Quick review (skip it):

In the formula from 8th grade, you figured out the length of the long side of the triangle

Sum of the short sides > long side

using this equation:

2d pythag

And in three dimensions:

3d pyth

This gives the length of the line between two opposite corners of a cube (it passes through the center).

Einstein said that time is a 4th dimension that you can move in and rotate into, but it's different from space in that you SUBTRACT (not add) the elapsed time from the total distance.


Note that you can convert elapsed time to a linear spatial distance by multiplying it by c: [Hours] x [miles]/[hour] = [miles].

A metric with a negative term is called a pseudometric. Einstein said that distance in 4D spacetime is like this. It is the E = MC^2 of special relativity.

This is called the "spacetime interval" metric. It is absolute distance (Lorentz-invariant), that is, everyone everywhere agrees on its length.

Because of this, scientists say that space is imaginary time. But multiply both sides by i, and you see that time is imaginary space also. The physical manifestation of that is, literally (and unfortunately, if you didn't already know it, because you won't want to believe it), negative length (distance).

It's why they call the null cone "null."

If you have a triangle with one side extended into ihe time direction (dimension), rotating it further into the time direction makes the hypotenuse shorter, not longer(!)

Believe it or not, this 4D rotation looks to us, in three dimensions, like acceleration. In fact, acceleration in any number of dimensions is actually rotation in n+1 dimensions.

Of course, I'm talking about the Lorentz transformation. Relativistic length contraction is actually just ordinary foreshortening, like the shadow of a rotating pencil.


Ignoring the pedantic dictionary definition involving the absolute value (use another term like "pseudometric distance" or "interval"), does the Einstein metric in fact say that elapsed time is negative spatial distance?


I imagine the answer is obviously "yes," but I want to establish it before I ask my next question, about the properties of null hypersurfaces. I didn't want to put all this in one question.

Ty, guys!

  • $\begingroup$ Hello! I have edited your question using MathJax (LaTeX) math typesetting. For future questions, you can refer to MathJax basic tutorial and quick reference. Thanks! $\endgroup$
    – jng224
    Feb 11 at 17:50
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    $\begingroup$ Sorry; I see ima have to learn MATHML! $\endgroup$
    – L Turner
    Feb 11 at 17:56
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    $\begingroup$ "negative spatial distance" is not a well-defined term. You should not be ascribing meaning to it. Also, MathJaX is not MathML. $\endgroup$ Feb 11 at 18:05
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    $\begingroup$ Yes, of course. But because it's subtracted from the sum of the squares of the spatial distance, it "behaves" like negative length. It's why they call the null cone "null." What you call it doesn't matter, I would think. $\endgroup$
    – L Turner
    Feb 11 at 18:23

2 Answers 2


does the Einstein metric in fact say that elapsed time is negative spatial distance?

No. Note that the question should be “does the Einstein metric in fact say that elapsed time squared is negative spatial distance squared?” But even with this correction the answer remains, no.

The usual interpretation is simply that $ds^2=-c^2 dt^2+dx^2+dy^2+dz^2$ identifies three different types of spacetime intervals.

$ds^2>0$ identifies spatial lengths which are measured with rulers.

$ds^2<0$ identifies times which are measured with clocks.

And $ds^2=0$ identifies light.

A time is not a negative spatial distance. They are both separate types of spacetime intervals. While different coordinate charts might split an interval into time and space differently, they all agree on the interval.

Similarly, the usual Pythagorean theorem does not say that a horizontal length is a vertical length. It says that while different coordinate charts may split a length into horizontal and vertical differently, they all agree on the length.

Because of this, scientists say that space is imaginary time.

While time is definitely not negative space, there are indeed some authors that historically have used the convention that time is an imaginary space. They would write coordinates like $(ic \ t,x,y,z)$ and then use the usual Euclidean metric to get $ds^2=(ic \ dt)^2 + dx^2 + dy^2 + dz^2 = - c^2 dt^2 + dx^2 + dy^2 + dz^2$.

This practice was largely abandoned quite some time ago. Instead of changing the coordinates, it is better to change the metric. Changing the metric instead of the coordinates has the following advantages:

  1. It geometrically is more insightful
  2. It physically matches key features of time
  3. It generalizes better to curved spacetime

The geometrical insight is that Euclidean distance is based on spheres. Meaning that the locus of points that are a given Euclidean distance from a center forms a sphere. In spacetime the locus of events that are a given interval from a center forms a hyperboloid. When you change the coordinates and use the old metric that becomes hidden. It seems that you are identifying a spherical locus in spacetime. It is better to recognize the geometry and have the formulas bring that to the front rather than hiding it.

The features of time that changing the metric captures are also related to those hyperboloids. For spacelike intervals, the locus is a hyperboloid of one sheet, meaning that it is possible to continuously transform between any events with a given spacelike separation. But for timelike intervals, the locus is a hyperboloid of two sheets, meaning that you cannot continuously transform between future pointing and past pointing events with a given timelike separation. This is a geometrical distinction between future and past, and it arises naturally with the metric approach.

Finally, in curved spacetime and with arbitrary coordinate systems the approach of simply multiplying time by $i$ will generally fail. Many coordinate systems do not have a time coordinate at all, and many will have off-diagonal terms that mix timelike and spacelike basis vectors. The metric modification approach generalizes nicely, but the coordinate modification approach gets discarded at that point.

  • $\begingroup$ >They are both separate types of spacetime intervals. No, they're the same type of spacetime interval because you can add them. it's just that one is negative. You can see this when you add the 2 types of distance to get the interval. Adding spatial distance increases the interval. Adding timelike distance (elapsed time) decreases the interval. $\endgroup$
    – L Turner
    Feb 11 at 20:42
  • $\begingroup$ They are different types, as one is measured with a clock and the other is measured with a ruler. $\endgroup$
    – Dale
    Feb 11 at 22:46
  • $\begingroup$ That's like saying that height and length are different things because you use a tape measure to measure length and a long pole to measure water depth. They're in the same units if you multiply by c. $\endgroup$
    – L Turner
    Feb 11 at 23:05
  • $\begingroup$ @Miss_Understands So what? If you multiply by $G/c^2$ then you can also express mass as length. This is called geometrized units. It doesn’t make mass the same thing as a spatial distance. Energy and torque have the same units even without multiplying by any factors. That doesn’t make them the same either. Your argument doesn’t hold. The sign difference remains and you are still neglecting the squares. Anyway, I am not going to argue. You asked and this is my answer. $\endgroup$
    – Dale
    Feb 12 at 0:58
  • $\begingroup$ Well, I still have questions, mainly about the sign difference, and I don't know what you mean by "neglecting the squares." A thing I DO know well from experience that when non-autistic people start to get angry, they are no longer useful for information so I have to shut up and get my answers elsewhere— In this case, the answer to why subtracting elapsed time from the space difference doesn't make time negative space. It just seems so obvious. $\endgroup$
    – L Turner
    Feb 12 at 4:13

No, elapsed time is not negative distance, and you are mistaken if you think the metric says otherwise. The spacetime interval considers two distances, namely the spatial distance between two events and the distance travelled by light in the temporal interval between them (which is the cdt term). You have to multiple time by a speed to get a distance, which is why time itself cannot be a distance.

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    $\begingroup$ >" You have to multiply time by a speed to get a distance" Yes, the speed is c. And the metric does that. you don't have to impose it. 2) I'm new here, and I learned at Reddit not to disagree with anyone. I'm not here to talk; i'm here to learn. I FINALLY found a place where everyone else is smarter than me, and I can NOT get thrown out of here. This is even better than college. $\endgroup$
    – L Turner
    Feb 13 at 11:19
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    $\begingroup$ Welcome to PSE! $\endgroup$ Feb 13 at 12:20

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