Energy-momentum tensor from a Lagrangian density?

  • #1
4
0

Homework Statement


I want to be able, for an arbitrary Lagrangian density of some field, to derive the energy-momentum tensor using Noether's theorem for translational symmetry.

I want to apply this to a specific instance but I am unsure of the approach.

Homework Equations


for a field undergoing a translational gauge transformation we also add a rotational gauge transformation to make it gauge invariant so we have the change in the vector field potential as
$$\delta A^\mu = \varepsilon^\mu (-\partial_\nu A^\mu + \partial^\mu A^\nu ) $$
for example the Klein-Gordon field has Lagrangian density
$$L = 1/2 \eta^{\mu\nu} \partial_\mu \phi \partial_\nu \phi - 1/2 m^2 \phi^2 $$

From Noether's theorem the current is
J^\mu = $$ \dfrac{\partial L}{ \partial (\partial_\mu A_\nu)} \delta A_\mu - \epsilon^\mu L$$

for the field

The Attempt at a Solution


My general procedure is to [/B]
substitute in the Langrangian density and change in the vector field into the expression for the Noether current leading to some expression I then intend on extracting the energy-momentum tensor out of using
$$ J^\mu = \varepsilon^\mu \left( T^{\mu\nu} \right) $$ as per my course notes.

I initially tried to re-arrange the Lagrangian by taking the metric tensor into the vector field to lower its indice to then allow the partial derivative to be resolved in the third equation but this is incorrect. I think this was incorrect because $$\eta^{\mu\nu}$$ cannot lower the index of $$A^\mu$$. I feel that I have the wrong approach.

Can anyone confirm my approach is incorrect and point me in the right direction?
 
Last edited:

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,370
975
$$\delta A^\mu = \varepsilon^\mu (-\partial_\nu A^\mu + \partial^\mu A^\nu ) $$
The indices in this equation do not make sense. Why not?
 
  • #3
4
0
I thought it was a rule that whenever there are repeated indices one should be up and one down. One of the indices should be down?
 
  • #4
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,370
975
I thought it was a rule that whenever there are repeated indices one should be up and one down. One of the indices should be down?
Also, free indices have to match up on both sides of an equation.
 
  • #5
4
0
Also, free indices have to match up on both sides of an equation.
So the factor of $$\varepsilon^\mu$$ is incorrect also
 
  • #6
4
0
Also, free indices have to match up on both sides of an equation.
I have since realised I was confusing the KG field as vector instead of scaler! I've now moved to correct this and have started like this
Lagrange density of KG
$$\mathcal{L} = \dfrac{1}{2}\left(\partial^\mu \phi\right) \left(\partial_\mu \phi \right) - \dfrac{1}{2}m^2\phi^2
$$

from taking some translation such that $$x \rightarrow x^\mu + \varepsilon^\mu $$ we have a change in the field given by $$
\phi \rightarrow \phi - \varepsilon^\mu \partial_\mu \phi $$

which gives a total derivative change in the Lagrange density and so all the equations of motion will be the same so the Lagrangian still applies all well and happy and because of the symmetry after this translation, Neother's theorem now also applies. We can then move to determin the conserved current with

$$\textbf{J}^\mu_{Noether} = \dfrac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi - \varepsilon^\mu\mathcal{L}$$

Now, the scaler field has changed by $$ \delta \phi = -\varepsilon^\mu \partial_\mu\phi $$
therefore, we have that
$$\textbf{J}^\mu_{Noether} = \dfrac{1}{2}\partial^\mu\phi\times -\varepsilon^\mu \partial_\mu\phi - \varepsilon^\mu \mathcal{L}$$
$$\textbf{J}^\mu_{Noether} = -\varepsilon^\mu\dfrac{1}{2}\partial^\mu\phi\partial_\mu\phi - \varepsilon^\mu \mathcal{L}$$
Now at this point I am lost because the notes I have just jump to the Tensor from here and just add some negative sign in front of the first term. Following this I get $$T^{\mu\nu} = \dfrac{1}{2}\partial^\mu\phi\partial_\mu\phi - \mathcal{L}$$
which is wrong but I don't see how Ive gone wrong
 
  • #7
nrqed
Science Advisor
Homework Helper
Gold Member
3,721
277
I have since realised I was confusing the KG field as vector instead of scaler! I've now moved to correct this and have started like this
Lagrange density of KG
$$\mathcal{L} = \dfrac{1}{2}\left(\partial^\mu \phi\right) \left(\partial_\mu \phi \right) - \dfrac{1}{2}m^2\phi^2
$$

from taking some translation such that $$x \rightarrow x^\mu + \varepsilon^\mu $$ we have a change in the field given by $$
\phi \rightarrow \phi - \varepsilon^\mu \partial_\mu \phi $$

which gives a total derivative change in the Lagrange density and so all the equations of motion will be the same so the Lagrangian still applies all well and happy and because of the symmetry after this translation, Neother's theorem now also applies. We can then move to determin the conserved current with

$$\textbf{J}^\mu_{Noether} = \dfrac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi - \varepsilon^\mu\mathcal{L}$$

Now, the scaler field has changed by $$ \delta \phi = -\varepsilon^\mu \partial_\mu\phi $$
therefore, we have that
$$\textbf{J}^\mu_{Noether} = \dfrac{1}{2}\partial^\mu\phi\times -\varepsilon^\mu \partial_\mu\phi - \varepsilon^\mu \mathcal{L}$$
Watch out, it is important to not use the same symbol for indices that are unrelated. Here the indices on ##\varepsilon^\mu \partial_\mu\phi## are unrelated to the index on ##\partial^\mu\phi##, so you should use different indices, since you use ##\mu## on the left side, you have to keep ##\mu## on the ## \partial^\mu\phi##, so use a different index on ##\varepsilon^\mu \partial_\mu\phi##.
 

Related Threads on Energy-momentum tensor from a Lagrangian density?

Replies
7
Views
3K
Replies
2
Views
952
Replies
16
Views
880
Replies
30
Views
994
Replies
16
Views
1K
Replies
15
Views
3K
  • Last Post
Replies
7
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
Top