Understanding the Stress-Energy Tensor in Special Relativity

[fab13]

TL;DR

I try to understand a demonstration in special relativity about the continuity equation
deduced from the stress-energy tensor

Hello,

I try to understand how to get the last relation below $(3)$ ( from stress energy tensor in special relativity - Wikipedia ).

This is to say that the divergence of the tensor in the brackets is $0$. Indeed, with this, we define the stress-energy tensor:
$T^{\mu \nu} \equiv \frac{\partial \mathcal{L}}{\partial\left(\partial_{\mu} \phi_{\alpha}\right)} \partial^{\nu} \phi_{\alpha}-g^{\mu \nu} \mathcal{L}\quad(1)$
By construction it has the property that
$\partial_{\mu} T^{\mu \nu}=0\quad(2)$

Note that this divergenceless property of this tensor is equivalent to four continuity equations. That is, fields have at least four sets of quantities that obey the continuity equation.

As an example, it can be seen that $T_{0}^{0}$ is the energy density of the system and that it is thus possible to obtain the Hamiltonian density from the stress-energy tensor.

Indeed, since this is the case, observing that $\partial_{\mu} T^{\mu 0}=0,$ we then have :

$\frac{\partial \mathcal{H}}{\partial t}+\nabla \cdot\left(\frac{\partial \mathcal{L}}{\partial \nabla \phi_{\alpha}} \dot{\phi}_{\alpha}\right)=0\quad(3)$

We can then conclude that the terms of $\frac{\partial \mathcal{L}}{\partial \nabla \phi_{\alpha}} \dot{\phi}_{\alpha}$ represent the energy flux density of the system.

I understand how to get equation $(1)$ but I don't grasp how to make appear the gradient operator in the dot product and the divergence operator in the bottom member nabla in equation $(3)$, i.e $\partial \nabla \phi_{\alpha}$.

Could anyone could help me to know how to introduce these 2 operators from equation $(1)$ ?

Regards

 

[strangerep]

Starting from $\,\partial_{\mu} T^{\mu 0}=0$,$$0 ~=~ \partial_0 T^{00} + \partial_k T^{k0}
~=~ \partial_0 {\mathcal H} ~+~ \partial_k \left( \frac{\partial \mathcal{L}}{\partial(\partial_k \phi_\alpha)} \, \partial^0 \phi_{\alpha}\right) ~-~ \xcancel{g^{k0}} \, \mathcal{L} ~.$$ Is that a sufficient clue?

 

[fab13]

@strangerep

Thanks for your help. I am just troubled by the writting of $\nabla$ in bottom term , i.e :

$\partial_k \left( \frac{\partial \mathcal{L}}{\partial(\partial_k \phi_\alpha)} \, \partial^0 \phi_{\alpha}\right)$.

If I expand this expression (without Einstein convention), we have :

$\partial_1 \left( \frac{\partial \mathcal{L}}{\partial(\partial_1 \phi_\alpha)} \, \partial^0 \phi_{\alpha}\right)+\partial_2 \left( \frac{\partial \mathcal{L}}{\partial(\partial_2 \phi_\alpha)} \, \partial^0 \phi_{\alpha}\right)+ \partial_3 \left( \frac{\partial \mathcal{L}}{\partial(\partial_3 \phi_\alpha)} \, \partial^0 \phi_{\alpha}\right)$.

I thought initially that $\nabla\,\cdot$ was a dot product where $\nabla$ is the gradient vector :

$\begin{align} \begin{bmatrix} \partial_1\\ \partial_2 \\ \partial_3 \end{bmatrix} \end{align} \cdot$

whereas this seems to be rather the divergence operator, doesn't it ?

If this is the case, then :

Is the other $\nabla$ in bottom member $\left(\frac{\partial \mathcal{L}}{\partial \nabla \phi_{\alpha}} \dot{\phi}_{\alpha}\right)$ not abusive ? since we don't sum over $k$ in this expression :

$\partial(\partial_k \phi_\alpha)\neq \partial \nabla \phi_{\alpha}$ as it is expressed above where I don't use Einstein's convention

Do you agree ?

 

[strangerep]

@strangerep
I am just troubled by the writting of $\nabla$ in bottom term , i.e :

$\partial_k \left( \frac{\partial \mathcal{L}}{\partial(\partial_k \phi_\alpha)} \, \partial^0 \phi_{\alpha}\right)$.

I consider it an abuse of notation, and never use it myself. For almost everything in physics I use explicit indices and the summation convention, since the meaning is much clearer that way (IMHO).

[...] Is the other $\nabla$ in bottom member $\left(\frac{\partial \mathcal{L}}{\partial \nabla \phi_{\alpha}} \dot{\phi}_{\alpha}\right)$ not abusive ? since we don't sum over $k$ in this expression :

$\partial(\partial_k \phi_\alpha)\neq \partial \nabla \phi_{\alpha}$ as it is expressed above where I don't use Einstein's convention

You'd have to write something like $$\partial(\partial_k \phi_\alpha) ~=~ \partial \Big( \nabla \phi_{\alpha}\Big)_k ~,$$but it's still a bit confusing.

This is just one of those situations where you've got to mentally translate what's really going on, which is best done with explicit index notation. It's easy enough when you get the hang of it.

 

[fab13]

ok, you reassure me since I wondered really why this nabla was appearing. Regards