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Error in books of conformal field theory?

+ 5 like - 0 dislike
42 views

If you look at the book Conformal Field Theory (by Philippe Francesco, Pierre Mathieu and David Senechal) or the lecture notes Applied Conformal Field Theory (by Paul Ginsparg), and many other places: The conformal transformation is defined as the subset of coordinate transformations under which the metric changes as follows, $$ g_{\mu \nu}'(x') = \Omega(x) g_{\mu \nu}(x) $$ under a coordinate change $x'=x'(x)$ where, $$ g_{\mu \nu}'(x')=\frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} g_{\alpha \beta}(x). $$ But I think the first equation should be: $$ g_{\mu \nu}'(x') = \Omega(x') g_{\mu \nu}(x'). $$ Although this looks like a typo, but it is consistant in many lectures/books. So I am little confused. Am I thinking it wrong? Books/Lectures are probably thinking that in the end of the transformation, they are relabeling $x'$ as $x$. But then that should be explicitly mentioned.

Derivation of conformal killing vector equation using book's version of the equation: Consider an infinitesimal change, $$ x'^{\mu} = x^{\mu} + \epsilon \xi^{\mu}(x) $$ and, $$ \Omega(x) = 1+\epsilon \omega(x) $$ Then the equation according to the books: $$ g_{\mu \nu}'(x') = \Omega(x) g_{\mu \nu}(x) $$ gives, $$ g_{\mu \nu}'(x') = g_{\mu \nu}(x) + \epsilon \omega(x) g_{\mu \nu}(x) $$ Also we can expand, $$ g_{\mu \nu}(x') = g_{\mu \nu}(x) + \epsilon \xi^{\alpha} \partial_{\alpha} g_{\mu \nu}(x) $$ The Lie derivative of the metric, $$ \mathcal{L}g=\lim_{\epsilon \to 0} \left( {g'_{\mu\nu}(x')-g_{\mu\nu}(x') \over \epsilon} \right) = \omega(x) g_{\mu \nu}(x) -\xi^{\alpha} \partial_{\alpha} g_{\mu \nu}(x) $$ Which implies, $$ -(\xi_{\mu;\nu}+\xi_{\nu;\mu})=\omega(x) g_{\mu \nu}(x) -\xi^{\alpha} \partial_{\alpha} g_{\mu \nu}(x) $$ as the killing vector equation. Where as if you consider the equation: $$ g_{\mu \nu}'(x') = \Omega(x') g_{\mu \nu}(x') $$ Then you get following similar steps, $$ -(\xi_{\mu;\nu}+\xi_{\nu;\mu})=\omega(x) g_{\mu \nu}(x) $$ Then which Killing vector equation is correct?

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Nilanjan
asked Nov 27, 2013 in Theoretical Physics by Nilanjan (25 points) [ no revision ]
No. The books are correct.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Prahar
To be clear, what is being said is that the two metrics are related by a conformal factor at the spacetime point. This spacetime point is labelled x in one coordinate system and x' in another.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Prahar
Hi Prahar, Please have a look at my source of confusion. I have included that in my question.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Nilanjan
conformal space-time transformations and Weyl transformations are different things (see physics.stackexchange.com/a/38189/6389)

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Christoph
@nilanjan the first one is correct (the version of the book) upto a sign of the last term.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Prahar
@nilanjan - sorry. I'm on my phone else I would have written out a longer clearer answer.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Prahar
@Prahar Some authors also use the second definition when $x'$ is an active transformation.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user10001

3 Answers

+ 4 like - 0 dislike

The books are correct. The statement is a definite relation that is being imposed between the 'old' metric structure and the transformed one, for the transformation to be conformal.

The equation you're unhappy about, $$ g_{\mu \nu}'(x') = \Omega(x) g_{\mu \nu}(x) $$ states that the transformed metric $g_{\mu \nu}'$ at the transformed point $x'$ can be obtained from the old metric $g_{\mu \nu}$ evaluated at the untransformed point $x$ by simply multiplying it, as a matrix by a scalar quantity $\Omega$ which is a function of the spacetime point in question.

This is distinct from the general change-of-coordinates transformation of the metric, which involves multiplying from left and right with the transformation jacobian: the inner structure of the metric is not being meddled with, which keeps the angles constant.

The old metric $g_{\mu \nu}$ is naturally a function of the old position, $x$. Because $x'$ is a function of $x$ and viceversa, writing $g_{\mu \nu}(x')$ is not formally incorrect, but it is misleading and should be avoided.

The scalar factor $\Omega$ is a function of the spacetime point in question and can be written as an explicit function of either the old or the transformed coordinates, as long as one is consistent. One would usually, though, keep the $x$s to the right of the equal sign, and the $x'$s to the left.

The point is that $x'$ is a function of $x$, and vice versa, and of course only one of the two is ever independent. Thus, you could write the equation as $g_{\mu \nu}'(x') = \Omega(x(x')) g_{\mu \nu}(x(x')) $, where $x'$ is the only independent variable, or $g_{\mu \nu}'(x'(x)) = \Omega(x) g_{\mu \nu}(x) $, where $x$ is independent. Both variants are quite ugly. Thus we usually write in a simpler notation with primed quantities on one side related to unprimed ones on the other, in the understanding that only one set of the two is ever independent but that they are, of course, completely equivalent. To a symmetric situation we respond, where possible, with symmetric notation.


Regarding your edit, I can't really say where you're getting tripped up, particularly as your definition of the Lie derivative is rather at odds with the more standard, more abstract ones. For an infinitesimal transformation that's "anchored" at the old point $x$, I would strongly urge you to not work in the transformed plane. Bring transformed quantities back to the untransformed plane and work with them there, instead of the other way. If you're careful, both ways should be equivalent, but it's a lot easier to mess up in the transformed plane, because it's "moving".

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user Emilio Pisanty
answered Nov 27, 2013 by Emilio Pisanty (520 points) [ no revision ]
+ 4 like - 0 dislike

First equation refers to the passive view of coordinate transformations while the second is the active view.

Let $M$ be a manifold with metric $g$ which in local coordinates $x$ is written as $g_{\alpha\beta}(x)dx^{\alpha}\otimes dx^{\beta}$. Let $\phi:M\rightarrow M$ be a differentiable function and let $g'=\phi^*g$ be the pullback of the metric $g$, which in the local coordinate $x$ is written as $g'_{\alpha\beta}(x)dx^{\alpha}\otimes dx^{\beta}$. Then in the active view map $\phi$ will be said to be a conformal transformation if it is invetible with a smooth inverse, and if for some positive function $\Omega$ on $M$ we have $g'=\Omega g$. Or in local coordinates $g'_{\alpha\beta}(x)= \Omega(x)g_{\alpha\beta}(x)$.

On the other hand in the passive view we are just looking at the same metric $g$ in different coordinates $x$ and $x'$. If in coordinates $x'$ metric $g$ has the form $g'_{\alpha\beta}(x')dx'^{\alpha}\otimes dx'^{\beta}$ then the coordinate change $x'=f(x)$ is said to be conformal if for some positive function $\Omega(x)$, we have $g'_{\alpha\beta}(x')=\Omega(x)g_{\alpha\beta}(x)$. Here it should be noted that $x$ and $x'=f(x)$ refer to the same underlying point $p$ of the manifold $M$ (as Prahar has remarked above).

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user10001
answered Nov 27, 2013 by user10001 (635 points) [ no revision ]
It is probably worthwhile to note that these two definitions of conformal are not equivalent. One compares the metric at different points and one compares them at the same point.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user NowIGetToLearnWhatAHeadIs
@NowIGetToLearnWhatAHeadIs Yes these definitions are not equivalent. In active view two different metrics (namely the given metric and its pullback) are being compared at the same point; while in the passive view the coefficients of the same metric in two different coordinates are being compared (again at the same point).

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user user10001
You said it better.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user NowIGetToLearnWhatAHeadIs
+ 3 like - 0 dislike

Let's look at an example. Let's consider $0+1$ dimensions. Our manifold will be $M=\mathbb{R}$ and the coordinate system we will use will map the coordinate $x \in \mathbb{R}$ to the point $p \in M$ according to the rule $p(x^a) = x^0$. Now suppose the metric in this coordinate system has coordinates $g_{00}=x^0$ in this coordinate system.

Now we would like to consider what happens if we do a conformal transformation. We define our conformal transformation by $x'^0(x) =(x^0)^2$.

Let's compute $\frac{\partial x'^\mu}{\partial x^\alpha}$. We find it is given by $\frac{\partial x'^0}{\partial x^0}(x)=2x^0.$ Thus $\frac{\partial x^0}{\partial x'^0}(x')=\frac{1}{2x^0}= \frac{1}{2\sqrt{x'^0}}.$

Now let's compute the metric in the new coordinates. We get $g'_{00}(x') = \frac{\partial x^0}{\partial x'^0}(x')\frac{\partial x^0}{\partial x'^0}(x')g_{00}(x(x')) = \frac{1}{4 x'^0}x^0(x') =\frac{1}{4 \sqrt{ x'^0}} $

Now what do we expect the conformal factor between the two metrics to be? In other words, if I am stading at a point $p \in M$ and I look at my metric in the $x$ coordinate system, then by what factor does the metric increase when I look at it instead in the $x'$ coordinate system (while standing at the same point $p$). I would expect it to be $\frac{1}{4 x'^0} = \frac{1}{4 (x^0)^2}$, since this is the square of the jacobian of the coordinate transformation, which should tell you how the metric scales at a given point.

Let's see what we actually get when we use the two definitions.

Let's start with the book's definition. $g'_{00}(x') = \Omega(x) g_{00}(x)$. We plug in $g_{00}(x)=x^0=\sqrt{x'^0}$ and $g'_{00}(x')=\frac{1}{4 \sqrt{ x'^0}} $. Comparing these, we find $\Omega(x')=\frac{1}{4 x'^0}$. This is what we expected.

Now let's compute it using your formula. Your formula is $g'_{00}(x') = \Omega(x') g_{00}(x')$ Do we get the same thing? Well $g'_{00}(x')=\frac{1}{4 \sqrt{ x'^0}} $ still, but if we evaluate $g_{00}$ using the coordinate $x'$ instead of $x$ we get something different. We get $g_{00}(x')=x'^0$. Thus our conformal factor is now $\Omega(x')=\frac{1}{4 (x'^0)^{3/2}}$. This is not what we expect. We conclude that this formula does not give a good definition for what a conformal transformation, and the formula above does.

The main point is that the metrics to be compared should correspond to the same point $p \in M$, but in your definition you evaluate the metric $g$ in the unprimed coordinates at the coordinate value $x=x'(x)$, so if say $x^0=2$ (so $x'^0=4$), then $g_{\mu \nu}(x')=g_{\mu \nu}(4)$ which is the value of the metric at the point $p \in M$ corresponding to $x^0=4$. So your formula compares $g'_{00}(4)$ with $g_{00}(4)$, when it should compare $g'_{00}(4)$ with $g_{00}(2)$, since the second combination corresponds to the same point in $M$.

Now presumably you already knew all this and by $g_{\mu \nu} (x')$ you really meant $g_{\mu \nu} (x(x'))$, in which case your formula is as good as the formula in the books and its just a matter of whether you want $g_{\mu \nu} (x)$ or $g_{\mu \nu} (x')$ to be a shorthand for $g_{\mu \nu} (x(x'))$.

This post imported from StackExchange Physics at 2015-03-30 13:56 (UTC), posted by SE-user NowIGetToLearnWhatAHeadIs
answered Nov 27, 2013 by NowIGetToLearnWhatAHeadIs (80 points) [ no revision ]

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