Mixing
Question on $B$ unit vector in frenet frame
$begingroup$
Given a regular curve $a(s) in mathbb{R}^3$, we can attach a moving frame given by three orthogonal unit vectors, $T(s), N(s), B(s)$.
Now, my question is does $B'(s)$ (the derivative of the $B(s)$ unit vector, which we obtained from taking $T(s) times N(s)$ only depend on the it's projection with respect to $N(s)$?
That is to say, $B'(s)= tau N$, where $tau$ is the torsion of the curve. I follow my books argument, and it seems to be technical rather than intuitive. To me, it seems like the rate of change of $B(s)$ should also depend on it's projection against $T$ also. It'd be possible for the rotation of the $B(s)$ vector to be completely in the $BT$ plane (the rectifying plane?), and if this is true, $B'(s) neq 0$ but none of this change in position of $B(s)$ that it's derative represents will be in the direction of the $N$ vector. Am I making myself clear on why I'm confused?
differential-geometry
asked Jan 23 at 2:57
Math is hardMath is hard
822211
$endgroup$
add a comment |
$begingroup$
Given a regular curve $a(s) in mathbb{R}^3$, we can attach a moving frame given by three orthogonal unit vectors, $T(s), N(s), B(s)$.
Now, my question is does $B'(s)$ (the derivative of the $B(s)$ unit vector, which we obtained from taking $T(s) times N(s)$ only depend on the it's projection with respect to $N(s)$?
That is to say, $B'(s)= tau N$, where $tau$ is the torsion of the curve. I follow my books argument, and it seems to be technical rather than intuitive. To me, it seems like the rate of change of $B(s)$ should also depend on it's projection against $T$ also. It'd be possible for the rotation of the $B(s)$ vector to be completely in the $BT$ plane (the rectifying plane?), and if this is true, $B'(s) neq 0$ but none of this change in position of $B(s)$ that it's derative represents will be in the direction of the $N$ vector. Am I making myself clear on why I'm confused?
differential-geometry
asked Jan 23 at 2:57
Math is hardMath is hard
822211
$endgroup$
add a comment |
$begingroup$
Given a regular curve $a(s) in mathbb{R}^3$, we can attach a moving frame given by three orthogonal unit vectors, $T(s), N(s), B(s)$.
Now, my question is does $B'(s)$ (the derivative of the $B(s)$ unit vector, which we obtained from taking $T(s) times N(s)$ only depend on the it's projection with respect to $N(s)$?
That is to say, $B'(s)= tau N$, where $tau$ is the torsion of the curve. I follow my books argument, and it seems to be technical rather than intuitive. To me, it seems like the rate of change of $B(s)$ should also depend on it's projection against $T$ also. It'd be possible for the rotation of the $B(s)$ vector to be completely in the $BT$ plane (the rectifying plane?), and if this is true, $B'(s) neq 0$ but none of this change in position of $B(s)$ that it's derative represents will be in the direction of the $N$ vector. Am I making myself clear on why I'm confused?
differential-geometry
asked Jan 23 at 2:57
Math is hardMath is hard
822211
$endgroup$
Given a regular curve $a(s) in mathbb{R}^3$, we can attach a moving frame given by three orthogonal unit vectors, $T(s), N(s), B(s)$.
Now, my question is does $B'(s)$ (the derivative of the $B(s)$ unit vector, which we obtained from taking $T(s) times N(s)$ only depend on the it's projection with respect to $N(s)$?
That is to say, $B'(s)= tau N$, where $tau$ is the torsion of the curve. I follow my books argument, and it seems to be technical rather than intuitive. To me, it seems like the rate of change of $B(s)$ should also depend on it's projection against $T$ also. It'd be possible for the rotation of the $B(s)$ vector to be completely in the $BT$ plane (the rectifying plane?), and if this is true, $B'(s) neq 0$ but none of this change in position of $B(s)$ that it's derative represents will be in the direction of the $N$ vector. Am I making myself clear on why I'm confused?
differential-geometry
differential-geometry
asked Jan 23 at 2:57
Math is hardMath is hard
822211
asked Jan 23 at 2:57
Math is hardMath is hard
822211
asked Jan 23 at 2:57
Math is hardMath is hard
822211
asked Jan 23 at 2:57
Math is hardMath is hard
822211
asked Jan 23 at 2:57
Math is hardMath is hard
822211
822211
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In order for two unit vectors to stay perpendicular, the second must turn away from the first at precisely the same rate as the first turns toward the second. (Mathematically, this is saying $B'cdot T = -T'cdot B$.) You can prove this by the product rule, but it's also quite clear if you just rotate your hand, keeping the thumb and forefinger perpendicular.
Since by definition $T$ turns only toward $N$ (i.e., it rotates in the osculating plane, instantaneously), it doesn't turn at all toward $B$ and so $B$ can't turn at all toward $T$.
answered Jan 23 at 21:00
Ted ShifrinTed Shifrin
64.2k44692
$endgroup$
$begingroup$
Thanks for the response, i'm honored!
$endgroup$
– Math is hard
Jan 24 at 2:43
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: "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%2f3084019%2fquestion-on-b-unit-vector-in-frenet-frame%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
$begingroup$
In order for two unit vectors to stay perpendicular, the second must turn away from the first at precisely the same rate as the first turns toward the second. (Mathematically, this is saying $B'cdot T = -T'cdot B$.) You can prove this by the product rule, but it's also quite clear if you just rotate your hand, keeping the thumb and forefinger perpendicular.
Since by definition $T$ turns only toward $N$ (i.e., it rotates in the osculating plane, instantaneously), it doesn't turn at all toward $B$ and so $B$ can't turn at all toward $T$.
answered Jan 23 at 21:00
Ted ShifrinTed Shifrin
64.2k44692
$endgroup$
$begingroup$
Thanks for the response, i'm honored!
$endgroup$
– Math is hard
Jan 24 at 2:43
add a comment |
$begingroup$
In order for two unit vectors to stay perpendicular, the second must turn away from the first at precisely the same rate as the first turns toward the second. (Mathematically, this is saying $B'cdot T = -T'cdot B$.) You can prove this by the product rule, but it's also quite clear if you just rotate your hand, keeping the thumb and forefinger perpendicular.
Since by definition $T$ turns only toward $N$ (i.e., it rotates in the osculating plane, instantaneously), it doesn't turn at all toward $B$ and so $B$ can't turn at all toward $T$.
answered Jan 23 at 21:00
Ted ShifrinTed Shifrin
64.2k44692
$endgroup$
$begingroup$
Thanks for the response, i'm honored!
$endgroup$
– Math is hard
Jan 24 at 2:43
add a comment |
$begingroup$
In order for two unit vectors to stay perpendicular, the second must turn away from the first at precisely the same rate as the first turns toward the second. (Mathematically, this is saying $B'cdot T = -T'cdot B$.) You can prove this by the product rule, but it's also quite clear if you just rotate your hand, keeping the thumb and forefinger perpendicular.
Since by definition $T$ turns only toward $N$ (i.e., it rotates in the osculating plane, instantaneously), it doesn't turn at all toward $B$ and so $B$ can't turn at all toward $T$.
answered Jan 23 at 21:00
Ted ShifrinTed Shifrin
64.2k44692
$endgroup$
In order for two unit vectors to stay perpendicular, the second must turn away from the first at precisely the same rate as the first turns toward the second. (Mathematically, this is saying $B'cdot T = -T'cdot B$.) You can prove this by the product rule, but it's also quite clear if you just rotate your hand, keeping the thumb and forefinger perpendicular.
Since by definition $T$ turns only toward $N$ (i.e., it rotates in the osculating plane, instantaneously), it doesn't turn at all toward $B$ and so $B$ can't turn at all toward $T$.
answered Jan 23 at 21:00
Ted ShifrinTed Shifrin
64.2k44692
answered Jan 23 at 21:00
Ted ShifrinTed Shifrin
64.2k44692
answered Jan 23 at 21:00
Ted ShifrinTed Shifrin
64.2k44692
answered Jan 23 at 21:00
Ted ShifrinTed Shifrin
64.2k44692
64.2k44692
$begingroup$
Thanks for the response, i'm honored!
$endgroup$
– Math is hard
Jan 24 at 2:43
add a comment |
$begingroup$
Thanks for the response, i'm honored!
$endgroup$
– Math is hard
Jan 24 at 2:43
$begingroup$
Thanks for the response, i'm honored!
$endgroup$
– Math is hard
Jan 24 at 2:43
$begingroup$
Thanks for the response, i'm honored!
$endgroup$
– Math is hard
Jan 24 at 2:43
add a comment |
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%2f3084019%2fquestion-on-b-unit-vector-in-frenet-frame%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
