# Poincaré Group element of pure spacetime translation

If we make a spacetime translation of the coordinates of a event $$x^{\mu}$$ such that $$x' ^{\mu} = x ^{\mu} + a^{\mu}$$, the element $$\eta _{\mu \nu} x'^\mu x' ^\nu$$. Must be invariant :

$$$$\eta _{\mu \nu} x'^\mu x' ^\nu = \eta _{\mu \nu} (x^\mu + a^ \mu )(x ^\nu + a^\nu)$$$$

I'm having trouble to show that. Can someone show how to proceed?

• where did you read that $\eta_{\mu \nu} x_\mu'x_\nu'$ must be invariant? source? – AccidentalFourierTransform Jan 20 at 2:38
• I've read that the poincaré transformations leave the spacetime interval between two events invariant, so I honk that this expression must be the spacetime interval between $x^\mu$ and the spacetime origin (0,0,0,0). – Lil'Gravity Jan 20 at 2:42
• Can u explain my mistake? – Lil'Gravity Jan 20 at 2:42
• No, it is the interval between an event at $x_A$ and another at $x_B$. The coordinates of both events undergo the translation. – G. Smith Jan 20 at 2:44
• So must be $\eta _{\mu \nu} (x_B - x_A)^\mu (x_B - x_A)^\nu$ to consider an invariant? – Lil'Gravity Jan 20 at 2:49

You are confusing points of a manifold and tangent vectors to that manifold. Vectors encode displacements. In general manifolds they encode infinitesimal displacement, while in Minkowski spacetime (or any $$\mathbb{R}^d$$ more generally) they may encode finite displacements.

Now, the metric tensor is a bilinear function of two vectors, not two points. So if $$x^\mu,y^\mu$$ are points in Minkowski spacetime and you see $$\eta_{\mu\nu}x^\mu y^\nu$$ you must understand this $$\eta$$ being applied not to the points, but to the vectors from the origin all the way to the points. This distinction may seem excessive formality but it is really important, since in the case of a general manifold you cannot identify points and vectors easily as it happens here.

Getting back to your question, the point is that in Minkowski spacetime vectors can be seem as just finite intervals, so that a vector becomes a difference of two points: $$v^\mu = x_B^\mu-x_A^\mu.\tag{1}$$

Now a translation is an operation on points. It shifts every point by the same translation vector $$\xi^\mu$$. So the point $$x^\mu$$ gets transformed to $${x'}^\mu = x^\mu +\xi^\mu$$. Since it affects all points equally we have $${x_B'}^\mu=x_B^\mu+\xi^\mu,\quad {x_A'}^\mu=x_A^\mu+\xi^\mu.\tag{2}$$

This transformation trivially preserves the Minkowski metric because vectors, in the sense of intervals as in (1), are obviously left unaltered:

$${v'}^\mu={x_B'}^\mu-{x_A'}^\mu=(x_B^\mu+\xi^\mu)-(x_A^\mu+\xi^\mu)=x_B^\mu-x_A^\mu=v^\mu.$$

• I understood this point of translations, thanks! – Lil'Gravity Jan 20 at 3:20