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Is there a notion of rotating an object about its time axis? I'm not sure if this question totally makes sense, but it seems intuitive to me that an object with dimensions in the three spatial directions and the time dimension (so an object paired with an interval of time) could be rotated about the time axis, just as it could be rotated about its "$x$-axis", "$y$-axis", or "$z$-axis". Is this intuition misguided? If it is possible, what would it look like to perform such a rotation?

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    $\begingroup$ I don't think there are objects per se in a spacetime diagram. There are events, which only exist at one instant in time, and there are world lines of objects. A world line represents the entire history of some object, past and future. $\endgroup$ Commented May 2, 2023 at 13:02
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    $\begingroup$ Imagining a 2D spatial case and time as a third dimension with time simply slotted in as the 'missing' third dimension. Are spatial dimensions defined as being normal to the time axis, or are they intrinsically so and my brain is too small to see that? $\endgroup$ Commented May 3, 2023 at 22:14
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    $\begingroup$ Regardless of physicality, I got curious what it'd look like, so here's a Python script that swaps the Y and time axis of a given GIF. And here's the album with a couple of very simple GIFs and their time rotated versions. $\endgroup$
    – BoppreH
    Commented May 5, 2023 at 17:42

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This a great question, and leads to some interesting ideas.

Firstly, the notion of a "rotation axis" is restricted to three dimensions. In more than three dimensions the rotation axis concept has to be replaced by a specification of the plane in which the rotation occurs. In 3d a rotation about the $z$ axis is the same thing as a rotation in the $x$-$y$ plane. The $z$ axis is chosen as "the axis" because it is the unique direction perpendicular to the $x$-$y$ plane. In higher dimensions there is more than one perpendicular direction and so the "axis" notion is no longer useful, but the choice of plane still works.

Secondly, once we have four dimensions, one of which is time, there is another complication: the new 4d space is not Euclidean but rather is "Minkowskian". A Minkowski-space "rotation" in the $t$-$x$ plane is same thing as a Lorentz boost in the $x$ direction. It's not a normal rotation because of the minus sign in the Minkowski metric. This sign changes the $\sin \theta$ and $\cos \theta$ of the usual rotation into $\pm\sinh s$ and $\cosh s$ where $s$, called the rapidity, is the analogue of the rotation angle $\theta$.

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    $\begingroup$ Would it be accurate to say that rotating in the $t$-$x$ plane amounts to changing the velocity of the object along the $x$ axis? $\endgroup$ Commented May 3, 2023 at 16:46
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    $\begingroup$ @TannerSwett: That’s a fine intuition. The Lorentz boost is the change of coordinates corresponding to a difference in (coordinate) velocity between a reference and an observer. As this answer points out, it’s not a rotation in the same way as a purely spatial plane—in effect, you can change the slope of the time axis, relative to the original frame, but the parameters are constrained by the speed of light: no degree of hyperbolic rotation can “flip” an object in spacetime, only increasingly contract lengths & dilate times (by the same factor). $\endgroup$
    – Jon Purdy
    Commented May 3, 2023 at 19:46
  • $\begingroup$ So, in keeping with OP's final question If it is possible, what would it look like to perform such a rotation?: if some aliens with advanced timespace manipulation technology could indeed perform such a Minkowski-space "rotation" within the $t-x$ plane, what would be the visible effect to an observer? $\endgroup$
    – AnoE
    Commented May 4, 2023 at 13:59
  • $\begingroup$ @AnoE The object would be boosted to a velocity $v= c \tanh s$. $\endgroup$
    – mike stone
    Commented May 4, 2023 at 14:07
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    $\begingroup$ So literally just an acceleration, in everyday terms? @mikestone $\endgroup$
    – AnoE
    Commented May 4, 2023 at 14:22
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Is there a notion of rotating an object about its time axis? I'm not sure if this question totally makes sense, but it seems intuitive to me that an object with dimensions in the three spatial directions and the time direction (so an object paired with an interval of time) could be rotated about the time axis, just as it could be rotated about its "x-axis", "y-axis", or "z-axis". Is this intuition misguided?

The idea of "rotating" implies that at one time the object has some orientation with respect to the spatial coordinate axes and at another time the object has some different orientation with respect to the spatial coordinate axes. So there is a before rotation and an after rotation.

Spacetime has all time, all future and all past. So there cannot be a before rotation or an after rotation in spacetime because there is no before or after, all time is already included in spacetime. An object is simply one unified 4D shape in spacetime embedded in a 4D manifold.

Instead, in spacetime a spatial rotation is a "twist" along the length of the object. The orientation of the cross-section with respect to the spatial axes is different at different points along the time dimension. This is similar to the sculpture below, where the vertical direction represents the time axis.

Twisted wood sculpture by Michael Semsch.

So similarly, the orientation of the cross-section with respect to the time axis could be different at different points along the time dimension. This would be a bend in the spacetime shape and would correspond to an acceleration of the object.

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    $\begingroup$ I think this answer nicely illustrates the spacetime representation of a physical motion--a continuous rotation of an object in 3D space that occurs over a finite span of time. But I'm not sure that that was what the OP was asking about. Seems like the OP was asking about a transform in 4D spacetime that... Well, I'm not sure I even know the right words to describe it, never mind the math. But, it seems like Mike Stone's answer comes closer to what OP was asking. $\endgroup$ Commented May 2, 2023 at 14:13
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    $\begingroup$ Indeed, I attempted to find the words to ask about what it would be like to take a 3-dimensional (or possibly 4-dimensional) subspace of spacetime and rotate it about the time axis, which Mike Stone addressed. This is an interesting perspective nonetheless. $\endgroup$
    – lanerogers
    Commented May 3, 2023 at 14:48
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There can be such a notion, and it ties in with the notion of curvature of spacetime. Curvature of spacetime is measured by parallel transport of a 4 vector around a small square. If spacetime is curved, the square doesn't come back to the same point and the 4 vector changes direction and/or magnitude.

You can see the same thing in a curved 2D space, such as the surface of the Earth. Consider a square with sides 1 m long oriented parallel to meridian and longitude lines. Start with a 2D vector at the southeast corner pointing north. Move it north, then west, then south, then east back to the start. The change is too small to see. The surface of the earth is very nearly flat.

Try again with a smaller sphere, where the distance from the equator to the north pole is 1 m. The square now looks different. It goes from the equator north to the north pole. It makes a 90$^o$ turn and goes south to the equator, then east along the equator, then north back to the north pole. If you follow the direction of the vector, you see that it has rotated 90$^o$ in its journey.

You can make similar loops in spacetime. Since you cannot travel backward in time, it is more convenient to go halfway around the square in opposite directions. The square can have sides parallel to the 3 space axes and/or the time axis. The sides $x$ and $t$ are of equal length if $x = ct$.

General Relativity tells us that mass curves spacetime. Time travels more slowly in a gravitational well than in space far from a planet. See Why can't I do this to get infinite energy?.

Let us choose a square that starts above a black hole. One side is spatial, toward the center. The next is time - you just wait a bit. For one half-trip, you wait $t$ and then find a point $x$ below you. For the other, you find the point $x$ below you, and wait there. Since time travels more slowly at the lower point, you arrive at two different events.

To see how a vector changes, you define a vector by placing a rock at the start and seeing how far it moves in a short time. The rock both the time and space component of the vector are different if you wait then move vs move then wait.

See this for more on visualizing curvature of spacetime - A new way to visualize General Relativity

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  • $\begingroup$ Re, "...cannot travel backward in time...The square can have sides parallel to the 3 space axes and/or the time axis." If you are proposing a physical experiment to actually measure the curvature of spacetime at some particular place and time, then it won't do to have any corners of the square separated by space-like intervals. That pretty well rules out having any of its sides be parallel to a pure-space axis. $\endgroup$ Commented May 2, 2023 at 23:10
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Your intuition is interesting, but rotating an object about the time axis isn't quite analogous to rotating an object in spatial dimensions. In special relativity, however, there is a concept that somewhat resembles your idea called Lorentz transformation. Lorentz transformations can be seen as "rotations" in a four-dimensional spacetime, which includes three spatial dimensions and one time dimension.

In Minkowski spacetime, the time and spatial coordinates are combined into a four-vector called the spacetime interval. A Lorentz transformation relates the coordinates of an event in one inertial frame to the coordinates of the same event in another inertial frame, which is moving with a constant velocity relative to the first frame. This transformation mixes the time and spatial coordinates, similar to how rotations mix the spatial coordinates in three-dimensional Euclidean space.

For a Lorentz transformation along the $x$-axis with a relative velocity $v$, the transformation equations are:

\begin{align} t' &= \gamma \left( t - \frac{v x}{c^2} \right) \\ x' &= \gamma \left( x - vt \right) \\ y' &= y \\ z' &= z \end{align}

where $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$ is the Lorentz factor, $c$ is the speed of light, and $(t, x, y, z)$ and $(t', x', y', z')$ are the spacetime coordinates in the unprimed and primed frames, respectively.

Although Lorentz transformations can be seen as a kind of rotation in spacetime, they don't have the same geometric interpretation as spatial rotations. They involve hyperbolic functions, rather than the trigonometric functions used in spatial rotations, and their effects are not visually intuitive in the same way.

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