Geodesic Deviation Derivation
$begingroup$
From Section 5.8 in Woodhouse's General Relativity:
Let $Y^a(tau)$ be a vector field. Its covariant derivative $DY^b$ along $omega$ is defined by the following equivalent expressions,
$$DY^b = V^atriangledown_a Y^b$$
Why have we chosen to add the $V^a$ to the definition?
The four-velocities of the particles form a vector field $V^a$. Because the individual particle worldlines are geodesic,
$$V^btriangledown_b V^a = DV^a = frac{dV^a}{dtau} + Gamma_{bc}^a V^bV^c = 0$$
1) Why do we care that $V^btriangledown_b V^a = 0$?
2) how do we get the last equality?
differential-geometry tensors geodesic general-relativity
asked Dec 21 '18 at 13:57
PermianPermian
2,2631135
$endgroup$
add a comment |
$begingroup$
From Section 5.8 in Woodhouse's General Relativity:
Let $Y^a(tau)$ be a vector field. Its covariant derivative $DY^b$ along $omega$ is defined by the following equivalent expressions,
$$DY^b = V^atriangledown_a Y^b$$
Why have we chosen to add the $V^a$ to the definition?
The four-velocities of the particles form a vector field $V^a$. Because the individual particle worldlines are geodesic,
$$V^btriangledown_b V^a = DV^a = frac{dV^a}{dtau} + Gamma_{bc}^a V^bV^c = 0$$
1) Why do we care that $V^btriangledown_b V^a = 0$?
2) how do we get the last equality?
differential-geometry tensors geodesic general-relativity
asked Dec 21 '18 at 13:57
PermianPermian
2,2631135
$endgroup$
$begingroup$
The whole section 5.8 feels very skippy
$endgroup$
– Permian
Dec 21 '18 at 13:59
add a comment |
$begingroup$
From Section 5.8 in Woodhouse's General Relativity:
Let $Y^a(tau)$ be a vector field. Its covariant derivative $DY^b$ along $omega$ is defined by the following equivalent expressions,
$$DY^b = V^atriangledown_a Y^b$$
Why have we chosen to add the $V^a$ to the definition?
The four-velocities of the particles form a vector field $V^a$. Because the individual particle worldlines are geodesic,
$$V^btriangledown_b V^a = DV^a = frac{dV^a}{dtau} + Gamma_{bc}^a V^bV^c = 0$$
1) Why do we care that $V^btriangledown_b V^a = 0$?
2) how do we get the last equality?
differential-geometry tensors geodesic general-relativity
asked Dec 21 '18 at 13:57
PermianPermian
2,2631135
$endgroup$
From Section 5.8 in Woodhouse's General Relativity:
Let $Y^a(tau)$ be a vector field. Its covariant derivative $DY^b$ along $omega$ is defined by the following equivalent expressions,
$$DY^b = V^atriangledown_a Y^b$$
Why have we chosen to add the $V^a$ to the definition?
The four-velocities of the particles form a vector field $V^a$. Because the individual particle worldlines are geodesic,
$$V^btriangledown_b V^a = DV^a = frac{dV^a}{dtau} + Gamma_{bc}^a V^bV^c = 0$$
1) Why do we care that $V^btriangledown_b V^a = 0$?
2) how do we get the last equality?
differential-geometry tensors geodesic general-relativity
differential-geometry tensors geodesic general-relativity
asked Dec 21 '18 at 13:57
PermianPermian
2,2631135
asked Dec 21 '18 at 13:57
PermianPermian
2,2631135
asked Dec 21 '18 at 13:57
PermianPermian
2,2631135
asked Dec 21 '18 at 13:57
PermianPermian
2,2631135
asked Dec 21 '18 at 13:57
PermianPermian
2,2631135
2,2631135
$begingroup$
The whole section 5.8 feels very skippy
$endgroup$
– Permian
Dec 21 '18 at 13:59
add a comment |
$begingroup$
The whole section 5.8 feels very skippy
$endgroup$
– Permian
Dec 21 '18 at 13:59
$begingroup$
The whole section 5.8 feels very skippy
$endgroup$
– Permian
Dec 21 '18 at 13:59
$begingroup$
The whole section 5.8 feels very skippy
$endgroup$
– Permian
Dec 21 '18 at 13:59
add a comment |
0
active
oldest
votes
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: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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%2fmath.stackexchange.com%2fquestions%2f3048516%2fgeodesic-deviation-derivation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics 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%2fmath.stackexchange.com%2fquestions%2f3048516%2fgeodesic-deviation-derivation%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
$begingroup$
The whole section 5.8 feels very skippy
$endgroup$
– Permian
Dec 21 '18 at 13:59