Basic introduction to gravitation as curved spacetime

  • #1
cianfa72
1,246
155
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.
 
Last edited:

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
21,996
12,906
How did your daughter respond to this explanation?
 
  • #3
cianfa72
1,246
155
How did your daughter respond to this explanation?
She seems to have grasped it :wink:. 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).
 
Last edited:
  • #4
cianfa72
1,246
155
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 ?
 
  • #5
haushofer
Science Advisor
Insights Author
2,701
1,156
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.
 
  • Like
Likes PhDeezNutz and ergospherical
  • #6
pervect
Staff Emeritus
Science Advisor
Insights Author
10,187
1,351
She seems to have grasped it :wink:. 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.
 
  • #7
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,721
15,330
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.
 
  • #8
cianfa72
1,246
155
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.
 
Last edited:
  • #9
ergospherical
877
1,203
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"?
 
  • #10
cianfa72
1,246
155
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' ?
 
Last edited:
  • #11
martinbn
Science Advisor
3,047
1,382
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".
 
  • #12
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,721
15,330
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?
 
  • #13
cianfa72
1,246
155
Is the ball wearing a wristwatch?
Yes, it is.
 
  • #14
ergospherical
877
1,203
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.
 
  • Like
Likes vanhees71 and PeroK
  • #15
cianfa72
1,246
155
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).
 
  • #16
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,721
15,330
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.
 
  • #17
martinbn
Science Advisor
3,047
1,382
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.
 
  • Like
Likes vanhees71 and ergospherical
  • #18
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,721
15,330
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?
 
  • Like
Likes vanhees71 and ergospherical
  • #19
cianfa72
1,246
155
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 🤔
 
  • #20
cianfa72
1,246
155
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.
 
  • #21
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
21,996
12,906
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.
 
  • #22
cianfa72
1,246
155
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).
 
  • #23
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,721
15,330
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.
 
  • #24
cianfa72
1,246
155
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).
 
  • #25
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
21,996
12,906
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.
 
  • #26
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,721
15,330
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.
 
  • #27
cianfa72
1,246
155
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.
 
  • #28
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,721
15,330
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.
 
  • #29
cianfa72
1,246
155
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.
 
  • #30
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,721
15,330
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.
 
  • #31
cianfa72
1,246
155
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.
 
  • #32
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,721
15,330
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!
 
  • #33
cianfa72
1,246
155
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.
 
  • #34
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
21,996
12,906
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}##.
 
  • #35
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,721
15,330
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.
 

Suggested for: Basic introduction to gravitation as curved spacetime

  • Last Post
Replies
2
Views
315
Replies
7
Views
777
  • Last Post
Replies
21
Views
848
Replies
141
Views
3K
  • Last Post
Replies
1
Views
611
Replies
7
Views
778
  • Last Post
Replies
6
Views
478
  • Last Post
2
Replies
61
Views
2K
Replies
2
Views
305
Replies
27
Views
893
Top