The Belinfante_rosenfeld tensor

[Michael_1812]

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 S^μ_γλ = T_λγ - T_γλ

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

T^γλ

= (T_γλ + T_λγ)/2 + (T_γλ -T_λγ)/2

= T^γλ_B + (T_γλ - T_λγ)/2

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

T^μγ = T^γλB - ∂_μ S^μ_γλ/2

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

T^γλ_B = T^γλ + ∂_μ(S^γλμ + S^λγμ - S^μλγ)/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!

Michael

 

[Michael_1812]

Further to my question.

I certainly understanding that T_γλ^B should not necessarily be defined as (T^γλ + T^λγ)/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 T_γλ^B is defined not as (T^γλ + T^λγ)/2 but in a more complex way, via

(T^γλ + T^λγ)/2 = T_γλ^B - (∂_μS^λγμ + ∂_μS^γλμ )/2

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

Nor is it clear to me why the simpler definition of T_γλ^B as (T^γλ + T^λγ)/2 would fail to warrant conservation.

 

[samalkhaiat]

Further to my question.

I certainly understanding that T_γλ^B should not necessarily be defined as (T^γλ + T^λγ)/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 T_γλ^B is defined not as (T^γλ + T^λγ)/2 but in a more complex way, via

(T^γλ + T^λγ)/2 = T_γλ^B - (∂_μS^λγμ + ∂_μS^γλμ )/2

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

Nor is it clear to me why the simpler definition of T_γλ^B as (T^γλ + T^λγ)/2 would fail to warrant conservation.

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

$$T_{ \nu \mu } = T_{ \mu \nu } + \partial^{ \sigma } S_{ \sigma \mu \nu }, \ \ (1)$$
where
$$S_{ \sigma \mu \nu } = - S_{ \sigma \nu \mu }.$$

So, let us write $S_{ \sigma \mu \nu }$ as

$$S_{ \sigma \mu \nu } = F_{ \sigma \mu \nu } - F_{ \sigma \nu \mu }, \ \ (2)$$

and try to determine $F_{ \sigma \mu \nu }$ 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 $\partial \cdot T = 0$. 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
$$\partial^{ \mu } \partial^{ \sigma } F_{ \sigma \mu \nu } = 0.$$
This will be the case if
$$F_{ \sigma \mu \nu } = - F_{ \mu \sigma \nu }. \ \ \ (3)$$

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

Sam

 

[Michael_1812]

Dear Sam,
Many thanks! I agree with your derivation.

You have built such an F that the sum T_γλ + ∂^μ F_μγλ is both conserved and symmetrical.

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

T_γλ

= (T_γλ + T_λγ)/2 + (T_γλ -T_λγ)/2

= (T_γλ + T_λγ)/2 - ∂^μ S_μγλ/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 S_μγλ/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 S_μγλ/2 is not conserved?

Many thanks,

Michael

 

[samalkhaiat]

Dear Sam,
Many thanks! I agree with your derivation.

You have built such an F that the sum T_γλ + ∂^μ F_μγλ is both conserved and symmetrical.

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

T_γλ

= (T_γλ + T_λγ)/2 + (T_γλ -T_λγ)/2

= (T_γλ + T_λγ)/2 - ∂^μ S_μγλ/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 S_μγλ/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 S_μγλ/2 is not conserved?

Many thanks,

Michael

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 $\partial$. The $\partial^{ \mu } \partial^{ \sigma } S_{ \sigma \mu \nu }$ is not zero because S is not antisymmetric in $( \sigma \mu )$. For this reason, your symmtrized T is not conserved.

 

[Michael_1812]

Dear Sam,

Many thanks for your time and help !

Now I understand this.

Michael