I What, exactly, are invariants?

[Freixas]

That's simple. Assume the object is accelerating in the positive x direction.

One question first.

In Gamma, I plot constant acceleration using a formula from https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html. Assuming $c = 1$, then $t = \sqrt{x^2 + 2x / a}$. Is this the same as proper acceleration?

You gave the proper acceleration for the rear as $1/x_r$. If I substitute this for $a$, I get $x^2 - t^2 = -2xx_r$, instead of $x^2 - t^2 = x_r^2$. I'm not sure how to relate the formulas.

 

[PeterDonis]

Is this the same as proper acceleration?

The $a$ on the web page you linked to is the proper acceleration of the rocket, yes.

I'm not sure how to relate the formulas.

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).

If you plug those two formulas for $d$ into the equation from the web page, you should get the same formulas for $t$ that you get if you take the two equations for the hyperbolas that I gave you and solve them for $t$ (i.e., rearrange them algebraically to give you a formula for $t$).

 

[Freixas]

If you plug those two formulas for into the equation from the web page, you should get the same formulas for that you get if you take the two equations for the hyperbolas that I gave you and solve them for (i.e., rearrange them algebraically to give you a formula for ).

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.

 

[robphy]

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".

Only rapidities make sense

And rapidities are generalized angles.

Sigh, yes, I don't deny that, but why must one draw this analogy which confuses beginners of the subject?

Do you introduce hyperbolic-trigonometric methods to find components of [4-]vectors,
akin to using [circular-]trigonometric methods with forces on a free-body diagram?
(I think few beginning students will use a dot-product or [Pythagorean] invariant methods.)

I think $E=\frac{ADJ}{HYP}=m\cosh\theta$ and $p=\frac{OPP}{HYP}=m\sinh\theta$ [ and $v=\frac{OPP}{ADJ}$]
is more intuitive and more familiar
than $E=\gamma m$ and $p=\gamma mv$,
not to mention $E=\frac{m}{\sqrt{1-v^2}}$ and $p=\frac{mv}{\sqrt{1-v^2}}$.

(Many collision problems in energy-momentum space
can be solved using methods similar to those
used in statics [equilibrium] problems on a free-body diagram.)

 

[vanhees71]

I introduce of course hyperbolic functions and rapidities, but I don't call them "trigonometric". This name is everywhere in the literature reserved for the functions sin, cos, tan, etc. Of course, I introduce Minkowski space with its indefinite fundamental form first, before drawing any diagrams.

 

[robphy]

I introduce of course hyperbolic functions and rapidities, but I don't call them "trigonometric". This name is everywhere in the literature reserved for the functions sin, cos, tan, etc. Of course, I introduce Minkowski space with its indefinite fundamental form first, before drawing any diagrams.

Yes, I know you introduce the hyperbolic functions.
But do use them (like the circular-trigonometric functions) to solve problems involving triangles,
e.g. finding components of vectors in the given frame or in another frame.
Do you use the terms opposite, adjacent, and hypotenuse?

 

[vanhees71]

Of course not. I use vector algebra to solve kinematical problems. The greatest progress in the understanding of geometry since Euclid was the introduction of analytical methods by Descartes et al.

 

[robphy]

Of course not. I use vector algebra to solve kinematical problems. The greatest progress in the understanding of geometry since Euclid was the introduction of analytical methods by Descartes et al.

So,

• in a first introduction to free-body diagrams [for introductory students],
  do you use vector-algebra and dot-products to solve (say) an inclined-plane problem?
• in a first introduction to relativity [for introductory students],
  do you use vector-algebra and dot-products to solve (say) a time-dilation problem?

 

[vanhees71]

For the inclined plane I use sometimes a free-body diagram, and there of course Euclidean geometry and language is adequate. For me the liberation from free-body diagrams by using generalized coordinates and Hamilton's principle was a great relief ;-)).

For the kinematical effects in SR I use the Lorentz transformation, which I introduce from the two postulates, motivating the Minkowski product (or rather the corresponding quadratic form). Only then I introduce Minkowski diagram for illustrative purposes making clear from the first moment on that there are no Euclidean notions left on the corresponding "paper plane".

 

[robphy]

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).

So which is it (or is it both)?

An important feature of elapsed “proper time” (“wristwatch time”) is that it is a function of the timelike-path between the endpoint-events.
So, the difference of two readings of the wristwatch worn by the astronaut measures the elapsed proper time for that astronaut.

(To me, “Interval” suggests “magnitude of a displacement”, which is associated with a straight path. So I avoid it in this context.)

“Proper distance” as an analogue of proper time is of limited value for physics. While proper time can measured by a wristwatch along a timelike curve (the worldline of the wristwatch), a general spacelike curve doesn’t have much physical meaning, e.g, those spacelike curves that are not on a constant time slice associated with an observer.

We have examples of spacelike curves that are not achronal…. so there could be a causal curve joining two events on a “curve with an everywhere-spacelike tangent-vector.”

However “proper time along a worldline with time measured by a wristwatch” is a good analogy to “distance along a path in Euclidean space with distance measured by an odometer “.

 

[Freixas]

I strongly suggest taking the time to draw a spacetime diagram of the above.

It turns out I had already done this in #113. However, I combined the two approaches and here is the result:

First, I drew a red and green line using your formulas, with $x_r = 1$ and $x_f =2$, stepping through rest time from 0 to 10 in steps of 0.1. Here is the Gamma code:

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; }

Then I created observers whose acceleration was 1/1 and 1/2. I drew the former in blue, but didn't draw the latter. The blue line overlays the red, as I would expect.

I located events on the rear worldline at intervals of 0.1 rest time. I found the comoving frame for this event and transformed the point (1, 0) in the comoving frame back to the rest frame. I connected the event to the transformed point. The right endpoint of this line falls on the green line generated using your formulas.

I then used the formulas on the rocket page to calculate tau for each worldline at each of the lines connecting the two curves. I had formerly eyeballed that the front clock would have more time than the rear clock and this seems to be the case.

The lines represent the lines of simultaneity for the rear worldline, so to an observer at the rear endpoint, everything on the line has the same time value. But the observer on the front endpoint would not agree. Here's what I originally said:

If an object is under acceleration does it still have a rest frame (and thus a proper length)? If the endpoints are accelerating uniformly, then its "proper length" is not invariant. If we insist on maintaining an invariant "proper length", then clocks at the endpoints of the length must be moving at different velocities, and so cannot be in the same rest frame.

It looks like the phrase "cannot be in the same rest frame" is incorrect. I should have said something like "the two ends of the object will not agree on a common rest frame (except at the start)."

Would this mean that if someone at the front of an accelerating object shined a light at someone at the rear, the light would be blue-shifted?

 

[PeterDonis]

The lines represent the lines of simultaneity for the rear worldline, so to an observer at the rear endpoint, everything on the line has the same time value. But the observer on the front endpoint would not agree.

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.

In other words, for any given line of simultaneity, you can construct an inertial frame for which that line is the $x'$ axis (this frame is what I referred to as the momentarily comoving inertial frame), and both observers will be at rest in that inertial frame at the events on their worldlines that intersect the line of simultaneity you chose. But you will have to adjust both observers' clocks by different amounts to make them (momentarily) synchronized in that inertial frame.

What this means is that if you try to construct a single non inertial frame that includes both observers (which is what Rindler coordinates is), the metric coefficient $g_{00}$ in this frame will depend on spatial position; it will not be constant.

I should have said something like "the two ends of the object will not agree on a common rest frame (except at the start)."

No, that's not correct either. See above.

Another invariant way of showing that there is a "common rest frame" for both observers is to look at round-trip light travel times. If a light signal bounces back and forth between the two observers, each observer will measure its round-trip travel time to be constant, indicating that the distance between the two observers is not changing. The actual time they measure will be different for the two observers (the rear observer will measure a smaller round-trip travel time than the front), but for both observers, the time will not change.

Would this mean that if someone at the front of an accelerating object shined a light at someone at the rear, the light would be blue-shifted?

Yes.

 

[Freixas]

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.

Ok, I partially understand this, although I can see the gaps in my knowledge.

Let's select three points: $(0, 0)$, $(x_r, 0)$, $(x_f, 0)$. We'll say we are viewing these points in an instantaneous rest frame: a comoving frame. If we boost these points to view them relative to any other frame, $(0, 0)$ stays where it is and the other two trace hyperbolic paths. Lines connecting the points remain colinear regardless of the boost.

The boosted views are just different views of the original setup. The lines connecting the points represent proper lengths as well as common lines of simultaneity.

But that's boosting. If we create worldlines that match the boosted hyperbolic paths, we
can pick points that can be "reverse boosted" back to the original setup. So any worldlines set up this way will have points that maintain proper length and simultaneity.

Boosting the $(x_r, 0)$ and $(x_f, 0)$ points and connecting them with a line yields the same line as finding the comoving frame of the hyperbolic worldlines (from either end). To find the line for the comoving frame, I pick an event on the worldline, find the tangent, and then find the line mirroring the tangent around a 45 degree line that goes through that event. Why this is the same line as we get by boosting the original two points (a Lorentz transformation of each point using a common relative velocity) requires some mathematical magic than I don't know. For the moment, I can accept that these are equivalent methods.

Another missing gap is why creating worldlines that match boosted points results in proper acceleration (that it results in acceleration is not a mystery).

Then there's the mystery of how two observers can maintain simultaneity with each other at every instant yet have different elapsed time. If observers move inertially and have different clock rates, they also have different lines of simultaneity.

Because "simultaneity" is a little woo-woo in the sense that the selection of a line of simultaneity is arbitrary, I thought I would look for things that could be directly observed, which is why I mentioned the blue shift.

So here are some of the other freaky things that I think happen in an accelerating ship that maintains its proper length:

• An observer looking toward the front sees a blue shift, indicating relative motion (in the past) toward the observer. Since the blue shift increases with time, the observer could conclude that the front of the ship is approaching at ever-greater velocity.
• Looking toward the rear, the effects are reversed: there is an increasing red shift, showing that the rear of the ship in receding at an ever-greater rate.
• On the other hand, you mentioned that the round-trip travel time of light would be constant for observers at each end, so they would each conclude that the other end of the ship remained a constant distance away. I gather that the frequency of the reflected light would be unchanged.
• If an observer at the rear of the ship pulls out a tape measure and begins to lay it down while walking toward the front, the length reading on the tape will be the same as when the ship was at rest and will not change at any time during acceleration. Any other equivalent method will work, too: the observer could count steps moving from one end to the other and compare the count to one obtained moving in the opposite direction.

To repeat, I'm not doing anything other than highlighting the gaps in my understanding. I understand some bits, but not others.

 

[PeterDonis]

Let's select three points: $(0, 0)$, $(x_r, 0)$, $(x_f, 0)$. We'll say we are viewing these points in an instantaneous rest frame: a comoving frame

Yes, that's obvious from the coordinates you give.

If we boost these points to view them relative to any other frame, $(0, 0)$ stays where it is and the other two trace hyperbolic paths.

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.

Lines connecting the points remain colinear regardless of the boost.

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.

The boosted views are just different views of the original setup.

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.)

The lines connecting the points represent proper lengths as well as common lines of simultaneity.

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.

If we create worldlines that match the boosted hyperbolic paths, we
can pick points that can be "reverse boosted" back to the original setup. So any worldlines set up this way will have points that maintain proper length and simultaneity.

See above regarding "passive" vs. "active" transformations.

Boosting the $(x_r, 0)$ and $(x_f, 0)$ points and connecting them with a line yields the same line as finding the comoving frame of the hyperbolic worldlines (from either end).

Yes.

To find the line for the comoving frame, I pick an event on the worldline, find the tangent, and then find the line mirroring the tangent around a 45 degree line that goes through that event.

Yes.

Why this is the same line as we get by boosting the original two points (a Lorentz transformation of each point using a common relative velocity) requires some mathematical magic than I don't know.

It has to do with the hyperbolas being integral curves of the boost Killing vector field.

Another missing gap is why creating worldlines that match boosted points results in proper acceleration (that it results in acceleration is not a mystery).

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.

Then there's the mystery of how two observers can maintain simultaneity with each other at every instant yet have different elapsed time.

This isn't a mystery. It's a simple consequence of spacetime geometry and the geometry of the hyperbolas.

If observers move inertially and have different clock rates, they also have different lines of simultaneity.

Obviously this is false since we have just gone to considerable lengths to describe a counterexample, so why are you asserting it?

Because "simultaneity" is a little woo-woo in the sense that the selection of a line of simultaneity is arbitrary

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.

• An observer looking toward the front sees a blue shift

Yes.

• indicating relative motion (in the past) toward the observer.

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.)

• the blue shift increases with time

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.

• you mentioned that the round-trip travel time of light would be constant for observers at each end,

Yes.

• so they would each conclude that the other end of the ship remained a constant distance away. I gather that the frequency of the reflected light would be unchanged.

Yes.

• If an observer at the rear of the ship pulls out a tape measure and begins to lay it down while walking toward the front, the length reading on the tape will be the same as when the ship was at rest and will not change at any time during acceleration.

Assuming that the walking is done slowly enough, yes.

 

[Freixas]

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.

Yes.

(These two different ways of viewing the action of a Lorentz boost are sometimes referred to as "passive" vs. "active" transformations in the literature.)

Interesting.

This isn't a mystery. It's a simple consequence of spacetime geometry and the geometry of the hyperbolas.

It's a mystery to me. I've verified that this is true with Gamma. But that's different from understanding it.

If observers move inertially and have different clock rates, they also have different lines of simultaneity.

Obviously this is false since we have just gone to considerable lengths to describe a counterexample, so why are you asserting it?

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.

From the viewpoint of an inertial frame, yes. But not from the viewpoint of the non-inertial frame (Rindler coordinates).

Interesting.

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.

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.

Assuming that the walking is done slowly enough, yes.

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.

The tape that is unwound has a relative speed of 0 and so it has no length contraction relative to the ship. The tape on the unwinding spool has a high relative velocity with small fluctuations, and so is length contracted. At the point where it unwinds onto the ship, it would experience a rapid change of velocity and so its length would rapidly expand. Would this cause a problem to the person unwinding the spool? I don't know.

In picturing the worldlines of points of the tape and of the person carrying it, I don't see a problem, but since you required walking slowly, there must be.

Thanks for all the cross-checking and the explanations!

 

[Freixas]

An important feature of elapsed “proper time” (“wristwatch time”) is that it is a function of the timelike-path between the endpoint-events. So, the difference of two readings of the wristwatch worn by the astronaut measures the elapsed proper time for that astronaut.

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.

People might seem to understand "proper time", because problems often have a context that implies that one is to calculate a duration. To date, I've thought it could be used in both senses--the time at some instant and a duration.

 

[robphy]

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.

I carry a ruler. The "measurement I read from it" is my measurement of a length.

I carry a wristwatch. The "measurement I read from it" (between two readings of it) is my measurement of the elapsed time that I and my wristwatch experienced... this is my elapsed proper time
or more-descriptively "wristwatch time"
(and everyone will agree that this is the elapsed time that I and my wristwatch experienced).

Practically any technical term is ambiguous... that's why it's a technical term.

I think part of the problem is that the word "proper" is what is ambiguous.
https://en.wikipedia.org/wiki/Proper_time is an invariant.
https://en.wikipedia.org/wiki/Proper_acceleration is an invariant
https://en.wikipedia.org/wiki/Proper_velocity is not invariant, it is relative to an observer.
It might be difficult for a novice to use "proper" properly.

 

[Freixas]

I carry a wristwatch. The "measurement I read from it" (between two readings of it) is my measurement of the elapsed time that I and my wristwatch experienced... this is my elapsed proper time (and everyone will agree that this is the elapsed time that I and my wristwatch experienced).

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.

They adopted the term thinking it made things easier for beginners to to understand. I found it confusing. "Proper time" is only a little better, but it's the most commonly used term and so it's required learning. "Proper duration" is crystal clear and is the term I would prefer, but physicists aren't going to adopt my terminology. If you really like wristwatches, "wristwatch duration" would work. "Stopwatch time" might also work.

 

[PeterDonis]

I thought we had been talking about observers who were accelerating. Observers moving at constant velocity

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).

 

[PeterDonis]

It's a mystery to me.

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.

 

[PeterDonis]

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.

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.

 

[PeterDonis]

At the point where it unwinds onto the ship, it would experience a rapid change of velocity and so its length would rapidly expand.

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.

 

[Ibix]

Then there's the mystery of how two observers can maintain simultaneity with each other at every instant yet have different elapsed time. If observers move inertially and have different clock rates, they also have different lines of simultaneity.

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.

Here's a statement about Euclidean geometry: If we have straight lines, lines perpendicular to those lines are only parallel if the original lines were parallel. The equivalent in Minkowski geometry is: If we move inertially, we only share simultaneity planes orthogonal to our worldlines if we have the same velocity.

Euclid: If we have circles, lines perpendicular to those circles are only parallel if the circles are concentric, and the arc length between one radial line and another is different on different circles. Minkowski: if we follow hyperbolic worldlines we share simultaneity only if our paths are concentric, and the proper time between one shared simultaneity plane and the next is different along different hyperbolae.

 

[robphy]

Then there's the mystery of how two observers can maintain simultaneity with each other at every instant yet have different elapsed time. If observers move inertially and have different clock rates, they also have different lines of simultaneity.

Since our astronaut is now non-inertial,
durations measured with radar are now different from
durations measured with a wristwatch.

 

[Freixas]

As I said, it's a simple consequence of spacetime geometry and the geometry of the hyperbolas.

Those "nested" hyperbolae you've drawn are the Minkowski equivalent of concentric circles

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).