# Lorentz-invariance of step function

I was reading about the Lorentz invariant integration measure $\int \frac{d^3k}{2E_K}$, and ways to prove that this was Lorentz invariant. Many of the proofs I have read use the step function (or Heaviside function) $\theta(k^0)$, and claim that this is obviously Lorentz-invariant (e.g Alex Nelson's notes). Other sources claim that there is some requirement that the Lorentz transformations have determinant = 1, or that $k^\mu$ is timelike. What are the actual conditions on the vector $k$ for the step function to be lorentz-invariant?

• Can you link to a specific proof, please? Taken out of context this leaves too much to guess, at least for my taste. The question for a quality proof is, of course, legitimate. – CuriousOne Apr 27 '15 at 20:48
• @CuriousOne quite a good proof can be found here physics.ucsd.edu/students/courses/fall2009/physics130b/PS_3.pdf – Jack Richings Apr 27 '15 at 20:54
• – Heterotic Apr 28 '15 at 9:48
• – Qmechanic Jul 20 '16 at 1:20

Proper, orthochronous Lorentz transformations can't change the sign of a $x^0$ component in a time-like four-vector - $dx^\mu dx_\mu$ and $\text{sign}\,x^0$ are both Lorentz invariants if $x_\mu$ is time-like. Because $\text{sign}\,x^0$ is Lorentz invariant, the step function $\theta(x^0)$ is Lorentz invariant - the condition is that $x$ is time-like.

A general Lorentz transformation that wasn't orthochronous could violate causality by making one observer see time-like events occurring in a different order.

• Why proper, orthochronous Lorentz transformations can't change the sign of $x^0$ of a time-like vector? – Louis Yang Oct 12 '15 at 23:35
• @LouisYang See my answer – Selene Routley Jul 20 '16 at 5:21

I'd like to add to Innisfree's answer, which depends on the orthochronous property of certain Lorentz transformations, so as to clear up the follow-up question to his/her answer::

Why proper, orthochronous Lorentz transformations can't change the sign of $$x^0$$ of a time-like vector?

First let's clear up some terminology. An orthochronous Lorentz transformation means, by definition, one that doesn't reverse the sign of the time component $$x^0$$ of a timelike vector $$X$$, i.e. one for which $${x^0}^2-{x^1}^2-{x^2}^2-{x^3}^2 > 0$$. It can reverse the $$x^0$$ component if this inequality is not fulfilled. So this nonreversal is something that is a by-definition property of orthochronous transformations, so, in these terms, the question has no physical content. But a question that does have physical consequence, and whose answer is crucial to Innisfree's arguent, is that:

The identity-connected component of the Lorentz group $$SO(1,\,3)$$ comprises only orthochronous transformations

Physically, what this means is that no finite number of finite rapidity boosts and rotations can reverse the sign of $$x_0$$. This is what allows us to salvage a sound notion of causality from special relativity even though simulteneity is observer-dependent, as long as we postulate that supraluminal signalling is impossible.

The proof of the statement is the following simple argument.

Let's suppose that $$(t,\,x,\,y,\,x)$$ is an event in an observer's future such that $$t^2 - x^2 - y^2 - z^2=Q^2 > 0$$.

Then, under the action of any Lorentz transformation, this event's co-ordinates become $$(t^\prime,\,x^\prime,\,y^\prime,\,x^\prime)$$ where:

$${t^\prime}^2 = Q^2 + {x^\prime}^2 + {y^\prime}^2 + {z^\prime}^2\tag{1}$$

where $$Q$$ (the Minkowski length of the vector) is constant and therefore:

$$t = \pm\sqrt{Q^2 + {x^\prime}^2 + {y^\prime}^2 + {z^\prime}^2}\tag{2}$$

which means that, as $$t,\,x,\,y,\,z$$ are varied through a Lorentz transformation, either $$t \geq +Q$$ or $$t\leq-Q$$, and $$t$$ is excluded from the interval nonzero length interval $$(-|Q|,\,|Q|)$$.

But if the event co-ordinates $$X^\prime$$ are acted on by a Lorentz transformation from the Lorentz group's-identity connected component, this means there is a continuous path of the form:

$$X_s:\mathbb{R}\to\mathbb{R}^{1+3};\;X_s(s) = \prod_{k=0}^N\exp(s\,K_k)\,X;\;X_k\in\mathfrak{so}(1,\,3)\quad X_s(1) = X^\prime\tag{3}$$

linking $$X$$ to $$X^\prime$$, so the time co-ordinate must vary continuously along this path. Therefore, it cannot jump the excluded interval $$(-|Q|,\,|Q|)$$ and so therefore, the negative root of (2) is ruled out and we conclude that the time component must stay positive. Thus all members of the Lorentz group's identity connected component are orthochronous and no finite number of finite rapidity boosts and rotations (i.e. the $$\exp(K_k)$$ in (3) at $$s=1$$) can reverse the time order of timelike-separated spacetime events.

The idea for this little argument comes from the the technique in Schreier's original 1926 proof of Schreier's theorem that any discrete normal subgroup of a Lie group is needfully contained within the Lie group's center. See Schreier, O. "Abstrakte kontinuierliche Gruppen", Abhandlungen aus dem Mathematischen Seminar der Universitat Hamburg, 4: pp15-32, 1926

Thanks WetSavannaAnimal aka Rod Vance for the rigorous proof. Here I would like to add a short answer in a different perspective.

An orthochronous proper Lorentz transformation can always be split into a boost and a rotation. The rotation part doesn’t change the $k^0$ component. A boost on $k^{0}$ has the form $$k^{0\prime}=\gamma\left(k^{0}-\vec{v}\cdot\vec{k}\right).$$ Since for time-like or light-like vectors, $\left|k^{0}\right|\ge\left|\vec{k}\right|$, $\left|v\right|<1$, and $\gamma>0$, boosts preserve the sign of $k^{0}$. Thus, $\theta\left(k^{0}\right)$ is Lorentz invariant for time-like and light-like vectors.