Spacetime question

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Alan Grayson

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Sep 6, 2024, 4:03:25 AM9/6/24
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Do the 4-tuple labels on spacetime, (ct, x, y, z), form a vector space? I was told on good authority that the answer is negative, but now I have grave doubts of that conclusion. AG

smitra

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Sep 6, 2024, 8:26:14 AM9/6/24
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Yes, but then a vector space without a norm. No inner product is
defined, so from a pure math point of view, you are free to define any
arbitrary inner product that satisfies the axioms for it. Alan the
physicist will prefer that inner product that is dictated by the
relevant physics, so this has to take the form of the indefinite Lorentz
inner product in a locally free-falling coordinate system. This means
that you must also consider the metric, so the distance between
infinitesimally separated points. So, labels are arbitrary and hey are
then meaningless, unless you also specify what the distance is between a
point at [ct, x, y, z] and at [c(t+dt), x+dx, y+dy, z+dz].

Saibal
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Alan Grayson

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Sep 6, 2024, 8:34:26 AM9/6/24
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I think we can use the usual metric in relativity, ds^2, with the minus sign in from of dt^2. AG

Brent Meeker

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Sep 6, 2024, 2:51:23 PM9/6/24
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That's a representation of a vector in 4-space in a certain coordinate system.

Brent



On 9/6/2024 1:03 AM, Alan Grayson wrote:
Do the 4-tuple labels on spacetime, (ct, x, y, z), form a vector space? I was told on good authority that the answer is negative, but now I have grave doubts of that conclusion. AG --
You received this message because you are subscribed to the Google Groups "Everything List" group.
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Alan Grayson

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Sep 6, 2024, 3:26:13 PM9/6/24
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What exactly is your point? AG

Brent Meeker

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Sep 6, 2024, 3:53:52 PM9/6/24
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You mean this one?

Schwarzschild Metric

Brent

Alan Grayson

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Sep 6, 2024, 4:03:08 PM9/6/24
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I meant   ds^2 = -dt^2 + dx^2 + dy^2 + dz^2

Alan Grayson

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Sep 6, 2024, 4:05:37 PM9/6/24
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Correction:   ds^2 = -(cdt)^2 + dx^2 + dy^2 + dz^2

Brent Meeker

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Sep 6, 2024, 4:24:05 PM9/6/24
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It's not a "vector space", it's just a representation of one component of a vector space.  Something you could have learned by a quick internet search or picking up a text book, instead of trolling this email group.

Brent
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Alan Grayson

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Sep 6, 2024, 7:38:57 PM9/6/24
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The set of all such components satisfies the rules of a vector space, so it sure seems like a vector space! I seriously doubt that this specific question is covered in your huge reference. BTW, since when is asking a question "trolling"? Maybe you need to consult a dictionary? AG
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Alan Grayson

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Sep 7, 2024, 3:11:55 AM9/7/24
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I just googled "Is spacetime a vector space?", and the results were uniformly positive. So, I think you should be grateful for my "trolling", since it yielded a truth you should cherish! AG

Alan Grayson

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Sep 7, 2024, 6:47:09 PM9/7/24
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Was I "trolling" when I told you about the Tao? You might be among the third type of man mentioned, laughing when you have zero contact with the Tao. As for spacetime as a vector space, maybe you deny it in the context of a private language, which is often a problem when conversing with "experts". AG
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