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The Belinfante_rosenfeld tensor

  1. Jan 5, 2014 #1
    Hi guys,

    Can anyone please help me to grasp a minor detail in the derivation of the Belinfante-Rosenfeld version of the Stress-Energy Tensor (SET) ?

    To save type, I refer to the wiki webpage http://en.wikipedia.org/wiki/Belinfante–Rosenfeld_stress–energy_tensor

    Using the Noether Theorem, it is indeed easy to arrive at the conservation of the tensor M.

    And yes, therefrom we indeed obtain

    μ Sμγλ = Tλγ - Tγλ

    Now, the canonical SET can be expressed as a sum of its symmetrical and antisymmetrical parts:


    = (Tγλ + Tλγ)/2 + (Tγλ -Tλγ)/2

    = TγλB + (Tγλ - Tλγ)/2

    As we have just seen, the antisymmetric part can be expressed through the spin tensor, whence we obtain:

    Tμγ = TγλB - ∂μ Sμγλ/2

    However, this is not what we see in the article in Wikipedia. There, two more terms are present:

    TγλB = Tγλ + ∂μ(Sγλμ + Sλγμ - Sμλγ)/2

    I don't think this is in error, because I saw those extra two terms also in the Relativity book by M. Gasperini /which is a good book, except that sometimes the author skips parts of the proof, clearly overestimating the abilities of an average reader/.

    Could someone please tell me how the two extra terms have shown up in the above formula?

    Many thanks!

    Last edited: Jan 5, 2014
  2. jcsd
  3. Jan 5, 2014 #2
    Further to my question.

    I certainly understanding that TγλB should not necessarily be defined as (Tγλ + Tλγ)/2 .

    We are always free to add a full divergence. And we should add it, to make sure that the result be conserved

    Therefore it is possible that TγλB is defined not as (Tγλ + Tλγ)/2 but in a more complex way, via

    (Tγλ + Tλγ)/2 = TγλB - (∂μSλγμ + ∂μSγλμ )/2

    But then what is the advantage of this definition? It is not apparently obvious to me that it guarantees conservation of TγλB.

    Nor is it clear to me why the simpler definition of TγλB as (Tγλ + Tλγ)/2 would fail to warrant conservation.
    Last edited: Jan 5, 2014
  4. Jan 6, 2014 #3


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    Start from the necessary and sufficient condition of Lorentz invariance. That is

    T_{ \nu \mu } = T_{ \mu \nu } + \partial^{ \sigma } S_{ \sigma \mu \nu }, \ \ (1)
    [tex]S_{ \sigma \mu \nu } = - S_{ \sigma \nu \mu }.[/tex]

    So, let us write [itex]S_{ \sigma \mu \nu }[/itex] as

    [tex]S_{ \sigma \mu \nu } = F_{ \sigma \mu \nu } - F_{ \sigma \nu \mu }, \ \ (2)[/tex]

    and try to determine [itex]F_{ \sigma \mu \nu }[/itex] later. Inserting (2) in (1), we find
    T_{ \mu \nu } + \partial^{ \sigma } F_{ \sigma \mu \nu } = T_{ \nu \mu } + \partial^{ \sigma } F_{ \sigma \nu \mu }.
    Clearly, this is nothing but the statement that the object
    \mathcal{ T }_{ \mu \nu } \equiv T_{ \mu \nu } + \partial^{ \sigma } F_{ \sigma \mu \nu },
    is symmetric. Next, translation invariance implies [itex]\partial \cdot T = 0[/itex]. Thus
    \partial^{ \mu } \mathcal{ T }_{ \mu \nu } = \partial^{ \mu } \partial^{ \sigma } F_{ \sigma \mu \nu }.
    So, to preserve translation invariance (the two tensors must be physically equivalent), we must have
    [tex] \partial^{ \mu } \partial^{ \sigma } F_{ \sigma \mu \nu } = 0.[/tex]
    This will be the case if
    [tex]F_{ \sigma \mu \nu } = - F_{ \mu \sigma \nu }. \ \ \ (3)[/tex]

    Now the two tensorial equations (2) and (3) can be solved. The solution is
    F_{ \sigma \mu \nu } = \frac{ 1 }{ 2 } ( S_{ \sigma \mu \nu } - S_{ \mu \sigma \nu } + S_{ \nu \mu \sigma } ).

  5. Jan 6, 2014 #4
    Dear Sam,
    Many thanks! I agree with your derivation.

    You have built such an F that the sum Tγλ + ∂μ Fμγλ is both conserved and symmetrical.

    Now, suppose I go a different way and extract directly the symmetrical part out of the canonical SET:


    = (Tγλ + Tλγ)/2 + (Tγλ -Tλγ)/2

    = (Tγλ + Tλγ)/2 - ∂μ Sμγλ/2

    In this case, the half-sum will be symmetric.

    Why then wouldn't we simply appoint the symmetric half-sum to play the role of the new SET ?

    The total T is conserved, and I presume that the antisymmetric part - ∂μ Sμγλ/2 is conserved also (is it not??)

    Then the symmetric half-sum must be conserved too.

    Or am I wrong, and the antisymmetric part - ∂μ Sμγλ/2 is not conserved?

    Many thanks,

    Last edited: Jan 6, 2014
  6. Jan 7, 2014 #5


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    I made it clear why we need to construct the F-tensor. We need an object that is antisymmetric in the FIRST TWO indeces so that it vanishes when operating with the second [itex]\partial[/itex]. The [itex]\partial^{ \mu } \partial^{ \sigma } S_{ \sigma \mu \nu }[/itex] is not zero because S is not antisymmetric in [itex]( \sigma \mu )[/itex]. For this reason, your symmtrized T is not conserved.
  7. Jan 7, 2014 #6
    Dear Sam,

    Many thanks for your time and help !!

    Now I understand this.

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