I Why is action considered a scalar in physics?

[Adams2020]

TL;DR Summary

Why is the action a scalar? Please explain.

Why is the action a scalar? Please explain.

 

[Dale]

Why would it be a vector?

 

[Adams2020]

Why would it be a vector?

I don't know. This is the question our professor asked us without explanation to think. What is the reason that it is a vector?

 

[anorlunda]

Is energy a vector?

 

[Dale]

I don't know. This is the question our professor asked us without explanation to think. What is the reason that it is a vector?

If your professor wanted you to think then maybe you should think a little about it and write your thoughts. Sometimes it is easier to think what would happen if it were not a scalar. Action is the basis of "the principle of least action", so how would that work if action were a vector?

 

[Adams2020]

Is energy a vector?

no

 

[wrobel]

I understand what the professor means but for me the question still sounds strange or at least too philosophically.
A scalar is a kind of a tensor on a manifold. The scalar is a function of the manifold with values in $\mathbb{R}$. In this sense the Action is not a scalar. The Action is a function of definite functional space with values in $\mathbb{R}$. The Action is not defined on the manifold. It is defined on functions. Reasonable question could sound as "is the Lagrangian a scalar?" "on which manifold is the Lagrangian defined?"

 

[vanhees71]

Summary:: Why is the action a scalar? Please explain.

Why is the action a scalar? Please explain.

I guess your professor gave a bit more of context before asking the question. I can only guess, what he might be after. First of all you need to specify in which sense the action is a scalar, i.e., with respect to which transformations it should be invariant.

If you describe Newtonian mechanics, the equations of motion for the full system must be Galilei invariant. For this to be true it's sufficient that the action is an invariant under all Galilei transformations.

That's of course NOT true for the standard Lagrangian,
$$L=T-V=\sum_k \frac{m_k}{2} \dot{\vec{x}}_k^2 + \sum_{j<k} V(|\vec{x}_j-\vec{x}_k|),$$
when considering Galilei boosts. On the other hand it's known that the equations of motion are Galilei invariant. So there must be a weaker condition on the action sufficient to yield Galilei-invariant equations of motion. What's this weaker condition also applying to Galilei boosts?

 

[mpresic3]

It is hard to think about the question posed by your professor without knowing the context in which he raised it. I have been looking at general relativity lately and I can begin to see the context that wrobel puts his answer in, but this may not be where your professor was going.

He may likely be asking, if a observer in a rotating or translating frame formulates the lagrangian, and solves for the equations of motion, will the observer get the same lagrangian and motion when the observer goes back to the original frame?