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, 20241 yr by 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

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, 20241 yr by Genady

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.

21 minutes ago, joigus said:

at this point in your life

The advantage is, no deadlines.