Suppose we have the vector space $\mathbb{R}^4$ and the Lorentz's transformation $f:\mathbb{R}^4\to\mathbb{R}^4$. Consider a inner product $g$ given by: $$g(x,y)=x^1y^1+x^2y^2+x^3y^3-c^2t^1t^2$$ for all $$x=(x^1,x^2,x^3,t^1)$$ and $$y=(y^1,y^2,y^3,t^2)$$ in $\mathbb{R}^4$.

  • How Minkowski concluded this expression?

  • Why we need that $$g(f(x),f(y))=g(x,y),\ \forall x,y\in\mathbb{R}^4~?$$

  • Can we define another invariant scalar product in all inertial system?

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    $\begingroup$ 2. is because the inner product should be the same for any observer, and applying a Lorentz-transformation (your $f$) is equivalent to a change of the observer. (And shouldn't it be $(c)^2$?) $\endgroup$ Dec 30, 2015 at 16:26

1 Answer 1


Regarding question 1. : Our task is to come up with an invariant scalar product in all inertial systems.

For easiness, I'll set $c=1$ and work in one dimension but the conclusions apply for higher dimensions.

We know that any two inertial frames $S'$ and $S$ are related via lorentz transformation as follows

$\begin{bmatrix}x'\\t'\end{bmatrix}=\gamma \begin{bmatrix}1 & -v\\-v & 1\end{bmatrix} \begin{bmatrix}x\\t\end{bmatrix}$

Where $$\gamma=\dfrac{1}{\sqrt{1-v^2}}$$

In a more compact form, the former matrix equation can be rewritten as

$$\mathbf{\chi'}= \Lambda \chi$$

Usually a scalar product can be written like this $$\mathbf { A^T \eta A}$$

Where $\mathbf{A^T}$ is the transpose of matrix $\mathbf A$, and $\mathbf {\eta}$ is a 2 by 2 matrix.

The invariant product we look for then is $$\mathbf {\chi^T\eta \chi=\chi'^T\eta \chi'}$$

We know all the terms in the equation except $\mathbf \eta$, so our task is to find it so that we are able to find out what the invariant scalar product is.

Given that the coordinate systems are related by $\mathbf{\chi'= \Lambda \chi}$.

If we applied Lorentz transformation for $$\chi'^T\eta \chi'$$ we get $$\mathbf{\chi^T \Lambda^T \eta \Lambda \chi}$$

Therefore for $\mathbf {\chi^T \eta \chi=\chi'^T\eta \chi'}$ to be satisfied it must be the case that $$\mathbf \Lambda^T \eta \Lambda=\eta$$

Solving this equation we get $$\eta = \begin{bmatrix}1 & 0\\0 & -1\end{bmatrix} $$

Therefore the inner product is given by

$$\mathbf {\chi^T\eta \chi}=\begin{bmatrix}x & t\end{bmatrix} \begin{bmatrix}1 & 0\\0 & -1\end{bmatrix} \begin{bmatrix}x\\t\end{bmatrix}=x^2 -t^2$$

Which is the Minkowski metric.

On question 3: Any Four-vector $\mathbf V$ that transforms under lorentz transformation, that is

$$\mathbf{V'=\Lambda V}$$

Has an invariant that is associated with it. Since the spacetime coordinates transform under LT, they're Four-vectors with the spacetime interval as the invariant scalar product associated with them, which is $x^2-t^2=x'^2-t'^2$ in one dimension. There are other four vectors like the Four-momentum or Four-current which have an invariant scalar product associated with them.

  • $\begingroup$ How exactly do you solve $\mathbf \Lambda^T \eta \mathbf\Lambda=\eta$? I am stuck on that, because your $\eta$ can be scaled and so is not the only solution. Can't see at the moment why there could not be even other solutions. $\endgroup$ Dec 30, 2015 at 17:23
  • $\begingroup$ @GyroGearloose Indeed $\mathbf \Lambda^T \eta \mathbf\Lambda=\eta$ has no unique solution, its general solution is given by $$\eta = \begin{bmatrix}n & 0\\0 & -n\end{bmatrix}$$ for all $n$ that is real. The solution $n=0$ is excluded for being trivial, since any arbitrary transformation matrix $A$ satisfy it, therefore the zero matrix metric does not uniquely specify Lorentz transformation. Actually you can choose any $n$(except $0$), but then your inner product will be given by $n(x^2-t^2)$. Obviously $n=1$ is the most natural and convenient to work with. $\endgroup$
    – Omar Nagib
    Dec 30, 2015 at 21:48
  • $\begingroup$ @GyroGearloose But the physics is the same if you choose any other $n$. $\endgroup$
    – Omar Nagib
    Dec 30, 2015 at 21:49
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    $\begingroup$ I see, $\Lambda^T=\Lambda$, so $\eta=\Lambda^T\eta\Lambda=\Lambda\eta\Lambda$. If $v$ is an eigenvector of $\Lambda$ with $v\Lambda=\alpha v$ then $v\Lambda\eta\Lambda=v\eta$ and $v\Lambda\eta\Lambda=\alpha v\eta\Lambda=v\eta$. Thus $v\eta$ is an eigenvector of $\Lambda$ and thus $\eta$ maps eigenvectors onto eigenvectors of $\Lambda$. So, up to scaling (or permutation), not much of a choice. $\endgroup$ Dec 30, 2015 at 22:06

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