I suppose you could call it an "arc interval", but somehow that doesn't sound right!vanhees71 said:
I also call it "proper time" to give it its physical interpretation.vanhees71 said:
I think that is a good approach. Your students should still get the benefits and should “easily” avoid the risks. Maybe they get the benefits slightly slower than a less cautious approach, but what is an extra week or two on concepts like these?vanhees71 said:
I'm not sure what counts as "overemphasizing".vanhees71 said:
This is a good example of a concept that gets generalized. Well before I started doing relativity I had to generalize the notion of "orthogonal" when I learned the Fourier transform and found out that cos functions formed an orthogonal basis for functions. There doesn't even exist any word to distinguish this notion of orthogonality from the less general notion of orthogonality as a 90 degree angle. But nevertheless the presentation of this new concept of orthogonality was sufficiently clear that I didn't get confused.robphy said:
There are some ways of introducing relativistic concepts without the use of coordinates, but these don't seem to be widely used. Bondi k-calculus is one method where the introduction of coordinates can be delayed (you can work with proper time only, to start with), and (as far as I can remember) Geroch's General Relativity from A to B also emphasises geometry and de-emphasises coordinates.martinbn said:
I mean Minkowski's analysis first enabled the formulation of the theory using tensors in the pseudo-Euclidean affine manifold, which indeed describes the special-relativistic spacetime model most adequately, and the great benefit is the "manifestly covariant formalism", which enables us to find physical laws that are compatible with this spacetime model. For me the main feature is also "geometry", but not in the sense that it is in many respects analogous to Euclidean affine space but that it provides the Poincare group as its symmetry group (i.e., ##\text{ISO}(1,3)^{\uparrow}## as the Lie group of the symmetry-transformation group that is connected smoothly with the identity). That is of course also analogous to the Euclidean case (the closest analogy would be ##\text{SO}(4)##).robphy said:
I never avoid four-vectors and -tensors. To the contrary, I introduce them as soon as possible, even before I draw the first Minkowski diagram. Many introductory textbooks avoid four-vectors in the beginning and make SR more complicated than necessary.robphy said:
Exactly. That's why for SR I start with Einstein's two postulates and introduce the four-vector and -tensor formalism with the Lorentzian fundamental form first. Then, specializing one-selves to two-dynamical point-particle motion, I introduce the Minkowski diagrams in a time-space plane, and then you don't come to the idea to misinterpret a Minkowski diagram by drawing any Euclidean circles in it but the hyperbolae (space-like and light-like as well as their "degenerate" case of the lightlike light-cone) to construct the correct "unit tics" on the axes of different inertial frames. You also don't introduce "angles" but "rapidities" in this plane (of course there's a great analogy between the Euclidean "angles" and the Minkowskian "rapidities", but they are clearly not the same).robphy said:
). I'm just musing on the problems with learning (and teaching) relativity. Their are a lot of barriers and no one approach will work for everyone. It sounds like you've all taken unique paths.If it is under the right kind of acceleration (the technical name for which is "Born rigid acceleration"), yes.Freixas said:
First, do you mean "invariant" or "constant"? An object's proper length can be invariant--the same in all frames--but still change as we move along the object's worldline. I'm going to assume you actually meant "constant", which means, not only that we can identify an appropriate invariant (which will turn out to be arc length along a particular spacelike curve), but that this invariant remains the same as we move along the object's worldline (so there is actually a whole series of spacelike curves, each describing the proper length of the object at successive instants of the object's proper time).Freixas said:
Different velocities in what frame?Freixas said:
Yes, they can, in two senses:Freixas said:
Freixas said:I'm not actually asking for answers to any questions posed here (although I suspect I will hear some
Freixas said:
Freixas said:
In any technical discussion, one has to learn the vocabulary and the definitions to fully participate in the discussion. For me, rather than just vague words, having a mathematical definition helps, particularly ones I can draw.Freixas said:
"proper time" is defined by Minkowski as "eigenzeit" to be one's own time.Freixas said:
That works well because it is easy to connect the 2nd postulate to the fundamental form.vanhees71 said:
I just think you can do all of that without needing to say that angles in spacetime make no sense.vanhees71 said:
That was one option. Under that option, the object's proper length increases (i.e. the string breaks).PeterDonis said:
Hmm... I tried to model a case where the proper length of an accelerating object remained the same (using Gamma). Using the viewpoint of the initial rest frame, I assigned a constant acceleration to the rear endpoint. I then calculated the position of the front endpoint so that, adjusting for the length contraction given the speed of the rear endpoint, the length of the object was constant. The front endpoint does not move as far as the rear one, so it's velocity must be less, which implies that the front clock moves faster than the rear.PeterDonis said:
PeterDonis said:
robphy said:
robphy said:
I meant to comment on this. "Wristwatch time" for me means that I look at my watch and it displays a value. But if Wikipedia is to trusted ("Proper distance is analogous to proper time"), "proper time" is a duration, an interval between two wristwatch numbers (in the same way that distance is measured between two spatial coordinates).robphy said:
That's simple. Assume the object is accelerating in the positive ##x## direction. The front and rear of the object, in any inertial frame, will have worldlines that are concentric hyperbolas, of the form ##x^2 - t^2 = x_r^2## (for the rear) and ##x^2 - t^2 = x_f^2## (for the front), where ##x_r < x_f##. The proper accelerations of the two ends will be ##c^2 / x_r## (for the rear) and ##c^2 / x_f## (for the front). And the "surfaces of constant time" for the object--the surfaces in which the line segments of constant proper length lie--are simply straight lines through the origin of the inertial frame, with gradually increasing slopes, i.e., lines of the form ##t = k x##, where ##0 \le k \lt 1## (the ##k = 0## line is just the line ##t = 0##, which we can take as the instant at which the acceleration begins).Freixas said:
I simply don't introduce angles in an (1+1)D Minkowski diagram. Only rapidities make sense and they have a geometrical meaning as the areas in connection with hyperbolas as discussed above although this area visualization doesn't help much, at least for me. The areas make more sense when working with light-cone coordinates aka @robphy 's "rotated graph paper".Dale said:
And rapidities are generalized angles.vanhees71 said:
I think you are right that it is not necessary. I don’t object to that.vanhees71 said:
One question first.PeterDonis said:
The ##a## on the web page you linked to is the proper acceleration of the rocket, yes.Freixas said:
Note that on the web page you linked to, it says ##d##, not ##x##. ##d## is the distance traveled from the starting point (in the inertial rest frame of the starting point). For the two hyperbolas describing the front and rear of the object, the starting points are ##x_r## (for the rear) and ##x_f## (for the front), in the coordinates of the inertial rest frame of the starting point. So for those two hyperbolas, you would have ##d = ( x - x_r )## (for the rear) and ##d = ( x - x_f )## (for the front).Freixas said:
Thanks. I'll get back to you on the diagram. I've drawn it, but want to think about this a bit more before posting or commenting further.PeterDonis said:
vanhees71 said:
Dale said:
vanhees71 said:
Yes, I know you introduce the hyperbolic functions.vanhees71 said:
So,vanhees71 said:
Freixas said:
It turns out I had already done this in #113. However, I combined the two approaches and here is the result:PeterDonis said:
xr = 1;
xf = 2;
lastR = (xr, 0);
lastF = (xf, 0);
for t = 0 to 10 step .1 {
r = (sqrt(xr^2 + t^2), t);
f = (sqrt(xf^2 + t^2), t);
path [path lastR, r], style: "color: red";
path [path lastF, f], style: "color: green";
lastR = r;
lastF = f;
}Freixas said:
Careful. The lines of simultaneity are lines of simultaneity for both observers--for both observers, everywhere on the line does have the same time value. But the elapsed proper time between two particular lines of simultaneity is not the same for both observers; the front observer has more elapsed proper time than the rear.Freixas said:
No, that's not correct either. See above.Freixas said:
Yes.Freixas said:
Ok, I partially understand this, although I can see the gaps in my knowledge.PeterDonis said:
Yes, that's obvious from the coordinates you give.Freixas said:
Yes, that's one way of looking at it. In somewhat more technical language, the hyperbolas are integral curves of the boost Killing vector field.Freixas said:
I'm not sure what you mean here. A single line is always collinear with itself. Perhaps you mean to say that the three points always lie along a single line regardless of the boost.Freixas said:
One can view them that way. But one can also view the hyperbolas as the worldlines of two accelerating observers who remain at rest relative to each other, with the proper distance between them remaining constant, and the boost moves the two observers along their respective worldlines, i.e., it represents their time evolution. (These two different ways of viewing the action of a Lorentz boost are sometimes referred to as "passive" vs. "active" transformations in the literature.)Freixas said:
More precisely, the arc lengths along the lines between the points are the proper lengths between them; the lines themselves, as a whole, are common lines of simultaneity.Freixas said:
See above regarding "passive" vs. "active" transformations.Freixas said:
Yes.Freixas said:
Yes.Freixas said:
It has to do with the hyperbolas being integral curves of the boost Killing vector field.Freixas said:
Proper acceleration is the same thing, geometrically, as path curvature. Since we are working in flat Minkowski spacetime, any curve that is not a straight line in an inertial frame has nonzero path curvature. So in this special case, proper acceleration and coordinate acceleration are in perfect correspondence.Freixas said:
This isn't a mystery. It's a simple consequence of spacetime geometry and the geometry of the hyperbolas.Freixas said:
Obviously this is false since we have just gone to considerable lengths to describe a counterexample, so why are you asserting it?Freixas said:
In general, yes, but you have described a particular method of selecting lines of simultaneity: pick lines that are orthogonal (in the Minkowski sense) to the hyperbolic worldlines at every point. This is a common method of defining lines of simultaneity, and it often picks out a set of lines of simultaneity with useful properties.Freixas said:
Yes.Freixas said:
From the viewpoint of an inertial frame, yes. But not from the viewpoint of the non-inertial frame (Rindler coordinates). From the viewpoint of that frame, the blueshift is analogous to "gravitational" blueshift. (In fact, Einstein's derivation of this result was one of his early results from the equivalence principle.)Freixas said:
No, it doesn't. The blueshift is constant. So is the redshift observed by the front observer, with respect to light signals from the rear observer.Freixas said:
Yes.Freixas said:
Yes.Freixas said:
Assuming that the walking is done slowly enough, yes.Freixas said:
Yes.PeterDonis said:
Interesting.PeterDonis said:
It's a mystery to me. I've verified that this is true with Gamma. But that's different from understanding it.PeterDonis said:
Freixas said:
Hmm... I thought we had been talking about observers who were accelerating. Observers moving at constant velocity appear (to me) to have different lines of simultaneity except when they are moving at the same velocity.PeterDonis said:
Interesting.PeterDonis said:
Ok, I think I see my mistake. I thought that since the starting velocity was 0, there would be no red/blue shift at the start so that it must increase with time. But light doesn't travel instantly, so by the time the light from an endpoint at the start of the worldline reaches the other end, the receiving end is moving. My initial assumption was wrong.PeterDonis said:
Clearly, there are problems that are way beyond my skills, but you just inspired another one. Let's simplify by stating that a ship is moving inertially (I don't think it matters). Assume a measuring tape is anchored to one end of a ship and wound on a spool. Someone picks up the spool and moves at a high relative velocity toward the other end.PeterDonis said:
Thanks for clarifying. The point remains that the term "wristwatch time" is ambiguous. When you already know what it means, the ambiguity might not be obvious. No one uses the term "proper duration", but it seems unambiguous.robphy said:
Freixas said:
Yes, I understand what you mean. But I carry a wristwatch. Right now, it reads 7:55 AM. Unless I have a stopwatch mode and activate it, my wristwatch doesn't tell me elapsed time. When Taylor/Wheeler introduced and then used the term "wristwatch time" in their book, I thought they meant that most natural interpretation, not the interpretation that required two readings and a calculation, so some of what they wrote just didn't make much sense to me.robphy said:
Ah, sorry, I missed that you switched to talking about inertial observers. Yes, if you use the "orthogonal" method to define lines of simultaneity, inertial observers in relative motion will have different lines of simultaneity (they will not be parallel).Freixas said:
As I said, it's a simple consequence of spacetime geometry and the geometry of the hyperbolas. The hyperbolas have the same asymptotes (the lines ##t= x## and ##t = -x## in an inertial frame), so they are "concentric" in a sense similar to concentric circles in Euclidean space, and the lines of simultaneity are "radial lines" from the "center" (the origin of the inertial frame) in a sense similar to radial lines in Euclidean space. Concentric circles in Euclidean space will be orthogonal to the same set of radial lines; similarly, "concentric" hyperbolas in Minkowski spacetime will be orthogonal to the same set of "radial" lines--i.e., they will have the same set of lines of simultaneity.Freixas said:
Yes. That is basically the argument Einstein originally used to show that there must be a "gravitational" blueshift (or redshift in the opposite direction) in an accelerating elevator.Freixas said:
It's not just that. The tape might deform in this process, because it has only a finite material strength, elasticity, etc., and those might be exceeded during the process. This means the distance between the markings on it when the process is finished might not be the same as it was when the tape was originally manufactured and wound up. Specifying that the unwinding is done very slowly is a way of avoiding all such issues, so that we can be reasonably sure the tape doesn't deform and the lengths it shows after the process are the same as the ones it was originally manufactured with.Freixas said:
Those "nested" hyperbolae you've drawn are the Minkowski equivalent of concentric circles, in the sense that the "distance" to the center, ##\sqrt{x^2-c^2t^2}##, is constant along each one, just as ##\sqrt{x^2+y^2}## is constant along a circle.Freixas said:
Freixas said:
PeterDonis said:
Ok, I think I get it using the analogy to concentric circles. Concentric circles have different circumferences. A line sweeping from the common center to a point on the other circle will intersect the inner circle. The line is orthogonal to the tangents of both points (common simultaneity), but the distance swept through the outer circle is more than is swept through the inner circle (different elapsed time).Ibix said:
