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?

  • $\begingroup$ What source has $x^\mu = (t,-x)$? $\endgroup$
    – G. Smith
    Commented Jan 19, 2021 at 20:37
  • $\begingroup$ 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. $\endgroup$ Commented Jan 19, 2021 at 22:28
  • $\begingroup$ This might be helpful, or the bottom of p 11 here. $\endgroup$ Commented Jan 19, 2021 at 22:48
  • $\begingroup$ 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. $\endgroup$
    – A. Map
    Commented Jan 20, 2021 at 1:53

1 Answer 1


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.

  • $\begingroup$ But the Poincare Group, or the inhomogeneous Lorentz are the set of transformations in Minkowski space consisting of all translations and proper Lorentz transformations. $\endgroup$
    – A. Map
    Commented Jan 20, 2021 at 1:57
  • $\begingroup$ @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? $\endgroup$
    – J. Murray
    Commented Jan 20, 2021 at 2:40
  • $\begingroup$ 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. $\endgroup$
    – A. Map
    Commented Jan 20, 2021 at 12:16
  • $\begingroup$ @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. $\endgroup$
    – J. Murray
    Commented Jan 20, 2021 at 12:21

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