39
$\begingroup$

I'm struggling to understand the relativity of simultaneity and position.

If my conception and birth are separated by time but not space, a frame of reference in which my birth and conception are simultaneous should exist right?

If another observer moves in the opposite direction, will he see my birth before my conception?

$\endgroup$
  • 7
    $\begingroup$ "a frame of reference...should exist" --- have you attempted to write down thst frame? $\endgroup$ – WillO Mar 4 at 17:07
  • 1
    $\begingroup$ Probably an accelerated frame of reference... $\endgroup$ – Rob Jeffries Mar 4 at 17:27
  • 2
    $\begingroup$ I notice that people are voting to put this on hold; personally I think it's a reasonable question, although it could benefit from a bit more research effort. If this does get put on hold (which is a reasonable thing to happen, if a fifth person thinks it should be done), perhaps we could open a discussion on meta to better understand the reasons. $\endgroup$ – David Z Mar 4 at 20:18
  • 2
    $\begingroup$ So, I see a bunch of people saying that this isn't possible in the answers, but wouldn't it be possible if your mother managed to leave her light cone (hopped on an FTL rocket, flew through a wormhole, etc) prior to giving birth to him? $\endgroup$ – nick012000 Mar 5 at 3:55
  • 12
    $\begingroup$ @nick012000 I don't think travelling through a wormhole is recommended when you're pregnant. $\endgroup$ – rubenvb Mar 5 at 15:30
95
$\begingroup$

Suppose we take the spacetime point of your conception as the origin, $(t=0, x=0)$, then the spacetime point for your birth would be $(t=T, x=uT)$. The time $T$ is approximately $9$ months, and we are writing the spatial position of your birth as $x=uT$ where $u$ is a velocity. The velocity $u$ can be any value from zero (i.e. born in the same spot as conception) up to $c$ (because your mother can't move faster than light).

Now we'll use the Lorentz transformations to find out how these events appear for an observer moving at a speed $v$ relative to you. The transformations are:

$$ t' = \gamma \left( t - \frac{vx}{c^2} \right ) $$

$$ x' = \gamma \left( x - vt \right) $$

though actually we'll only be using the first equation as we're only interested in the time. Putting $(0,0)$ into the equation for $t'$ gives us $t'=0$ so the clocks of the observer and your mother both read zero at the moment of your conception. Now feeding the position of your birth $(T,uT)$ into the equation for $t'$ we get:

$$ t' = \gamma \left( T - \frac{vuT}{c^2} \right ) $$

For you to be born before you were conceived we need $t'\lt 0$ and that gives us:

$$ T \lt \frac{vuT}{c^2} $$

or:

$$ vu \gt c^2 $$

We know that the observer's velocity $v$ cannot be greater than $c$, and your mother's velocity $u$ cannot be greater than $c$, so this inequality can never be satisfied. That is, there is no frame in which you were born before you were conceived.

The rule is that two events that are timelike separated, i.e. their separation in space is less than their separation in time times $c$, can never change order. All observers will agree on which event was first. For the order to change the events have to be spacelike separated. In this case this would mean $uT \gt cT$ i.e. your mother would have to have moved at a speed $u$ faster than light between your conception and birth.

$\endgroup$
  • 2
    $\begingroup$ Xe isn't pre-supposing it, M. Fischler. That these are not separated in space is specified in the question. $\endgroup$ – JdeBP Mar 4 at 23:39
  • 48
    $\begingroup$ @MarkFischler it's normal for your mother to be present at both your conception and your birth. My mother assures me that this was the case. $\endgroup$ – John Rennie Mar 5 at 7:54
  • 17
    $\begingroup$ @Mars That would be a good answer if this were puzzling.stackexchange.com. However, since it's Physics, it is safe to assume any unstated circumstance is in a reasonable, typical configuration. $\endgroup$ – Oscar Bravo Mar 5 at 9:46
  • 7
    $\begingroup$ @MarkFischler there is always a reference frame in which the event of conception and the event of birth differ only in time coordinate. $\endgroup$ – Danijel Mar 5 at 10:49
  • 8
    $\begingroup$ @Mars test tube babies aka in-vitro fertilisation makes no difference to the argument unless we have glassware that can travel faster than light. $\endgroup$ – bdsl Mar 5 at 21:36
27
$\begingroup$

Simultaneity is relative, but causality is always preserved in moving from one reference frame to another. In no reference frame can you be born prior to being conceived or be born at the same time that you're conceived.

$\endgroup$
  • 9
    $\begingroup$ There are reference frames where you can be born prior to being conceived: any frame traveling sufficiently faster than light relative to you. The fact that we've never observed such a reference frame doesn't mean we can't describe it. $\endgroup$ – Mark Mar 4 at 20:14
  • $\begingroup$ @Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)? $\endgroup$ – Fax Mar 5 at 10:15
  • 5
    $\begingroup$ @Fax JR's answer suggests no such thing, it compares the product of two speeds (the speed of the observer v and the speed of the mother u) with c*c and then makes the inequality impossible by v < c ^ u < c. If either velocity was sufficiently faster than c, the inequality may hold. $\endgroup$ – DonFusili Mar 5 at 11:32
  • $\begingroup$ @DonFusili I missed that, thanks. So a superluminal mother would experience the events in reverse order. Does that mean a luminal mother would perceive the events as being simultaneous? $\endgroup$ – Fax Mar 5 at 12:46
  • 2
    $\begingroup$ @Fax, a superluminal mother would perceive the events as happening in the normal order. It's the observer who would perceive them happening in the reverse order. $\endgroup$ – Mark Mar 5 at 19:51
14
$\begingroup$

Another way to recognize that causality is preserved is to notice that for events to have ambiguous time order (i.e. you could switch your experience of their order with a Lorentz boost), they must be space-like separated. If two events happen at the same location, their time order is unambiguous. No Lorentz transformation could switch them.

$\endgroup$
  • $\begingroup$ I don't understand this answer as causality and unambiguous order should be here even if the mother had moved. Am I right? $\endgroup$ – Alchimista Mar 6 at 17:41
  • 1
    $\begingroup$ @Alchimista yes, but since the mother presumably moves slower than the speed of light, the events will still be light-like separated. $\endgroup$ – Gilbert Mar 6 at 17:49
  • $\begingroup$ Thanks. This was already clear. Might be that causality is rooted in my mind. I see useful this kind of light like separation etc when I already know that events do not originate from the same thing/cause. I admit that relativity still confuse me though not in this current question. I have to think on something which is not light- like separated, but I would see it as two independent events like me typing this and a firecracker exploding somewhere else. Could you give me a scenario (with no stress, if you have time). $\endgroup$ – Alchimista Mar 7 at 8:40
  • $\begingroup$ @Alchimista any two events which happen simultaneously from your perspective, but at different places, will easily satisfy the space-like separation condition to have ambiguous time order. An example would be firecrackers going off simultaneously to your left and right. Then if an observer had been moving (relative to your rest frame) in the left or right direction, they would have observed one or the other firecracker go off first. But this is okay since it’s intuitive that neither firecracker could have been the “cause” of the other in this case. $\endgroup$ – Gilbert Mar 7 at 12:11
  • $\begingroup$ Super. I think sometimes dictionary is more confusing than of help. $\endgroup$ – Alchimista Mar 7 at 15:41
8
$\begingroup$

There are coordinate systems in which your birth preceded your conception. However, special relativity deals only with coordinate systems that can be related through translations, rotations, and transformations that are known as Lorentz transformation. Translations correspond to changing where the origin of the coordinate system is, rotations correspond to changing what directions the axes point, and a Lorentz transformation corresponds to changing what is considered "at rest". General relativity is more complicated, but those complications don't affect the answer to this question. Your conception and birth are what's known as timelike-separated events. For timelike-separated events, you can't switch the order through any of the standard transformations of relativity. So technically there are frames of references where your birth occurred before your conception, but those don't correspond to the frame of reference of any physical object, or possible physical object, under current understanding of relativity.

$\endgroup$

protected by David Z Mar 4 at 21:14

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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