B Basic introduction to gravitation as curved spacetime

[cianfa72]

TL;DR Summary

Basic introduction to gravitation as curved spacetime

Hi,

my daughter saw my MTW copy on the desk and she asked me about the picture with the apple in front. To introduce her to the idea of gravitation as curved spacetime I answered like this:

Consider you (A) and a your friend (B) at two different spots on a garden each with a firecracker. Take a ball with a wristwatch attached to it. Throw the ball when your firecracker explode (the launch event at A) and consider all possible ball's paths going to your friend that happen to arrive just when her firecracker explode (the arrive event at B).

Now the actual path the ball takes from those two events is that that maximize the reading of the ball attached wristwatch.

What do you think about, could such an introduction make sense ? Thank you.

 

[vanhees71]

How did your daughter respond to this explanation?

 

[cianfa72]

How did your daughter respond to this explanation?

She seems to have grasped it . I tried to tell her (nine years old) that there is actually no 'magnet' inside the Earth (that was her first understanding of why all bodies seem to be 'attracted' from it).

 

[cianfa72]

By the way, she saw the 'elastic rubber sheet' model from school some time ago. I spend much time to convince her that it is actually a wrong mental 'picture' of what is really going on.

In your opinion, which is an easy and effective way to introduce such concepts ?

 

[haushofer]

The elastic sheet rubber model.

It has its drawbacks, as many simple analogies for complex physics, but it grasps the paradigm shift from force to geometry well.

 

[pervect]

She seems to have grasped it . I tried to tell her (nine years old) that there is actually no 'magnet' inside the Earth (that was her first understanding of why all bodies seem to be 'attracted' from it).

The explanation of Hamilton's principle was correct, but I have my doubts how helpful it was to someone only 9 years old, unless she has a surprisingly good background in special relativity.

The rationale is as follows. I believe one needs to understand the principle of maximal aging in flat space-time, before attempting to understand it in curved space-time. Newtonian approaches, since they don't have any differential aging, aren't adequate to motivate the principle of maximal aging even in flat space-time.

With a sufficient knowledge of special relativity to understand differential aging, and also to be able to draw space-time diagrams, I think the approach based on maximal aging can be understood. One can start to explain the consequences of drawing space-time diagrms (which represent the abstract idea of space-time) on curved surfaces. But without the background in SR to start, I don't see the approach working.

 

[PeroK]

Summary:: Basic introduction to gravitation as curved spacetime

Hi,

my daughter saw my MTW copy on the desk and she asked me about the picture with the apple in front. To introduce her to the idea of gravitation as curved spacetime I answered like this:

There is a series of Wired videos where topics are explained in five different levels - from child to PhD student. Here's the one on gravity:

https://www.wired.com/video/watch/5...xplains-one-concept-in-5-levels-of-difficulty

She starts explaining gravity to an eight-year-old. I'm not sure that you can do much better than this.

 

[cianfa72]

The explanation of Hamilton's principle was correct, but I have my doubts how helpful it was to someone only 9 years old, unless she has a surprisingly good background in special relativity.

The rationale is as follows. I believe one needs to understand the principle of maximal aging in flat space-time, before attempting to understand it in curved space-time.

I take it as the Hamilton's principle of maximal aging applies in SR as well (no gravity at all). That's true.

Just to be clear the point is that if we do the same 'experiment' in the OP in flat spacetime then the path of maximal wristwatch aging is in space a straight line with euclidean properties (yes, we know its worldline in spacetime is actually a timelike geodesic of Minkowski geometry).

The same 'experiment' done in curved spacetime (as in OP) shows that the path of maximal aging involves a change of gravitational potential since we know, let me say, that time at higher height w.r.t. the Earth surface 'runs faster' (see for example Feynman Lecture 42 - 8).

So, IMO, the take-home message is that near Earth the 'time curvature' of spacetime is actually much more noticeable than the spatial one.

 

[ergospherical]

So, IMO, the take-home message is that near Earth the 'time curvature' of spacetime is actually much more noticeable than the spatial one.

What do you mean by time curvature and spatial curvature?

That being, curves, 2-surfaces, 3-surfaces, spacetime, have on them defined various measures of curvature, but you need to explicitly specify the (sub-)manifold to talk about it (not just "time", "space", etc.)

How do you define "much more noticeable"?

 

[cianfa72]

What do you mean by time curvature and spatial curvature?

That being, curves, 2-surfaces, 3-surfaces, spacetime, have on them defined various measures of curvature, but you need to explicitly specify the (sub-)manifold to talk about it (not just "time", "space", etc.)

Yes, that's true. To be more precise maybe we should describe the OP experiment using Schwarzschild static spacetime geometry in Schwarzschild chart.

So, using Schwarzschild chart we can describe the OP experiment as follows (to take it simple we drop the $\theta$ coordinate):

Consider you (A) and a your friend (B) described by worldlines having the same constant coordiante radius $r$ and different even if constant coordinate $\phi$ (they are basically the Schwarzschild hovering observers at fixed coordiante radius $r$). Take a ball with a wristwatch attached to it.

Throw the ball when your firecracker explode (the launch event with coordinates $(r,\phi_a,t_A)$ ) and consider all possible ball's paths going to your friend that happen to arrive just when her firecracker explode (the arrive event at B with coordinates $(r,\phi_B,t_B)$ ).

Now we know the actual path the (free) ball takes in spacetime is the Schwarzschild timelike geodesic connecting those two events. From a physical point of view it is the spacetime path maximizing the wristwatch aging.

From this kind of experiment that does not involve any geodesic deviation between nearby timelike geodesics, can we actually infer any 'spacetime curvature' ?

 

[martinbn]

What do you mean by time curvature and spatial curvature?

That being, curves, 2-surfaces, 3-surfaces, spacetime, have on them defined various measures of curvature, but you need to explicitly specify the (sub-)manifold to talk about it (not just "time", "space", etc.)

How do you define "much more noticeable"?

For a B level thread just saying "time", "space", etc. might be the best one can do. It is definitely better than "explicitly specifying the submanifold".

 

[PeroK]

Yes, that's true. To be more precise maybe we should describe the OP experiment using Schwarzschild static spacetime geometry in Schwarzschild chart.

So, using Schwarzschild chart we can describe the OP experiment as follows (to take it simple we drop the $\theta$ coordinate):

Consider you (A) and a your friend (B) described by worldlines having the same constant coordiante radius $r$ and different even if constant coordinate $\phi$ (they are basically the Schwarzschild hovering observers at fixed coordiante radius $r$). Take a ball with a wristwatch attached to it.

Throw the ball when your firecracker explode (the launch event with coordinates $(r,\phi_a,t_A)$ ) and consider all possible ball's paths going to your friend that happen to arrive just when her firecracker explode (the arrive event at B with coordinates $(r,\phi_B,t_B)$).

Now we know the actual path the (free) ball takes in spacetime is the Schwarzschild timelike geodesic connecting those two events. From a physical point of view it is the spacetime path maximizing the wristwatch aging.

From this kind of experiment that does not involve any geodesic deviation between nearby timelike geodesics, can we actually infer any 'spacetime curvature' ?

Is the ball wearing a wristwatch?

 

[cianfa72]

Is the ball wearing a wristwatch?

Yes, it is.

 

[ergospherical]

For a B level thread just saying "time", "space", etc. might be the best one can do. It is definitely better than "explicitly specifying the submanifold".

But the issue is that I don’t know what curvature of time means! It could be many things.

 

[cianfa72]

But the issue is that I don’t know what curvature of time means!

You're right. Mine was a tentative to describe, as Feynman did, why the ball has to increase its height w.r.t. the Earth surface along its path between the two given events.

I believe it is the main take-home of the OP experiment. We cannot use Newtonian mechanics to explain why the ball's attached wristwatch ages that way and in the context of SR the ball's path would be different as well.

As in my post #10, I am not sure we can actually infer spacetime curvature by that experiment since it does not involve at all the geodesic deviation between nearby timelike geodesics (i.e. tidal gravity).

 

[PeroK]

I believe it is the main take-home of the OP experiment. We cannot use Newtonian mechanics to explain why the ball's attached wristwatch ages that way and in the context of SR the ball's path would be different as well.

Locally, in an experiment like a ball dropped a few metres above the ground, everything can be explained in terms of Newtonian mechanics. The Earth's surface locally experiences a real, upward force, hence proper upward acceleration. The ball in free-fall experiences no force and no proper acceleration. It is moving inertially. And, of course, $F = ma$ applies.

To explain Newtonian gravitation globally requires gravity as a real force.

Given that everyone's experience of gravity is the acceleration of falling objects, appealing to the notion that wristwatches are doing something different hardly seems a good introduction - as no one can have much intuition about that. Perhaps after a course in SR, that might mean something, but not for a child or beginner.

 

[martinbn]

But the issue is that I don’t know what curvature of time means! It could be many things.

I agree, but the other issue is that a high school kid most likely doesn't know what a manifold is.

 

[PeroK]

I agree, but the other issue is that a high school kid most likely doesn't know what a manifold is.

Nor Minkowsi geometry, as mentioned in post #8. The simple fact is that it is impossible to adequately or sensibly explain GR to a nine-year old. Newtonian gravity, as in the Wired video, is possible. How is a nine-year-old going to makes sense of gravitational time dilation as a motivation for free-fall acceleration?

 

[cianfa72]

How is a nine-year-old going to makes sense of gravitational time dilation as a motivation for free-fall acceleration?

Remember, my point was to try to add some explanation about the picture with the apple in front of my MTW copy

 

[cianfa72]

Locally, in an experiment like a ball dropped a few meters above the ground, everything can be explained in terms of Newtonian mechanics. The Earth's surface locally experiences a real, upward force, hence proper upward acceleration. The ball in free-fall experiences no force and no proper acceleration. It is moving inertially. And, of course, $F = ma$ applies.

So, locally in an experiment like a ball dropped a few meters above the ground, since the Earth's surface undergoes an upward proper acceleration, the Newton law $F=ma$ actually applies to it and not to the ball that moves inertially.

To explain Newtonian gravitation globally requires gravity as a real force.

As in the local case, see above.

 

[vanhees71]

Well, I think it makes much more sense to explain to a 9y-old first the idea of Newtonian gravitation and the equivalence principle. This is already a challenge! It's not even easy for students in the mechanics 1 lecture.

One first has to explain what an inertial reference frame is (Newton's first postulate). Then you can introduce the notion of inertial mass and force (Newton's second postulate) and finally the action-reaction principle (Newton's third law). That sets the minimal framework for discussing (Newtonian) gravity and the great discovery of the equivalence principle. You can start with the gravitational force on bodies on Earth in the usual approximation $\vec{F}_g=m \vec{g}$ with $\vec{g}=\text{const}$. This already includes the (weak) equivalence principle, i.e., the proportionality of the gravitational force of a body on Earth to its inertial mass. It should be made clear that this is an amazing discovery, which is the specialty of the gravitational force of all the forces in nature. It implies that a reference frame which is freely falling in a homogeneous gravitational field, realizes an inertial frame.

That's of course the heuristic way setting Einstein on track to discover general relativity, but it relies of course heavily on special relativity to argue that from starting with free-falling reference systems as (local) inertial reference frames one comes finally to the conclusion that gravity is described by a 4D curved spacetime manifold and that free-falling bodies move on geodesics in spacetime.

 

[cianfa72]

That's of course the heuristic way setting Einstein on track to discover general relativity, but it relies of course heavily on special relativity to argue that from starting with free-falling reference systems as (local) inertial reference frames one comes finally to the conclusion that gravity is described by a 4D curved spacetime manifold and that free-falling bodies move on geodesics in spacetime.

Thinking about it, IMO, we cannot infer the 4D spacetime curvature if we do not consider experiments that show the geodesic deviation (i.e. tidal gravity).

 

[PeroK]

So, locally in an experiment like a ball dropped a few meters above the ground, since the Earth's surface undergoes an upward proper acceleration, the Newton law $F=ma$ actually applies to it and not to the ball that moves inertially.

Precisely.

As in the local case, see above.

Not at all. All the projectile motion problems of elementary mechanics can equally be done by considering the Earth's surface as an accelerating reference frame and zero gravitational force on the projectile. I.e. gravity can be treated as a fictitious force in the same way as the centrifugal force in a rotating reference frame.

That all works locally. The difficulties arise when you look at a global scenario or where the gravitational force varies with distance and is no longer approximately constant.

This is a way to explain why Newton's gravity must be updated. In a nutshell, curved spacetime is the additional factor that allows all these locally valid analyses to be joined together into a globally valid solution.

If I had to explain GR to high school students that might be my approach. Push Newtonian gravity as far as it will go, then introduce curved spacetime non-mathematically as the global solution. The missing piece in the jigsaw.

 

[cianfa72]

All the projectile motion problems of elementary mechanics can equally be done by considering the Earth's surface as an accelerating reference frame and zero gravitational force on the projectile. I.e. gravity can be treated as a fictitious force in the same way as the centrifugal force in a rotating reference frame.

That all works locally. The difficulties arise when you look at a global scenario or where the gravitational force varies with distance and is no longer approximately constant.

This is a way to explain why Newton's gravity must be updated. In a nutshell, curved spacetime is the additional factor that allows all these locally valid analyses to be joined together into a globally valid solution.

ok, so your point is that locally in the Earth's rest frame (an accelerating not inertial rest frame) there exist a "field" responsible of the fictitious forces on all free bodies (e.g. on the projectile).

 

[vanhees71]

Well, I prefer the interaction/field picture for gravity anyway. It's more intuitive than to just jump to the geometric interpretation, which can be derived from the field point of view, as demonstrated in the Feynman lectures on gravitation. That's of course also not suitable at the high-school level.

 

[PeroK]

ok, so your point is that locally in the Earth's rest frame (an accelerating not inertial rest frame) there exist a "field" responsible of the fictitious forces on all free bodies (e.g. on the projectile).

Simpler than that. The Earth's surface locally is an accelerating reference frame. No different from the accelerating elevator in free space. In a sense, the constant gravitational field disappears.

The kinematics of a local projectile on Earth are equivalent to the kinematics of a projectile in an accelerating box in free space.

Moreover, if you have an accelerating vehicle in classical mechanics, the simplest solution is to add the negative of its acceleration vectorially to that of gravity and have a single resultant "gravity". Then proceed as though that is gravity. That's all pure classical mechanics and does not require more than classical kinematics.

PS what we know as gravity has a centrifugal element caused by the Earth's rotation. It's all the equivalence principle, whether you know it or not.

 

[cianfa72]

Simpler than that. The Earth's surface locally is an accelerating reference frame. No different from the accelerating elevator in free space. In a sense, the constant gravitational field disappears.

The kinematics of a local projectile on Earth are equivalent to the kinematics of a projectile in an accelerating box in free space.

That's true. However from the point of view of the inside of the accelerating box in free space there exist a constant field acting on all free bodies including the (free) projectile.

 

[PeroK]

That's true. However from the point of view of the inside of the accelerating box in free space there exist a constant field acting on the free projectile.

It's simply an accelerating box - there is no need to invoke a field.

 

[cianfa72]

It's simply an accelerating box - there is no need to invoke a field.

Yes, but that does mean you have to do the classic kinematics analysis from an 'external' not accelerating reference frame, I believe.

 

[PeroK]

Yes, but that does mean you have to do the classic kinematics analysis from an 'external' not accelerating reference frame, I believe.

If you insist that a fictitious force on a projectile must be due to a force field (perhaps you've watched too many epsiodes of Star Trek!) then I can't stop you. It doesn't mean that I have to adopt the notion of a force field. I can leave field theory out of things until I really need it.

 

[cianfa72]

If you insist that a fictitious force on a projectile must be due to a force field (perhaps you've watched too many epsiodes of Star Trek!) then I can't stop you. It doesn't mean that I have to adopt the notion of a force field. I can leave field theory out of things until I really need it.

Sorry, maybe I've not grasp it. From the inside of the accelerating box (image really an accelerating box without any window) how do you explain the parabolic motion of the projectile ? Thank you.

 

[PeroK]

Sorry, maybe I've not grasp it. From the inside of the accelerating box (image really an accelerating box without any window) how do you explain the parabolic motion of the projectile ? Thank you.

An object on the floor of the box can feel the real force accelerating it. You know, therefore, the box is an accelerating reference frame, so you imagine a fictitious force acting on the projectile and apply Newton's laws. The explanation is that you know the real force is acting on the box. You know the fictitious force is not real, but that doesn't stop you using it.

If you say that a force implies a field, then we need a fictitious force field. But, that's not mandated by Newton's laws. There's no Newton's fourth law: all forces arise from fields, fictional or otherwise!

 

[cianfa72]

If you say that a force implies a field, then we need a fictitious force field. But, that's not mandated by Newton's laws. There's no Newton's fourth law: all forces arise from fields, fictional or otherwise!

ok, so your point is simply that inside the accelerating box there is a fictitious force acting on all bodies (e.g. on the projectile) without introducing any field it arises from.

 

[vanhees71]

The "fictitious forces" of course come from the acceleration calculated in the non-inertial frame. Then you bring them to the right-hand side where they add to the force (in Newtonian mechanics of course).

Take the most simple case of a constantly accelerated reference frame. If $\vec{x}$ is the position vector in the inertial frame then
$$\vec{x}'=\vec{x}-\frac{\vec{a}}{2} t^2$$
is the position vector wrt. the accelerated frame. Then the EoM. in the inertial frame of course is
$$m \ddot{\vec{x}}=\vec{F}.$$
In the non-inertial frame that reads
$$m \ddot{\vec{x}'}+m \vec{a}=\vec{F} \; \Rightarrow \; m \ddot{\vec{x}}'=\vec{F}-m \vec{a}.$$
So the observer in the non-inertial frame, interpreting his acceleration as if he was in an inertial frame, finds an additional fictitious (or rather inertial) force $\vec{F}_{\text{inert}}=-m \vec{a}$.

 

[PeroK]

ok, so your point is simply that inside the accelerating box there is a fictitious force acting on all bodies (e.g. on the projectile) without introducing any field it arises from.

Yes, exactly.

 

[pervect]

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.

 

[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).

 

[PeterDonis]

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.

 

[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.

 

[PeterDonis]

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".

 

[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).

 

[PeterDonis]

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.

 

[PeterDonis]

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.

 

[pervect]

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.

 

[vanhees71]

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.

 

[A.T.]

In your opinion, which is an easy and effective way to introduce such concepts ?

Like this way to introduce it:


It's from this book (scroll down for free download links):
https://archive.org/details/L.EpsteinRelativityVisualizedelemTxt1994Insight/mode/2up

 

[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.

 

[DrGreg]

... 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.