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Improved energy-momentum tensor changing dilation operator

  1. Mar 27, 2015 #1
    I'm trying to show that [itex]\int d^3x \,x^\mu \left(\partial_\mu \partial_0-g_{\mu 0} \partial^2 \right)\phi^2(x)=0 [/itex]. This term represents an addition to a component of the energy-momentum tensor [itex]\theta_{\mu 0} [/itex] of a scalar field and I want to show that this does not change the dilation operator [itex]D=\int d^3 x \, x^\mu \theta_{\mu 0} [/itex]

    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. jcsd
  3. Mar 30, 2015 #2

    samalkhaiat

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    Science Advisor

    Oh, you are very confused about this stuff. Okay, I will be naughty :smile: 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. :wink:
     
  4. Mar 31, 2015 #3

    samalkhaiat

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    Science Advisor

    Okay, let’s start from the basics. Under infinitesimal scale transformation, a generic field [itex]\varphi ( x )[/itex] transforms according to [tex]\delta^{*} \varphi ( x ) \equiv \bar{\varphi} ( \bar{x} ) - \varphi ( x ) = \Delta \ \varphi ( x ) ,[/tex] where [itex]\Delta[/itex] is the scaling dimension of the field. The canonical Noether current associated with this transformation (the dilatation current) is [tex]D_{c}^{\mu} = \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \varphi )} \Delta \varphi + T^{\mu \nu} \ x_{\nu} ,[/tex] where [tex]T^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \varphi )} \partial^{\nu} \varphi - \eta^{\mu \nu} \mathcal{L} ,[/tex] is the canonical energy-momentum tensor. For a Lorentz scalar field, we have [itex]\Delta = 1[/itex] and [tex] \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \varphi )} = \partial^{\mu} \varphi .[/tex] Therefore, the canonical dilatation current becomes [tex]D_{c}^{\sigma} = T^{\sigma \tau} \ x_{\tau} + \varphi \ \partial^{\sigma} \varphi , \ \ \ \ \ (1)[/tex] and the associated scaling generator is [tex]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) .[/tex] Now, let’s play with the second term in Eq(1). First, you can check that [tex]\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} .[/tex] Using the relation [itex]\eta_{\rho \tau} = \partial_{\rho} x_{\tau}[/itex] , we find [tex]\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} .[/tex] Finally, with help of the identity [itex]( \partial X ) Y = \partial ( X Y ) - X ( \partial Y )[/itex], we rewrite the above equation in the form [tex]\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)[/tex] Now, substituting (2) in (1) we get [tex]D_{c}^{\sigma} + \partial_{\rho} X^{\sigma \rho} = \theta^{\sigma \tau} x_{\tau} , \ \ \ \ (3)[/tex] where [tex]X^{\sigma \rho} = - X^{\rho \sigma} \equiv \frac{1}{6} \ ( x^{\sigma} \partial^{\rho} - x^{\rho} \partial^{\sigma} ) \varphi^{2} ,[/tex] and [tex]\theta^{\sigma \tau} = T^{\sigma \tau} + \frac{1}{6} \ ( \eta^{\sigma \tau} \partial^{2} - \partial^{\sigma} \partial^{\tau} ) \varphi^{2} . \ \ \ (4)[/tex] Now, let us examine the left-hand-side of Eq(3) by writing [tex]D^{\sigma} \equiv D_{c}^{\sigma} + \partial_{\rho} X^{\sigma \rho} .[/tex] 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 [itex]\partial_{\sigma} D^{\sigma} = \partial_{\sigma} D_{c}^{\sigma}[/itex] and, since [tex]\int d^{3} x \ \partial_{\rho} X^{\rho 0} = \int d^{3} x \ \partial_{j} X^{j 0} = 0 ,[/tex] we find [tex]\int d^{3} x \ D^{0} ( x ) = \int d^{3} x \ D_{c}^{0} ( x ) = D .[/tex] Thus, [itex]D^{\mu}[/itex] is a legitimate dilatation current and, therefore, can be used instead of the canonical Noether current [itex]D_{c}^{\mu}[/itex]. This implies that the tensor [itex]\theta^{\mu \nu}[/itex] 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 [tex]\theta^{\mu \nu} = \theta^{\nu \mu} , \ \ \partial_{\mu} \theta^{\mu \nu} = \partial_{\mu} T^{\mu \nu} = 0 .[/tex] And, more importantly, the added piece in (4) does not change the Poincare’ generators [tex]P^{\mu} = \int d^{3} x \ \theta^{0 \mu} = \int d^{3} x \ T^{0 \mu} ,[/tex] [tex]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} ) .[/tex] So, given [itex]\theta^{\mu \nu}[/itex] we can rewrite dilatation current in the following compact form [tex]D^{\sigma} = \theta^{\sigma \tau} x_{\tau} .[/tex] So, [tex]\partial_{\mu} D^{\mu} = \theta^{\mu}_{\mu} .[/tex] Thus, for scale-invariant scalar field theory, the new energy-momentum tensor is traceless [itex]\theta^{\mu}_{\mu} = 0[/itex]. I believe you have realized now that your question is a trivial one as you can see that [tex]\int d^{3} x \ \theta^{0 \tau} \ x_{\tau} = \int d^{3} x \ ( T^{0 \tau} \ x_{\tau} + \varphi \ \partial^{0} \varphi ) = D .[/tex]

    Sam
     
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