Where was the symmetry?

[Genady]

Here are steps of derivation of energy-momentum conservation:

Consider a shift of the field ϕ by a constant 4-vector ξ :

(1) ϕ(x)→ϕ(x+ξ)=ϕ(x)+ξν∂νϕ(x)+...

The infinitesimal transformation makes

(2) δϕδξν=∂νϕ

and

(3) δLδξν=∂νL

Using the E-L equations, the variation of Lagrangian is

(4) δL[ϕ,∂μϕ]∂ξν=∂μ(∂L∂(∂μϕ)δϕδξν)

Using (2) and (3),

(5) ∂νL=∂μ(∂L∂(∂μϕ)∂νϕ)

or equivalently

(6) ∂μ(∂L∂(∂μϕ)∂νϕ−gμνL)=0

The conclusion is, "The four symmetries have produced four Noether currents, one for each ν :

(7) Tμν=∂L∂(∂μϕ)∂νϕ−gμνL

all of which are conserved: ∂μTμν=0 ."

My question:

where in this derivation the assumption was used that the transformation is a symmetry?

P.S. I am sorry that LaTex is so buggy here. I don't have a willing power to do this again. Ignore. Bye. 

Edited February 27 by Genady

[Genady]

Rather than trying to fix the OP, I've prepared the text elsewhere and just post its image:

A minor correction: the equation (4) above should rather be

[Genady]

I think, I got it.

The symmetry validates the equation (3), because this equation makes the variation of Lagrangian a total derivative, and this makes the variation of action vanish:

 

IOW, without the symmetry, we can't go from (4) to (5).

Edited February 27 by Genady

[joigus]

1 hour ago, Genady said:

I think, I got it.

The symmetry validates the equation (3), because this equation makes the variation of Lagrangian a total derivative, and this makes the variation of action vanish:

 

IOW, without the symmetry, we can't go from (4) to (5).

Exactly! In fact, that's how you define a symmetry in the action, as a total divergence does not change the "surface" (hypersurface) terms t=t₁ and t=t₂.

You must apply Stoke's theorem first. Perhaps you did, but I didn't have time to check.

It is amazing that you're doing this stuff at this point in your life.

[Genady]

21 minutes ago, joigus said:

at this point in your life

The advantage is, no deadlines.