Saturday, January 19, 2008 ... Deutsch/Español/Related posts from blogosphere

Numerical BFSS matrix model & black holes

Press releases can occassionally be useful.

Japanese KEK
has pointed out a work of the kind I always wanted to do.
Jun Nishimura et al. (Phys Rev Lett)
have numerically calculated the properties of non-supersymmetric D0-like black holes in the maximally supersymmetric BFSS matrix model. The most difficult part is the strong coupling that is relevant for the Schwarzschild black holes which is what the D0-branes become.



They have used powerful computers and flexible algorithms to optimize their calculation and the resulting energy-temperature relation agrees with gravity even at strong coupling and even though the agreement cannot be guaranteed by any supersymmetric non-renormalization theorems because the whole setup breaks supersymmetry. Of course, I have never had any doubts that it would work but it is cool that one can actually do it. They can now literally calculate how the black holes are composed out of stringy objects.



Their calculation is not a lattice gauge theory approach but rather a sophisticated Fourier expansion over time with the Polyakov lines used as the key order parameter whose nonzero vev at all temperatures shows that supersymmetry removes the phase transition, as expected (in contrast to a purely bosonic model).




I completely missed the paper when it appeared in the summer - otherwise it would have been included in 2007 in theoretical physics.

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reader Anonymous said...

Lubos,

I was wondering if you could help figure out a puzzle.

Various authors have put forth the idea that space should be viewed as a collection of finite volumed points.

I have a problem with this picture because if it were true, for a massive object; when it moved in a circle, it would move along a segmented path.

It would seem to me that an object could appear moving in a circle at the speed of light, when in reality it was moving at less than the speed of light (since the distance around the segmented circle is less than the circle itself)

Now I know this is a simple geometric view of things; but as I develop my understanding of strings I still haven't found any discussion of this, and I can't seem to find anyone smart enough to explain it. Is it purely a misconception? Is there an easy explanation that I haven't found?

Since your one of the few people in physics that seem interested in truth and not fairy tales, I was hoping you could shed some light on this; warmest regards.


reader Anonymous said...

Lubos,

I think I figured this one out, but any confirmation of my conclusions, or revelation about my stupidity would be appreciated.

It seems that the answer is that the object in question would occasionally have to jump to points outside the apparent circular path, this would give it the opportunity to "self correct" the distance traveled along the segments.


reader Anonymous said...

oh yeah, and the path would look like a tube and not a circle...go string theory


reader Luboš Motl said...

Dear Hal S, I think that you are absolutely correct and yours is one of several arguments to show that space cannot be separated into finite volumes in this way. It cannot be discretized.

One would get contradictions with continuous symmetries, the limiting of light, Lorentz invariance, rotational symmetry, and vanishing of the entropy of the vacuum.

At very short distances, the geometry as we know it and this new effect is associated with a new distance scale - analogous to the length of the minimum volumes in the toy model you mention.

That's why we can talk about the discretization at the popular level. However, at the literal level, the discretized picture is physically inconsistent. It must still be possible for physics to explain that space is kind of continuous, in order for all the continuous things to work.

Let me offer an analogy. Phase space - whose coordinates are "x" and momentum "p" - is a smooth 2D space. In quantum mechanics, it is effectively clumped into volumes of size of Planck's constant h, because of quantum mechanics. But it is done in such a fine way that nothing about the concepts you can derive in the continuous setup - such as continuous symmetries - ever fails.

There are people who imagine that the popular picture of the discretization is literally true but these people just don't understand physics well.


reader Anonymous said...

Thanks Lubos!

I might have more questions in the future but out of courtesy and respect I will try to limit them in quantity and to simple logical contradictions like this one.

Thanks again!

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