Motivation for mass term in Lagrangians

In summary: The reason for including the mass term is that it is a necessary condition for the Klein-Gordon equation to hold. In other words, without it the equation would not be valid. Additionally, the mass term is also a self-interaction term.
  • #1
Frank Castle
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In field theory a typical Lagrangian (density) for a "free (scalar) field" ##\phi(x)## is of the form $$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi -\frac{1}{2}m^{2}\phi^{2}$$ where ##m## is a parameter that we identify with the mass of the field ##\phi(x)##.

My question is, what is the motivation for including this mass term? Is it simply that in doing so one can reproduce the Klein-Gordon equation or are there some other physical arguments? Furthermore, is it in some sense a "self-interaction" term? Classically the Lagrangian of a free particle would just have a kinetic term so how does this Lagrangian describe a free field?
 
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  • #2
If you do not include the mass term, you cannot describe a massive field and there is nothing forbidding its presence.

You can choose whether to include it in the free lagrangian or not. If you do it becomes part of your propagator, if you don't it becomes a (rather boring) self-energy contribution. Both approaches give the same results in the end.

Also, this is not the Lagrangian of a particle, it is the Lagrangian of a field. There are several classical situations that would lead to such a Lagrangian.
 
  • #3
Orodruin said:
If you do not include the mass term, you cannot describe a massive field and there is nothing forbidding its presence.

I was just wondering if there's any motivation for why one considers a Lagrangian of this form? Is it simply because we know that a relativistic massive field should satisfy the Klein-Gordon equation and so we want to construct a Lagrangian such that the Euler-Lagrange equations reproduce it?!

Also, is the reason why free-field Lagrangians are always constructed to be quadratic because this ensures that the corresponding EOM are linear and as such the solutions satisfy the superposition principle, i.e. they are non-interacting (intuitively they "pass through" each other)?!
 

1. What is the significance of the mass term in Lagrangians?

The mass term in Lagrangians represents the mass of a particle in a physical system. It is a crucial component in determining the dynamics and behavior of the system.

2. How is the mass term related to the Higgs field?

The mass term is related to the Higgs field through the Higgs mechanism, which explains how particles acquire mass through interactions with the Higgs field. The mass term is a result of this interaction and is responsible for the mass of particles in the Standard Model of particle physics.

3. Can the mass term be negative?

No, the mass term in Lagrangians cannot be negative as it represents a physical property of a particle and must have a positive value in order to have a physical interpretation.

4. How does the mass term affect the stability of a system?

The mass term can affect the stability of a system by determining the energy levels and frequencies of the particles in the system. A larger mass term can lead to a more stable system, while a smaller mass term can result in a less stable system.

5. Can the mass term be modified or manipulated?

Yes, in certain theoretical models, the mass term can be modified or manipulated to account for different phenomena or to explain discrepancies in experimental data. However, these modifications must still be consistent with the overall principles of the theory and be physically meaningful.

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