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Wednesday, October 3, 2012

gravity and dimensionality

would the description of gravity as a metric theory work in any number of dimensions?

bonus question: can you show without the rule of de l'Hospital or Taylor expansion that $\sin(x)/x=1$ for $x\rightarrow0$?

Yes. (Page 184 of Hobson, Lasenby and Efstathiou ...)

* In 2+1 dimensions or less, there are as many field equations as independent components of the curvature tensor, so in empty space the curvature tensor must vanish. I.e. No gravitational fields in empty space! * 3 + 1 = our universe. * 4 + 1 dimensions: Kaluza and Klein showed that the extra equations that appear when you add an extra dimension to GR are equivalent to the Maxwell equations for electromagnetism. This started physicists thinking about unifying the forces using higher dimensions, leading eventually to string theory and the like.

Sin(x) can be expressed as a generalized continued fraction of the shape: a0+b0/(a1+b1/(a2+b2/(a3+b3/ (...)))) , (based on recurrence formulas) with a0 = 0, b0 = x, a1 = 1 and the whole rest being a function of x, which goes to zero for x->0 as all the nested denominators stay finite numbers > 0. So we get: x / (1+ latter function). Thus, deviding by x yields: 1 / (1 + a function that is 0 in the limit).

There's a very intuitive derivation... from the unit circle you can read off that $\sin(x)\leq x\leq\tan(x)$, which you can divide by $\sin(x)$ and take the inverse, switching the direction of the less-or-equal-signs, yielding $1\geq\sin(x)/x\geq\cos(x)$. If you consider now the limit $x\rightarrow0$ you'll obtain the relation $1\geq\lim_{x\rightarrow0} \sin(x)/x\geq1$ which proves the limit.

mmmh, that's the point: you need the result that $\lim_{x\rightarrow0}\sin(x)/x=1$ for getting the derivative of $\sin(x)$, so you can't really use the Taylor-expansion to prove the relation...

While it's true there are no dynamical degrees of freedom in 2+1 gravity, you can do some cool stuff with conical singularities: each massive particle (that is, a source of gravity) is represented by such a conical singularity (we did some visualization here); you can even formulate a Kepler problem (e.g. here), and people have been using this weird gravity as a simple toy model for quantum gravity for some time now.

An additional consideration perhaps, in regards to differentiation and smoothness (i.e. infinite orders of differentiation), is that 'the number of ways in which to smooth a manifold, is greatest in 3 or fewer dimensions'.

Yes. (Page 184 of Hobson, Lasenby and Efstathiou ...)

ReplyDelete* In 2+1 dimensions or less, there are as many field equations as independent components of the curvature tensor, so in empty space the curvature tensor must vanish. I.e. No gravitational fields in empty space!

* 3 + 1 = our universe.

* 4 + 1 dimensions: Kaluza and Klein showed that the extra equations that appear when you add an extra dimension to GR are equivalent to the Maxwell equations for electromagnetism. This started physicists thinking about unifying the forces using higher dimensions, leading eventually to string theory and the like.

In[1]:= Limit[Sin[x]/x, x -> 0]

ReplyDeleteOut[1]= 1

Sin(x) can be expressed as a generalized continued fraction of the shape:

ReplyDeletea0+b0/(a1+b1/(a2+b2/(a3+b3/ (...)))) ,

(based on recurrence formulas) with a0 = 0, b0 = x, a1 = 1 and the whole rest being a function of x, which goes to zero for x->0 as all the nested denominators stay finite numbers > 0. So we get: x / (1+ latter function). Thus, deviding by x yields: 1 / (1 + a function that is 0 in the limit).

There's a very intuitive derivation... from the unit circle you can read off that $\sin(x)\leq x\leq\tan(x)$, which you can divide by $\sin(x)$ and take the inverse, switching the direction of the less-or-equal-signs, yielding $1\geq\sin(x)/x\geq\cos(x)$. If you consider now the limit $x\rightarrow0$ you'll obtain the relation $1\geq\lim_{x\rightarrow0} \sin(x)/x\geq1$ which proves the limit.

ReplyDeleteTaylor expand Sin(x) about zero.

ReplyDeletemmmh, that's the point: you need the result that $\lim_{x\rightarrow0}\sin(x)/x=1$ for getting the derivative of $\sin(x)$, so you can't really use the Taylor-expansion to prove the relation...

ReplyDeleteCouldn't you just get the derivatives of the sine from its complex exponential form and then proceed with l'Hôpital / Taylor series?

DeleteWhile it's true there are no dynamical degrees of freedom in 2+1 gravity, you can do some cool stuff with conical singularities: each massive particle (that is, a source of gravity) is represented by such a conical singularity (we did some visualization here); you can even formulate a Kepler problem (e.g. here), and people have been using this weird gravity as a simple toy model for quantum gravity for some time now.

ReplyDeleteAn additional consideration perhaps, in regards to differentiation and smoothness (i.e. infinite orders of differentiation), is that 'the number of ways in which to smooth a manifold, is greatest in 3 or fewer dimensions'.

ReplyDelete