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












15












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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?










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$endgroup$








  • 3




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










  • $begingroup$
    Probably an accelerated frame of reference...
    $endgroup$
    – Rob Jeffries
    18 hours ago










  • $begingroup$
    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
    15 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
    7 hours ago


















15












$begingroup$


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











$endgroup$








  • 3




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










  • $begingroup$
    Probably an accelerated frame of reference...
    $endgroup$
    – Rob Jeffries
    18 hours ago










  • $begingroup$
    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
    15 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
    7 hours ago
















15












15








15


4



$begingroup$


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 17 hours ago









SmarthBansal

655423




655423










asked 19 hours ago









IchVerlorenIchVerloren

153112




153112








  • 3




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










  • $begingroup$
    Probably an accelerated frame of reference...
    $endgroup$
    – Rob Jeffries
    18 hours ago










  • $begingroup$
    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
    15 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
    7 hours ago
















  • 3




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










  • $begingroup$
    Probably an accelerated frame of reference...
    $endgroup$
    – Rob Jeffries
    18 hours ago










  • $begingroup$
    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
    15 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
    7 hours ago










3




3




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




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












$begingroup$
Probably an accelerated frame of reference...
$endgroup$
– Rob Jeffries
18 hours ago




$begingroup$
Probably an accelerated frame of reference...
$endgroup$
– Rob Jeffries
18 hours ago












$begingroup$
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
15 hours ago




$begingroup$
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
15 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
7 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
7 hours ago












4 Answers
4






active

oldest

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47












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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
    13 hours ago






  • 1




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






  • 2




    $begingroup$
    @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
    3 hours ago










  • $begingroup$
    Test tube babies?
    $endgroup$
    – Mars
    3 hours ago






  • 2




    $begingroup$
    @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
    1 hour ago



















20












$begingroup$

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











$endgroup$









  • 5




    $begingroup$
    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
    15 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
    1 hour 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
    4 mins ago



















10












$begingroup$

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









$endgroup$





















    5












    $begingroup$

    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 14 hours ago



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      4 Answers
      4






      active

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      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      47












      $begingroup$

      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
        13 hours ago






      • 1




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






      • 2




        $begingroup$
        @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
        3 hours ago










      • $begingroup$
        Test tube babies?
        $endgroup$
        – Mars
        3 hours ago






      • 2




        $begingroup$
        @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
        1 hour ago
















      47












      $begingroup$

      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
        13 hours ago






      • 1




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






      • 2




        $begingroup$
        @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
        3 hours ago










      • $begingroup$
        Test tube babies?
        $endgroup$
        – Mars
        3 hours ago






      • 2




        $begingroup$
        @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
        1 hour ago














      47












      47








      47





      $begingroup$

      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 17 hours ago

























      answered 19 hours ago









      John RennieJohn Rennie

      276k44551795




      276k44551795








      • 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
        13 hours ago






      • 1




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






      • 2




        $begingroup$
        @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
        3 hours ago










      • $begingroup$
        Test tube babies?
        $endgroup$
        – Mars
        3 hours ago






      • 2




        $begingroup$
        @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
        1 hour 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
        13 hours ago






      • 1




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






      • 2




        $begingroup$
        @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
        3 hours ago










      • $begingroup$
        Test tube babies?
        $endgroup$
        – Mars
        3 hours ago






      • 2




        $begingroup$
        @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
        1 hour 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
      13 hours ago




      $begingroup$
      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
      13 hours ago




      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
      11 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
      11 hours ago




      2




      2




      $begingroup$
      @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
      3 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
      3 hours ago












      $begingroup$
      Test tube babies?
      $endgroup$
      – Mars
      3 hours ago




      $begingroup$
      Test tube babies?
      $endgroup$
      – Mars
      3 hours ago




      2




      2




      $begingroup$
      @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
      1 hour 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
      1 hour ago











      20












      $begingroup$

      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











      $endgroup$









      • 5




        $begingroup$
        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
        15 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
        1 hour 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
        4 mins ago
















      20












      $begingroup$

      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











      $endgroup$









      • 5




        $begingroup$
        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
        15 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
        1 hour 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
        4 mins ago














      20












      20








      20





      $begingroup$

      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 18 hours ago

























      answered 19 hours ago









      PiKindOfGuyPiKindOfGuy

      601519




      601519








      • 5




        $begingroup$
        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
        15 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
        1 hour 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
        4 mins ago














      • 5




        $begingroup$
        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
        15 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
        1 hour 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
        4 mins ago








      5




      5




      $begingroup$
      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
      15 hours ago




      $begingroup$
      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
      15 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
      1 hour 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
      1 hour 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
      4 mins 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
      4 mins ago











      10












      $begingroup$

      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









      $endgroup$


















        10












        $begingroup$

        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









        $endgroup$
















          10












          10








          10





          $begingroup$

          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 19 hours ago









          GilbertGilbert

          5,070818




          5,070818























              5












              $begingroup$

              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









              $endgroup$


















                5












                $begingroup$

                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









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  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 18 hours ago









                  AcccumulationAcccumulation

                  2,626312




                  2,626312

















                      protected by David Z 14 hours ago



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