Get the equation of motion given a Lagrangian density

[JD_PM]

Homework Statement

Given the Lagrangian density for the real vector field $\phi^{\alpha} (x)$



$$ \mathcal{L} = - \frac 1 2 (\partial_{\alpha} \phi_{\beta}) (\partial^{\alpha} \phi^{\beta}) + \frac 1 2 (\partial_{\alpha} \phi^{\alpha}) (\partial_{\beta} \phi^{\beta}) + \frac{\mu^2}{2} \phi_{\alpha} \phi^{\alpha}$$



a) Show that the equation of motion is given by



$$\Big( \eta_{\alpha \beta} (\partial^{k} \partial_{k} + \mu^2) - \partial_{\alpha} \partial_{\beta} \Big) \phi^{\beta} = 0$$



b) Show that $\partial_{\alpha} \phi^{\alpha} = 0$


*Source: QFT book by Franz Mandl and Graham Shaw; second edition

Relevant Equations

$$ \mathcal{L} = - \frac 1 2 (\partial_{\alpha} \phi_{\beta}) (\partial^{\alpha} \phi^{\beta}) + \frac 1 2 (\partial_{\alpha} \phi^{\alpha}) (\partial_{\beta} \phi^{\beta}) + \frac{\mu^2}{2} \phi_{\alpha} \phi^{\alpha}$$

a)

Alright here we have to use Euler-Lagrange equation

$$\partial_{\alpha} \Big( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} \Big) - \frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$

Let's focus on the term $\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )}$

I know that

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( - \frac 1 2 (\partial_{\alpha} \phi_{\beta}) (\partial^{\alpha} \phi^{\beta}) \Big) = - \partial^{\alpha} \phi^{\beta}$$

After reviewing my work, I think that my mistake has to be in computing the term

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( \frac 1 2 (\partial_{\alpha} \phi^{\alpha}) (\partial_{\beta} \phi^{\beta}) \Big)$$

What I have tried:

Note I have swapped $\alpha \rightarrow \gamma$ and $\beta \rightarrow b$ in the original Lagrangian

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( \frac 1 2 (\partial_{\gamma} \phi^{\gamma}) (\partial_{b} \phi^{b}) \Big) = \frac 1 2 \eta^{\gamma c} \partial_{b} \phi^{b} \frac{\partial(\partial_{\gamma} \phi_{c})}{\partial (\partial_{\mu} A_{\nu})} + \frac 1 2 \eta^{b d} \partial_{\gamma} \phi^{\gamma} \frac{\partial(\partial_{b} \phi_{d})}{\partial (\partial_{\mu} A_{\nu})} = \frac 1 2 \eta^{\gamma c} \partial_{b} \phi^{b} \delta_{\gamma}^{\mu} \delta_{c}^{\nu} + \frac 1 2 \eta^{b d} \partial_{\gamma} \phi^{\gamma} \delta_{b}^{\mu} \delta_{d}^{\nu} = \frac 1 2( \partial_{b} \phi^b + \partial_{\gamma} \phi^{\gamma}) \eta^{\mu \nu}$$

Mmm but I do not see where I got wrong...

 

[strangerep]

I think you should review Orodruin's insight about how to use indices and implicit summation. Your index usage is all over the place. You also seem to swap $A$ for $\phi$ in random places.

$$\partial_{\alpha} \Big( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} \Big) - \frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$

Free indices need to match in all terms.

I know that $$\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( - \frac 1 2 (\partial_{\alpha} \phi_{\beta}) (\partial^{\alpha} \phi^{\beta}) \Big) = - \partial^{\alpha} \phi^{\beta}$$

You have $\alpha,\beta$ as summation indices on the LHS, but also as free indices on both sides. Use a different pair of symbols for the free indices.

The rest of your problems seem likewise due to inconsistent use of indices.

 

[JD_PM]

Oops I should have checked it better, there are many mistakes.

Let me correct them.

a)

Alright here we have to use Euler-Lagrange equation

$$\partial_{\alpha} \Big( \frac{\partial \mathcal{L}}{\partial(\partial_{\alpha} \phi_{\beta})} \Big) - \frac{\partial \mathcal{L}}{\partial \phi_{\beta}} = 0$$

Let's focus on the term $\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )}$

I know that

$$\frac{\partial}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( - \frac 1 2 (\partial_{\gamma} \phi_{b}) (\partial^{\gamma} \phi^{b}) \Big) = - \partial^{\alpha} \phi^{\beta} \ \ \ \ (1)$$

After reviewing my work, I think that my mistake has to be in computing the term

$$\frac{\partial}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( \frac 1 2 (\partial_{\gamma} \phi^{\gamma}) (\partial_{b} \phi^{b}) \Big)$$

What I have tried:

Note I have swapped $\alpha \rightarrow \gamma$ and $\beta \rightarrow b$ in the original Lagrangian

$$\frac{\partial }{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( \frac 1 2 (\partial_{\gamma} \phi^{\gamma}) (\partial_{b} \phi^{b}) \Big) = \frac 1 2 \eta^{\gamma c} \partial_{b} \phi^{b} \frac{\partial(\partial_{\gamma} \phi_{c})}{\partial (\partial_{\alpha} \phi_{\beta})} + \frac 1 2 \eta^{b d} \partial_{\gamma} \phi^{\gamma} \frac{\partial(\partial_{b} \phi_{d})}{\partial (\partial_{\alpha} \phi_{\beta})} = \frac 1 2 \eta^{\gamma c} \partial_{b} \phi^{b} \delta_{\gamma}^{\alpha} \delta_{c}^{\beta} + \frac 1 2 \eta^{b d} \partial_{\gamma} \phi^{\gamma} \delta_{b}^{\alpha} \delta_{d}^{\beta} = \frac 1 2( \partial_{b} \phi^b + \partial_{\gamma} \phi^{\gamma}) \eta^{\alpha \beta} \ \ \ \ (2)$$

Mmm but I do not see where I got wrong...

Hopefully now it is clear.

 

[JD_PM]

For the sake of completeness, let's show the whole computation for a)

We already know that

$$\frac{\partial}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( - \frac 1 2 (\partial_{\gamma} \phi_{b}) (\partial^{\gamma} \phi^{b}) \Big) = - \partial^{\alpha} \phi^{\beta} \ \ \ \ (1)$$

And

$$\frac{\partial }{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( \frac 1 2 (\partial_{\gamma} \phi^{\gamma}) (\partial_{b} \phi^{b}) \Big) = \frac 1 2( \partial_{b} \phi^b + \partial_{\gamma} \phi^{\gamma}) \eta^{\alpha \beta} \ \ \ \ (2)$$

We're left to show the computation for $\frac{\partial \mathcal{L}}{\partial \phi_{\beta}}$

$$\frac{\partial \mathcal{L}}{\partial \phi_{\beta}} = \frac{\partial}{\partial \phi_{\beta}} \Big( \phi_{\gamma} \phi^{\gamma} \Big) = \delta_{\gamma}^{\beta} \phi^{\gamma} + \eta^{\gamma e} \phi_{\gamma} \frac{\partial \phi_e}{\partial \phi_{\beta}} = \delta_{\gamma}^{\beta} \phi^{\gamma} + \eta^{\gamma e} \phi_{\gamma} \delta_e^{\beta} = \phi^{\beta} + \phi^{\beta} = 2 \phi^{\beta} \ \ \ \ (3)$$

Thus the equation of motion I get is

$$- \partial^{\alpha} \partial_{\alpha} \phi^{\beta} + \frac 1 2 \partial^{\beta} ( \partial_b \phi^b + \partial_{\gamma} \phi^{\gamma}) - \mu^2 \phi^{\beta} = 0$$

As $b$ and $\gamma$ are dummy indices, we can label them as we wish; let me label them $k$. Thus we can simplify the equation of motion:

$$- \partial^{\alpha} \partial_{\alpha} \phi^{\beta} + \partial^{\beta} ( \partial_k \phi^k) - \mu^2 \phi^{\beta} = 0$$

But this is not the provided solution

$$\Big( \eta_{\alpha \beta} (\partial^{k} \partial_{k} + \mu^2) - \partial_{\alpha} \partial_{\beta} \Big) \phi^{\beta} = 0$$

Mmm where did I get wrong?

Thank you.

 

[Gaussian97]

It's correct, the two equations $- \partial^{\alpha} \partial_{\alpha} \phi^{\beta} + \partial^{\beta} ( \partial_k \phi^k) - \mu^2 \phi^{\beta} = 0$ and $\Big( \eta_{\alpha \beta} (\partial^{k} \partial_{k} + \mu^2) - \partial_{\alpha} \partial_{\beta} \Big) \phi^{\beta} = 0$ are equivalent. To realize that first of all notice that the solution has $\alpha$ as a free index, while your equation has $\beta$, so I would recommend you to relabel your indices to make $\alpha$ the free index. Once you have done this, try to write all your $\phi$ with index $\beta$ contravariant, after this manipulations you may see clearer that these two equations are the same.

 

[JD_PM]

I would recommend you to relabel your indices to make $\alpha$ the free index. Once you have done this, try to write all your $\phi$ with index $\beta$ contravariant, after this manipulations you may see clearer that these two equations are the same.

Mmm alright let's check it out explicitly.

Let's manipulate a bit:

$$- \partial^{\gamma} \partial_{\gamma} \phi^{\alpha} + \partial^{\alpha} ( \partial_k \phi^k) - \mu^2 \phi^{\alpha} = 0$$

$$- \partial^{\gamma} \partial_{\gamma} \phi^{\alpha} + \eta^{\alpha f} \partial_{f} ( \partial_k \phi^k) - \mu^2 \phi^{\alpha} = 0$$

Which is equivalent to

$$- \partial^{\gamma} \partial_{\gamma} \phi_{\alpha} + \eta_{\alpha f} \partial^{f} ( \partial_k \phi^k) - \mu^2 \phi_{\alpha} = 0$$

Then we finally get

$$- \partial^{\gamma} \partial_{\gamma} \phi_{\alpha} + \partial_{\alpha} ( \partial_k \phi^k) - \mu^2 \phi_{\alpha} = 0$$

Swapping $\gamma \rightarrow k$ and $k \rightarrow \beta$ in the above equation:

$$- \partial^{k} \partial_{k} \phi_{\alpha} + \partial_{\alpha} ( \partial_{\beta} \phi^{\beta}) - \mu^2 \phi_{\alpha} = 0$$

Which is the provided solution! Thanks Gaussian97!

Soon I'll get to b)

 

[JD_PM]

Mmm but how can we show that $\partial_{\alpha} \phi^{\alpha} = 0$?

I am used to be asked to show that $\partial_{\mu} T^{\mu \nu} = 0$, where $T^{\mu \nu}$ is a tensor. Here we simply plug the tensor formula and start manipulating, as I did here: https://www.physicsforums.com/threads/computing-an-energy-momentum-tensor-given-a-lagrangian.985353/

But in this exercise we do not have an explicit formula for $\phi^{\alpha}$, so that method does not work...

 

[Gaussian97]

Well, note that your equation $- \partial^{\kappa} \partial_{\kappa} \phi_{\alpha} + \partial_{\alpha} ( \partial_{\beta} \phi^{\beta}) - \mu^2 \phi_{\alpha} = 0$ has a free index, while the equation you want $\partial_\alpha \phi^\alpha=0$ has no free index.
With that in mind is always a good idea to take your equation an try to contract the free index with some tensor.

PD: In the literature is a very established convention that all the indices that run between 0 and 3 are written with Greek letters, while the Latin letters are reserved for indices that run only between 1 and 3. I encourage you to follow this convention.

 

[JD_PM]

With that in mind is always a good idea to take your equation an try to contract the free index with some tensor.

Oh so do you mean doing the following?

$$- \partial^{\kappa} \partial_{\kappa} \phi_{\epsilon} T^{\epsilon \alpha} + \partial_{\epsilon} T^{\epsilon \alpha}( \partial_{\beta} \phi^{\beta}) - \mu^2 \phi_{\epsilon} T^{\epsilon \alpha}= 0$$

I do not see why this leads to $\partial_{\alpha} \phi^{\alpha} = 0$ though...

 

[Gaussian97]

Well, you still have the same problem as before, your equation has a free index $\alpha$, while the equation you want to get has no free index... so sure contracting with a rank 2 tensor will not help you in this. Worse, now you have a dependence on some unknown tensor $T$, so of course, this cannot be the path.

Your LHS is a tensor of rank 1, and you want something without indexes (a tensor of rank 0), so you need to contract with another tensor of rank 1. Furthermore, you don't want any dependence in an arbitrary tensor, so how many tensors of rank 1 do you know?

 

[JD_PM]

Well, you still have the same problem as before, your equation has a free index $\alpha$, while the equation you want to get has no free index... so sure contracting with a rank 2 tensor will not help you in this. Worse, now you have a dependence on some unknown tensor $T$, so of course, this cannot be the path.

Your LHS is a tensor of rank 1, and you want something without indexes (a tensor of rank 0), so you need to contract with another tensor of rank 1. Furthermore, you don't want any dependence in an arbitrary tensor, so how many tensors of rank 1 do you know?

Ahhh so we have

$$- \partial^{\kappa} \partial_{\kappa} \phi_{\alpha} \phi^{\alpha} + \partial_{\alpha} \phi^{\alpha}( \partial_{\beta} \phi^{\beta}) - \mu^2 \phi_{\alpha} \phi^{\alpha}= 0$$

$$\Big( - \partial^{\kappa} \partial_{\kappa} - \mu^2 \Big) \phi_{\alpha} \phi^{\alpha} + \partial_{\alpha} \phi^{\alpha} (\partial_{\beta} \phi^{\beta}) = 0$$

Mmm but how to continue?

 

[Gaussian97]

Well, ok, one possible rank 1 tensor that you have is $\phi^\alpha$, but it's not the only one, there's another rank 1 tensor that you can use to contract your expression.

PD: This is not important in order to solve the problem, but your contraction with $\phi^\alpha$ is wrong, the second term should be $\phi^\alpha \left(\partial_\alpha \partial_\beta \phi^\beta\right)$, not $\left(\partial_\alpha \phi^\alpha\right)\left(\partial_\beta\phi^\beta\right)$.

 

[JD_PM]

Ahhh I see the trick! It is about using $-\partial^{\kappa} \partial_{\kappa} \phi_{\alpha} + \partial_{\alpha} ( \partial_{\beta} \phi^{\beta}) - \mu^2 \phi_{\alpha} = 0$ at certain point.

We proceed as follows

$$-\partial^{\kappa} \partial_{\kappa} \phi_{\alpha} \phi^{\alpha} + \phi^{\alpha} \partial_{\alpha} ( \partial_{\beta} \phi^{\beta}) - \mu^2 \phi_{\alpha} \phi^{\alpha}= 0$$

Which can be rearranged as follows

$$\Big( -\partial^{\kappa} \partial_{\kappa} \phi_{\alpha} + \partial_{\alpha} ( \partial_{\beta} \phi^{\beta}) - \mu^2 \phi_{\alpha} \Big) \phi^{\alpha} = 0$$

But how does the above equation justify that $\partial_{\alpha} \phi^{\alpha} = 0$ ?

 

[Gaussian97]

Well, you've just done the same as in the previous post (except that now you have contract correctly) I said that, although the equation you obtain is a valid one, this is not the way to prove that $\partial_\alpha \phi^\alpha=0$.
Let's return to the post

Your LHS is a tensor of rank 1, and you want something without indexes (a tensor of rank 0), so you need to contract with another tensor of rank 1. Furthermore, you don't want any dependence in an arbitrary tensor, so how many tensors of rank 1 do you know?

In the equation
$$\left( \eta_{\alpha \beta} (\partial^{\kappa} \partial_{\kappa} + \mu^2) - \partial_{\alpha} \partial_{\beta} \right) \phi^{\beta} = 0 $$
there are two rank-1 tensors involved, one is $\phi^\beta$, which is the other one?

 

[JD_PM]

although the equation you obtain is a valid one, this is not the way to prove that $\partial_\alpha \phi^\alpha=0$.

Alright.

In the equation
$$\left( \eta_{\alpha \beta} (\partial^{k} \partial_{k} + \mu^2) - \partial_{\alpha} \partial_{\beta} \right) \phi^{\beta} = 0 $$
there are two rank-1 tensors involved, one is $\phi^\beta$, which is the other one?

$\eta_{\alpha \beta} \phi^{\beta} = \phi_{\alpha}$ but I am afraid I still do not see your point.

 

[Gaussian97]

Well, for me $\phi_\alpha$ and $\phi^\alpha$ are essentially the same tensor. What else, aside from $\phi$ carries a single Lorentz index in the equation
$$\left( \eta_{\alpha \beta} (\partial^{\kappa} \partial_{\kappa} + \mu^2) - \partial_{\alpha} \partial_{\beta} \right) \phi^{\beta} = 0 $$
?

 

[JD_PM]

Do you mean $\partial_{\alpha}$?

 

[Gaussian97]

Yes, exactly, now try to contract the equation $\left( \eta_{\alpha \beta} (\partial^{\kappa} \partial_{\kappa} + \mu^2) - \partial_{\alpha} \partial_{\beta} \right) \phi^{\beta} = 0$ with $\partial^{\alpha}$, and see what conclusions can you extract from here.

 

[JD_PM]

Alright I got it!

$$\partial^{\alpha} \phi_{\alpha} ( \partial^{k} \partial_{k} + \mu^2) - \partial^{\alpha} \partial_{\alpha} (\partial_{\beta} \phi^{\beta}) = 0$$

$$\Big( \partial^{k} \partial_{k} + \mu^2 - \partial^{k} \partial_{k} \Big) \partial_{\beta} \phi^{\beta} = 0$$

$$\mu^2 \partial_{\alpha} \phi^{\alpha} = 0$$

Assuming that $\mu \neq 0$ we indeed get

$$\partial_{\alpha} \phi^{\alpha} = 0$$