Improved energy-momentum tensor changing dilation operator

[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.

 

[samalkhaiat]

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.

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
www.physicsforums.com/showthread.php?t=172461
If, by tomorrow, you don’t get any replies, I post a detailed solution for you.

 

[samalkhaiat]

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 $$\int d^{3} x \ \theta^{0 \tau} \ x_{\tau} = \int d^{3} x \ ( T^{0 \tau} \ x_{\tau} + \varphi \ \partial^{0} \varphi ) = D .$$

Sam