General Relativity

[Perpetual Student]

The equations would be correct only in that one coordinate system. So they wouldn't correctly predict experimental results unless that coordinate system happened to coincide precisely with the one relevant to the experiment.

So the answer to your question is "yes" - either that, or you'd be smart enough to realize that the equations simply don't apply except in one special frame.

If we choose a specific coordinate system and then write out the components of Einstein's equations in those coordinates -- as you suggest -- can you give an example of how and why experimental results might go astray in a different coordinate system?

 

[Perpetual Student]

...
Actually deriving the EFEs is more useful than just giving the historical info.

Another good source of GR education is Leonard Susskind's YouTube lecures on Modern Physics concentrating on General Relativity (recorded September 22, 2008 at Stanford University).

Thanks, Reality Check, for that wonderful suggestion. I have plodded my way half way through lecture 4 at this point. The lectures are amazingly clear and entertaining. What a fabulous tool these lectures on YouTube are! I can pause the lecture to look up and better understand some fact and then continue from that same point.
For example, Susskind makes no effort to explain the difference between covariance and contravariance, choosing instead to focus on the manipulation of the tensor indices involved. So, I paused the play, did a little investigation to better understand the geometric significance of those terms and went back to where i was in the lecture. What a contrast to the effort it would have taken to do that when I was in a classroom!

 

[Vorpal]

It's certainly true that the equation as written indicates that GR is generally covariant, because the equation above is true and looks exactly the same in all coordinate systems. For that to be possible requires tensors, or something like them.

I don't disagree with this, but I would add that it also requires there to not be any fixed fields, regardless of whether or not they're expressible as tensors. Which the EFE certainly doesn't have.

I don't understand the distinction you're making here. It sounds like you're saying (to translate to my language) that GR allows gravity waves, while Nordstrom's theory doesn't. Said another way, that specifying the stress-energy tensor in a region doesn't uniquely specify the metric.

In Nordström's theory, vacuum has vanishing Weyl curvature but only fixes the Ricci scalar to zero. It's still possible to have gravitational waves that consist of traceless Ricci curvature, unlike in GTR where they're pure Weyl. They're scalar rather than spin-2 as well.

... If you were to specify the appropriate boundary conditions on those degrees of freedom - meaning in the case of GR that you specify the metric and some derivatives on a Cauchy surface - you do fully specify the solution, including the geometry. I don't see how that's different from (say) including a scalar field in your theory, and then having to specify its initial conditions.

But then under Nordström's theory, if you drop the conformally flat requirement altogether, then some of those degrees of freedom don't have anything to do with gravity or local matter distribution. I don't see how you'd get a unique solution up to diffeomorphism from the same kind of initial conditions--not without supplementing the theory with something else, rather.

To me, any theory specified by an action that's an integral of a scalar density over spacetime is generally covariant. Do you disagree with that?

Not per se, no: I'll just say that what Einstein called "general covariance" would be, in more modern terms, (I think) something like "general covariance + background independence."

 

[sol invictus]

I don't disagree with this, but I would add that it also requires there to not be any fixed fields, regardless of whether or not they're expressible as tensors. Which the EFE certainly doesn't have.

I'm not sure about that. Even in a generally covariant theory like GR, you can always expand around a fixed background metric - and if you kept every term in the expansion, that theory would be equivalent to the original.

In Nordström's theory, vacuum has vanishing Weyl curvature but only fixes the Ricci scalar to zero. It's still possible to have gravitational waves that consist of traceless Ricci curvature, unlike in GTR where they're pure Weyl. They're scalar rather than spin-2 as well.

I see, OK.

But then under Nordström's theory, if you drop the conformally flat requirement altogether, then some of those degrees of freedom don't have anything to do with gravity or local matter distribution. I don't see how you'd get a unique solution up to diffeomorphism from the same kind of initial conditions--not without supplementing the theory with something else, rather.

Sure - if you drop the conformally flat requirement. But if you don't, what's the relevant difference?

Not per se, no: I'll just say that what Einstein called "general covariance" would be, in more modern terms, (I think) something like "general covariance + background independence."

Perhaps so - but the utility of background independence has always been a little vague to me. Roughly it's supposed to mean that the solutions determine the background, rather than the other way around. But in most theories only some field configurations are solutions. So what if your theory determines one background uniquely? Is that bad?

Moreover, generally you can expand about any background, and - if you keep enough of the theory - find other backgrounds. So how do you distinguish between "background" and "fluctuations" in general?

 

[sol invictus]

If we choose a specific coordinate system and then write out the components of Einstein's equations in those coordinates -- as you suggest -- can you give an example of how and why experimental results might go astray in a different coordinate system?

Let's take an easier example: Newton II (F=ma) in two dimensions. In Cartesian coordinates in 2D, F=(F_x, F_y) and the acceleration vector a=(x'',y'') (' means derivative with respect to time).

Now, suppose we change from (x,y) to polar coordinates (r,θ). If we naively tried to use the same equation, we might write F=(F_x, F_y)=ma=m(r'',θ''), or maybe F=(F_r, F_θ)=m(r'',θ''). But r and θ don't have the same dimensions, so we certainly can't set the thing on the right hand side equal to F. If you did that, you'd get nonsense.

To find the correct equation, you have to transform Newton's equation correctly into the new coordinates, and it will have a different expression in terms of those coordinates than it did in the old ones. Another example would be to choose a rotating system of coordinates. Those can be Cartesian if you like, but Newton's equation will be different (there will be "fictitious forces").

 

[Perpetual Student]

Let's take an easier example: Newton II (F=ma) in two dimensions. In Cartesian coordinates in 2D, F=(F_x, F_y) and the acceleration vector a=(x'',y'') (' means derivative with respect to time).

Now, suppose we change from (x,y) to polar coordinates (r,θ). If we naively tried to use the same equation, we might write F=(F_x, F_y)=ma=m(r'',θ''), or maybe F=(F_r, F_θ)=m(r'',θ''). But r and θ don't have the same dimensions, so we certainly can't set the thing on the right hand side equal to F. If you did that, you'd get nonsense.

To find the correct equation, you have to transform Newton's equation correctly into the new coordinates, and it will have a different expression in terms of those coordinates than it did in the old ones. Another example would be to choose a rotating system of coordinates. Those can be Cartesian if you like, but Newton's equation will be different (there will be "fictitious forces").

How obvious! Thanks.

 

[sol invictus]

How obvious! Thanks.

Everything's obvious in retrospect .

It seems that a big part of Einstein's contribution in this area was to figure out how to write equations using a notation (tensors) that smoothly and correctly handles such coordinate transformations.

Once you know how to do that, it's very easy to see which equations make sense and which don't (at least assuming there's no preferred choice of coordinates), and which are Lorentz invariant (Lorentz transformations are a special sub-class of coordinate transforms). Einstein's theory of GR follows almost uniquely from that - it's essentially the only thing you can write for the metric that follows from a scalar action (so it satisfies the above) and has a dominant effect at long distances.

 

[Perpetual Student]

I have a trivial off topic question that I hope can be answered by a simple statement:
When using c = 1, in any equation involving spacetime, is it assumed the spacial units are approximately 3x10⁸ meters? In other words, would x = 3 mean x = 9x10⁸ meters?

 

[sol invictus]

I have a trivial off topic question that I hope can be answered by a simple statement:
When using c = 1, in any equation involving spacetime, is it assumed the spacial units are approximately 3x10⁸ meters? In other words, would x = 3 mean x = 9x10⁸ meters?

No. "Setting c=1" simply means you measure all speeds in units of 3x10⁸m/s. So of s=1/2 is the speed of something, that thing is moving at 3x10⁸/2 m/s.

But it doesn't say anything about lengths (although it does mean that lengths and time intervals can have the same units). x=3 (assuming x is a position) is simply meaningless, unless some additional convention is in play.

 

[Perpetual Student]

No. "Setting c=1" simply means you measure all speeds in units of 3x10⁸m/s. So of s=1/2 is the speed of something, that thing is moving at 3x10⁸/2 m/s.

But it doesn't say anything about lengths (although it does mean that lengths and time intervals can have the same units). x=3 (assuming x is a position) is simply meaningless, unless some additional convention is in play.

Sorry, x should be replaced by Δx in my question. If a unit of x is not 3x10⁸ meters, what does it mean for lengths and time intervals to have the same units where c = 1?

 

[sol invictus]

Sorry, x should be replaced by Δx in my question. If a unit of x is not 3x10⁸ meters, what does it mean for lengths and time intervals to have the same units where c = 1?

Just that you can describe distance by the time it would take light to travel that distance (that's the idea behind lightyears, lightseconds, etc.).

 

[Perpetual Student]

So when I consider (proper time) ΔΤ² = Δt² - Δx² - Δy² - Δz², with c = 1, if t is in seconds, are the units of x,y and z necessarily 3x10⁸ meters?

 

[DazzaD]

So when I consider (proper time) ΔΤ² = Δt² - Δx² - Δy² - Δz², with c = 1, if t is in seconds, are the units of x,y and z necessarily 3x10⁸ meters?

If you choose to set c=1 and G=1 then that still allows you to deal with the other units as you wish....

(except when you want to come out of the "natural units" session perhaps you may want to insert factors of c and c^2 into your formulae as appropriate).

http://en.wikipedia.org/wiki/Natural_units

 

[H'ethetheth]

In Nordström's theory, vacuum has vanishing Weyl curvature but only fixes the Ricci scalar to zero. It's still possible to have gravitational waves that consist of traceless Ricci curvature, unlike in GTR where they're pure Weyl. They're scalar rather than spin-2 as well.

So I wanted to see what this discussion had gotten up to, and then I saw this sentence.

Funny how a sentence can become completely unintelligible when you don't know what any of the terms mean.

 

[sol invictus]

So when I consider (proper time) ΔΤ² = Δt² - Δx² - Δy² - Δz², with c = 1, if t is in seconds, are the units of x,y and z necessarily 3x10⁸ meters?

If you've decided to measure t in seconds, then yes, with c=1 you'll be using lightseconds to measure distance.

It's perhaps easier just to think of x and t as having the same units, and speeds/velocities as being dimensionless. When you get to the end of the calculation, you multiply by the necessary power of c to restore standard units.

 

[Perpetual Student]

Another question for anyone willing to help (thanks in advance):

Is there any reasonably intuitive geometric interpretation of the Christoffel symbol, Γ^δ_αβ , by which the metric is multiplied and added to the derivative of the tensor, when obtaining a covariant derivative of a tensor? Since the covariant derivative is also a tensor, is it a function of the geometry of the coordinates of the original tensor? Does that question make any sense?

 

[W.D.Clinger]

Another question for anyone willing to help (thanks in advance):

sol invictus or Vorpal or many others could do a better job of answering your questions, but I'll try.

Is there any reasonably intuitive geometric interpretation of the Christoffel symbol, Γ^δ_αβ , by which the metric is multiplied and added to the derivative of the tensor, when obtaining a covariant derivative of a tensor?

Yes. The caption for Misner/Thorne/Wheeler Figure 8.3, on page 212, starts out "The why of connection coefficients, schematically portrayed." That figure shows a great-circle geodesic for an aviator flying from Peking to Vancouver, and relates the Christoffel symbols to the geodesic equation in the latitude/longitude basis.

If you don't have a copy of MTW handy, the most similar description I could find online is Example 8 (and the following text) at
http://www.lightandmatter.com/html_books/genrel/ch05/ch05.html

Since the covariant derivative is also a tensor, is it a function of the geometry of the coordinates of the original tensor? Does that question make any sense?

The covariant derivative of a tensor is a particular function of the original tensor, and that particular function is determined by the geometry.

The coordinate-dependent components of the covariant derivative of a tensor are functions of the coordinates of the original tensor, and those particular functions are determined by the geometry and by the coordinate system.

By the way, my Google search turned up what appears to be a nice set of lecture notes by Matthias Blau. Its first chapter parallels a lot of the discussion we've seen scattered over several threads, and it might help to see a more systematic and less noisy treatment of that material.

 

[Perpetual Student]

While the derivative of any function (even a tensor) should be reasonably intuitive, the covariant derivative is not so easy to fathom. The extra Γ term (with its three partial derivative terms) is a too complex (for me) to intuit. Section 5.6.2 of the material by Benjamin Crowell that you linked seems to be helpful. This is challenging stuff! Thanks.

 

[Perpetual Student]

I have (what I hope is) a simple yes/no question. I have seen it described that the Einstein field equations, which are represented in his tensor equation, as being ten partial differential equations. So, is the number of equations ten because the four dimensional metric tensor is symmetric? Can I assume it follows that the stress-energy tensor is also symmetric?

 

[W.D.Clinger]

I have (what I hope is) a simple yes/no question. I have seen it described that the Einstein field equations, which are represented in his tensor equation, as being ten partial differential equations. So, is the number of equations ten because the four dimensional metric tensor is symmetric?

Yes.

Can I assume it follows that the stress-energy tensor is also symmetric?

Yes.