Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The conserved charge associated with the Lorentz transfomation of a scalar field is given by $Q^{\alpha\beta}=\int d^3x\frac{1}{2}(x^\alpha T^{0\beta}-x^\beta T^{0\alpha})$. The quantities $Q^{ij}$ is recognized as the components of orbital angular momentum but what physical quantities does $Q^{0i}$ correspond to for $i=1,2,3$? This cannot be intrinsic angular momentum or spin because we are considering a scalar field. What is the significance of these quantities?

share|improve this question

1 Answer 1

It is the the boost conserved quantity. The factor $1/2$ is not used. To have an idea of its meaning you should write down its conservation law. Start form the definition, where $t= x^0$ (I assume $c=1$):

$$B^i(t) := t\int_{\mathbb R^3}d^3x T^{0i}- \int_{\mathbb R^3}d^3x x^i T^{00}\:, \quad i=1,2,3 \:.$$

The conservation law of $B^i$ is $\vec{B}(t)= \vec{B}(0)$, namely:

$$t \vec{P}= \int_{\mathbb R^3}d^3x T^{00}(t, \vec{x}) \:\vec{x}\:-\int_{\mathbb R^3}d^3x T^{00}(0, \vec{x})\:\vec{x}\:$$

differentiating in $t$ and taking into account that $\vec{P}$ is costant:

$$\vec{P} = \frac{d}{dt} \int_{\mathbb R^3}d^3x T^{00}(t, \vec{x}) \:\:\vec{x}\:\:$$

that is, in turn:

$$\vec{P} = M\frac{d}{dt}\left( \frac{1}{M}\int_{\mathbb R^3}d^3x T^{00}(t, \vec{x})\:\:\vec{x}\:\:\right) \quad (1)$$

Where I have introduced the total energy of the system in the considered reference frame (which is a conserved quantity as well):

$$M = \int_{\mathbb R^3}d^3x T^{00}\:.$$

(1) is nothing but the relativistic version of the classic theorem of center of mass. The position of the center of mass is:

$${\vec X}(t) := \frac{1}{M}\int_{\mathbb R^3}d^3x T^{00}(t, \vec{x})\:\:\vec{x}\:\:$$

The product of the total energy and the velocity of the center of mass is the total $3$-momentum. Now the right notion of "mass" to be used to define the "center of mass" is the temporal component of the $4$-momentum, i.e. the "relativistic mass".

ADDENDUM. The same result exists in classical mechanics too for a system made of some points of matter. In that case the invariance group is Galileo's one. The theorem of center of mass arises in the same way when considering the boost associated with the symmetry with respect to the pure Galileian transformations, i.e. $t \to t':=t$, $\vec{x}_i \to \vec{x}_i' = \vec{x}_i + t\vec{V}$. The difference is that, in classical mechanics the total mass $M$ is the sum of the masses of the points of the systems, in relativity, instead for a system of non interacting material points, the relevant mass includes the energies, since it is the temporal component of the total $4$-momentum. So this conservation law includes the relation $E=mc^2$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.