So we have a graph of space vs. time, space being $y$, time being $x$. What's the unit for the distance between them?

$$\sqrt{(y m)^2+(x s)^2}$$


$$\sqrt{\frac{(3m)^2}{(3 m)^2+(4s)^2}+\frac{(4s)^2}{{(3 m)^2+(4s)^2}}}=\sqrt{\frac{(3m)^2+(4s)^2}{(3 m)^2+(4s)^2}}=1$$

To me it seems simple, the answer is $\sqrt{m^2+s^2}$. But I'm wondering if the $\sqrt{m^2+s^2}$ is something specific. Is it energy or something like that? Also what is that unitless 1?

  • $\begingroup$ One multiplies the "time" unit by $c$ to get a "distance" unit... $\endgroup$ – Kyle Kanos Mar 28 '17 at 20:52

So what's special about distances in 3D is that they are invariant under translations and rotations.

In 4D spacetime, we usually care about a generalization of "distance" which is invariant under space-translations, space-rotations, time-translations, and "Lorentz boosts." This generalization of distance is as follows: let $\Delta P = P_1 - P_0$ be the difference symbol between two events that we'll call 0 and 1, so $\Delta x^2 = (x_1 - x_0)^2$ for example.

You are accustomed to the distance $s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$; in 4D we generalize this to the "spacetime interval" $$I = c^2 \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2.$$Please note, first, these minus signs. These minus signs are very important, because the central idea is that when something happens somewhere and somewhen in relativity, the light that announces that event to the world expands outward as a thin bubble growing uniformly at speed $c$ in all directions. The "distance" from this bubble to the event which emitted it is just $I=0.$ While it is obvious that rotations and translations aren't going to change uniformly expanding bubbles, it is somewhat more surprising that there is this simple account of "boosts" which also doesn't. But there is, and it comes from this simple fact that the boosts preserve $I$ which means that these expanding bubbles $I=0$ get mapped to other expanding bubbles $I=0.$ In other words our usual idea is that when a train is moving with speed $v$ and we move in the same direction with speed $u$ it is now moving relative to us with speed $v - u,$ that is not quite how these Lorentz boosts work. As long as $v$ is small that's fine; but as $v$ gets larger we notice that its speed is larger and larger than $v - u$ until eventually at $v = c$ we find out that its new speed is $c$, the same as its old speed, no matter what $u$ is.

These minus signs also mean that the interval can either be negative or positive. If the interval is positive then $\sqrt{I}/c$ is called a "proper time" between the two events. Everyone agrees on the proper time because everyone agrees on $I$ and $c$. But not everyone measures the time between the two events as the proper time; that interpretation belongs specifically to the set of reference frames that think both events happened at the same place, $\Delta x = \Delta y = \Delta z = 0.$ So as this light-bubble is expanding outward announcing some event, the bubble itself is $I=0$, the points inside the bubble have seen the event already and have $I > 0$, the points outside the bubble have not yet seen the event and have $I < 0.$ For the $I > 0$ points, someone moving less than the speed of light, without accelerating or anything, happened to have coordinates which saw both events happen at the same place. Such a spaceship thinks they were separated by the proper time. Everyone else thinks the time takes longer, by a "time dilation factor" $\gamma = 1/\sqrt{1-(v/c)^2},$ where $v$ is the speed that they think those spaceships are going.

Finally for the $I < 0$ case the quantity $\sqrt{-I}$ is known as the proper distance between the two events, and as you can imagine, it is measured by someone who thinks they both happened at the same time, $\Delta t = 0$. You are probably used to thinking of two things as being objectively either simultaneous or not, but it turns out that in relativity, if $I < 0$ there is legitimate disagreement about which one came first. Things that are objectively separated in time ($I>0$) are only subjectively separated by space and things that are objectively separated by space ($I<0$) are only subjectively separated in time.

Okay, that is your tutorial on relativity. How does this answer your question?

Well, $\text{1 s} + \text{1 m}$ is on its face wrong, you cannot add a second to a meter. Similarly $\text{1 s}^2 + \text{1 m}^2$ is wrong. Notice that in all of these expressions you see $c^2 \Delta t^2,$ the time is being converted into a distance by multiplying by the speed of light. Indeed, we can do this explicitly, defining $w = c t$ and then $I = \Delta w^2 - \Delta x^2 - \Delta y^2 - \Delta z^2.$

But in another sense it doesn't matter that much. After multiplying by $c$ one second corresponds exactly to a certain distance, one light-second, and we can just pretend that this distance is what we mean when we write $\text{1 s}$ and everything works just fine.

  • $\begingroup$ How do you do math in 4d without a cross product? The cross product only exists in R^3 and R^7. $\endgroup$ – user1698948 Mar 31 '17 at 15:10
  • $\begingroup$ @user1698948 So, if you ask what observer-independent thing the cross product is in 3D, it linearly maps two vectors to a vector, or equivalently 3 vectors to a scalar, and it is totally antisymmetric in its three inputs. That makes it a [0, 3] antisymmetric tensor called the orientation tensor, just like the dot product is secretly a [0, 2] symmetric tensor called the metric tensor. It turns out that in N dimensions the metric is still valence [0, 2] but the orientation has valence [0, N]. That's the only change. $\endgroup$ – CR Drost Mar 31 '17 at 18:35
  • $\begingroup$ So the "cross product" still exists in 4D, but it gives you a 4x4 matrix which in the appropriate representation is antisymmetric. Any antisymmetric 4x4 matrix field has a representation as two 3D vector fields, in fact the electric and magnetic fields turn out to be such a pair. $\endgroup$ – CR Drost Mar 31 '17 at 18:39
  • $\begingroup$ The "E-field" that you get turns out to be just the cross product of the two vector parts of the two 4-vector fields, the "B-field" is a little more subtle, and sort of says that there is no Lorentz force on either of the two original 4-vectors if they are interpreted as charge-current 4-vectors. $\endgroup$ – CR Drost Mar 31 '17 at 18:50
  • $\begingroup$ I'm pretty sure the $$\sqrt{(x m)^2+(y m)^2+(z m)^2+(t s)^2}$$ is already in the denominator if we observe any undamped sine wave. I could be wrong. Sin and cos functions are great, but don't they make us forgetful of the dimensional analysis? Any time you touch a unit vector the units end up in the denominator. $\endgroup$ – user1698948 Apr 1 '17 at 3:11

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