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According to Wikipedia, the mass $M$ is one of the Casimir invariants of the Galilean group. Casimir invariants of a group are made out of the generators, and they commute with all the generators of the group. For example, the Casimir invariant of the group $SU(2)$ is $J^2$ which is made out of $J_1, J_2, J_3$ as $$J^2:=J_1^2+J_2^2+J_3^2.\tag{1}$$ Another example is a Casimir invariant of the Poincare group $P^\mu P_\mu$ which is made out of $P_0, P_1,P_2, P_2$ as $$P^\mu P_\mu:=-P_0^2+P_1^2+P_2^2+P_3^2.\tag{2}$$

In the same manner, can we write $M$ as a function of the generators Galilean group?

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  • $\begingroup$ It is not qudratic. $\endgroup$
    – DanielC
    Jul 26, 2021 at 7:04
  • $\begingroup$ @DanielC Can you elaborate? I am asking how to write M using the generators of the Galilean group. $\endgroup$ Jul 26, 2021 at 7:54
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    $\begingroup$ See also this and links within: physics.stackexchange.com/q/103441 $\endgroup$
    – DanielC
    Jul 26, 2021 at 10:30

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  1. No, the mass operator $M$ is the central charge operator in the central extension [known as the Bargmann algebra (BA)] of the Galilean algebra (GA).

  2. In other words, the mass operator $M$ is by definition not part of the GA. [The momenta and Galilean boosts commute by definition within the GA.]

  3. If we define a Casimir invariant of a Lie algebra as a central element of its universal enveloping algebra (UEA), then the mass operator $M$ is a Casimir invariant for the BA but not for the GA.

    See also this related Phys.SE post.

  4. The above bolsters the following point:

    The natural non-relativistic Lie algebra in Newtonian mechanics is the BA, not the GA!

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