Differentiating Lagrangian in Tensor Notation

I guess one equation, since you're just taking the trace. But it's better to be consistent throughout and keep all the indices Greek.I think this might be right actually, but I'm still uneasy about the double covariant derivative product; what does that actually represent physically?It's just the wave operator acting on the field. The equation is saying that the wave operator acting on the field is proportional to the field itself.
  • #1
JackDP
7
0

Homework Statement


Hi all, I'm trying to learn how to manipulate tensors and in particular to differentiate expressions. I was looking at a Lagrangian density and trying to apply the Euler-Lagrange equations to it.

Homework Equations


Lagrangian density:
[tex]\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{1}{2}\mu^2 \phi^{\alpha} \phi_{\alpha}[/tex]

Euler-Lagrange:
[tex]\frac{\partial \mathcal{L}}{\partial \phi^i} = \partial^k \frac{\partial \mathcal{L}}{\partial \phi^{i,k}}[/tex]

The Attempt at a Solution


I have attempted to differentiation the expression several times; I can compute [tex]\frac{\partial \mathcal{L}}{\partial \phi^i}[/tex] with no problems and can compute [tex]\frac{\partial \mathcal{L}}{\partial \phi^{i,k}}[/tex] for the first and third terms.

However, I just cannot figure out how to differentiate the middle term. My attempt:

[tex]\mathcal{L}_2 = \frac{1}{2} \partial_{\alpha} \phi^{\alpha} \partial_{\beta} \phi^{\beta}
= \frac{1}{2} g_{\alpha \lambda} g_{\beta \sigma} \partial^{\lambda} \phi^{\alpha} \partial^{\sigma} \phi^{\beta}[/tex]
Hence
[tex] \frac{\partial \mathcal{L}_2}{\partial \phi^{i,k}} =
\frac{1}{2} g_{\alpha \lambda} g_{\beta \sigma} \left(
\delta_k^{\lambda} \delta_i^{\alpha} \partial^{\sigma} \phi^{\beta} +
\delta_k^{\sigma} \delta_i^{\beta} \partial^{\lambda} \phi^{\alpha}
\right)
= \frac{1}{2} \left(
g_{i k} \partial_{\beta} \phi^{\beta} +
g_{i k} \partial_{\alpha} \phi^{\alpha}
\right)
= g_{i k} \phi_i \phi^i
[/tex]

So as you can see, I have somehow picked up this additional factor of the metric. I'm not sure what to do with it, or where I have gone wrong!

Best wishes,
J
 
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  • #2
The final step should be ##g_{ik}\partial_j\phi^j##. But other than the ##\partial## that got swapped for a ##\phi## (typo?) and the choice to call the summation index i (the same as one of the free indices), I don't see anything wrong with what you did. Are you sure that the answer isn't supposed to be ##g_{ik}\partial_j\phi^j##?

I think it's pretty natural for the metric to show up there. If the metric is diagonal, the expression you're differentiating with respect to ##\partial^i\phi^k## contains no factors of the form ##\partial^l\phi^m## with ##l\neq m##, so the derivative should contain a factor that's zero when ##i\neq k##.
 
  • Like
Likes JackDP
  • #3
Fredrik said:
The final step should be ##g_{ik}\partial_j\phi^j##. But other than the ##\partial## that got swapped for a ##\phi## (typo?) and the choice to call the summation index i (the same as one of the free indices), I don't see anything wrong with what you did. Are you sure that the answer isn't supposed to be ##g_{ik}\partial_j\phi^j##?

I think it's pretty natural for the metric to show up there. If the metric is diagonal, the expression you're differentiating with respect to ##\partial^i\phi^k## contains no factors of the form ##\partial^l\phi^m## with ##l\neq m##, so the derivative should contain a factor that's zero when ##i\neq k##.
Oh, yes, that is a typo. It is meant to be a ##\partial## not a ##\phi##. My decision to call the sum index ##i## was just because the derivative was taken with respect to ##\phi^{i,j}## but this might be bad practice. I'm not familiar with the conventions people use when using tensors because I've really only been doing it a few days. I actually couldn't find anything online through google about doing this kind of thing, which is what led me to ask here!

As for the answer being "wrong", I suppose it might not be – I just didn't expect to have a metric in the result, since the next step is to calculate ##\partial^k## of the answer as per the E-L equation. With the metric term, this would give

[tex]\partial^k g_{ik}\partial_j \phi^j = \partial_i \partial_j \phi^j[/tex]. Is this a legitimate expression?

Taken with the other terms, the final E-L would then be:
[tex]g_{i j} \mu^2 \phi^j = \partial^k (-\partial_k \phi_i + g_{ik} \partial_j \phi^j) = - \square \phi_i + \partial_i \partial_j \phi^j = - g_{i j} \square \phi^j + \partial_i \partial_j \phi^j[/tex]
or, rearranging
[tex][g_{i j}(\mu^2 + \square) - \partial_i \partial_j]\phi^j = 0[/tex]

I think this might be right actually, but I'm still uneasy about the double covariant derivative product; what does that actually represent physically?
 
  • #4
I found a simpler way to get the result you got. We have
$$\mathcal L_2=\frac 1 2 \partial_\alpha\phi^\alpha \partial_\beta\phi^\beta =\frac 1 2(\partial_\alpha\phi^\alpha)^2.$$ So by the chain rule, we have
$$\frac{\partial\mathcal L_2}{\partial(\partial_\mu\phi^\nu)}=\partial_\alpha\phi^\alpha \frac{\partial}{\partial(\partial_\mu\phi^\nu)}\partial_\beta\phi^\beta =\partial_\alpha\phi^\alpha\delta^\mu_\nu.$$ My left-hand side is however different from yours. If I just change your Latin indices to Greek, your left-hand side is ##\frac{\partial\mathcal L_2}{\partial(\partial^\mu\phi^\nu)}##. We get
$$\frac{\partial\mathcal L_2}{\partial(\partial^\mu\phi^\nu)}=g_{\mu\rho} \frac{\partial\mathcal L_2}{\partial(\partial_\rho\phi^\nu)}=g_{\mu\rho} \partial_\alpha\phi^\alpha\delta^\rho_\nu =g_{\mu\nu}\partial_\alpha\phi^\alpha.$$ I would write the Euler-Lagrange equation for ##\mathcal L_2## as
$$0=\frac{\partial L_2}{\partial \phi^\nu}-\partial_\mu\left(\frac{\partial\mathcal L_2}{\partial(\partial_\mu\phi^\nu)}\right) =-\partial_\mu\left(\partial_\alpha\phi^\alpha \delta^\mu_\nu\right) =-\delta^\mu_\nu\partial_\mu \partial_\alpha\phi^\alpha =-\partial_\nu\partial_\alpha\phi^\alpha.$$

JackDP said:
My decision to call the sum index ##i## was just because the derivative was taken with respect to ##\phi^{i,j}## but this might be bad practice.
Yes, that's actually the reason why you can't use i or j as the summation index, but any other symbol will do.

You should also start following the convention to have Latin indices run from 1 to 3 and Greek indices from 0 to 3. For example ##A^i_i## and ##A^\mu_\mu## don't mean the same thing. The former means ##A^1_1+A^2_2+A^3_3## and the latter means ##A^0_0+A^1_1+A^2_2+A^3_3##.

JackDP said:
With the metric term, this would give

[tex]\partial^k g_{ik}\partial_j \phi^j = \partial_i \partial_j \phi^j[/tex]. Is this a legitimate expression?
Yes. But the indices should be Greek throughout the calculation. Otherwise your sums will be interpreted as having three terms rather than four, and your result will be interpreted as three equations rather than four.
 
Last edited:

Related to Differentiating Lagrangian in Tensor Notation

1. What is the Lagrangian in Tensor Notation?

The Lagrangian, in tensor notation, is a mathematical function used in the study of mechanics to describe the dynamics of a system. It is defined as the difference between the kinetic and potential energy of a system, and is commonly used in the field of theoretical physics.

2. How is the Lagrangian differentiated in Tensor Notation?

The Lagrangian is differentiated in tensor notation using the Einstein summation convention, where repeated indices are summed over. This notation allows for a more compact and simplified representation of the equations involved in differentiating the Lagrangian.

3. What is the importance of Tensor Notation in differentiating the Lagrangian?

Tensor notation is important in differentiating the Lagrangian because it allows for a concise and elegant representation of the equations involved. It also simplifies the process of differentiating complex equations and helps to avoid errors in calculations.

4. What are the benefits of using Tensor Notation in Lagrangian mechanics?

Using tensor notation in Lagrangian mechanics has several benefits, such as simplifying the equations involved, making them easier to solve and manipulate. It also allows for a more general and efficient way of representing physical systems, making it a useful tool in theoretical physics.

5. Are there any limitations to using Tensor Notation in differentiating the Lagrangian?

While tensor notation is a powerful tool in differentiating the Lagrangian, it can be challenging for beginners to understand and apply. It also has limitations in representing certain types of physical systems, such as those involving non-inertial frames of reference. In these cases, alternative methods may be necessary.

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