Symmetry Condition for Scaling a Lagrangian?

1. Sep 28, 2014

Xenosum

1. The problem statement, all variables and given/known data

Take the action

$$S = \int d^4x \frac{1}{2} \left( \partial_{\mu}\phi(x)\partial^{\mu}\phi(x) - m^2\phi^2(x) - g\phi(x)^p \right) ,$$

and consider the following transformations:

$$x^{\mu} \rightarrow x^{'\mu} = \lambda x^{\mu}$$
$$\phi(x) \rightarrow \phi^{'}(x) = \lambda^{-D}\phi(\lambda^{-1} x) .$$

For what values of $m$ and $p$ is this a symmetry?

2. Relevant equations

N/A

3. The attempt at a solution

This seems like an elementary calculus problem, but I'm having trouble. Under the condition that the derivative terms remain invariant, we want

$$\frac{\partial}{\partial x^{\mu}} \phi(x) \rightarrow \frac{\partial}{\partial(\lambda^{-1}x^{\mu})}\phi(\lambda^{-1}x),$$

which is satisfied when $D = -2$. From the action we calculate the equation of motion and get

$$\partial_{\mu}\partial^{\mu} \phi(x) + m^2\phi(x) + gp\phi^{p-1}(x) = 0.$$

In the primed frame, the Euler-Lagrange equations are taken with respect to the primed coordinates and the equation of motion reads

$$\frac{\partial}{\partial(\lambda x^{\mu})} \left( \partial^{'}_{\mu} \phi^{'}(x) \right) + m^2\phi^{'}(x) + gp\phi^{'}(x)^{p-1} = 0.$$

Invoking the condition that $D = -2$, the $\partial^{'}_{\mu}\phi^{'}(x)$ simply becomes $\frac{\partial}{\partial\alpha}\phi(\alpha)$, where $\alpha = \lambda^{-1}x$. The remaining derivative in the first time requires an extra factor of $\lambda^{-2}$ for it to be evaluated over $\alpha$ as well, and the entire equation becomes

$$\frac{\partial}{\partial\alpha_{\mu}}\frac{\partial}{\partial\alpha^{\mu}}\phi(\alpha) + \lambda^{4}m^2\phi(\alpha) + \lambda^{2p}gp\phi(\alpha)^{p-1} = 0.$$

In order for the transformation to be a symmetry, we want $\lambda$ to equal unity, and we find that $m^2 = m^2/\lambda^4$. But this is some sort of recursion relation and doesn't really make any sense.

I'm also skeptical about whether or not the equation I have written for the EOM in the primed frame is actually a symmetry when $\lambda = 1$ because the derivatives aren't even taken with respect to the primed frame's coordinates. If it is, though, I'm kind of lost. (Incidentally it's also possible that there was a typo in the problem; I don't see any reason to define the transformation such that $x$ transforms by a different factor than the argument of $\phi(x)$...?)

Help appreciated, thanks!

2. Sep 30, 2014

vanhees71

What you should calculate to check, e.g., the kinetic term is
$$\partial_{\mu}' \phi'(x').$$
Then plug it into the Lagrangian and check if the corresponding piece of the action is invariant. The same you do for all the other terms. In this way you can determine the values for your parameter space, $D$, $m$, and $p$ for which the action is invariant under the given scaling transformation.

3. Sep 30, 2014

vanhees71

What you should calculate to check, e.g., the kinetic term is
$$\partial_{\mu}' \phi'(x').$$
Then plug it into the Lagrangian and check if the corresponding piece of the action is invariant. The same you do for all the other terms. In this way you can determine the values for your parameter space, $D$, $m$, and $p$ for which the action is invariant under the given scaling transformation.

4. Sep 30, 2014

Xenosum

I see, thanks, but I'm a bit thrown off because the condition of the mass term to remain invariant is

$$m^2\phi(x) = m^2\phi^{'}(x^{'}) = m^2\lambda^{-D}\phi(x)$$

So that $D=0$. However the condition for the derivative term to remain invariant requires a different $D$ as shown above (actually in my derivations above I took the field in the primed coordinates as $\phi^{'}(x) = \lambda^{-D}\phi(\lambda^{-1}x)$, but I guess I should actually have used $\phi^{'}(x^{'})$? In this case the lambdas would cancel and it makes sense to define the transformations this way. In this case we get $D=-1$ for the derivative term to remain invariant; still a contradiction.)

5. Sep 30, 2014

vanhees71

That's correct! Now you must argue with the physics. The mass term is not essential to the theory, but you must have the kinetic term for the field. So first we have to investigate the kinetic term to determine $D$. Now you must take into account that not the Lagrangian, but the action should be invariant. So you need
$$\mathrm{d}^4 x' (\partial_{\mu}' \phi(x')) (\partial'{}^{\mu} \phi'(x') = \mathrm{d}^4 x (\partial_{\mu} \phi(x)) (\partial^{\mu} \phi(x).$$
So you start from the left-hand side of this equation and express everything in terms of the unprimed quantities. At the very end you set it equal to the right-hand side of this equation, which determins $D$.

A hint is, that this can be solved by dimensional analysis. With $\hbar=c=1$ the action is dimension less. From this it follows that the Lagrangian is of dimension $\text{length}^{-4}$, because $\mathrm{d}^4 x$ is of dimension $\text{length}^4$. Now you can "count" of which length dimension $\phi$ must be, and from this you conclude, what's $D$. In the same way you can count the length dimension of the other terms in the Lagrangian. After this idea is established, it's very simple to decide, that all the constants in the Lagrangian must be dimensionless in order to make the action dilation invariant. This lets you very easily decide what has to happen to the mass of the particle and which power of $\phi$ has to occur in the interaction-potential term.

Finally, you intuitively understand why then in the quantized theory this dilation symmetry is broken anyway: It's because you must necessarily introduce a momentum scale when renormalizing the theory when calculating loop diagrams in perturbation theory.

6. Sep 30, 2014

Xenosum

Ah, I forgot to include the differentials in calculating the invariance of the kinetic term! Thanks!

Though, it still seems like the same problem arises. When I expand out

$$\mathrm{d}^4 x' (\partial_{\mu}' \phi(x')) (\partial'{}^{\mu} \phi'(x')) = \mathrm{d}^4 x (\partial_{\mu} \phi(x)) (\partial^{\mu} \phi(x)),$$

I get

$$\frac{\lambda^{2}}{\lambda^{2D}} \mathrm{d}^4 x (\partial_{\mu} \phi(x)) (\partial^{\mu} \phi(x)) = \mathrm{d}^4 x (\partial_{\mu} \phi(x)) (\partial^{\mu} \phi(x)),$$

so that $D = 1$. With this value for $D$ we can expand out the action and obtain

$$S^{'} = \int \frac{\mathrm{d}^4x}{2} \left( \partial_{\mu}\phi(x)\partial^{\mu}\phi(x) - \lambda^{2}m^2\phi(x)^2 - g\lambda^{4-p}\phi(x)^p \right).$$

In order for it to remain invariant, then, we again we find that the $m$ term must satisfy some sort of recursion relation: $m^2 = \frac{m^2}{\lambda^2}$. Am I missing something obvious?

7. Sep 30, 2014

Xenosum

Well, I suppose it's possible that this transformation is only a symmetry for a massless scalar field field, $m=0$...

I don't know why I didn't think of this at the outset (derp) but in any case there was a number of things wrong with my procedure (most importantly, it's invariance in the action, not the lagrangian, that gives a symmetry), and thanks for pointing them out!

8. Sep 30, 2014

Xenosum

Okay but I'm also a little bit concerned about the transformations I have written down. In Peskin and Schroeder, it's claimed that

"We can describe the infinitesimal translation

$$x^{\mu} \rightarrow x^{\mu} - a^{\mu}$$

alternatively as a transformation of the field configuration

$$\phi(x) \rightarrow \phi(x+a) = \phi(x) + a^{\mu}\partial_{\mu}\phi(x)."$$

But why is this an alternative description? The wording is weird because doesn't it imply that $x$ and the argument of $\phi(x)$ transform independently? Which would mean that I shouldn't take $\phi(x) \rightarrow \phi^{'}(x^{'})$ but rather simply $\phi(x) \rightarrow \phi^{'}(x)$? It seems to me like it should read 'the infinitesimal translation requires a transformation of the field configuration...'? Or not?