Orthochronous condition - Lorentz transformations

Question

I'm trying to learn by myself some special relativity. By reading online I've come across the fact that Lorentz transformations are rotations on a 4D spacetime with a Minkowski metric. A rotation $\Lambda$ must preserve lengths, therefore, for a generic 4-vector $\vec{x}$: $$ \Vert\Lambda \vec{x}\Vert = \Vert\vec{x}\Vert \Rightarrow \left( \Lambda^\alpha_\mu x^\mu \right) \left( \Lambda^\beta_\nu x^\nu \right) \eta_{\alpha \beta} = x^\mu x^\nu \eta_{\mu \nu} \Rightarrow \Lambda^\alpha_\mu \Lambda^\beta_\nu \eta_{\alpha \beta} = \eta_{\mu \nu} \Rightarrow \Lambda^T \eta \Lambda = \eta,$$ which is the condition that allows $\Lambda \in O(1, 3)$. Now it is necessary to exclude spatial reflections, therefore it is necessary to force $\mathrm{det}\Lambda = 1$: $$ \Lambda \in SO(1, 3) \subset O(1, 3).$$ I've also read that it is necessary to exclude time reflections, which swap the past and the present. The condition which must be forced is $\Lambda^0_0 \geq 0$, the orthochronous condition, so that $\Lambda \in SO^+(1,3)$.

Now I find it difficult to understand why that condition excludes time reflections. I've tried to interpret the condition like this: applying $\Lambda$ to a 4-vector $\vec{x}$ produces a 4-vector $\vec{x}'$ whose time component should be: $$ x'^0 = \Lambda^0_\mu x^\mu = \Lambda^0_0 ct + \Lambda^0_1 x + \Lambda^0_1 y + \Lambda^0_1 z,$$ I get that if $\Lambda^0_0 < 0$ the sign in front of $ct$ changes, but it is not guaranteed that the whole $x'^0$ is actually negative. Could someone please explain it to me?

Answer 1

The correct condition for an orthochronous Lorentz transformation is that it does not change the sign of the time component for a timelike vector. For a spacelike vector, the sign of the time component is allowed to change.

Consider the problem of two points $x_{a}$ and $x_{b}$ with a spacelike separation between them. The first thing Einstein showed in his first relativity paper was that simultaneity is relative. One observer will the see $x_{a}$ and $x_{b}$ has occurring at the same time, but another (moving) observer will not. Depending on which direction the second observer is moving, they may see either $x_{a}$ or $x_{b}$ as happening first. The frames of the two observers are related by a proper, orthochronous Lorentz transformation, but this transformation does not preserve the sign of the time component of $x_{a}-x_{b}$—because $x_{a}-x_{b}$ is spacelike.

However, for a timelike vector $$|x^{0}|>|\vec{x}|=\sqrt{x^{j}x^{j}}=\left[(x^{1})^{2}+(x^{2})^{2}+(x^{3})^{2}\right]^{1/2},$$ and for an orthochronous Lorentz boost, with $\Lambda_{0}{}^{0}=\gamma$ and $\Lambda_{j}{}^{0}=\gamma\beta_{j}$, $$|\Lambda_{0}{}^{0}|>|\Lambda_{j}{}^{0}x^{j}|=\gamma\left|\vec{\beta}\cdot\vec{x}\right|.$$ Together, these ensure that the contributions to $x'^{0}$ coming from the spatial components of $x^{\mu}$ are always smaller in magnitude than the contribution from $x^{0}$, so the sign of $$x'^{0}=\Lambda_{\mu}{}^{0}x^{\mu}=\Lambda_{0}{}^{0}+\Lambda_{j}{}^{0}x^{j}$$ will always be the same as the sign of the original $x^{0}$.

(One additional note: The conditions for a proper, orthochronous Lorentz transformation is that $\det\Lambda=1$ and $\Lambda_{0}{}^{0}>0$. However, $\det\Lambda=1$ does not, on its own, mean that the transformation is proper—that is, that it preserves parity. The real condition for a proper Lorentz transformation is that the determinant of the spatial part of $\Lambda$ should be positive. This is evident from looking at the case of $\Lambda=\operatorname{diag}[-1,-1,-1,-1],$ which inverts both space and time, yet has $\det\Lambda=1$.)

Answer 2

After reading online, I want to try to answer to my own question: I've been searching deeper and I've found this article (pages 11-13) that features a proof for the orthochronous condition without referring the entries of the Lorentz transformation $\Lambda$. Even if this kind of approach is more general I find the Buzz's one to be more intuitive.

The author starts from the condition

$$ \Lambda^T \eta \Lambda = \eta $$

that in tensor notation becomes

$$ \Lambda^\alpha_\mu \Lambda^\beta_\nu \eta_{\alpha \beta} = \eta_{\mu \nu} \tag{1}$$

but also

$$ \Lambda^\alpha_\mu \Lambda^\beta_\nu \eta_{\alpha \beta} = \eta_{\mu \nu} \Rightarrow \Lambda^\alpha_\mu \Lambda^\beta_\nu \eta_{\alpha \beta} \eta^{\alpha \mu} \eta^{\beta \nu} = \eta_{\mu \nu} \eta^{\alpha \mu} \eta^{\beta \nu} \Rightarrow \Lambda_\mu^\alpha \Lambda_\nu^\beta \delta^\mu_\beta \eta^{\beta \nu} = \delta_\nu^\alpha \eta^{\beta \nu} $$

$$ \Lambda_\mu^\alpha \Lambda_\nu^\beta \eta^{\mu \nu} = \eta^{\alpha \beta} \tag{2} $$

Now, it is possible to retrieve a useful relation from $(1)$, setting $\mu = \nu = 0$:

$$ \left( \Lambda_0^0 \right)^2 - \sum_{i = 1}^3 \left( \Lambda_0^i \right)^2 = 1 \tag{1.1}$$

Doing the same for $(2)$, and setting $\alpha = \beta = 0$

$$ \left( \Lambda_0^0 \right)^2 - \sum_{i = 1}^3 \left( \Lambda^0_i \right)^2 = 1 \tag{2.1}$$

Now, consider a generic $4$-vector $\vec{x}$, by applying $\Lambda$ on it, it is obtained a new $4$-vector $\vec{x}'$ whose components are

$$ x'^\mu = \Lambda^\mu_\alpha x^\alpha $$

In particular, the time component $x'^0$

$$ x'^0 = \Lambda^0_\alpha x^\alpha = \Lambda^0_0 x^0 + \sum_{i=1}^3 \Lambda^0_i x^i \tag{3} $$

Now the first step to derive the orthochronous condition is to build the following inequality from the sum of the transformed spatial terms $\sum_{i=1}^3 \Lambda^0_i x^i$ of $\vec{x}$ inside $x'^0$

$$ \left( \sum_{i=1}^3 \Lambda^0_i x^i \right)^2 \leq \sum_{i=1}^3 \left( \Lambda^0_i \right)^2 \sum_{j=1}^3 \left( x^j \right)^2 $$

Now it is possible to sobstitute $(2.1)$ in place of $\sum_{i=1}^3 \left( \Lambda^0_i \right)^2$

$$ \left( \sum_{i=1}^3 \Lambda^0_i x^i \right)^2 \leq \left[ \left( \Lambda^0_0 \right)^2 - 1 \right] \sum_{j=1}^3 \left( x^j \right)^2 \tag{4}$$

Now, as Buzz said, the orthochronous condition applies to timelike $4$-vectors. The most intuitive way for me to understand why only the time component of timelike $4$-vectors matters, relies on the following interpretation of spacetime vectors

A timelike vector connects two events that are causally connected, that is the second event is in the light cone of the first event. A spacelike vector connects two events that are causally disconnected, that is the second event is outside the light cone of the first event.

From Vagelford's answer on this question. Timelike $4$-vectors lie inside the lightcone of a first event $O$. Suppose a timelike $4$-vector connects $O$ to a second event $A$. This means that if some kind of information (i.e. sound, EM waves...) propagates from $O$ to $A$, it is possible that it reaches $A$ before $A$ happens with a velocity $v = \beta c < c$, and hence it is possible that it influences the event $A$. The defining feature of a timelike $4$-vector is indeed the following

$$ || \vec{x} ||^2 = x^\mu x_\nu = x^\mu x^\nu \eta_{\nu \mu} = \left( c t \right)^2 - || \vec{r} ||^2 > 0 \Longrightarrow \left( c t \right)^2 > || \vec{r} ||^2 $$

Where $ \vec{r} = \left( x, y, z \right) $. Therefore timelike $4$-vectors can causally connect to events, that is the first event $O$ is the cause that influences the second event $A$. Thanks to the possible causality, in principle, it is sensible the notion of reversing the time flow, that is, switching the cause and the effect, using a Lorentz transformation $\Lambda$.

On the other hand, spacelike $4$-vectors lie outside the lightcone of a first event $O$ . This means that, given a spacelike $4$-vector that connects $O$ to a second event $B$ in order for them to be causally connected it is necessary that some kind of information, originated in $O$, reaches $B$ before $B$ happens, and this implies that the same information would need to propagate faster than $c$, which, in turn, violates one of the postulate of special relativity. The defining feature of a spacelike $4$-vector is indeed the following

$$ || \vec{x} ||^2 = x^\mu x_\nu = x^\mu x^\nu \eta_{\nu \mu} = \left( c t \right)^2 - || \vec{r} ||^2 < 0 \Longrightarrow \left( c t \right)^2 < || \vec{r} ||^2 $$

There is another way of interpreting timelike and spacelike $4$-vectors, suggested by zeldredge's answer to this question

Spacelike separation means that there exists a reference frame where the two events occur simultaneously, but in different places. Timelike separation means that there exists a reference frame where the two events occur at the same place, but at different times. Lightlike means that, well, light could travel between those points.

and by Mayou36's answer to the same post

[...]

time-like: if you are fast enough, you can be at (think spatial, like "at the festival") event a and at event b, it is only a "matter of time" until you see the second event

space-like: the two events are too far apart (in space). You cannot see both of them together, no matter how fast you are. As soon as event a happened and you go as fast as possible, event b will have happened before you arrive there.

[...]

So a space-like separation makes any causal relation between the two events impossible, i.e. one cannot cause or influence the other$^1$.

$^1$as mentioned, a common cause (an event that is time-like to both events) can still make the two correlated.

This means that, to ensure that a Lorentz transformation $\Lambda$ doesn't reflect time, swapping the past and the present, it is necessary to force that for a timelike $4$-vector, after the transformation, the past remains the past and the future remains the future (this is what the author of the paper means with the phrase "mapping the future on itself" (and the past on itself), in proposition 1.2).

The verse in which the time is flowing is encoded by the sign of the time component of the aforementioned $4$-vector, therefore, for a timelike $4$-vector $\vec{x}'$, $x'^0 = \Lambda^0_\alpha x^\alpha$ must have the same sign of $x^0$. For a timelike $4$-vector

$$ \left( x^0 \right)^2 > \sum_{i=1}^3 \left( x^i \right)^2 $$

Applying the transitive property in $(4)$

$$ \left( \sum_{i=1}^3 \Lambda^0_i x^i \right)^2 \leq \left[ \left( \Lambda^0_0 \right)^2 - 1 \right] \sum_{j=1}^3 \left( x^j \right)^2 \leq \left[ \left( \Lambda^0_0 \right)^2 - 1 \right] \left( x^0 \right)^2 \Longrightarrow $$

$$ \left( \sum_{i=1}^3 \Lambda^0_i x^i \right)^2 \leq \left[ \left( \Lambda^0_0 \right)^2 - 1 \right] \left( x^0 \right)^2 = \left( \Lambda^0_0 \right)^2 \left( x^0 \right)^2 - \left( x^0 \right)^2 \Longrightarrow $$

$$ \left( \sum_{i=1}^3 \Lambda^0_i x^i \right)^2 \leq \left( \Lambda^0_0 x^0 \right)^2 - \left( x^0 \right)^2 \leq \left( \Lambda^0_0 x^0 \right)^2 \Longrightarrow $$

$$ \left( \sum_{i=1}^3 \Lambda^0_i x^i \right)^2 \leq \left( \Lambda^0_0 x^0 \right)^2 \tag{5} $$

Note that, from $(3)$

$$ x'^0 = \Lambda^0_0 x^0 + \sum_{i=1}^3 \Lambda^0_i x^i \Longrightarrow \sum_{i=1}^3 \Lambda^0_i x^i = x'^0 - \Lambda^0_0 x^0 $$

Sobstituting in $(5)$

$$ \left( x'^0 - \Lambda^0_0 x^0 \right)^2 \leq \left( \Lambda^0_0 x^0 \right)^2 \Longrightarrow $$

$$ \left( x'^0 \right)^2 - 2 x'^0 \Lambda^0_0 x^0 + \left( \Lambda^0_0 x^0 \right)^2 \leq \left( \Lambda^0_0 x^0 \right)^2 \Longrightarrow x'^0 \left( x'^0 - 2 \Lambda^0_0 x^0 \right) \leq 0 \tag{6} $$

Now the orthochronous contition is both necessary and sufficient not to reflect time:

$$ \mathrm{sgn}\left( x'^0 \right) = \mathrm{sgn}\left( x^0 \right) \Longleftrightarrow \textrm{"orthochronous condition on } \Lambda \textrm{"}$$

Less formally: the past remains the past and the future remains the future if and only if the orthochronous condition is enforced on $\Lambda$.

1. Proof that: $ \mathrm{sgn}\left( x'^0 \right) = \mathrm{sgn}\left( x^0 \right) \Longrightarrow \Lambda^0_0 > 0$ :

   a. The future remains the future: $x^0 > 0$ (hence, for hypothesis, $ \mathrm{sgn}\left( x'^0 \right) = \mathrm{sgn}\left( x^0 \right) \Longrightarrow x'^0 > 0 $). From $(6)$: $$ x'^0 - 2 \Lambda^0_0 x^0 \leq 0 \overset{x^0, x'^0 > 0}{\Longrightarrow} -2\Lambda^0_0 < 0 \Longrightarrow \Lambda^0_0 > 0 $$

   b. The past remains the past: $x^0 < 0$. The proof is similar.

2. Proof that: $ \Lambda^0_0 > 0 \Longrightarrow \mathrm{sgn}\left( x'^0 \right) = \mathrm{sgn}\left( x^0 \right) $ :

   a. If $x'^0 > 0$: $$ x'^0 - 2 \Lambda^0_0 x^0 \leq 0 \overset{x^0, x'^0 > 0}{\Longrightarrow} -2\Lambda^0_0 x^0 < 0 \Longrightarrow x^0 < 0 \Longrightarrow \mathrm{sgn}\left( x'^0 \right) = \mathrm{sgn}\left( x^0 \right) $$

   b. if $x'^0 < 0$: the proof is similar.

Therefore the orthochronous condition $\Lambda^0_0 > 0$ is proven. $\blacksquare$

Note that, from $(1.1)$ - however the same could be derived from $(1.2))$ -

$$ \left\vert \Lambda^0_0 \right\vert = \sqrt{1 + \sum_{i=1}^3 \left( \Lambda_i^0 \right)} \Longleftrightarrow \Lambda^0_0 > 0$$

Then

$$ \Lambda^0_0 = \sqrt{1 + \sum_{i=1}^3 \left( \Lambda_i^0 \right)} \Longrightarrow \Lambda^0_0 \geq 1 $$

which is a stricter form for the orthochronous condition.

At this point it is possible to conclude that

$$ \left.\begin{matrix} \left.\begin{matrix} \Lambda^T \eta \Lambda = \eta \Rightarrow \Lambda \in O(1,3) \textrm{, rotations and reflections}\\ \mathrm{det} \Lambda = +1 \textrm{, proper condition: no spatial reflections} \end{matrix}\right\} \Rightarrow \Lambda \in SO(1,3)\\ \\ \Lambda^0_0 \geq 1 \textrm{, orthochronous condition: no time reflections} \end{matrix}\right\} \Rightarrow \Lambda \in SO^+(1,3) $$

Let me know if there are some errors. Hope this answer could help in the future!