Sergei Winitzki's GR Book: Examining Errors in Conformal Transformations

[bcrowell]

Sergei Winitzki has a very nice copylefted GR book: http://sites.google.com/site/winitzki/index/topics-in-general-relativity (direct PDF link: http://sites.google.com/site/winitzki/index/topics-in-general-relativity/GR_course.pdf?attredirects=0&d=1 ) I'm puzzling over his treatment of conformal transformations.

On p. 85, near the bottom left corner of the page, he has
\tilde{r}(i^o)=\lim_{r\rightarrow\infty}e^\lambda r^2=\lim_{v\rightarrow -\infty, u\rightarrow\infty}f'(u)f'(v)\frac{(u-v)^2}{4}.
Comparing with the expressions near the top of the column, it seems to me that the e^\lambda should clearly be e^{2\lambda}. I emailed him about this, hoping that I wasn't just making some dumb mistake.

After this, there is another thing that seems wrong to me, but considering that my understanding of GR is far inferior to his, I'm a little hesitant to shoot off another email. He says suppose that f'(s) goes like s^-n for large s; then the limit in the third expression above should approach zero if n\ge 2. But shouldn't the condition be n\ge 1??

If I'm right about the second error, then it's a more substantial hole in his analysis. Basically you want i^o to be pointlike (a 2-sphere of radius 0) and points on scri+ to be non-pointlike (each such point represents a 2-sphere of finite radius). The best motivation I know of for wanting these to be this way is that it allows you to make a tiling of Penrose diagrams to cover the static Einstein universe. (Is there some more fundamental reason?) So the condition on i^o requires n\ge 1, and the condition on scri+ forces n\le 2. But then this leaves a whole range of possible exponents between 1 and 2 (contrary to Winitzki's analysis, which fixes the exponent uniquely at 2). Does this seem right?

One would then need some other reason to pick n=2. I think that if you want f' to be analytic and positive-definite, then that does rule out n=1, or any other value besides n=2. Does this make sense?

 

[arkajad]

Surely it should be e^{2\lambda}. As for n, for n>1 the limit is zero, for n<1 the limit is infinite. For n=1 the limit will probably depend on the direction (speed) you approach the infinity.

 

[George Jones]

On p. 85, near the bottom left corner of the page, he has
\tilde{r}(i^o)=\lim_{r\rightarrow\infty}e^\lambda r^2=\lim_{v\rightarrow -\infty, u\rightarrow\infty}f'(u)f'(v)\frac{(u-v)^2}{4}.

After this, there is another thing that seems wrong to me, but considering that my understanding of GR is far inferior to his, I'm a little hesitant to shoot off another email. He says suppose that f'(s) goes like s^-n for large s; then the limit in the third expression above should approach zero if n\ge 2. But shouldn't the condition be n\ge 1??

As for n, for n>1 the limit is zero, for n<1 the limit is infinite. For n=1 the limit will probably depend on the direction (speed) you approach the infinity.

Suppose \left( u , v \right) = \left( e^t , t \right) and f'\left(s\right) = s^{-n}. Then,

<br /> u^{-n} v^{-n} \frac{\left( u - v \right)^2}{4} = e^{-nt} t^{-n} \frac{\left( e^t - t \right)^2}{4}.<br />

 

[arkajad]

Suppose \left( u , v \right) = \left( e^t , t \right) and f&#039;\left(s\right) = s^{-n}.

Small correction: In Winitzki's text he wants:

"... corresponding to r\rightarrow\infty\quad (u\rightarrow\infty,\, v\rightarrow -\infty)."

 

[George Jones]

In Winitzki's notes he writes:

"... corresponding to r\rightarrow\infty\quad (u\rightarrow\infty,\, v\rightarrow -\infty).

Oops. I didn't see the minus sign in -\infty at the bottom of the limit above. And I just got my eyes checked and new glasses a couple of months ago.

 

[George Jones]

Suppose \left( u , v \right) = \left( e^t , -t \right) and f&#039;\left(s\right) = s^{-n}. Then,

<br /> u^{-n} v^{-n} \frac{\left( u - v \right)^2}{4} = \left(-1\right)^{-n} e^{-nt} t^{-n} \frac{\left( e^t + t \right)^2}{4}.<br />

[edit]The (-1)^(-n) is not nice, but I think that this suggests that n >= 2 is needed[/edit]

 

[arkajad]

[edit]The (-1)^(-n) is not nice, but I think that this suggests that n >= 2 is needed[/edit]

Winitzki has

e^{2\lambda}=f&#039;(u)f&#039;(v)

Therefore the product f&#039;(u)f&#039;(v) should always be positive. Anyway the result goes to zero for any n&gt;1. Since Wintzki probably assumes n to be an integer (though I do not know why?), it starts with n=2 or higher.

And I just got my eyes checked and new glasses a couple of months ago.

I have them self-checked and have about 20 pairs of different glasses, each for about $3.

 

[bcrowell]

Thanks very much, arkajad and George Jones, for your replies!

When Winitzki says f' "tends to zero as s^-n," he could mean (a) that f' has limiting behavior that is that same as that of s^-n, up to a constant factor of proportionality, with a relative error that approaches zero. On the other hand, he could mean (b) that f' is bounded above by something of the form k|s^-n|, and that n is the largest real number such that you can make a bound like this. In case b, we could have possibilities like f'(s)=1/|s|, so that the sign is not an issue. We would probably like f' to be C^∞, but then we can still make up possibilities like f&#039;(s)=1/\sqrt{1+s^2}, which is not analytic.

There is also the question of whether he intends the limit to be independent of the direction from which you approach i^o. In the following paragraph, he talks about approaching a particular point on \mathscr{I}+ along a particular light ray. If you approach i^o along a spacelike curve, I think only n\ge 1 is required. George's <br /> \left( u , v \right) = \left( e^t , -t \right)<br /> is timelike.

Trying to put all this together, here's what I get. He has two physical requirements, one on i^o and one on \mathscr{I}+. The latter seems to be all that is really required. If you want every point on \mathscr{I}+ to represent a 2-sphere with a nonzero, finite radius, then this should certainly be true in the case where you approach that point along a lightlike geodesic, and therefore we need n=2 exactly. It seems like we would also like the same finite-r limit if we approach a point on \mathscr{I}+ along any other curve, and I don't know if this holds or not. Once we've fixed n=2, it also follows that r=0 at i^o, regardless of the curve along which the limit is taken.

-Ben