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I often read from textbooks that in relativity, space and time are treated on an equal footing. What do authors mean when they say this?

Are there any examples that show space and time are treated on an equal footing? Conversely, what examples show that space and time are not treated on an equal footing?

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Putting space and time on the same footing means to treat time as another dimension in addition to the other three physical dimensions. In the context of relativity, time is treated as another dimension (but within this idea of Spacetime space and time are not the same).

In classical Newtonian physics, space is treated within the ideas of three dimensional space. In this approach, time is absolute, as oppose to relativity.

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    $\begingroup$ Can I understand it as in Newtonian physics, when transforming coordinates, only the space coordinates are transformed while time coordinate is not (i.e. is absolute). While in relativity, when transforming coordinates, both space and time coordinates are transformed "equally" using Lorentz transformations. $\endgroup$ – TaeNyFan Nov 13 '20 at 8:32
  • $\begingroup$ Correct. Well put. $\endgroup$ – Dr jh Nov 13 '20 at 8:50
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After some thought, this is what I understand:

In Newtonian physics, a particle's path can be specified by $x^i(t)$ where the time $t$ can be seen as an independent parameter. The space coordinates $x^i(t)$ are dependent variables that depend on $t$. We thus say that space and time are not treated on an equal footing.

In relativity, a particle's worldline is specified by $x^\mu(\lambda)$ where $\lambda$ is an independent parameter (often taken as the particle's proper time). Both space and time coordinates $x^\mu(t)$ are dependent variables that depend on $\lambda$. We thus say space and time are treated on an equal footing.

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  • $\begingroup$ This isn’t a relevant distinction. For example, when solving the two-body problem in Newtonian gravity, it is common to parameterize the solution so that $x$, $y$, and $t$ are all functions of a parameter along the orbit called the eccentric anomaly. (See Wikipedia.) This doesn’t make anything about the problem relativistic. It just makes it analytically solvable. $\endgroup$ – G. Smith Nov 13 '20 at 18:10
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    $\begingroup$ Therefore this answer is not helpful in understanding the unification of space and time into spacetime, despite its many upvotes. $\endgroup$ – G. Smith Nov 13 '20 at 18:14
  • $\begingroup$ I think this answer is just a rephrase of Dr jh's answer. It just means that time is treated as another ordinary dimension in relativity, instead of being a the absolute time in Newtonian physics. $\endgroup$ – TaeNyFan Nov 14 '20 at 8:40
  • $\begingroup$ But I can see what you mean. This definitely is a flawed description for the unification of spacetime. But I find that it helps a bit in understanding why time is no longer so "special" in relativity. $\endgroup$ – TaeNyFan Nov 14 '20 at 8:42
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I often read from textbooks that in relativity, space and time are treated on an equal footing. What do authors mean when they say this?

I actually give a brilliant aid to understand what does it mean? It's called surveyors parable introduced by Tayloe and Wheeler. Suppose a town has daytime surveyors, who have the North star. These notions differ, of course, since the magnetic north is not the direction to the North pole. Suppose, further that both groups measure north/south distances in miles and east/west distances in meters, with both being measured from the town center. How does one go about comparing the measurements of the two groups?

With our knowledge of Euclidean geometry, we see how to do this: convert miles to meters (or vice versa). Distances computed with the Pythagorean theorem do not depend on which group does the surveying. Finally, it is easily seen that the 'daytime' coordinate can be obtained from the 'nighttime' coordinate by a simple rotation. The geometry of this situation is therefore described, in which the $x$ and $y$ directions correspond to geographic east and north, respectively, and the $x'$ and $y'$ directions correspond to the magnetic east and north, respectively. If the surveyors measure $x$ and $x'$ in meters, and $y$ and $y'$ in miles, and if they do not understand how to convert between the two, communication between the two groups will not be easy.

Applying the lesson of special relativity, we should measure both time and space in the same units. How do we measure distance in second? That's easy: simply multiply by $c$. This has the effect of setting $c=1$ since the number of seconds traveled by light in 1 sec is precisely 1.

Are there any examples that show space and time are treated on an equal footing? Conversely, what examples show that space and time are not treated on an equal footing?

In Newtonian mechanics, We never care about time, it's flowing at a constant speed. At the same rate for all the observers in the universe.

While If you look at special relativity you always find that time is one of the coordinates just like the other 3 coordinates are. Like the following

According to Newtonian mechanics, If a rod is moving with some speed, all observers in the inertial frame will agree on the length of the rod. But If you work out the space-time diagram for the moving rod. It will look like :

enter image description here

As you can see for the moving observer the length of the rod gets contracted. (It needs some elementary knowledge to appreciate it). Drawing the same for Newtonian nothing will change. As time and space, axis remains parallel to rest obersevers frame.

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  • $\begingroup$ "Suppose a town has daytime surveyors, who have the North star. These notions differ" I think you might want to review this. It's not really sensible for at least two reasons. $\endgroup$ – JimmyJames Nov 13 '20 at 16:54
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    $\begingroup$ I think a section was accidentally deleted. I think it should be: "Suppose a town has daytime surveyors, who have compasses they can see in the sun, and nighttime surveyors, who have the north star." $\endgroup$ – user253751 Nov 13 '20 at 16:55
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In Special Relativity, there is the invariant interval defined as $$\Delta s^2=c^2\Delta t^2-\Delta x^2$$ (for relative motion in the x-direction only). Here $\Delta t$ and $\Delta x$ are the difference in t and x for two events in some frame of reference. It has the same value in any other inertial reference frame using that frame's coordinates t' and x' to describe the same two events.

As t and x both appear in the equation in a similar fashion, one might say that time and space are being treated on an equal footing.

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I think the short answer is that Special Relativity doesn't "treat time and space on equal footing". In Newtonian mechanics, Galilean relativity is assumed. To switch between two different reference frames you simply have to rotate/shift your spatial coordinates.

In Special Relativity, the transformation between reference frames is more complicated. Firstly, the transformation is based on the relative velocity between the reference frames. Secondly, not only are rotation and shifting of space required, but also, in general, contraction and expansion of space (along the direction of relative motion) and time.

So, as to SR "treating time and space on the same footing", it is true that they are both treated as things that have be transformed when switching reference frames. However, the ways in which they are transformed for a particular change-of-reference are always different.

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