + dX^2)+r^2 d\Omega^2$$

There is a typo in the Kruskal metric here. Eq1.9 should read:

$$(Eq1.9b) \hskip 1 cm \frac{32 M^3 e^{-r/2M}}{r}(-dT^2 + dX^2) + r^2 d \Omega^2$$

REf: http://en.wikipedia.org/wiki/Kruskal-Szekeres_coordinates

We see that the new expression is perfectly finite in the new variables X,T at r=2M (which is at X=T), removing the coordinate singularity at the event horizon (r=2M, or X=T).

However, a singularity remains at r=0. We know we can't eliminate that because the curvature scalar diverges.

In short, a "simple" (it's simple with computer algebra, anyway) variable substitution eliminates the singularity at the event horizon.

 

[yuiop]

Part 2

I had to split my original post into two because the latex was not displaying properly.

This is the second post where Pervect attacks the problem from a slightly different angle, but I am still missing the trick where he obtains the differentials.

Phase 1: The Kruskal metric - the gory details.

Piggybacking a bit on a previous post
https://www.physicsforums.com/showpost.php?p=1019121&postcount=67

we consider the line element

$$(Eq2.1) \hskip 1 cm ds^2 = \frac{32 M^3 e^{r/2M}}{r}(-dT^2 + dX^2) + r^2 d \Omega^2$$

Again, the typo in the Kruskal metric is still here. Eq2.1 should read:

$$(Eq2.1b) \hskip 1 cm \frac{32 M^3 e^{-r/2M}}{r}(-dT^2 + dX^2) + r^2 d \Omega^2$$

REf: http://en.wikipedia.org/wiki/Kruskal-Szekeres_coordinates

where r = r(X,T) = 2M*LambertW((X^2-T^2)/e) + 2M

The LambertW(x) is defined for the domain x>=-1/e, and is a monotonically increasing function that satisfies

LambertW(x)*exp(LambertW(x)) = x

(Note that LambertW(-1/e) = -1)

also
(Eq2.2) d (LambertW(x))/dx = LambertW(x) /[x (1+LambertW(x))]

This definition is equivalent to the perhaps conceptually simpler implicit definition of r(X,T) given by the relationship

(Eq2.3) (r/2M - 1)* exp(r/2M) = X^2 - T^2

The first defintion is convenient for computer algebra, though, which is a practical necessity for what follows. It *could* in principle be done by hand, by someone sufficiently dedicated, I suppose, but would be extremely time consuming.

We can confirm by direct (computer-aided) calculation that the Ricci tensor for the for line element Eq2.1 is zero.

Thus the Kruskal metric represents a vacuum solution for Einstein's field equations.

The range of coordinates for which we have a non-singular metric is just X^2 - T^2 > -1 (equality is not allowed). This is equivalent to r > 0.

Now consider the diffeomorphism

(X,T) -> (r,t)

where r = r(X,T) = 2M*LambertW((X^2-T^2)/e) + 2M
and t = 2M ln[(X+T)/(X-T)]

These are obtained from the Kruskal coordinates as defined in the earlier post with reference to Eq 1.2 and Eq1.3

Applying this diffeomorphism to the well-behaved Kruskal metric above yields the apparently singular Schwarzschild metric, i.e line element:

$$(Eq2.4) \hskip 1 cm ds^2 = dr^2 / (1-2*M/r) + dt^2 (-1 + 2M/r) + r^2 d\Omega^2$$

This can be demonstrated by substituting

(Eq2.5) dr = (dr/dX)*dX + (dr/dT)*dT
(Eq2.6) dt = (dt/dX)*dX + (dt/dT)*dT

Should that be:
(Eq2.5b) 2*dr = (dr/dX)*dX + (dr/dT)*dT
(Eq2.6b) 2*dt = (dt/dX)*dX + (dt/dT)*dT

or are Eq2.5 and 2.6 correct and that is just a quirk of how partial differential equations work?

into the line element of the Schwarzschild metric Eq2.4 and finding that it reduces to

$$(Eq2.7) \hskip 1 cm ds^2 = 16\,{\frac { \left( {{\it dX}}^{2}-{{\it dT}}^{2} \right) {M}^{2}{\it LambertW} \left( \left( {X}^{2}-{T}^{2} \right) {e^{-1}} \right) }{ \left( {\it LambertW} \left( \left( {X}^{2}-{T}^{2} \right) {e^{-1}} \right) +1 \right) \left( {X}^{2}-{T}^{2} \right) }}$$

which can be further simplified to the line element (Eq2.1) by using the expression for X^2-T^2 previously given.

I have no problem in converting the last equation Eq2.7 into the Schwarzschild line element Eq2.1 but I am not sure how Pervect obtains the equations for the differentials Eq2.5 and Eq2.6 and what they evaluate to individually.

I tried this method using maple 12:

Start with the simplified radial motion only Schwarzshild metric:

$$dS^2 = \left(1-\frac{2m}{r} \right)^{-1} (dr)^2 - \left(1-\frac{2m}{r} \right)(dt)^2$$
and then squaring Eq2.5 and Eq2.6 to obtain expressions for dr^2 and dt^2 that can be directly inserted into the Schwarzschild metric to obtain:

$$dS^2 = \left(1-\frac{2m}{r} \right)^{-1}\left(\frac{dX}{\frac{d}{dr}X}+\frac{dT}{\frac{d}{dr}T}\right)^2 - \left(1-\frac{2m}{r} \right)*\left(\frac{dX}{\frac{d}{dt}X}+\frac{dT}{\frac{d}{dt}T}\right)^2$$

which after simplification eventually spits out:

$$\frac{32 m^3 e^{(-r/2M)}(-dT^2 + dX^2)}{r}* \frac{1}{(cosh(t/4M)^2-1)cosh(t/4m)^2}$$

This is basically the Kruskal metric but has additional terms with the variable t in it. How do I eliminate the t variable?

This diffeomorphism is non-singular only for (X^2 - T^2) > 0, i.e. X>T, which is equivalent to r > 2M

We can thus see that the Schwarzschild metric has a non-singular diffeomorphism to the Kruskal metric [1] in the region (X^2 - T^2) > 0, which is equivalent to the region (r > 2M), a region which is a subset of the entire manifold spanned by the Kruskal metric.

It is the ill-behavior of t(X,T) that leads to apparent ill-behavior in the Schwarzschild metric at r=2M.

Thus, as promised, we have found a diffeomorphism from a subset of the manifold covered by Kruskal coordinates to the Schwarzschild metric.

We can map the interior part of the Schwarzschild solution (0<r<2M) into the Kruskal metric as well, by the same set of diffeomorphisms, as long as we omit the event horizon (r=2M) itself.

We can furthermore see that the ill-behavior of the Schwarzschild metric was induced by the ill-behavior of the diffeomorphism that defined the 't' coordinate. The diffeomorphism that defines the 'r' coordinate remains well-behaved, which is why coordinates such as Eddington-Finklestein which use the r coordinate and an additional coordinate (usually T+X/2 or T-X/2) are possible.

Obviously I am missing a trick somewhere. Can anyone enlighten me?

Maybe this should have been posted in the Relativity forum, but I am only interested in the mathematical aspects that seem to require the knowledge of a expert in differential equations so I posted it here.

 

[HallsofIvy]

I am trying to understand a derivation posted by Pervect a long time ago (I suspect Pevect is no longer active) that involves differentiation and I was hoping someone here could fill in some of the steps to make it clearer.

The original posts by Pervect are here:

https://www.physicsforums.com/newreply.php?do=newreply&p=1019121

https://www.physicsforums.com/showpost.php?p=1020560&postcount=5


I have taken the liberty of posting a cut down version of Pervects posts here. I have added equation numbers and comment. The comments in blue are just notes and the comments in magenta are the places where I am stuck and need some more explanation.

OK, this is one part I need help with. How does Pervect obtain dr and dt in Eq1.6 and Eq1.7?
Is he using ordinary or partial differential equations?
Is his result correct?

He is not using differential equations here but just the definition of "differential"

From
$$X+ T= (\frac{r}{2m}-1)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}$$
take the differential of both sides. On the right, we need to apply the product rule to a product of three terms: Since the first and second are functions of r and the third term a function of t,we basically get d(f(r)g(r)h(t))= f'(r)g(r)h(t)dr+ f(r)g'(r)h(t)dr+ f(r)g(r)h'(r)dt.

More specifically,
$$dX+ dT= \frac{1}{2}\left(\frac{r}{2m}-1\right)^{-1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}dr$$
$$+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{r}{4m}}dr$$
$$+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{r}{4m}}dt$$

Do the same with the "X- T" equation and solve the two equations for dr and dt.

 

[cristo]

For the next part: eqns 2.5 and 2.6 are correct, and just arise from the definition of a differential: $$dr=\frac{\partial r}{\partial X}dX+\frac{\partial r}{\partial T} dT$$

Your last question: I'm not sure what you're asking. One can evaluate the above, since one has an expression of r in terms of X and T, so can take the derivatives and plug into the above.

 

[yuiop]

He is not using differential equations here but just the definition of "differential"

Hi,

thanks for the help and sorry for the delay responding. I have spent the afternoon trying to learn differential calculus, product rules and multivariate calculus from the internet. I had no idea there was so much to it I can not honestly say I understand 10 percent of it but I hope I have picked up enough to address this problem. I found the following resources very helpful:

Multivariate calculus
Rules of calculus
Calculus - Wikibook
Product rule - Wikipedia

Back to the problem in hand.

From
$$X+ T= (\frac{r}{2m}-1)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}$$
take the differential of both sides. On the right, we need to apply the product rule to a product of three terms: Since the first and second are functions of r and the third term a function of t,we basically get d(f(r)g(r)h(t))= f'(r)g(r)h(t)dr+ f(r)g'(r)h(t)dr+ f(r)g(r)h'(r)dt.

More specifically,
$$dX+ dT= \frac{1}{2}\left(\frac{r}{2m}-1\right)^{-1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}dr$$
$$+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{r}{4m}}dr$$
$$+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{r}{4m}}dt$$

Applying the product rule to (X+T) with respect to r:

$$X+ T= (\frac{r}{2m}-1)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}$$

I get:

$$\begin{align*} dX+ dT &= \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{-1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}(dr)&\\ &+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}(dr)&\\ &+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}(dt)& (Eq 4.1) \end{align*}$$

Do the same with the "X- T" equation and solve the two equations for dr and dt.

O.K.

$$X- T= (\frac{r}{2m}-1)^{1/2}e^{\frac{r}{4m}}e^{\frac{-t}{4m}}$$

$$\begin{align*} dX- dT &= \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{-1/2}e^{\frac{r}{4m}}e^{\frac{-t}{4m}}(dr)&\\ &+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{-t}{4m}}(dr)&\\ &- \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{-t}{4m}}(dt)& (Eq 4.2) \end{align*}$$

Solve (Eq 4.1) for $$e^{\frac{t}{4m}}$$ :

$$e^{\frac{t}{4m}} = \frac{4m(dX+dT)}{e^{\frac{r}{4m}}\left[ dr\left(\left(\frac{r}{2m}-1\right)^{1/2}+\left(\frac{r}{2m}-1\right)^{-1/2}\right) + dt\left(\frac{r}{2m}-1\right)^{1/2}\right] }}$$

and do the same for (Eq 4.2):

$$e^{\frac{t}{4m}} = \frac{e^{\frac{r}{4m}}\left[ dr\left(\left(\frac{r}{2m}-1\right)^{1/2}+\left(\frac{r}{2m}-1\right)^{-1/2}\right) - dt\left(\frac{r}{2m}-1\right)^{1/2}\right] }{4m(dX-dT)}$$

Since both the above equations are equal to $$e^{\frac{t}{4m}}$$ they can be equated to each other to eliminate the pesky t term and by applying the difference of squares rule the following is obtained:

$$16m^2(dX^2-dT^2) = e^{\frac{r}{2m}}\left[dr^2\left(\left(\frac{r}{2m}-1\right)^{1/2}+\left(\frac{r}{2m}-1\right)^{-1/2}\right)^2 -dt^2 \left(\frac{r}{2m}-1\right) \right]$$

By inspection it is obvious that the last equation is close to a solution and this suggests a shortcut that avoids having to find dr and dt individually.

Multiply both sides by $$\frac{2m }{ r} e^{\frac{-r}{2m}}$$ :

$$\frac{32m^3 e^{\frac{-r}{2m}} }{r} (dX^2-dT^2) = dr^2\left(1-\frac{2m}{r}\right)^{-1} -dt^2 \left(1-\frac{2m}{r}\right)$$

which is the Kruskal-Szekeres metric on the left hand side and the Schwazschild metric on the right hand side.

 

[yuiop]

For the next part: eqns 2.5 and 2.6 are correct, and just arise from the definition of a differential: $$dr=\frac{\partial r}{\partial X}dX+\frac{\partial r}{\partial T} dT$$

Your last question: I'm not sure what you're asking. One can evaluate the above, since one has an expression of r in terms of X and T, so can take the derivatives and plug into the above.

Hi cristo,

could you flesh this out a bit and explain why this form is used here in this particular example or point to some online reference material? I'm afraid I'm still not "getting it".

Is it because the cooordinate equations for R and T are being treated as parametric equations with respect to r and t for the purposes of differentiation?

I think we have one solution thanks to help from HallsofIvy but I never feel comfortable with a solution until I can get it from at least two different angles. Sorry if I being obtuse about all this but differentiation (at this level of complexity) is all fairly new to me.


<EDIT>
This is how it seems to me but I readily admit I am probably out of my depth here.

Starting with:

$$(X +T ) = (X+T)$$

Differentiating both sides with respect to r:

$$\frac{d}{dr} (X+T) = \frac{\partial X}{\partial r} + \frac{\partial T}{\partial r}$$

$$(dX + dT) = \left( \frac{\partial X}{\partial r} + \frac{\partial T}{\partial r} \right) dr$$

$$\frac{ (dX + dT) }{\left( \frac{\partial X}{\partial r} + \frac{\partial T}{\partial r} \right)} = dr$$

$$\partial r \frac{ ( dX + dT) }{\left( \partial X + \partial T \right)} = dr$$

$$dr = \frac{ \partial r }{\left( \partial X + \partial T \right)}dX + \frac{ \partial r }{\left( \partial X + \partial T \right)}dT$$

...then again..maybe you can not throw differential symbols around like that..

 

[yuiop]

For the next part: eqns 2.5 and 2.6 are correct, and just arise from the definition of a differential: $$dr=\frac{\partial r}{\partial X}dX+\frac{\partial r}{\partial T} dT$$

Your last question: I'm not sure what you're asking. One can evaluate the above, since one has an expression of r in terms of X and T, so can take the derivatives and plug into the above.

OK, I have finally figured out what you were trying to tell me here. Sorry for being so slow I have posted a few corrections and a complete solution below which I hope will clear up any confusion for anyone hoping to learn anything from this thread. Thanks again Christo (and HallsofIvy) for the pointers in the right direction


I now realize I was on completely the wrong track in my last post and that post should be disregarded... I had the wrong end of the stick

This is the correct way to do it:

Starting with the definition of the Kruskal-Szekeres coordinates:

$$X=e^{(r/4M)}(r/2M-1)^{1/2}cosh(t/4M)$$
$$T=e^{(r/4M)}(r/2M-1)^{1/2}sinh(t/4M)$$

(there was a typo in post #1)

Solving these two equations for r gives:

$$r = 2M LambertW((X^2-T^2)/e) + 2M$$

Note: The lambertW function is sometimes know as the ProductLog or Omega function. Solving for r with complicated expressions like this is best done with mathematical software such as this free online one here: http://www.quickmath.com/

Now dr is found by finding the sum of the partial differential of (2M*LambertW((X^2-T^2)/e) + 2M) with respect to X plus the partial differential of (2M*LambertW((X^2-T^2)/e) + 2M) with respect to T using the product rule

$$dr=\frac{\partial r}{\partial X}dX+\frac{\partial r}{\partial T} dT$$

(Exactly as cristo said)

$$dr=\frac{4M LambertW((X^2-T^2)/e)(XdX-TdT)}{(LambertW((X^2-T^2)/e)+1)(X^2-T^2) }$$


The first two coordinate equations can also be solved for t in terms of X and T to give:

$$t = 2M ln[(X+T)/(X-T)]$$

and as before

$$dt=\frac{\partial t}{\partial X}dX+\frac{\partial t}{\partial T} dT$$

$$dt = \frac{4m(XdT-TdX)}{(X^2-T^2) }$$

Now that expressions for dr and dt in terms of X and T have been obtained they can be plugged into the Schwarzschild metric

$$dS^2 = \left(1-\frac{2M}{r} \right)^{-1} (dr)^2 - \left(1-\frac{2M}{r} \right)(dt)^2$$

to obtain:

$$dS^2 = \frac{16M^2 (dX^2-dT^2) LambertW((X^2-T^2)/e)}{(LambertW((X^2-T^2)/e)+1)(X^2-T^2)}$$

which is the expression originally obtained by Pervect. (Anyone know what happened to Pervect? I miss his knowledgeable contributions to PF )

By substituting (r/2m-1)exp(r/2m) for (x^2-T^2) and (r/2m-1) for LambertW(X^2-T^2)/e) into the last equation as suggested by Pervect, the Kruskal-Szekeres metric is obtained:

$$dS^2 = \frac{32 M^3 e^{-r/2M}}{r}(-dT^2 + dX^2)$$