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How can I derive the Euler-Lagrange equations valid in the field of special relativity? Specifically, consider a scalar field.

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Short answer: take the action for your (special relativistic) Lagrangian and use variational principle... I might consider writing a detailed answer for this, but first let us know what kind of system are you dealing with. At the very least, whether it has finite or infinite number of degrees of freedom (i.e. particles vs. fields). –  Marek Dec 6 '10 at 16:16
    
I want to consider a scalar field $\Phi(x)$ –  Andrea Amoretti Dec 6 '10 at 16:25
    
Hint: Particles travel along paths of maximum proper time. –  Matt Dec 6 '10 at 16:35
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@Matt: there is no proper time if the particle is massless (e.g. photon). It's better to consider general parameter of the path and let the theory posses gauge reparametrization freedom. It gives more insight into the problem and makes the quantization easier (if one ever wanted to do it). –  Marek Dec 6 '10 at 16:40
    
I edited your question a little (I hope in a way that agrees with what you want). If your intentions with this question were different, please provide more details. –  Marek Dec 6 '10 at 17:27

1 Answer 1

up vote 13 down vote accepted

General approach

First recall that Euler-Lagrange equations are conditions for the vanishing of the variation of action $S$. For a scalar field $\Phi$ with Lagrangian density $\mathcal L$ on some open subset U we have

$$S[\Phi] = \int_U {\mathcal L}(\Phi(x), \partial^{\mu}\Phi(x)) {\rm d}^4 x$$

Consider a variation of the field in direction $\chi$ and compute

$$S[\Phi + \varepsilon \chi] = \int_M {\mathcal L}(\Phi(x) + \varepsilon \chi(x), \partial^{\mu}(\Phi(x) + \varepsilon \chi(x))) {\rm d}^4 x$$ Then using Taylor expansion $$S[\Phi + \varepsilon \chi] - S[\Phi] = \int_U \left[ \varepsilon \chi(x) {\partial{\mathcal L} \over \partial \Phi}(\Phi(x), \partial^{\mu}\Phi(x)) + \varepsilon (\partial^{\mu} \chi(x)) {\partial{\mathcal L} \over \partial (\partial^{\mu} \Phi)}(\Phi(x), \partial^{\mu}\Phi(x)) + O(\varepsilon^2) \right] {\rm d}^4 x$$

Using integration by parts on the second term (assuming $\chi$ vanishes on $\partial U$), diving by $\varepsilon$ on both sides and letting $\varepsilon \to 0$ this becomes a variation in direction $\chi$

$$\delta S [\Phi][\chi] = \int_U \chi(x) \left[ {\partial{\mathcal L} \over \partial \Phi}(\Phi(x), \partial^{\mu}\Phi(x)) - \partial^{\mu}\left( {\partial{\mathcal L} \over \partial (\partial^{\mu} \Phi)}(\Phi(x), \partial^{\mu}\Phi(x))\right) \right] {\rm d}^4 x$$

By requiring variations in all directions equal zero we obtain

$$ {\partial{\mathcal L} \over \partial \Phi} - \partial^{\mu}\left( {\partial{\mathcal L} \over \partial (\partial^{\mu} \Phi)}\right) = 0 $$

(arguments the same as always, so omitted).

Massive scalar field example

Consider Lagrangian density $${\mathcal L} = {1 \over 2}\eta_{\mu \nu} \partial^{\mu} \Phi \partial^{\nu} \Phi - {1 \over 2} m^2 \Phi^2 $$ By using the E-L equations we have just derived we obtain Klein-Gordon equation.

$$ \eta_{\mu \nu} \partial^{\mu} \partial^{\nu} \Phi + m^2 \Phi = \square \Phi + m^2 \Phi = 0$$

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Lagrangian density? Interesting; in all my study if Lagrangian mechanics I've never encountered that term. We tend to just say "Lagrangian" or "Lagrangian operator". Not sure if they're the same. –  Noldorin Dec 6 '10 at 21:24
    
@Noldorin: The duality is the same as between energy and energy density. The former deals with whole system and the latter with local distribution. Speaking about density only makes sense when you are dealing with fields. So I suppose you were only dealing with particles in your study? Although note that physicists often abuse the terminology and call (and even write) Lagrangian density just Lagrangian. –  Marek Dec 6 '10 at 21:34
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@Noldorin: maybe one thing worth pointing out is that action for particles is just integral over time. But for fields it is integral over whole space-time. You can rewrite that as an integral of the Lagrangian over time and the Lagrangian itself would then become integral over Lagrangian density over space. –  Marek Dec 6 '10 at 21:37
    
@Marek: Probably. The principle application of the Lagrangian in classical mechanics is too particles, so yeah. –  Noldorin Dec 6 '10 at 22:07
    
@Noldorin: I don't think I've ever heard of the Lagrangian density outside of field theory, so if you haven't studied QFT it's no surprise that you haven't heard of it. –  David Z Dec 7 '10 at 4:37

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