Does a wristwatch measure imaginary time?

1. Jan 22, 2006

DiamondGeezer

This is from "Exploring Black Holes" by Taylor and Wheeler. It's a very good book but I struggle not with the math, but the explanations (sometimes)

On page B-13 is a frame called "Metric for the Rain Frame", which is a transformation of the Schwarzschild Metric from "bookkeeper coordinates" to shell coordinates to "rain coordinates"

All well and good.

Here is the final equation for the rain frame:

$$d \tau^2 = (1-\frac {2M}{r}) dt^2 - 2( \frac{2M}{r})^\frac{1}{2} dt_{rain} dr -dr^2 -r^2 d \phi ^2$$

The text continues:

1. Yes, the metric has got rid of the infinity at $$r=2M$$ but
when r < 2M all of the parts of the right hand side must be negative because $$1 - \frac {2M}{r} < 0$$ ​
this means that $$d \tau^2 < 0$$ which means that $$d \tau$$ is imaginary

2. When r is slightly greater than 2M, then in order for the wristwatch of the plunger to record real time then

$$(1-\frac {2M}{r}) dt^2 > 2( \frac{2M}{r})^\frac{1}{2} dt_{rain} dr -dr^2 -r^2 d \phi ^2$$

Now I interpret part (2) to mean that regardless of initial conditions, any infalling object or person must cross normally to the event horizon.

But (1) puzzles me. There may be no "jolt" but how does a wristwatch measure imaginary time?

Perhaps the hands bend at $$90^ \circ$$ to the plane of the watchface...

Last edited: Jan 22, 2006
2. Jan 22, 2006

George Jones

Staff Emeritus
I don't have either the book or my notes on the book with me right now, and I forget the exact context of this, but I can still make a couple of general comments.

First, along the wordline of any observer inside the event horizon, $dr < 0$, i.e., $r$ must decease along the wordline of any observer inside the horizon. Consequently, not all the terms are negative.

Secondly, if the metric does become negative, this means that it's measuring proper space, not imaginary proper time.

Regards,
George

3. Jan 22, 2006

robphy

My books are still in boxes. So, I can't get to my copy now.
Is $$d\tau^2$$ in (1) measured along an observer's worldline?

4. Jan 22, 2006

pervect

Staff Emeritus
This appears to be a variant of Eddington-Finklestein coordinates. However, I'm not clear on the details, not having the book you cite.

Ingoing EF coordiantes have a metric that looks like this:

$$ds^2 = -(1-2M/r) dV^2 + 2 dV dr + r^2(d\theta^2 + \mathrm{sin^2} \theta d \phi^2)$$

here dt has been removed entirely, replaced by the ingoing EF coordinate V which I suspect is your "rain" coordinate, leaving r, theta, and phi as the other coordinates.

I suspect that your metric is actually a mixed metric, having five differentials (r, theta, t, V, and phi), though you don't show a theta term (?).

Anyway, in the ingoing EF coordinate system, V is a constant for radially infalling light, which is why it might be called a 'rain' coordinate. If someone beamed numbered pulses of light at regular intervalsl from infinity directly towards the black hole (i.e. having the light travel a strictly raidal path), the number of the pulse you are currently receiving would be the V coordinate.

As far as r and t inside the event horizon go, r becomes timelike (a time coordinate), and t becomes a space coordinate, which is probably why you are getting imaginary time. Your time coordinate is really now a space coordinate. You can see this directly from the usual form of the Schwarzschild metric.

5. Jan 22, 2006

George Jones

Staff Emeritus
These are Painleve-Gullstrand coordinates. See here. I believe they're also discussed in Poisson's book.

Regards,
George

6. Jan 22, 2006

DiamondGeezer

The so-called "rain frame" is the frame of an observer free-falling into a spherical non-rotating black hole.

The metric is a variant of the Schwarzschild metric.

I'm not saying that Taylor and Wheeler are wrong. I have a little theory that both space and time beyond the event horizon are imaginary.

It's that Taylor and Wheeler claims that nothing unusual happens when you cross the event horizon, but even their modified metric shows that something weird happens.

7. Jan 22, 2006

pervect

Staff Emeritus
Someone looking at their wristwatch is always going to read a real number, not an imaginary or complex one.

While you may be attached to your own theories, try thinking about the traditional explanation (r being timelike inside the event horizon, and t being spacelike) to see if you can understand that point of view.

8. Jan 23, 2006

DiamondGeezer

Ok, lets go with that. Are you saying that time and space swap roles inside the event horizon? That a wristwatch is recording r and not t ?

9. Jan 23, 2006

pervect

Staff Emeritus
Yes, odd as it may seem, the Schwarzschild coordinate 'r' is timelike inside the event horizon. This is why one cannot avoid the singularity with rockets - no matter how hard you thrust, time advances on, and the singularity at the center of the black hole is not some distance away, but some time (in the future) away.

MTW has a rather nice quote on this, I will try and see if I can dig it up.

10. Jan 23, 2006

George Jones

Staff Emeritus
I am now looking at the book.

You left out the subscript "rain" on $t$ the $dt^2$ term. This led to confusion in this thread.

From page B-12: "Instead we used the time $t_{rain}$ measured on the wristwatch of a single in-falling rain observer."

So, along the worldline of an infalling observer, $t_{rain}$ is a measure of proper time $\tau$. Consider a radially infalling observer, so that $d\phi = 0$ and $d\tau = dt_{rain}$.

What do you get when you when you substitute these into the metric. You should end up with equation [2] on page B-6. Note that everything is consistent, and that along this wordline, $d\tau^{2} > 0$.

Regards,
George

11. Jan 27, 2006

DiamondGeezer

That would be good. Thanks

12. Jan 28, 2006

pervect

Staff Emeritus
OK, I found it - it's on pg 823. (MTW is a huge book, and it's index isn't the best :-().

Last edited: Jan 28, 2006