Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm writing a paper for a Philosophy of Science course about GR/SR and I'm wondering if I can (1) characterize the curvature of spacetime as invariant and (2) argue that this is what Einstein referred to in 1920 when he said "space without ether is unthinkable."

I take (1) from Gauss' proof that the curvature of 2-surfaces has one invariant which seems to be an intrinsic quality of the space (i.e. non-reference-frame-dependent). Shown by:

$K=\frac{(\nabla_{2}\nabla_{1}-\nabla_{1}\nabla_{2})e_{1},e_{2}}{det(g)}$ where $\nabla_{i}=\nabla_{e}$ is the covariant derivative and $g$ is the metric tensor.

And (2) from a rather terse paper (which I don't fully understand) found here.

I would just like to know if I'm completely off-base as even though I have the math background, my physics knowledge is spotty at best.

share|improve this question
If you have a star in some region then near that star you have some curvature but far away you have zero curvature. Also from the mathematical point of view when you say invariant you have to specify under which transformation –  Barefeg Feb 27 '13 at 20:05
That Gaussian curvature only applies to 2-surfaces, so it doesn't seem fit for a paper on GR. The curvature invariants in GR are built out of the Riemann tensor. Depending on your background, you can maybe work with this. –  Vibert Feb 27 '13 at 21:01
I've read some sections of that paper, but didn't really understand much of it. At best, it seems that we're talking about some nomenclature here - even without matter particles, Einstein seems to argue that spacetime still has (geometric) properties. Curvature invariants do measure this (you can have a black hole at the origin and measure a non-zero Ricci scalar far away), but it's a tad vague at to me. –  Vibert Feb 27 '13 at 21:07
Invariant under what? –  alexarvanitakis Feb 27 '13 at 22:56
Any scalar quantity is invariant ("scalar" = contraction of tensors, not components of tensors) with respect to coordinate changes. But why do you seek a scalar? The metric as a whole is invariant (some people use the phrase "transforms like a tensor") and is found everywhere in space. That is, Einstein could very well be saying a rank 2 tensor field exists everywhere in space, even in the vacuum between the stars. –  Chris White Feb 27 '13 at 23:15

2 Answers 2

I'm wondering if I can (1) characterize the curvature of spacetime as invariant

Spacetime is attributed a curvature tensor field. You plug a point into the field, say (t,x,y,z), and it returns the value of the Riemann curvature tensor at that point. From the Riemann tensor, you can construct a number of scalar invariants: the Ricci scalar $R=g^{\mu \nu} R^\sigma_{~\mu \sigma \nu}$ (this is the most frequently used curvature invariant in GR), the Kretschmann scalar $K=R_{\sigma \mu \lambda \nu} R^{\sigma \mu \lambda \nu}$, and a few others like $g^{\lambda \mu} g^{\gamma \nu} R^\sigma_{~\mu \sigma \nu}R^\sigma_{~\lambda \sigma \gamma}$. Each of these is invariant in the sense that they do not depend on your choice of coordinates. That does not mean that they have the same value at every point.

and (2) argue that this is what Einstein referred to in 1920 when he said "space without ether is unthinkable."

That seems like untestable nonsense to me. It looks as if he didn't like the fact that spacetime without matter didn't make much sense in the context of GR, so he wanted to invent some imaginary "stuff" that would give spacetime meaning even in the absence of matter fields. I don't think there are any modern theories concerning "ether" that are taken seriously by the physics community.

share|improve this answer
Neither Einstein (I'm certain) nor the author (I think) of that paper had in mind any physical, fluid substance or light-propagating medium when using the word "ether." The word has many meanings, and here could be taken to be closely akin to what today is known as a "field" in physics - a quantity or set of quantities for every point in space. –  Chris White Feb 27 '13 at 23:12
That makes more sense. I tend to shy away from anything that mentions "ether" because 9/10 times it ends up being crackpot nonsense. –  elfmotat Feb 27 '13 at 23:16
the author has left us, but I think the simplest answer is that ether was supposed to be a universal inertial reference frame against which everything is moving, and that is the ether disproved experimentally. Any medium which respects Lorenz transformations can be thought of as "ether" in a "metaphysical" way, a medium/framework against which motion can be gauged. the QFT vacuum for example, populated by virtual pairs and respecting Lorenz invariance. It is not a univeral inertial frame. –  anna v Mar 1 '13 at 5:14

I have no idea what the spec is for your assignment but if you want to talk about physics you should think about referencing papers written by the physicists to support your main arguments if you want it to have more weight in my view. That paper you linked seemed a bit wishy-washy at first glance (though I should admit I only skimmed through) I also do not know about the quote or the Gauss' equation you have. I will openly admit that this topic is not something I know very much about (yet) before I continue but hopefully I can still help.

I believe that Einstein had actually related the distribution of mass to the curvature of space time in his field equations as shown in this equation:


$R_{\mu\upsilon}$ as the Ricci curvature tensor. $R$ as the local curvature. $T_{\mu\upsilon}$ as the matter tensor specifying the distribution of 4-momentum

Schwarzschild found the following solutions to this equation for outside a spherical mass in 1916:

$$e(r)=1+\frac{2\Phi}{c^2}$$ $$f(r)=\left(1+\frac{2\Phi}{c^2}\right)^{-1}$$

With $\Phi(r)=-\frac{Gm}{r}$

These solutions have no time dependency.

If I have understood things correctly here, then this should help support the argument you are trying to make.

share|improve this answer

protected by Qmechanic Mar 3 '13 at 18:55

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?