Put into more words, to my knowledge we see 3 dimensions of space moving through a 4th dimension of time. My thought then is that if there was a hypothetical creature/device that could see time via replacing one of the spatial axes with time, let´s say X. So essentially instead of seeing the X, Y, Z coordinates through all the unique time T coordinates, they´d see all the T, Y, Z coordinates through a string of X coordinates. The kind of foundation of this question is then would the observed physics differ between the two observers due to time having some distinction from space I don´t fully understand or would they observe the same physics implying that time is only significant due to our biology assigning it significance. For reference I understand that they won´t exactly see the same thing as particles would then be placed in their different spatial and time coordinates, I´m just curious if the physics would be the same, kind of like a principle of relativity but for observing unique dimensions, if I swap the X and Y axes on a graph the laws of physics still apply, is the same true for swapping the T and X axes?
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2It's a bit difficult to answer without being a bit more specific about your question. For example, what if I decided to just start calling my X variable "time", and my T variable "first space dimension" (as in, I relabel my axes, but don't change any of my physics models). This would be a silly relabeling, but by construction nothing 'physical' changes, as long as I remember the relabeling procedure. – QCD_IS_GOOD Mar 14 '23 at 02:56
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1For some background info on this see What is time, does it flow, and if so what defines its direction? – John Rennie Mar 14 '23 at 06:48
1 Answers
Time is geometrically distinct from space. The metric (in units where c=1) in a local inertial frame is $$ds^2=-dt^2+dx^2+dy^2+dz^2$$ Notice that time has the opposite sign as space. Time is also distinguished by the fact that there is only one time dimension while there are three space dimensions. Those two facts together accounts for all of the differences between time and space that are not related to entropy/thermodynamics.
EDIT: So what physical features does this $ds^2$ encode?
First, if $ds^2<0$ then the interval is timelike, meaning that it will be physically measured with a clock. Conversely, if $0<ds^2$ then the interval is spacelike, meaning that it will be physically measured with a ruler.
Second, because there are multiple positive-signature coordinates it is possible to make a closed loop in space. You can draw a circle at an instant of time that loops back on itself and is everywhere spacelike (measured with a ruler not a clock).
Third, because there is a single negative-signature coordinate it is not possible to make a closed loop in time. Any closed loop will necessarily be spacelike somewhere.
Fourth, related to the third, there is a geometric distinction between future and past. You can rotate a "left" vector into a "right" vector through a series of small rotations, but you cannot rotate a "future" vector into a "past" vector through a series of small rotations. Surfaces defined by a constant $ds^2$ form hyperboloids of one sheet for $0<ds^2$ and hyperboloids of two sheets (future and past) for $ds^2<0$.
Finally, the invariance of $ds^2$ directly encodes Einstein's 2nd postulate. For $ds^2=0$ we get $$dt^2=dx^2+dy^2+dz^2$$ which is the equation of a sphere whose radius is expanding at a speed of $1=c$.
So, what you can do is to swap your coordinate labels so that you have $$ds^2= dt^2 - dx^2 + dy^2 + dz^2$$ Then $x$ is time and $t$ is one of the three spatial dimensions. Intervals in $x$ are measured with clocks and there is a future $x$ and a past $x$ direction. Physics doesn't care about our coordinate labels.
What you cannot do is to change the sign so that all are positive $$ds^2 \ne dt^2+dx^2+dy^2 + dz^2$$ This would mean that you have only space and no time. There is no future or past and no quantity is measured with a clock.
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I understand why c is included as it translates the units from seconds to meters, but out of curiosity why is it negative, if I´m not mistaken the spacetime interval is simply calculating the distance between two points which adds all of the squares of every positional value together. Because of this I´m curious as to why time is denoted as subtracted from the total rather than added. – JohnIsBueno Mar 14 '23 at 02:42
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1The c is only needed if the units are such that c is not equal to 1. By scaling your units so that c = 1, you can always remove it from consideration (note that this puts all achievable speeds between 0 and 1, which is $\beta$ in Lorentz factor formulas). The metric Dale put above is the metric of Minkowski space. I agree with Dale's conclusion that it makes space and time geometrically distinct. Phrases like "space and time are the same in relativity" are always found in pop science but they are wrong. The metric gives space and time opposite signs (either -+++ as above or +---). – Poisson Aerohead Mar 14 '23 at 06:17
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@JohnIsBueno : A beam of light travels (according to me) at speed 1, covering a distance dx in time dt. Therefore dt = dx. According to relativity, you (using different coordinates) must agree that the speed is 1.In other words, whenever dx=dt, we must also have dx'=dt', where x', t' is an alternative set of allowable coordinates. The easiest way to guarantee this is to require that all allowable coordinate transformations preserve the value of $dt^2-dx^2$ (so that if I get zero for that quantity, so must you). – WillO Mar 14 '23 at 06:36
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@JohnIsBueno I made a pretty big modification to the answer that I hope clarifies sufficiently – Dale Mar 14 '23 at 13:51
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Your edit should be in every book on relativity. Very compact and clear. – StephenG - Help Ukraine Mar 14 '23 at 16:16
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