# Transformation of a function

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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?

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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 $$x' = g(x)$$

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and ##f'##?

Higgsono
and ##f'##?

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

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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?

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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.

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