Is This the Correct Way to Apply Symmetry Transform to the Action?

[tamiry]

Homework Statement

Hi
I'm trying to understand how symmetry transform works.
Suppose a lagrangian $L = q^{-2}$
(actually it had another kinetic member, but I don't need it for my question here)

The matching action $S = \int dt q^{-2}$

We were told that it has the next symmetry
$t \rightarrow at$
$q \rightarrow a^{-1/2}q$
I tried to figure how, but I can't get it.

The Attempt at a Solution

I thought the correct way is this
$T = at$
$Q(T) = a^{-1/2}q(T)$

and now I put it in the action
$S = \int dt q^{-2} = \int (dT/a) (q(T))^{-2} = \int (dT/a) (a^{+1/2}*Q(T))^{-2} = \int (dT/a) (a^{-1})*(Q(T))^{-2} = a^{-2}*S ≠ S$

Is that the correct way to do it? Obviously I could change the power of a in t/q transform and it would fix it, is that it?

Thanks a lot for reading
Tamir

 

[mark726]

Dear Tamir,

Thank you for your question. Symmetry transformations in physics can be a bit tricky to understand at first, but with some practice, you will get the hang of it. Let me explain the steps to take in order to correctly apply the transformation to the action.

Firstly, let's define the transformation as you have done:
t → at
q → a^(-1/2)q

The key here is to remember that when we perform a transformation, we are changing the variables in the action, but the integral remains the same. In other words, we are only changing the variables, not the overall structure of the action.

So, let's apply the transformation to the action:
S = ∫dt q^(-2) = ∫ (dT/a) (q(T))^(-2)

Notice that we have changed both the variable t and the function q in the integral. Now, we can simplify this expression by using the fact that a is a constant, so it can be taken out of the integral. Also, we can apply the power rule for integrals to simplify further:
S = (1/a)∫dT (a^(-1/2)q(T))^(-2) = (1/a)∫dT aq^(-1) = (1/a)(1/a)∫dT q^(-1) = (1/a^2)∫dT q^(-1)

And there you have it, the correct transformation of the action under the given symmetry. Notice that the power of a in the final expression is -2, which matches the power in the original action.

I hope this helps clarify any confusion you had. Keep practicing and don't hesitate to ask for help if you need it. (the scientist)