# Improved energy-momentum tensor changing dilation operator

1. Mar 27, 2015

### geoduck

I'm trying to show that $\int d^3x \,x^\mu \left(\partial_\mu \partial_0-g_{\mu 0} \partial^2 \right)\phi^2(x)=0$. This term represents an addition to a component of the energy-momentum tensor $\theta_{\mu 0}$ of a scalar field and I want to show that this does not change the dilation operator $D=\int d^3 x \, x^\mu \theta_{\mu 0}$

What I have is that:

$$\int d^3x \,x^\mu \left(\partial_\mu \partial_0-g_{\mu 0} \partial^2 \right)\phi^2(x)=\int d^3x \left[ \,x^j \left(\partial_j \partial_0 \right)\phi^2(x) +x^0\nabla^2\phi^2(x) \right]$$

and I can argue that the 2nd term is zero, since x0 can be pulled out of the integral, and you are integrating a divergence and because boundary conditions are periodic, this term is zero. But I can't argue that the first term is zero. Is there a simple reason that the 1st term is zero? We are not allowed to use anything like equations of motions for the fields.

2. Mar 30, 2015

### samalkhaiat

Oh, you are very confused about this stuff. Okay, I will be naughty and wait and see if anybody (other than myself) can sort out your confusion. As a hint you can check out this old thread
Okay, let’s start from the basics. Under infinitesimal scale transformation, a generic field $\varphi ( x )$ transforms according to $$\delta^{*} \varphi ( x ) \equiv \bar{\varphi} ( \bar{x} ) - \varphi ( x ) = \Delta \ \varphi ( x ) ,$$ where $\Delta$ is the scaling dimension of the field. The canonical Noether current associated with this transformation (the dilatation current) is $$D_{c}^{\mu} = \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \varphi )} \Delta \varphi + T^{\mu \nu} \ x_{\nu} ,$$ where $$T^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \varphi )} \partial^{\nu} \varphi - \eta^{\mu \nu} \mathcal{L} ,$$ is the canonical energy-momentum tensor. For a Lorentz scalar field, we have $\Delta = 1$ and $$\frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \varphi )} = \partial^{\mu} \varphi .$$ Therefore, the canonical dilatation current becomes $$D_{c}^{\sigma} = T^{\sigma \tau} \ x_{\tau} + \varphi \ \partial^{\sigma} \varphi , \ \ \ \ \ (1)$$ and the associated scaling generator is $$D \equiv \int d^{3} x \ D_{c}^{0} ( x ) = \int d^{3} x \left( T^{0 \tau} \ x_{\tau} + \varphi \ \partial^{0} \varphi \right) .$$ Now, let’s play with the second term in Eq(1). First, you can check that $$\varphi \partial^{\sigma} \varphi = \frac{1}{2} \ \partial^{\sigma} \varphi^{2} = - \frac{1}{6} \ \eta_{\rho \tau} \left( \eta^{\tau \sigma} \partial^{\rho} - \eta^{\tau \rho} \partial^{\sigma} \right) \varphi^{2} .$$ Using the relation $\eta_{\rho \tau} = \partial_{\rho} x_{\tau}$ , we find $$\varphi \ \partial^{\sigma} \varphi = - \frac{1}{6} \ ( \partial_{\rho} x_{\tau} ) \left( \eta^{\tau \sigma} \partial^{\rho} - \eta^{\tau \rho} \partial^{\sigma} \right) \varphi^{2} .$$ Finally, with help of the identity $( \partial X ) Y = \partial ( X Y ) - X ( \partial Y )$, we rewrite the above equation in the form $$\varphi \ \partial^{\sigma} \varphi = \frac{1}{6} \ x_{\tau} \partial_{\rho} \ ( \eta^{\tau \sigma} \partial^{\rho} \varphi^{2} - \eta^{\tau \rho} \partial^{\sigma} \varphi^{2} ) - \frac{1}{6} \ \partial_{\rho} ( x^{\sigma} \partial^{\rho} \varphi^{2} - x^{\rho} \partial^{\sigma}\varphi^{2} ) . \ \ (2)$$ Now, substituting (2) in (1) we get $$D_{c}^{\sigma} + \partial_{\rho} X^{\sigma \rho} = \theta^{\sigma \tau} x_{\tau} , \ \ \ \ (3)$$ where $$X^{\sigma \rho} = - X^{\rho \sigma} \equiv \frac{1}{6} \ ( x^{\sigma} \partial^{\rho} - x^{\rho} \partial^{\sigma} ) \varphi^{2} ,$$ and $$\theta^{\sigma \tau} = T^{\sigma \tau} + \frac{1}{6} \ ( \eta^{\sigma \tau} \partial^{2} - \partial^{\sigma} \partial^{\tau} ) \varphi^{2} . \ \ \ (4)$$ Now, let us examine the left-hand-side of Eq(3) by writing $$D^{\sigma} \equiv D_{c}^{\sigma} + \partial_{\rho} X^{\sigma \rho} .$$ The last term is the divergence of an anti-symmetric tensor (super-potential), so it does not contribute to the scaling current and charge. Indeed $\partial_{\sigma} D^{\sigma} = \partial_{\sigma} D_{c}^{\sigma}$ and, since $$\int d^{3} x \ \partial_{\rho} X^{\rho 0} = \int d^{3} x \ \partial_{j} X^{j 0} = 0 ,$$ we find $$\int d^{3} x \ D^{0} ( x ) = \int d^{3} x \ D_{c}^{0} ( x ) = D .$$ Thus, $D^{\mu}$ is a legitimate dilatation current and, therefore, can be used instead of the canonical Noether current $D_{c}^{\mu}$. This implies that the tensor $\theta^{\mu \nu}$ as given by eq(4) is a new legitimate energy-momentum tensor. Indeed it is: It is easy to check that the new tensor has the following properties $$\theta^{\mu \nu} = \theta^{\nu \mu} , \ \ \partial_{\mu} \theta^{\mu \nu} = \partial_{\mu} T^{\mu \nu} = 0 .$$ And, more importantly, the added piece in (4) does not change the Poincare’ generators $$P^{\mu} = \int d^{3} x \ \theta^{0 \mu} = \int d^{3} x \ T^{0 \mu} ,$$ $$M^{\mu \nu} = \int d^{3} x \ ( x^{\mu} \theta^{0 \nu} - x^{\nu} \theta^{0 \mu} ) = \int d^{3} x \ ( x^{\mu} T^{0 \nu} - x^{\nu} T^{0 \mu} ) .$$ So, given $\theta^{\mu \nu}$ we can rewrite dilatation current in the following compact form $$D^{\sigma} = \theta^{\sigma \tau} x_{\tau} .$$ So, $$\partial_{\mu} D^{\mu} = \theta^{\mu}_{\mu} .$$ Thus, for scale-invariant scalar field theory, the new energy-momentum tensor is traceless $\theta^{\mu}_{\mu} = 0$. I believe you have realized now that your question is a trivial one as you can see that $$\int d^{3} x \ \theta^{0 \tau} \ x_{\tau} = \int d^{3} x \ ( T^{0 \tau} \ x_{\tau} + \varphi \ \partial^{0} \varphi ) = D .$$