Get the equation of motion given a Lagrangian density

In summary, the conversation discusses the use of Euler-Lagrange equations in a specific problem and focuses on computing a specific term. The expert summarizer provides a step-by-step summary of the conversation, including the equations used and the mistakes made in the computation. The final computation is also provided, along with a comparison to the provided solution.
  • #1
JD_PM
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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...
 
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  • #2
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.

JD_PM said:
$$\partial_{\alpha} \Big( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} \Big) - \frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$
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.
 
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  • #3
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.
 
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  • #4
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.
 
  • #5
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.
 
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  • #6
Gaussian97 said:
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! :biggrin:

Soon I'll get to b)
 
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  • #7
  • #8
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.
 
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  • #9
Gaussian97 said:
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...
 
  • #10
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?
 
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  • #11
Gaussian97 said:
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?
 
  • #12
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)##.
 
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  • #13
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## ?
 
  • #14
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
Gaussian97 said:
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?
 
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  • #15
Gaussian97 said:
although the equation you obtain is a valid one, this is not the way to prove that ##\partial_\alpha \phi^\alpha=0##.

Alright.

Gaussian97 said:
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.
 
  • #16
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
$$\left( \eta_{\alpha \beta} (\partial^{\kappa} \partial_{\kappa} + \mu^2) - \partial_{\alpha} \partial_{\beta} \right) \phi^{\beta} = 0 $$
?
 
  • #17
Do you mean ##\partial_{\alpha}##?
 
  • #18
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.
 
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  • #19
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

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

Related to Get the equation of motion given a Lagrangian density

What is a Lagrangian density?

A Lagrangian density is a mathematical function used in theoretical physics to describe the dynamics of a physical system. It is derived from the Lagrangian, which is a function that describes the total energy of a system in terms of its generalized coordinates and their time derivatives.

What is the equation of motion?

The equation of motion is a mathematical expression that describes how a physical system will evolve over time. It is derived from the Lagrangian density and takes into account all the forces acting on the system.

How is the equation of motion derived from the Lagrangian density?

The equation of motion is derived using the Euler-Lagrange equations, which are a set of differential equations that relate the Lagrangian density to the motion of the system. These equations take into account the principle of least action, which states that the actual path taken by a system is the one that minimizes the action (a measure of the system's energy) over a given time interval.

What are the advantages of using the Lagrangian density to derive the equation of motion?

Using the Lagrangian density to derive the equation of motion allows for a more elegant and concise description of a physical system. It also allows for the incorporation of constraints and symmetries in the system, making it a more powerful and flexible tool for analyzing complex systems.

Can the Lagrangian density be used for all physical systems?

Yes, the Lagrangian density can be used to describe the dynamics of all physical systems, including classical mechanics, electromagnetism, and quantum mechanics. It is a fundamental concept in theoretical physics and has been successfully applied to a wide range of physical phenomena.

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