Relativity Anomaly

[Qfwfq]

As length contraction would demand, sigma is smaller where t has nonzero value.

Actually, that isn't how length contraction is deduced. For one thing, [imath]dt[/imath] doesn't approach zero in the low velocity limit. The slower you go, the longer it takes you! The faster you go, the less it takes you and, for a non-zero [imath]dx[/imath], [imath]dt[/imath] approaches zero for infinite velocity!

 

OTOH, in the timelike case, since [imath]dx[/imath] is proportional to velocity for a given [imath]dt[/imath], it would seem you deduce time contraction for velocity other than zero.

 

In any case this isn't the point for geodesics but I'll give you a chance to try sorting it out better cuz the librarians are kicking me out of here. It is all a bit more subtle than you're supposing, though.

[modest]

Actually, that isn't how length contraction is deduced. For one thing, [imath]dt[/imath] doesn't approach zero in the low velocity limit. The slower you go, the longer it takes you! The faster you go, the less it takes you and, for a non-zero [imath]dx[/imath], [imath]dt[/imath] approaches zero for infinite velocity!

 

I don't understand at all. The slower you go, the longer it takes you to do what, exactly? You can't move between events with space-like separation.

 

I was meaning that,

the two red events have the greatest spatial separation (dt between them is zero). Length contraction would demand that the other two lines have less spatial separation (where dt is not zero). As dt approaches dx it would seem spatial separation approaches zero (as does temporal separation for that matter).

 

It makes sense to me that the inertial path between time-like events is the greatest proper time. It does not make sense to me that a geodesic between events with spacelike separation represents the least spatial separation, because, it would seem in Minkowski's metric the red line is longer than the gray line:

 

 

Is the red line not longest? Is it not a geodesic?

 

~modest

[Qfwfq]

The slower you go, the longer it takes you to do what, exactly?

To get there, of course! And if you go rilly rilly slow, I mean even slower than c, you go between timelike separated points!

 

At least, that was because I thought you were confusing different intervals with the coordinates of different observers, especially as you mentioned length contraction. :doh: But now from your last post I get a clue to what you meant.

 

...the two red events have the greatest spatial separation (dt between them is zero). Length contraction would demand that the other two lines have less spatial separation (where dt is not zero). As dt approaches dx it would seem spatial separation approaches zero (as does temporal separation for that matter).

Usually spatial separation is understood to mean [imath]\Delta x[/imath] (or given by Pythagoras) and so the three intervals in the figure have the same spatial separation. Length contraction is a tricky matter and, although related to where your difficulty lies, isn't really the point for answering the question. What's more relevant is:

It makes sense to me that the inertial path between time-like events is the greatest proper time. It does not make sense to me that a geodesic between events with spacelike separation represents the least spatial separation, because, it would seem in Minkowski's metric the red line is longer than the gray line:

You are considering a variation that is purely [imath]\delta t(s)[/imath] when instead only the [imath]\delta x_i(s)[/imath] or [imath]\delta x(s)[/imath], [imath]\delta y(s)[/imath] and [imath]\delta z(s)[/imath] are relevant.

 

Down to earth, take a thread and pull it taut; you are quite obviously minimizing the length of thread between two given points of space. Each of these and each point of the thread is a world line in spacetime, not a single point of it. As you stare at the thread you see successive points of each world line. It makes no sense to consider [imath]\delta t(s)[/imath] for a given observer and this goes for the particle trajectory as well.

 

The Minkowski metric has the same eigenvalue for the three spatial coordinates; space is isotropic. Spacetime is not, the eigenvalue for the time coordinate is distinct and has opposite sign. For GR of course we must consider the equivalence principle and that of general covariance, with the less often mentioned prescription of coordinate transformations being a congruence (maintaining the number of positive and negative eigenvalues of the metric). So, the [imath]\delta x_i(s)[/imath] variations around a spacelike curve between the same two points of space will increase the integral in any case. Given the opposite signs of the eigenvalues, they decrease the integral for a timelike curve.

 

Formally, the geodesic can be defined in differential geometry by a variational principle, with the integral being in any case stationary; this can mean maximum, minimum or perhaps even (in freaky cases) something analogous to a horizontal inflection point (also called saddle point).

[modest]

Usually spatial separation is understood to mean [imath]\Delta x[/imath] (or given by Pythagoras) and so the three intervals in the figure have the same spatial separation.

 

:eek: Pythagoras in the Minkowski metric? But, the line element in Minkowski space would make that highly irregular, would it not?

 

Length contraction is a tricky matter and, although related to where your difficulty lies, isn't really the point for answering the question. What's more relevant is:You are considering a variation that is purely [imath]\delta t(s)[/imath] when instead only the [imath]\delta x_i(s)[/imath] or [imath]\delta x(s)[/imath], [imath]\delta y(s)[/imath] and [imath]\delta z(s)[/imath] are relevant.

 

How could only the spatial dimensions be relevant :confused: Perhaps, let me back up. Here is the quote I'm troubled by:

A geodesic between two events could also be described as the curve joining those two events which has the maximum possible length in time — for a timelike curve —

or the minimum possible length in space — for a spacelike curve.

Geodesic (general relativity) - Wikipedia, the free encyclopedia

The bolded part doesn't seem to make sense either physically or in the Minkowski metric. It seems as if the geodesic should be the path of greatest spatial separation between space-like separated events.

 

In the way that a clock can be taken on a path to measure the proper time, let's say that a ruler is taken along on a path to measure the proper distance.

At event A, Alice measures AC directly at 2 lightyears. The spatial length of the red path is 2 lightyears. My understanding is that she can measure this spatial length in such a way (her ruler gives the proper distance) because both events are simultaneous in her frame. At point A Tom also measures AB with his ruler. In his frame the events are simultaneous. His velocity relative to Alice is 0.5c and the proper distance he finds between events is 0.866 lightyears. In the Minkowski metric proper spatial distance is given by,

[math]d \sigma = \sqrt{d x^2 - d t^2}[/math]

which gives,

[math]d \sigma = \sqrt{1^2 - 0.5^2} = 0.866[/math]

We could alternatively use the Lorentz transformations by considering AB coincides with an object Alice measures at 1 ly in length and Tom (going .5c relative to Alice) finds,

[math]L = L_0\sqrt{1-v^2/c^2} = 0.866 \ ly[/math]

It would seem the spatial distance AB is inescapably 0.866 ly. Using Pythagoras for this would make no physical sense to me.

 

At event P Tom changes direction such that B and C are simultaneous at which point he measures BC finding again [math]\sigma[/math] = 0.866 with his comoving and rigid ruler. AB + BC = 1.732. The proper spatial distance of the path ABC looks to me like 1.732 lightyears.

 

If the red path depicted above is 2 lightyears while the gray path is 1.732 lightyears as I believe it must be and the red path is the geodesic (which I assume) then the geodesic cannot be the least spatial distance between space-like separated events.

 

Down to earth, take a thread and pull it taut; you are quite obviously minimizing the length of thread between two given points of space. Each of these and each point of the thread is a world line in spacetime, not a single point of it. As you stare at the thread you see successive points of each world line. It makes no sense to consider [imath]\delta t(s)[/imath] for a given observer and this goes for the particle trajectory as well.

 

Yes, but a string curved in space (curved in the x-y plane) is not the same as a path curved in spacetime in the x-t plane. Oh... for f**ks sake!

 

:doh: :doh: :doh:

 

Ok, I got it... where the quote says "for a spacelike curve" I was taking that to mean curved into the time dimension—the curve being on the x-t plane. The quote intends the curve to be in the x-y-z plane with no curvature in time. That's why you're saying that distance is given by the Pythagorean theorem :doh: Oh, good Lord

 

Ok, clearly the gray line is longer than the red line:

where the line is curved into the y direction. Yeah, I was taking "spacelike curve" to mean a curve between spacelike events... Ok, sorry, and thank you for the help, Q.

 

~modest :doh:

[Qfwfq]

Indeed it is perhaps a bit misleading, by saying "for a spacelike curve", I might even try to find a better wording for that wiki since folks rely on it so much.

 

Edit: actually I realized the trouble isn't those words, they aren't wrong although, as the article was, they can suggest considering [imath]\delta t[/imath] so I just specified about the variations and changed a few other words.

[modest]

I might even try to find a better wording for that wiki since folks rely on it so much.

 

I think you did find better wording which would have avoided confusion at least in my case, but it looks like one of the editors restored the previous version. :shrug:

 

~modest

[Qfwfq]

Just goes to show. Trust wiki! :phones: