What is the meaning of invariance for an equation f=0?

[FrederikPhysics]

Hey. When talking about invariance of a function f under some transformation T we mean that T(f)=f. But what is meant by invariance of an equation f=0? As far as I can see it makes sense to call an equation invariant when the transformed equation T(f)=T(0) is equivalent to the original equation f=0, or maybe just if T(f)=T(0) implies f=0.

To be specific, I am asking because the source free Yang-Mills equation D^μF_μν=0 is said to be invariant under gauge transformations and i am wondering what exactly is meant by this. When preforming the gauge transformation U we obtain UD^μF_μνU^†=0 which is equivalent to the original equation but not the same as the original equation.

 

[jedishrfu]

Here's some general info on gauge transformations, later in the article is a discussion on Yang-Mills:

https://en.wikipedia.org/wiki/Gauge_theory

and here's a more personalized discussion using E&M theory:

https://www.phy.duke.edu/~rgb/Class/Electrodynamics/Electrodynamics/node30.html

and a Leonard Susskind video on it in the context of Relativity:

http://theoreticalminimum.com/courses/special-relativity/2008/spring/lecture-7

 

[fresh_42]

Hey. When talking about invariance of a function f under some transformation T we mean that T(f)=f. But what is meant by invariance of an equation f=0?

Not necessarily. It can also mean $f \circ T = f$, that is a transformation of coordinates leaves the relationship intact. One should always consider the context and how it is meant. $T \circ f = f$ should better be noted as $f$ is an eigenvector of the transformation operator $T$.

As far as I can see it makes sense to call an equation invariant when the transformed equation T(f)=T(0) is equivalent to the original equation f=0, or maybe just if T(f)=T(0) implies f=0.

This would only mean, that $T$ is injective. It's not called invariant.

To be specific, I am asking because the source free Yang-Mills equation D^μF_μν=0 is said to be invariant under gauge transformations and i am wondering what exactly is meant by this. When preforming the gauge transformation U we obtain UD^μF_μνU^†=0 which is equivalent to the original equation but not the same as the original equation.

Have you heard of Noether's theorem?

 

[jedishrfu]

Have you heard of Noether's theorem?

and the punch line: Noether heard of it.

 

[fresh_42]

and the punch line: Noether heard of it.

Noether heard of it, but it took Little (aka F. Klein) to be published!

 

[FrederikPhysics]

The Lagrange function generating these Yang-Mills equations are gauge invariant L=L', but as mentioned the Yang-Mills equations themselves transform according to the adjoint representation (D_μF^μν)'=UD_μF^μνU^† which does not seem invariant to me, still people say that the are invariant under gauge transformations. My question is, in what sense are they invariant?

Yes I have heard about Nöethers theorem I do not see how it relates. For every continuous symmetry of the action there exists a continuity equation.

 

[fresh_42]

The Lagrange function generating these Yang-Mills equations are gauge invariant L=L', but as mentioned the Yang-Mills equations themselves transform according to the adjoint representation (D_μF^μν)'=UD_μF^μνU^† which does not seem invariant to me, still people say that the are invariant under gauge transformations. My question is, in what sense are they invariant?

Yes I have heard about Nöethers theorem I do not see how it relates. For every continuous symmetry of the action there exists a continuity equation.

Noether (without "ö") defined as an invariant of a Lie group for her theorem the existence of a relation
$$P(x,u,\frac{\partial u}{\partial x},\frac{\partial^2 u}{\partial x^2},\ldots) = P(y,v,\frac{\partial v}{\partial y},\frac{\partial^2 v}{\partial y^2},\ldots)$$
which shows, that it is a statement about the variables, coordinates.

 

[FrederikPhysics]

Thank you for your answer. If i understand your post the statement of invariance is a statement of the form of an equation independent of variables. But then is not every equation involving only the gauge field A_μ and coordinates invariant under gauge transformations since the gauge transformation of the equation makes A_μ→A_μ', and so the transformed equation has the same form as the original with A_μ' in place of A_μ as variable?
Please go into details.