Symmetries & Lie groups in physics

Question

This is not a homework, neither it is any exercise. It is my understanding of $U(1)$ symmetry. I would request if anybody can please correct me on any one of the following understandings:

1. The bottom line of particle physics is Lagrangian density which is a generalization of classical mechanics, $L= V-T$.

2. We come across lot of symmetries in nature which are invariant under any changes made. Like an equilateral triangle is more symmetric than other triangles...... For a symmetry, if the object is rotated or flipped, the appearance will remain unchanged.

3. But when we label a symmetry, we could differentiate it which is something called 'symmetry breaking'.

4. $U(1)$ is a Lie group. Now, typically if a circle is rotated, flipped, it appears the same, called Lie group.

5. There are different type of symmetries like (i) Translational --> Invariance of the laws of physics under any translation as pointed out in Noether's theorem (ii) Rotational -->Whichever direction you rotate it is the same (iii) Time symmetry --> the laws of physics are eternally unchanging (iv) Boost symmetry

6. The rotational and boost symmetry are grouped under Lorentz group. The translational, rotational and boost symmetry together make up the Poincare group.

7. Local symmetry is not a symmetry in physical spacetime.

8. When Einstein applied special relativity to electromagnetism, he found electromagnetic 4 potential-- the four vector (1 time like, and 3 space like, which relates the electric scalar potential and magnetic vector potential). The physicists were happy, as the electromagentic 4 potential appears in the Lagrangian density.

9. But the value of the electromagnetic 4 potential can be changed. The Electric field $E(t,x)$ and the Magnetic field, $B(t,x)$ can be expressed in scalar potential and vector field.The term gauge invariance refers to the property that a whole class of scalar and vector potentials, related by so-called gauge transformations, describe the same electric and magnetic fields. As a consequence, the dynamics of the electromagnetic fields and the dynamics of a charged system in a electromagnetic background do not depend on the choice of the representative $(A_0(t,x),A(t,x))$.

10. $U(1)$ being a circle. When the charged particle move across the $U(1)$ plane, its' mass, kinetic energy does not depend on the position of the particle. It's value depend on the rate at which it circles the plane.

11. This is the $U(1)$ symmetry of the standard model. If we rotate $U(1)$ at any angles, the Lagrangian density will remain unchanged.

Please do correct me where I am wrong.

Thanks.

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Shounak Bhattacharya

Comments

The item (4) is too vague. A Lie group is just an infinite-dimensional and smooth group. Some of them, the $SO(N)$, can be realised as groups that leave spheres $S^{N-1}$ invariant, but that's not the way you should think of Lie groups in general.

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Vibert

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Comment to the question (v1): Similar to another Phys.SE question by OP the title mentions susy despite the body of text does not.

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Qmechanic

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And as to (9): a charged particle doesn't "move across the $U(1)$ plane" since there is no $U(1)$ plane. The $U(1)$ is some abstract space that has nothing to do with spacetime. Similarly there is no "rate at which the particle traverses $U(1)$."

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Vibert

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@Vibert: Lie groups are not infinite-dimensional.

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user MBN

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@MBN: sorry, of course I meant finite but nonzero. Too late to edit, however.

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Vibert

Answer 1

You have a few mistakes:

(3) I find the words label and differentiate odd. A symmetry could be broken, either spontaneously e.g. Higgs mechanism, or explicitly.

(4) is a poor definition of a Lie group... as mentioned in the comments, a Lie group is a group that is also a differentiable manifold.

(7) a local symmetry is continuous symmetry for which the continuous parameter that parameterizes the symmetry is a function of space and time. cf. a global symmetry, in which the continuous parameter is constant. For a theory with kinetic terms (e.g. Standard Model), local symmetries are possible only if the local symmetries are gauge symmetries.

(10) As per the comments, the $U(1)$ is an internal symmetry - a particle doesn't move around the $U(1)$ line (parametrized by only one continuous parameter) in the ordinary sense.

Please, somebody feel free to edit this answer to include more corrections or expand mine.

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user innisfree

Comments

About （7）, in a field theory which contains kinetic term, the local symmetry must be gauged. In that case, the local symmetry is not a symmetry.

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Craig Thone

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@CraigThone, thanks, I've incorporated your comment.

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user innisfree

Answer 2

1. Right, but note that the Lagrangian Density $\mathcal L\,$ is different from the Lagrangian $L$, in that $L=\iiint\mathcal L\mathrm{d}x^3$.\

   Also, you are incorrect to state that the Lagrangian Density is fundamental to "particle physics", whereas the Lagrangian is fundamental to "Classical Mechanics". Actually, the Lagrangian is used in pre-relativistic physics, where space and time are considered to be very different entities, almost as if time is an unnatural, construct, or a construct introduced purely for the sake of formalism.

   However, when it comes to relativistic physics, it is hard completely unnatural to use the Lagrangian, as the (sort of fundamental, due to the principle of least action or the integrand of the feynman's path integral) action is the integral of the Lagrangian over time; it's unnatural in relativistic physics that time is treated so differently from space (why is it the integral over time only?), so one introduces the Lagrangian Density, such that his integral over space is the Lagrangian, so that it's integral over spacetime is the Action, so you could sort of think; that the Lagrangian Density is more fundamental when one considers relativistic physics.

2. Right,; more generally, if an object or construct $A$ is unchanged under transformation $T$, then $A$ is symmetric with respect to $T$.

3. Symmetry breaking is something like breaking a symmetry group into different symmetry groups.

4. I don't see what you're trying to say here, . A lie group refers to a group which is also a differentiable manifold.

5. Right, but No - ether's theorem is more general than that. It also applies to rotaions; for example, rotational invariance is equivalent to conservation of angular momentum.

6. Right.

7. As said by innisfree (23389).

8. Yes, it's something to be quite happy about.

9. Right.

10. As said by innisfree (23389).

11. No, $U(1)$ does' not roatate, the $\psi$ is rotated by $U(1)$.

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On abstract plane

While it is true that the $U(1)$ bundle is rather abstract, the geometric interpretation of electromagnetism is that it is the curvature of the $U(1)$ bundle, like how the geometric interpretation of gravity is that it is the curvature of spacetime.

On rotating fields

Multiplying the field $\psi$ by a certain matrix $P$ in the group $U(1)$, along with gauge transforming the partial derivative, leaves the theory unchanged. I.e., $P\psi$ leaves the theory unchanged.

Comments

"The U(1) is some abstract space that has nothing to do with spacetime. Similarly there is no "rate at which the particle traverses U(1)" This is confusing. If anybody can please explain.

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Shounak Bhattacharya

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In the 10th.point you have told that the psi rotates, not the U(1) plane. So, if you could please explain this point.

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Shounak Bhattacharya

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@ShounakBhattacharya: Re:1; yes;. Re:2; No, A lie group just means a group which is also a differentiablke manifold; Re:Abstract; I'll add that into my answer. Re:10; I'll add that into my answer; .

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Dimensio1n0

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Well, are you speaking of the Kaluza-Klein or the Rainich-Misner-Wheeler theory?

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Shounak Bhattacharya

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@ShounakBhattacharya: Neither. QFT/QED.

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Dimensio1n0

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@ShounakBhattacharya: What?

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Dimensio1n0

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So (1) Lagrangian is more used in classical mechanics. In relativistic physics it is used, so that the integral over spacetime gives a better picture.

This post imported from StackExchange Physics at 2014-03-07 16:40 (UCT), posted by SE-user Shounak Bhattacharya