What is the curvature of this metric?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite
1












This is a follow up to a question about a four dimensional geometry that's suggested by data I've collected.



If I have a surface defined by:$$ds^2=-(phi^2t^2)dt^2+dx^2+dy^2+dz^2$$where $phi$ is a constant with units of $km$ $s^-2$, t is time ($s$), x, y and z are spatial coordiantes (in $km$), how would I calculate the curvature of the spatial surface?



Is the curvature zero (that is, does this metric describe perfectly flat space)?






differential-geometry differential-topology







share|cite|improve this question























  • I don't understand what your formula is supposed to mean. Usually, the notation $ds^2$ stands for a Riemannian metric, that is, a positive definite symmetric $2$-tensor field. But $dt^4$ is a $4$-tensor, so it doesn't make sense to add it to a bunch of $2$-tensors. And I have no idea how that formula is supposed to define a surface.
    – Jack Lee
    Jan 3 at 17:57










  • @JackLee - So sorry, I've updated the question to include the units of the constant. Hopefully it makes more sense now.
    – Donald Airey
    Jan 3 at 18:11










  • It still doesn't make mathematical sense, for the same reason I mentioned above. Maybe it would help if you explain exactly how this equation is supposed to define a surface.
    – Jack Lee
    Jan 3 at 18:13










  • Why doesn't it make sense? The $phi^2dt^4$ expression is a 2-tensors (that is, it has units of $km^2$, like the rest of the equation.)
    – Donald Airey
    Jan 3 at 18:16











  • You should reference the question which lead to this. Having that context makes answering this question (or pointing out mistakes in the way you formulated it) more likely.
    – MvG
    Jan 3 at 18:23














up vote
2
down vote

favorite
1












This is a follow up to a question about a four dimensional geometry that's suggested by data I've collected.



If I have a surface defined by:$$ds^2=-(phi^2t^2)dt^2+dx^2+dy^2+dz^2$$where $phi$ is a constant with units of $km$ $s^-2$, t is time ($s$), x, y and z are spatial coordiantes (in $km$), how would I calculate the curvature of the spatial surface?



Is the curvature zero (that is, does this metric describe perfectly flat space)?






differential-geometry differential-topology







share|cite|improve this question























  • I don't understand what your formula is supposed to mean. Usually, the notation $ds^2$ stands for a Riemannian metric, that is, a positive definite symmetric $2$-tensor field. But $dt^4$ is a $4$-tensor, so it doesn't make sense to add it to a bunch of $2$-tensors. And I have no idea how that formula is supposed to define a surface.
    – Jack Lee
    Jan 3 at 17:57










  • @JackLee - So sorry, I've updated the question to include the units of the constant. Hopefully it makes more sense now.
    – Donald Airey
    Jan 3 at 18:11










  • It still doesn't make mathematical sense, for the same reason I mentioned above. Maybe it would help if you explain exactly how this equation is supposed to define a surface.
    – Jack Lee
    Jan 3 at 18:13










  • Why doesn't it make sense? The $phi^2dt^4$ expression is a 2-tensors (that is, it has units of $km^2$, like the rest of the equation.)
    – Donald Airey
    Jan 3 at 18:16











  • You should reference the question which lead to this. Having that context makes answering this question (or pointing out mistakes in the way you formulated it) more likely.
    – MvG
    Jan 3 at 18:23












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





This is a follow up to a question about a four dimensional geometry that's suggested by data I've collected.



If I have a surface defined by:$$ds^2=-(phi^2t^2)dt^2+dx^2+dy^2+dz^2$$where $phi$ is a constant with units of $km$ $s^-2$, t is time ($s$), x, y and z are spatial coordiantes (in $km$), how would I calculate the curvature of the spatial surface?



Is the curvature zero (that is, does this metric describe perfectly flat space)?






differential-geometry differential-topology







share|cite|improve this question















This is a follow up to a question about a four dimensional geometry that's suggested by data I've collected.



If I have a surface defined by:$$ds^2=-(phi^2t^2)dt^2+dx^2+dy^2+dz^2$$where $phi$ is a constant with units of $km$ $s^-2$, t is time ($s$), x, y and z are spatial coordiantes (in $km$), how would I calculate the curvature of the spatial surface?



Is the curvature zero (that is, does this metric describe perfectly flat space)?






differential-geometry differential-topology




differential-geometry differential-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 31 at 1:16

























asked Jan 3 at 13:46









Donald Airey

5510




5510











  • I don't understand what your formula is supposed to mean. Usually, the notation $ds^2$ stands for a Riemannian metric, that is, a positive definite symmetric $2$-tensor field. But $dt^4$ is a $4$-tensor, so it doesn't make sense to add it to a bunch of $2$-tensors. And I have no idea how that formula is supposed to define a surface.
    – Jack Lee
    Jan 3 at 17:57










  • @JackLee - So sorry, I've updated the question to include the units of the constant. Hopefully it makes more sense now.
    – Donald Airey
    Jan 3 at 18:11










  • It still doesn't make mathematical sense, for the same reason I mentioned above. Maybe it would help if you explain exactly how this equation is supposed to define a surface.
    – Jack Lee
    Jan 3 at 18:13










  • Why doesn't it make sense? The $phi^2dt^4$ expression is a 2-tensors (that is, it has units of $km^2$, like the rest of the equation.)
    – Donald Airey
    Jan 3 at 18:16











  • You should reference the question which lead to this. Having that context makes answering this question (or pointing out mistakes in the way you formulated it) more likely.
    – MvG
    Jan 3 at 18:23
















  • I don't understand what your formula is supposed to mean. Usually, the notation $ds^2$ stands for a Riemannian metric, that is, a positive definite symmetric $2$-tensor field. But $dt^4$ is a $4$-tensor, so it doesn't make sense to add it to a bunch of $2$-tensors. And I have no idea how that formula is supposed to define a surface.
    – Jack Lee
    Jan 3 at 17:57










  • @JackLee - So sorry, I've updated the question to include the units of the constant. Hopefully it makes more sense now.
    – Donald Airey
    Jan 3 at 18:11










  • It still doesn't make mathematical sense, for the same reason I mentioned above. Maybe it would help if you explain exactly how this equation is supposed to define a surface.
    – Jack Lee
    Jan 3 at 18:13










  • Why doesn't it make sense? The $phi^2dt^4$ expression is a 2-tensors (that is, it has units of $km^2$, like the rest of the equation.)
    – Donald Airey
    Jan 3 at 18:16











  • You should reference the question which lead to this. Having that context makes answering this question (or pointing out mistakes in the way you formulated it) more likely.
    – MvG
    Jan 3 at 18:23















I don't understand what your formula is supposed to mean. Usually, the notation $ds^2$ stands for a Riemannian metric, that is, a positive definite symmetric $2$-tensor field. But $dt^4$ is a $4$-tensor, so it doesn't make sense to add it to a bunch of $2$-tensors. And I have no idea how that formula is supposed to define a surface.
– Jack Lee
Jan 3 at 17:57




I don't understand what your formula is supposed to mean. Usually, the notation $ds^2$ stands for a Riemannian metric, that is, a positive definite symmetric $2$-tensor field. But $dt^4$ is a $4$-tensor, so it doesn't make sense to add it to a bunch of $2$-tensors. And I have no idea how that formula is supposed to define a surface.
– Jack Lee
Jan 3 at 17:57












@JackLee - So sorry, I've updated the question to include the units of the constant. Hopefully it makes more sense now.
– Donald Airey
Jan 3 at 18:11




@JackLee - So sorry, I've updated the question to include the units of the constant. Hopefully it makes more sense now.
– Donald Airey
Jan 3 at 18:11












It still doesn't make mathematical sense, for the same reason I mentioned above. Maybe it would help if you explain exactly how this equation is supposed to define a surface.
– Jack Lee
Jan 3 at 18:13




It still doesn't make mathematical sense, for the same reason I mentioned above. Maybe it would help if you explain exactly how this equation is supposed to define a surface.
– Jack Lee
Jan 3 at 18:13












Why doesn't it make sense? The $phi^2dt^4$ expression is a 2-tensors (that is, it has units of $km^2$, like the rest of the equation.)
– Donald Airey
Jan 3 at 18:16





Why doesn't it make sense? The $phi^2dt^4$ expression is a 2-tensors (that is, it has units of $km^2$, like the rest of the equation.)
– Donald Airey
Jan 3 at 18:16













You should reference the question which lead to this. Having that context makes answering this question (or pointing out mistakes in the way you formulated it) more likely.
– MvG
Jan 3 at 18:23




You should reference the question which lead to this. Having that context makes answering this question (or pointing out mistakes in the way you formulated it) more likely.
– MvG
Jan 3 at 18:23















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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2590224%2fwhat-is-the-curvature-of-this-metric%23new-answer', 'question_page');

);

Post as a guest






















active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2590224%2fwhat-is-the-curvature-of-this-metric%23new-answer', 'question_page');

);

Post as a guest
































































-

這個網誌中的熱門文章

How to combine Bézier curves to a surface?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
閱讀完整內容

Mutual Information Always Non-negative

圖片
Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
閱讀完整內容

Why am i infinitely getting the same tweet with the Twitter Search API?

圖片
up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
閱讀完整內容