# Units of spacetime in Minkowski metric

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1. Oct 14, 2013

In the equation
ds2=dx2+dy2+dz2-c2*dt2
the units on the RHS are units of distance squared. But it would seem that units for a spacetime metric should somehow be in units which incorporate both space and time units.
Undoubtedly this is an elementary question, but one has to start somewhere. Thanks.

Last edited: Oct 14, 2013
2. Oct 14, 2013

### ghwellsjr

That's actually the formula for one version of the Spacetime Interval (except you left off the squared term on "c") and it's actually the square of distance (just so we don't have to worry about taking the square root of a negative number). And it does incorporate both time and space but since they are actually different, we convert time to a distance that light travels in a given time.

3. Oct 14, 2013

### dextercioby

The speed of light is a conversion factor in terms of units. You must pass somehow from seconds to meters and do that with something measured in meters/second. The second postulate of SR forces this conversion factor to be exactly c (and not c+/- 2%).

4. Oct 14, 2013

### dauto

It is also possible to write the space-time interval as dτ2 = dt2 - 1/c2[dx2 + dy2 + dz2]

5. Oct 14, 2013

thank you, ghwellsjr and dextercioby. Yes, I have corrected the formula to having the c-squared and that we are looking at distance squared. That two sides of the equation should have the same units, and that the c2 accomplishes this is clear, and that both space and time are implicitly taken into account in the calculation of ds2 is also clear. However, this still bypasses my question: with this mechanism, we end up with ds2 having the units of, for example, m2. Yes, in a calculation this works, but spacetime is not a spatial distance and should hence not have the units of spatial distance, just as it would be odd to express a measurement of mass in units of electric charge: the two are fundamentally different. My own guess is either that
(a) there is some sort of implicit isomorphism at work here, so that we have a distance-squared
(dx2+dy2+dz2-c2*dt2) which relates to the measurement of a distance in a structure in space, and that this structure of space is isomorphic to a structure in spacetime, in which the units of the metric are no longer purely spatial.
Or,
(b) once the units have been all set to the same units, then these quantities are really unitless, merely computing comparative lengths of the space, the time, and the space-time vectors.

But I don't know whether either of these guesses are viable. I would be grateful for further enlightenment.

6. Oct 14, 2013

PS the reply of dauto just came in as I sent in my reply. Thanks, dauto; yes, using proper time (tau) has both sides of the equation expressed in time-squared units, but this then transforms my question from
(A) why spacetime-squared is in terms of spatial-distance-units squared, to
(B) why spacetime-squared is in terms of time-units squared.
Just as spacetime is different to space, it is also different to time. Again, both time and space are taken into account in the calculation of either mode of expression, but in neither does spacetime have its proper units distinct from space alone or time alone.
My dubious guesses (a) and (b) would also apply here.

7. Oct 14, 2013

### ghwellsjr

If you start with your version of the spacetime interval and it evaluates to a positive number, then you can take the squareroot and you will have a value that is purely spacial. It's called a spacelike spacetime interval. If it's negative, then you can recalculate it using the form that dauto provided, take its squareroot and you will have a value that purely temporal. That one is called a timelike spacetime interval. If the value is non-zero, it's always either one or the other, never both. If it's zero, it's neither and it's called a null spacetime interval.

8. Oct 14, 2013

thanks, ghwellsjr. Interesting. I shall work further on this, starting from your explanation.

9. Oct 15, 2013

### phyti

The evolution of the interval.

From the SR 1905 paper, the (equality) expression for the invariant interval between two events is

1. x12+ x22+ x32= (ct)2.

Nice and simple, with obvious meaning.
Minkowski, using complex notation, modifies the ct variable to ict. This allows a general form for the interval, that (mathematically) represents ct as a 4th dimension when expressed as

2. s2 = x12 + x22 + x32 + x42.

More complex, allowing for misinterpretation, such as ‘moving in time’.
The s term can have ± values or = 0. Using a Minkowski diagram, the sign of the values only indicates a speed <, >, or = to c, i.e. time like, space like, or null (photon) world lines.

Spacetime is a measure of spatial units. If you examine the light clock, each tick is counting multiples of light motion within the clock. If you could unfold this distance into a straight line, it would represent ct. The Minkowski diagram is a plot of speed, vt/ct, or v/c, and is not a 2D plane.

10. Oct 15, 2013

Thanks, phyti. Your remark that it is ct, not t, that is a separate dimension is enlightening, as was your comment that spacetime is really a spatial measurement (implicitly: for a spacelike spacetime separation; mutatis mutandi for timelike spacetime measurements). Also enlightening was my attempt to give a counterexample; I kept finding myself guilty of assuming either simultaneity or motionlessness, both of which are unprovable. This helped not only my understanding but also my gut reaction against spacetime not having some units of its own.

11. Oct 15, 2013

### ghwellsjr

Where does that occur in the SR 1905 paper? I couldn't find it or any discussion about the spacetime interval or about it being invariant.

I did find a similar equation in the middle of article 3 but it concerned the formula for the expansion of a spherical wave and had nothing to do with the invariant interval between two events. Einstein's use is nice and simple, with obvious meaning, but I don't see how it could be true for any two events, even if one of those events is at the origin.

12. Oct 15, 2013

### The_Duck

In special relativity, you should ideally measure distance and time in the same units, and you should also use these units for the spacetime interval. Note that "spacetime" is not a quantity, or a unit--what we are speaking of is called the spacetime *interval*. "Interval" here is supposed to mean "distance"--the spacetime interval defines a sort of distance between any two events, and it is perfectly fine to measure this distance in meters. The spacetime interval has the same units as distance and time for the same reason that all distances in the 2D Euclidean plane have the same units as distances along the x axis.

13. Oct 15, 2013

Thanks, The_Duck. A few comments, sentence by sentence:
Yes, phyti's post also emphasizes this point.
Why isn't it a quantity? If it is measured or calculated, it is a quantity, no? In this case, although it is not a unit, it is measured in units, no?
If you already have the two axes in distance units (as clarified by phyti), then fine. I was starting from the assumption that one of the axes in spacetime was time, in temporal units, and the other axis was space, in spatial units (as is usually the case in popular descriptions of spacetime), in which case the analogy to the 2D Euclidean plane would no longer work, since you have curves measured in the same units as those of the x axis only if the units of the y axis are the same as those of the x axis; however, if the axes are of different quantities measured in different units, this no longer works.

14. Oct 15, 2013

### Staff: Mentor

As far as the mathematics of space-time are concerned, there is no distinction between the units of time-like intervals and space-like intervals.

We humans don't experience these intervals the same way, but that affects neither the mathematical structure nor the correspondence between that structure and what we do experience (in part because Minkowski mathematics does recognize a difference between time-like intervals and space-like intervals, it just doesn't need a different unit of measure to do so).

Measuring intervals on the t-axis in seconds and measuring intervals on the spatial axes in meters is like measuring distances on the y-axis in inches and distances on the x-axis in feet... It obscures without being any more correct.

15. Oct 15, 2013

### yuiop

I agree with ghwellsjr. This expression is not the invariant interval between two events.
Using x2=x3=0 and rearranging yields:

$\frac{\Delta x1}{\Delta t} = c^2$

which is obviously only true for a light particle. If this is in the 1905 SR paper, then it was in the context of the speed of light being invariant.

16. Oct 15, 2013

Nugatory, thanks and a further question. If spatial and temporal units are considered essentially equivalent , would that make velocity (and hence the constant relating the two units) essentially unitless?

17. Oct 18, 2013

### phyti

For this discussion, it should just be "interval" as in line 1. The expression you mentioned in par 3, is the one noted. The purpose was to compare (1) and (2) (post 9), showing the misinterpretation of (2) with "time" dressed up as a literal dimension. The classification of interval types, seems redundant, when they are so obvious in Minkowski diagrams. Mass moves at <c, light moves at c, and nothing known moves >c.
This idea of a time dimension just perpetuates the philosophical/scientific debates about the nature of time. More is learned about time by studying clocks and how they have been used. The conclusion is the same throughout history until now. Time is an alias for spatial motion, the earth for millenia, and light for the last century.

From 'The Meaning of Relativity', Albert Einstein, 1956:
page 1.
"The experiences of an individual appear to us arranged in a series of events; in this series the single events which we remember appear to be ordered according to the criteria of "earlier" and "later", which cannot be analysed further. There exists, therefore, for the individual, an I-time, or subjective time."
page 31.
"The non-divisibility of the four-dimensional continuum of events does not at all, however, involve the equivalence of the space coordinates with the time coordinate."
page 32.
"Finally, with Minkowski, we introduce in place of the real time co-ordinate l=ct, the imaginary time co-ordinate..."

18. Oct 18, 2013

### ghwellsjr

But couldn't you just as easily claim that the spatial dimensions are misinterpretations of the time coordinate as per dauto's post:

After all, the international standard for the meter is no longer based on a rigid rod or on a fraction of the size of the earth but rather on the distance that light travels during a specified interval of time. Apparently, the world's best scientists have become convinced that there is no such thing as a rigid rod. Why wouldn't you conclude based on that and since there has been so much debate concerning Length Contraction of so-called rigid rods that there is no legitimate way to construct a spatial coordinate system based on rigid rods but only on the measurement of distances using clocks to determine how far light has traveled during a specified time interval.

So why wouldn't you conclude your remarks with "Distance is an alias for light travel time and motion is just a fraction of the speed of light (or something similar).

19. Oct 22, 2013

### phyti

You can rearrange x=ct, in different ways, depending on what variable is unknown, but that does not define them. It defines a relation among the variables.

If light travels a fixed distance, we can set a clock to run at different rates, and assign various “times” for the same distance. The distance is constant, the time is a convention, and thus variable.

They knew this long before 1900, when they discovered that heated matter expanded.

That was the point, that the “time” corresponded to a distance travelled by light.

Since lc and td scale length and time for the moving observer, by the same scale, 1/γ, the spatial coordinate system preserves relations between “time” and “space”, as if rigid rods exist, i.e. relative to that observer, the unit of time and length are constant.

We already do that with light second, light year, etc., and expressing (speed) v as .6c for example.

The light clock counts units of light motion (t) that are simultaneous with the object motion. The numbers t1 & t2 are only events corresponding with the start and end of the motion, just as the marks on a ruler correspond to the ends of an object.

I can’t regard “time” as something fundamental which causes events. Observing and recording events using a clock (a standard event generator), and comparing to previous events is an operational definition of “time”, an accounting method. In its primitive form, a criminal in a prison marking a line on the wall for each day.

The Minkowski diagram shows explicitly, vertical ct axis (light motion) vs horizontal x axis (object motion), v/c, an apples to apples comparison.

20. Oct 22, 2013

### ghwellsjr

Do you regard "distance" as something fundamental which causes events?