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The Hamiltonian is the energy, which is just one component of a four-vector and therefore not Lorentz invariant.

The Lagrangian is the Legendre transform of the Hamiltonian and I was wondering if there is some good reason why we get through the Legendre transform something invariant?

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The Lagrangian is what is integrated over spacetime in the action, i.e. has to be a 4-form. As such, it is necessarily a (pseudo-)scalar under Lorentz transformations.

When wondering about Lorentz transformations and such, the Hamiltonian is, as a non-Lorentz-covariant object, not a good starting point, by the way. It is often better to start with the Lagrangian that makes the Lorentz covariance of the theory manifest.

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    $\begingroup$ Everything you say is correct, but unfortunately does not answer my question. I already wrote in my question that the Hamiltonian isn't invariant and I'm aware of the fact that for example in QFT we use the Lagrangian instead. The connection between the two is that the Lagrangian is the Legrendre transform of the Hamiltonian. My question was/is: Why does this work? In other words: Why do we get something invariant (the Lagrangian) from something non-invariant (the Hamiltonian) through the Legendre transform? $\endgroup$ – Tim Dec 2 '14 at 7:11
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    $\begingroup$ @Tim: Ah, I see. I'll think about it a bit, if I don't come up with something, I'll delete this. $\endgroup$ – ACuriousMind Dec 2 '14 at 9:17
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    $\begingroup$ @Tim That's the wrong way of looking at it. Here's a better way: As ACuriousMind's answer explains, the Lagrangian must be Lorentz invariant. Once you take its Legendre transform to get a Hamiltonian, this breaks Lorentz symmetry. Put another way: The only class of interesting Hamiltonians are the ones that are Legendre transforms of Lorentz invariant Lagrangians. $\endgroup$ – jwimberley Jan 15 '15 at 14:10
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    $\begingroup$ @Tim: I think almost everyone would say that the Lagrangian is the fundamental thing. When we do time-dependent Legendre tranforms to it, it should be hardly surprising that we get non-covariant stuff out. And the legendre transform of a general Hamiltonian is totally not Lorentz covariant. $\endgroup$ – Jerry Schirmer Jan 15 '15 at 14:24
  • $\begingroup$ @ACuriousMind can you elaborate, please? What is a 4-form? $\endgroup$ – Michael Angelo Jan 23 '17 at 15:40
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The previous answer is very good, but I think can be simplified a bit.

In particle mechanics, the Lagrangian $L$ is $$L = p\dot{q} - H$$ So let's look at this in special relativity. We get, with $p = \gamma m v$, $\dot{q} = v$, and $H = E = \gamma mc^{2}$, $$L = \gamma mv^{2} - \gamma mc^{2} = - \gamma mc^{2}(1 - (v/c)^{2}) = - mc^{2}/\gamma$$ It's not that $L$ is a scalar (which is what I thought originally), but that $\int L\,dt$ is a scalar. And this is easy, because $$\int L\,dt = -\int (mc^{2}/\gamma)\,dt = - \int mc^{2}\,d\tau$$ where $\tau$ is the proper time. This integral is clearly invariant, as we should wish for the action.

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We will here give our interpretation of OP's question (v4).

  1. We assume that OP's definition of Lorentz covariance is that the equations of motion (eom) of the theory is Lorentz covariant.

  2. We will assume that the theory has an action principle, and that the eoms are the Euler-Lagrange (EL) equations.

  3. One may prove that Lorentz invariance of the action implies Lorentz covariance of the EL eqs., cf. e.g. this Phys.SE post.

  4. The implication (3) does in principle not hold in the other direction, but in practice Lorentz covariant EL eqs. arise from a Lorentz invariant action principle.

Putting these facts together show that it is natural to expect that the action to be Lorentz invariant for a Lorentz covariant theory, cf. definition (1).

  1. Next, we will assume that the Legendre transformation is well-defined.

  2. Also we will assume that the Legendre transformation is an involution, i.e. performing the Legendre transformation twice gets us back to the starting point.

In particular, if OP starts from a Lorentz covariant (but not necessarily manifestly$^1$ Lorentz covariant) Hamiltonian formulation, this means that the Hamiltonian eoms are Lorentz covariant, cf. definition (1). The Hamiltonian$^2$ $H(q,p)$ itself is of course not Lorentz invariant, but the temporal component of a four-vector, as OP correctly writes. Points 2-4 now motivate that the Hamiltonian action $$S_H[q,p]~=~\int \!dt ~(p_i\dot{q}^i-H(q,p))$$ is Lorentz invariant. It follows that the Lagrangian action $S[q]$ is also Lorentz invariant.

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$^1$ For manifestly Lorentz covariant Hamiltonian formulations, see e.g. my Phys.SE answer here.

$^2$ The following argument can be extended to field theory.

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  • $\begingroup$ Example: Scalar field theory. Lagrangian density ${\cal L}= \frac{1}{2} \phi_{,\mu}\eta^{\mu\nu}\phi_{,\nu} -V(\phi)$ with Minkowski sign convention $(+,-,-,-)$. Let $\pi^{\mu}:=\frac{\partial {\cal L}}{\partial\phi_{,\mu}}=\eta^{\mu\lambda} \phi_{,\lambda}$. EL eqs. $d_{\mu} \pi^{\mu}=\Box\phi \approx -V^{\prime}(\phi)$. The $\pi^0$ field is the conjugate/canonical momentum field. It is the temporal component of a 4-vector. Non-zero Poisson brackets $\{\phi({\bf x},t),\pi^0({\bf y},t)\}_{PB}=\delta^3({\bf x}-{\bf y})$. $\endgroup$ – Qmechanic Jan 26 '15 at 14:49
  • $\begingroup$ Hamilton's eom. $\dot{\phi}\approx \pi^0$ and $\dot{\pi}^0\approx \nabla^2\phi-V^{\prime}(\phi)$. Canonical stress-energy-momentum tensor $T^{\mu}{}_{\nu} = \pi^{\mu} \phi_{,\nu}-\delta^{\mu}_{\nu} {\cal L}$. Noether's theorem from translation symmetry $d_{\mu}T^{\mu}{}_{\nu}\approx 0$. 4-momentum density ${\cal P}_{\nu}=T^0{}_{\nu}$. The Hamiltonian density ${\cal P}^0=T^{00}=\frac{1}{2}(\pi^0)^2 +\frac{1}{2}(\nabla\phi)^2 +V(\phi)$ is the $00$-component of a symmetric $(2,0)$ tensor. ${\cal P}_{i}=\pi^0\phi_{,i}$. 4-momentum $P_{\nu}(t)=\int d^3x~{\cal P}_{\nu}({\bf x},t)$. $\endgroup$ – Qmechanic Jan 26 '15 at 15:46
  • $\begingroup$ Example: Free point particle. Lagrangian $L=-\frac{m_0c^2}{\gamma}$ with Minkowski sign convention $(-,+,+,+)$. Here $\gamma := \left(1-\left(\frac{\dot{x}}{c}\right)^2\right)^{-\frac{1}{2}} $ $ = \sqrt{\left(\frac{p}{m_0c}\right)^2+1}$, where $3$-momentum $p_i:=\frac{\partial L}{\partial \dot{x}^i}=\gamma m_0\dot{x}^i$. Hamiltonian $p^0:=p_i\dot{x}^i-L = \gamma m_0c^2$ is the $0$-component of a $4$-vector. $\endgroup$ – Qmechanic Jan 26 '15 at 23:56

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