# Is there an orthogonal time dimension in Minkowski spacetime?

Space dimensions are orthogonal one to each other. But what about time in the Minkowski diagram?

At first sight, time seems to be orthogonal to space. But we have to consider that each Minkowski diagram is an observer's diagram. Thus, an object that is not moving with regard to the observer will describe a vertical worldline in the Minkowski diagram of the observer. However, from the point of view of another observer (inertial frame) the worldline described by the object might be inclined. Even the worldline of the observer himself which is the y-axis might be sloped from the point of view of other observers.

The conclusion seems to be that time is not orthogonal to space. Or is there an error?

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Time is orthogonal to space dimensions in drawings only. In real life, when you accelerate in your car its tires do not draw a curve on the asphalt. If the car leaves marks, they are just straight lines. Because time is not like other dimensions, and when you draw it as orthogonal it's only in order to visualize the change (of velocity for instance) better. – bright magus Jun 3 '14 at 19:09
These are just equations - no less no more. And time dilatation and distance contraction that result from them contradict the constancy of light. If Δx′=Δx/γ and Δt′=γΔt then c=x/t and c=x'/t' cannot be both correct. Because time and distance change under transforms in inverse proportions (if t'>t then x'<x), while the above requires that x/t=x'/t' which means direct proportions. These are the traps of mathematics applied to physics without real understanding of the problem. Therefore your equations prove nothing. – bright magus Jun 3 '14 at 19:22
@brightmagus: I do not agree with you that there is no orthogonality at all, because for the observer there is true orthogonality in his own Minkowki diagram. – Moonraker Jun 3 '14 at 19:33
@neuneck: What you are saying is that time is orthogonal from an observer's point of view, but not in an objective, universal sense. Each time when I am stating that time is orthogonal I must admit that this is a very relative statement, and that the application of any Lorentz boost may invalidate it. The rotation formulas you mentioned are characteristic for orthogonal systems, and rotations do not harm orthogonality. – Moonraker Jun 3 '14 at 19:34
@Moonraker Time is orthogonal from every observer's point of view. Observe that there is no $t'$ dependence in $x'$, even after the transformation. It is just that going from one observer to the other will mix time and space. – Neuneck Jun 3 '14 at 19:36

Time is orthogonal to space. Check that $(1, 0, 0, 0) \cdot (0, 1, 0, 0) = 0$ and likewise for all other spacelike unit vectors, where $\cdot$ represents the Lorentz invariant scalar product and I put time in the 0th coordinate.

The inclination originates in the Lorentz transformation that you use to go from one observer's point of view to another one's. Perfoming a boost creates an $x$ dependence in $t'$ and a $t$ dependence in $x'$: $$t' = \gamma \left(t - \frac{v}{c^2} x\right), x' = \gamma (x - v t)$$

This also happens in regular euclidean space, e.g. when perfoming a rotation about the $z$ axis, giving $$x' = \cos\theta \ x - \sin\theta\ y,\quad y' = \sin\theta\ x + \cos\theta\ y$$ So after a rotation a previousely vertical line will look inclined as well.

Edit: Observe that there is no t′ dependence in x′, even after the transformation. So, while time is always orthogonal to space, an object at rest for observer $O$ (say the observer's nose) will not be at rest for an observer $O'$ that moves away from observer $O$ at a speed $v \neq 0$.

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In your formulas t' and x' are depending on v, the relative velocity of the observer which is depending on t and x, and the 2 formulas of dependence you mention are not the same. By consequence, as long as v=0, there may be (and there is) orthogonality. But as soon as v >0 orthogonality disappears. – Moonraker Jun 3 '14 at 20:13
@Moonraker $\vec{t}=(1,0,0,0)$ is orthogonal to $\vec{x}=(0,1,0,0)$, and $\vec{t}'=(\gamma t,-\gamma v,0,0)$ is orthogonal to $\vec{x}'=(-\gamma v,\gamma,0,0)$. Sure $\vec{t}'$ is not orthogonal to $\vec{x}$, but that's not a particularly interesting comparison, since those are basis vectors drawn from different coordinate systems. – Chris White Jun 3 '14 at 20:25
@Chris White: I agree. But that means that orthogonality is relative. If an object is moving in 3D space any observer will agree that the 3D space the object is moving in is orthogonal. This absolute vision does apparently not exist in 4D-spacetime. – Moonraker Jun 3 '14 at 20:35
@Moonraker I think we're talking cross purposes. Realize that I could take normal Newtonian 3D space and describe it with a non-orthogonal coordinate system too. There is no sense in which Newtonian space is more orthogonal than Minkowski spacetime, because orthogonality is not a property of the vector space but rather it is a relation between two specific vectors. – Chris White Jun 3 '14 at 20:40
@Moonraker also note that Lorentz transformations will keep orthogonality between purely spacelike vectors intact (actually even between fully general four-vectors), i.e. if $x \cdot y = 0 \Leftrightarrow x' \cdot y' = 0$! Therefore "orthogonality" is observer independent, since if one observer finds two four-vectors to be orthogonal, all other observers will find them to be orthogonal in their own coordinate systems as well. – Neuneck Jun 3 '14 at 21:01

Any timelike vector has a 3-dimensional subspace orthogonal to it, orthogonal under the Minkowski inner product. Such a subspace is spanned by vectors that are spacelike under the inner product, and all linear combinations of those vectors are spacelike.

Any timelike vector might be the direction a massive object follows at some point on its worldline. The 3-dimensional subspace orthogonal to that four-velocity is what an observer following that trajectory would instantaneously consider "space".

An observer's "time" is always orthogonal to his "space", and further, all observers would agree that that is so.

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