# I Applying Euler-Lagrange to (real) Klein-Gordon Lagrangian

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1. Jan 24, 2017

### joebentley10

I'm currently studying Quantum Field Theory and I have a confusion about some mathematics in page 30 of Mandl's Quantum Field Theory (Wiley 2010).

Here is a screenshot of the relevant part: https://www.dropbox.com/s/fsjnb3kmvmgc9p2/Screenshot 2017-01-24 18.10.10.png?dl=0

My issue is in finding the equations of motion and showing that they are the Klein-Gordon equation. So first we apply the Euler-Lagrange equation:

$$\frac{\partial}{\partial x^\alpha} \left(\frac{\partial \mathcal{L}}{\partial \phi_{,\alpha}} \right) - \frac{\partial \mathcal{L}}{\partial \phi} = 0$$

where $\mathcal{L}$ is the Lagrangian density and $\phi_{,\alpha} \equiv \frac{\partial \phi}{\partial x^\alpha}$

My confusion arises in the first term of the Euler-Lagrange equation when we substitute in our Lagrangian density. The answer suggests we should find the first term equal to $\Box \phi \equiv \partial_\alpha \partial^\alpha \phi$. If we substitute in for the first term we get:

$$\frac{\partial}{\partial x^\alpha} \left(\frac{\partial}{\partial \phi_{,\alpha}} \left[ \frac{1}{2} \phi_{,\alpha} \phi_,^{\,\alpha} \right]\right)$$

for which we expect the term in $( \dots )$ to equal $\frac{\partial \phi}{\partial x_\alpha} \equiv \partial^\alpha \phi$ so that the answer is $\Box \phi \equiv \partial_\alpha \partial^\alpha \phi$.

I can see how we can use the metric to raise one of the indices before differentiation, so we get a "product rule" which cancels the factor of 2, but then I'm still confused as to why the answer is contravariant (i.e. why the alpha is upstairs) rather than covariant (downstairs). We are differentiate with respect to the derivative of the field, and somehow this gives us a contravariant derivative? I'm not sure about the mathematical justification of this.

If anyone has any mathematical intuition to give me I would be very grateful, when my lecturer went through this it was very handwavey, basically just "a derivative of a derivative makes the index go upstairs", but that doesn't satisfy me mathematically.

2. Jan 24, 2017

### dextercioby

1st rule: Respect the Einstein convention. In your equation alpha is written 4 times, it should be appearing only once (if free) or two times If summed after.
2nd rule: Don't mix partial derivatives shorthand notation. You either use the comma/semi-colon convention throughout, or at all.
3rd rule: Use the metric tensor to place indices in an expression involving derivatives either all upstairs or all downstairs. Then you will automatically discover why the end result has the index in that position.

3. Jan 24, 2017

### joebentley10

Yes sorry about the 1st. About the 2nd it is used like that in Mandl and my QFT course but I understand that it is confusing. I'd usually use the partial derivatives shorthand, but I wanted it to make sense relative to the screenshot.

So in the 3rd. For example if I lower the second derivative w.r.t phi I get,

$$\frac{\partial}{\partial x^\alpha} \left(\frac{\partial}{\partial \phi_{,\alpha}} \left[ \frac{1}{2} g^{\mu\nu} \phi_{,\nu} \phi_{,\mu} \right]\right)$$

and then I differentiate w.r.t $\phi_{,\alpha}$ using the product rule. Well the metric doesn't depend on the field derivative so that's okay, but I still differentiate with respect to the derivative to get,

$$\frac{\partial}{\partial x^\alpha} \left(\frac{\partial}{\partial \phi_{,\alpha}} \left[ \frac{1}{2} g^{\mu\nu} \phi_{,\nu} \phi_{,\mu} \right]\right) = \frac{\partial}{\partial x^\alpha} \left(\phi_{,\alpha}\right)$$

i.e. the derivative is still downstairs which is illegal. I could use the metric to bring them both upstairs instead but I can't differentiate w.r.t that using a downstairs derivative, can I?

EDIT: One sec I think my mistake is neglecting to apply the metric, i.e. I just seem to let it disappear after the differentiation. I will try it again

4. Jan 24, 2017

### joebentley10

Yep indeed it did work after properly applying the metric. Thank you dextercioby!

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