2
$\begingroup$

From Woodhouse's General Relativity:

If $A$ is the origin and $B$ is a nearby event with coordinates $dt, dx, dy, dz$, then,

$$ds^2 = dt^2 - dx^2 - dy^2 - dz^2$$ is the same in all local inertial coordinate systems with origin A.

Timelike separation. If $ds^2 >0$, then $ds$ is the time from $A$ to $B$ on a clock travelling between the two events in free-fall

Spacelike separation. If $ds^2 < 0$, then $ds^2=-D^2$, where $D$ is the distance from $A$ to $B$ measured in a frame in free-fall in which $A$ and $B$ are simultaneous.

1) If we are talking about timelike separation why given that $ds^2 >0$ is $ds$ is the time from $A$ to $B$? I cannot see how this works in the equation above

2) If given that $ds^2 < 0$, I get how we can say $ds^2=-D^2$ but why are we now able to say that this is a distance (and not a time)? How can we discard considering the $dt^2$ in the above equation?

I have tried to think about this by writing

$$C^2 = dt^2 - dx^2 - dy^2 - dz^2$$ and

$$-C^2 = dt^2 - dx^2 - dy^2 - dz^2$$

but I cannot get my head around this.

$\endgroup$
0
2
$\begingroup$

I'll try to re-describe the same ideas in a different way. This isn't meant to be a quick answer to the question; rather, this is meant to be a resource to help build some intuition.

In this answer, the word "frame" is not used. That's because "frame" might carry connotations of something rigid, something defined by "axes." This answer is expressed using the more general concept of a coordinate system, which doesn't rely on anything like axes or straight lines.

Coordinates are arbitrary labels for the points in spacetime. They should assign a unique 4-tuple of numbers $(w,x,y,z)$ to each point, and they should do this in a smooth way, but otherwise they are arbitrary. Any worldline (curve in spacetime) can be described by giving the four coordinates as functions of some other parameter $\lambda$ that runs along the worldline. Examples will be shown below, after the general principles.

Mathematically, coordinate systems and worldlines are defined without the help of any geometric concepts like time, distance, timelike, or spacelike, and without the help of any dynamic concepts like free-fall. Geometry (including time) and free-fall are both defined instead by the metric. A convenient way to specify the metric is by specifying the line element. The line element takes any $\lambda$-parameterized worldline as input and returns a single function $G(\lambda)$ as output. In special relativity, the line element can be expressed as $$ G(\lambda) = \dot w^2 - (\dot x^2+\dot y^2+\dot z^2) \tag{1} $$ where a dot denotes a derivative with respect to the parameter $\lambda$ along the given worldline. The worldline is called

  • timelike wherever $G(\lambda)>0$,

  • spacelike wherever $G(\lambda)<0$,

  • lightlike wherever $G(\lambda)=0$.

A worldline is called causal if it is either timelike or lightlike. The causality principle says that only a causal worldline can represent the history of a physical object. Proper time is defined only along such a worldline. Given any causal worldline, its proper time $\tau(\lambda)$ is defined by the condition $$ \dot\tau^2 = G(\lambda) \geq 0. \tag{2} $$ This tells us how the proper time $\tau$ progresses along the worldline as a function of the parameter $\lambda$.

In hindsight, now that the line element (1) has been specified, we see that a wordline cannot be timelike unless $w$ changes monotonically along the worldline. In this sense, we can think of $w$ as a "timelike" coordinate — but it is still just a coordinate. Proper time is given by equation (2), and this is what an object actually experiences as time. Proper time is specific to the given worldline, and it is invariant under changes of the coordinate system.

If the quantity (1) is negative, then we have a spacelike worldline. Proper time is undefined along such a worldline. Physical objects, including clocks, cannot move according to such a worldline, so we shouldn't expect to have any invariant notion of the progression of time along such a worldline. What we have instead for such a worldline is proper distance $\ell$, given by the condition $$ \dot\ell^2 = -G(\lambda) > 0. \tag{3} $$

Two points in spacetime are said to be "timelike separated" if they can be connected to each other by some timelike worldline, and they are said to be "spacelike separated" if they cannot be connected to each other by any causal worldline. The concept of "spacelike separated" events is an extension of the concept of "simultaneous" events. Spacelike-separated events cannot be time-ordered in any invariant way.

By the way, even if two points are timelike separated (which means that one of the points is unambiguously in the future of the other), they can still be connectd to each other by a spacelike worldline. The following pair of examples illustrates this.


Example 1

Choose constants $A,B,C$ and consider the worldline given by $$ w(\lambda)=A\lambda \hskip1cm x(\lambda)=B\lambda+C \hskip1cm y(\lambda)=0 \hskip1cm z(\lambda)=0. \tag{4} $$ Then $$ \dot w = A \hskip1cm \dot x = B \hskip1cm \dot y = 0 \hskip1cm \dot z = 0, \tag{5} $$ so $G(\lambda)=A^2-B^2$, which is independent of $\lambda$ in this simple example. This worldline is:

  • timelike if $A^2>B^2$, and then equation (2) gives $\tau(\lambda) = \sqrt{A^2-B^2}\,\lambda$ for the proper time along this worldline.

  • spacelike if $A^2<B^2$, and then equation (3) gives $\ell(\lambda) = \sqrt{B^2-A^2}\,\lambda$ for the proper distance along this worldline.

  • lightlike if $A^2=B^2$, and then the proper time and proper distance are both zero along this worldline.

For the special metric defined by (1), a timelike (or lightlike) worldline correpsonds to free-fall if and only if the derivatives $(\dot w,\dot x,\dot y,\dot z)$ are all proportional to each other. In particular:

  • The worldline defined by (4) represents free-fall if $A^2\geq B^2$.

  • If $A^2<B^2$, then it does not represent any physically possible motion.


Example 2

Consider the worldline defined by \begin{align} w(\lambda) &= \lambda + \lambda^3 \\ x(\lambda) &= \cos(\beta\lambda + \beta\lambda^3) \\ y(\lambda) &= \sin(\beta\lambda + \beta\lambda^3) \\ z(\lambda) &= 0. \tag{6} \end{align} where $\beta$ is a constant. For each value of $\lambda$, these equations specify the coordinates of one point in the four-dimensional manifold, so they define a worldline. Plug (6) into (1) to get $$ G(\lambda)=(1-\beta^2)(1+3\lambda^2)^2. \tag{7} $$

  • If $\beta^2<1$, then this worldline is timelike, and then equation (2) says that its proper time is given by $$ \tau(\lambda)=\sqrt{1-\beta^2}\,(\lambda+\lambda^3). \tag{8} $$ This tells us how the proper time $\tau$ progresses along the worldline as a function of the parameter $\lambda$. Equation (8) is independent of the coordinates, as it should be; proper time is invariant under coordinate transformations.

  • If $\beta^2>1$, then this worldline is spacelike. Notice, though, that this spacelike worldline passes through all of the points $(w,x,y,z)=(2\pi\,n/\beta,1,0,0)$ for all integers $n$, and these points are also all contained in the timelike worldline (4) with $A=1$, $B=0$, and $C=1$. (The parameters $\lambda$ of the two worldlines are not the same; the symbol $\lambda$ was recycled.) This shows that the worldline (6) with $\beta^2>1$ is an example of a spacelike worldline that connects some timelike-separated points.

$\endgroup$
7
$\begingroup$

Time-like intervals are a small distance and a long time. For example one event is here and now. A second event is 1 m away and 1 sec later.

If you travel at ~1 m/s in the right direction, you can pass through both events. In that frame, the separation is purely time: here and now, here and ~1 sec later. Any time-like separation can be reduced to a pure time separation this way.

It does get a little counter intuitive at higher speeds. You can ignore the time dilation and space contraction at 1 m/s. For a separation of .8 light-sec and 1 sec, someone could still travel through both events. But you would have to calculate the time and distance separation the traveler saw.

Light-like separations are separated equally by time and distance. For example, event is here and now. The other is 1 light-second away and 1 second later. You cannot travel the speed of light, so you cannot reduce this to a pure time separation. A beam of light can pass through both events.

A space-like separation is a long distance and a short time. An example is here and now, and an event 1 light sec away, and 0.8 sec from now. Nothing can go through both events.

But here it gets really counter intuitive. We are used to the idea that I see two places at different times, while a traveler that passes through both sees them as the same place at different times.

We are not used to the idea that a traveler can choose a velocity that makes the two events simultaneous. But time does work like that. Any space-like separation can be reduced to a purely spatial separation, two simultaneous events, by choosing the right velocity.

$\endgroup$
3
  • $\begingroup$ Im still confused with how we can go from $t$ to $s$ to $\tau$. Potentially I cannot see this mathematically as well. $\endgroup$ – Trajan Dec 19 '18 at 17:12
  • $\begingroup$ Part of it may be notation. There is more than one convention for saying the same thing in relativity. Sometimes people use ds2 for the interval between events, and sometimes dτ2. Sometimes they may choose ds2 for space like intervals and dτ2 for time like intervals, sometimes not. $\endgroup$ – mmesser314 Dec 19 '18 at 21:24
  • $\begingroup$ Because of time dilation and spatial contraction, You measure different time and space intervals in different frames. But everybody gets the same ds2. For a time like interval, if you choose the frame where the two events are in the same place, dx2=dy2=dz2=0, and ds2=dt2. In this case, the interval is called the proper time. $\endgroup$ – mmesser314 Dec 19 '18 at 21:24
2
$\begingroup$

This is simply because the interval $ds^2$ is invariant, and so we can calculate it in any frame we want. So if $ds^2$ is timelike, we can go a system $(t', x', y', z')$ in which the world line from A to B is at rest, and this coordinate system we have

$$ds^2 = dt'^2,$$

since the spatial displacement is zero. Similarly, if the separation is spacelike, we can go to a system in which A and B are at the same time, so $dt'$ disappears from the interval.

$\endgroup$
3
  • $\begingroup$ Im still confused with how we can go from $t$ to $s$ to $\tau$. Potentially I cannot see this mathematically as well. $\endgroup$ – Trajan Dec 19 '18 at 17:12
  • $\begingroup$ @Permian could you give a bit more detail? Do you know the definitions of $s$ and $\tau$? $\endgroup$ – Javier Dec 19 '18 at 18:44
  • $\begingroup$ I get that $tau$ is the proper time and but $s$ seems to be set by the first equation in my question. Incidentally, I am not sure how the right hand side arises (as it was given by definition in my book)? $\endgroup$ – Trajan Dec 19 '18 at 20:22
1
$\begingroup$

For a precise appreciation you must take into account that timelike intervals and spacelike intervals have different foundations. Time is not space, and the symmetry of time and space is limited to Lorentz symmetry. By consequence, you must appreciate timelike intervals and spacelike intervals separately:

Timelike intervals are time intervals because of the phenomena of proper time and time dilation.

In contrast, the consideration of spacelike intervals as space intervals is a mere model which satisfies the assumption that spacetime is a manifold, but there is not the same kind of strict derivation as for proper time.

1.Timelike intervals: The particle worldlines are timelike intervals, and they consist of proper time intervals dτ. That means, in the reference frame of the particle, the intervals are intervals of its own time (proper time dτ). Also, according to the reference frame of the particle, the particle is not moving at all, its spatial coordinates are always null coordinates because it is its own reference frame.

In contrast, observers are observing a moving particle. The observed time coordinate is dt, corresponding to the "time-dilated proper time". As the particle is moving from the point of view of all observers with other reference frames, there is a spatial coordinate.

That means inversely, that from a given time and space coordinate of the worldline of a particle you can retrieve the corresponding proper time of the particle. As timelike intervals correspond to the proper time of a worldline, there is no problem to understand that timelike intervals have time character.

2.The situation is different and more complicated for spacelike intervals. The speed limit of spacetime is c, and for this reason there are no spacelike worldlines. This implies also that time dilation cannot work for spacelike worldlines which explained above 1) the fusion between time and space coordinates.

On the other hand, distances are measured by 3D space intervals, not by 4D spacelike intervals. The distance between Sun and Earth is always (approximately) 8 light minutes, even if we consider the Sun 4 minutes ago and the Earth now. We may call this distance "8 light minutes + 4 minutes" or "space interval + time interval", but we do not get a simple distance as space interval.

As you outlined in your question: squared spacelike intervals are negative. Thus we get imaginary results for space distances of spacelike intervals. Misner/ Thorne/ Wheeler, Gravitation, tried to repair this flaw by the means of the signature (-,+,+,+), but until today this concept has not found general acceptance, and especially in particle physics this concept does not seem to correspond to experimental needs.

So, as a consequence, the consideration of spacelike intervals as distances is a mere model, and this model has not the same physical foundation as timelike intervals. If you follow the signature (-,+,+,+), this flaw is hidden by the fact that squared spacelike intervals are positive and spacelike intervals are real in the same way as 3D distances. According to this model, spacelike intervals may be considered as "space distances which are corrected by a time coordinate". But exactly such a correction may be physically justified for timelike intervals as shown above 1), but for spacelike intervals there is no such justification.

In summary, as soon as you abandon the signature (-,+,+,+), your question is arising. And the answer in short is that a model is used in order to satisfy human intuition, not more and not less.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.