# Energy-momentum tensor from a Lagrangian density?

• Kyri_Phys
In summary, 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$$L = 1/2 \eta^{\mu\nu} \partial_\mu \phi \partial_\nu \phi - 1/2 m^2 \phi^2$$
Kyri_Phys

## 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:
Kyri_Phys said:
$$\delta A^\mu = \varepsilon^\mu (-\partial_\nu A^\mu + \partial^\mu A^\nu )$$

The indices in this equation do not make sense. Why not?

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?

Kyri_Phys said:
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.

George Jones said:
Also, free indices have to match up on both sides of an equation.
So the factor of $$\varepsilon^\mu$$ is incorrect also

George Jones said:
Also, free indices have to match up on both sides of an equation.
I have since realized 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 I've gone wrong

Kyri_Phys said:
I have since realized 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##.

## 1. What is the energy-momentum tensor?

The energy-momentum tensor is a mathematical quantity used in theoretical physics to describe the distribution of energy and momentum in a system. It is a symmetric tensor that represents the density and flux of energy and momentum in space and time.

## 2. How is the energy-momentum tensor derived from a Lagrangian density?

The energy-momentum tensor is derived from the Lagrangian density through the use of Noether's theorem, which states that for every continuous symmetry of a system, there exists a corresponding conserved quantity. In the case of translation invariance, the conserved quantity is the energy-momentum tensor.

## 3. What information does the energy-momentum tensor provide?

The energy-momentum tensor provides information about the energy and momentum content of a physical system. It can also be used to calculate the stress and pressure within a system, as well as the rate of change of energy and momentum.

## 4. What is the significance of the energy-momentum tensor in physics?

The energy-momentum tensor is an important concept in physics, as it allows for the conservation of energy and momentum to be understood and applied in a wide range of physical systems. It is also a crucial component in various field theories, such as general relativity and quantum field theory.

## 5. How is the energy-momentum tensor used in practical applications?

The energy-momentum tensor is used in a variety of practical applications, such as in the study of fluid dynamics, astrophysics, and particle physics. It is also used in the development of models and simulations for energy and momentum transfer in various systems, as well as in the analysis and prediction of physical phenomena.

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