Spacetime Curvature: Exploring Eddington's Experiment

[Carpe Physicum]

TL;DR Summary

Further details on what ST curvature really means

Here's what I read in a closed thread asking about the meaning of spacetime curvature:

The curvature of space-time is a mathematical curvature, and should not be taken to mean that space and time are somehow curved through each other or through some as yet undiscovered dimension in the way a two dimensional space can be curved into the third dimension in a three-dimensional world.

So does that mean if we were able to move to some spot "way above the earth/sun/distant galaxy" where we could watch light and trace it from a distant galaxy (which is otherwise hidden from us) pass around (?) the sun and hit our telescope, it wouldn't literally curve around the sun as is depicted in so many laymen's physics books? (I'm describing the Eddington experiment of course.)

 

[Dale]

The curvature of spacetime is more than just mathematical. It is physical. In curved spacetime the rules of flat geometry fail. Triangles can have interior angles different from 180 degrees. Straight parallel lines can intersect. Lines can continuously curve away from each other without getting further apart. Etc.

All of those non-flat geometrical features can be physically measured, so the claim that it is just math is incorrect.

 

[ergospherical]

Curvature of spacetime is related to the failure of vectors to return to themselves after being subjected to a parallel displacement around an infinitesimal closed contour. In other words if a vector $X^{\mu}$ is parallel transported ($t^{\nu} \nabla_{\nu} X^{\mu} = 0$) around a closed curve $\mathcal{C} = \partial \mathcal{S}$ of tangent $t^{\mu}$ then upon returning to its original position it has changed by\begin{align*}
\oint \delta X^{\mu} = -\oint_{\mathcal{C}} dx^{\rho} \Gamma^{\mu}_{\nu \rho} X^{\nu} &= -\int_{\mathcal{S}} dS^{\sigma \rho} \frac{\partial}{\partial x^{\sigma}} (\Gamma^{\mu}_{\nu \rho} X^{\nu}) \\ \\

&= -\frac{1}{2} \int_{\mathcal{S}} dS^{\rho \sigma} \left( \frac{\partial}{\partial x^{\rho}}(\Gamma^{\mu}_{\nu \sigma} X^{\nu}) - \frac{\partial}{\partial x^{\sigma}}(\Gamma^{\mu}_{\nu \rho} X^{\nu}) \right) \\ \\

&= -\frac{1}{2} \int_{\mathcal{S}} dS^{\rho \sigma} \left( \frac{\partial \Gamma^{\mu}_{\nu \sigma}}{\partial x^{\rho}}X^{\nu} - \frac{\partial \Gamma^{\mu}_{\nu \rho}}{\partial x^{\sigma}}X^{\nu} + \Gamma^{\mu}_{\nu \sigma} \frac{\partial X^{\nu}}{\partial x^{\rho}} - \Gamma^{\mu}_{\nu \rho} \frac{\partial X^{\nu}}{\partial x^{\sigma}} \right) \\

&= -\frac{1}{2} \int_{\mathcal{S}} dS^{\rho \sigma} \left( \frac{\partial \Gamma^{\mu}_{\nu \sigma}}{\partial x^{\rho}} - \frac{\partial \Gamma^{\mu}_{\nu \rho}}{\partial x^{\sigma}} - \Gamma^{\mu}_{\alpha \sigma} \Gamma^{\alpha}_{\nu \rho} + \Gamma^{\mu}_{\alpha \rho} \Gamma^{\alpha}_{\nu \sigma} \right)X^{\nu}
\end{align*}since $\delta X^{\mu} = -dx^{\rho} \Gamma^{\mu}_{\nu \rho} X^{\nu}$ under a parallel translation and consequently $\dfrac{\partial X^{\mu}}{\partial x^{\rho}} = - \Gamma^{\mu}_{\nu \rho} X^{\nu}$. If the loop is infinitesimal then this goes over to \begin{align*}
\oint \delta X^{\mu} = - \frac{1}{2} R^{\mu}_{\nu \rho\sigma} X^{\nu} \delta S^{\rho \sigma}
\end{align*}where $\delta S^{\rho \sigma}$ is the infinitesimal area enclosed by the loop and $R^{\mu}_{\nu \rho \sigma} := \dfrac{\partial \Gamma^{\mu}_{\nu \sigma}}{\partial x^{\rho}} - \dfrac{\partial \Gamma^{\mu}_{\nu \rho}}{\partial x^{\sigma}} - \Gamma^{\mu}_{\alpha \sigma} \Gamma^{\alpha}_{\nu \rho} + \Gamma^{\mu}_{\alpha \rho} \Gamma^{\alpha}_{\nu \sigma}$ is the curvature tensor.

 

[Dale]

But again, it is not just mathematical. This is the way the measurable geometry of the universe works.

 

[PeroK]

Summary:: Further details on what ST curvature really means

Here's what I read in a closed thread asking about the meaning of spacetime curvature:

So does that mean if we were able to move to some spot "way above the earth/sun/distant galaxy" where we could watch light and trace it from a distant galaxy (which is otherwise hidden from us) pass around (?) the sun and hit our telescope, it wouldn't literally curve around the sun as is depicted in so many laymen's physics books? (I'm describing the Eddington experiment of course.)

In my opinion, the quotation is ambiguous.

If your mathematical model is curved, then it makes sense to say the physical spacetime is curved. What curvature means mathematically is the important point.

The key point here is that you don't have to think of a curved manifold as being embedded in a higher dimensional space.

 

[Ibix]

Summary:: Further details on what ST curvature really means

Here's what I read in a closed thread asking about the meaning of spacetime curvature:

So does that mean if we were able to move to some spot "way above the earth/sun/distant galaxy" where we could watch light and trace it from a distant galaxy (which is otherwise hidden from us) pass around (?) the sun and hit our telescope, it wouldn't literally curve around the sun as is depicted in so many laymen's physics books? (I'm describing the Eddington experiment of course.)

I'm not sure of the context of the quote you provided (link?), but I suspect it's talking about the "dent in a rubber sheet" model of gravity which is highly misleading. Spacetime is not a flat sheet with dents in it floating in some higher dimensional space. The curvature that we talk about in relativity (so called "intrinsic curvature") isn't the same thing as the curvature depicted by that diagram ("extrinsic curvature").

However, to answer your question, if you traced out a light path through space near the Sun you would find it curved. The curvature of spacetime is a measurable physical phenomenon with measurable effects. It's just not a dent in a rubber sheet.

 

[ergospherical]

It's probably better to say that curvature is a mathematical concept with physical consequences. So for example, the relative acceleration of two neighbouring geodesics $\gamma(s=0)$ and $\gamma(s=1)$ with deviation vector $\xi^a = \partial x^a / \partial s$ is\begin{align*}
\dfrac{D^2 \xi^a}{dt^2} &= u^{c} \nabla_c (u^b \nabla_b \xi^a) \\

&= u^{c} \nabla_c (\xi^b \nabla_b u^a) \\

&= (u^c\nabla_c \xi^b)(\nabla_b u^a ) + u^c \xi^b \nabla_c \nabla_b u^a \\

&= (\xi^c \nabla_c u^b)(\nabla_b u^a) + u^c \xi^b(\nabla_b \nabla_c u^a - R^{a}_{dbc} u^d) \\

&= (\xi^c \nabla_c u^b)(\nabla_b u^a) + \xi^b\nabla_b(u^c \nabla_c u^a) - (\xi^c\nabla_c u^b)(\nabla_b u^a) - u^c \xi^b u^d R^{a}_{dbc} \\

&= -u^c \xi^b u^d R^{a}_{dbc}
\end{align*}which is non-zero in the presence of curvature (non-vanishing $R^{a}_{bcd}$).

 

[Dale]

It's probably better to say that curvature is a mathematical concept with physical consequences.

I for one don’t think that is better. The idea that a mathematical concept has physical consequences is repulsive to me. I think Max Tegemark would agree with your statement, but few others.

 

[ergospherical]

Yeah maybe, it's a question better left to the philosophers. I didn't mean it literally though - more that curvature is a property of a manifold, and its presence results in observable effects such as the relative acceleration of nearby geodesics (and failure of initially parallel geodesics to remain parallel, etc.).

 

[Dale]

I didn't mean it literally though - more that curvature is a property of a manifold, and its presence results in observable effects such as the relative acceleration of nearby geodesics (and failure of initially parallel geodesics to remain parallel, etc.).

Yes, I like that much better.

 

[PeroK]

The idea that a mathematical concept has physical consequences is repulsive to me.

I wonder whether "repugnant" would be a better word in this context?

 

[vanhees71]

Yeah maybe, it's a question better left to the philosophers. I didn't mean it literally though - more that curvature is a property of a manifold, and its presence results in observable effects such as the relative acceleration of nearby geodesics (and failure of initially parallel geodesics to remain parallel, etc.).

Of course, if we have a mathematical model of some aspect of the physical world, it tells us the consequences for observation, and thus you can test these consequences of the mathematical model. In fact, when Einstein discovered possible extensions of special relativity to describe also the gravitational interaction, he was eager to find all kinds of possible phenomena being testable by observation.

The only known deviation from Newtonian gravitation was an unexplained tiny part of the perihelion shift of the planet Mercury, and even this was first rather thought to be a hint for another hitherto unknown planet, which however couldn't be found given the constraints of the observed deviation.

The other prediction Einstein came up with was very early was the deflection of light by the Sun. His first value of the deflection angle was however wrong by a factor of 2, because he used some predecessor theory of GR which was not yet right, giving only half of the deflection angle than the correct one. In a sense it was fortunate for Einstein that the solar-eclipse expedition of 1914 had to be cancelled, because then he'd been proven wrong (provided Freundlich et al had been successful to accurately measure the deflection). The correct prediction using the complete GR of 1915 was confirmed by Eddington et al using data from the eclipse expedition of 1919 making Einstein a celebrity over night.

Another predicted effect was the gravitational red shift of light, and the famous Einstein tower in Potsdam was built to measure it, but at this time they failed to measure the tiny effect of corresponding shifts of spectral lines of the Sun light.

 

[Carpe Physicum]

Here's the link, it's post 4: https://www.physicsforums.com/threads/what-does-curvature-of-spacetime-really-mean.196359/

I guess that agrees with what folks have posted here?? It probably gets into speculation to ask why it's NOT a curvature into a fourth spatial dimension.

 

[Dale]

Here's the link, it's post 4: https://www.physicsforums.com/threads/what-does-curvature-of-spacetime-really-mean.196359/

View attachment 288138
I guess that agrees with what folks have posted here?? It probably gets into speculation to ask why it's NOT a curvature into a fourth spatial dimension.

I have not heard of that book before, so I cannot vouch for its accuracy. I agree with the fact that it is not necessary to think of "some as yet undiscovered dimension", but disagree with the "mathematical curvature" bit.

The surface of the Earth is not just mathematically curved, it is physically curved. If you start at the north pole, walk straight south to the equator, turn 90 deg, walk straight east 1/4 of the globe, turn 90 deg, and walk straight north to the pole then you will find you are facing 90 deg off from where you started. Thus, the interior angles of this triangle sum to 270 deg, not 180 deg. That is how we can know that the surface of the Earth is physically curved, it does not follow the geometrical rules of flat surfaces. This physical curvature can be physically measured with physical protractors and other physical measuring devices.

Similarly the physical curvature of spacetime can be measured with protractors, accelerometers, rulers, clocks, etc. This is the physical geometry of the universe as actually measured by physical devices in the universe.

 

[PeroK]

Here's the link, it's post 4: https://www.physicsforums.com/threads/what-does-curvature-of-spacetime-really-mean.196359/

View attachment 288138
I guess that agrees with what folks have posted here?? It probably gets into speculation to ask why it's NOT a curvature into a fourth spatial dimension.

Spacetime is a curved 4D manifold. To some extent you have to understand the mathematics of manifolds to appreciate why curvature does not imply embedding in a higher dimension.

 

[Carpe Physicum]

By comparison, shooting a bullet at the Earth from a distant galaxy around a giant magnetic sun would also curve the bullets path. But that curving would be different from what we're talking about here because...?

 

[Dale]

By comparison, shooting a bullet at the Earth from a distant galaxy around a giant magnetic sun would also curve the bullets path. But that curving would be different from what we're talking about here because...?

Because that curvature can be measured with an accelerometer, meaning that the bullet's path through spacetime is curved in a physically measurable sense. In the curvature of spacetime we are talking about what happens to straight lines, which can be identified by the fact that an attached accelerometer reads 0 along the whole path.

 

[Carpe Physicum]

Hmm, that sounds like What's the difference between hot and cold? Well, hot is the one that melts ice cubes. The question is, WHY does hot melt ice cubes. Point being, what is it about the magnetic case that's different from the gravitational case? Maybe that's the big question - why does gravity seem to be categorically different from the other forces.

 

[Dale]

Hmm, that sounds like What's the difference between hot and cold? Well, hot is the one that melts ice cubes.

I would suggest that if you think that is what it sounds like then you haven't thought about it carefully. That is ok, most people have not. But don't dismiss the concept so readily.

Here is an insights article that explains the core elements of these geometrical concepts:
https://www.physicsforums.com/insights/understanding-general-relativity-view-gravity-earth/

Point being, what is it about the magnetic case that's different from the gravitational case? Maybe that's the big question - why does gravity seem to be categorically different from the other forces.

What is different is that magnetic forces produce an extrinsic curvature of a worldline while gravitation produces an intrinsic curvature of spacetime. Why they behave that way instead of some other way is a philosophical/religious question that we don't get into here.

 

[Carpe Physicum]

I would suggest that if you think that is what it sounds like then you haven't thought about it carefully. That is ok, most people have not. But don't dismiss the concept so readily.

Here is an insights article that explains the core elements of these geometrical concepts:
https://www.physicsforums.com/insights/understanding-general-relativity-view-gravity-earth/What is different is that magnetic forces produce an extrinsic curvature of a worldline while gravitation produces an intrinsic curvature of spacetime. Why they behave that way instead of some other way is a philosophical/religious question that we don't get into here.

Good references thanks. I think I get it, extrinsic meaning within the "framework" so to speak provided by s-t, vs gravity which affects the framework/s-t itself. As always I appreciate folks' patience with us curious laymen.

 

[PeterDonis]

By comparison, shooting a bullet at the Earth from a distant galaxy around a giant magnetic sun would also curve the bullets path.

Curvature of paths is something different from curvature of spacetime. Also, curvature of paths in spacetime is different from curvature of paths in space.

If you throw a rock on Earth, its path through space looks curved, but its path through spacetime is straight, because it's freely falling. The same applies to a light ray passing near the Sun; its path through space looks curved, but its path through spacetime is straight.

Curvature of spacetime, physically, is tidal gravity. That is, if you have two freely falling objects that, at some instant of time, are at rest relative to each other, they will not stay at rest relative to each other (whereas in the flat spacetime of special relativity, they would). So, for example, if I have two rocks that, at some instant of time, are at rest relative to each other and at rest relative to the Earth, high above the Earth, at different altitudes, they will not stay at rest relative to each other. From an Earth observer's point of view, the lower rock falls faster because it is closer to the Earth so the Earth's gravity pulls it down more strongly. But from a curved spacetime point of view, both rocks are following straight paths through spacetime, but spacetime itself is curved, so the two rocks' paths through spacetime, which started out parallel, don't stay parallel.

 

[martinbn]

Hmm, that sounds like What's the difference between hot and cold? Well, hot is the one that melts ice cubes. The question is, WHY does hot melt ice cubes. Point being, what is it about the magnetic case that's different from the gravitational case? Maybe that's the big question - why does gravity seem to be categorically different from the other forces.

Because gravity affects everything the same way. With other forces you can use different test objects to see what is due to the force and what to the spacetime geometry. With gravity you cannot do that. Thats why it is different and why it is connected to the geometry of spacetime.

 

[vanhees71]

Here's the link, it's post 4: https://www.physicsforums.com/threads/what-does-curvature-of-spacetime-really-mean.196359/

View attachment 288138
I guess that agrees with what folks have posted here?? It probably gets into speculation to ask why it's NOT a curvature into a fourth spatial dimension.

Well, yes. I don't think that this passage is very clear. Particularly the claim that cuvature were only "mathematical" is highly misleading. The curvature of spacetime is observable and has been observed. One example is the famous gravity-B experiment demonstrating "frame dragging" which I'd consider as the most direct proof of (a) the curvature of spacetime and (b) that spacetime is part of the dynamics of all the fields describing gravity, matter, and the other known interactions (strong and electroweak), i.e., it demonstrates that the interpretation of the field describing gravitation within Einstein's GR as the geometry of a pseudo-Riemannian 4D manifold is correct. The beauty of this is that together with the (Einsteinian) equivalence principle it's almost inevitable if you start with a massless spin-2 field to describe gravity. For this approach see "The Feynman Lectures on Gravitation".

The important point is that the curvature of a Riemannian (or pseudo-Riemannian as in GR) manifold needs not to be interpreted as embedded in some higher-dimensional affine (flat) space. Though in our intuition we know curvature from embedded two-dimensional "surfaces" in Euclidean affine space, there is no necessity to consider such a surface as embedded in that way, but you can describe all its geometrical properties only using the corresponding Riemannian manifold. This goes back to Gauss's famous work on "non-Euclidean geometries". Later this has been extended by Riemann, and for our physical purposes also the general characterization of differentiable manifolds, affine manifolds, Riemannian and pseudo-Riemannian manifolds, etc. in terms of symmetries is very important, but all these mathematical descriptions do not consider the manifold as embedded in some higher-dimensional flat affine space, which wouldn't have any physical meaning anyway. To our present knowledge there are 4 spacetime dimensions. There's no empircal evidence for any extradimensions which are sometimes used in model building (the oldest example is Kaluza-Klein theory).

 

[vanhees71]

Because that curvature can be measured with an accelerometer, meaning that the bullet's path through spacetime is curved in a physically measurable sense. In the curvature of spacetime we are talking about what happens to straight lines, which can be identified by the fact that an attached accelerometer reads 0 along the whole path.

I'd expect such a bullet to fly on a geodesic in spacetime, because it's freely falling. There are no other interactions than gravity acting on it after realeased. So it's not accelerated and an implemented accelerometer would read 0.

 

[Dale]

I'd expect such a bullet to fly on a geodesic in spacetime, because it's freely falling. There are no other interactions than gravity acting on it after realeased. So it's not accelerated and an implemented accelerometer would read 0.

I assumed from the wording that they intended the bullet to be subject to a real magnetic force, not free falling. E.g. a steel jacketed bullet or an iron slug

 

[vanhees71]

Sure, then you have a real force, and the spacetime trajectory is no longer geodesic.

 

[cianfa72]

Similarly the physical curvature of spacetime can be measured with protractors, accelerometers, rulers, clocks, etc. This is the physical geometry of the universe as actually measured by physical devices in the universe.

Can you give us an example of measuring the angles of a geodesic triangle 'built' in spacetime? Thanks.

 

[DrGreg]

Can you give an example of measuring the angles of a geodesic triangle 'built' in spacetime? Thanks.

See https://www.physicsforums.com/threa...ciple-suggests-curvature.1002557/post-6487001 for one example where all sides of the triangle are timelike.

 

[cianfa72]

See https://www.physicsforums.com/threa...ciple-suggests-curvature.1002557/post-6487001 for one example where all sides of the triangle are timelike.

Reading that post, my understanding is that the relative rapidity $\psi$ between Carol and Alice is basically the same as measured at the common event (Carol catches up with Alice) in both locally inertial frames (namely the local inertial frame in which Alice is at rest and the other one in which Carol is at rest).

The rapidity has got a geometrical significance as the 'angle' between the 4-velocity of Carol and Alice at the common event (4-velocity at each point along a worldline is really the worldline's tangent vector -- an element of the tangent space at the given event).

Does it make sense ? Thanks.

 

[DrGreg]

Reading that post, my understanding is that the relative rapidity $\psi$ between Carol and Alice is basically the same as measured at the common event (Carol catches up with Alice) in both locally inertial frames (namely the local inertial frame in which Alice is at rest and the other one in which Carol is at rest).

The rapidity has got a geometrical significance as the 'angle' between the 4-velocity of Carol and Alice at the common event (4-velocity at each point along a worldline is really the worldline's tangent vector -- an element of the tangent space at the given event).

Does it make sense ? Thanks.

Yes. "Angle between worldlines" means angle between tangent vectors (=4-velocities when the lines are timelike) at the event of intersection. In curved spacetime it makes sense to compare velocities (independently of coordinate system or other imposed convention) only where worldlines intersect.