I have a lot of trouble finding out what the rules are for doing algebra and calculus with 4-vectors. This example shall illustrate one of my problems:

The Lagrangian for a real scalar field is

$$\mathcal{L}=\frac{1}{2}\eta^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi-\frac{1}{2}m^2\phi^2.$$ When trying to solve the Euler-Lagrange equation I do not know how to evaluate $(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)})$. These are the ideas I have:

1. $$(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)})=\frac{1}{2}\eta^{\mu\nu}\partial_{\nu}\phi=\frac{1}{2}\partial^{\mu}\phi.$$

2. The Lagrangian can be written as $\mathcal{L}=\frac{1}{2}\eta^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi-\frac{1}{2}m^2\phi^2=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^2\phi^2$, so


3. But the Lagrangian can also be written as $\mathcal{L}=\frac{1}{2}\partial^{\nu}\phi\partial_{\nu}\phi-\frac{1}{2}m^2\phi^2$, so in this case how do I evaluate it? Do I change $(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)})$ to $(\frac{\partial\mathcal{L}}{\partial(\partial_{\nu}\phi)})$?

4. We can write the Lagrangian as $\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^2\phi^2=\frac{1}{2}(\partial_{\mu}\phi)^2-\frac{1}{2}m^2\phi^2$, so in this case we have


None of these are correct $(\partial^{\mu}\phi)$! How do I get the correct answer? What am I doing wrong? Are there any resources where they can help me specifically with such problems?


You seem to be having some problems with index notation and the Einstein summation convention, so I recommend brushing up on those.

Firstly, the $\mu$ in $\mathcal{L}$ is a dummy index, while the $\mu$ in $\partial/\partial\left(\partial_\mu\phi\right)$ is a live index. You cannot write both of them as $\mu$ or else you will run into problems. For example, if you are going to use $\mu$ in $\partial/\partial\left(\partial_\mu\phi\right)$, you should change the dummy indices in $\mathcal{L}$ to something like $$\mathcal{L}=\frac{1}{2}\eta^{\rho\lambda}\partial_{\rho}\phi\partial_{\lambda}\phi-\frac{1}{2}m^2\phi^2$$

Secondly, $\partial^\mu \phi$ is not independent of $\partial_\mu\phi$, so you cannot treat it as a constant in your second point. In addition, in your fourth point, you cannot possibly have $\partial_\mu\phi \partial^\mu\phi = \left(\partial_\mu\phi\right)^2$ as you have a dummy index on the left but a live index on the right.

Lastly, the derivative of one component with respect to another component of the same object is a delta function. For example, $$\frac{\partial v_1}{\partial v_\mu} = \delta^1_\mu$$ since components are linearly independent. Then you can apply this to the expanded expression $\partial_\mu\phi \partial^\mu\phi = -\partial_0\phi \partial_0\phi+\partial_1\phi \partial_1\phi+\partial_2\phi \partial_2\phi+\partial_3\phi \partial_3\phi$. Alternatively, if you are confident enough, you could also apply the product rule directly to $\eta^{\rho\lambda}\partial_{\rho}\phi\partial_{\lambda}\phi$.

Hope this clears up (at least some of) your confusion.

  • 1
    $\begingroup$ Explicitly, by the product rule$$\tfrac{\partial}{\partial(\partial_\mu\phi)}(\tfrac12\eta^{\rho\lambda}\partial_\rho\phi\partial_\lambda\phi)=\tfrac12\eta^{\rho\lambda}(\delta_\rho^\mu\partial_\lambda\phi+\partial_\rho\phi\delta_\lambda^\mu)=\tfrac12(\eta^{\mu\lambda}\partial_\lambda\phi+\eta^{\rho\mu}\partial_\rho\phi)=\partial^\mu\phi.$$ $\endgroup$
    – J.G.
    Aug 22 at 9:00
  • $\begingroup$ Your answer and the comment by J.G. helped a lot. I can now see how to get the correct answer. However I've got one more question. I know what $\partial_{\mu}x^{\mu}$ is, as you said $\mu$ here is a dummy index so I have to expand it. But what is $\partial_{\mu}\phi$? Is it the 4-vector $(\partial_{t}\phi,\partial_{x}\phi,\partial_{y}\phi,\partial_{}\phi)$ or just a representation for the 4 possible partial derivatives? My textbook defines $\partial_{\mu}$ as the operator $(\partial_{t},\partial_{x},\partial_{y},\partial_{z})$. $\endgroup$ Aug 22 at 9:40
  • $\begingroup$ @ColourfulSpacetime It is a representation for the 4 possible partial derivatives, one in each coordinate direction. The collection of partial derivatives form the components of the covector field (or one-form) $\text{d}\phi = \partial_\mu\phi \text{d}x^\mu$. Note that this only works for scalars; for tensors you'll need to use the covariant derivative. $\endgroup$ Aug 22 at 9:47

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