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Definition of observer and time measured by different observers in general relativity
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$begingroup$
An observer in general relativity is defined as a future directed timelike worldline
begin{align*}
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
end{align*}
together with an orthonormal basis $e_a(lambda) in T_{gamma(lambda)}M$ where $e_0(lambda)= v_{gamma, gamma(lambda)}$ and
begin{align}
g_{gamma(lambda)}(e_a(lambda),e_b(lambda))=eta_{ab}~. qquad (1)
end{align}
Here, $v_{gamma, gamma(lambda)}$ is the velocity of the worldline $gamma$ at the point $gamma(lambda)in M$ and $g$ is the metric tensor field on $M$. The time measured by the clock carried by this observer between events $lambda_0, lambda_1$ is defined as
begin{align}
tau_gamma = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)})}~.
end{align}
However,
begin{align}
g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)}) = g_{gamma(lambda)}(e_0(lambda),e_0(lambda))=1 qquad (2)
end{align}
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.
This is all standard definition. Suppose, we have another observer $delta$:
begin{align*}
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
end{align*}
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
begin{align}
tau_delta = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{delta(lambda)}(v_{delta, delta(lambda)},v_{delta, delta(lambda)})}~.
end{align}
From equations (1) and (2), we get $tau_gamma = tau_delta$ and this will be true for all observers measuring time between $lambda_0, lambda_1$.
However, I know that my conclusion is wrong. Can you point out where I went astray?
Edit: I am trying to make the situation I am referring to clearer.
$gamma(lambda)$ and $delta(lambda)$ meeting at points $p,q in M$">
(Sorry for this huge picture. I wanted to make it smaller, but couldn't figure out how to go about it.)
This picture shows the two observers $gamma$ and $delta$ defined above. Both worldlines are parameterised by the same parameter $lambda$. This need not be the case but I choose it to convey my point. I wish to determine the proper time measured by observers $gamma$ and $delta$ between events $p$ and $q$ in the spacetime manifold $M$.
begin{align}
p =& gamma(lambda_1) = delta(lambda_1) \
q =& gamma(lambda_2) = delta(lambda2)
end{align}
This scenario is possible, isn't it? I do not see why $tau_gamma$ and $tau_delta$ need to be the same. (In fact, in the twin paradox, for example, we see this explicitly.) However, from equation (1) and the deduction above, it follows that $tau_gamma = tau_delta$. This is my confusion.
Note From the definition of an observer in general relativity, the observer worldline seems to be always parameterised by its propertime. But the propertime measured by two observers between the same two events need not be the same, isn't it?
general-relativity differential-geometry observers
edited 15 hours ago
Qmechanic♦
107k121991239
asked yesterday
damaihatidamaihati
684
$endgroup$
add a comment |
$begingroup$
An observer in general relativity is defined as a future directed timelike worldline
begin{align*}
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
end{align*}
together with an orthonormal basis $e_a(lambda) in T_{gamma(lambda)}M$ where $e_0(lambda)= v_{gamma, gamma(lambda)}$ and
begin{align}
g_{gamma(lambda)}(e_a(lambda),e_b(lambda))=eta_{ab}~. qquad (1)
end{align}
Here, $v_{gamma, gamma(lambda)}$ is the velocity of the worldline $gamma$ at the point $gamma(lambda)in M$ and $g$ is the metric tensor field on $M$. The time measured by the clock carried by this observer between events $lambda_0, lambda_1$ is defined as
begin{align}
tau_gamma = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)})}~.
end{align}
However,
begin{align}
g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)}) = g_{gamma(lambda)}(e_0(lambda),e_0(lambda))=1 qquad (2)
end{align}
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.
This is all standard definition. Suppose, we have another observer $delta$:
begin{align*}
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
end{align*}
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
begin{align}
tau_delta = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{delta(lambda)}(v_{delta, delta(lambda)},v_{delta, delta(lambda)})}~.
end{align}
From equations (1) and (2), we get $tau_gamma = tau_delta$ and this will be true for all observers measuring time between $lambda_0, lambda_1$.
However, I know that my conclusion is wrong. Can you point out where I went astray?
Edit: I am trying to make the situation I am referring to clearer.
$gamma(lambda)$ and $delta(lambda)$ meeting at points $p,q in M$">
(Sorry for this huge picture. I wanted to make it smaller, but couldn't figure out how to go about it.)
This picture shows the two observers $gamma$ and $delta$ defined above. Both worldlines are parameterised by the same parameter $lambda$. This need not be the case but I choose it to convey my point. I wish to determine the proper time measured by observers $gamma$ and $delta$ between events $p$ and $q$ in the spacetime manifold $M$.
begin{align}
p =& gamma(lambda_1) = delta(lambda_1) \
q =& gamma(lambda_2) = delta(lambda2)
end{align}
This scenario is possible, isn't it? I do not see why $tau_gamma$ and $tau_delta$ need to be the same. (In fact, in the twin paradox, for example, we see this explicitly.) However, from equation (1) and the deduction above, it follows that $tau_gamma = tau_delta$. This is my confusion.
Note From the definition of an observer in general relativity, the observer worldline seems to be always parameterised by its propertime. But the propertime measured by two observers between the same two events need not be the same, isn't it?
general-relativity differential-geometry observers
edited 15 hours ago
Qmechanic♦
107k121991239
asked yesterday
damaihatidamaihati
684
$endgroup$
$begingroup$
It doesn't make sense to say that two worldlines have the same parameter. You can have the values of the parameter coincide at one crossing point if you want, but nothing guarantees that this will happen when the curves cross again (which is the only place where it makes sense to say the the proper times are the same).
$endgroup$
– Javier
16 hours ago
$begingroup$
@Javier Then what is your opinion on the twin paradox? Let us work in some chart $(U,x)$. The worldline $gamma: (0,1) to M$ such that $gamma_{(x)}^i=(lambda,0,0,0)^i$ and $delta:(0,1) to M$ such that $delta_{(x)}^i=(lambda, alpha lambda, 0,0)^i$ for $lambda leq 1/2$ while $delta_{(x)}^i=(lambda, (1-lambda)alpha, 0,0)^i$ for $lambda > 1/2$. Here, $alpha in (0,1)$. Both the worldlines lie in $U subset M$. So, these two curves have the same parameter $lambda$ but the maps $gamma$ and $delta$ are defined differently. The parameters coincide in the beginning and end.
$endgroup$
– damaihati
13 hours ago
$begingroup$
That is a perfectly good parameterization and it will let you calculate the time difference of the twin paradox, but the second curve is not parametrized by proper time. That's not a problem.
$endgroup$
– Javier
12 hours ago
add a comment |
$begingroup$
An observer in general relativity is defined as a future directed timelike worldline
begin{align*}
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
end{align*}
together with an orthonormal basis $e_a(lambda) in T_{gamma(lambda)}M$ where $e_0(lambda)= v_{gamma, gamma(lambda)}$ and
begin{align}
g_{gamma(lambda)}(e_a(lambda),e_b(lambda))=eta_{ab}~. qquad (1)
end{align}
Here, $v_{gamma, gamma(lambda)}$ is the velocity of the worldline $gamma$ at the point $gamma(lambda)in M$ and $g$ is the metric tensor field on $M$. The time measured by the clock carried by this observer between events $lambda_0, lambda_1$ is defined as
begin{align}
tau_gamma = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)})}~.
end{align}
However,
begin{align}
g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)}) = g_{gamma(lambda)}(e_0(lambda),e_0(lambda))=1 qquad (2)
end{align}
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.
This is all standard definition. Suppose, we have another observer $delta$:
begin{align*}
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
end{align*}
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
begin{align}
tau_delta = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{delta(lambda)}(v_{delta, delta(lambda)},v_{delta, delta(lambda)})}~.
end{align}
From equations (1) and (2), we get $tau_gamma = tau_delta$ and this will be true for all observers measuring time between $lambda_0, lambda_1$.
However, I know that my conclusion is wrong. Can you point out where I went astray?
Edit: I am trying to make the situation I am referring to clearer.
$gamma(lambda)$ and $delta(lambda)$ meeting at points $p,q in M$">
(Sorry for this huge picture. I wanted to make it smaller, but couldn't figure out how to go about it.)
This picture shows the two observers $gamma$ and $delta$ defined above. Both worldlines are parameterised by the same parameter $lambda$. This need not be the case but I choose it to convey my point. I wish to determine the proper time measured by observers $gamma$ and $delta$ between events $p$ and $q$ in the spacetime manifold $M$.
begin{align}
p =& gamma(lambda_1) = delta(lambda_1) \
q =& gamma(lambda_2) = delta(lambda2)
end{align}
This scenario is possible, isn't it? I do not see why $tau_gamma$ and $tau_delta$ need to be the same. (In fact, in the twin paradox, for example, we see this explicitly.) However, from equation (1) and the deduction above, it follows that $tau_gamma = tau_delta$. This is my confusion.
Note From the definition of an observer in general relativity, the observer worldline seems to be always parameterised by its propertime. But the propertime measured by two observers between the same two events need not be the same, isn't it?
general-relativity differential-geometry observers
edited 15 hours ago
Qmechanic♦
107k121991239
asked yesterday
damaihatidamaihati
684
$endgroup$
An observer in general relativity is defined as a future directed timelike worldline
begin{align*}
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
end{align*}
together with an orthonormal basis $e_a(lambda) in T_{gamma(lambda)}M$ where $e_0(lambda)= v_{gamma, gamma(lambda)}$ and
begin{align}
g_{gamma(lambda)}(e_a(lambda),e_b(lambda))=eta_{ab}~. qquad (1)
end{align}
Here, $v_{gamma, gamma(lambda)}$ is the velocity of the worldline $gamma$ at the point $gamma(lambda)in M$ and $g$ is the metric tensor field on $M$. The time measured by the clock carried by this observer between events $lambda_0, lambda_1$ is defined as
begin{align}
tau_gamma = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)})}~.
end{align}
However,
begin{align}
g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)}) = g_{gamma(lambda)}(e_0(lambda),e_0(lambda))=1 qquad (2)
end{align}
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.
This is all standard definition. Suppose, we have another observer $delta$:
begin{align*}
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
end{align*}
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
begin{align}
tau_delta = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{delta(lambda)}(v_{delta, delta(lambda)},v_{delta, delta(lambda)})}~.
end{align}
From equations (1) and (2), we get $tau_gamma = tau_delta$ and this will be true for all observers measuring time between $lambda_0, lambda_1$.
However, I know that my conclusion is wrong. Can you point out where I went astray?
Edit: I am trying to make the situation I am referring to clearer.
$gamma(lambda)$ and $delta(lambda)$ meeting at points $p,q in M$">
(Sorry for this huge picture. I wanted to make it smaller, but couldn't figure out how to go about it.)
This picture shows the two observers $gamma$ and $delta$ defined above. Both worldlines are parameterised by the same parameter $lambda$. This need not be the case but I choose it to convey my point. I wish to determine the proper time measured by observers $gamma$ and $delta$ between events $p$ and $q$ in the spacetime manifold $M$.
begin{align}
p =& gamma(lambda_1) = delta(lambda_1) \
q =& gamma(lambda_2) = delta(lambda2)
end{align}
This scenario is possible, isn't it? I do not see why $tau_gamma$ and $tau_delta$ need to be the same. (In fact, in the twin paradox, for example, we see this explicitly.) However, from equation (1) and the deduction above, it follows that $tau_gamma = tau_delta$. This is my confusion.
Note From the definition of an observer in general relativity, the observer worldline seems to be always parameterised by its propertime. But the propertime measured by two observers between the same two events need not be the same, isn't it?
general-relativity differential-geometry observers
general-relativity differential-geometry observers
edited 15 hours ago
Qmechanic♦
107k121991239
asked yesterday
damaihatidamaihati
684
edited 15 hours ago
Qmechanic♦
107k121991239
asked yesterday
damaihatidamaihati
684
edited 15 hours ago
Qmechanic♦
107k121991239
edited 15 hours ago
Qmechanic♦
107k121991239
edited 15 hours ago
Qmechanic♦
107k121991239
107k121991239
asked yesterday
damaihatidamaihati
684
asked yesterday
damaihatidamaihati
684
asked yesterday
damaihatidamaihati
684
684
$begingroup$
It doesn't make sense to say that two worldlines have the same parameter. You can have the values of the parameter coincide at one crossing point if you want, but nothing guarantees that this will happen when the curves cross again (which is the only place where it makes sense to say the the proper times are the same).
$endgroup$
– Javier
16 hours ago
$begingroup$
@Javier Then what is your opinion on the twin paradox? Let us work in some chart $(U,x)$. The worldline $gamma: (0,1) to M$ such that $gamma_{(x)}^i=(lambda,0,0,0)^i$ and $delta:(0,1) to M$ such that $delta_{(x)}^i=(lambda, alpha lambda, 0,0)^i$ for $lambda leq 1/2$ while $delta_{(x)}^i=(lambda, (1-lambda)alpha, 0,0)^i$ for $lambda > 1/2$. Here, $alpha in (0,1)$. Both the worldlines lie in $U subset M$. So, these two curves have the same parameter $lambda$ but the maps $gamma$ and $delta$ are defined differently. The parameters coincide in the beginning and end.
$endgroup$
– damaihati
13 hours ago
$begingroup$
That is a perfectly good parameterization and it will let you calculate the time difference of the twin paradox, but the second curve is not parametrized by proper time. That's not a problem.
$endgroup$
– Javier
12 hours ago
add a comment |
$begingroup$
It doesn't make sense to say that two worldlines have the same parameter. You can have the values of the parameter coincide at one crossing point if you want, but nothing guarantees that this will happen when the curves cross again (which is the only place where it makes sense to say the the proper times are the same).
$endgroup$
– Javier
16 hours ago
$begingroup$
@Javier Then what is your opinion on the twin paradox? Let us work in some chart $(U,x)$. The worldline $gamma: (0,1) to M$ such that $gamma_{(x)}^i=(lambda,0,0,0)^i$ and $delta:(0,1) to M$ such that $delta_{(x)}^i=(lambda, alpha lambda, 0,0)^i$ for $lambda leq 1/2$ while $delta_{(x)}^i=(lambda, (1-lambda)alpha, 0,0)^i$ for $lambda > 1/2$. Here, $alpha in (0,1)$. Both the worldlines lie in $U subset M$. So, these two curves have the same parameter $lambda$ but the maps $gamma$ and $delta$ are defined differently. The parameters coincide in the beginning and end.
$endgroup$
– damaihati
13 hours ago
$begingroup$
That is a perfectly good parameterization and it will let you calculate the time difference of the twin paradox, but the second curve is not parametrized by proper time. That's not a problem.
$endgroup$
– Javier
12 hours ago
$begingroup$
It doesn't make sense to say that two worldlines have the same parameter. You can have the values of the parameter coincide at one crossing point if you want, but nothing guarantees that this will happen when the curves cross again (which is the only place where it makes sense to say the the proper times are the same).
$endgroup$
– Javier
16 hours ago
$begingroup$
It doesn't make sense to say that two worldlines have the same parameter. You can have the values of the parameter coincide at one crossing point if you want, but nothing guarantees that this will happen when the curves cross again (which is the only place where it makes sense to say the the proper times are the same).
$endgroup$
– Javier
16 hours ago
$begingroup$
@Javier Then what is your opinion on the twin paradox? Let us work in some chart $(U,x)$. The worldline $gamma: (0,1) to M$ such that $gamma_{(x)}^i=(lambda,0,0,0)^i$ and $delta:(0,1) to M$ such that $delta_{(x)}^i=(lambda, alpha lambda, 0,0)^i$ for $lambda leq 1/2$ while $delta_{(x)}^i=(lambda, (1-lambda)alpha, 0,0)^i$ for $lambda > 1/2$. Here, $alpha in (0,1)$. Both the worldlines lie in $U subset M$. So, these two curves have the same parameter $lambda$ but the maps $gamma$ and $delta$ are defined differently. The parameters coincide in the beginning and end.
$endgroup$
– damaihati
13 hours ago
$begingroup$
@Javier Then what is your opinion on the twin paradox? Let us work in some chart $(U,x)$. The worldline $gamma: (0,1) to M$ such that $gamma_{(x)}^i=(lambda,0,0,0)^i$ and $delta:(0,1) to M$ such that $delta_{(x)}^i=(lambda, alpha lambda, 0,0)^i$ for $lambda leq 1/2$ while $delta_{(x)}^i=(lambda, (1-lambda)alpha, 0,0)^i$ for $lambda > 1/2$. Here, $alpha in (0,1)$. Both the worldlines lie in $U subset M$. So, these two curves have the same parameter $lambda$ but the maps $gamma$ and $delta$ are defined differently. The parameters coincide in the beginning and end.
$endgroup$
– damaihati
13 hours ago
$begingroup$
That is a perfectly good parameterization and it will let you calculate the time difference of the twin paradox, but the second curve is not parametrized by proper time. That's not a problem.
$endgroup$
– Javier
12 hours ago
$begingroup$
That is a perfectly good parameterization and it will let you calculate the time difference of the twin paradox, but the second curve is not parametrized by proper time. That's not a problem.
$endgroup$
– Javier
12 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your conclusion is correct, because what you are doing by saying that $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tilde{lambda}$ of the curve $gamma$ that have $g(v_{gamma,gamma(tilde{lambda})},v_{gamma,gamma(tilde{lambda})}) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.
Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!
As concerns your edit:
When you state
$$p = gamma(lambda_1) = delta(lambda_1)$$
$$q = gamma(lambda_2) = delta(lambda_2)$$
and impose the normalisation of the tangents $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = g(v_{delta,delta(lambda)},v_{delta,delta(lambda)}) = 1$, you are NOT choosing two general curves (worldlines) passing through $p,q$. Instead, you are choosing two curves for which the same amount of proper time passes between $p,q$.
Perhaps you should consider a concrete example. Choose a space-time such as Minkowski and two randomly chosen curves $gamma, delta$ on it that have an arbitrary parametrization $gamma(tilde{lambda}_gamma), delta(tilde{lambda}_delta)$ and that meet at some events $p,q$. Now find the proper-time parametrisation $lambda$ along the entire curves $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = g(v_{delta,delta(lambda)},v_{delta,delta(lambda)}) = 1$, this can be recast as two first-order differential equations for the functions $lambda(tilde{lambda}_{gamma,delta})$. Each of the two differential equations have a unique solution up to a single integration constant. By setting $p = gamma(lambda_1) = delta(lambda_1)$, you specify both of these integration constants. So, you do not have any freedom to set the same condition for $lambda_2$ at $q$ because there is no freedom in the solution left and the value of $lambda$ at $q$ will be generally different for each of the curves. $blacksquare$
answered yesterday
VoidVoid
10.7k1757
$endgroup$
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
yesterday
$begingroup$
@Javier Yes, the limits need not be the same but I have made this choice. If they were different, $tau_gamma$ and $tau_delta$ would be different even if the two worldlines had the same parameterisation.
$endgroup$
– damaihati
19 hours ago
add a comment |
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1 Answer
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$begingroup$
Your conclusion is correct, because what you are doing by saying that $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tilde{lambda}$ of the curve $gamma$ that have $g(v_{gamma,gamma(tilde{lambda})},v_{gamma,gamma(tilde{lambda})}) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.
Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!
As concerns your edit:
When you state
$$p = gamma(lambda_1) = delta(lambda_1)$$
$$q = gamma(lambda_2) = delta(lambda_2)$$
and impose the normalisation of the tangents $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = g(v_{delta,delta(lambda)},v_{delta,delta(lambda)}) = 1$, you are NOT choosing two general curves (worldlines) passing through $p,q$. Instead, you are choosing two curves for which the same amount of proper time passes between $p,q$.
Perhaps you should consider a concrete example. Choose a space-time such as Minkowski and two randomly chosen curves $gamma, delta$ on it that have an arbitrary parametrization $gamma(tilde{lambda}_gamma), delta(tilde{lambda}_delta)$ and that meet at some events $p,q$. Now find the proper-time parametrisation $lambda$ along the entire curves $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = g(v_{delta,delta(lambda)},v_{delta,delta(lambda)}) = 1$, this can be recast as two first-order differential equations for the functions $lambda(tilde{lambda}_{gamma,delta})$. Each of the two differential equations have a unique solution up to a single integration constant. By setting $p = gamma(lambda_1) = delta(lambda_1)$, you specify both of these integration constants. So, you do not have any freedom to set the same condition for $lambda_2$ at $q$ because there is no freedom in the solution left and the value of $lambda$ at $q$ will be generally different for each of the curves. $blacksquare$
answered yesterday
VoidVoid
10.7k1757
$endgroup$
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
yesterday
$begingroup$
@Javier Yes, the limits need not be the same but I have made this choice. If they were different, $tau_gamma$ and $tau_delta$ would be different even if the two worldlines had the same parameterisation.
$endgroup$
– damaihati
19 hours ago
add a comment |
$begingroup$
Your conclusion is correct, because what you are doing by saying that $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tilde{lambda}$ of the curve $gamma$ that have $g(v_{gamma,gamma(tilde{lambda})},v_{gamma,gamma(tilde{lambda})}) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.
Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!
As concerns your edit:
When you state
$$p = gamma(lambda_1) = delta(lambda_1)$$
$$q = gamma(lambda_2) = delta(lambda_2)$$
and impose the normalisation of the tangents $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = g(v_{delta,delta(lambda)},v_{delta,delta(lambda)}) = 1$, you are NOT choosing two general curves (worldlines) passing through $p,q$. Instead, you are choosing two curves for which the same amount of proper time passes between $p,q$.
Perhaps you should consider a concrete example. Choose a space-time such as Minkowski and two randomly chosen curves $gamma, delta$ on it that have an arbitrary parametrization $gamma(tilde{lambda}_gamma), delta(tilde{lambda}_delta)$ and that meet at some events $p,q$. Now find the proper-time parametrisation $lambda$ along the entire curves $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = g(v_{delta,delta(lambda)},v_{delta,delta(lambda)}) = 1$, this can be recast as two first-order differential equations for the functions $lambda(tilde{lambda}_{gamma,delta})$. Each of the two differential equations have a unique solution up to a single integration constant. By setting $p = gamma(lambda_1) = delta(lambda_1)$, you specify both of these integration constants. So, you do not have any freedom to set the same condition for $lambda_2$ at $q$ because there is no freedom in the solution left and the value of $lambda$ at $q$ will be generally different for each of the curves. $blacksquare$
answered yesterday
VoidVoid
10.7k1757
$endgroup$
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
yesterday
$begingroup$
@Javier Yes, the limits need not be the same but I have made this choice. If they were different, $tau_gamma$ and $tau_delta$ would be different even if the two worldlines had the same parameterisation.
$endgroup$
– damaihati
19 hours ago
add a comment |
$begingroup$
Your conclusion is correct, because what you are doing by saying that $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tilde{lambda}$ of the curve $gamma$ that have $g(v_{gamma,gamma(tilde{lambda})},v_{gamma,gamma(tilde{lambda})}) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.
Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!
As concerns your edit:
When you state
$$p = gamma(lambda_1) = delta(lambda_1)$$
$$q = gamma(lambda_2) = delta(lambda_2)$$
and impose the normalisation of the tangents $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = g(v_{delta,delta(lambda)},v_{delta,delta(lambda)}) = 1$, you are NOT choosing two general curves (worldlines) passing through $p,q$. Instead, you are choosing two curves for which the same amount of proper time passes between $p,q$.
Perhaps you should consider a concrete example. Choose a space-time such as Minkowski and two randomly chosen curves $gamma, delta$ on it that have an arbitrary parametrization $gamma(tilde{lambda}_gamma), delta(tilde{lambda}_delta)$ and that meet at some events $p,q$. Now find the proper-time parametrisation $lambda$ along the entire curves $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = g(v_{delta,delta(lambda)},v_{delta,delta(lambda)}) = 1$, this can be recast as two first-order differential equations for the functions $lambda(tilde{lambda}_{gamma,delta})$. Each of the two differential equations have a unique solution up to a single integration constant. By setting $p = gamma(lambda_1) = delta(lambda_1)$, you specify both of these integration constants. So, you do not have any freedom to set the same condition for $lambda_2$ at $q$ because there is no freedom in the solution left and the value of $lambda$ at $q$ will be generally different for each of the curves. $blacksquare$
answered yesterday
VoidVoid
10.7k1757
$endgroup$
Your conclusion is correct, because what you are doing by saying that $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tilde{lambda}$ of the curve $gamma$ that have $g(v_{gamma,gamma(tilde{lambda})},v_{gamma,gamma(tilde{lambda})}) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.
Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!
As concerns your edit:
When you state
$$p = gamma(lambda_1) = delta(lambda_1)$$
$$q = gamma(lambda_2) = delta(lambda_2)$$
and impose the normalisation of the tangents $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = g(v_{delta,delta(lambda)},v_{delta,delta(lambda)}) = 1$, you are NOT choosing two general curves (worldlines) passing through $p,q$. Instead, you are choosing two curves for which the same amount of proper time passes between $p,q$.
Perhaps you should consider a concrete example. Choose a space-time such as Minkowski and two randomly chosen curves $gamma, delta$ on it that have an arbitrary parametrization $gamma(tilde{lambda}_gamma), delta(tilde{lambda}_delta)$ and that meet at some events $p,q$. Now find the proper-time parametrisation $lambda$ along the entire curves $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = g(v_{delta,delta(lambda)},v_{delta,delta(lambda)}) = 1$, this can be recast as two first-order differential equations for the functions $lambda(tilde{lambda}_{gamma,delta})$. Each of the two differential equations have a unique solution up to a single integration constant. By setting $p = gamma(lambda_1) = delta(lambda_1)$, you specify both of these integration constants. So, you do not have any freedom to set the same condition for $lambda_2$ at $q$ because there is no freedom in the solution left and the value of $lambda$ at $q$ will be generally different for each of the curves. $blacksquare$
answered yesterday
VoidVoid
10.7k1757
edited 15 hours ago
answered yesterday
VoidVoid
10.7k1757
answered yesterday
VoidVoid
10.7k1757
answered yesterday
VoidVoid
10.7k1757
10.7k1757
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
yesterday
$begingroup$
@Javier Yes, the limits need not be the same but I have made this choice. If they were different, $tau_gamma$ and $tau_delta$ would be different even if the two worldlines had the same parameterisation.
$endgroup$
– damaihati
19 hours ago
add a comment |
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
yesterday
$begingroup$
@Javier Yes, the limits need not be the same but I have made this choice. If they were different, $tau_gamma$ and $tau_delta$ would be different even if the two worldlines had the same parameterisation.
$endgroup$
– damaihati
19 hours ago
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
yesterday
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
yesterday
$begingroup$
@Javier Yes, the limits need not be the same but I have made this choice. If they were different, $tau_gamma$ and $tau_delta$ would be different even if the two worldlines had the same parameterisation.
$endgroup$
– damaihati
19 hours ago
$begingroup$
@Javier Yes, the limits need not be the same but I have made this choice. If they were different, $tau_gamma$ and $tau_delta$ would be different even if the two worldlines had the same parameterisation.
$endgroup$
– damaihati
19 hours ago
add a comment |
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$begingroup$
It doesn't make sense to say that two worldlines have the same parameter. You can have the values of the parameter coincide at one crossing point if you want, but nothing guarantees that this will happen when the curves cross again (which is the only place where it makes sense to say the the proper times are the same).
$endgroup$
– Javier
16 hours ago
$begingroup$
@Javier Then what is your opinion on the twin paradox? Let us work in some chart $(U,x)$. The worldline $gamma: (0,1) to M$ such that $gamma_{(x)}^i=(lambda,0,0,0)^i$ and $delta:(0,1) to M$ such that $delta_{(x)}^i=(lambda, alpha lambda, 0,0)^i$ for $lambda leq 1/2$ while $delta_{(x)}^i=(lambda, (1-lambda)alpha, 0,0)^i$ for $lambda > 1/2$. Here, $alpha in (0,1)$. Both the worldlines lie in $U subset M$. So, these two curves have the same parameter $lambda$ but the maps $gamma$ and $delta$ are defined differently. The parameters coincide in the beginning and end.
$endgroup$
– damaihati
13 hours ago
$begingroup$
That is a perfectly good parameterization and it will let you calculate the time difference of the twin paradox, but the second curve is not parametrized by proper time. That's not a problem.
$endgroup$
– Javier
12 hours ago