definition of observer and time measured by different observers in general relativityWhy do clocks measure arc-length?Are there more distinctive names of “null curves” with certain additional properties (absence of “chord curves”)?An argument that massive particles don't redshift?How to determine the three-velocity measured by a single observer?Global symmetries of spacetime and general covarianceHow 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?How one uses the definition of observers in General Relativity?How do we measure distances in the FLRW metric?
how to check a propriety using r studio
How much of data wrangling is a data scientist's job?
expand `ifthenelse` immediately
Arrow those variables!
Get value of a counter
Theorems that impeded progress
What's that red-plus icon near a text?
How do I deal with an unproductive colleague in a small company?
Maximum likelihood parameters deviate from posterior distributions
Why doesn't H₄O²⁺ exist?
Can I make popcorn with any corn?
How old can references or sources in a thesis be?
Can a Cauchy sequence converge for one metric while not converging for another?
Alternative to sending password over mail?
Java Casting: Java 11 throws LambdaConversionException while 1.8 does not
How is it possible to have an ability score that is less than 3?
Why is consensus so controversial in Britain?
Can you really stack all of this on an Opportunity Attack?
Roll the carpet
Why "Having chlorophyll without photosynthesis is actually very dangerous" and "like living with a bomb"?
Two films in a tank, only one comes out with a development error – why?
How to source a part of a file
Languages that we cannot (dis)prove to be Context-Free
Linear Path Optimization with Two Dependent Variables
definition of observer and time measured by different observers in general relativity
Why do clocks measure arc-length?Are there more distinctive names of “null curves” with certain additional properties (absence of “chord curves”)?An argument that massive particles don't redshift?How to determine the three-velocity measured by a single observer?Global symmetries of spacetime and general covarianceHow 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?How one uses the definition of observers in General Relativity?How do we measure distances in the FLRW metric?
$begingroup$
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_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
However,
beginalign
g_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = g_gamma(lambda)(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_delta(lambda)(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?
general-relativity observers
$endgroup$
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_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
However,
beginalign
g_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = g_gamma(lambda)(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_delta(lambda)(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?
general-relativity observers
$endgroup$
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_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
However,
beginalign
g_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = g_gamma(lambda)(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_delta(lambda)(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?
general-relativity observers
$endgroup$
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_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
However,
beginalign
g_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = g_gamma(lambda)(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_delta(lambda)(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?
general-relativity observers
general-relativity observers
asked 8 hours ago
damaihatidamaihati
683
683
add a comment |
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 $tildelambda$ 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!
$endgroup$
$begingroup$
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
2 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
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
1 Answer
1
active
oldest
votes
1 Answer
1
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!
$endgroup$
$begingroup$
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
2 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 $tildelambda$ 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!
$endgroup$
$begingroup$
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
2 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 $tildelambda$ 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!
$endgroup$
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!
answered 8 hours ago
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
2 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
2 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
2 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
2 hours ago
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