# Different weights for time and spatial derivative in Lagrangian Density

I'm new to QFT and trying to understand the form of the Lagrangian densitys used. As a simple model you often see a Lagrangian density of the form $${\mathcal L} = \frac{1}{2} \partial_j \phi_n \partial^j \phi_n + \frac{1}{2}m^2 \phi_n \phi_n.\tag{1}$$ The two terms corresponding to the time and spatial derivatives are often refered to as the kinetic and gradient energy. I see a strong analogy to the Lagrangian density of a vibrating string: $$\mathcal{L} = \frac{\mu}{2} \left(\frac{\partial \phi}{\partial t}\right)^2 - \frac{E}{2} \left(\frac{\partial \phi}{\partial x} \right)^2.\tag{2}$$ Now the question that arraises to me is why there are different factors before the time and spatial derivatives in the second Lagrangian density but not in the first one. I guess the answer is that the first Lagrangian density has to be relativisticaly covariant while the second one isn't. But then my question is how can you treat fields with Lagrangian densitys like the second one in a covariant manner?

I guess the answer is that the first Lagrangian density has to be relativisticaly covariant while the second one isn't.

Yep, that's why.

But then my question is how can you treat fields with Lagrangian densitys like the second one in a covariant manner?

By introducing a nontrivial coupling tensor: $${\mathcal L} = \frac{1}{2} M_{\alpha\beta} \partial_\alpha \phi \partial^\beta \phi + \frac{1}{2}m^2 \phi \phi, \tag{1'}$$ with $$M_{\alpha\beta}=M_{\beta\alpha}$$ a covariant tensor. Generically, this tensor will be diagonal in some frames (like the original frame in which the formulation $$\frac{\mu}{2} \left({\partial \phi}/{\partial t}\right)^2 - \frac{E}{2} \left({\partial \phi}/{\partial x} \right)^2$$ is formulated), but it will acquire cross-terms if you boost away from those frames unless $$M_{\alpha\beta} = K g_{\alpha\beta}$$ is proportional to the metric $$g_{\alpha\beta}$$ that's used to define those boosts.

• Thanks, so the answer is actually pretty simple. Do you know if there are examples where this kind of Lagrangian Densitys with that cuppling tensor is actually used in a Quantum Field Theory? – Jürgen Oct 18 '18 at 16:44
• @Pisanty One more question: Could I construct a Lagrangian density this way that has no gradient energy by setting the coupling tensor in a frame to (1, 0, 0, 0)? When changing frame then of course I would get some nontrivial coupling of the gradients, but in the frame where the tensor is diagonal only the time derivative would occur. Is this possible or do I miss something? – Jürgen Jan 29 at 1:33
• Sure, there's nothing stopping you from doing that, with the consequences that you note. – Emilio Pisanty Jan 29 at 13:25
• @Pisanty ok thanks for the clarification, I do understand it now. I guess the reason that normaly one only considers a trivial $M_{\alpha \beta} = \text{diag}(-1, 1, 1, 1)$ is that one tries to build the simplest possible Lagrangian that meets all given requirements. Do you know an example where Lagrangians like $\text{(1')}$ with nontrivial $M_{\alpha \beta}$ are considered in applications of QFT? – Jürgen Jan 29 at 16:19
• No, I have no idea of whether people who do QFT in practice find those things useful. – Emilio Pisanty Jan 29 at 16:30

FWIW, if $$\mu$$ & $$E$$ in eq. (2) are two positive constants, they lead to a characteristic speed $$c=\sqrt{E/\mu}$$. They can be scaled away in (2) via appropriate scaling of the $$t$$ & $$x$$ variables to reach (1).

• That's an interesting point. I guess this rescaling would then correspond to the frame where the coupling tensor in Emilio's answer is diagonal with the scaling factors as diagonal elements, would it? And the covariance of the equation would then be manifest inspite of the different scaling factors for the t and x variables because of the transformation propertys of the coupling tensor. Is that right? – Jürgen Oct 18 '18 at 16:52
• @Jürgen No. If it's in that form, then (even before the re-scaling) it's already in the frame where $M_{\alpha\beta}$ is diagonal - the diagonality just means that there are no terms in $\partial^2\phi/\partial x\partial t$. – Emilio Pisanty Oct 18 '18 at 17:05
• As for whether the re-scaling in this answer is physical - it kind of depends on how married you are to the initial natural metric $\mathrm{diag}(-1,1,1,1)$ on the frame where you have the form $(2)$. If you have no pre-conceived ideas about what boosts are acceptable and how they look like, then the rescaling in this answer brings the lagrangian to a canonical form which specifies what boosts leave it invariant. On the other hand, if you already have a natural metric, then the re-scaling is a 'trick' but it doesn't change the fact that your lagrangian is fundamentally different to form $(1)$. – Emilio Pisanty Oct 18 '18 at 17:08
• @Pisanty Do I understand your first point correctly that it means if you don't have an initial natural metric you can take $\text{diag}(-\mu/E, 1, 1, 1)$ as metric and this would result in $\text{(1)}$ and $\text{(2)}$ beeing the same (in a hypothetical universe with this metric)? – Jürgen Jan 29 at 16:08