Is there a frame of reference in which I was born before I was conceived?How does time dilation work without...

How can I be pwned if I'm not registered on the compromised site?

Forward slip vs side slip

Why can't we make a perpetual motion machine by using a magnet to pull up a piece of metal, then letting it fall back down?

How to lift/raise/repair a segment of concrete slab?

Is there a frame of reference in which I was born before I was conceived?

Skis versus snow shoes - when to choose which for travelling the backcountry?

Citing contemporaneous (interlaced?) preprints

Erro: incompatible type for argument 1 of 'printf'

Non-Italian European mafias in USA?

Levi-Civita symbol: 3D matrix

Is divide-by-zero a security vulnerability?

In iTunes 12 on macOS, how can I reset the skip count of a song?

How can I handle a player who pre-plans arguments about my rulings on RAW?

Why do members of Congress in committee hearings ask witnesses the same question multiple times?

Why adding the article "the" when it is not needed?

What happened to QGIS 2.x

What are all the squawk codes?

Didactic impediments of using simplified versions

Dystopian novel where telepathic humans live under a dome

Does an unattuned Frost Brand weapon still glow in freezing temperatures?

Giving a talk in my old university, how prominently should I tell students my salary?

What type of investment is best suited for a 1-year investment on a down payment?

Why is s'abonner reflexive?

A bug in Excel? Conditional formatting for marking duplicates also highlights unique value



Is there a frame of reference in which I was born before I was conceived?


How does time dilation work without a privileged reference frame?Perceived direction of light emitted in moving reference frameAbsoluteness of Simultaneity?Is this an inertial frame of reference in relativistic context?Meaning and logic of Einstein's train thought experimentHow to interpret Hermann Minkowski's comments on the construction of spacetime“We certainly cannot have observers in the same reference frame disagree on whether clocks are synchronized or not”-is this true?Special relativity - loss of simultaneity - Is that real?Do Events Conditional to Simultaneity Occur in Every Reference Frame?Is Relativity of simultaneity just a flaw in perception?













29












$begingroup$


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?










share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    "a frame of reference...should exist" --- have you attempted to write down thst frame?
    $endgroup$
    – WillO
    yesterday










  • $begingroup$
    Probably an accelerated frame of reference...
    $endgroup$
    – Rob Jeffries
    yesterday






  • 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
    yesterday






  • 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
    18 hours ago








  • 3




    $begingroup$
    @nick012000 I don't think travelling through a wormhole is recommended when you're pregnant.
    $endgroup$
    – rubenvb
    7 hours ago
















29












$begingroup$


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?










share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    "a frame of reference...should exist" --- have you attempted to write down thst frame?
    $endgroup$
    – WillO
    yesterday










  • $begingroup$
    Probably an accelerated frame of reference...
    $endgroup$
    – Rob Jeffries
    yesterday






  • 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
    yesterday






  • 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
    18 hours ago








  • 3




    $begingroup$
    @nick012000 I don't think travelling through a wormhole is recommended when you're pregnant.
    $endgroup$
    – rubenvb
    7 hours ago














29












29








29


6



$begingroup$


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?










share|cite|improve this question











$endgroup$




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?







special-relativity spacetime inertial-frames observers causality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









SmarthBansal

655423




655423










asked yesterday









IchVerlorenIchVerloren

326212




326212








  • 5




    $begingroup$
    "a frame of reference...should exist" --- have you attempted to write down thst frame?
    $endgroup$
    – WillO
    yesterday










  • $begingroup$
    Probably an accelerated frame of reference...
    $endgroup$
    – Rob Jeffries
    yesterday






  • 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
    yesterday






  • 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
    18 hours ago








  • 3




    $begingroup$
    @nick012000 I don't think travelling through a wormhole is recommended when you're pregnant.
    $endgroup$
    – rubenvb
    7 hours ago














  • 5




    $begingroup$
    "a frame of reference...should exist" --- have you attempted to write down thst frame?
    $endgroup$
    – WillO
    yesterday










  • $begingroup$
    Probably an accelerated frame of reference...
    $endgroup$
    – Rob Jeffries
    yesterday






  • 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
    yesterday






  • 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
    18 hours ago








  • 3




    $begingroup$
    @nick012000 I don't think travelling through a wormhole is recommended when you're pregnant.
    $endgroup$
    – rubenvb
    7 hours ago








5




5




$begingroup$
"a frame of reference...should exist" --- have you attempted to write down thst frame?
$endgroup$
– WillO
yesterday




$begingroup$
"a frame of reference...should exist" --- have you attempted to write down thst frame?
$endgroup$
– WillO
yesterday












$begingroup$
Probably an accelerated frame of reference...
$endgroup$
– Rob Jeffries
yesterday




$begingroup$
Probably an accelerated frame of reference...
$endgroup$
– Rob Jeffries
yesterday




2




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
yesterday




$begingroup$
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
yesterday




2




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
18 hours ago






$begingroup$
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
18 hours ago






3




3




$begingroup$
@nick012000 I don't think travelling through a wormhole is recommended when you're pregnant.
$endgroup$
– rubenvb
7 hours ago




$begingroup$
@nick012000 I don't think travelling through a wormhole is recommended when you're pregnant.
$endgroup$
– rubenvb
7 hours ago










4 Answers
4






active

oldest

votes


















68












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You are pre-supposing that the event C "you are conceived" occurs in the same place or on the same physical particle (your mother M) as the event B "you are born." But say the laws of nature were that you are conceived in Brooklyn at the same instant in which (in some frame) M is in Manhattan. Then there would be a frame in which B happens before C.
    $endgroup$
    – Mark Fischler
    yesterday






  • 1




    $begingroup$
    Xe isn't pre-supposing it, M. Fischler. That these are not separated in space is specified in the question.
    $endgroup$
    – JdeBP
    23 hours ago






  • 19




    $begingroup$
    @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
    14 hours ago






  • 13




    $begingroup$
    @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
    12 hours ago






  • 2




    $begingroup$
    @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
    11 hours ago



















21












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 6




    $begingroup$
    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
    yesterday










  • $begingroup$
    @Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)?
    $endgroup$
    – Fax
    12 hours ago








  • 2




    $begingroup$
    @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
    11 hours ago










  • $begingroup$
    @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
    9 hours ago










  • $begingroup$
    @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
    2 hours ago



















11












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$












      protected by David Z yesterday



      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?














      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      68












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        You are pre-supposing that the event C "you are conceived" occurs in the same place or on the same physical particle (your mother M) as the event B "you are born." But say the laws of nature were that you are conceived in Brooklyn at the same instant in which (in some frame) M is in Manhattan. Then there would be a frame in which B happens before C.
        $endgroup$
        – Mark Fischler
        yesterday






      • 1




        $begingroup$
        Xe isn't pre-supposing it, M. Fischler. That these are not separated in space is specified in the question.
        $endgroup$
        – JdeBP
        23 hours ago






      • 19




        $begingroup$
        @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
        14 hours ago






      • 13




        $begingroup$
        @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
        12 hours ago






      • 2




        $begingroup$
        @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
        11 hours ago
















      68












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        You are pre-supposing that the event C "you are conceived" occurs in the same place or on the same physical particle (your mother M) as the event B "you are born." But say the laws of nature were that you are conceived in Brooklyn at the same instant in which (in some frame) M is in Manhattan. Then there would be a frame in which B happens before C.
        $endgroup$
        – Mark Fischler
        yesterday






      • 1




        $begingroup$
        Xe isn't pre-supposing it, M. Fischler. That these are not separated in space is specified in the question.
        $endgroup$
        – JdeBP
        23 hours ago






      • 19




        $begingroup$
        @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
        14 hours ago






      • 13




        $begingroup$
        @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
        12 hours ago






      • 2




        $begingroup$
        @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
        11 hours ago














      68












      68








      68





      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$



      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.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited yesterday

























      answered yesterday









      John RennieJohn Rennie

      276k44551796




      276k44551796








      • 1




        $begingroup$
        You are pre-supposing that the event C "you are conceived" occurs in the same place or on the same physical particle (your mother M) as the event B "you are born." But say the laws of nature were that you are conceived in Brooklyn at the same instant in which (in some frame) M is in Manhattan. Then there would be a frame in which B happens before C.
        $endgroup$
        – Mark Fischler
        yesterday






      • 1




        $begingroup$
        Xe isn't pre-supposing it, M. Fischler. That these are not separated in space is specified in the question.
        $endgroup$
        – JdeBP
        23 hours ago






      • 19




        $begingroup$
        @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
        14 hours ago






      • 13




        $begingroup$
        @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
        12 hours ago






      • 2




        $begingroup$
        @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
        11 hours ago














      • 1




        $begingroup$
        You are pre-supposing that the event C "you are conceived" occurs in the same place or on the same physical particle (your mother M) as the event B "you are born." But say the laws of nature were that you are conceived in Brooklyn at the same instant in which (in some frame) M is in Manhattan. Then there would be a frame in which B happens before C.
        $endgroup$
        – Mark Fischler
        yesterday






      • 1




        $begingroup$
        Xe isn't pre-supposing it, M. Fischler. That these are not separated in space is specified in the question.
        $endgroup$
        – JdeBP
        23 hours ago






      • 19




        $begingroup$
        @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
        14 hours ago






      • 13




        $begingroup$
        @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
        12 hours ago






      • 2




        $begingroup$
        @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
        11 hours ago








      1




      1




      $begingroup$
      You are pre-supposing that the event C "you are conceived" occurs in the same place or on the same physical particle (your mother M) as the event B "you are born." But say the laws of nature were that you are conceived in Brooklyn at the same instant in which (in some frame) M is in Manhattan. Then there would be a frame in which B happens before C.
      $endgroup$
      – Mark Fischler
      yesterday




      $begingroup$
      You are pre-supposing that the event C "you are conceived" occurs in the same place or on the same physical particle (your mother M) as the event B "you are born." But say the laws of nature were that you are conceived in Brooklyn at the same instant in which (in some frame) M is in Manhattan. Then there would be a frame in which B happens before C.
      $endgroup$
      – Mark Fischler
      yesterday




      1




      1




      $begingroup$
      Xe isn't pre-supposing it, M. Fischler. That these are not separated in space is specified in the question.
      $endgroup$
      – JdeBP
      23 hours ago




      $begingroup$
      Xe isn't pre-supposing it, M. Fischler. That these are not separated in space is specified in the question.
      $endgroup$
      – JdeBP
      23 hours ago




      19




      19




      $begingroup$
      @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
      14 hours ago




      $begingroup$
      @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
      14 hours ago




      13




      13




      $begingroup$
      @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
      12 hours ago




      $begingroup$
      @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
      12 hours ago




      2




      2




      $begingroup$
      @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
      11 hours ago




      $begingroup$
      @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
      11 hours ago











      21












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$









      • 6




        $begingroup$
        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
        yesterday










      • $begingroup$
        @Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)?
        $endgroup$
        – Fax
        12 hours ago








      • 2




        $begingroup$
        @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
        11 hours ago










      • $begingroup$
        @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
        9 hours ago










      • $begingroup$
        @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
        2 hours ago
















      21












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$









      • 6




        $begingroup$
        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
        yesterday










      • $begingroup$
        @Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)?
        $endgroup$
        – Fax
        12 hours ago








      • 2




        $begingroup$
        @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
        11 hours ago










      • $begingroup$
        @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
        9 hours ago










      • $begingroup$
        @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
        2 hours ago














      21












      21








      21





      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$



      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.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited yesterday

























      answered yesterday









      PiKindOfGuyPiKindOfGuy

      613520




      613520








      • 6




        $begingroup$
        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
        yesterday










      • $begingroup$
        @Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)?
        $endgroup$
        – Fax
        12 hours ago








      • 2




        $begingroup$
        @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
        11 hours ago










      • $begingroup$
        @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
        9 hours ago










      • $begingroup$
        @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
        2 hours ago














      • 6




        $begingroup$
        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
        yesterday










      • $begingroup$
        @Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)?
        $endgroup$
        – Fax
        12 hours ago








      • 2




        $begingroup$
        @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
        11 hours ago










      • $begingroup$
        @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
        9 hours ago










      • $begingroup$
        @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
        2 hours ago








      6




      6




      $begingroup$
      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
      yesterday




      $begingroup$
      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
      yesterday












      $begingroup$
      @Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)?
      $endgroup$
      – Fax
      12 hours ago






      $begingroup$
      @Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)?
      $endgroup$
      – Fax
      12 hours ago






      2




      2




      $begingroup$
      @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
      11 hours ago




      $begingroup$
      @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
      11 hours ago












      $begingroup$
      @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
      9 hours ago




      $begingroup$
      @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
      9 hours ago












      $begingroup$
      @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
      2 hours ago




      $begingroup$
      @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
      2 hours ago











      11












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$


















        11












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          11












          11








          11





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          GilbertGilbert

          5,080818




          5,080818























              6












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$


















                6












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  AcccumulationAcccumulation

                  2,636312




                  2,636312

















                      protected by David Z yesterday



                      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?



                      Popular posts from this blog

                      El tren de la libertad Índice Antecedentes "Porque yo decido" Desarrollo de la...

                      Castillo d'Acher Características Menú de navegación

                      Connecting two nodes from the same mother node horizontallyTikZ: What EXACTLY does the the |- notation for...