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I've been trying to get to grips with SpaceTime.

As I understand it, we move at a set rate through spacetime. Any increase in our rate of travel through space results in a decrease in our rate of travel through time (via standard vector maths, pythagoras)

Where I get confused is the implication for our perception of time. Do we perceive time as the magnitude of our movement through spacetime, and the speed of someone stationary as the 'time' component of our vector through spacetime? Wouldn't this mean that, as a diagonal line is longer than a straight one of the same height, that things around us should slow down as we move more quickly?

enter image description here

Say this triangle is our vector in Spacetime. If $a$ is the 'time' component of spacetime, and $b$ is the 'space' component, does that mean that the $c$ component is of fixed magnitude? Do we perceive things around us moving at a rate of $a/c$? (ie, slower than they occur now)

I know other questions about spacetime have been asked, but I can't find one that explains this aspect.

Qmechanic
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Witnaaay
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    The relation in spacetime isn't the pythagorean one. it is instead: $$ds^{s} = - dt^{2} + dx^{2} + dy^{2} + dz^{2} $$ The minus sign changes a lot. – Zo the Relativist Jan 29 '15 at 06:37
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    @JerrySchirmer "The minus sign changes a lot." (LOL at your understatement!). I like to think of it as shattering the transitivity of closeness asserted by the triangle inequality: this lack of transitivity is what gives the twin paradox legs (see the Bert Mendelson quote in my answer) – Selene Routley Jan 29 '15 at 07:38

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You've hit on one of the fundamental "weirdnesses" of the Minkowski "metric". Its name "metric" is a bit misleading if you're a mathematician: it does not fulfil two out of the of the three axioms of the distance function defining a metric (in the topologist's words) space. There are null vectors, i.e. nonzero vectors $X$ for which $\langle X,\,X\rangle=0$ and it is NOT subadditive i.e. it does not fulfill the triangle identity. So the intuition you have from Euclidean (and general metric) space that the sum of the lengths of two of a triangle's sides must be longer than the third alone is WRONG!

The names "metric", "inner product" and "norm" in connexion with Minkowski spacetime reflects (1) the "structural" likeness that these operations as algebraic operations have to genuine metrics, inner products and norms (written on a page, they look a great deal like the genuine ones) and (2) the fact that many of the main theorems of Riemannian geometry also hold when we replace genuine inner products with nondegenerate ones (i.e. the matrix representing the two form is nonsingular) such as the Minkowski "inner product".

I recall many years ago reading Bert Mendelson in his wonderful introductory undergrad textbook "Topology" making the memorable and intuitive statement:

".... the [triangle] inequality $\mathrm{d}(x,\,z)\leq \mathrm{d}(x,\,y) + \mathrm{d}(y,\,z)$ may be thought of as asserting the transitivity of [the] closeness [relationship]; that is, if $x$ is close to $y$ and $y$ is close to $z$, then $z$ is close to $x$." (Chapter 2, Section 2, p 31 in my edition)

This wonderful statement - this transitivity - is precisely what is shattered in the twin paradox. The spacefaring twin at the halfway point ($y$) in his journey can be close to his beginning point ($x$) (he ages very little - the path has only short proper time), and at the end of his journey at ($z$) he can be near to his halfway point (again, he ages very little coming back), but ($z$) in spacetime is far, far away from ($x$): the Earthbound twin is now a decrepit, bent up old man.

PS if you study topology and they try to get you to read Munkres, I highly recommend Bert Mendelson's book to be read together with the heavier Munkres.

Selene Routley
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  • Ah, so its non-Euclidean Space. Thats frustrating.

    Okay, I'll have to do some reading on the equivalent vector formula in Minkowski Space.

     – Witnaaay Jan 29 '15 at 06:46
  • @Witnaaay Frustrating, but interesting. It is this property that allows for the universal speed (observed to be the same for all observers) $c$. Also, some people make a big deal out of "imaginary" time so that the "metric" looks Euclidean: this has its uses in some cases (it's called Wick rotation) but in SR and classical GTR it's simply unhelpful and misleading. The co-ordinates are real, so that so-called signature (the fact that there is one minus sign) is nontrivial and cannot be spirited away by co-ordinate transformations. Once you add the extra dimensions in complexification, it ... – Selene Routley Jan 29 '15 at 06:56
  • ... signature, which has physical consequences, loses its meaning: you can spirit signature away in complexified spaces. – Selene Routley Jan 29 '15 at 06:57
  • @Witnaaay See also the wonderful Bert Mendelson quote that I read many years ago: it sheds light on the twin paradox. – Selene Routley Jan 29 '15 at 07:14
  • It's sort of like you swap c for a in the Pythagorean relationship. I like your point on transitivity - you have to wonder about the usefulness of philosophy when such obvious concepts prove to be incomplete.

    Thanks for the answer

     – Witnaaay Jan 29 '15 at 08:15
  • @Witnaaay It's not philosophy: the metric axioms (zero distance between two points implies the points are the same, symmetry of distance and subadditivity) hold in some mathematical systems and not others. They belong, if you like, to the taxonomy of mathematical systems and geometries. – Selene Routley Jan 29 '15 at 08:52
  • Actually you're right that Minkowski "metric" is not a metric in the language of metric spaces. But it's a pseudo-Riemannian metric. I was wondering about this some time ago when first read the definition of metric space, see my question on Math.SE. – Ruslan Jan 29 '15 at 13:00
  • WetSavannaAnimal aka Rod Vance:"[...] Minkowski "metric". Its name "metric" is a bit misleading [...]: it does not fulfil [...]" -- non-negativity, pos. definiteness (identity of indiscernibles), and subadditivity (triangle inequality). Right. So: How do mathematicians call such general "functions of pairs of a set" in the literature ??; without requiring "smoothness", or "flatness", but possibly: $$\forall x,y,z:0\gt s^2[~x,y~]\ge s^2[~y,z~]\gt s^2[~x,z~]\implies$$ $$\sqrt{-s^2[~x,y~]}+\sqrt{-s^2[~y,z~]}\le\sqrt{-s^2[~x,z~]}$$. – user12262 Jan 29 '15 at 21:58
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    @user12262 Well they are all billinear functionals of a pair of members of the set or billinear forms Such objects can be degenerate (see Wiki "degenerate billinear form". Names pseudo/Minkowski-metric, pseudo/Minkowski norm and pseudo-Riemannian manifold specifically betoken nontrivially signatured forms. As I said, this is useful because many of the big theorems of Riemannian geometry also hold if the tangent bundle is kitted with a general (i.e. possibly degenerate) billinear form. For instance .... – Selene Routley Jan 29 '15 at 23:51
  • @user12262 ..., this one, which lets us choose the unique Levi-Civita connexion to absorb the torsion into the curvature. I'm not criticising the usage: it's perfectly OK within GR and SR because, after a while everyone knows what is being spoken of, so no-one is going to make a mistake. It's just that it grates a bit at first to see a "metric" stripped of most of the important properties topologists understand by the word and still being called such. The OP, just like I did, tripped up on meeting it for this reason. – Selene Routley Jan 29 '15 at 23:56
  • WetSavannaAnimal aka Rod Vance: "it grates a bit at first to see a "metric" stripped" -- Yes, this exactly reflects my experience and present state of understanding: Looking at the "Wikipedia table of available generalizations" and missing the appropriate entry "xyz-metric space". "Well they are all billinear functionals of a pair of members of the set or billinear forms" -- I'm still skeptical about the hint of "linear" in the name ... http://en.wikipedia.org/wiki/Degenerate_form -- Cool. News to me. Thx. – user12262 Jan 30 '15 at 00:29
  • WetSavannaAnimal aka Rod Vance: p.s. Well, since that what I've been specificly asking about (i.e. including the inequality above), and what you've more or less been getting at, too, is apparently not yet listed in the Wikipedia table of available generalizations of the "metric space" notion, may I suggest (after a quick check with Google, and with a nod to J. L. Synge): "chrono-metric space". – user12262 Jan 30 '15 at 06:36
  • p.p.s. Once more for completeness: May I suggest to call the pair $$(X, s) \text{ a } {\mathbf{ \text{ chrono-metric space }}}$$ if $X$ is a set and $s$ is a map $s : X \times X \rightarrow \mathbb R$ such that $$ s[~x, x~] = 0 \quad {\text{ (indiscernability of the identical)}} $$ and $$\forall x, y, z \in X : 0 \ge s[~x, y~] \ge s[~y, z~] \ge s[~x, z~] \implies s[~x, y~] + s[~y, z~] \ge s[~x, z~] \quad {\text{ (general chronology).}}$$ – user12262 Jan 30 '15 at 07:06
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Any increase in our rate of travel through space results in a decrease in our rate of travel through time (via standard vector maths, pythagoras)

Movement through space is not absolute; there are an infinity of relatively moving reference frames from which your 'mixture' of (rate of) space and time displacement differs.

Do we perceive time as the magnitude of our movement through spacetime, and the speed of someone stationary as the 'time' component of our vector through spacetime?

You are at rest with respect to yourself; in this reference frame, your are stationary and others are moving relative to you.

The fact that relatively moving others see you as moving through space does not affect your perception of time.

Alfred Centauri
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