How to generate or guess an invariant quantity?

Question

What are the methods for guessing the invariant quantity:

$$x^2 - t^2$$

under Lorentz transformation

$$x'=\gamma(x-vt)$$ $$t'=\gamma\left(t - {vx}\right)$$

where c = 1.

Answer 1

It's helpful to consider the Euclidean version of your question.

• What is the simple step-by-step procedure for "discovering" this particular invariant quantity $$x^2+y^2$$ using the vector method?
  
  Let $\vec V=(x,y)$.
  Given the dot-product $g_e(\vec A, \vec B)=A_xB_x+A_yB_y$, we have $$g_e(\vec V, \vec V)=x^2+y^2. $$

• So, by analogy, let $\tilde S=(t,x/c)$ be a vector in (1+1)-Minkowski spacetime
  with Minkowski dot-product $g_m(\tilde A,\tilde B)=A_tB_t- A_xB_x$. Thus $$g_m(\tilde S, \tilde S)=t^2-(x/c)^2. $$

Now, you may have a different but related question:
Why is the Minkowski dot-product defined as that?
That's follows from the Relativity Principle and the Speed of Light Principle.

Rather than think of "switching a sign", I prefer Bondi's motivation, which I describe here at https://physics.stackexchange.com/a/508251/148184 in response to Minkowski Metric Signature .

A possible follow-up... about generating invariants... first consider how one does it in the Euclidean case (with its Euclidean metric).

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update:

Vector and tensor algebra is an efficient way to generate scalars, if you can form the appropriate quantities.

With a vector alone $V^a$, you can't do anything.

You need additional structure, like a covector $\omega_a$ or a tensor like $Z_{ab}$. Then you can form $V^a \omega_a$ and $Z_{ab}V^aV^b$. For example, the dot product via a metric tensor $g_{ab}$ [a symmetric tensor] is an example of a $Z_{ab}$.

With two vectors and a metric, you can form dot products among them $$g_{ab}V^aV^b ,\qquad g_{ab}P^aP^b,\qquad g_{ab}V^aP^b .$$ If $g_{ab}$ is the Minkowski metric, $V^a$ is the 4-velocity of an observer, and $P^a$ is the 4-momentum of a particle, then the scalars above are the square-norm (1) [in the +--- signature], the square of the invariant mass of the particle $m^2$, and the energy of the particle according to the observer.

Answer 2

The invariance can be generated by a clever algebraic term exchange between x and t during transformation. From this backward:

Because - signs are required at different times to properly balance the term exchange, intuitively other quantities having only + terms or odd powers are not likely to work as invariants.