# What, exactly, are invariants?

• I
• Freixas
In summary, the conversation discusses various aspects of Special Relativity (S.R.) basics, including the concept of invariants and their role in reflecting actual physics. The conversation also touches on the idea of observables being invariants, and the potential for certain invariants to depend on frame of reference and simultaneity conventions. The conversation also includes a simple problem involving two observers meeting and the concept of a lattice as a way to measure events. The conversation concludes with a discussion of different types of invariants and the important distinction between invariants and observables.
Freixas
I have a number of questions about some S.R. basics. The answer to one question may affect the next, so I will try to focus on one question at a time. My first post was on simultaneity conventions. This one is about invariants.

@PeterDonis has been a big influence on how I try to approach S.R. What follows is a summary of some basic rules he gave me (any errors in paraphrasing are mine, of course):
• If something is an invariant, it reflects actual physics.
• Things that are not invariant are artifacts of a choice of coordinates, but can still be useful in calculating invariants.
• Any analysis that is coordinate dependent is going to tell you some things that have nothing to do with the actual physics, but are artifacts of your choice of coordinates.
Clearly, I need to ensure that I fully understand S.R. invariants.

I’ve found definitions similar to this: Relativistic invariance (or Lorentz invariance) means "the same regardless of frame of reference".

I think observables are invariants. The definition I’ve found for an observable is “a physical quantity that can be measured.”

Are events invariants? Some definitions are that an event is a “point in spacetime”. Others say an event matches a physical occurrence with a point in spacetime. A physical occurrence sounds like an observable, although some physical occurrences might not be measurable. Time and space, however, are coordinates that are generally not thought of as invariants.

Bob and Alice meet. This sounds like an event and even sounds observable, but I’m not clear what we would measure.

I came up with an invariant-qualification scheme that will lead to some further questions.
• Type I invariants have no qualifiers. I would like to include events in this class. A less problematic example would be c, the speed of light (the two-way speed, anyway).
• Type II variants are qualified only by their frame of reference. The only examples I have are proper time and proper location—the term “proper” identifies that these are measured relative to the inertial frame in which the observer and any relevant objects are at rest.
• Type III invariants require a frame of reference and a simultaneity convention.
The type III invariants are problematic as the group includes things such as lengths and time intervals which (if not proper) are generally considered not invariant. (Does anyone say “variant” instead of “not invariant”?)

Let’s consider a simple problem: Two observers meet. One is at rest; the other is moving at 80%c relative to the first. How long (by the first observer’s clock) after their meeting will the second observer be 4 LY away (a proper length relative to the first)?

We can use a 3D lattice as per Taylor/Wheeler, with observation equipment and clocks at each lattice junction. When a lattice point 4 LY from the first observer notes the presence of the second, it marks the time. This information can later be downloaded.

The meeting of the second observer with this lattice point is an event; let’s call it A. The lattice will also contain other events whose time matches this event’s time. One of those events will match a lattice point with the first observer; let’s call it B.

The clock time for event A is an invariant, but to say B occurs at the same time as A seems like an artifact of the coordinate system. But both things depend on the same (arbitrary) simultaneity convention used to initialize the lattice’s clocks. Also, that the two events occurred at the same time is literally what the lattice tells us.

In this example, a single time value is associated with event B. We only get in trouble if we try to match this time to a time occurring elsewhere. There are some type III invariants, such as lengths, velocities, and clock rate comparisons, which always require matching the time for two points separated by some distance. But we could imagine, for example, a ship passing a planet and then having the observed length of the ship posted on a huge sign. This seems to transform a length into an observable and thus an invariant.

I am clearly hazy as to whether what I’ve called type III invariants are actually invariants at all or only under certain conditions.

P.S. I'm sure there are other types of invariants. I just focused on the basic ones I'm familiar with.

Freixas said:
• If something is an invariant, it reflects actual physics.
• Things that are not invariant are artifacts of a choice of coordinates, but can still be useful in calculating invariants.
• Any analysis that is coordinate dependent is going to tell you some things that have nothing to do with the actual physics, but are artifacts of your choice of coordinates.
These are all fine.

Freixas said:
I think observables are invariants. The definition I’ve found for an observable is “a physical quantity that can be measured.”
Yes. More precisely, any quantity that is observable--that can be directly observed and measured--must be represented by an invariant in the math. Unfortunately, many presentations of relativity obfuscate this by choosing coordinates in which many invariants of interest are numerically equal to particular coordinate values.

Freixas said:
Are events invariants?
In the most general sense of "invariant", yes--meaning, the sense that refers to geometric objects themselves, as well as particular observable properties that they have. Geometrically, events are points in spacetime at which particular things happen--for example, "lightning strikes a particular point on the embankment next to a train track on which Einstein's train is moving to the right" is a happening that takes place at a particular point in spacetime, and that point is an event. Something like the energy delivered by the lightning strike would be a particular observable property associated with that event.

Freixas said:
Time and space, however, are coordinates that are generally not thought of as invariants.
That depends on how you define "time" and "space". "Time" can mean proper time along some particular worldline, which is an invariant. "Space" can mean the proper length along some particular spacelike curve (although this would more likely be called "distance"), which is also an invariant.

Freixas said:
Bob and Alice meet. This sounds like an event and even sounds observable, but I’m not clear what we would measure.
"Bob and Alice meet" is an observable happening. The point in spacetime at which this happening takes place is an event.

Freixas said:
I came up with an invariant-qualification scheme that will lead to some further questions.
• Type I invariants have no qualifiers. I would like to include events in this class. A less problematic example would be c, the speed of light (the two-way speed, anyway).
• Type II variants are qualified only by their frame of reference. The only examples I have are proper time and proper location—the term “proper” identifies that these are measured relative to the inertial frame in which the observer and any relevant objects are at rest.
• Type III invariants require a frame of reference and a simultaneity convention.
Unfortunately, this is wrong, because invariants don't require any frame of reference. Proper time is not measured relative to any frame; it's measured by a clock following the chosen worldline. "Proper location" isn't a term I'm familiar with in relativity, but "proper length" and "proper distance" are, and both are normally used to refer to invariants that don't require any frame of reference.

Freixas said:
Let’s consider a simple problem: Two observers meet. One is at rest; the other is moving at 80%c relative to the first. How long (by the first observer’s clock) after their meeting will the second observer be 4 LY away
Normally one would say this depends on whose frame we choose, or more specifically what simultaneity convention we choose (it should be obvious that a simultaneity convention must be involved since you are asking for a time by the first observer's clock of an event on the second observer's worldline where the second observer is not co-located with the first).

Freixas said:
(a proper length relative to the first)?
No, it's one of at least two possible proper lengths, depending on which spacelike curve you choose to measure along:

(1) There is a spacelike curve with proper length 4 LY that goes from the first observer's worldline to the second observer's worldline, and is orthogonal to the first observer's worldline. This curve intersects the first observer's worldline at an event with proper time t1 by the first observer's clock.

(2) There is a spacelike curve with proper length 4 LY that goes from the first observer's worldline to the second observer's worldline, and is orthogonal to the second observer's worldline. This curve intersects the first observer's worldline at an event with proper time t2 by the first observer's clock.

Most relativity texts would describe (1) as using the simultaneity convention of the first observer's rest frame, and (2) as using the simultaneity convention of the second observer's rest frame.

Freixas said:
We can use a 3D lattice as per Taylor/Wheeler, with observation equipment and clocks at each lattice junction.
This is equivalent to adopting a simultaneity convention, and, as above, there are at least two choices of that.

Freixas said:
The clock time for event A is an invariant, but to say B occurs at the same time as A seems like an artifact of the coordinate system.
More precisely, it's an artifact of the simultaneity convention chosen.

Freixas said:
But both things depend on the same (arbitrary) simultaneity convention used to initialize the lattice’s clocks.
No. The clock time along a single worldline doesn't depend on any simultaneity convention. Only claims that two events on different worldlines occurred at the same time do.

Freixas said:
that the two events occurred at the same time is literally what the lattice tells us.
Well, of course, because you used a particular simultaneity convention to build the lattice.

Freixas said:
There are some type III invariants, such as lengths, velocities, and clock rate comparisons, which always require matching the time for two points separated by some distance.
No, they don't. They require picking out particular geometric objects--particular worldlines, particular spacelike curves, etc. A person might have different reasons for picking out particular geometric objects, and those reasons, for example "following a particular simultaneity convention", might depend on artifacts of coordinates or other conventions. But a particular person's reason for picking a particular geometric object is not the same thing as the invariants associated with that geometric object. The proper length along a particular spacelike curve between two events is an invariant regardless of why we picked out that particular curve and those particular events.

vanhees71, topsquark, Orodruin and 1 other person
I find it's worth drawing a direct analogy between Minkowski geometry and Euclidean geometry. Invariants are things you can talk about without referencing facts not in evidence.

So the length of a line segment drawn on a piece of paper is an invariant - I can point to the line and (beside a text on Euclidean geometry) that's all you need. Similarly proper time - the interval along a worldline between two specified events - is an invariant.

The angle of a line is not an invariant, because what do I mean by "angle"? Angle between the line and... what? As soon as I say "the angle between this line and that one", or even "the angle between this line and the direction I called ##x##", then that is an invariant. Similarly, an object's rapidity is not an invariant until I specify a worldline that I am using as "at rest". Then everyone can measure the rapidity of the object with respect to my "rest" worldline and will agree my value.

That's kind of what goes wrong in OP's distance example. Defining the relative velocity as 0.8c works fine - it specifies the angle between the worldlines. Asking about time by their clocks means to measure "distance" along their worldlines after the crossing point - again, fine. But talking about "the distance between them" fails because the distance from an event on one worldline to the other worldline is ill-defined. What it probably means is the distance from point on A's worldline to B's worldline along a line perpendicular to A's worldline - but it doesn't say that last bit. As soon as that last bit is added the problem is entirely specified in terms of angles and lines and we're fine.

The complexity in Minkowski geometry is that all worldlines are timelike. That's different from Euclidean geometry where we can draw a real physical line pointing in any direction. This property makes proper time easy to talk about because the worldlines are physical objects and we can talk about "distances" along those lines between events. But there are no spacelike lines. Those we have to specify ourselves, which we usually do either by specifying a straight line between two events or by specifying orthogonality with a timelike line.

robphy, vanhees71 and topsquark
Freixas said:
I think observables are invariants. The definition I’ve found for an observable is “a physical quantity that can be measured.”
PeterDonis said:
Yes. More precisely, any quantity that is observable--that can be directly observed and measured--must be represented by an invariant in the math.
@Freixas this point is important and it deserves some detail.

Suppose that we have some free particle of mass ##m## and two inertial frames, primed and unprimed. There is a device which measures the total energy of the particle, and that device is at rest in the unprimed frame.

Using units where c=1, the particle has a four-momentum ##\mathbf p = m \mathbf u## where ##\mathbf u## is the particle’s four-velocity. The energy measuring device has a four-velocity of ##\mathbf v##

What is important to understand is that the four-vectors, ##\mathbf p##, ##\mathbf v##, and ##\mathbf u## are geometric objects. Different reference frames agree on the geometric object itself.

What they disagree on are the components of the four-vectors. So they will say $$\mathbf p = \sum_{\mu=0}^{3} p^\mu \mathbf{ e}_\mu = \sum_{\mu=0}^{3} p’^\mu \mathbf{ e’}_\mu$$ where the ##p^\mu\ne p’^\mu## are the components of the vector with respect to the basis vectors ##\mathbf{ e}## and ##\mathbf{ e’}## for the unprimed and primed frames respectively.

The different frames agree on the ##\mathbf p## geometric object but not on the ##p^\mu## components because they are using different basis vectors. However, as long the basis vectors, ##\mathbf{e}##, have been specified then we can reconstruct the geometric four-vector ##\mathbf p## from the components ##p^\mu##. So we will often write ##p^\mu## referring to ##\mathbf p## when the basis vectors have been identified, e.g. by explicitly identifying the reference frame.

This is why very often you will see old members tell new members that they need to specify the reference frame for something to have meaning. Without the basis vectors we cannot reconstruct the geometric object from the components.

Now, with that background, the energy of the particle in the unprimed frame is ##p^0\ne p’^0## which is not equal to the energy in the unprimed frame. The two frames disagree on the energy of the particle because the energy is a single component of ##\mathbf p## and they are using different basis vectors.

However, when the device measures the energy the result is ##\mathbf{p}\cdot \mathbf{v}=p^0## which all frames do agree on. So different frames do not agree that ##p^0## is the energy, all frames do agree that ##\mathbf{p}\cdot \mathbf{v}## is what the device measures the energy to be.

Hopefully that subtlety is clear. The energy (a component, ##p_0##) is frame variant, but the detector's measurement of the energy (a scalar, ##\mathbf{p}\cdot\mathbf{v}##) is invariant.

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member 728827, dextercioby, vanhees71 and 3 others
There is a lot to think about in the responses. It may be best to deal with one topics in a response. Here I will discuss proper time and length.

The terms don't seem analogous; time is to duration as location is to length. I think the term "proper time" is usually used to mean "proper duration" or maybe "proper clock rate." Time, in the sense of reading a time value of a clock, depends on knowing the arbitrary origin point of the clock's time and so doesn't seem like an invariant--until you specify the origin.

In Spacetime Physics, Taylor/Wheeler like to talk about "wristwatch time", which has the same problem--I can change the time on my wristwatch at any time. If the clock is accurate, however, the durations it measures are probably what people mean by "proper time."

So the terms "proper time" and "proper length" have an inherent confusion, at least when the words are interpreted with their regular meanings.

Qualifying the terms by a worldline rather than a rest frame sounds good. It probably accommodates accelerated movement better, although I always thought of an accelerated observer as always in a rest frame, even if that frame is different in the next instant. Therefore, it seemed that if I said that a particular distance was a proper distance relative to an observer, that was sufficient to be both accurate and brief.

Speaking of which, is there a shorter way of qualifying a length than saying "as measured on a spacelike curve orthogonal to the first observer's worldline and going from the first observer's worldline to the second's"? I just searched Taylor/Wheeler's book and the word "orthogonal" does not appear anywhere.

Here's a sentence that does appear in the book: "Let ##L_0## be the proper length of the rod when measured at rest." Shouldn't this be "Let ##L_0## be the proper length of the rod," or "Let ##L_0## be the length of the rod when measured at rest"? I'm pretty sure that the concept of "proper length" being the rest length of an object (meaning the length of an object as measured by an observer at rest with respect to the object), is a common way of introducing proper lengths.

Thanks for the precise definition. How would you rephrase the problem so that it is both accurate and brief? If it's not yet clear, I meant option (1) of the two you listed.
Freixas said:
Two observers meet. One is at rest; the other is moving at 80%c relative to the first. How long (by the first observer’s clock) after their meeting will the second observer be 4 LY away (a proper length relative to the first)?

In my problem, the observation of the time value on the clock at event A make the time value an invariant; that's clear. All observers, regardless of their motion, would observe the same time value.

But to calculate the value of this invariant (to "do" physics), I need to know that the lattice is at rest relative to observer A and I need to know how the lattice's clocks were set. Without these two things, I cannot calculate the invariant time value at event A.

Thus, I came up with type III invariants. But this classification cannot be right because invariants cannot be frame-dependent (I suspect they cannot also be dependent on a simultaneity convention). So I go in a circle.

One possible way around this problem: the 80%c velocity implies both a relative frame and a simultaneity convention. If we assume the lattice must be set using the same convention, maybe we need no further assumptions to calculate the answer. In other words, for any two inertial frames for the two observers and any simultaneity convention (applied consistently), the velocity leads us to the final answer. I'll have to think about this.

@Dale's response may address this question; unfortunately, as soon he gets into four-momentum, four-vectors, and basis vectors, he has zoomed over my math horizon. The paper mentioned by @Demystifier might be equally relevant, but it's equally difficult to follow.

Freixas said:
unfortunately, as soon he gets into four-momentum, four-vectors, and basis vectors, he has zoomed over my math horizon
You should pause then, and update your math horizon. The four-vectors are the simplest way to understand invariants. It will be easier to learn the math that naturally expresses these concepts than it will be to learn these concepts without the math. A little effort now will make this thing you are interested in easier, as well as many related future questions that you will have. It is by far worth the effort, particularly as you have an immediate application/need for it.

topsquark, vanhees71 and berkeman
Indeed, four-vectors and tensors make special relativity much easier to formulate and understand. Imho it's bad didactics to avoid them. E.g., it's much more simple to use Einstein's two postulates (special principle of relativity + "constancy" of the speed of light) and first introduce the affine Minkowski space and only then derive the Lorentz transformations between inertial frames than to first go through the cumbersome gedanken experiments a la Einstein (1905).

On the other hand, of course, one should also discuss the physics coming out from the analysis in Minkowski space (relativity of simultaneity, time dilation, length contraction and all that), but it's much easier to explain, having the four-vector formalism at hand than with the (1+3)-formalism a la Einstein.

dextercioby, topsquark, Dale and 1 other person
vanhees71 said:
first introduce the affine Minkowski space and only then derive the Lorentz transformations between inertial frames than to first go through the cumbersome gedanken experiments a la Einstein (1905).
This cannot be overstated. For some reason many seem to insist on teaching relativity the way it was discovered by Einstein more than a century ago. The modern understanding of relativity and the didactics have developed significantly since then.

dextercioby, Vanadium 50, topsquark and 3 others

Orodruin said:
This cannot be overstated. For some reason many seem to insist on teaching relativity the way it was discovered by Einstein more than a century ago. The modern understanding of relativity and the didactics have developed significantly since then.

I blame Einstein's initial reactions to the mathematical formulations. (TL;DR: Math is hard.)
"Since the mathematicians have invaded the theory of relativity, I do not understand it myself anymore."
(In A. Sommerfeld "To Albert Einstein's Seventieth Birthday" in Paul A. Schilpp (ed.) Albert Einstein, Philosopher-Scientist, Evanston, 1949.
https://www.maa.org/press/periodicals/convergence/quotations/einstein-Albert-1879-1955-21 )
Abraham Pais also reported that before 1912 Einstein told V. Bergmann that
he regarded the transcription of his theory into tensor form as "überflüssige Gelehrsamkeit" (superfluous learnedness).
(in Pais, Abraham, Subtle is the Lord. The Science and Life of Albert Einstein, 1982, Oxford: Oxford University Press, p. 152.)

quote from the footnote of page 2 of
Galina Weinstein
"Max Born, Albert Einstein and Hermann Minkowski's Space-Time Formalism of Special Relativity"
https://arxiv.org/abs/1210.6929

And now some quotes from a spacetime-diagram proponent: JL Synge. Bolding mine.
"We are not embarking on a programme of "graphical relativity".
Our space-time diagrams are to be used as a mathematician or physicist uses rough sketches,
rather than as an architect or engineer uses blueprints.
The diagrams are to serve as guides for the mind.
Anyone who studies relativity without understanding
how to use simple space-time diagrams
is as much inhibited as a student of
functions of a complex variable who
does not understand the Argand diagram."

- J.L. Synge in Relativity: The Special Theory (1956), p. 63
"The basic idea is to present the essentials of relativity from the
Minkowskian point of view, that is, in terms of the geometry of space-time.
This geometrical approach is used to some extent in all expositions of relativity,
but I have emphasised it more than is customary,
because it is to me (and I think to many others) the key which unlocks many mysteries.
My ambition has been to make space-time a real workshop for physicists,
and not a museum visited occasionally with a feeling of awe. "

[snip]

"To understand a subject, one must tear it apart
and reconstruct it in a form intellectually satisfying to oneself,
and that (in the view of the differences between individual minds)
is likely to be different from the original form.
This new synthesis is of course not an individual effort;
it is the result of much reading and of countless informal discussions,
but for it one must in the end take individual responsibility.
Therefore, I apologise, if apology is necessary,
for departing from certain traditional approaches which seemed to me unclear,
and for insisting that the time has come in relativity
to abandon an historical order and to present the subject as a completed whole,
completed, that is, in its essentials.
In this age of specialisation, history is best left to the historians."

- J.L. Synge in Relativity: The Special Theory (1956), p. vii

"I have described the spirit of this book as ironical, but it is also geometrical.
Perhaps the two things go together, for a simple space-time diagram
will often bring out the inner meaning of a mass of calculations.
Surely one of the reasons why the general theory of relativity
remains a mystery to so many physicists is that
they do not realize how easy it is to form a qualitative geometrical image of what is going on.
It is in fact easier to deal with space-time diagrams, which remain fixed,
than with the kinematical pictures of Newtonian mechanics."

- J.L. Synge in Relativity: The General Theory (1960), p. viii

Freixas
Ibix said:
I find it's worth drawing a direct analogy between Minkowski geometry and Euclidean geometry. Invariants are things you can talk about without referencing facts not in evidence.

So the length of a line segment drawn on a piece of paper is an invariant - I can point to the line and (beside a text on Euclidean geometry) that's all you need. Similarly proper time - the interval along a worldline between two specified events - is an invariant.

The angle of a line is not an invariant, because what do I mean by "angle"? Angle between the line and... what? As soon as I say "the angle between this line and that one", or even "the angle between this line and the direction I called ##x##", then that is an invariant. Similarly, an object's rapidity is not an invariant until I specify a worldline that I am using as "at rest". Then everyone can measure the rapidity of the object with respect to my "rest" worldline and will agree my value.
I would add that there should also be a geometric figure that implies the choice of metric
(for example, a unit circle or a unit hyperbola... or a "square" or rhombus?)
I am thinking along the lines of the "geometrical constructions with only a restricted set of tools".

Note that a vector operation ##\vec C=\vec A + \vec B## works equally well
in Euclidean geometry, Minkowski spacetime geometry, and Galilean spacetime geometry
(which are all affine spaces).
Vectors need not have a "magnitude"-function.. all they need are vector-addition and scalar multiplication.
They have a notion of parallelism, but not orthogonality.
There is a limited sense of length along the same line (from scalar multiplication).. e.g. bisection of a segment.

"Lengths" (i.e. comparing lengths in different directions) and "angles"
[and dot-products] are determined by the choice [specification] of a "unit circle".

To get a better handle on invariants,
as @Ibix says, think about Euclidean geometry.
What can one say universally
without laying down a coordinate grid or
independently of what coordinate grids are laid down?
For example, the intersection of two lines is an invariant.

In terms of a calculation,
what algebraic operations can be done with the structures you have?
With a vector space, vector-addition and scalar-multiplication. are available.
But a dot-product or a metric tensor is needed to (say) form a scalar from vectors.

In my opinion, many questions posed for Minkowski spacetime geometry
can also be posed for Euclidean geometry.
One distinction is that we often take our Euclidean tools for granted.
But we should rightly ask for operational definitions
of corresponding geometric constructions in Minkowski spacetime.

Freixas, Ibix and topsquark
Galilean spacetime is not an affine space but a fiber bundle! You just have a stack of 3D Euclidean affine space along the "time axis".

Nevertheless, indeed geometry plays a tremendous role in all of physics, but it's geometry understood in a wide sense as introduced by Riemann and, particularly, Klein with his "Erlangen Program". Imho. the key for an understanding of modern physics is its relation with the theory of symmetries. The physical laws take the form they take because of the symmetries of the underlying mathematical descriptions, starting with the spacetime models but also the symmetries underlying the Standard Model of elementary particle physics (both the fundamental local gauge symmetries of the strong (QCD) and electroweak interactions (QFD) and the "accidental" (approximate) symmetries like chiral symmetry of the light-quark sector) and their relations with conservation laws. From the corresponding Lie groups and algebras you understand also, how the operator algebras used in quantum mechanics (non-relativistic) as well as relativistic quantum field theory come about. In a sense it's all symmetry.

One should also note that Einstein changed his attitude towards math quite drastically, when he became aware of the usefulness of particularly these geometrical concepts (tensor analysis) in formulating his general theory of relativity, which cannot be formulated in any other way at all.

topsquark
vanhees71 said:
Galilean spacetime is not an affine space but a fiber bundle! You just have a stack of 3D Euclidean affine space along the "time axis".

Again?
( New Shimmer: https://www.nbc.com/saturday-night-live/video/shimmer-floor-wax/n8625
https://snltranscripts.jt.org/75/75ishimmer.phtml
It's a floor wax AND a dessert topping! )

In short: it's both a fiber bundle and an affine space.

Old discussions

I see now there is a disconnect in terminology...
for me, I am assuming we have the Law of Inertia: [timelike] lines as geodesics.

All bolding below is mine.

Here's Penrose
Penrose said:
(The Road to Reality 17.2; p.387)
Galilean spacetime ##\cal G## is not a product space ##E^1\times E^3##, it is a fiber bundle with base space ##E^1## and fiber ##E^3##! In a fiber bundle, there is no pointwise identification between
one fiber and the next; nevertheless the fibers fit together to form a connected whole.

(17.3; p. 388)
This ‘bundle’ picture of spacetime is all very well, but how are we to express the dynamics of Galileo–Newton in terms of it?
[snip]
In fact, in our Galilean spacetime,
world lines must always be cross-sections of the Galilean bundle; see §15.3.[17.2] and Fig. 17.3.
The notion of ‘uniform and in a straight line’, in ordinary spatial terms (an inertial motion), is interpreted simply as ‘straight’, in spacetime terms. Thus, the Galilean bundle ##\cal G## must have a structure that encodes the notion of ‘straightness’ of world lines. One way of saying this is to assert that ##\cal G## is an affine space (§14.1) in which the affine structure, when restricted to individual ##E^3## fibers, agrees with the Euclidean affine structure of each ##E^3##.

(robphy: interesting passage, p. 389)
Yet another way is to assert that the Galilean spacetime, considered as a manifold, possesses a connection which has both vanishing curvature and vanishing torsion (which is quite different from it possessing a bundle connection, when considered as a bundle over ##E^1##).[17.3] In fact, this third point of view is the most satisfactory, as it allows for the generalizations that we shall be needing in §§17.5,9 in order to describe
gravitation in accordance with Einstein’s ideas.

(14.1, p. 292)
However, a vector space is itself a very special type of space, and something much more general is needed for the mathematics of much of modern physics. Even Euclid’s ancient geometry is not a vector space, because a vector space has to have a particular distinguished point, namely the origin (given by the zero vector), whereas in Euclidean geometry every point is on an equal footing. In fact, Euclidean space is an example of what is called an affine space. An affine space is like a vector space but we ‘forget’ the origin; in effect, it is a space in which there is a consistent notion of parallelogram. As soon as we specify a particular point as origin this allows us to define vector addition by the ‘parallelogram law’.

Here is Trautman ( https://en.wikipedia.org/wiki/Andrzej_Trautman )
http://trautman.fuw.edu.pl/publications/scientific-articles.html
Trautman said:
#86 in the list above
Theory of gravitation, pp. 179–198 in: The Physicist’s Conception of Nature, ed. by J. Mehra, 1973.
http://trautman.fuw.edu.pl/publications/Papers-in-pdf/30_Andrzej_Trautman.pdf

(p.186)
In Newtonian physics, space-time E may be represented as a product T X S in
many ways; none of these representations are natural in a sense, which can be easily
related to the notation of natural transformations. Space is relative because there is
no absolute method of ascertaining whether or not two non-simultaneous events
happen at the same place. In other words, there is no natural horizontal slicing of E;
there is only a vertical f‌ibring corresponding to the projection ##\pi: E\rightarrow T ##
which associates to any event ##p \in E## the corresponding instant of time ##\tau=pi(p)##; or, time is absolute.

The last example provides us with the essential set-theoretic ingredients of a bundle...

(p.187)
I. The Galilean category ##\cal Gal## has Galilean spaces as objects.
A Galilean space is an aff‌ine space (E, V, +) endowed with a bilinear map ...

(diagram from p. 190)
Trautman said:
Spacetime and Gravitation (by Kopczynski and Trautman)
http://trautman.fuw.edu.pl/publicat...d_gravitation_Kopczynski_Trautman.pdf#page=30

(p. 20, Ch 3: Galilean Spacetime)
(p. 26)
Aff‌ine space.
An af‌fine space is a pair (E, V), where E is a set and V a vector space, with a mapping...

(p. 31)
We shall now def‌ine the geometrical structure of Galilean spacetime in a more mathematical manner. Galilean spacetime is a four-tuplet ##(E, V, \tau, h)##.
Here (E, V) is a four-dimensional aff‌ine space and ##\tau## is a form on ##V##, ...

In short: Galilean Spacetime is both a fiber bundle
and (when we have the Law of Inertia) an affine space.

topsquark and vanhees71
robphy said:
Again?
( New Shimmer: https://www.nbc.com/saturday-night-live/video/shimmer-floor-wax/n8625
https://snltranscripts.jt.org/75/75ishimmer.phtml
It's a floor wax AND a dessert topping! )

In short: it's both a fiber bundle and an affine space.

Old discussions

I see now there is a disconnect in terminology...
for me, I am assuming we have the Law of Inertia: [timelike] lines as geodesics.

All bolding below is mine.

Here's Penrose

Here is Trautman ( https://en.wikipedia.org/wiki/Andrzej_Trautman )
http://trautman.fuw.edu.pl/publications/scientific-articles.html
In short: Galilean Spacetime is both a fiber bundle
and (when we have the Law of Inertia) an affine space.
Or, in the world of internet memes …

topsquark and vanhees71
I'd like to focus on one point:

PeterDonis said:
Unfortunately, this is wrong, because invariants don't require any frame of reference.

I was looking at a thread about proper length and saw this (back from 2016):

vanhees71 said:
Defining something in a special frame of reference, well-defined by the physical situation is an invariant definition.

@vanhees71 statement is similar to my thinking about type III invariants. For example, the length of an object is not an invariant, but relative to a given frame and using a given simultaneity convention, the length is an invariant.

@Ibix says something that sounds similar to @vanhees71:

Ibix said:
The angle of a line is not an invariant, because what do I mean by "angle"? Angle between the line and... what? As soon as I say "the angle between this line and that one", or even "the angle between this line and the direction I called x", then that is an invariant.

Is there a difference between what @vanhees71 said and what I said?

I think the most useful thing I said was about "facts not in evidence". If there are two lines and you can point to them, everyone will agree about the angle (in Euclidean space) or "angle" (in Minkowski spacetime) between them. All the relevant facts are there. If you have a frame of reference, that's just a set of lines. As long as you tell me what those lines are I can measure things with respect to that frame even if it's not convenient for me to do so. All the relevant facts are there.

This is why I think your classification scheme is misconceived. All there is is lines and intersections and lengths and angles between lines. We sometimes call some of those lines "timelike axes of a reference frame", but they're still just lines that you either tell me about or don't. Classifying things according to which bits of information you haven't told me or how directly useful to me a particular invariant would be seems pointless to me.

Where things stop being invariant is just where you stop specifying all the information. "Speed" is not invariant because you haven't told me what "at rest" is. Add that information and we have an invariant, whether it is a useful invariant to me or not.

Dale
Ibix said:
Where things stop being invariant is just where you stop specifying all the information. "Speed" is not invariant because you haven't told me what "at rest" is. Add that information and we have an invariant, whether it is a useful invariant to me or not.

There were a few different things I've been trying to get a handle on. Due to Peter's influence, I was trying to approach physics problems as starting with invariants and ending with invariants. In the process of getting from the starting point to the ending point, coordinates might prove useful (apparently, there are ways solving problems purely with invariants), but the endpoints should be invariants.

To do that, I have to be able to identify an invariant. In many sources, especially sources written for beginners, something like "speed" is said to be not invariant, full stop, no exceptions (as in the Taylor/Wheeler book). But, according to you (and others), that appears to be incorrect. Properly qualified, speed can be an invariant.

Listing what information is required to make something an invariant is a separate issue, although it would seem useful to know.

If I want to start a problem with invariants, I should provide all the information needed to ensure that all values are actually invariants. In your example for "speed", you state that I should tell you what "at rest" is if I want it to be invariant. For some problems and some things, I might also need to tell you the simultaneity convention used (i.e. the length of a moving object depends on the frame relative to which it's measured and the simultaneity convention used as per https://sites.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/significance_conv_sim/index.html).

Ibix said:
All there is is lines and intersections and lengths and angles between lines. We sometimes call some of those lines "timelike axes of a reference frame", but they're still just lines that you either tell me about or don't.

I'm clearly missing the point you're making here. Lengths and angles seem like coordinates, not invariants. Could you present a simple physics problem using just the elements you list above? That might make things clearer.

Freixas said:
I have to be able to identify an invariant.
There are only two ways to identify an invariant:

1) transform it and show that it does not change under an arbitrary coordinate transform

2) write it in a manifestly invariant form

The second way is far easier, but requires updating your “math horizon” as I recommended in post 8. Based on your stated goals, this is really the path you should pursue.

PeroK, vanhees71 and PeterDonis
Dale said:
There are only two ways to identify an invariant:

1) transform it and show that it does not change under an arbitrary coordinate transform

2) write it in a manifestly invariant form

The second way is far easier, but requires updating your “math horizon” as I recommended in post 8. Based on your stated goals, this is really the path you should pursue.
Got it. Thanks.

Could you, by any chance, rewrite the problem I initially proposed in the OP using the math approach you recommend? Feel free to modify the problem to remove any ambiguities. I'm hoping that this is a trivial exercise, so that I'm not asking for a lot of work.

So that you don't have to scroll, here's the problem for reference.

Freixas said:
Two observers meet. One is at rest; the other is moving at 80%c relative to the first. How long (by the first observer’s clock) after their meeting will the second observer be 4 LY away (a proper length relative to the first)?

I thought I had clearly specified the 4 LY distance, but apparently not. My intent was that if the first observer had a ruler 4 LY long that he was holding (thus at rest relative to him) and lined up with observer 2's path, then the second observer would be 4 LY away when he is colocated with the far end of the ruler.

I don't believe the answer to the question is actually an invariant (the answer that I think is invariant is the time on a clock at the end of the ruler at the second observer's arrival; to determine this time, it may be necessary to provide the simultaneity convention used). It would be useful if you could demonstrate how your math approach would show that it is or isn't invariant.

The notion of “invariant” is not unique to relativity. It’s found in ordinary Euclidean geometry.

If you want to understand invariants,
I think you should first do it in a familiar setting… presumably, Euclidean geometry.

Once you understand invariants in that context, then move on to do it in relativity.

To do them simultaneously makes it difficult (in my opinion)
to distinguish things that are about “being invariant”
from things that are about “modeling kinematics in spacetime”.

An "x-component of a vector ##\vec a##", namely, ##a_x## (without reference to an axis) is not invariant... because rotating the graph paper leads to different x-components

A dot product of two vectors ##\vec a\cdot \vec b## is an invariant… since its value is the same independent of the orientation of the graph paper. It is a property of those two vectors and the [chosen] dot-product (because of the operations needed to construct the scalar), but not a property of any particular choice of orientation of axes.

An x-component of a vector with a specification of an axis is an invariant ##\vec a \cdot \hat O_x## (since it's a dot product). It's a property of the vector of interest ##\vec a##, a property of the measurer ##\hat O_x##, and the [chosen] dot product. Yes, it was what was measured by that axis (which all orientations of graph paper will confirm)... but it's not just about the vector of interest ##\vec a##.

PeroK and Freixas
Freixas said:
Two observers meet. One is at rest; the other is moving at 80%c relative to the first. How long (by the first observer’s clock) after their meeting will the second observer be 4 LY away (a proper length relative to the first)?
The last part is based on a notion of simultaneity so it is frame variant, but (although it is not clearly stated) it seems that you intend this to be in the frame of observer ##A##.

So I would say: the worldline of observer ##A## is $$r_A=(t_A,x_A)=(t,0)$$ and the worldline of observer ##B## is $$r_B=(t_B,x_B)=(t,0.8 \ t)$$ find ##t## when ##x_B-x_A=4##

Freixas
Dale said:
So I would say ... etc. ...
So you simplified this from a 4-vector to a 2-vector (assuming I am using the right terminology)? This is fine; I just want to make sure I have it right. I assume I can look up 4-vectors and just ignore the y and z terms to try to decode what you wrote.

Freixas said:
So you simplified this from a 4-vector to a 2-vector (assuming I am using the right terminology)? This is fine; I just want to make sure I have it right. I assume I can look up 4-vectors and just ignore the y and z terms to try to decode what you wrote.
Yes, the y and z terms were just 0.

Freixas
Freixas said:
Is there a difference between what @vanhees71 said and what I said?
Yes. Notice the precise wording that @vanhees71 used: "a special frame of reference, well-defined by the physical situation". In other words, the "frame of reference" he is talking about is made up of actual physical observables. And, as has already been pointed out, actual physical observables are invariants.

vanhees71
Freixas said:
To do that, I have to be able to identify an invariant.
And that means you need to read carefully and not just take what you read at face value. For example:

Freixas said:
In many sources, especially sources written for beginners, something like "speed" is said to be not invariant, full stop, no exceptions (as in the Taylor/Wheeler book). But, according to you (and others), that appears to be incorrect. Properly qualified, speed can be an invariant.
The word "speed" is vague ordinary language. You have to look behind that and ask what "speed" means in the math as the term is used in that particular source. Source that say "speed" is not invariant are talking about coordinate speed. Sources that say "speed" is an invariant are talking about relative speed (also known as "angle in spacetime between two timelike worldlines at the point where they intersect).

Freixas said:
Lengths and angles seem like coordinates, not invariants.
Read off a length from a ruler. That's not a coordinate; it's a direct physical observable, i.e., an invariant.

Read off an angle from a protractor. That's not a coordinate; it's a direct physical observable, i.e., an invariant.

You not only need to read carefully, you need to think carefully, in order to understand this topic properly.

Freixas
Freixas said:
To do that, I have to be able to identify an invariant. In many sources, especially sources written for beginners, something like "speed" is said to be not invariant, full stop, no exceptions (as in the Taylor/Wheeler book). But, according to you (and others), that appears to be incorrect. Properly qualified, speed can be an invariant.
Speed relative to some specified state of rest is an invariant. Take driving a car. Oncoming vehicles, pedestrians, cars you overtake, and a copper on the roadside will all give different numbers for your speed - and that's what's meant by "speed is relative". But all those people could each be equipped with a radar gun, and those guns must read some value that everyone can (in principle) see regardless of their choice of frame. So whatever it is they are measuring must be an invariant, or the other people would not be able to understand why one person's radar gun is showing what it is. That is, the speed relative to each radar gun must be an invariant. Ultimately, this is because the radar guns measure the "angle" between worldlines (known as the rapidity) on a funny hyperbolic scale.

It really is much easier to think of SR in terms of geometry and invariants. The thinking change is quite profound, though, because familiar physical quantities (like speed) often turn out to be less convenient to use in the model than more abstract quantities like ##\gamma## or rapidity. @Dale is correct that this is all much easier with four vectors and tensors. If you are familiar with three vectors, four vectors are only a tiny bit harder and are very powerful.

Last edited:
Freixas and vanhees71
Ibix said:
,It really is much easier to think of SR in terms of geometry and invariants. The thinking change is quite profound, though, because familiar physical quantities (like speed) often turn out to be less convenient to use in the model than more abstract quantities like ##\gamma## or rapidity. @Dale is correct that this is all much easier with four vectors and tensors. If you are familiar with three vectors, four vectors are only a tiny bit harder and are very powerful.
While vectors and tensors are great for some things (especially efficient calculation),
they may not be great in some other things
(like giving a geometric intuition of what the tensor calculation is saying).
So, I'd suggest that one also use spacetime-vectors on a planar (1+1)-Spacetime diagram, together with its trigonometry… since it’s not too much harder than Euclidean geometry, when properly guided in it.
The angle between two given vectors is an invariant.Since relativity is about the "geometry of spacetime",
I think it's good to be able to connect with the "geometry" visually,
and not just [vector- and tensor-]algebraically or "as a set of components that transforms as ...".

(We can [in principle] solve high-school geometry problems vectorially and tensorially.
But could we easily and intuitively explain the geometry that is involved from those vectors and tensors,
or from transformations (rotations)?
Many situations in relativity can be treated geometrically, but we sadly don't seem to do so.

Dale
robphy said:
So, I'd suggest that one also use spacetime-vectors on a planar (1+1)-Spacetime diagram, together with its trigonometry… since it’s not too much harder than Euclidean geometry, when properly guided in it.
Yes - I completely agree, and if memory serves, @Freixas has written a program to draw spacetime diagrams already, suggesting some degree of familiarity. Building on that with the basics of four vectors and inner products is, I would suggest, worthwhile.

robphy said:
So, I'd suggest that one also use spacetime-vectors on a planar (1+1)-Spacetime diagram, together with its trigonometry

Yes. I generally try to figure out everything geometrically. However, there are some concepts that aren't about geometry, such as Peter's point below:

PeterDonis said:
Unfortunately, many presentations of relativity obfuscate this by choosing coordinates in which many invariants of interest are numerically equal to particular coordinate values.

If I look at a spacetime diagram, I still have to be careful about what is an invariant and what is not. I have a more detailed response in mind, but it will have to wait a bit.

Ibix said:
Yes - I completely agree, and if memory serves, @Freixas has written a program to draw spacetime diagrams already, suggesting some degree of familiarity. Building on that with the basics of four vectors and inner products is, I would suggest, worthwhile.

Your memory is correct. I'll have to see if I can translate four vectors and inner products to geometrical constructs. My initial skimming of the literature hasn't highlighted the geometrical connections, but I haven't spent much time looking.

Freixas said:
If I look at a spacetime diagram, I still have to be careful about what is an invariant and what is not.
Lengths along of curves and angles between curves where they intersect. In other words, geometric invariants.

PeterDonis said:
And that means you need to read carefully and not just take what you read at face value.
Here is the longer response I promised.

At least to start, I'm going to use the Taylor/Wheeler lattice as the equivalent of training wheels (or of @Ibix's radar gun). What can be observed using the lattice will be an invariant.

In order to do predictive physics (where I have some set of initial conditions and want to predict some ending conditions), I need to state a problem so that values are relative to the lattice, which defines the rest frame. I also need to define how the lattice's clocks are initialized (I could initialize them in any of a number of ways, but let's limit the choices to ones that implement "valid" simultaneity conventions).

If I want to start and end with invariants, I need to begin with initial conditions defined in terms of observable events. Relative speed might be an "angle in spacetime between two timelike worldlines at the point where they intersect," but I don't know how to observe that. Instead, I might state that, say, Alice is moving inertially and she was spotted at coordinate ##c_1## and time ##t_1## and coordinate ##c_2## at time ##t_2##. Alternately, I could create a special lattice observer whose job is to measure Alice's relative velocity and display it on a billboard for all to see. Since I'm talking about initial conditions, I could provide the number seen on the billboard.

The problem in the OP asked about the elapsed time in one location based on an event happening elsewhere. Using the billboard method, I could place a special observer in a position such that light from both locations reach it at the same time. If the lattice's clocks were initialized using Einsteinian synchronization, then we could place the observer midway between the two events; if some other synchronization was used, the location might be closer to one event or the other. The observer could then view an event occurring in one place, view the time on a clock elsewhere on the lattice, and display the result on a billboard.

This is a slow and crude technique, but it establishes the simultaneity of two events as an invariant physical observable. This is one of the pieces I was missing.

Once I can do this, I can use the same technique for other things that depend on a simultaneity convention: length, speed, and clock rates. In each case, I need to understand how someone can use the raw event data captured by the lattice to measure these quantities. If I go through the mental exercise, then I can verify that I start and end with invariants.

The process may be cumbersome, but at least I can understand it.

I notice that everyone seems to drop the simultaneity convention requirement. Yes, I gather that if nothing is mentioned, Einsteinian synchronization is universally assumed. But no one can predict the billboard numbers displayed unless the calculation method takes the lattice's simultaneity convention into account, so I keep including it.

PeterDonis said:
Lengths along of curves and angles between curves where they intersect. In other words, geometric invariants.
Here are two Minkowski spacetime diagrams. Both show the same worldlines relative to different frames, but the lengths (on the diagrams anyway--I realize the proper lengths are invariant) and the angles are not obviously invariant. I'm not sure how looking at a diagram clarifies what is and isn't an invariant.

robphy said:
While vectors and tensors are great for some things (especially efficient calculation),
they may not be great in some other things
(like giving a geometric intuition of what the tensor calculation is saying).
Well, you always have to physically understand what your model is saying. Manifestly covariant formulations have the advantage that they provide you with models that are at least not contradicting the very foundations of relativity, i.e., the spacetime model it is based on. Also I've the impression that geometry is overemphasized (e.g., in endless discussions of Minkowski diagrams which are not so simple to understand as one might think).
robphy said:
So, I'd suggest that one also use spacetime-vectors on a planar (1+1)-Spacetime diagram, together with its trigonometry… since it’s not too much harder than Euclidean geometry, when properly guided in it.
The angle between two given vectors is an invariant.
It's not too much harder than Euclidean geometry. What's hard is to give up ones Euclidean interpretation of the "paper plane"!
robphy said:
Since relativity is about the "geometry of spacetime",
I think it's good to be able to connect with the "geometry" visually,
and not just [vector- and tensor-]algebraically or "as a set of components that transforms as ...".
Relativity provides a spacetime model, and that's of course "geometric", but then you have to go further and understand the "geometry" in the sense of Riemann, Klein, et al. i.e., providing a symmetry group, which can be used to find the dynamical laws of Nature, which can be compared with observations. E.g., you can argue that gravitation is just spacetime curvature, but at the same time it's an interaction as the other fundamental interactions too. One should also have in mind the approach as given in, e.g., Weinberg's book of 1971 or Feynman's Lectures on Gravitation and not stick only to the standard geometric interpretation. Particularly it becomes clear that "general covariance" is a "local gauge symmetry", which is very important. One of the problems Einstein had to derive his field equations was that this important feature was not clear to him and his collaborators and that's why it took him about 10 years to finally get GR.
robphy said:
(We can [in principle] solve high-school geometry problems vectorially and tensorially.
But could we easily and intuitively explain the geometry that is involved from those vectors and tensors,
or from transformations (rotations)?
Many situations in relativity can be treated geometrically, but we sadly don't seem to do so.
For me the analytical approach to geometry via vectors was a revelation. Unfortunately we did not learn about tensors but simple transformations like rotations we learned about. Most situations in relativity can be treated with great advantage using manifestly covariant tensor (or spinor) formulations, which of course is also geometry but without the obstacle of easily misunderstood drawings in a non-Euclidean plane ;-).

vanhees71 said:
Well, you always have to physically understand what your model is saying. Manifestly covariant formulations have the advantage that they provide you with models that are at least not contradicting the very foundations of relativity, i.e., the spacetime model it is based on. Also I've the impression that geometry is overemphasized (e.g., in endless discussions of Minkowski diagrams which are not so simple to understand as one might think).

As you say, one has to understand physically what is happening.
My point is that vectors and tensors are great, but not enough.
Although they provide and display the invariance of relativity,
especially in complicated calculations,
it's not so easy to connect them to one's physical and mathematical intuition.

That's why I said (new red bolding mine)
"one also use spacetime-vectors on a planar (1+1)-Spacetime diagram".
That can help with the visualization of the geometry

Even with vectors, tensors, and spacetime diagrams,
that's not enough for physical intuition.

As an undergrad, I had 3 courses in relativity using texts by Skinner, Lawden, and Landau.
(I also had Taylor-Wheeler and Misner-Thorne-Wheeler as side references.)
I'm certainly not claiming to have learned anywhere-near-everything from these books.
I could do enough of the work to get A's in them, but I didn't feel I "got it".

In grad school, learning more tensorial methods and the abstract-index notation from Wald's text helped.
But it wasn't until I learned about operational definitions of distance and time measurements via radar-methods on a spacetime diagram from Geroch, did things finally click for me. For me, I could now tie together the verbiage of introductory texts, the notations of vectors and tensors in coordinate form and in abstract-index form, and physics connected to observation using light signals and clocks (which is more relativistic in spirit than rods and clocks).

On my own in grad school, I stumbled upon Yaglom's "A simple non-Euclidean geometry and its physical basis", which introduced me to Klein and the Cayley-Klein geometries. I also tumbled upon Schouten's "Ricci Calculus" and "Tensor Analysis for Physicists", which introduced me to visualizing tensors.

Ideally, one really should try to be fluent in (and fluid in, in the sense of "being able to inter-connect") all of these methods.

My $0.03. vanhees71 and Dale robphy said: My$0.03.
Hmm, inflation hits PF also?

For me, the spacetime diagrams and the concept of four-vectors built my intuition. But I always use explicit four-vector or tensor-based math to check my intuition. I think that it is important to have both. As you said "also". Sometimes our intuition is wrong and needs to be retrained. So it is necessary to have a solid foundation that you can trust to get the right answer even when intuition fails.

robphy

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