# Basic introduction to gravitation as curved spacetime

• B
Staff Emeritus
The principle of maximal aging, aka Hamilton's principle, is rather nice because it's independent of the choice of frame of reference.

In an inertial frame of reference in flat space-time, the path of maximal aging will appear to be a straight line in that frame. In an accelerating frame , such as that of an accelerating spaceship, the same path through space-time will appear to be curved in space. One can explain why the curved path has maximal aging in an accelerated frame by invoking what is usually called "gravitational time dilation", which interacts with the usual velocity-dependent "time dilation" in special relativity to generate the path which satisfies the condition of maximal aging. Different frames have different "explanations" with various degrees of complexity, but everyone agrees on the end result.

The "straightforwards" approach is to compute the total elapsed time along any timelike path, then use variational principles to find the differential equation that represents the path of maximal aging.

Taylor wrote a fair number of papers on the principle of maximal aging and the related principle of "least action", some discussion and bibliographical references are available on his website https://www.eftaylor.com/leastaction.html.

It is worth noting that in some cases, one needs to use the principle of extremal or stationary aging, rather than a true maximum, but that's probably best left for another thread.

Sagittarius A-Star, cianfa72 and vanhees71
cianfa72
The principle of maximal aging, aka Hamilton's principle, is rather nice because it's independent of the choice of frame of reference.

In an accelerating frame, such as that of an accelerating spaceship, the same path through space-time will appear to be curved in space.

The "straightforwards" approach is to compute the total elapsed time along any timelike path, then use variational principles to find the differential equation that represents the path of maximal aging.
In the accelerating spaceship in flat spacetime there is no curvature at all. So a non-local approach is actually needed to introduce curved spacetime. IMO we need to consider a non-local experiment involving geodesic deviation (i.e tidal gravity).

Mentor
In the accelerating spaceship in flat spacetime there is no curvature at all.
No spacetime curvature. But the worldline of the spaceship is curved; it has nonzero proper acceleration, and proper acceleration is path curvature. Which means that in a non-inertial frame in which the spaceship is at rest, the paths of freely falling objects will appear curved. That is what @pervect was referring to.

vanhees71 and cianfa72
cianfa72
No spacetime curvature. But the worldline of the spaceship is curved; it has nonzero proper acceleration, and proper acceleration is path curvature. Which means that in a non-inertial frame in which the spaceship is at rest, the paths of freely falling objects will appear curved.
Yes that's true for the paths of free falling objects in that non-inertial frame. However it does not imply non-zero spacetime curvature, I believe.

Mentor
it does not imply non-zero spacetime curvature, I believe.
No, it doesn't, and @pervect was not claiming it did. The curvature he was talking about was not spacetime curvature. But in your previous post that I responded to, you didn't say "no spacetime curvature". You said "no curvature".

vanhees71 and cianfa72
cianfa72
But in your previous post that I responded to, you didn't say "no spacetime curvature". You said "no curvature".
Ah ok, you are right.

cianfa72
So, to bring out the spacetime curvature I suggest the following:

Repeat again the experiment in the OP this time throwing two balls in the same (spatial) direction and with the same velocity from different heights w.r.t. the Earth surface (the two timelike geodesics from different heights start parallel in spacetime).

You can see that they will arrive at their destinations (her friend at location B at different heights) with slightly different velocities (the two timelike geodesics are no longer parallel in spacetime).

Of course if we do this experiment a few meters apart we do not see any difference, but in principle I believe is a good 'test' for geodesic deviation (i.e. spacetime curvature).

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Mentor
So, to bring out the spacetime curvature I suggest the following:

Repeat again the experiment in the OP this time throwing two balls in the same (spatial) direction and with the same velocity from different heights w.r.t. the Earth surface (the two timelike geodesics from different heights start parallel in spacetime).

You can see that they will arrive at their destinations (her friend at location B at different heights) with slightly different velocities (the two timelike geodesics are no longer parallel in spacetime).
I assume that the destination heights are different from the starting heights, correct? In other words, the two balls start at heights ##A_1## and ##A_2##, both launched upward with velocity ##v##, and end at heights ##B_1## and ##B_2##, where ##B_1 > A_1##, ##B_2 > A_2##, but ##B_1 - A_1 = B_2 - A_2##? If that's the case, yes, their velocities at arrival will be different, but the setup is rather complicated.

A simpler way to set up this kind of geodesic deviation experiment is to drop two balls in free fall, both from rest at slightly different heights, and keep track of their separation; it will change, indicating that their geodesics, which were initially parallel (at rest relative to each other) no longer are (since if they stayed at rest relative to each other their separation would not change).

cianfa72
I assume that the destination heights are different from the starting heights, correct? In other words, the two balls start at heights ##A_1## and ##A_2##, both launched upward with velocity ##v##, and end at heights ##B_1## and ##B_2##, where ##B_1 > A_1##, ##B_2 > A_2##, but ##B_1 - A_1 = B_2 - A_2##?
Yes, exactly.

drop two balls in free fall, both from rest at slightly different heights
In order to verify that they are actually at rest with each other, they can exchange light signals between them and check that the round-trip time does not change over time (excuse the pun ), I believe.

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Mentor
In order to verify that they are actually at rest with each other, they can exchange light signals between them and check that the round-trip time does not change over time (excuse the pun ), I believe.
Yes.

Staff Emeritus
In the accelerating spaceship in flat spacetime there is no curvature at all. So a non-local approach is actually needed to introduce curved spacetime. IMO we need to consider a non-local experiment involving geodesic deviation (i.e tidal gravity).

The Riemann curvature tensor for the space-time of an accelerated spaceship is zero, just as it is for an inertial frame of reference. This is because the Riemann curvature tensor is coordinate independent. This implies that it only depends on the geometry of the space-time, if it's zero in one coordinate system, it's zero in all coordinate systems. In the language of components, if the Riemann is zero, all components are zero.

Now, we often use the example of an accelerated spaceship to describe "gravity" via the equivalence principle, this is potentially confusing. The confusion is in the lay language, not the mathematics. Unfortunately, the lay language is less precise than the math.

The mathematical entity associated with the "gravity" of an accelerating spaceship is not the Riemann curvature tensor. I would say in this case it is the set of Christoffel symbols, which match the conceptual idea that whatever "gravity" is, it can be non-zero in an accelerating spaceship but zero in free-fall for the exact same space-time. There aren't really any clear references that I am aware of that go into this, unfortunately.

A good analogy here is polar coordinates in the plane, vs cartesian coordinates. In Cartesian coordinates, the Christoffel symbols are all zero, in polar coordinates, they are not. The coordinates don't matter to the geometry of the plane - at the level of geometry, a plane is a plane, the coordinates are just a human convention we use to identify points in it.

Unfortunately, if we use the idea of "gravity" as being the Christoffel symbols, we can't talk about the physics in terms that are independent of the observer. Einstein made remarks to this effect, I don't recall the exact quote, and a keyword search for "It is not good to" didn't dig up the quote :(.

MTW talks about this a little bit, I would summarize their remarks as there being a lot of mathematical quantities associated with "gravity", and that when we use the generic term in lay language, it can be confusing.

Unfortunately, the alternative to such broad use of terms is to be very precise and formal. This is unfortunately both hard to do and also doesn't reach people without the necessary background.

The takeaway here, for me at least, is that when someone is talking about "gravity", one needs to figure out which mathematical entity that the author is referring to, as there are several possibilities.

cianfa72, vanhees71 and PeroK
Gold Member
2022 Award
Indeed that's the difference between a true gravitational field and an accelerated frame of reference in special relavity: A true gravitational field has a non-vanishing curvature tensor, the Minkowski space is flat, no matter in which coordinates/reference frames you describe it.

cianfa72
cianfa72
Which means that in a non-inertial frame in which the spaceship is at rest, the paths of freely falling objects will appear curved.
Just to be more specific, as non-inertial frame in which the spaceship is at rest consider the following:

Imagine a latticework of rods and clocks built inside the spaceship at rest each other (each of them has a non zero proper acceleration). This set of clocks defines a non-geodesic timelike congruence in the spacetime region involved in the experiment. Assign fixed values for the spatial coordinates of each of them and take the proper time measured from each of them as the coordinate time of the chart being defined. The coordinate chart (aka reference frame) defined this way is just one of the possibile reference frames in which the spaceship as whole (i.e. its worldtube) is at rest.

Is that correct ? Thanks.

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Gold Member
... take the proper time measured from each of them as the coordinate time of the chart being defined. The coordinate chart (aka reference frame) defined this way is just one of the possibile reference frames in which the spaceship as whole (i.e. its worldtube) is at rest.
That does define a coordinate chart, provided you also specify how to initially synchronise all the clocks. Unfortunately the clocks won't remain synchronised (by the same criteria), because of "(pseudo-) gravitational time dilation".

A better solution is to redefine coordinate time as a constant multiple of proper time (a different constant for each clock) so as to maintain synchronisation. That's the principle for Rindler coordinates.

PeterDonis and vanhees71
cianfa72
That does define a coordinate chart, provided you also specify how to initially synchronise all the clocks. Unfortunately the clocks won't remain synchronised (by the same criteria), because of "(pseudo-) gravitational time dilation".
Sorry, even if you do not specify how to initially synchronize clocks they still define a coordinate chart -- namely it smoothly maps events of that spacetime region to ##\mathbb R^4## with no singularities.

A better solution is to redefine coordinate time as a constant multiple of proper time (a different constant for each clock) so as to maintain synchronisation.
Do you mean 'adjust' the rate of standard clocks depending on their location in order to maintain synchronization when initially synchronized ?

Gold Member
Sorry, even if you do not specify how to initially synchronize clocks they still define a coordinate chart -- namely it smoothly maps events of that spacetime region to ##\mathbb R^4## with no singularities.
Well, if you don't specify how to synchronise, you haven't fully defined the chart. But they can be synced any "smooth" way you like. (Any two "nearby" events deemed to be simultaneous must be spacelike-separated.)
Do you mean 'adjust' the rate of standard clocks depending on their location in order to maintain synchronization when initially synchronized ?
That's one way to do it in practice, yes.

cianfa72
Well, if you don't specify how to synchronize, you haven't fully defined the chart. But they can be synced any "smooth" way you like. (Any two "nearby" events deemed to be simultaneous must be spacelike-separated.)
I assumed the following as reference system (actually the mathematical map it defines from spacetime events to ##\mathbb R^4##) from Landau "The Classic Theory of Field - vol 2" sect 82.

In the latter (SR) we meant by a reference system a set of bodies at rest relative to one another in unchanging relative positions. Such systems of bodies do not exist in the presence of a variable gravitational field,
and for the exact determination of the position of a particle in space we must, strictly speaking, have an infinite number of bodies which fill all the space like some sort of "medium". Such a system of bodies with arbitrarily running clocks fixed on them constitutes a reference system in the general theory of relativity.

IMO there is no requirement about clock's rates as well as the procedure employed to synchronize them.

vanhees71
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Of course not, since in a general spacetime you cannot synchronize clocks at different positions.

cianfa72
cianfa72
Of course not, since in a general spacetime you cannot synchronize clocks at different positions.
Not sure to grasp it: since in GR we cannot synchronize clocks at different positions, there is actually no requirement from the point of view of clock synchronization in order to define a coordinate chart, don't you ?

vanhees71
Gold Member
2022 Award
Yes, sure. I was confirming what you said in #53.

cianfa72
Gold Member
IMO there is no requirement about clock's rates as well as the procedure employed to synchronize them.
You are right. The only point I was trying to make was that, to analyse a particular problem, you need to know how one clock relates to another clock, either by specifying what it is or by experimental measurement.

And after initial syncing there's no requirement to remain in sync. In many spacetimes, it's impossible.

But in a particular scenario it may be convenient to choose a coordinate system where there is a specific kind of synchronisation that is maintained.

cianfa72
cianfa72
you need to know how one clock relates to another clock, either by specifying what it is or by experimental measurement.
Could you be more specific ? Thank you.

trainman2001
That video explaining the concept of gravity to five levels of experience was terrific! Still left with more questions than answers, as were the folks in the video.