General Relativity

[Farsight]

Perpetual Student: don't think that general relativity is the problem here. Instead the problem is what some people say about general relativity. They're often mathematicians rather than physicists and cosmologists, and broadly speaking, they take the phrase "all coordinate systems are equally valid" and put a spin on it that delivers a different meaning altogether.

To appreciate this, take a very very simple case where we have no gravity at all, where we have 20 static objects in our universe, and we are not concerned about time, merely distance. You can adopt a coordinate system with an origin of your choice, with X Y and Z directions of your choice, and with units of your choice. I can make totally different choices, and our coordinate systems are equally valid.

Let's now change the scenario such that one object is the Earth, and along with another 8 objects called the planets, it orbits a tenth object called the Sun. The other 10 objects are very distant stars. You are free to use any coordinate system of your choice, and it's just as valid as mine. But just because it's a valid coordinate system, it doesn't mean the the Sun goes round the Earth. In similar vein you are free to use a local map of your town as the basis of a coordinate system, but that doesn't mean the Earth doesn't spin.

I agree with Zig's sentiment re the CMB rest frame. This allows you to gauge your motion through the universe. It doesn't provide an "absolute reference frame" because when you're in a black box you can't see it, and so you still can't tell whether you're moving. But we study the universe, we look outside that black box, and that CMB rest frame tells us important information about motion within our universe. And we note that Einstein did not assert that the universe revolves around Phobos just because "all coordinate systems are equally valid".

 

[DazzaD]

Minor point but possibly important..

Generally speaking cosmologists ARE mathematicians.

In most universities the cosmology "department" is normally located within the mathematics department.

There is a very sensible reason for this.

We need to distinguish between "the laws of physics appear to be the same in all frames of reference" from "what I am calculating here doesnt tally with my everyday sense of appreciation of the world around me".

There is no requirement for the second statement to happen.

 

[alexi_drago]

This is very likely a really stupid question but if the earth was the centre and everything is rotating around it then does that mean that everything (roughly) at neptune and beyond is travelling faster than c?

 

[sol invictus]

This is very likely a really stupid question but if the earth was the centre and everything is rotating around it then does that mean that everything (roughly) at neptune and beyond is travelling faster than c?

The coordinate speed of objects at large enough distance will certainly exceed c. But that doesn't pose any problem - you can transform to that frame, and it makes correct predictions. Rotating frames are not inertial, and it's only in inertial frames that c is the speed limit.

 

[Perpetual Student]

The coordinate speed of objects at large enough distance will certainly exceed c. But that doesn't pose any problem - you can transform to that frame, and it makes correct predictions. Rotating frames are not inertial, and it's only in inertial frames that c is the speed limit.

So, with Princeton NJ as my stationary place, how would I calculate the kinetic energy of some very distant object traveling in a huge circle around me at many times c? If I use:

[latex] E = \dfrac{mc^2}{\sqrt{1-\dfrac{v^2}{c^2}}}[/latex]

I seem to be in a lot of trouble, with an imaginary number in the denominator.

 

[Farsight]

Don't bother. It's pseudoscience garbage. And there's a lot of it about.

 

[Almo]

If you believe in commonsense and logic, you will immediately throw away your Relativity.

Incorrect. General Relativity has been shown to explain things we can't explain otherwise.

 

[Mashuna]

[qimg]http://t0.gstatic.com/images?q=tbn:ANd9GcQRkSoOrCyjOUbVaRk6ofcld1ooqA6UfH_75ymkVl2efge0JavUjVHuSyJ5sQ[/qimg]

This thread should answer your question.

 

[sol invictus]

So, with Princeton NJ as my stationary place, how would I calculate the kinetic energy of some very distant object traveling in a huge circle around me at many times c? If I use:

[latex] E = \dfrac{mc^2}{\sqrt{1-\dfrac{v^2}{c^2}}}[/latex]

I seem to be in a lot of trouble, with an imaginary number in the denominator.

You have to use the correct expression for the coordinate. Here's a simpler example of your "trouble" - spherical coordinates. An object near the origin can have an angular coordinate theta that changes arbitrarily rapidly with time. If you tried to use the time derivative of theta for "v", you'd get in just as much trouble (more actually, since it has the wrong units).

 

[Perpetual Student]

You have to use the correct expression for the coordinate.
/.../

Yes...? And that is...?

 

[Perpetual Student]

If you relax what you mean by "preferred", then GR actually does provide something that may suit your purposes, and you actually alluded to it. And that's the co-moving reference frame of the universe, which we can observe by watching the CMB. This reference frame isn't preferred in the sense that the laws of physics are any different in this frame from any other frame. But it is still a unique reference frame in terms of a number of observable details of the universe, such as the CMB being essentially isotropic. In more tangible terms than the CMB, the co-moving reference frame is the reference frame in which mater is (on average) stationary within the universe. Local measurements can't distinguish this reference frame from any other reference frame (so again, the laws of physics are no different), but we're not confined to local measurements, and large-scale measurements (like the CMB) are sensitive to it. So if you want a reference frame on which you can hang a sense of place without everything becoming seemingly completely arbitrary, well, the co-moving reference frame can serve that purpose perfectly well.

I don't think I have been saying anything very different from the above. However, going one small step further to say that I would like to view the universe as it actually is -- with the CMB stationary -- seems to provoke accusations of GR anathema.
I have accepted that GR allows us to view and examine the universe or any part of it using any frame of reference -- and all these analyses would be perfectly accurate and valid.
However, when I ask myself, "what is the universe?" describing it as some vast complexity gyrating around Princeton, NJ does not seem to be a very good answer. The universe is not that! It is the immense structure that we all know and love with the CMB stationary! With that thought, I can sleep better.

 

[sol invictus]

Yes...? And that is...?

It depends on the coordinates. For spherical coordinates it would be something like [latex]$v=\partial x/\partial t \rightarrow r \partial \theta/\partial t$[/latex]. Even though [latex]$ \partial \theta/\partial t$[/latex] can be arbitrarily large near the origin r=0, in all physical solutions the factor of r renders the velocity finite (and less than c).

Similarly, in any other coordinate system there will be some expression for the energy that depends on the analog of v. The equations of motion in those coordinates will guarantee that the energy can never become infinite or imaginary, just as they guarantee that the derivative of the Cartesian coordinate position of a particle with respect to time never exceeds c.

All of that holds true precisely because the laws of physics are invariant (or covariant, really) under coordinate transformations. If you get shot with an infinite energy bullet, it hurts a lot. Getting hurt a lot is a physical outcome. If infinite energy bullets (and the corresponding outcome) are forbidden in Cartesian coordinates, then they are forbidden in all other coordinates as well.

 

[Perpetual Student]

It depends on the coordinates. For spherical coordinates it would be something like [latex]$v=\partial x/\partial t \rightarrow r \partial \theta/\partial t$[/latex]. Even though [latex]$ \partial \theta/\partial t$[/latex] can be arbitrarily large near the origin r=0, in all physical solutions the factor of r renders the velocity finite (and less than c).

Similarly, in any other coordinate system there will be some expression for the energy that depends on the analog of v. The equations of motion in those coordinates will guarantee that the energy can never become infinite or imaginary, just as they guarantee that the derivative of the Cartesian coordinate position of a particle with respect to time never exceeds c.

All of that holds true precisely because the laws of physics are invariant (or covariant, really) under coordinate transformations. If you get shot with an infinite energy bullet, it hurts a lot. Getting hurt a lot is a physical outcome. If infinite energy bullets (and the corresponding outcome) are forbidden in Cartesian coordinates, then they are forbidden in all other coordinates as well.

I guess I'm missing something again. If r is something like a billion light years, how can we avoid
[latex] r\partial \theta/\partial t$ \rightarrow \partial x/\partial t = v[/latex]
from exceeding c? What does using spherical coordinates have to do with anything, since v = rω and a sufficiently large r will result in v exceeding c for any ω?

 

[Astrodude]

So, with Princeton NJ as my stationary place, how would I calculate the kinetic energy of some very distant object traveling in a huge circle around me at many times c? If I use:

[latex] E = \dfrac{mc^2}{\sqrt{1-\dfrac{v^2}{c^2}}}[/latex]

I seem to be in a lot of trouble, with an imaginary number in the denominator.

You'd have to calculate the velocity relative to a truly inertial reference frame, which doesn't exist in general relativity, if you were to use that formula. That formula is an approximation for real space as it only applies to Euclidian metrics. Princeton, NJ is inert relative to the Earth(hopefully), but it is not inert relative to an observer somewhere on another planet like Mars. General relativity avoids this by dealing with all frames of reference with a tensor to describe all motion in terms of curved spacetime. Accurate calculation of the kinetic energy would require calculating the geodesic deviation of your frame with respect to the object which should eventually give:

[latex] E_k = mc^2({\sqrt{\dfrac{g_t_t}{g_t_t+g_s_sv^2}}-1})[/latex]

With g being the metric tensor.

 

[sol invictus]

I guess I'm missing something again. If r is something like a billion light years, how can we avoid
[latex] r\partial \theta/\partial t$ \rightarrow \partial x/\partial t = v[/latex]
from exceeding c?

The same thing that forbids it from happening in Cartesian coordinates - the equations of motion in those coordinates that follow from the underlying laws of physics.

What does using spherical coordinates have to do with anything, since v = rω and a sufficiently large r will result in v exceeding c for any ω?

It's just an example that shows why the question is naive. "v" has to be defined with the coordinates you're using, and the expression for the energy may also take a different form in different coordinates.

The point is, if the laws of physics forbid physical quantities from being infinite or imaginary in Cartesian coordinates, they forbid them from being infinite or imaginary in every other coordinate system too. So there can't possibly be anything to worry about.

 

[Astrodude]

OK, similarly, from the perspective of pure mathematics, there can be no preferred solution to a quadratic -- all solutions are equally valid. Consider that all solutions may not be meaningful for some real situation that is being modeled but by choosing a preferred solution to a quadratic we are not rejecting quadratic equations. Can we not treat GR in the same way?

General relativity usually does this. The invariant differential metric of the space time is usually given in quadratic form as:
[latex]ds^2={g^{\mu \nu}dx_{\mu}dx_\nu=ds'^2[/latex]

However, the metric is ds and not ds^2 and for calculation purposes only the positive square root is used, a line of thinking similar to your own on quadratics. So physicists have definitely heard your argument about general relativity.

Personally, I agree with Mendel Sachs and George Raetz and suspect that this disregard is scientifically incorrect and consider the quaternion reformulation of general relativity to be the most accurate description of gravitation and spacetime. I think the main reasons people reject it are because 'sol_invictus' line of thinking about imaginary quantities leads to a false interpretation of the formalism, and that the Mach principle is required to give a quaternion formalism any meaning.

 

[Ziggurat]

I don't think I have been saying anything very different from the above. However, going one small step further to say that I would like to view the universe as it actually is -- with the CMB stationary -- seems to provoke accusations of GR anathema.

That's because "as it actually is" doesn't mean anything in this context. The co-moving reference frame is very convenient, and it suits our intuition, but the universe isn't any less real in any other coordinate system. And as long as you handle the math correctly, all these different coordinate systems will make exactly the same predictions. So how can one reference frame be any less real than any other reference frame if they all describe reality equally well? We can prefer one over another for a whole host of reasons (a particular calculation becomes easier, the equations of motion look simpler, it matches our intuition better, or whatever), but these are our own human preferences. The universe cares not one bit about any of them.

 

[Perpetual Student]

That's because "as it actually is" doesn't mean anything in this context. The co-moving reference frame is very convenient, and it suits our intuition, but the universe isn't any less real in any other coordinate system. And as long as you handle the math correctly, all these different coordinate systems will make exactly the same predictions. So how can one reference frame be any less real than any other reference frame if they all describe reality equally well? We can prefer one over another for a whole host of reasons (a particular calculation becomes easier, the equations of motion look simpler, it matches our intuition better, or whatever), but these are our own human preferences. The universe cares not one bit about any of them.

We have been over this many times in this thread and some others, but I'll state the case on more time in an attempt to achieve some clarity.
Let's say the solar system is the whole universe to simplify the model a little. Now, GR will allow us to examine and describe this universe from the perspective of any frame we can imagine: Phobos, Titan, Princeton, or any molecule on or in any object we might choose. The universe does not care one iota; it is the same universe with the same physics every time. Some of these frames are very convenient for some particular purpose.
However, for me it is very compelling that the mathematics is dizzyingly complex for the whole universe when we choose any of these frames but simplifies strikingly when we have the sun in the "middle" with the planets orbiting with the moons orbiting around the planets, etc. That tells me that GR is not telling the whole story. The heliocentric essence of the solar system is "missed" by GR. There simply must be more to it all than GR is capable of telling us! That latter comment gets a rousing Bronx cheer from physicists because they see GR as a complete and final description of the universe. I remain skeptical -- and, yes. it is only a mere layman's skepticism and of little consequence.

 

[Perpetual Student]

You'd have to calculate the velocity relative to a truly inertial reference frame, which doesn't exist in general relativity, if you were to use that formula. That formula is an approximation for real space as it only applies to Euclidian metrics. Princeton, NJ is inert relative to the Earth(hopefully), but it is not inert relative to an observer somewhere on another planet like Mars. General relativity avoids this by dealing with all frames of reference with a tensor to describe all motion in terms of curved spacetime. Accurate calculation of the kinetic energy would require calculating the geodesic deviation of your frame with respect to the object which should eventually give:

[latex] E_k = mc^2({\sqrt{\dfrac{g_t_t}{g_t_t+g_s_sv^2}}-1})[/latex]

With g being the metric tensor.

Thanks for trying. As sol invictus and others have said, it would be easier to discuss GR if I had a handle on tensor calculus. Sadly, my MS in mathematics (some 45 years ago -- yikes!) included only a cursory introduction to tensors -- not much beyond defining them -- so as much as I have tried with this 72 year old brain, I have not made much headway. If anyone can recommend a book like "tensor calculus for dummies" I might give it a try again.

 

[Vorpal]

However, for me it is very compelling that the mathematics is dizzyingly complex for the whole universe when we choose any of these frames but simplifies strikingly when we have the sun in the "middle" with the planets orbiting with the moons orbiting around the planets, etc. That tells me that GR is not telling the whole story.

Again, YMMV, but I have the completely opposite reactions to those situations.

Let's step back from the universe at large and do something way more basic: elementary geometry. Let's say you have a piece of paper with some geometric figures drawn on it. Lines, circles, maybe other conic sections... whatever. You're trying to figure out the answers to some questions about them. Knowing that analytic geometry is good for something, you lay down some Cartesian coordinates and go to town.

Have conic sections? Some Cartesian axes make them a bit complicated, while others very simple (depending on how they align on the major/minor axes). Have a circle? Maybe polar would work better. Etc.

And yet all these coordinates you can lay down are not the thing you're investigating. Coordinates are not geometry. They're just a book-keeping device. Those figures on that paper are not changed one bit by your choice of coordinates. None of their properties change. All the intrinsic questions you can ask about them have the same answers.

I really can't imagine the the universe to be different. So to add to how sol invictus answered your previous question of "why?", your view seem to me horrifyingly anthropocentric. It's really is like saying English is the universe's language because you personally are more conversant in it.

Or in the above analogy, are you going to tell me there the one true coordinate system for those figures on that paper?