Fun Problem: Field Theory on S2!

**BenTheMan** · 10-30-08, 02:21 PM

Ok, I worked this problem out the other night. It's kind of cute, and straightforward for a scalar field.

Consider field theory on $\mathbb{R}^4\times S^2$ . Given the six dimensional Klein-Gordon equation for a massless scalar field:

$\box_6 \Phi = 0$ ,

derive the Kaluza-Klein mode expansion.

This problem is pretty easy, so let's complicate it! Define an operation on the sphere such that $\theta \rightarrow \theta + \pi$ , where $\theta$ is the azimuthal angle. What does this do to the sphere? What happens to the points at the north and south pole of the sphere?

Next, define a parity operation on the six dimensional scalar field, such that

$\mathcal{P}: \Phi(x,\theta,\phi)\mapsto\Phi(x,\theta+\pi,\phi) = \pm \Phi(x,\theta,\phi)$

Now what do the mode expansions look like, given that $\Phi$ can be either even or odd under the operation?

For extra credit, work out the normalizations of the scalar field so that it is canonically normalized.

**rpenner** · 10-30-08, 03:22 PM

I assume the Kaluza-Klein mode expansion of the solution is
$\sum_{\ell,|m| \le \ell} Y_{\ell}^{m}(\phi,\theta) e^{i\mathbf{p}\cdot\mathbf{x}} = \sum_{\ell,|m| \le \ell} \sqrt{{(2\ell+1)\over 4\pi}{(\ell-m)!\over (\ell+m)!}} P_{\ell}^{m}(\cos \phi) e^{i \ell \theta} e^{i\mathbf{p}\cdot\mathbf{x}}$

Let's call your operation the Ben Twist. It seems to rotate the the sphere on it's polar axis by one half revolution. Solutions (even at the north and south pole) pick up a phase factor of $(-1)^\ell$ .

So the parity is even for even $\ell$ and odd otherwise.

Or I could be completely off base since I never took a graduate physics course.

**BenTheMan** · 10-30-08, 04:17 PM

Anyway, the "Ben Twist" is actually an orbifold---something about which AN knows quite a bit

It's not a proper manifold, because you have these singular points (conical singularities) called fixed points. These are points which are unaffected (i.e. map to themselves) by the orbifold action.

The poles are fixed points in this construction, I think, because they are poorly defined by the spherical coordinate system. You might ask whether the fixed points are still fixed if you put a different coordinate system on the sphere---I think the answer is yes, because any coordinates will have places where they don't work.

**rpenner** · 10-30-08, 05:22 PM

I'm lost on terminology. I thought the orbifold was the quotient space not the operation -- since BenTwist is an action faithful to the non-trivial element of Z_2 (such that $\mathrm{BenTwist}^2 = I$ ). So, I think that's written as $\mathbb{R}^4 \times (S^2/\mathbb{Z}_2)$ -- Seeing as I don't know how to make the mass spectrum out of this six- dimensional Kaluza-Klein theory or even the basics of orbifolds, I'll shut up now.

**BenTheMan** · 10-30-08, 08:13 PM

I thought the orbifold was the quotient space not the operation

Ahh yes. We typically call the operation "orbifolding", which probably pisses QuarkHead and Guest off (no offense!).

Seeing as I don't know how to make the mass spectrum out of this six- dimensional Kaluza-Klein theory or even the basics of orbifolds, I'll shut up now.

Would you like to learn?

**rpenner** · 10-30-08, 10:11 PM

Originally Posted by BenTheMan

Would you like to learn?

If you have the time. I haven't internalized most of any of the quantum field theory textbooks on my shelf, and only Kaku mentions our friend Kazula-Klein.

**BenTheMan** · 10-31-08, 01:18 AM

Sure it's easy.

Presumably you know how to do separation of variables?

Let's start with an easier example. Suppose you have a scalar field on a something like $\mathbb{R}^4\times S^1$ . The five dimensional Klein-Gordon Equation for a massless scalar looks like

$\box_5 \Phi(x,y) = 0$ .

Explicitly, we can write this as

$\box_4 \Phi(x,y) - \frac{\partial^2}{\partial y^2} \Phi(x,y) = 0$ .

Now take an ansatz for $\Phi(x,y) = \sum \phi_n(x) f_n(y)$ .

What kinds of functions should f(y) be?

**BenTheMan** · 10-31-08, 01:20 AM

Also, this is a bit tricky---where does the minus sign in $\box_4 \Phi - \partial_y\partial^y \Phi = 0$ come from?

(This one ALWAYS screws me up)

**rpenner** · 10-31-08, 02:23 AM

Originally Posted by BenTheMan

Sure it's easy.

Presumably you know how to do separation of variables?

Presumably. If it's 100-years old in math undergrad materials, I should have a grip on it.

Originally Posted by BenTheMan

Let's start with an easier example. Suppose you have a scalar field on a something like $\mathbb{R}^4\times S^1$ . The five dimensional Klein-Gordon Equation for a massless scalar looks like

$\box_5 \Phi(x,y) = 0$ .

Explicitly, we can write this as

$\box_4 \Phi(x,y) - \frac{\partial^2}{\partial y^2} \Phi(x,y) = 0$ .

Or to the very slow among us:
$\frac{\partial^2}{\partial x_0^2} \Phi(x,y) - \nabla^2 \Phi(x,y) - \frac{\partial^2}{\partial y^2} \Phi(x,y) = \frac{\partial^2}{\partial x_0^2} \Phi(x,y) - \frac{\partial^2}{\partial x_1^2} \Phi(x,y) - \frac{\partial^2}{\partial x_2^2} \Phi(x,y) - \frac{\partial^2}{\partial x_3^2} \Phi(x,y) - \frac{\partial^2}{\partial y^2} \Phi(x,y) = 0$

Originally Posted by BenTheMan

Also, this is a bit tricky---where does the minus sign in $\box_4 \Phi - \partial_y\partial^y \Phi = 0$ come from?

(This one ALWAYS screws me up)

That would come from the space-like metric convention of your author/professor/research team, which I surmise is +---.

Originally Posted by BenTheMan

Now take an ansatz for $\Phi(x,y) = \sum \phi_n(x) f_n(y)$ .

What kinds of functions should f(y) be?

Since I know the solutions for $\box_4 \phi = 0$ are $e^{i E x_0 - i \mathbf{p} \cdot \mathbf{x}} = e^{i (E x_0 - p_1 x_1 - p_2 x_2 - p_3 x_3)} = e^{i E x_0}e^{-i p_1 x_1}e^{-i p_2 x_2}e^{-i p_3 x_3}$ I surmise that a good candidate would be $f_n(y) = e^{-i \, \mathrm{something} \, y}$ but by the requirement that y be a coordinate that closes $S^1$ (a circle), then we need our convention to close and so we get quantification. Rescale y to the range $\left[ 0, 2\pi \right)$ and we get $f_n(y) = e^{-iny}$ . (Or if y is a actual spacial distance, then there is another distance r such that y/r is radians and we get $f_n(y) = e^{-iny/r}$ ).

Since $\frac{\partial^2}{\partial y^2} g(x)f_n(y) = -(\frac{n}{r})^2 g(x)f_n(y)$ it follows that for fixed n, a solution to $\box_5 \Phi(x,y) = 0$ looks a lot like $(\box_4 + \mu^2) \Phi(x,y) = 0$ .

**BenTheMan** · 10-31-08, 03:12 AM

Originally Posted by rpenner

That would come from the space-like metric convention of your author/professor/research team, which I surmise is +---.

Yes! This minus sign always kills me.

Since $\frac{\partial^2}{\partial y^2} g(x)f_n(y) = -(\frac{n}{r})^2 g(x)f_n(y)$ it follows that for fixed n, a solution to $\box_5 \Phi(x,y) = 0$ looks a lot like $(\box_4 + \mu^2) \Phi(x,y) = 0$ .

And that's all there is to it! Welcome to the illustrious world of Kaluza Klein modes.

**rpenner** · 10-31-08, 04:15 AM

Checking.
$-\nabla^2 Y_{\ell}^{m}(\phi, \theta) \\ = <br /> - {{1}\over{r^2 \sin \phi}} {{\partial}\over{\partial \phi}} ( \sin \phi {{\partial}\over{\partial \phi}} Y_{\ell}^{m}(\phi, \theta) )<br /> - {{1}\over{r^2 \sin^2 \phi}} {{\partial^2}\over{\partial \theta^2}} Y_{\ell}^{m}(\phi, \theta) \\ = {{\ell(\ell+1)}\over{r^2}} Y_{\ell}^{m}(\phi, \theta) = \mu^2 Y_{\ell}^{m}(\phi, \theta)$ , right?

**AlphaNumeric** · 10-31-08, 04:39 AM

Originally Posted by BenTheMan

Yes! This minus sign always kills me.

Cliff Burgess refers to the +--- metric as 'the wrong metric'. Doesn't give proper Euclidean space when you do a Wick rotation on it.

Originally Posted by BenTheMan

And that's all there is to it! Welcome to the illustrious world of Kaluza Klein modes.

Plus, if you're doing any kind of vague phenomenology with KK modes you just set n=0, since all n>0 are masses of order the Planck mass

**BenTheMan** · 10-31-08, 11:37 AM

Originally Posted by AlphaNumeric

CPlus, if you're doing any kind of vague phenomenology with KK modes you just set n=0, since all n>0 are masses of order the Planck mass

Well, not always. Suppose you have a dimension that's [url=http://arxiv.org/abs/0805.4186]bigger than the string scale...[/QUOTE]

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