Symmetry breaking Lagrangian

[Genady]

This is a multi-step exercise. It would be very helpful if somebody could check my step(s) as I go. @joigus, I'm sure it is a child play for you.

I'd like to make sure that I've derived correctly the equation of motion for this Lagrangian:

\[\mathcal L=- \frac 1 2 \phi \Box \phi + \frac 1 2 m^2 \phi^2 - \frac {\lambda} {4!} \phi^4\]

The EL equation:

\[\frac {\partial \mathcal L} {\partial \phi} + \Box \frac {\partial \mathcal L} {\partial (\Box \phi)} = 0\]

The equation of motion:

\[\Box \phi - \frac 1 2 m^2 + \frac {\lambda} {3!} \phi^3 = 0\]

 

How is it?

P.S. As edit LaTex does not work, I add a typo correction here. The equation of motion is rather

\[\Box \phi - m^2 \phi + \frac {\lambda} {3!} \phi^3 = 0\]

[joigus]

12 hours ago, Genady said:

This is a multi-step exercise. It would be very helpful if somebody could check my step(s) as I go. @joigus, I'm sure it is a child play for you.

I'd like to make sure that I've derived correctly the equation of motion for this Lagrangian:

 

L=−12ϕ□ϕ+12m2ϕ2−λ4!ϕ4

 

The EL equation:

 

∂L∂ϕ+□∂L∂(□ϕ)=0

 

The equation of motion:

 

□ϕ−12m2+λ3!ϕ3=0

 

 

How is it?

P.S. As edit LaTex does not work, I add a typo correction here. The equation of motion is rather

 

□ϕ−m2ϕ+λ3!ϕ3=0

 

Your eq. of motion (once corrected) looks fine. Most people prefer to write (1/2)(grad)phi(grad)phi instead of -(1/2)phi(grad)²phi, but they differ in just a total divergence, so they are equivalent (lead to the same equations of motion).

By grad² I mean the D'Alembert operator. I'll check in more detail later, I would have to do it in my head now and I could miss a sign. This is a famous equation.

[Genady]

Q: How many constants \(c\) are there so that \(\phi(x)=c\) is a solution to the equation of motion?

A: Three. \(c=0\) and two solutions for \(c^2= \frac {3!} {\lambda} m^2\)

Q: Which solution has the lowest energy (the ground state)?

A: The potential energy from the Lagrangian is \[\frac {\lambda} {4!} \phi^4 - \frac 1 2 m^2 \phi^2\]It is the lowest for the non-zero \(c\): \(- \frac{3!m^4} {4 \lambda}\).

[RobertSmart]

Your derivation of the equation of motion from the given Lagrangian appears to be mostly correct, but there's a small mistake in your final result. Let's go through it step by step:

Given Lagrangian:

4L=−21ϕ□ϕ+21m2ϕ2−4!λϕ4

The Euler-Lagrange equation is:

∂ϕ∂L+□(∂(□ϕ)∂L)=0

Taking the derivatives with respect to ϕ and 3∂ϕ∂L=−□ϕ+m2ϕ−3!λϕ3∂(□ϕ)∂L=−21

Now, plugging these derivatives into the Euler-Lagrange equation: −□ϕ+m2ϕ−3!λϕ3+□(−21)=0

Simplifying: −□ϕ+m2ϕ−3!λϕ3+21□=0

Rearranging terms: □ϕ−m2ϕ+3!λϕ3=0

So, the corrected equation of motion should be: □ϕ−m2ϕ+3!λϕ3=0

This is the correct equation of motion derived from the given Lagrangian.

If you want to know more about lagrang theorem,I will suggest you to visit link removed once.

[joigus]

Ok. Yes, @RobertSmart is right. There is a little mistake in the constants. Let me display my calculation in detail, because his Latex seems to have been messed up by the compiling engine or whatever and I seem to find a small discrepancy with him.

Your Lagrangian,

\[ \mathscr{\mathcal{L}}=-\frac{1}{2}\phi\Box\phi+\frac{1}{2}m^{2}\phi^{2}-\frac{\lambda}{4!}\phi^{4} \]

I prefer to write with an index notation, which is more convenient for variational derivatives:

\[ \mathscr{\mathcal{L}}=-\frac{1}{2}\phi\left.\phi^{,\mu}\right._{,\mu}+\frac{1}{2}m^{2}\phi^{2}-\frac{\lambda}{4!}\phi^{4} \]

As we have no dependence on first order derivatives,

\[ \frac{\partial\mathscr{\mathcal{L}}}{\partial\phi_{,\mu}}=0 \]

we get as the only Euler-Lagrange equation,

\[ \frac{\partial\mathscr{\mathcal{L}}}{\partial\phi}=-\frac{1}{2}\left.\phi^{,\mu}\right._{,\mu}+m^{2}\phi-\frac{\lambda}{3!}\phi^{3}=0 \]

Or,

\[ -\frac{1}{2}\Box\phi+m^{2}\phi-\frac{\lambda}{3!}\phi^{3}=0 \]

Or a bit more streamlined,

\[ \Box\phi-2m^{2}\phi+\frac{\lambda}{3}\phi^{3}=0 \]

Sorry I didn't get around to it sooner.

Paraphrasing Sir Humphrey Appleby:

Is that finally final?

I hope so.

PS: BTW, this is a simplified symmetry-breaking Lagrangian. The real thing in the SM is a complex SU(2)-symmetric multiplet \( \left(\phi_{1},\phi_{2},\phi_{3},\phi_{4}\right) \).

Edited February 29 by joigus
Latex editing

[Genady]

Unfortunately, I can't read this post:

1 hour ago, RobertSmart said:

Given Lagrangian:

4L=−21ϕ□ϕ+21m2ϕ2−4!λϕ4

The Euler-Lagrange equation is:

∂ϕ∂L+□(∂(□ϕ)∂L)=0

and I don't know how his result is different from mine, but it seems that his EL equation is the same as mine,

On 2/27/2024 at 6:36 PM, Genady said:

The EL equation:

<<<<<

which is different from

36 minutes ago, joigus said:

we get as the only Euler-Lagrange equation,

<<<<<<<

I disagree with the latter. We need to use the generalized EL equation, which I have already derived in this exercise:

and got the answer compatible with this:

(Euler–Lagrange equation - Wikipedia)

 

Edited February 29 by Genady

[joigus]

12 minutes ago, Genady said:

Unfortunately, I can't read this post:

and I don't know how his result is different from mine, but it seems that his EL equation is the same as mine,

<<<<<

which is different from

<<<<<<<

I disagree with the latter. We need to use the generalized EL equation, which I have already derived in this exercise:

and got the answer compatible with this:

(Euler–Lagrange equation - Wikipedia)

 

Yes, you're right. The unusual writing of the Lagrangian set me off. Sorry. That is indeed the way to generalise to higher-order derivatives. I've proven it many times, but now I had just a couple of minutes and I screwed up. There's just a coefficient difference. 

Later.