# How to define the components of the Poincare group?

I know that the Poincare group/inhomogeneous Lorentz group can be defined as:

$$x^\mu = (t,-x) \\ t \rightarrow t^\prime = \gamma x + \delta t + b^0 \\ x \rightarrow x^\prime = \alpha x + \beta t + b^1 \\ x^{\mu \prime} = R(b) L x$$ and that it has to be invariant under Minkowski Metric $$ds^2 = c^2dt^2 - dx^2 = {ds^2}^\prime = c^2{dt^2}^\prime - {dx^2}^\prime$$

Usually I could use $$x^\prime = 0 \\ x= vt$$ and use this in the metric and get the components of the Lorentz transformation, but I am not sure that using translation works too.

Can please someone try to explain to me how to get the Lorentz transformation assuming translation?

• What source has $x^\mu = (t,-x)$? Jan 19, 2021 at 20:37
• Near duplicate. Few books write the ten 5x5 matrices of the Poincare group, but most decent QFT books write the infinitesimal realizations. Wu-Ki Tung's group theory book certainly does. The linked question illustrates how translations are expressed in matrix form. Jan 19, 2021 at 22:28
• This might be helpful, or the bottom of p 11 here. Jan 19, 2021 at 22:48
• Whoa, the Wu-Ki Tung book are really good. But what I want to show is that $$\alpha, \beta, \gamma, \delta$$ are the Lorentz transformation components. Which is just said in the book. $${x}^\prime = \Lambda ^\mu_\nu x^\nu + b^\mu.$$ With lambda beeing the proper Lorentz transformations. But I cant get this result. Jan 20, 2021 at 1:53

Lorentz transformations relate two reference frames whose origins coincide. If there is a spacetime translation involved, then those two frames are not related via Lorentz transformation.

Starting from an inhomogeneous Lorentz transformation $$x'^\mu = \Lambda^\mu_{\ \ \nu} x^\nu + b^\mu$$, consider an infinitesimal displacement $$dx^\mu$$. In the transformed coordinates, the inhomogeneous term falls away and you are left with $$dx'^\mu = \Lambda^\mu_{\ \ \nu} dx^\nu$$, since

$$x'^\mu-y'^\mu = (\Lambda^\mu_{\ \ \nu} x^\nu + b^\mu) - (\Lambda^\mu_{\ \ \nu}y^\nu + b^\mu) = \Lambda^\mu_{\ \ \nu}(x^\mu-y^\mu)$$

From there, you can proceed as usual to derive the form of $$\Lambda$$ from physical considerations.

• But the Poincare Group, or the inhomogeneous Lorentz are the set of transformations in Minkowski space consisting of all translations and proper Lorentz transformations. Jan 20, 2021 at 1:57
• @A.Map When you say “Lorentz transformation” without any other precondition, that will be interpreted as a proper Lorentz transformation. The general transformation which includes translations cannot be written in the form $\Lambda^\mu_{\ \ \nu}x^\nu$. They can be realized as 5x5 matrices as referenced by Cosmas in the comments, is that what you’re looking for? Jan 20, 2021 at 2:40
• They can be written in the form $${x}^\prime = \Lambda^\mu_\nu x^\nu + b^\mu$$, right? If it is correct, I want to show that the $$\Lambda$$ term is from the Lorentz transfomartion. Jan 20, 2021 at 12:16
• @A.Map Oh, I see. I've updated my answer. The key is to consider displacements, not absolute coordinates, in which case the inhomogeneous term is canceled out. Jan 20, 2021 at 12:21