Definition of observer and time measured by different observers in general relativity2019 Community Moderator Election Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar Manara 2019 Moderator Election Q&A - QuestionnaireWhy do clocks measure arc-length?Do observers at rest at different positions in a gravitational field see the Universe expand at different ratesHow to determine the three-velocity measured by a single observer?Variation of products of Riemann tensor $delta (sqrt-g RR epsilon epsilon)$How a reference frame relates to observers and charts?When the $x^0$ coordinate represents time in GR?How to make sense of this definition of a reference frame?What is the “reference frame of a particle” in General Relativity?How one uses the definition of observers in General Relativity?Expanding a summation of covariant derivatives
Contradiction:Maximum Power Transfer and High resistance of load
Where can I find how to tex symbols for different fonts?
AI positioning circles within an arc at equal distances and heights
PIC mathematical operations weird problem
What helicopter has the most rotor blades?
My admission is revoked after accepting the admission offer
Putting Ant-Man on house arrest
I preordered a game on my Xbox while on the home screen of my friend's account. Which of us owns the game?
What is /etc/mtab in Linux?
Raising a bilingual kid. When should we introduce the majority language?
What do you call an IPA symbol that lacks a name (e.g. ɲ)?
What was Apollo 13's "Little Jolt" after MECO?
Are there existing rules/lore for MTG planeswalkers?
All ASCII characters with a given bit count
Mechanism of the formation of peracetic acid
What to do with someone that cheated their way though university and a PhD program?
yticklabels on the right side of yaxis
using NDEigensystem to solve the Mathieu equation
Is it OK if I do not take the receipt in Germany?
Retract an already submitted recommendation letter (written for an undergrad student)
Did war bonds have better investment alternatives during WWII?
Arriving in Atlanta after US Preclearance in Dublin. Will I go through TSA security in Atlanta to transfer to a connecting flight?
Why didn't the Space Shuttle bounce back into space many times as possible so that it loose lot of kinetic energy over there?
/bin/ls sorts differently than just ls
Definition of observer and time measured by different observers in general relativity
2019 Community Moderator Election
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manara
2019 Moderator Election Q&A - QuestionnaireWhy do clocks measure arc-length?Do observers at rest at different positions in a gravitational field see the Universe expand at different ratesHow to determine the three-velocity measured by a single observer?Variation of products of Riemann tensor $delta (sqrt-g RR epsilon epsilon)$How a reference frame relates to observers and charts?When the $x^0$ coordinate represents time in GR?How to make sense of this definition of a reference frame?What is the “reference frame of a particle” in General Relativity?How one uses the definition of observers in General Relativity?Expanding a summation of covariant derivatives
$begingroup$
An observer in general relativity is defined as a future directed timelike worldline
beginalign*
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
endalign*
together with an orthonormal basis $e_a(lambda) in T_gamma(lambda)M$ where $e_0(lambda)= v_gamma, gamma(lambda)$ and
beginalign
g(gamma(lambda))(e_a(lambda),e_b(lambda))=eta_ab~. qquad (1)
endalign
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
beginalign
tau_gamma = int_lambda_0^lambda_1 dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
However,
beginalign
g(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = g(e_0(lambda),e_0(lambda))=1 qquad (2)
endalign
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.
This is all standard definition. Suppose, we have another observer $delta$:
beginalign*
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
endalign*
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
beginalign
tau_delta = int_lambda_0^lambda_1 dlambda sqrtg(v_delta, delta(lambda),v_delta, delta(lambda))~.
endalign
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 $M$.
beginalign
p =& gamma(lambda_1) = delta(lambda_1) \
q =& gamma(lambda_2) = delta(lambda2)
endalign
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
asked Apr 5 at 11:44
damaihatidamaihati
704
$endgroup$
add a comment |
$begingroup$
An observer in general relativity is defined as a future directed timelike worldline
beginalign*
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
endalign*
together with an orthonormal basis $e_a(lambda) in T_gamma(lambda)M$ where $e_0(lambda)= v_gamma, gamma(lambda)$ and
beginalign
g(gamma(lambda))(e_a(lambda),e_b(lambda))=eta_ab~. qquad (1)
endalign
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
beginalign
tau_gamma = int_lambda_0^lambda_1 dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
However,
beginalign
g(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = g(e_0(lambda),e_0(lambda))=1 qquad (2)
endalign
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.
This is all standard definition. Suppose, we have another observer $delta$:
beginalign*
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
endalign*
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
beginalign
tau_delta = int_lambda_0^lambda_1 dlambda sqrtg(v_delta, delta(lambda),v_delta, delta(lambda))~.
endalign
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 $M$.
beginalign
p =& gamma(lambda_1) = delta(lambda_1) \
q =& gamma(lambda_2) = delta(lambda2)
endalign
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
asked Apr 5 at 11:44
damaihatidamaihati
704
$endgroup$
1
$begingroup$
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
Apr 6 at 12:01
$begingroup$
@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
Apr 6 at 14:59
1
$begingroup$
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
Apr 6 at 16:06
add a comment |
$begingroup$
An observer in general relativity is defined as a future directed timelike worldline
beginalign*
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
endalign*
together with an orthonormal basis $e_a(lambda) in T_gamma(lambda)M$ where $e_0(lambda)= v_gamma, gamma(lambda)$ and
beginalign
g(gamma(lambda))(e_a(lambda),e_b(lambda))=eta_ab~. qquad (1)
endalign
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
beginalign
tau_gamma = int_lambda_0^lambda_1 dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
However,
beginalign
g(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = g(e_0(lambda),e_0(lambda))=1 qquad (2)
endalign
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.
This is all standard definition. Suppose, we have another observer $delta$:
beginalign*
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
endalign*
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
beginalign
tau_delta = int_lambda_0^lambda_1 dlambda sqrtg(v_delta, delta(lambda),v_delta, delta(lambda))~.
endalign
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 $M$.
beginalign
p =& gamma(lambda_1) = delta(lambda_1) \
q =& gamma(lambda_2) = delta(lambda2)
endalign
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
asked Apr 5 at 11:44
damaihatidamaihati
704
$endgroup$
An observer in general relativity is defined as a future directed timelike worldline
beginalign*
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
endalign*
together with an orthonormal basis $e_a(lambda) in T_gamma(lambda)M$ where $e_0(lambda)= v_gamma, gamma(lambda)$ and
beginalign
g(gamma(lambda))(e_a(lambda),e_b(lambda))=eta_ab~. qquad (1)
endalign
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
beginalign
tau_gamma = int_lambda_0^lambda_1 dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
However,
beginalign
g(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = g(e_0(lambda),e_0(lambda))=1 qquad (2)
endalign
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.
This is all standard definition. Suppose, we have another observer $delta$:
beginalign*
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
endalign*
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
beginalign
tau_delta = int_lambda_0^lambda_1 dlambda sqrtg(v_delta, delta(lambda),v_delta, delta(lambda))~.
endalign
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 $M$.
beginalign
p =& gamma(lambda_1) = delta(lambda_1) \
q =& gamma(lambda_2) = delta(lambda2)
endalign
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
asked Apr 5 at 11:44
damaihatidamaihati
704
asked Apr 5 at 11:44
damaihatidamaihati
704
edited Apr 10 at 9:22
damaihati
asked Apr 5 at 11:44
damaihatidamaihati
704
asked Apr 5 at 11:44
damaihatidamaihati
704
asked Apr 5 at 11:44
damaihatidamaihati
704
704
1
$begingroup$
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
Apr 6 at 12:01
$begingroup$
@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
Apr 6 at 14:59
1
$begingroup$
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
Apr 6 at 16:06
add a comment |
1
$begingroup$
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
Apr 6 at 12:01
$begingroup$
@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
Apr 6 at 14:59
1
$begingroup$
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
Apr 6 at 16:06
1
1
$begingroup$
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
Apr 6 at 12:01
$begingroup$
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
Apr 6 at 12:01
$begingroup$
@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
Apr 6 at 14:59
$begingroup$
@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
Apr 6 at 14:59
1
1
$begingroup$
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
Apr 6 at 16:06
$begingroup$
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
Apr 6 at 16:06
add a comment |
2 Answers
2
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 $tildelambda$ of the curve $gamma$ that have $g(v_gamma,gamma(tildelambda),v_gamma,gamma(tildelambda)) neq 1$ 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 $gamma(tildelambda_gamma), delta(tildelambda_delta)$ 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 $lambda(tildelambda_gamma,delta)$. 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 Apr 5 at 11:52
VoidVoid
11.3k11959
$endgroup$
1
$begingroup$
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
Apr 5 at 18:10
$begingroup$
@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
Apr 6 at 9:52
add a comment |
$begingroup$
I am just answering my question in a way that is more transparent to me. There is no extra information here compared to the accepted answer or comments therein.
The definition of an observer $(gamma, e)$ is given in the OP. By definition, the worldline of the observer is a curve with unit speed i.e.,
beginalign
g(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = 1~. qquad (1)
endalign
This implies that the curve is parameterised by its arc length parameter. The proper time that an observer $(gamma,e)$ measures between two events $p=gamma(lambda=a)$ and $q=gamma(lambda=b)$ in spacetime is,
beginalign
tau_gamma &= int_a^b dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) \
&= int_a^b dlambda~.
endalign
So, any two observers whose arc length (or proper time) parameters increase by the same amount between the events $p$ and $q$ will measure the same proper time interval. (Naturally, observers whose arc length parameters change by different amounts between $p$ and $q$ will measure different proper time intervals.)
Note: Suppose, we relax the requirement of eq. (1) of having a unit speed worldline on the observer. So, we are allowed to reparameterise the observer worldline by some arbitrary parameter. Now, the length of a curve $gamma$,
beginalign
L[gamma] = int_a^b dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
is invariant under reparameterisation. This means that, given a smooth curve
beginalign
gamma:I to M
endalign
if
beginalign
sigma:tildeI to I
endalign
is smooth, bijective and increasing then
beginalign
L[gamma]=L[gamma circ sigma]~.
endalign
Owing to this, the proper time interval between the spacetime events $p$ and $q$ will be same also along the reparameterised curve $(gamma circ sigma)$.
answered Apr 10 at 9:19
damaihatidamaihati
704
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "151"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f470705%2fdefinition-of-observer-and-time-measured-by-different-observers-in-general-relat%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 $tildelambda$ of the curve $gamma$ that have $g(v_gamma,gamma(tildelambda),v_gamma,gamma(tildelambda)) neq 1$ 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 $gamma(tildelambda_gamma), delta(tildelambda_delta)$ 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 $lambda(tildelambda_gamma,delta)$. 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 Apr 5 at 11:52
VoidVoid
11.3k11959
$endgroup$
1
$begingroup$
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
Apr 5 at 18:10
$begingroup$
@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
Apr 6 at 9:52
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 $tildelambda$ of the curve $gamma$ that have $g(v_gamma,gamma(tildelambda),v_gamma,gamma(tildelambda)) neq 1$ 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 $gamma(tildelambda_gamma), delta(tildelambda_delta)$ 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 $lambda(tildelambda_gamma,delta)$. 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 Apr 5 at 11:52
VoidVoid
11.3k11959
$endgroup$
1
$begingroup$
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
Apr 5 at 18:10
$begingroup$
@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
Apr 6 at 9:52
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 $tildelambda$ of the curve $gamma$ that have $g(v_gamma,gamma(tildelambda),v_gamma,gamma(tildelambda)) neq 1$ 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 $gamma(tildelambda_gamma), delta(tildelambda_delta)$ 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 $lambda(tildelambda_gamma,delta)$. 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 Apr 5 at 11:52
VoidVoid
11.3k11959
$endgroup$
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 $tildelambda$ of the curve $gamma$ that have $g(v_gamma,gamma(tildelambda),v_gamma,gamma(tildelambda)) neq 1$ 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 $gamma(tildelambda_gamma), delta(tildelambda_delta)$ 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 $lambda(tildelambda_gamma,delta)$. 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 Apr 5 at 11:52
VoidVoid
11.3k11959
edited Apr 6 at 13:22
answered Apr 5 at 11:52
VoidVoid
11.3k11959
answered Apr 5 at 11:52
VoidVoid
11.3k11959
answered Apr 5 at 11:52
VoidVoid
11.3k11959
11.3k11959
1
$begingroup$
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
Apr 5 at 18:10
$begingroup$
@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
Apr 6 at 9:52
add a comment |
1
$begingroup$
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
Apr 5 at 18:10
$begingroup$
@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
Apr 6 at 9:52
1
1
$begingroup$
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
Apr 5 at 18:10
$begingroup$
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
Apr 5 at 18:10
$begingroup$
@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
Apr 6 at 9:52
$begingroup$
@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
Apr 6 at 9:52
add a comment |
$begingroup$
I am just answering my question in a way that is more transparent to me. There is no extra information here compared to the accepted answer or comments therein.
The definition of an observer $(gamma, e)$ is given in the OP. By definition, the worldline of the observer is a curve with unit speed i.e.,
beginalign
g(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = 1~. qquad (1)
endalign
This implies that the curve is parameterised by its arc length parameter. The proper time that an observer $(gamma,e)$ measures between two events $p=gamma(lambda=a)$ and $q=gamma(lambda=b)$ in spacetime is,
beginalign
tau_gamma &= int_a^b dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) \
&= int_a^b dlambda~.
endalign
So, any two observers whose arc length (or proper time) parameters increase by the same amount between the events $p$ and $q$ will measure the same proper time interval. (Naturally, observers whose arc length parameters change by different amounts between $p$ and $q$ will measure different proper time intervals.)
Note: Suppose, we relax the requirement of eq. (1) of having a unit speed worldline on the observer. So, we are allowed to reparameterise the observer worldline by some arbitrary parameter. Now, the length of a curve $gamma$,
beginalign
L[gamma] = int_a^b dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
is invariant under reparameterisation. This means that, given a smooth curve
beginalign
gamma:I to M
endalign
if
beginalign
sigma:tildeI to I
endalign
is smooth, bijective and increasing then
beginalign
L[gamma]=L[gamma circ sigma]~.
endalign
Owing to this, the proper time interval between the spacetime events $p$ and $q$ will be same also along the reparameterised curve $(gamma circ sigma)$.
answered Apr 10 at 9:19
damaihatidamaihati
704
$endgroup$
add a comment |
$begingroup$
I am just answering my question in a way that is more transparent to me. There is no extra information here compared to the accepted answer or comments therein.
The definition of an observer $(gamma, e)$ is given in the OP. By definition, the worldline of the observer is a curve with unit speed i.e.,
beginalign
g(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = 1~. qquad (1)
endalign
This implies that the curve is parameterised by its arc length parameter. The proper time that an observer $(gamma,e)$ measures between two events $p=gamma(lambda=a)$ and $q=gamma(lambda=b)$ in spacetime is,
beginalign
tau_gamma &= int_a^b dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) \
&= int_a^b dlambda~.
endalign
So, any two observers whose arc length (or proper time) parameters increase by the same amount between the events $p$ and $q$ will measure the same proper time interval. (Naturally, observers whose arc length parameters change by different amounts between $p$ and $q$ will measure different proper time intervals.)
Note: Suppose, we relax the requirement of eq. (1) of having a unit speed worldline on the observer. So, we are allowed to reparameterise the observer worldline by some arbitrary parameter. Now, the length of a curve $gamma$,
beginalign
L[gamma] = int_a^b dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
is invariant under reparameterisation. This means that, given a smooth curve
beginalign
gamma:I to M
endalign
if
beginalign
sigma:tildeI to I
endalign
is smooth, bijective and increasing then
beginalign
L[gamma]=L[gamma circ sigma]~.
endalign
Owing to this, the proper time interval between the spacetime events $p$ and $q$ will be same also along the reparameterised curve $(gamma circ sigma)$.
answered Apr 10 at 9:19
damaihatidamaihati
704
$endgroup$
add a comment |
$begingroup$
I am just answering my question in a way that is more transparent to me. There is no extra information here compared to the accepted answer or comments therein.
The definition of an observer $(gamma, e)$ is given in the OP. By definition, the worldline of the observer is a curve with unit speed i.e.,
beginalign
g(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = 1~. qquad (1)
endalign
This implies that the curve is parameterised by its arc length parameter. The proper time that an observer $(gamma,e)$ measures between two events $p=gamma(lambda=a)$ and $q=gamma(lambda=b)$ in spacetime is,
beginalign
tau_gamma &= int_a^b dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) \
&= int_a^b dlambda~.
endalign
So, any two observers whose arc length (or proper time) parameters increase by the same amount between the events $p$ and $q$ will measure the same proper time interval. (Naturally, observers whose arc length parameters change by different amounts between $p$ and $q$ will measure different proper time intervals.)
Note: Suppose, we relax the requirement of eq. (1) of having a unit speed worldline on the observer. So, we are allowed to reparameterise the observer worldline by some arbitrary parameter. Now, the length of a curve $gamma$,
beginalign
L[gamma] = int_a^b dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
is invariant under reparameterisation. This means that, given a smooth curve
beginalign
gamma:I to M
endalign
if
beginalign
sigma:tildeI to I
endalign
is smooth, bijective and increasing then
beginalign
L[gamma]=L[gamma circ sigma]~.
endalign
Owing to this, the proper time interval between the spacetime events $p$ and $q$ will be same also along the reparameterised curve $(gamma circ sigma)$.
answered Apr 10 at 9:19
damaihatidamaihati
704
$endgroup$
I am just answering my question in a way that is more transparent to me. There is no extra information here compared to the accepted answer or comments therein.
The definition of an observer $(gamma, e)$ is given in the OP. By definition, the worldline of the observer is a curve with unit speed i.e.,
beginalign
g(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = 1~. qquad (1)
endalign
This implies that the curve is parameterised by its arc length parameter. The proper time that an observer $(gamma,e)$ measures between two events $p=gamma(lambda=a)$ and $q=gamma(lambda=b)$ in spacetime is,
beginalign
tau_gamma &= int_a^b dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) \
&= int_a^b dlambda~.
endalign
So, any two observers whose arc length (or proper time) parameters increase by the same amount between the events $p$ and $q$ will measure the same proper time interval. (Naturally, observers whose arc length parameters change by different amounts between $p$ and $q$ will measure different proper time intervals.)
Note: Suppose, we relax the requirement of eq. (1) of having a unit speed worldline on the observer. So, we are allowed to reparameterise the observer worldline by some arbitrary parameter. Now, the length of a curve $gamma$,
beginalign
L[gamma] = int_a^b dlambda sqrtg(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
is invariant under reparameterisation. This means that, given a smooth curve
beginalign
gamma:I to M
endalign
if
beginalign
sigma:tildeI to I
endalign
is smooth, bijective and increasing then
beginalign
L[gamma]=L[gamma circ sigma]~.
endalign
Owing to this, the proper time interval between the spacetime events $p$ and $q$ will be same also along the reparameterised curve $(gamma circ sigma)$.
answered Apr 10 at 9:19
damaihatidamaihati
704
answered Apr 10 at 9:19
damaihatidamaihati
704
answered Apr 10 at 9:19
damaihatidamaihati
704
answered Apr 10 at 9:19
damaihatidamaihati
704
704
add a comment |
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f470705%2fdefinition-of-observer-and-time-measured-by-different-observers-in-general-relat%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
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
Apr 6 at 12:01
$begingroup$
@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
Apr 6 at 14:59
1
$begingroup$
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
Apr 6 at 16:06