Distance between two points in spacetime

[dhumidifier]

So I'm new to the forum and to relativity as well and I am confused as to how the equation for an invariant distance between two points in spacetime is derived, specifically where subtraction comes from

s^2=srqrt[((ct)^2)-x^2)]

i understand most of that but just not why its subtraction when the regular equation for distance between two points is addition

 

[Matterwave]

If you define it with the plus sign, then that quantity is no longer a scalar, meaning it's not the same in every frame. The fact that S^2 is the same in every inertial frame only works with the minus sign.

 

[WannabeNewton]

Well I'm not sure if this is the reason why really but metric tensors in relativity are defined on pseudo - Riemannian manifolds where the metric signature is indefinite (so it won't always be positive - definite) thus allowing for space - like, time - like, and null - like intervals on the manifold.

 

[JesseM]

The equation you have is slightly off, if you have s^2 on the left side then you don't need a square root on the right (if you just had s on the left side then you'd need that square root). Also, it would be more common to write either s^2=x^2-(ct)^2 (which for points with a spacelike separation is the proper distance between them), or s^2=t^2-(x/c)^2 (which for points with a timelike separation is the proper time between them), not s^2=(ct)^2-x^2.

Anyway, the reason that equation is used is that if you look at the Lorentz transformation, you'll see this is a frame-invariant quantity (i.e if you find the coordinates of two events in different frames, although the dx and dt between them changes, t^2-(x/c)^2 remains unchanged), while t^2+(x/c)^2 would not be frame-invariant so it doesn't hold the same interest for physicists. Also it happens that for events with a timelike separation, if you have an inertial clock with both events on its worldline, sqrt[t^2-(x/c)^2] is actually the "proper time" elapsed on the clock between those events.

 

[dhumidifier]

So would it be fair to say that S it is not measured as a distance but as an interval in order to account for the temporal distance between the two points as well as the spatial? and because its measured as an interval that's why there is subtraction instead of addition?

 

[bcrowell]

The reason we like the Pythagorean x^2+y^2 in the Euclidean plane is that when you rotate things, it stays the same.

In the (t,x) plane, the corresponding idea is that when you do a Lorentz transformation, the speed of light stays the same. A ray of light has ds=0, and x^2-t^2=0 has to stay the same.

 

[bobc2]

So I'm new to the forum and to relativity as well and I am confused as to how the equation for an invariant distance between two points in spacetime is derived, specifically where subtraction comes from

s^2=(ct)^2-x^2

i understand most of that but just not why its subtraction when the regular equation for distance between two points is addition

Here is a quick direct derivation of your equation using the Pythagorean theorm on the right triangle sketched below. We use a symmetric spacetime diagram so that line lengths on the computer screen are calibrated the same for the blue coordinate system and the red coordinate system. We diagram the case for two rockets moving in opposite directions at relativistic speeds with respect to the black coordinates. I continued on with the algebra to derive the rotation transformation for time. The sketches on the left show graphically the time dilation (upper left) and length contraction (lower left).

You can see that you do in fact start out with the positive sign that you expected (starting with the Pythagorean theorem for distances in 4-dimensional space), but then solve for the S^2 you were referring to.

 

[GrayGhost]

So I'm new to the forum and to relativity as well and I am confused as to how the equation for an invariant distance between two points in spacetime is derived, specifically where subtraction comes from

s^2=srqrt[((ct)^2)-x^2)]

I understand most of that but just not why its subtraction when the regular equation for distance between two points is addition

You might want to draft a quick 2-space diagram of this, and it'll all be clear ...

Imagine a 2-space diagram, with vertical axis y and horizontal axis x (with time implied). From the origin, imagine a spherical light pulse is emitted (at t=0) from an emitter moving along +x at speed v, which expands at c from the origin. Imagine a system X,Y (with time Tau implied) whose origin is always the moving emitter, X and x always colinear with Y and y always parallel. The expanding EM-sphere is always intersecting the moving Y-axis as it goes, with further and further intercepts. At some subsequent time t wrt emission, you consider the expanding EM's intersection with the Y-axis. There exists a single ray of the expanding lightsphere that departed the origin and intersects that moving Y-axis at time t. The ray travels a distance ct thru x,y space, a slanted path thru quadrant 1. That same ray that intersects upon Y "appears per the emitter" to always have traveled vertically along his Y-axis, the emitter always assuming itself stationary. Call that distance Y=cTau. It turns out that this distance (cTau) is actually s of the invariant spacetime interval eqn. So since the emitter moves at vt along +x, then by Pathagorus's theorom at time t ...

(ct)² = (vt)²+y² ... <- note no minus sign
y² = (ct)²-(vt)²​

but x=vt, so ...

y² = (ct)²-x²​

There is no length contraction wrt axes orthogonal to the direction of motion, so the length Y=y ...

Y² = (ct)²-x²​

And per the moving emitter's own POV, Y = cTau, so ...

(cTau)² = (ct)²-x²​

It should be pointed out here that this equation relates the time (Tau) of one system (X,Y,Tau) to space and time of the other system (x,y,t), under an invariant c context. If everything is of one system (say x,y,t or X,Y,Tau), then there would be no minus sign. The minus sign only appears out of order because the variables are not of the same single system, but rather of 2 differing systems related together ...

So while it is true that ...

(ct)² = x²+y² ... <- note no minus sign​

It is also true that ...

(ct)² = x²+y²​

(ct)² = x²+Y²​

Y² = (ct)²-x²​

(cTau)² = (ct)²-x²​

[I]s[/I]² = (ct)²-x² ... <- note the minus sign​

... given s=cTau, and it is.

s is numerically equivalent to time Tau, and it is derived from a length wrt the y-axis of the other system (x,y,z,t) ... and this is why So the above eqn relates

EDIT: The remainder of this post was re-edited for amplification of key related points ...

If you draft the figure as I recommended, (while reading thru this) you will see geometrically why it is that the minus sign must come into play. The spacetime interval's length s is numerically equivalent to Tau, and s is derived from a length upon the other system's y-axis. So the minus sign arises in the spacetime interval's eqn because variables of 2 different systems are being related to one another, versus variables all of the same system. You'll note that in the 2-space figure you draft, time is NOT depicted as an axis in either system (it's only implied). Therefore, the direction of each time-vector is not presented, however its magnitude wrt the defined interval was indeed determinable by considering light's velocity wrt non-contracted spatial axes orthogonal to the direction of propagation. So here's a question for you ...

What are the relative orientations of the 2 time vectors within the 4d spacetime continuum?

Although we are not casually aware of it, a clue is inherent within the 2-space figure. Note the direction of the ray in system x,y which is slanted thru quadrant 1 (+x,+y). Then note the direction of the same ray in system X,Y which is always colinear (vertically) along +Y. The angle (theta) that the ray makes wrt the y-axis in system x,y is of importance here. Since Tau was derived from Y, and Y derived from y, then the lightpath is a common reference wrt the y,Y axes. Given a relation exists between the lightpath-and-y and the lightpath-and-Y, that relation may also relate the relative orientation of each system's time-axis (ie arrow-of-time), t and Tau. And, it does. As it turns out, the time vectors are angularly rotated wrt one another by theta/2. The reason it is half of theta, is because the lightpath must always bisect the time-axis and spatial-axis-of-motion, which maintains light's speed at 1-unit-of-space per 1-unit-of-time (ie normalized to unity) per the Minkowski model. So the moving "spatial and time axes" share equally in this bisection, hence theta/2. In-a-nutshell, the relative angular orientation of the 2 systems (within the 4-space continuum) are related by the relative motion between the 2 systems "using light as the common reference". This "relative angular orientation differential" between the 2 systems not only produces their relative motion, but also all the relativistic effects. Because we perceive time differently than space, this orientation differential goes casually unbeknownst, but is deducable from the relativistic effects considered in collective.

One more point ... familiarizing oneself with Minkowski spacetime diagrams is the ticket in expediting a good understanding of the Special Theory of Relativity, IMO.

Hope this helps!

GrayGhost

 

[bobc2]

One more point ... familiarizing oneself with Minkowski spacetime diagrams is the ticket in expediting a good understanding of the Special Theory of Relativity, IMO.

Hope this helps!

GrayGhost

Excellent advice, GrayGhost.

 

[GrayGhost]

Excellent advice, GrayGhost.

Yeh, I thought you would like that :)

 

[robphy]

Here is a more operational argument... based on Geroch's book.
(It relies only on the t-x plane.)

On p. 81,
consider a radar-experiment to measure a remote event q, relative to an event p:
http://books.google.com/books?id=rwPDssnbHPEC&pg=PA81
It is argued in the pages following p.81 (in particular, p. 88-91
http://books.google.com/books?id=rwPDssnbHPEC&pg=PA91)
that the product of those two elapsed-times T1 and T2 is an invariant.
(Another observer through p would obtain two different elapsed times U1 and U2 in measuring q, but would find that (U1)(U2)=(T1)(T2).)

On p. 84,
http://books.google.com/books?id=rwPDssnbHPEC&pg=PA84
one can interpret spatial separation of q from p as
(dx) = c(T1+T2)/2 [(speed of light)*(half-the-round-trip time)]
and elapsed time of q from p
(dt) = (T1-T2)/2 .
By solving the system for T1 and T2, then forming the product (T1)(T2), you get the expression for s^2, up to a constant-of-convention.

 

[Naty1]

Crowell's post is the easiest shorthand to keep in mind when getting started.

So would it be fair to say that S it is not measured as a distance but as an interval in order to account for the temporal distance between the two points as well as the spatial?

ok

and because its measured as an interval that's why there is subtraction instead of addition?

not really.

There is no stand alone "inherent and obvious logic" that can be gleaned from staring at that interval formulation such as might correspond to the distance between two points on a ruler likely being a positive value. It makes no more or less "sense" than time and distance being variable while the speed of light is fixed for all observers.

You can also get some background here:

http://en.wikipedia.org/wiki/Spacetime_interval#Spacetime_intervals

 

[bobc2]

You might want to draft a quick 2-space diagram of this, and it'll all be clear ...

By the way, GrayGhost, that was a nifty presentation, using just two of our normal 3-D spatial coordinates (x1 and x2) without resorting to the use of the x4 coordinate.

s is numerically equivalent to time Tau, and it is derived from a length wrt the y-axis of the other system (x,y,z,t) ...
GrayGhost

Please don't think I'm trying to be picky on this point, but I question whether you would want to say that s is numerically equal to Tau (unless you were thinking of scaling c with a value of 1.0). I thought your derivation was nice the way it made it clear that

s = Y = c(Tau).

In other words, the time Tau, is just a parameter. Y = f(Tau) for anything moving along the Y axis in the moving system. and I like that because it's no different for the 4th spatial dimension, i.e., time is just a parameter in exactly the same way there.

 

[GrayGhost]

By the way, GrayGhost, that was a nifty presentation, using just two of our normal 3-D spatial coordinates (x1 and x2) without resorting to the use of the x4 coordinate.

Thanx. I probably could have worded it a little better, but it locked me out and I could not re-edit it further.

Please don't think I'm trying to be picky on this point, but I question whether you would want to say that s is numerically equal to Tau (unless you were thinking of scaling c with a value of 1.0).

No problem bobc2, and likewise.

Indeed, only if the figure is scaled in the Minkowskian way is that true. However it always remains true that s=cTau, so Tau = s/c = Y/c = y/c ... assuming both systems use the same units of measure. The right triangle will always exist on a 2-space figure with time implied.

I thought your derivation was nice the way it made it clear that

s = Y = c(Tau)​

In other words, the time Tau, is just a parameter. Y = f(Tau) for anything moving along the Y axis in the moving system. and I like that because it's no different for the 4th spatial dimension, i.e., time is just a parameter in exactly the same way there.

Thanx again. Also, y=Y=f(Tau)=cTau, and that's the important thing given the goal to relate 2 systems in relative motion.

The 2-space figure with time implied will always explain how the minus sign arises in the spacetime interval eqn, and so that question is laid to rest IMO. Many seem to ask that question, and usually depart still scratching their heads. Your Loedel figure works fine as drafted, which represents 4-space (while showing only 2-space). They are great figures in many respects. However, if your red X"₄ worldline is rotated slightly in either direction, it falls short of being applicable because a right triangle cannot be made given the blue X'₄'s line-of-simultaneity. IOWs, it works on Loedel figures, but not on non-Loedel figures and not or standard Minkowski spacetime diagrams. The 2-space figure with time implied always gets the job done, and the overall meaning is always clear.

IOWs, no one has a problem with this minus sign here ...

(ct)²=x²+y²
y²=(ct)²-x² <- note the minus sign​

But many generally do have a problem with the minus sign here ...

y²=(ct)²-x²
(cTau)²=(ct)²-x² <- note the minus sign​

and they have the problem, because they do not realize that cTau = y. That is, they always assume that ct (or cTau) must be a hypthenuse of a right triangle, which is true when considering one system unto itself, hence no minus sign should exist in the eqn when constructed in the normal Pythagorean way. However the spacetime interval relates 2 different systems, and since cTau = y, cTau (ie y) is equivalent to a smaller leg (ie y) of the right traingle in system x,y,z,t, ie the ct,x=vt,y right triangle ... hence the minus sign :)

GrayGhost

 

[bobc2]

Thanx again. Also, y=Y=f(Tau)=cTau, and that's the important thing given the goal to relate 2 systems in relative motion.

Exactly. Good point.

The 2-space figure with time implied will always explain how the minus sign arises in the spacetime interval eqn, and so that question is laid to rest IMO. Many seem to ask that question, and usually depart still scratching their heads.

Actually, I think your triangle, taken as representing distances only, shows how the negative sign comes in. It's no different than a right triangle drawn on a sheet of paper, where you label X, Y, and H (for hypotenuse), then solve for Y. It's all spatial, but you can always introduce time as a parameter if you wish to describe motion of something moving along the side as a function of time, i.e., Y(t), or Y(Tau).

Your Loedel figure works fine as drafted, which represents 4-space (while showing only 2-space). They are great figures in many respects. However, if your red X"₄ worldline is rotated slightly in either direction, it falls short of being applicable. IOWs, it works on Loedel figures, but not on non-Loedel figures and not or standard Minkowski spacetime diagrams.

Yes, you certainly need the symmetric (Loedel) spacetime diagram to make that derivation work. However, I've never thought of that as a special case two-observer motion, because I don't care what the speeds are for any two observers (moving at constant velocities), you can always find a coordinate system (my black reference coordinates) for which the two observers are moving away from each other with the same speeds. So, if you wish to rotate the red coordinate, then find a the new black coordinates (it is always a trivial effort to find them) such that the blue time coordinate rotates accordingly so as to maintain the symmetric diagram.

The 2-space figure with time implied always gets the job done, and the overall meaning is always clear.

Yes, I agree (with time as a parameter, understanding that Y = (c)(Tau) ). However, the symmetric diagram also shows explicitly the time dilation and length contraction. But again, your derivation was very nice and insightful.

My grad school Special Relativity prof used to have us do all the old chestnut SR problems numerically, then do them again with spacetime diagrams. He tolerated (up to a point) many philosophical and metaphysical arguments about the physical interpretation of spacetime and pointed out the possibility of the spatial 4-D interpretation (time as just a parameter) while at the same time pointing out some bazaar implications of going down that path (which I am still cautious about pursuing--other than for pedagogical reasons).

 

[GrayGhost]

Yes, I agree (with time as a parameter, understanding that Y = (c)(Tau) ). However, the symmetric diagram also shows explicitly the time dilation and length contraction. But again, your derivation was very nice and insightful.

Indeed they do, and they do a geat job at that.

Wrt Minkowski spacetime illustrations ...

Mastering SR from the OEMB alone, is a wonderful thing IMO. However, the Minkowski model makes mastering the details of the theory so much easier, that words cannot say. A picture paints a 1000 words, and especially the case of SR. Yet, if one has never derived the LTs in the very way Einstein did (via OEMB math w/o spacetime figures), then they are likely missing out on some very important things as well. Both approaches should be attempted, and in unison if at all possible.

Far as 2-space illustrations (with time implied) such as that I defined here prior, one need only consider said 2-space figure as a horizontal slice of the stationary observer's spacetime system ... where the spacetime figure is instead presented in 3 dimensions as x,y,t. x and y being a horizontal plane, and t vertical. That's when things really get clear. It's not that difficult to draft either. Then, presenting the moving observer's slanted plane-of-simultaneity does some wonderful things too, as it presents in 1 figure how axes orthogonal to the propagation axis are not contracted ... and the meaning seems just a little bit clearer, geometrically.

GrayGhost

 

[GrayGhost]

My grad school Special Relativity prof used to have us do all the old chestnut SR problems numerically, then do them again with spacetime diagrams. He tolerated (up to a point) many philosophical and metaphysical arguments about the physical interpretation of spacetime and pointed out the possibility of the spatial 4-D interpretation (time as just a parameter) while at the same time pointing out some bazaar implications of going down that path (which I am still cautious about pursuing--other than for pedagogical reasons).

I'd be rather interested in hearing the implications your professor spoke of, assuming you still recall them.

GrayGhost

 

[PAllen]

My grad school Special Relativity prof used to have us do all the old chestnut SR problems numerically, then do them again with spacetime diagrams. He tolerated (up to a point) many philosophical and metaphysical arguments about the physical interpretation of spacetime and pointed out the possibility of the spatial 4-D interpretation (time as just a parameter) while at the same time pointing out some bazaar implications of going down that path (which I am still cautious about pursuing--other than for pedagogical reasons).

You should realize that bazaar is marketplace of goods. Presumably you mean bizarre.

 

[bobc2]

You should realize that bazaar is marketplace of goods. Presumably you mean bizarre.

Thanks, PAllen. Stream of consciousness does not serve me well.

 

[bobc2]

I'd be rather interested in hearing the implications your professor spoke of, assuming you still recall them.

GrayGhost

If I got very far into some of that my post would quickly get out of the bounds of SR posting here. But, one of the things that comes up immediately with the concept of 4 spatial dimensions (the 4-D universe is all there, populated by 4-dimensional objects) is that you need to identify what it is that is moving along the 4th dimension at the speed of light. It can't be the material structural body of the observer, because that is a 4-dimensional object and is always all there frozen along with the "block universe" (of course you can easily calculate the length of the material worldline structure by just multiplying c times the age in seconds from birth to death).

An obvious choice for the moving entity is something that illicits consciouness--let's just say consciousness itself in a sense (you get the continuous sequence of 3-D cross-section conscious views of the 4-D universe). It's like watching a movie, except that instead of the film frames flying by your view, your consiousness flies along the worldline watching the movie presented by the 4-D bundle of neurons strung out along the 4th dimension.

But the consciousnesses of observers with non-parallel worldlines are sampling different cross-sections of the universe. Thus, having all these different moving 3-D consciousnesses means that any given observer, A, at any given instant could have a 3-D cross-section view populated by zombies. In other words, the consciousness of another observer, B, moving along his own nonparallel worldline has already moved hundreds of millions of miles beyond the worldline point that is included in A's cross-section view (time dilation). So, A is not really cohabitating with a real living person, B--it's just his body structure with no consciousness present in B, that is, in A's 3-D cross-section of the universe at that instant.

Appologies to the Mentors. This is probably already enough to lock the thread.

 

[GrayGhost]

bobc2,

I must admit, I was not expecting conciousness to come into play. And you're right, that won't fly in these parts. That will surely take awhile to digest. I find it very interesting that your professor of relativity engaged in that discussion at university. But then, I suppose many physics professors are also philosophy majors.

GrayGhost

 

[bobc2]

bobc2,

I must admit, I was not expecting conciousness to come into play. And you're right, that won't fly in these parts. That will surely take awhile to digest. I find it very interesting that your professor of relativity engaged in that discussion at university. But then, I suppose many physics professors are also philosophy majors.

GrayGhost

In grad school we were required to take a philosophy of science course over in the philosophy department. The prof there first had completed a physics PhD before turning philosopher. That guy was way too far for my taste. He's the guy that told us on the last day of class that "Physics never has and never will make a useful contribution to philosophy." He also told us that the room next door did not exist until he walked down the hall and opened the door. I think he was a disgruntled ex-physicist (maybe he couldn't get tenure or something).

 

[GrayGhost]

In grad school we were required to take a philosophy of science course over in the philosophy department. The prof there first had completed a physics PhD before turning philosopher. That guy was way too far for my taste. He's the guy that told us on the last day of class that "Physics never has and never will make a useful contribution to philosophy." He also told us that the room next door did not exist until he walked down the hall and opened the door. I think he was a disgruntled ex-physicist (maybe he couldn't get tenure or something).

lol, now that's good :) He likely made the course colorful, quite the opposite of typical professors!

GrayGhost