# I Constant field in the Lagrangian

1. Sep 16, 2018 at 2:03 PM

### kelly0303

Hello! If you have a Lagrangian (say of a scalar field) depending only on the field and its first derivative and you want to calculate the ground state configuration, is it necessary a constant value? I read about Spontaneous symmetry breaking having this Lagrangian $$L= \frac{1}{2}(\partial_\mu \phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4$$ and they say that to get the lowest energy configuration you set $\phi$ to be a constant and work from there. In the case of a particle it makes sense that a motionless particle would have less energy that a moving one at a point. but is this obvious for fields? Can't a complicate configuration bring the energy lower than a constant value? Thank you!

2. Sep 16, 2018 at 2:50 PM

### Orodruin

Staff Emeritus

3. Sep 16, 2018 at 2:57 PM

### kelly0303

Sorry I did a mistake, the term mass should be + not - (in order to get spontaneous symmetry breaking). So the energy density would be $$H= \frac{1}{2}(\partial_\mu \phi)^2-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$. Is it obvious that the field must be constant?

4. Sep 16, 2018 at 2:58 PM

### Orodruin

Staff Emeritus
No, that is not the correct energy density. Are you familiar with how the energy-momentum tensor is derived from the Lagrangian?

5. Sep 16, 2018 at 3:06 PM

### kelly0303

I am not sure I am. But we were told that the hamiltonian density is $H=T+V$ which corresponds to the energy density (when you integrate it you get the actual hamiltonian which gives the energy for well behaved potentials). Did I miss-understand that?

6. Sep 16, 2018 at 3:08 PM

### Orodruin

Staff Emeritus
You are misinterpreting what $T$ is in this case so you have the wrong Hamiltonian. I suggest you look at the full expression for the energy-momentum tensor and extract the time-time component, which is the energy density.

7. Sep 16, 2018 at 5:45 PM

### kelly0303

Oh, right! So I got this for the energy density: $$\frac{1}{2}\dot{\phi}^2-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ Is it obvious that $\phi$ has to be a constant?

8. Sep 16, 2018 at 11:34 PM

### Orodruin

Staff Emeritus
That is also wrong. You are now missing terms.

9. Sep 17, 2018 at 9:54 AM

### kelly0303

So the way I did it is to use $$T_{\mu\nu}=\frac{\partial L}{\partial(\partial_\mu\phi)}\partial_\nu\phi-g_{\mu\nu}L$$ which gives $$T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi-g_{\mu\nu}(\frac{1}{2}(\partial_\mu \phi)^2+\frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4)$$ $$T_{\mu\nu}=\frac{1}{2}\partial_\mu \phi\partial_\nu \phi-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ Now the energy density is given by $$T_{00}=\frac{1}{2}\partial_0 \phi\partial_0 \phi-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ Where is my mistake? Thank you!

10. Sep 17, 2018 at 11:54 AM

### Orodruin

Staff Emeritus
11. Sep 18, 2018 at 10:18 AM

### kelly0303

Ah, ok. So it should be like this: $$T_{00}=\frac{1}{2}\partial_0\phi\partial_0\phi+\frac{1}{2}( \partial_x\phi\partial_x\phi+\partial_y\phi\partial_y\phi+\partial_z\phi\partial_z\phi)-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ I hope this is correct now. But still, why is this a minimum for constant $\phi$? If the mass term would have been added and not subtracted, that would have made sense, but I am not sure I see it as obvious here? Thank you and sorry for taking so long!

12. Sep 18, 2018 at 10:47 AM

### Orodruin

Staff Emeritus
The shape of the potential is irrelevant. What is relevant are the derivative terms that are always non-negative and zero only if the field is constant.

13. Sep 18, 2018 at 3:30 PM

### kelly0303

But if you have something of the form $$g(x)=f'(x)^2+f(x)$$ the minimum of $$\int g(x)$$ is not necessary when $f=ct$

14. Sep 18, 2018 at 4:45 PM

### Orodruin

Staff Emeritus
That’s because $f(x)$ does not have a minimum. Your potential does.

15. Sep 18, 2018 at 5:32 PM

### kelly0303

Hmm that makes it more clear. However, I am still a bit confused. The potential energy minimum is not necessarily attained for a constant value of $\phi$, I mean $\int V(x)$. So can't it happen that for some non-constant value of $\phi$ the kinetic term will indeed get bigger (as it is non-zero anymore) but the potential gets down by a higher amount and hence $\int T+V$ it's smaller overall? Intuitively makes sense that the more you move the more energy you have, but mathematically is not really obvious to me. Can I derive this using calculus of variation, something like $\frac{\delta E}{\delta \phi}=0$ and from here infer that $\phi$ is a constant?

16. Sep 18, 2018 at 9:41 PM

### Orodruin

Staff Emeritus
If a field value is a minimum of the potential for one point, it is for all points. Hence, a constant does give you the minimal potential at every point.