4

I don't think it is clear to me what exactly is the physical meaning of the energy-momentum tensor being traceless.

Qmechanic
  • 201,751
PsycoPulcino
  • 639
  • 1
    Related: https://physics.stackexchange.com/q/162297/2451 – Qmechanic Jan 26 '17 at 12:13
  • It means the theory is invariant under conformal transformation. For example, Pure bosonic string theory has 2D conformal invariance. Therefore it is scale invariant. – Mass Jan 26 '17 at 12:29
  • 2
    What sort of "physical meaning" are you looking for? Is "the theory is dilatation invariant" a "physical meaning"? – ACuriousMind Jan 26 '17 at 14:05
  • Ok! That's clear. @ACuriousMind i was also wondering if it had something to do with the field being massless or massive – PsycoPulcino Jan 26 '17 at 14:56
  • 1
    @user129511 If the field was massive, the theory wouldn't have conformal invariance, as the mass introduces a scale – coconut Jan 26 '17 at 15:40
  • For clarity, please say whether you are asking about $T^\mu_{; \mu}$ or some other trace (since at the moment there is an answer which considers $T_{00} + T_{11} + T_{22} + T_{33}$ which is not an invariant.) – Andrew Steane Feb 21 '21 at 16:57

1 Answers1

-2

The physical significance of a traceless energy-momentum tensor or $\text{Tr}(T_{ab}) = 0$ means that the addition of the diagonal terms of the matrix is $0$.

Now, the energy momentum tensor carries its identity with:

$T_{00}$ = $\text{energy density}$, and ${T_{\alpha\alpha}}$ = $\text{pressure}$

And the sum of the diagonal terms is $0$. Or,

$$T_{00} + T_{11} + T_{22}+T_{22} = 0$$

as summation, means that internal energy of the system, and internal pressure, do not contribute to the bulk motion of the system, as they cancel out each other's contribution, which indeed is a very significant property of the conformal invariant object, where, if the property $Tr(T_{ab}) = 0$ holds, it holds true for any metric transformation of type

$$g_{ab} \rightarrow \Omega^2g_{ab}$$

Monopole
  • 3,454
prikarsartam
  • 130
  • 1
  • 10
  • What do you mean when you say that they don't contribute to the bulk motion of the system? In particular, 1. What do you mean by bulk motion? 2. $T_{\mu\nu}$ enters the equations of motion, not only $T$. So the individual diagonal elements do enter the equations of motion, right? –  May 10 '20 at 10:19
  • 1
    Are you sure this is the trace which was asked about? I was expecting $T^\mu_{;\mu}$. – Andrew Steane Feb 21 '21 at 17:01