Can function transformation result in a constant variation?

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Discussion Overview

The discussion revolves around the transformation of scalar functions and the implications for variations in the context of Lagrangian mechanics. Participants explore whether the transformation leads to a constant variation and the meaning of the notation used in these transformations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants question whether the transformation $$\delta f(x) = f'(x') - f(x)$$ results in a variation of zero, arguing that since $$f(x)$$ is a scalar, it follows that $$f'(x') = f(x)$$ by definition.
  • Others express confusion about the notation of the prime on the function, suggesting that it may not be necessary or could lead to misunderstanding.
  • One participant notes that the Lagrangian $$f'$$ has the same form as $$f$$ and is not merely a function obtained through coordinate transformation, but rather the same function applied to different coordinates.
  • There is a suggestion that the transformation could be expressed as $$\delta f = f(x') - f(x)$$ instead of using the prime notation, reflecting a preference for clarity in the representation of the transformation.
  • Some participants highlight the inconsistency in how different texts handle the notation, with some writing $$f'(x') = f'(x)$$, which one participant believes is technically inaccurate.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and implications of the prime notation in function transformations, with no consensus reached on the correct interpretation or representation of the variations.

Contextual Notes

Participants note potential limitations in understanding due to the ambiguity of the notation and the definitions involved in the transformations, as well as the context of Lagrangian mechanics.

Higgsono
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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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Higgsono said:
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?
 
PeroK said:
Do you have more context?
PeroK said:
Do you have more context?

Given a coordinate transformation $$x' = g(x)$$
 
and ##f'##?
 
PeroK said:
and ##f'##?

I don't know. But they always write a prime on the function as well.
 
Higgsono said:
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.
 
PeroK said:
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?
 
Higgsono said:
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'##.
 
PeroK said:
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.
 
  • #10
Higgsono said:
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.
 

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