Lack of independence between the spatial dimensions and time within space-time

Question

I am having a conceptual difficulty understanding the following issue regarding space-time: it is clear to me why a full description of coordinates requires three spatial dimensions plus time. However, Linear Algebra also teaches that dimensions needs to be independent from one other, such that my movement along one dimension has no bearing on my movement along another dimension. This does not seem to hold in space-time, given the effects of time dilation and space contraction, which is essentially a rescaling transformation of one dimension based on the object's velocity along another. This seems to preclude such independence between said dimensions. Can somebody clear up my confusion?

Answer 1

Linear Algebra also teaches that dimensions needs to be independent from one other, such that my movement along one dimension has no bearing on my movement along another dimension.

This is not correct, linear algebra does not teach this. What it actually says is that the different vectors that form a basis must be linearly independent from each other. That means that none of the basis vectors can be formed as a linear combination of the other basis vectors.

That is satisfied by spacetime. No amount of translation forward, left, or up can teanslate you to the future. Similarly, no combination of translation to the future, forward, or left can translate you up.

the effects of time dilation and space contraction, which is essentially a rescaling transformation of one dimension based on the object's velocity along another

Mathematically it is not a rescaling but a rotation. When you rotate ordinary spatial axes the projection of the new vectors onto the old ones can “contract” also.

Neither of the things that you mention really distinguish Minkowski spacetime from any ordinary vector space. The thing which does distinguish it is the fact that the distance between two distinct points can be zero.

Answer 2

First I’ll make a tangential remark:

it is clear to me why a full description of coordinates requires three spatial dimensions plus time.

Well actually you don’t need this, in the sense that for any $(1+n)$-dimensional Minkowski vector space $(V,g)$, I can always find a basis $\{\zeta_0,\dots,\zeta_n\}$ such that each $\zeta_i$ is a null-vector (i.e lightlike). The point about the $1$ time and $n$ space is that those vectors are in addition orthogonal with respect to $g$ (i.e a basis $\{e_0,\dots, e_n\}$ such that $g(e_a,e_b)=\text{diag}(-1,1,\dots, 1)$).

In everything that follows, I set speed of light to be $1$ for convenience.

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I think you may benefit from a more linear-algebraically precise description of the phenomena of time-dilation and length contraction. These phenomena are less about linear independence/dependence than about the difference in descriptions corresponding to projections relative to different direct sum decompositions of the same vector space $V$. What follows is a more detailed explanation.

First let’s just recall some basic linear algebra. Given any vector space $V$ over a field $\Bbb{F}$ (imagine $\Bbb{R}$ if that’s more comfortable), and any two subspaces $T,S\subset V$, we write $V=T\oplus S$ if

• $T\cap S=\{0\}$
• $T+S=V$.

Another equivalent way of saying these two conditions is that for every vector $v\in V$, there must exist unique vectors $t\in T$ and $s\in S$ such that $v=t+s$ (existence of $t,s$ is equivalent to the second bullet point, and uniqueness is equivalent to the first bullet point). Yet another way of saying this is that the mapping $\Phi:T\times S\to V$, $\Phi(t,s)=t+s$ is required to be a linear isomorphism (due to linearity, injectivity is equivalent to the first bullet point, while surjectivity is equivalent to the second bullet point).

We say that $V=T\oplus S$ is a direct sum decomposition of $V$ with (direct) summands $T$ and $S$. Now, given such a decomposition, we can define two projection maps, $P_T:V\to T$ and $P_S:V\to S$ such that for each $v\in V$, we have that $P_T(v)\in T$ and $P_S(v)\in S$ are the unique vectors such that $v=P_T(v)+P_S(v)$ (we know they exist and are unique because of the definition of direct sums described above). So, the point is that with a direct sum decomposition $V=T\oplus S$, we can uniquely decompose a vector $v\in V$ into two pieces $t+s$ lying in the respective subspaces.

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Ok so how do direct sums relate to physics and special relativity (for GR, you repeat everything I’m about to say at each tangent space)? Well, an inertial observer $\mathcal{O}$ simply amounts to a choice of a (future-directed) unit timelike vector $t_0$, i.e a vector $t_0\in V$ such that $g(t_0,t_0)=-1$. There are infinitely many such vectors. Given such a timelike vector $t_0$, you can consider the 1-dimensional subspace $T=\text{span}\{t_0\}$. From this 1-dimensional subspace, you can consider the $g$-orthogonal complement $S:=T^{\perp}=\{v\in V\,:\, g(v,t_0)=0\}$. You can prove that $S$ is $n$-dimensional and $V=T\oplus S$ and that every non-zero vector in $S$ is spacelike.

In more pedestrian terms, what this means is that given an observer $\mathcal{O}$ (i.e a choice of a unit timelike vector $t_0$; so $g(t_0,t_0)=-1$) one can decompose the Minkowski spacetime as $V=T\oplus S$ into a “purely temporal part”, $T$, and a “purely spatial part”, $S$; but keep in mind that this is all relative to the given observer $\mathcal{O}$. So, for example, for any real number $a\in\Bbb{R}$, the vector (event) $at_0\in T$ is what the particular observer $\mathcal{O}$ would describe as “the event which occurs after $a$ units of purely moving through time”. In particular the event $t_0$ is described as “one second afterwards”. An event $s\in S$ is what the observer $\mathcal{O}$ would describe as being a “purely spatial displacement $s$ away”. And more generally, given any vector $v\in V$ (i.e a general event) we can decompose it uniquely as $v=P_T(v)+P_S(v)=at_0+P_S(v)$ for some unique number $a\in\Bbb{R}$ (because $T$ is 1-dimensional). The observer $\mathcal{O}$ would then say that this event occurs “$a$ units of time later, and a spatial separation of $s=P_S(v)$ away from my origin”.

Now, suppose we introduce a new inertial observer $\tilde{\mathcal{O}}$ to the game. This observer can be described by a unit timelike vector $\tilde{t}_0\in V$. This observer then creates their own direct sum decomposition $V=\tilde{T}\oplus (\tilde{T})^{\perp}\equiv \tilde{T}\oplus \tilde{S}$.

Now, we shall attempt to relate the two descriptions. Consider the vector (event) $\tilde{t}_0$. Observer $\tilde{\mathcal{O}}$ would call this event as “one second afterwards”. However, for our original observer $\mathcal{O}$, he would first decompose $\tilde{t}_0=P_{T}(\tilde{t}_0)+P_{S}(\tilde{t}_0)=at_0+P_{S}(\tilde{t}_0)\in T\oplus S$ for some unique number $a\in\Bbb{R}$. So, observer $\mathcal{O}$ would describe the event $\tilde{t}_0$ as “the event which occurs after $a$ units of time and a spatial displacement $s=P_{S}(\tilde{t}_0)$ away from my origin”. So, in general, $a\neq 1$, which means that although we’re talking about one and the same event (namely $\tilde{t}_0$) the observer $\mathcal{O}$ regards this as “one second later”, whereas the original observer $\mathcal{O}$ regards this as “$a$ seconds later (and a certain spatial displacement away)”. It is this discrepancy which is at the heart of the time-dilation phenomenon. Said another way: we’re taking the same event, and projecting it onto two different sets of axes corresponding to the two observers. So of course there’s no reason for the respective projections to be equal. In this light, time-dilation and length contraction aren’t really that fancy (we’re taking the same thing and “chopping it up differently”).

For the sake of completeness I’ll also define the concept of relative velocity in this language. With notation as above, (and assuming we actually have different observers, i.e $t_0\neq \tilde{t}_0$) given the decomposition $\tilde{t}_0=at_0+s$ for some unique $a\in\Bbb{R}$ and $s\in S$, we define the vector $v_0:=\frac{s}{a}\in S$ to be the (spatial, $3$-) velocity of observer $\tilde{\mathcal{O}}$ relative to observer $\mathcal{O}$.

One can then show (but I won’t here) that $a=\frac{1}{\sqrt{1-g(v_0,v_0)}}$, i.e $a=\frac{1}{\sqrt{1-\|v_0\|^2}}>1$, which as you can recognize, is exactly the time-dilation factor. You can describe length contraction similarly. So, the tldr is that you’re taking the same vector space $V$ and considering two different direct sum decompositions $T\oplus S$ vs $\tilde{T}\oplus\tilde{S}$, and “translating” between the two descriptions; what counts as “purely temporal movement” for one observer becomes “time and space movement” for a different observer. But keep in mind that the notions of time and space for both observers separately are independent (this is the meaning of the first bullet point in the direct sum definition), but when you try to relate one observer to another, that’s when things get mixed up. Again, when you think of “chopping things up differently”, it doesn’t sound as complicated.

Answer 3

This answer doesn't approach the existing one in rigor and detail, but at a conceptual level:

In 3D geometry and vector analysis, you have a coordinate basis, say $(x, y, z)$, and you can choose a new coordinate basis by rotating this to another direction $(x', y', z')$. Then the equations of linear algebra tell you how you can write the coordinates of a vector $\vec r$ in terms of $(x, y, z)$ or convert it to $(x', y', z')$. And the $x'$ component will be a linear combination of $ax+by+cz$.

The Lorentz transformations of special relativity are exactly the same thing, except in 4D and with hyperbolic instead of Cartesian coordinates. An object's proper time $\tau$ or proper length $l$ is (the magnitude of) an invariant vector like $\vec r$ in 3D space. The $(t,x,y,z)$ and $(t',x',y',z')$ are the components of different coordinate bases that correspond to rotations of an orthogonal coordinate system.