How does the Klein-Gordon Lagrangian relate to the equations of motion?

In summary, the conversation discusses the application of the Euler-Lagrange equations to a Klein-Gordin Lagrangian density in order to find the equations of motion. The conversation also touches on the use of dummy indices and the product rule in taking derivatives. Eventually, the conversation leads to a discussion on varying the action directly and using Stokes' Theorem to simplify the variation of the action.
  • #1
nylonsmile
8
0
Hi, I hope I put this in the right place!

I'm having trouble with some of the calculus in moving from the Klein-Gordin Lagrangian density to the equations of motion. The density is:

[tex]
L = \frac{1}{2}\left[ (\partial_μ\phi)(\partial^\mu \phi) - m^2\phi ^2 \right]
[/tex]

Now, to apply the Euler-Lagrange equations one needs to find:


[tex]
\frac{\partial L}{\partial(\partial_\mu\phi)}
[/tex]


Which to me, looked like it could be:
[tex]
\frac{\partial L}{\partial(\partial_\mu\phi)} = \frac{1}{2}\partial^\mu\phi
[/tex]

But that gives the wrong equation of motion - the half shouldn't be there. I guess my mistake is that this can sort of be thought of as being like:

[tex]
L = \frac{1}{2}\left[ (\partial_μ\phi)^2 - m^2\phi ^2 \right]
[/tex]

Which works out fine, but I'm just not quite sure what's happening here. What is the best way to think of this? In particular, how does the positioning of μ change how to think of it. This seems really basic but I'm pretty lost! Can anyone help me understand how it works?

Thanks!
 
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  • #2
You should write the Lagrangian as

[tex]\mathcal{L} = \frac12 \Big( \eta^{\mu\nu} \partial_\mu \phi \partial_\nu \phi - m^2 \phi^2 \Big)[/tex]
 
  • #3
Wow that makes so much sense! So the upstairs index is just a short-hand that allows you to avoid writing the metric explicitly. Thank you!
 
  • #4
Actually, I'm still not completely on board. When I take the derivative, I use the product rule? And that's where the factor of 2 comes from?

I think maybe the dummy indices are what are confusing me. If I think of it as:

[tex]
\mathcal{L} = \frac{1}{2}\left(\eta^{ab}\partial_a\phi\partial_b\phi - m^2\phi^2\right)
[/tex]

Then I get it better.
 
  • #5
Exact same doubt here.
 
  • #6
nylonsmile said:
Actually, I'm still not completely on board. When I take the derivative, I use the product rule? And that's where the factor of 2 comes from?

I think maybe the dummy indices are what are confusing me. If I think of it as:

[tex]
\mathcal{L} = \frac{1}{2}\left(\eta^{ab}\partial_a\phi\partial_b\phi - m^2\phi^2\right)
[/tex]

Then I get it better.

That's exactly right. You can't use the same index for the summation as for the derivative; one is a dummy, the other is free. With the form you've written it, the product rule sorts it all out.
 
  • #7
rsouza01 said:
Exact same doubt here.
[tex]\frac{\partial \mathcal{L}}{\partial (\partial_{a}\varphi)} = \frac{1}{2}(\eta^{bc}\delta^{a}_{b}\partial_{c}\varphi + \eta^{bc}\delta^{a}_{c}\partial_{b}\varphi) = \partial^{a}\varphi[/tex] so [tex]\partial_{a}(\frac{\partial \mathcal{L}}{\partial (\partial_{a}\varphi)}) - \frac{\partial \mathcal{L}}{\partial \varphi} = \partial_{a}\partial^{a}\varphi + m^{2}\varphi = 0[/tex].
 
  • #8
Should I understand [itex]\partial_{\mu}\partial_{\nu}[/itex] like [itex](\partial_{\mu})^{2}[/itex]?

Otherwise I'm kind of lost, feeling too stupid asking basic questions here... :cry:
 
  • #9
As you seem to have problems with the index convention, why don't you write out the sum explicitly to see what is going on?
 
  • #10
rsouza01 said:
Should I understand [itex]\partial_{\mu}\partial_{\nu}[/itex] like [itex](\partial_{\mu})^{2}[/itex]?

Otherwise I'm kind of lost, feeling too stupid asking basic questions here... :cry:
Don't feel stupid, the notation can be quite confusing at first (trust me it was insanely confusing for me when I first saw it). ##\partial_{\mu}\partial_{\nu}## isn't the same as ##\partial^{\mu}\partial_{\mu}##. Note that in the second expression, there is an implied summation over the ##\mu## index whereas in the first case there is no such implication. If we are given a Klein-Gordon field ##\varphi## propagating across a background Minkowski space-time, the effect of applying ##\partial^{\mu}\partial_{\mu}## can be seen by choosing a global inertial coordinate system ##(t,x,y,z)## for the background metric. We then have [tex]\partial^{\mu}\partial_{\mu}\varphi = \eta^{\mu\nu}\partial_{\mu}\partial_{\nu}\varphi = \eta^{tt}\partial^{2}_{t}\varphi + \eta^{xx}\partial^{2}_{x}\varphi + \eta^{yy}\partial^{2}_{y}\varphi + \eta^{zz}\partial^{2}_{z}\varphi = -\partial_{t}^{2}\varphi + \nabla^{2}\varphi[/tex]

On the other hand, ##\partial_{\mu}\partial_{\nu}\varphi## just represents any possible second partial derivative of the Klein-Gordin field e.g. it could be ##\partial_{t}\partial_{x}\varphi## or ##\partial_{y}\partial_{z}\varphi## depending on what you choose for the indies ##\mu,\nu##.
 
  • #11
I've already done, just like Griffiths did in Introduction to Elementary Particles, unnumbered equation between 11.12 and 11.13, and in this fashion I felt convinced. Problem is that I really wanted to understand this way with the indexes , which I think is more practical and usual.. (sorry my lame english).

Rodrigo
 
  • #12
Perhaps varying the action directly instead of using the E-L equations will make more sense to you:

[tex]S = \frac{1}{2} \int \left [ \eta^{ab} \partial_a \phi \partial_b \phi-m^2 \phi^2 \right ] d^4 x[/tex]

When varying the action, the metric is constant so:

[tex]\delta S = \frac{1}{2} \int \left [ \eta^{ab} \left [\partial_a \phi ~ \delta (\partial_b \phi)+ \partial_b \phi~ \delta (\partial_a \phi) \right ]-2m^2 \phi \delta \phi \right ] d^4 x[/tex]

The two terms involving variations of the gradient of the field are symmetric, i.e. [itex]\partial_a \phi ~ \delta (\partial_b \phi) = \partial_b \phi~ \delta (\partial_a \phi)[/itex]. Thus, they can be combined into a single term:

[tex]\delta S = \int \left [ \eta^{ab} \partial_a \phi ~ \delta (\partial_b \phi)-m^2 \phi \delta \phi \right ] d^4 x[/tex]

Next you can integrate the first term by parts (using the fact that partial derivatives and variations commute):

[tex]\int \eta^{ab} \partial_a \phi ~ \delta (\partial_b \phi) d^4 x= \int \partial_b (\eta^{ab} \partial_a \phi~ \delta \phi ) d^4 x - \int \partial_b (\eta^{ab} \partial_a \phi) \delta \phi d^4x[/tex]

The first integral is of a total derivative, which can be converted into a surface integral via Stokes' Theorem. Since the variation at the boundary is zero (by definition), the whole first integral is zero. The variation of the action becomes:

[tex]\delta S = -\int \left [ \eta^{ab} \partial_b \partial_a \phi +m^2 \phi \right ] \delta \phi d^4 x=0[/tex]

Since this holds for any [itex]\delta \phi[/itex], what you're left with is:

[tex]\eta^{ab} \partial_b \partial_a \phi + m^2 \phi = 0[/tex]
 

1. What is the Klein-Gordon Lagrangian?

The Klein-Gordon Lagrangian is a mathematical expression that describes the dynamics of a quantum field. It is used in quantum field theory to describe the behavior of scalar particles, such as the Higgs boson.

2. How is the Klein-Gordon Lagrangian derived?

The Klein-Gordon Lagrangian is derived from the Klein-Gordon equation, a relativistic wave equation that describes the behavior of spinless particles. It is obtained by applying the principles of Lagrangian mechanics to the equation.

3. What is the significance of the Klein-Gordon Lagrangian?

The Klein-Gordon Lagrangian is significant because it allows us to make predictions about the behavior of quantum fields and particles. It is an important tool in understanding the fundamental forces and particles of the universe.

4. How does the Klein-Gordon Lagrangian relate to other Lagrangians?

The Klein-Gordon Lagrangian is a special case of the Dirac Lagrangian, which describes the behavior of spin-half particles. It is also related to the Schrödinger Lagrangian, which describes the behavior of non-relativistic quantum systems.

5. Are there any limitations or criticisms of the Klein-Gordon Lagrangian?

One criticism of the Klein-Gordon Lagrangian is that it does not take into account the spin of particles, which is a fundamental property of many particles. It also does not fully describe the interactions between particles, as it only considers free particles. Additionally, it has been shown to lead to some mathematical inconsistencies in certain cases.

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