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Minimizing proper time

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1

$begingroup$

I've started studying general relativity course and now I have a question about proper time. Consider functional
$$S[x]=-int_A^B ds,$$
where $A$, $B$ are fixed points of the space-time and $ds^2=dt^2-dx^2$ (let our space-time be two-dimensional without loss of generality). Finding its minimum is equivalent to maximizing proper time $s=int_A^B ds$ and it a well-known fact that maximizers of proper time are straight lines, as it can be easily checked by writing Lagrange equation for $mathcalL(t,x,dotx)=-sqrt1-dotx^2$. So my question is: what are minimizers of proper time?

I've heard that there are, hm, lots of them, but I can't write down any. If one wants to minimize $s$, he takes lagrangian $mathcalL=sqrt1-dotx^2$ and he obtains that
$$ddotx=0,$$
so minimizers necessarily are straight lines. But we have already proved that they are maximizers, so I come to contradiction. Where am I wrong?

general-relativity special-relativity lagrangian-formalism variational-principle geodesics

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edited 8 hours ago

Qmechanic♦

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asked 9 hours ago

igor.tsutsurupaigor.tsutsurupa

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add a comment | 

1

$begingroup$

I've started studying general relativity course and now I have a question about proper time. Consider functional
$$S[x]=-int_A^B ds,$$
where $A$, $B$ are fixed points of the space-time and $ds^2=dt^2-dx^2$ (let our space-time be two-dimensional without loss of generality). Finding its minimum is equivalent to maximizing proper time $s=int_A^B ds$ and it a well-known fact that maximizers of proper time are straight lines, as it can be easily checked by writing Lagrange equation for $mathcalL(t,x,dotx)=-sqrt1-dotx^2$. So my question is: what are minimizers of proper time?

I've heard that there are, hm, lots of them, but I can't write down any. If one wants to minimize $s$, he takes lagrangian $mathcalL=sqrt1-dotx^2$ and he obtains that
$$ddotx=0,$$
so minimizers necessarily are straight lines. But we have already proved that they are maximizers, so I come to contradiction. Where am I wrong?

general-relativity special-relativity lagrangian-formalism variational-principle geodesics

share|cite|improve this question

edited 8 hours ago

Qmechanic♦

114k1313 gold badges224224 silver badges13461346 bronze badges

asked 9 hours ago

igor.tsutsurupaigor.tsutsurupa

622 bronze badges

New contributor

igor.tsutsurupa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I've started studying general relativity course and now I have a question about proper time. Consider functional
$$S[x]=-int_A^B ds,$$
where $A$, $B$ are fixed points of the space-time and $ds^2=dt^2-dx^2$ (let our space-time be two-dimensional without loss of generality). Finding its minimum is equivalent to maximizing proper time $s=int_A^B ds$ and it a well-known fact that maximizers of proper time are straight lines, as it can be easily checked by writing Lagrange equation for $mathcalL(t,x,dotx)=-sqrt1-dotx^2$. So my question is: what are minimizers of proper time?

I've heard that there are, hm, lots of them, but I can't write down any. If one wants to minimize $s$, he takes lagrangian $mathcalL=sqrt1-dotx^2$ and he obtains that
$$ddotx=0,$$
so minimizers necessarily are straight lines. But we have already proved that they are maximizers, so I come to contradiction. Where am I wrong?

general-relativity special-relativity lagrangian-formalism variational-principle geodesics

share|cite|improve this question

edited 8 hours ago

Qmechanic♦

114k1313 gold badges224224 silver badges13461346 bronze badges

asked 9 hours ago

igor.tsutsurupaigor.tsutsurupa

622 bronze badges

New contributor

igor.tsutsurupa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 8 hours ago

Qmechanic♦

114k1313 gold badges224224 silver badges13461346 bronze badges

asked 9 hours ago

igor.tsutsurupaigor.tsutsurupa

622 bronze badges

New contributor

igor.tsutsurupa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 8 hours ago

Qmechanic♦

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asked 9 hours ago

igor.tsutsurupaigor.tsutsurupa

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igor.tsutsurupa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 8 hours ago

Qmechanic♦

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edited 8 hours ago

Qmechanic♦

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asked 9 hours ago

igor.tsutsurupaigor.tsutsurupa

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igor.tsutsurupa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 9 hours ago

igor.tsutsurupaigor.tsutsurupa

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1 Answer
1

active

oldest

votes

5

$begingroup$

1. Given two (timelike separated) points in Minkowski space, there is a unique timelike curve that maximizes the proper time, namely the straight line, as OP already mentions. However there is no timelike curve that minimizes the proper time. Nevertheless proper time does have an infimum, namely zero. This is essentially because a massive point particle can always fly a bit closer to the speed of light without reaching it.



2. 
   

   For a general action functional, the Euler-Lagrange (EL) equations yield stationary configurations. Extremal configurations might not exist.

   
   
   

   Example. For a differentiable function $f:Ito mathbbR$ on an (open or closed) interval $I$, recall that a stationary point is neither a necessary nor a sufficient condition for an extremum for $f$. Similar statements are true in calculus of variations.

   
   

share|cite|improve this answer

edited 3 hours ago

answered 8 hours ago

Qmechanic♦Qmechanic

114k1313 gold badges224224 silver badges13461346 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  1. If there is no minimizer, is there a sequence of curves that converges to minimum? 2. It states that "If I am local extremum of the functional, then I am a solution to Euler-Lagrange equation", the same statement holds for a differentiable function of one variable.
  $endgroup$
  – igor.tsutsurupa
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I updated the answer.
  $endgroup$
  – Qmechanic♦
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  well, again: if a function $fcolon ItomathbbR$ is differentiable and $xin I$ is its extremum point, then it is stationary point. It is known as Fermat theorem. For Euler-Lagrange equation, if $x$ is local extremum of an action $int_t_A^t_BmathcalL(t,x,dotx)dt$, THEN it is a solution of EL-equation, that's the point. I use exactly this statement in my reasoning.
  $endgroup$
  – igor.tsutsurupa
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not necessarily if $x$ is an endpoint of $I$.
  $endgroup$
  – Qmechanic♦
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok, let's talk about open $I$. Anyway, the question concerns to the statement about EL-equation
  $endgroup$
  – igor.tsutsurupa
  4 hours ago
  

  
  
  
  

 | 
show 3 more comments

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5

$begingroup$

1. Given two (timelike separated) points in Minkowski space, there is a unique timelike curve that maximizes the proper time, namely the straight line, as OP already mentions. However there is no timelike curve that minimizes the proper time. Nevertheless proper time does have an infimum, namely zero. This is essentially because a massive point particle can always fly a bit closer to the speed of light without reaching it.



2. 
   

   For a general action functional, the Euler-Lagrange (EL) equations yield stationary configurations. Extremal configurations might not exist.

   
   
   

   Example. For a differentiable function $f:Ito mathbbR$ on an (open or closed) interval $I$, recall that a stationary point is neither a necessary nor a sufficient condition for an extremum for $f$. Similar statements are true in calculus of variations.

   
   

share|cite|improve this answer

edited 3 hours ago

answered 8 hours ago

Qmechanic♦Qmechanic

114k1313 gold badges224224 silver badges13461346 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  1. If there is no minimizer, is there a sequence of curves that converges to minimum? 2. It states that "If I am local extremum of the functional, then I am a solution to Euler-Lagrange equation", the same statement holds for a differentiable function of one variable.
  $endgroup$
  – igor.tsutsurupa
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I updated the answer.
  $endgroup$
  – Qmechanic♦
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  well, again: if a function $fcolon ItomathbbR$ is differentiable and $xin I$ is its extremum point, then it is stationary point. It is known as Fermat theorem. For Euler-Lagrange equation, if $x$ is local extremum of an action $int_t_A^t_BmathcalL(t,x,dotx)dt$, THEN it is a solution of EL-equation, that's the point. I use exactly this statement in my reasoning.
  $endgroup$
  – igor.tsutsurupa
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not necessarily if $x$ is an endpoint of $I$.
  $endgroup$
  – Qmechanic♦
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok, let's talk about open $I$. Anyway, the question concerns to the statement about EL-equation
  $endgroup$
  – igor.tsutsurupa
  4 hours ago
  

  
  
  
  

 | 
show 3 more comments

5

$begingroup$

1. Given two (timelike separated) points in Minkowski space, there is a unique timelike curve that maximizes the proper time, namely the straight line, as OP already mentions. However there is no timelike curve that minimizes the proper time. Nevertheless proper time does have an infimum, namely zero. This is essentially because a massive point particle can always fly a bit closer to the speed of light without reaching it.



2. 
   

   For a general action functional, the Euler-Lagrange (EL) equations yield stationary configurations. Extremal configurations might not exist.

   
   
   

   Example. For a differentiable function $f:Ito mathbbR$ on an (open or closed) interval $I$, recall that a stationary point is neither a necessary nor a sufficient condition for an extremum for $f$. Similar statements are true in calculus of variations.

   
   

share|cite|improve this answer

edited 3 hours ago

answered 8 hours ago

Qmechanic♦Qmechanic

114k1313 gold badges224224 silver badges13461346 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  1. If there is no minimizer, is there a sequence of curves that converges to minimum? 2. It states that "If I am local extremum of the functional, then I am a solution to Euler-Lagrange equation", the same statement holds for a differentiable function of one variable.
  $endgroup$
  – igor.tsutsurupa
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I updated the answer.
  $endgroup$
  – Qmechanic♦
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  well, again: if a function $fcolon ItomathbbR$ is differentiable and $xin I$ is its extremum point, then it is stationary point. It is known as Fermat theorem. For Euler-Lagrange equation, if $x$ is local extremum of an action $int_t_A^t_BmathcalL(t,x,dotx)dt$, THEN it is a solution of EL-equation, that's the point. I use exactly this statement in my reasoning.
  $endgroup$
  – igor.tsutsurupa
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not necessarily if $x$ is an endpoint of $I$.
  $endgroup$
  – Qmechanic♦
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok, let's talk about open $I$. Anyway, the question concerns to the statement about EL-equation
  $endgroup$
  – igor.tsutsurupa
  4 hours ago
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

1. Given two (timelike separated) points in Minkowski space, there is a unique timelike curve that maximizes the proper time, namely the straight line, as OP already mentions. However there is no timelike curve that minimizes the proper time. Nevertheless proper time does have an infimum, namely zero. This is essentially because a massive point particle can always fly a bit closer to the speed of light without reaching it.



2. 
   

   For a general action functional, the Euler-Lagrange (EL) equations yield stationary configurations. Extremal configurations might not exist.

   
   
   

   Example. For a differentiable function $f:Ito mathbbR$ on an (open or closed) interval $I$, recall that a stationary point is neither a necessary nor a sufficient condition for an extremum for $f$. Similar statements are true in calculus of variations.

   
   

share|cite|improve this answer

edited 3 hours ago

answered 8 hours ago

Qmechanic♦Qmechanic

114k1313 gold badges224224 silver badges13461346 bronze badges

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 8 hours ago

Qmechanic♦Qmechanic

114k1313 gold badges224224 silver badges13461346 bronze badges

answered 8 hours ago

Qmechanic♦Qmechanic

114k1313 gold badges224224 silver badges13461346 bronze badges

answered 8 hours ago

Qmechanic♦Qmechanic

114k1313 gold badges224224 silver badges13461346 bronze badges