Spacetime and limit c->infinity

[JustinLevy]

If we start with the Lorentz transformation
$$\begin{align*} ct' &= \gamma (ct - \beta x) \\ x' &= \gamma (x - \beta ct) \\ y' &= y \\ z' &= z \end{align*}$$

with the usual $$\beta = v/c, \gamma = 1/\sqrt{1-\beta^2}$$

and take the limit $$c \rightarrow \infty$$, then we get:
$$\begin{align*} t' &= t \\ x' &= x - v t \\ y' &= y \\ z' &= z \end{align*}$$

ie. we get the Galilean transformations.

But what of spacetime? Can we still discuss a spacetime manifold? If I look at the line element in an inertial frame:
$$ds^2 = g_{ab} dx^a dx^b = c^2 dt^2 - dx^2 - dy^2 - dz^2$$

taking $$c \rightarrow \infty$$ seems to render it as nonsense (or at least I am unsure how to interpret it).

Is there a way to have a Galilean spacetime?
And if we instead make Galilean symmetry just a local symmetry, can we obtain something like a "Galilean version" of GR?

 

[haushofer]

Yes, there is :) See this topic,

https://www.physicsforums.com/showthread.php?t=432933

You get two degenerate metrics, one for space and one for time. There does not exist a single non-degenerate metric which is kept invariant under the whole Galilei group. A quick way to see this also is to write down the Minkowski metric and its inverse explicitly with c in it and take c --> oo.

And indeed, if one makes Galilean symmetry local one obtains Newton-Cartan.

 

[JustinLevy]

You get two degenerate metrics, one for space and one for time.

When I hear degenerate, I think equal eigenvalues, which probably isn't what is meant there at all. So I'm sorry, but I don't understand what that means. Can you expound a bit?

And what do you mean by two metrics? The line element is written as a sum of two terms involving different metrics?

I think once I understand the terminology, that post you linked will hopefully clear up the rest.

 

[haushofer]

Degenerate means "having zero eigenvalues". The "time metric" has 3 eigenvectors with eigenvalue zero, and the spatial metric has 1 eigenvector with eigenvalue zero.

So your metric structure becomes simply degenerate. It's tempting to add the two metrics, but this results in a "metric" which is NOT kept invariant under the Galilei group; two separate degenerate metrics is the best you can do.

 

[arkajad]

You take the limit of the contravariant metric, not the covariant one.

 

[haushofer]

A "quick and dirty" way to see it is to regard the c --> oo of the metrics

$$c^{-2} \eta_{\mu\nu}$$

and

$$\eta^{\mu\nu}$$

:)

 

[Daverz]

Is there a way to have a Galilean spacetime?
And if we instead make Galilean symmetry just a local symmetry, can we obtain something like a "Galilean version" of GR?

Newtonian gravity can be "geometrized". There's no 3+1 spacetime metric, but there is a connection!

You might find chapter 12 of Misner, Thorne & Wheeler, "Newtonian Gravity in the Langauge of Curved Spacetime" to be of interest. This is also described in Penrose's Road to Reality.

 

[haushofer]

Yes, but this connection can be written in terms of two metrics; the ones which are described above. So Newton-Cartan can be described via a metrical structure, something which is not very clear from MTW.

 

[arkajad]

Moreover, the two-metric structure can be naturally "geometrically quantized" leading to Schrodinger's equation (with electromagnetic field) understood as a parallel transport equation in the bundle of Hilbert spaces over time.

 

[grav-universe]

One way of looking at it would be to take

∞^2 dt^2 - dx^2 - dy^2 - dz^2 = ∞^2 dt'^2 - dx'^2 - dy'^2 - dz'^2

dt^2 - (dx^2 + dy^2 + dz^2) / ∞^2 = dt'^2 - (dx'^2 + dy'^2 + dz'^2) / ∞^2

dt = dt'

and then

c^2 dt^2 - dx^2 - dy^2 - dz^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2, where dt = dt', so

c^2 dt^2 - dx^2 - dy^2 - dz^2 = c^2 dt^2 - dx'^2 - dy'^2 - dz'^2

dx^2 + dy^2 + dz^2 = dx'^2 + dy'^2 + dz'^2

 

[atyy]

Some details on haushofer's spatial and temporal metrics are given here

http://arxiv.org/abs/gr-qc/0506065

 

[grav-universe]

My last post was performed straight forward mathematically, but something still seems to be awry, since then we would have dx = dx'. It must have something to do with introducing infinities, but I can't tell exactly where it went astray yet.

 

[arkajad]

The quick and dirty method of haushofer is one one way. Another, less known, way is to start with a 5-dimensional space time. Usually it is assumed that one space-like dimension (Killing vector field for the 5D metric) is "hidden". That is the normal Kaluza-Klein Ansatz. If, instead, we assume that the hidden dimension is light-like, we end up with Einstein-Cartan "two-metrics" structure of our effective space-time. Not so quick, but also not so dirty?

 

[JustinLevy]

A "quick and dirty" way to see it is to regard the c --> oo of the metrics

$$c^{-2} \eta_{\mu\nu}$$

and

$$\eta^{\mu\nu}$$

:)

Some details on haushofer's spatial and temporal metrics are given here

http://arxiv.org/abs/gr-qc/0506065

Hmm... pg 45. discusses in more depth the two metrics haushofer refers to. But I'm still not sure what this means. We now have two line elements? Something like:
$$dT^2 = t_{ab} dx^a dx^b$$
$$dS^2 = h^{ab} dx_a dx_b$$

But I don't really understand what the symbols mean physically anymore. For an inertial frame we have:
$$t_{ab} = \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{matrix}\right), h^{ab} = \left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{matrix}\right)$$
Now if I have a contravariant displacement (in inertial frame components)
$$x^a = (c,u,v,w)$$
how do I use those interial frame metric components to get x in covariant form, $$x_a = (?,?,?,?)$$
It is hard to understand what these "not really metrics" are, and why it even still counts as spacetime.

It seems there is no spacetime. It is just like Newtonian mechanics where you only have space, and a parameter to describe evolution in this space.

 

[haushofer]

Well, one can introduce also

$$h_{ab}, \ \ \ T^{ab} = T^a T^b$$

with the following properties:

$$h^{ab}t_b = 0, \ \ \ h_{ab}T^b = 0, \ \ \ T^a t_a = 1$$

and finally

$$h^{ab}h_{bc} = \delta^a_c - T^a t_c$$

This last object works as an projection operator on the spatial hypersurfaces. Now one can split up any vector in a spatial and a temporal direction.

I'm not sure what you mean by "it seems there is no spacetime". That implies that you think that a spacetime should have a non-degenerate metric.

 

[haushofer]

Perhaps you like this article and the references therein:

http://arxiv.org/abs/1011.1145

 

[arkajad]

Now one can split up any vector in a spatial and a temporal direction.

Remark: Such a split depends on the state of the motion of the observer (or, on a frame of reference). It is not an intrinsic property of the Galilean space-time structure itself.

 

[haushofer]

Could you elaborate on that? Do you mean that the tensors you use for this splitting are only tensors under the Galilei group?

 

[arkajad]

A Galilean spacetime M is a fibre bundle over time. You may have a metric on the base (time metric), you may also have a Euclidean metric in each fibre (space at a given time) - they can be affine Eucldean spaces. But there is no identification of spaces at different time. In other words, even if the bundle is trivial, it is not naturally trivializable. If you think of a trivialization this defines a split and you can think of it as flow of a vector field in M - these are trajectories of an "observer". Change the observer, for instance using a Galilei transformation, and the trivilization will change: x'=x-vt. So the split will also change. This corresponds to the fact that we sometimes refer to Newton's spacetime as the product (absolute space and absolute time), and to Galilei spacetime as a trivial (i.e. trivializable) bundle but not intrinsically a product bundle.

 

[haushofer]

I'm more or less familiar with the notion of fibre bundles, but could you be more explicit on your phrase "there is no identification of spaces at different time"? I would really like to understand this better :)

What is your notion of "being trivializable"? As I understand a fibre bundle is trivial if it can be written as a product space globally, but how is this related to being "trivializable"? And what is the physical relevance of this to our Galilean spacetime?

 

[arkajad]

I'm more or less familiar with the notion of fibre bundles, but could you be more explicit on your phrase "there is no identification of spaces at different time"? I would really like to understand this better :)

Alright, will do it, but give me a little time, because I will have to prepare a short exposition. Of course it all can be found in the literature, but I will try to make it short and simple giving just geometric ideas that will illustrate the essence.
In the meantime you may like to have a look at http://arkadiusz-jadczyk.org/papers/9406204/topics.htm" - the section Galilean General Relativity - just to have an idea about what can be developed along these lines.

 

[haushofer]

Ok, I will definitively have a look there and wait for your answer! Thanks in advance!

 

[arkajad]

Ok. Here it goes. There are many ways in which the Galilei spacetime can be introduced, but the simplest one is via defining it in terms of its group of transformations, which is the inhomogeneous Galilei group. It consists of translations in space, translations in time, rotations in space and Galilei boosts. Altogether 3+1+3+3=10 parameters. Since rotations are not important for understanding its structure, let's restrict ourselves to 1 time and 1 space dimension. Then it is easy to draw a picture. In this case a element of the Galilei group is a linear transformation in $$\mathbf{R}^2$$ given by:

$$t'=t+t_0$$
$$x'=x+a-vt$$

Here $$a$$ is the translation in space parameter, $$t_0$$ - translation in space, and $$v$$ is the boost parameter.

Then the Galilei spacetime M is defined as a set (say, a 4-dimensional manifold) endowed with a family of global coordinate systems $$(x,t)$$ in such a way that the transitions between any two coordinate system from this family are given by Galilei transformations. So, we have a 10-parameter family of distinguished coordinate systems - we may associate with them terms such as "inertial frames" or "inertial observers". In our 2-dimensional case this is a 3-parameter family. I will refer to them simply as "frames"

Once this is done the next thing to do is to study its "geometry" i.e. according to Felix Klein, yo study the invariants of the group of transformations.

We notice that the form $$dt$$ is invariant.Even more, if two points $$P_1,P_2$$ have the same coordinate in one frame, they have the same coordinate in any other frame (of our distinguished family). We call the set of such points "space at a given time). We also notice that if two points $$P_1,P_2$$ have the same time coordinates, then the distance $$d(P_1,P_2)=(x(P_1)-x(P_2))^2$$ is an invariant. Therefore each space at a given time is an affine Euclidean space - with its Euclidean flat metric.

We notice that our transformations are linear. Therefore we can give M the structure of the affine space. Or, if you prefer, there is a distinguished globally flat affine connection on M characterized by the fact that its connection coefficients are globally zero in all frames. Both time form and space metric have covariant derivative zero with respect to this connection.

Now, let us have a look at this picture:

Blue lines are trajectories of points with a constant x of one frame, red lines - for another frame. We chose a point P at time t. It has red coordinates in one frame, blue coordinates in another frame. Here we neglect time translation - which is easy to take into account, but it will not add anything of importance. Verical lines are spaces at a fixed time. These are distinguished geometrically - defined in an invariant way. But there are no distinguished horizontal lines.

Each frame defines an isomorphism between M and $$\mathbf{R}\times\mathbf{R}$$, different frames define different isomorphisms. M can be represented as a product, but in many (10-parameter family for our 3-dimensional space and 1-dimensional time) ways, but they are all on equal footing. This is in contrast to Newton's time. For Newton (also Kant) there was also an "absolute space" - something like an "aether". Perhaps in reality there is such a thing, but here we have spacetime as an arena for studying, say, one or several particles in an otherwise empty space". Later we can study Galilean cosmology and find solutions of the corresponding field equations that break the full Galilei symmetry, but that is another subject.

 

[haushofer]

Thanks arkajad, I'll read it carefully tomorrow! :)