# General Relativity and the curvature of space: more space or less than flat?

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Can Gravitational time dilation be explained as a result of spacetime curvature?
Could you state that context?
Yes, one of the first results that you can derive from GR is gravitational time dilation. GR is a theory of spacetime curvature; as has already been said, there is nothing else to GR, really.

The context would be a particular experiment, such as Pound-Rebka:

https://en.wikipedia.org/wiki/Pound–Rebka_experiment

• vanhees71 and Martian2020
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Thank you for interesting info.
Just so to be sure (myself), spacetime (4d) is invariant for all observers?

I would say that it's clearer to say that the Lorentz interval is invariant for all observers.

The Lorentz interval in 4d space-time is rather similar to the distance in 3d Euclidean space, but the Lorentz interval is 4 dimensional. In space, the invariant distance in cartesian coordinates between two points is just dx^2 + dy2 + dz^2. Different observers may give different coordinates for the location of a point, but the distance between two points is independent of such coordinate choices.

In the flat space-time of special relativity, the invariant interval between two events, the Lorentz interval, is just dx^2 + dy^2 + dz^2 - dt^2. General relativity generalizes the above formula to an arbitrary quadratic form, a polynomial of degree 2 in four generalized coordinates.

If you look at two events in space-time, the distance between them (dx^2 + dy^2 + dz^2) is not invariant, it depends on the observer. "Lorentz contraction" is a specific example of the non-invariance of spatial intervals in special relativity. The time interval dt is also not invariant by itself. An example of this is the relativity of simultaneity - two events may have the same t coordinate so that dt=0 in some frame, but in a different frame, dt may, and in gneeral will, be nonzero.

In special relativity, the difference of the squares of the distance and the time IS an invariant quantity. And GR shares the same idea of the existence of an invariant interval, however the details of its computation are different / more complex.

Curvature is same? Why then is it called GR? Because different observers see/experience differently space (3d) and time (1d)?

I'm not quite understanding the question, sorry.

Related question that bothers me: do we draw geodesics in spacetime because of law of conservation of energy-momentum? In wiki (I know some say it is not authoritative source): https://en.wikipedia.org/wiki/Physical_theories_modified_by_general_relativity#Conservation_of_energy–momentum
"Unlike classical mechanics and special relativity, it is not usually possible to unambiguously define the total energy and momentum in general relativity, so the tensorial conservation laws are local statements only (see ADM energy, though)."

No. The wiki article you cite says as much, and I agree with it. So we can draw geodesics in space-time that lack the particular requirements that we need to have a conserved energy or momentum.

Rather than try and explain geodesics in GR, I'll settle for describing them in SR. In SR, two events are space-like separated if and only if there is some frame in which they are simultaneous. A space-like geodesic is just the shortest path between two space-like separated events in this frame where the two events are simultaneous.

SR also has timelike geodesics. In timelike geodesics one event always occurs before the other event, and an object can travel from one event to the other event, without exceeding or reaching the speed of light. In SR, timelike geodesics have the property of maximizing the length / duration of the connecting worldline. In the various twin paradoxes of SR, the "stay-at-home" twin is the one traveling along a timelike geodesic, and he'll be older than any twin who follows a non-geodesic path. This result does not generalize to GR, however.

There are also null geodesics, but I don't want to get into them.

Does above mean separately energy and momentum could be not conserved, but as a whole tensor remains same? So that energy-momentum tensor is conserved (when there is no acceleration) and that is the reason for geodesics in spacetime?

The energy-momentum tensor always has a local conservation law. This law states that the divergence of the energy-momentum tensor is zero. This is rather similar to some conservation laws in fluid dynamics. This conservation law is built into the structure of the GR.

While this local conservation law always exists, global conservation laws, such as the energy of a system, do not always exist.

The old sci.physics FAQ on this is pretty good. https://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

I'll quote briefly from it , the original is worth reading in its entirety though.

FAQ said:
Is Energy Conserved in General Relativity?

In special cases, yes. In general, it depends on what you mean by "energy", and what you mean by "conserved".

In flat spacetime (the backdrop for special relativity), you can phrase energy conservation in two ways: as a differential equation, or as an equation involving integrals (gory details below). The two formulations are mathematically equivalent. But when you try to generalize this to curved spacetimes (the arena for general relativity), this equivalence breaks down. The differential form extends with nary a hiccup; not so the integral form.

• vanhees71
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The way technology of humanity is now doing it (as I understand): atomic clocks are used to measure time and speed of light as traveled per unit of time of above clocks measure distance in space. What convention does this method use?

IT's rather the reverse - the constancy of the speed of light is used to define distance in terms of time. In the old days, distance used to be defined by a physical meter standard. The modern definition defines the speed of light to have a specific fixed value. The current definition of the SI meter is:

SI said:
The metre, symbol m, is the SI unit of length. It is defined by taking the fixed numerical value of the speed of light in vacuum c to be 299 792 458 when expressed in the unit m s–1, where the second is defined in terms of the caesium frequency  Cs.

BIPM is the group that handles our physical standards.

There are some other related conventions as well. One of the usual convention is the Einstein clock synchronization convention, which arises from isotropy.

https://en.wikipedia.org/w/index.php?title=Einstein_synchronisation&oldid=988269037

Times on the Earth have several conventions, TAI time is the convention relevant to atomic clocks. TAI time is used as the base for other time conventions, such as UTC time which you'll read from your wall clock.

https://en.wikipedia.org/w/index.php?title=International_Atomic_Time&oldid=981327586
https://en.wikipedia.org/w/index.php?title=Coordinated_Universal_Time&oldid=990427393

wiki said:
The current version of UTC is defined by International Telecommunication Union Recommendation (ITU-R TF.460-6), Standard-frequency and time-signal emissions, and is based on International Atomic Time (TAI) with leap seconds added at irregular intervals to compensate for the slowing of the Earth's rotation.

I'm not sure if I've answerd your question, it's a bit general.

• vanhees71
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A comment on "learning GR". GR is a very advanced topic, so this is an ambitious goal. It will also require a lot of background knowledge. A good place to start is special relativity.

Special relativity won't tell you anything about curvature, however. Curvature is a concept in geometry, in the case of General Relativity, the applicable theory of geometry is Riemannian and pseudo-Riemannian geometry.

Something simpler, that can tell you a little bit about curved geometries, is spherical trignometry and spherical geometry. In this geometry, great circles replace straight lines - great circles can also be regarded as geodesics of this geoemetry. It's a spatial geometry, not a space-time geometry but the surface of a sphere is a curved geometry, so it can give some basic insights into curvature.

It's also interesting in it's own right, and useful if one wants to navigate long distances on the Earth, since the Earth is almost spherical.

The applicability of curved geometries to space-time, rather than space, is a bit trickier. A useful idea is to introduce the space-time diagram of special relativity. Then one can imagine drawing these space-time diagrams with great circles on a sphere, rather than with straight lines on a plane, to get a tiny bit of insight into the sorts of effects one sees in curved space-time. Unfortunately, the technique can't handle anything higher than 1 spatial and 1 time dimension by it's very nature. One needs a more abstract treatment to handle more than this small number of dimensions, and it gets a lot more complicated.

• vanhees71 and PeroK
Martian2020
IT's rather the reverse - the constancy of the speed of light is used to define distance in terms of time.
Sorry for my long phrase where I actually wanted to tell the same, but apparently my wording was not perfect. Therefore you spent time explaining what I already knew.

What I really wanted answer to is:
you mentioned slicing spacetime depends on a convention. How could we (technology, humankind) measure time (atomic clocks) separately from space? It requires (per "problem" of slicing your mentioned) some convention how to do that. What is that convention?

Martian2020
I would say that it's clearer to say that the Lorentz interval is invariant for all observers.

The Lorentz interval in 4d space-time is rather similar to the distance in 3d Euclidean space, but the Lorentz interval is 4 dimensional.
I've watched recently videos by Sean Carroll (his lectures were recommended somewhere on that site). e.g. here:

he states that curvature of spacetime could be defined by Riemann tensor at each point (of spacetime). So let me rephrase my question on in-variance of spacetime.

Two observers in different points of spacetime, generally effected my external forces (accelerating). One "calculates" Riemann tensor at its' location (T11) and location of 2nd observer (T12). Similarly 2nd observer calculates tensor for itself (T22) and 1st observer (T21). Are T11=T21 and T12 = T22? Or maybe the notion of "calculating" tensor is ambiguous in that context?

Added: Posted the question and realized it requires observers to be on the same spacetime. Could we at least have that, as a result that they are in the same Universe?

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Mentor
he states that curvature of spacetime could be defined by Riemann tensor at each point (of spacetime).

Not "could be"; is.

Two observers in different points of spacetime

Are at different points of spacetime, so the Riemann tensors they measure could be different since the Riemann tensor can be different at different points of spacetime.

Posted the question and realized it requires observers to be on the same spacetime.

Obviously.

Could we at least have that, as a result that they are in the same Universe?

Obviously.

Staff Emeritus
Sorry for my long phrase where I actually wanted to tell the same, but apparently my wording was not perfect. Therefore you spent time explaining what I already knew.

What I really wanted answer to is:
you mentioned slicing spacetime depends on a convention. How could we (technology, humankind) measure time (atomic clocks) separately from space? It requires (per "problem" of slicing your mentioned) some convention how to do that. What is that convention?

The sort of time clocks measure has the technical name "proper time". It is a physical measurement, it does not require a convention, though it still requires standards. To measure proper time, a clock needs to be physically present at a starting event, and an ending event. The clock ticks as time progresses, one counts the number of ticks as the clock progress from the start to the finish event, and calls the result the proper time.

Note that when one measures the speed of light, it is not possible for the same clock to be present at both events. There are a lot of threads on this, I think we have one ongoing right now, though I haven't been following it so I'm not sure how helpful it will be.

Coordinate time is the sort of time that requires a convention relates to synchronizing spatially separated clocks. It arises when you have multiple clocks, and you want to compare their readings via some synchronization process. A usual setup for a coordinate system imagines a framework, in which there are a very large (even infinite) number of clocks, all of which are synchronized. Then the position in the framework defines the spatial coordinate, and the clock reading on the clock that one is physically at defines the coordinate time.

While what a single clock measures is physical, independent of the observer, the same is not true of the process of clock synchorinzation. This is one of the important issues arising from special relativity known as "the relativity of simultaneity". Frames in relative motion do not share a common notion of simultaneity, they do not share a common notion of how to synchronize clocks. There's a lot of discussion of this issue, which is a very common if not the most common source of confusion regarding special relativity.

The usual convention as to how to synchronize clocks is the "Einstein synchronization", see for instance https://en.wikipedia.org/w/index.php?title=Einstein_synchronisation&oldid=988269037. The conventional part of Einstein clock synchronization is specifying the frame of reference that one uses. There are some choices that are convenient, but no choice is mandated by the laws of nature, it's a human choice and hence a convention.

There's more that could be said, but this seems like a good time to stop.

While I hate to stop here, as there are some important things to say about some practical issues in timekeeping on the Earth, it'd be too confusing to go on further.

• vanhees71
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Two observers in different points of spacetime, generally effected my external forces (accelerating). One "calculates" Riemann tensor at its' location (T11) and location of 2nd observer (T12). Similarly 2nd observer calculates tensor for itself (T22) and 1st observer (T21). Are T11=T21 and T12 = T22? Or maybe the notion of "calculating" tensor is ambiguous in that context?

Added: Posted the question and realized it requires observers to be on the same spacetime. Could we at least have that, as a result that they are in the same Universe?
I think you are confused about a few things here.

First, there is the mathematical subject of Riemannian and pseudo-Riemannian geometry. This involves an infinite number of different geometrical configurations (manifolds or geometries). And, in any number of dimensions.

Second, if you confine your study to four-dimensional pseudo-Riemannian manifolds, then you are focusing on what may, in principle, be physically possible as a model for our universe. With the manifold representing spacetime.

Third, the curvature of spacetime is determined (through the Einstein Field Equations) by the distribution of matter (in fact the distribution of stress-energy). Matter tells space how to curve; space tell matter how to move.

Fourth, if you want to study a particular problem using GR, you need to decide first on the approriate stress-energy distribution (e.g. a single large star, a spinning black hole, two colliding black holes etc.), which determines the spacetime curvature you need to analyse.

(There is no useful sense in which you could have "two observers in different spacetimes".)

Fifth, you can use different coordinate systems to study any particular spacetime, but the choice of coordinates cannot affect the physical consequences.

(This is where a lack of knowldege of classical physics is hindering you, because the same applies there. Quantities like velocity, kinetic energy, electric current, magnetic field strength all vary from one reference frame to the other. But, the physical phenomena do not vary: did the cars crash?, did the charged particle go to detector A or detector B? etc. All these physical questions must have the same answer in all reference frames.)

Sixth, observers in GR make local measurements. You mustn't confuse mathematical tools with measurebale quantities. The Riemann tensor, for example, is not something you measure; it's a mathematical tool that is used to analyse the spacetime. Ultimately, all you really measure from distant events are lights signals (or gravitational waves). You may infer things from this, but if you think about it, all you measure is the light reaching you, ultimately.

Finally, to echo what @pervect said, GR is an advanced subject and you are at the point where you are drowning in a confusion of ideas: trying to ask more advanced questions without having a grasp of the basics.

For example, let's assume you want to study motion near a single large star (or planet). You have a specific spacetime curvature. You choose the best coordinate system for what you want to study. You study the problem using that coordinate system. You then use that analysis to predict the raw measurements made by any local observers. For example, you can predict the wavelength of light detected by an observer in a certain state of motion at a particular point in spacetime. Or, you could predict how much time elapses on a local clock while orbiting the star or planet; or while falling to the planet's surface.

GR demands a rigorous understanding of these basic ideas.

• vanhees71
Martian2020
Second, if you confine your study to four-dimensional pseudo-Riemannian manifolds, then you are focusing on what may, in principle, be physically possible as a model for our universe. With the manifold representing spacetime.
Third, the curvature of spacetime is determined (through the Einstein Field Equations) by the distribution of matter (in fact the distribution of stress-energy).
Fourth, if you want to study a particular problem using GR, you need to decide first on the approriate stress-energy distribution (e.g. a single large star, a spinning black hole, two colliding black holes etc.), which determines the spacetime curvature you need to analyse.
(There is no useful sense in which you could have "two observers in different spacetimes".)
Sixth, observers in GR make local measurements. You mustn't confuse mathematical tools with measurebale quantities. The Riemann tensor, for example, is not something you measure;
I said "calculate" tensor, not measure. I said two observers in different points of spacetime, not two spacetimes. I just was not sure we have singleton of a spacetime.
Some background so you understand my confusion and questions: I learned to apply formulas for translating from one inertial coordinate to another in SR. It gives time dilation and length contraction, difference in simultaneity, and I was content with the results and their application. I have not thought about spacetime as a whole.

Now I wanted to understand gravity and turned to GR. Both SR, GR are called "relativity". I know is SR time and space are relative.

I want to understand what is relative in GR? My thought was that spacetime is relative. Seems not as e.g. here answers suggest, now I see I did not formulate it as plainly as I see now (asking about invariance).

Is spacetime the same (single, not relative) for all our Universe, for all observers located in our Universe?

If yes (I expect the confirmation to above question), then we have matter "fixed" in spacetime in same positions for all observers in that spacetime. Correct?

If yes, then I can answer Yes to my previous post - Riemannian tensors defining how much vector would change from parallel transport in the vicinity of each point should be same for all observers. Then I can safely try to do some calculus for some star affecting the curvature etc.

Added: initially I wanted to visualize spacetime curvature of say near and in the Earth. And I asked the original question. I understand e.g. observers one on the surface and free-falling one would see different space and time. I now want to see if spacetime is single entity and therefore visualization of it is same for both observers.

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I now want to see if spacetime is single entity and therefore visualization of it is same for both observers.
Not sure if the "therefore" make sense here. Even if spacetime is a single entity, the visualization of it for different frames might differ.

See vor example the Rindler coordinates vs. a Minkowski diagram for an inertial frame.
https://en.wikipedia.org/wiki/Rindler_coordinates

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I want to understand what is relative in GR? My thought was that spacetime is relative. Seems not as e.g. here answers suggest, now I see I did not formulate it as plainly as I see now (asking about invariance).

Is spacetime the same (single, not relative) for all our Universe, for all observers located in our Universe?

"Relativity" is just a word. Let's call it "Einstein's Theory of Gravity". Now there is no "relativity" to worry about.

As far as GR is concerned, the universe is a unique manifold of events. Each event represents a unique point in spacetime. The spacetime manifold has geometry, defined by the metric tensor. At one level, you can fully describe our universe (or any part of it) without any recourse to observers or measurements.

You (only) need observers in order to test predictions. Because you need your theory to tell you something about what is ultimately measured.

One conceptual hurdle that you need not cross when you study SR is how to think about spacetime without invoking a suitable global IRF. You can learn SR and use the comfort blanket of always thinking about things in terms of IRF's. In GR you have no such luxury and you must confront the coordinate-independence of spacetime.

You also need to study the concepts of "invariance" and tensor transformation closely. Scalar quantities are the same in all coordinate systems, but vector and tensor quantities transform according to the appropriate transformation rules. This applies in SR also.

If you think you are at the right level, you could try these MIT lectures on GR. It is, however, a graduate physics course.

https://ocw.mit.edu/courses/physics...ction-and-the-geometric-viewpoint-on-physics/

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• Martian2020, vanhees71, Vanadium 50 and 1 other person
Martian2020
As far as GR is concerned, the universe is a unique manifold of events. Each event represents a unique point in spacetime. The spacetime manifold has geometry, defined by the metric tensor. At one level, you can fully describe our universe (or any part of it) without any recourse to observers or measurements.

You (only) need observers in order to test predictions. Because you need your theory to tell you something about what is ultimately measured.

One conceptual hurdle that you need not cross when you study SR is how to think about spacetime without invoking a suitable global IRF. You can learn SR and use the comfort blanket of always thinking about things in terms of IRF's. In GR you have no such luxury and you must confront the coordinate-independence of spacetime.
Yesterday (before reading this post) I spent much time thinking about flat space (SR), flat 2d space of like Minkowski diagram, with only 1d of space (empty sheet of paper). I thought ok, single spacetime, coordinates can have non-right angle in general.

But time direction is different from space one. I have one "observer" with direction toward future, then movement of second observer at angle of slightly less than light cone to the right. Then move to that new direction as starting point and draw another light cone and choose 3rd "observer" moving within the cone to the right. That way I can make full circle. And my 3rd observer would move in relation to 1st one outside of light cone for 1st one. That is as I know should not be in SR. I got puzzled how can I have single spacetime...

I'm still puzzled after reading your answer where you kind of confirmed there could be single spacetime: "In GR you have no such luxury and you must confront the coordinate-independence of spacetime." Can this be understood in general (w/out need for exact results) w/out calculus? For e.g. flat 1d space+time?

In the meanwhile I can try to listen to the lectures your recommended.

I think I've found a solution:
1. I imagine a plane.
2. Then scatter some events on it randomly. Fine as of now.
3. Add causal relationships. Then each two events of causal relationship can represent direction of time and space is at right angle. And with that coordinates check if other causal relationships are within light cones from cause to the result. Check for all pairs of events. If violated, then my set of events represent non-causal universe. If not then this is example of SR causal universe.

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Mentor
Can this be understood in general (w/out need for exact results) w/out calculus? For e.g. flat 1d space+time?
Yes, and indeed you need to have this down solid before you move on to more complicated situations. You’re trying to understand the geometric approach to special relativity, which is both the most elegant statement of the theory and essential before you can move on to curved spacetime and GR, but the one dimensional spacetime diagrams are still the starting point.

Taylor and Wheeler’s book ‘Spacetime Physics” is the best introduction that I know of.

Martian2020
but the one dimensional spacetime diagrams are still the starting point.
I've added solution to my puzzlement in the end of previous post. If you have time, please see if my understanding is correct there.

Mentor
move to that new direction as starting point and draw another light cone
The new light cone will still be at a 45 degree angle, just like the first one and just like the following one.

Martian2020
The new light cone will still be at a 45 degree angle, just like the first one and just like the following one.
Just so you see my addition: is it correct?
I think I've found a solution:
1. I imagine a plane.
2. Then scatter some events on it randomly. Fine as of now.
3. Add causal relationships. Then each two events of causal relationship can represent direction of time and space is at right angle. And with that coordinates check if other causal relationships are within light cones from cause to the result. Check for all pairs of events. If violated, then my set of events represent non-causal universe. If not then this is example of SR causal universe.

Mentor
is it correct?
I am not sure. I don’t really understand what you are saying.

In any inertial frame if ##ds^2=-dt^2+dx^2>0## then the two events are not causally related. Otherwise the one that occurs earlier could in principle cause the one that happens later.

Mentor
Then move to that new direction as starting point and draw another light cone and choose 3rd "observer" moving within the cone to the right. That way I can make full circle. And my 3rd observer would move in relation to 1st one outside of light cone for 1st one. That is as I know should not be in SR. I got puzzled how can I have single spacetime...
Draw the worldline of an object (Object A) that is at rest in whatever frame you’re using; it will be a vertical line going straight up the page. Next draw the worldline of an object (Object B) moving at .9c to the right relative to A; this will be a line that slants up and to the right, rising ten units for every nine units sideways. It’s easiest if you have the two worldlines intersect at the origin.

Pick a point, any point, on B’s worldline. Say that B emits a flash of light off to the left and off to the right at that point. Draw the path of those two flashes; they will of course form a lightcone because that’s the definition of a lightcone. Do remember that the slant of B’s worldline is irrelevant - one flash is moving to left at speed c and the other is moving to the right at speed c, no matter what B is doing. That is, all lightcones are always at a 45 degree angle.

Next draw the worldline of a third object, C, which is moving to the right at speed .9c relative to B and passes B at the moment the light flashes. But we’re still working in the frame in which A is at rest, so the slope of C’s worldline will be determined by C’s speed relative to A - you will have to use the relativistic velocity addition formula and you’ll find that the line is still inside the lightcone, a bit less than 45 degrees from the vertical.

Finally, start over with a fresh sheet of graph paper and draw the same exact situation, except using the frame in which B is at rest. A’s worldline now slants off and to the left, B’s worldline is straight up, and C’s worldline slants off to the right. You’ve just drawn the same spacetime in two different ways, with the angles on the paper distorted in different ways when you make a different choice of which set of x,t axes intersect at a 90 degree angle.

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I have one "observer" with direction toward future, then movement of second observer at angle of slightly less than light cone to the right.

If not then this is example of SR causal universe.
There is only one flat universe. In the same way that there is only one 3D Euclidean space. SR changed the way we understand the nature of spacetime, but there is still only one 4D flat spacetime.

Although much of SR is presented using "observers", the theory is really one of flat spacetime and inertial reference frames (related by Lorentz Transformations). Especially when studying the geometry of spacetime you do not need observers.

The "observer" in SR is often over-emphasised - to the point where some students wrongly believe it's all about the light signals received by each and every observer.

If we imagine an IRF, then we are doing something quite abstract - we are imagining that we can assign coordinates to every event in spacetime. This is much less physical than imagining what a single observer "sees". It's much better to think of an IFR as a grid (1D, 2D or 3D) of equally spaced observers, all at rest with respect to each other and all with synchronised clocks. All events are measured locally, by the nearest observer and the overall picture can only be pieced together by collating all those local observations/measurements.

And, as I mentioned previously, when you start studying GR, you must make a further abstraction to a coordinate-independent or reference-frame independent approach to curved spacetime. In particular, an observer does not have a global view of spacetime (only a local view, where SR applies).

• etotheipi
Martian2020
Do remember that the slant of B’s worldline is irrelevant - one flash is moving to left at speed c and the other is moving to the right at speed c, no matter what B is doing. That is, all lightcones are always at a 45 degree angle.
Thanx, but I do know how to draw Minkowski diagram. I used the name to imply I have 1d of space, flat sheet of paper representing flat spacetime. I want to draw events on spacetime w/out initial IRF to start with, as @PeroK said:
"One conceptual hurdle that you need not cross when you study SR is how to think about spacetime without invoking a suitable global IRF. You can learn SR and use the comfort blanket of always thinking about things in terms of IRF's. In GR you have no such luxury and you must confront the coordinate-independence of spacetime."
As I understood the quotation above, there is no global IRF to always draw cones at 45 angle with. But straight lines on 2d sheet represent list of event related by causality - object is at some point because it was in adjustant point just a moment ago in time and is moving inertially. If we draw light cones at 45% from some path, due to symmetry we should do the same for other line on spacetime. Otherwise there is one global time direction on spacetime. Maybe so, but then how to find it? I'm not certain what to do next.

Mentor
I want to draw events on spacetime w/out initial IRF to start with
When you draw a Minkowski diagram, you’re plotting the ##x,t## coordinates of various events on a piece of graph paper, so you need some frame to assign these coordinate values. You can choose some non-inertial frame (that is, one in which an object whose position coordinate is constant is not moving inertially) to assign these values and then you’ll get a diagram drawn using that non-inertial frame. One example would be the Rindler diagram you get by using the Rindler X and T coordinates for your axes, and labeling events by their Rindler coordinates. (Rindler coordinates are one in which we consider an accelerating object to be at rest and the rest of the universe to be accelerating away from it in the opposite direction).
But straight lines on 2d sheet represent list of event related by causality
Only if you’ve chosen coordinates such that when you plot the path of a flash of light, you get straight lines moving up and out at a 45 degree angle. The causality relationships at an event are determined by the paths of hypothetical light signals passing through that point; whether these paths follow straight lines is completely a matter of the coordinates you’re using.

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• PeroK
Martian2020
When you draw a Minkowski diagram, you’re plotting the ##x,t## coordinates ...
Only if you’ve chosen coordinates ...
I'm not sure why @PeroK liked this post. It is correct, as far as I can see for Minkowski. But in my post before I tried to expain I wanted to do what he said:
you must confront the coordinate-independence of spacetime.
I want to confront spacetime w/out coordinates, as @PeroK challenged me to do. Please help me to, if you can, explain your point w/out coordinates.
Coordinates are necessary so say point (1,3) etc. So say go from point A to point/event B we don't need coordinates, as far as I understand.

Mentor
want to confront spacetime w/out coordinates, as @PeroK challenged me to do. Please help me to, if you can, explain your point w/out coordinates.
You can confront spacetime without coordinates, but you can't draw spacetime diagrams without them. When you draw a diagram you are putting marks on a piece of paper, and the coordinates tell you where the marks go.

Confronting spacetime without coordinates doesn't mean you never use coordinates, it means that you have to learn to distinguish the things that are true no matter what cordinates we choose ("invariants" in the lingo) like the fact that causal relationships are determined by the path that a hypothetical flash of light would follow through spacetime ("lightlike geodesics" in the lingo) from the things that are coordinate dependent like the 45-degree straight line lightcones that appear in Minkowski diagrams.

• PeroK
Martian2020
You can confront spacetime without coordinates, but you can't draw spacetime diagrams without them. When you draw a diagram you are putting marks on a piece of paper, and the coordinates tell you where the marks go.

Confronting spacetime without coordinates doesn't mean you never use coordinates, it means that you have to learn to distinguish the things that are true no matter what cordinates we choose ("invariants" in the lingo) like the fact that causal relationships are determined by the path that a hypothetical flash of light would follow through spacetime ("lightlike geodesics" in the lingo) from the things that are coordinate dependent like the 45-degree straight line lightcones that appear in Minkowski diagrams.
Thank you. I think I'm understanding better now. Could you please clarify: invariants are e.g. causal relationships, that is clear. Are paths that a hypothetical flash of light would follow through spacetime invariants (I say paths, because light can go many directions, correct?)?

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Thank you. I think I'm understanding better now. Could you please clarify: invariants are e.g. causal relationships, that is clear. Are paths that a hypothetical flash of light would follow through spacetime invariants (I say paths, because light can go many directions, correct?)?
I would say that the path that a flash of light follows is invariant, yes. The coordinates used to describe the set of events on that path will vary from one coordinate system to the next. But the set of events on the path is the same.

However, the angle that a particular light pulse takes from its launching point or the angle at which it arrives at its detection point can vary depending on one's choice of reference frame (e.g. stellar abberation). The path is still an invariant. The angle that it takes when projected onto a spacelike snapshot is not.

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Thank you. I think I'm understanding better now. Could you please clarify: invariants are e.g. causal relationships, that is clear. Are paths that a hypothetical flash of light would follow through spacetime invariants (I say paths, because light can go many directions, correct?)?
A path through spacetime is not really what we mean by invariant. That's a defined set of points in the spacetime manifold. It's not easy to describe that path until you have chosen a coordinate system, but (and this is the key point), the path exists and is well-defined without being given a coordinate description.

Generally there are two types of path: timelike (followed by massive particles) and null (followed by light). And, there are general timelike and null paths and geodesic timelike and null paths, which are the natural paths that particles and light follow through spacetime. Massive particles can, of course, be forced off geodesic paths, but the path remains timelike. I'm not sure there's any way to force a light onto a null non-geodesic path(?)

There are clearly an infinitude of possible paths, but each particle or light ray can only take one path through spacetime (its worldline).

An invariant is something you calculate, like the length of a spacetime path between two events. Null paths have zero length in all coordinate systems and timelike paths have the same non-zero length in all coordinate systems. So, it's the length of the spacetime path that is invariant.

We don't really talk about the path itself being invariant.

• etotheipi and jbriggs444
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A path through spacetime is not really what we mean by invariant.

It can be. You say:

the path exists and is well-defined without being given a coordinate description

That's what "invariant" means, so yes, a path through spacetime would be an invariant.

What it would not be is what I would call a "local" invariant, i.e., an invariant defined at a single spacetime point. As you say, it's a set of spacetime points. But that set of points is the same no matter what coordinates you choose.

I'm not sure there's any way to force a light onto a null non-geodesic path(?)

There is: a waveguide or fiber optic cable are examples of things that can do this.

it's the length of the spacetime path that is invariant.

We don't really talk about the path itself being invariant.

It's true that the term "invariant" is more likely to be used to describe the arc length along the path than the path itself. However, I don't think that means it's wrong to describe the path itself as invariant; it's just a less common use of the term.

• PeroK