I Can function transformation result in a constant variation?

[Higgsono]

Given a scalar function, we consider the following transformation:

$$\delta f(x) = f'(x') - f(x) $$ Given a coordinate transformation $$x' = g(x)$$

But since $f(x)$ is a scalar isn't it true that $f'(x') = f(x)$

Then the variation is always zero? What am I missing?

 

[PeroK]

Given a scalar function, we consider the following transformation:

$$\delta f(x) = f'(x') - f(x) $$

But since $f(x)$ is a scalar isn't it true that $f'(x') = f(x)$

Then the variation is always zero? What am I missing?

Do you have more context?

 

[Higgsono]

Do you have more context?

Do you have more context?

Given a coordinate transformation $$x' = g(x)$$

 

[PeroK]

and $f'$?

 

[Higgsono]

and $f'$?

I don't know. But they always write a prime on the function as well.

 

[PeroK]

I don't know. But they always write a prime on the function as well.

If you don't know what it is, how can you calculate $\delta f(x)$.

My initial interpretation of $f'$ would agree with what you said in your OP. That $f'(x') = f(x)$ by definition.

 

[Higgsono]

If you don't know what it is, how can you calculate $\delta f(x)$.

My initial interpretation of $f'$ would agree with what you said in your OP. That $f'(x') = f(x)$ by definition.

f is a scalar field, and I am considering variations of this field in the Lagrangian. But if $$ f'(x') = f(x)$$ by definition. What is the point of varying the fields?

 

[PeroK]

f is a scalar field, and I am considering variations of this field in the Lagrangian. But if $$ f'(x') = f(x)$$ by definition. What is the point of varying the fields?

Ah, Lagrangian, you see, the magic word!

The Lagrangian $f'$ has the same form as the Lagrangian $f$. It's not the function obtained by a coordinate transformation. It's the same function as $f$ applied to the coordinates $x'$.

 

[Higgsono]

Ah, Lagrangian, you see, the magic word!

The Lagrangian $f'$ has the same form as the Lagrangian $f$. It's not the function obtained by a coordinate transformation. It's the same function as $f$ applied to the coordinates $x'$.

So if it has the same form, the transformation would be $$\delta f= f(x') - f(x)$$ intead of $$\delta f= f'(x') - f(x)$$

I'm always confused why they put a prime on the function when they consider coordinate transformations.

 

[PeroK]

So if it has the same form, the transformation would be $$\delta f= f(x') - f(x)$$ intead of $$\delta f= f'(x') - f(x)$$

Yes, that's how I tend to write things, although most books seem to prefer to put a prime on the Lagrangian as well. I would write:

$f'(x) = f(x') = f(g(x))$, assuming $x' = g(x)$

But, some texts write $f'(x') = f'(x)$, which (technically) I think is inaccurate.