# Klein-Gordon Lagrangian

#### nylonsmile

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:

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

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

$$\frac{\partial L}{\partial(\partial_\mu\phi)}$$

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

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:

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

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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#### Ben Niehoff

Gold Member
You should write the Lagrangian as

$$\mathcal{L} = \frac12 \Big( \eta^{\mu\nu} \partial_\mu \phi \partial_\nu \phi - m^2 \phi^2 \Big)$$

#### nylonsmile

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!

#### nylonsmile

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:

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

Then I get it better.

#### rsouza01

Exact same doubt here.

#### VantagePoint72

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:

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

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.

#### WannabeNewton

Exact same doubt here.
$$\frac{\partial \mathcal{L}}{\partial (\partial_{a}\varphi)} = \frac{1}{2}(\eta^{bc}\delta^{a}_{b}\partial_{c}\varphi + \eta^{bc}\delta^{a}_{c}\partial_{b}\varphi) = \partial^{a}\varphi$$ so $$\partial_{a}(\frac{\partial \mathcal{L}}{\partial (\partial_{a}\varphi)}) - \frac{\partial \mathcal{L}}{\partial \varphi} = \partial_{a}\partial^{a}\varphi + m^{2}\varphi = 0$$.

#### rsouza01

Should I understand $\partial_{\mu}\partial_{\nu}$ like $(\partial_{\mu})^{2}$?

Otherwise I'm kind of lost, feeling too stupid asking basic questions here...

#### DrDu

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?

#### WannabeNewton

Should I understand $\partial_{\mu}\partial_{\nu}$ like $(\partial_{\mu})^{2}$?

Otherwise I'm kind of lost, feeling too stupid asking basic questions here...
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 $$\partial^{\mu}\partial_{\mu}\varphi = \eta^{\mu\nu}\partial_{\mu}\partial_{\nu}\varphi = \eta^{tt}\partial^{2}_{t}\varphi + \eta^{xx}\partial^{2}_{x}\varphi + \eta^{yy}\partial^{2}_{y}\varphi + \eta^{zz}\partial^{2}_{z}\varphi = -\partial_{t}^{2}\varphi + \nabla^{2}\varphi$$

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 $\mu,\nu$.

#### rsouza01

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

#### elfmotat

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

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

When varying the action, the metric is constant so:

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

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

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

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

$$\int \eta^{ab} \partial_a \phi ~ \delta (\partial_b \phi) d^4 x= \int \partial_b (\eta^{ab} \partial_a \phi~ \delta \phi ) d^4 x - \int \partial_b (\eta^{ab} \partial_a \phi) \delta \phi d^4x$$

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:

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

Since this holds for any $\delta \phi$, what you're left with is:

$$\eta^{ab} \partial_b \partial_a \phi + m^2 \phi = 0$$