Is there a frame of reference in which I was born before I was conceived?
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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?
special-relativity spacetime inertial-frames observers causality
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add a comment |
$begingroup$
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?
special-relativity spacetime inertial-frames observers causality
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3
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"a frame of reference...should exist" --- have you attempted to write down thst frame?
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– WillO
18 hours ago
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Probably an accelerated frame of reference...
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– Rob Jeffries
18 hours ago
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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.
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– David Z♦
15 hours ago
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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?
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– nick012000
7 hours ago
add a comment |
$begingroup$
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?
special-relativity spacetime inertial-frames observers causality
$endgroup$
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
special-relativity spacetime inertial-frames observers causality
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
add a comment |
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
add a comment |
4 Answers
4
active
oldest
votes
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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.
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1
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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.
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– Mark Fischler
13 hours ago
1
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Xe isn't pre-supposing it, M. Fischler. That these are not separated in space is specified in the question.
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– JdeBP
11 hours ago
2
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@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.
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– John Rennie
3 hours ago
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Test tube babies?
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– Mars
3 hours ago
2
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@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.
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– Oscar Bravo
1 hour ago
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show 2 more comments
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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.
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5
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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.
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– Mark
15 hours ago
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@Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)?
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– Fax
1 hour ago
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@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.
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– DonFusili
4 mins ago
add a comment |
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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.
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add a comment |
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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.
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add a comment |
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4 Answers
4
active
oldest
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4 Answers
4
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
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.
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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.
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– 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.
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– JdeBP
11 hours ago
2
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@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.
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– 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.
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– Oscar Bravo
1 hour ago
|
show 2 more comments
$begingroup$
Suppose we take the spacetime point of your conception as the origin, $(t=0, x=0)$, then the spacetime point for your birth would be $(t=T, x=uT)$. The time $T$ is approximately $9$ months, and we are writing the spatial position of your birth as $x=uT$ where $u$ is a velocity. The velocity $u$ can be any value from zero (i.e. born in the same spot as conception) up to $c$ (because your mother can't move faster than light).
Now we'll use the Lorentz transformations to find out how these events appear for an observer moving at a speed $v$ relative to you. The transformations are:
$$ t' = gamma left( t - frac{vx}{c^2} right ) $$
$$ x' = gamma left( x - vt right) $$
though actually we'll only be using the first equation as we're only interested in the time. Putting $(0,0)$ into the equation for $t'$ gives us $t'=0$ so the clocks of the observer and your mother both read zero at the moment of your conception. Now feeding the position of your birth $(T,uT)$ into the equation for $t'$ we get:
$$ t' = gamma left( T - frac{vuT}{c^2} right ) $$
For you to be born before you were conceived we need $t'lt 0$ and that gives us:
$$ T lt frac{vuT}{c^2} $$
or:
$$ vu gt c^2 $$
We know that the observer's velocity $v$ cannot be greater than $c$, and your mother's velocity $u$ cannot be greater than $c$, so this inequality can never be satisfied. That is, there is no frame in which you were born before you were conceived.
The rule is that two events that are timelike separated, i.e. their separation in space is less than their separation in time times $c$, can never change order. All observers will agree on which event was first. For the order to change the events have to be spacelike separated. In this case this would mean $uT gt cT$ i.e. your mother would have to have moved at a speed $u$ faster than light between your conception and birth.
$endgroup$
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
|
show 2 more comments
$begingroup$
Suppose we take the spacetime point of your conception as the origin, $(t=0, x=0)$, then the spacetime point for your birth would be $(t=T, x=uT)$. The time $T$ is approximately $9$ months, and we are writing the spatial position of your birth as $x=uT$ where $u$ is a velocity. The velocity $u$ can be any value from zero (i.e. born in the same spot as conception) up to $c$ (because your mother can't move faster than light).
Now we'll use the Lorentz transformations to find out how these events appear for an observer moving at a speed $v$ relative to you. The transformations are:
$$ t' = gamma left( t - frac{vx}{c^2} right ) $$
$$ x' = gamma left( x - vt right) $$
though actually we'll only be using the first equation as we're only interested in the time. Putting $(0,0)$ into the equation for $t'$ gives us $t'=0$ so the clocks of the observer and your mother both read zero at the moment of your conception. Now feeding the position of your birth $(T,uT)$ into the equation for $t'$ we get:
$$ t' = gamma left( T - frac{vuT}{c^2} right ) $$
For you to be born before you were conceived we need $t'lt 0$ and that gives us:
$$ T lt frac{vuT}{c^2} $$
or:
$$ vu gt c^2 $$
We know that the observer's velocity $v$ cannot be greater than $c$, and your mother's velocity $u$ cannot be greater than $c$, so this inequality can never be satisfied. That is, there is no frame in which you were born before you were conceived.
The rule is that two events that are timelike separated, i.e. their separation in space is less than their separation in time times $c$, can never change order. All observers will agree on which event was first. For the order to change the events have to be spacelike separated. In this case this would mean $uT gt cT$ i.e. your mother would have to have moved at a speed $u$ faster than light between your conception and birth.
$endgroup$
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.
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.
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– 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
|
show 2 more comments
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
|
show 2 more comments
$begingroup$
Simultaneity is relative, but causality is always preserved in moving from one reference frame to another. In no reference frame can you be born prior to being conceived or be born at the same time that you're conceived.
$endgroup$
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
add a comment |
$begingroup$
Simultaneity is relative, but causality is always preserved in moving from one reference frame to another. In no reference frame can you be born prior to being conceived or be born at the same time that you're conceived.
$endgroup$
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
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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.
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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.
edited 18 hours ago
answered 19 hours ago
PiKindOfGuyPiKindOfGuy
601519
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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.
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– Mark
15 hours ago
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@Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)?
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– Fax
1 hour ago
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@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.
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– DonFusili
4 mins ago
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5
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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.
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– Mark
15 hours ago
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@Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)?
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– Fax
1 hour ago
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@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.
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– DonFusili
4 mins ago
5
5
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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.
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– Mark
15 hours ago
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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.
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– Mark
15 hours ago
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@Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)?
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– Fax
1 hour ago
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@Mark does "sufficently faster" mean faster than $c$, or faster than $c^2$ (as John Rennie's answer appears to suggest)?
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– Fax
1 hour ago
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@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.
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– DonFusili
4 mins ago
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@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.
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– DonFusili
4 mins ago
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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.
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add a comment |
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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.
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add a comment |
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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.
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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.
answered 19 hours ago
GilbertGilbert
5,070818
5,070818
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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.
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add a comment |
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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.
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add a comment |
$begingroup$
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.
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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.
answered 18 hours ago
AcccumulationAcccumulation
2,626312
2,626312
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protected by David Z♦ 14 hours ago
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?
3
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"a frame of reference...should exist" --- have you attempted to write down thst frame?
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– WillO
18 hours ago
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Probably an accelerated frame of reference...
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– Rob Jeffries
18 hours ago
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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.
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– David Z♦
15 hours ago
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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?
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– nick012000
7 hours ago