4
$\begingroup$

For example Schwarzschild metric, or Alcubierre metric, but using only intrinsic (natural, canonical, etc.) physical objects (like length, angles, etc.) for relations between natural objects on tangent structures. So not using anything like basis vectors, frames, charts, etc.

A metric tensor is just function of angles and absolute values of vectors. It is a section of the dual of the tensor product bundle of tangent bundle with itself. It has nothing to do with coordinates. So why can't it be written down without them?

$\endgroup$
  • 4
    $\begingroup$ More on coordinate-free constructions: physics.stackexchange.com/q/65084/2451 $\endgroup$ – Qmechanic Aug 1 at 12:13
  • $\begingroup$ You´ll have to introduce an observer relative to which the quantities are defined (distance, lenght, time intervall, ...). $\endgroup$ – Cham Aug 1 at 13:09
  • $\begingroup$ A sort of evidence: the most mathematically rigorous GR book I know, O'Neill, still introduces coordinates when defining particular spacetimes, so I don't think it can be done. $\endgroup$ – Javier Aug 1 at 15:01
  • $\begingroup$ can you provide a hypothetical example in math? $\endgroup$ – Paradoxy Aug 1 at 15:48
  • $\begingroup$ Can you do this for Euclidean space? I suspect you can not. $\endgroup$ – tfb Aug 5 at 13:04
5
+25
$\begingroup$

I'm not sure if this will precisely answer your question concerning "metrics".... but this might have the spirit of what you may be seeking.

Here's an overview of a coordinate-free derivation of the Schwarzschild solution by Robert Geroch.

[short answer: using symmetries specified by Killing vector fields, construct various scalar fields for use in the Einstein Fields Equations to obtain a set of differential equations for those scalar fields. After the solutions are obtained, the results can be expressed in coordinate-form, if desired.]

(sources:
General Relativity: 1972 Lecture Notes (Lecture Notes Series) (Volume 1)
Minkowski Institute Press; 1 edition (February 25, 2013) ISBN 978-0987987174
also
http://home.uchicago.edu/~geroch/Course%20Notes (latexed draft?)
http://www.gravity.psu.edu/links/general_relativity_notes.pdf (scan of original notes) )

Refer to the above for details.
Below I will quote some passages from the LaTeXed file and summarize some parts of the approach given. (Hopefully my transcriptions are accurate.)

Ch 25: The Schwarzschild Solution

Physically, the Schwarzschild solution represents the geometry of an “isolated, non rotating star, which has settled down to equilibrium”. What properties would we expect such a solution to have? Firstly, we would expect the solution to be static, i.e., we would expect to have a timelike, hypersurface-orthogonal Killing vector $t^a$. Secondly, we would expect the solution to be spherically symmetric, i.e., we would expect to have Killing vectors ${l_1}^a$, ${l_2}^a$, ${l_3}^a$ which are spacelike, linearly dependent, and have the commutation relations $$[{l_1},{l_2}]^a={l_3}^a\quad [{l_2},{l_3}]^a={l_1}^a\quad [{l_3},{l_1}]^a={l_2}^a\quad (79)$$ Finally, we would expect that the time-translations and rotations commute, i.e., we would expect to have additional commutation relations $$[t,{l_1}]^a=[t,{l_2}]^a=[t,{l_3}]^a=0\quad (80)$$ To summarize, we are concerned with space-time having four Killing vectors, with the commutation relations (79) and (80). For the matter composing the star, we take a fluid. Thus, we have the density $\rho$, pressure $p$, and (unit) velocity field $\eta^a$. Since the star is supposed to have “settled down to equilibrium”, we suppose that the fluid does not “move relative to static observers”, i,e., we take $\eta^a$ a multiple of $t^a$.

To summarize, the Schwarzschild solution is a space-time with four Killing vectors, $t^a$ (timelike, hypersurface-orthogonal), and ${l_1}^a$, ${l_2}^a$, ${l_3}^a$ (spacelike, linearly dependent), subject to (79) and (80), where the matter is a fluid with four-velocity field proportional to $t^a$. We now discuss the geometry of the Schwarzschild solution.

Then, Geroch proceeds as follows:

  1. Define a scalar field $\lambda=t^a t_a$. ($\lambda<0$ since $t^a$ is timelike [signature $(-+++)$])

  2. Write Ricci in terms of $\lambda$ using the hypersurface-orthogonality of $t^a$: $$R_{mb} t^m =\frac{1}{2}\lambda^{-2}t_b(\nabla^c \lambda \nabla_c \lambda) -\frac{1}{2}\lambda^{-1}t_b\nabla^2\lambda\quad (83)$$

  3. Use the Einstein field equations for a perfect fluid to introduce matter variables (in place of the Ricci terms) to obtain $$ R_{ab}=8\pi G\left[ -\lambda^{-1}(\rho+p)t_a t_b+\frac{1}{2}(\rho-p)g_{ab}\right] \quad(84) $$

$$ \lambda^{-1}\nabla^2\lambda-\lambda^{-2}(\nabla^c \lambda \nabla_c \lambda) =8\pi G(\rho+ 3p) \quad (85) $$ which "can be rewritten in the more suggestive form" $$\nabla^2 \left[\frac{1}{2}\ln(-\lambda)\right]=4\pi G(\rho+3p) \quad (86)$$

  1. Define a positive scalar field $r$ in spacetime as $$2r^2= l_1{}^a l_1{}_a + l_2{}^a l_2{}_a +l_3{}^a l_3{}_a\qquad (88) $$ which he describes "as a sort of 'radial distance from the center of the star' "

  2. Define the scalar field $\mu=(\nabla^a r)\nabla_a r$, where $\mu=1$ for flat space, and deviations of $\mu$ from 1 represent the "curvature of space"

Let us summarize the situation. We think of $r$ as a “radial coordinate”. We think of $\lambda$ and $\mu$ as "fields which describe the geometry of space-time." Since our space-time is static and spherically symmetric, we expect that everything of interest will be a function only of $r$....

The idea is to use Einstein’s equation to obtain a pair of ordinary differential equations on the functions $\lambda(r)$ and $\mu(r)$.

Eventually, for the region outside the star (so $\rho=0$, $p=0$), Geroch arrives at these $$\lambda''\mu -\frac{1}{2}\lambda^{-1} \mu(\lambda')^2+ \frac{1}{2}\lambda'\mu' +2\mu r^{-1} \lambda'=0\quad(94)$$

$$-\frac{1}{4}\lambda^{-1} \mu \lambda' \mu' -\mu \mu' r^{-1} + \frac{1}{4}\lambda^{-2} \mu^2(\lambda')^2 -\frac{1}{2}\lambda^{-1} \mu^2\lambda''=0\quad(95)$$ where $d/dr$ is denoted by a prime.

We have now obtained the ordinary differential equations we sought. What remains is to solve them. Eliminating $\lambda''$ between (94) and (95), we obtain simply $\lambda'/\lambda=\mu'/\mu$. So, $\lambda$ is a constant multiple of $\mu$. What multiple should we choose?
$\vdots$
[physical and mathematical arguments]
In Minkowski space, $\lambda=-1$ and $\mu=1$, which suggests $\lambda=-\mu$.
$\vdots$
Setting $\lambda=-\mu$ in (95)... the solution is $\lambda=a+b/r$
$\vdots$
We write $\lambda= −1+2GM/r$...
$\vdots$

It should now be clear that one can choose coordinates in which the metric for the Schwarzschild solution takes the well-known form $$−\left(1 −\frac{2GM}{r} \right) dt^2 + \left(1 −\frac{2GM}{r}\right)^{−1} dr^2 + r^2( d\theta^2 + sin^2 \theta d\phi^2)$$ The $\theta$ and $\phi$ are "angular coordinates", while the scalar field $r$ becomes a "radial coordinate".

...so coordinates are introduced at the last step.

$\endgroup$
0
$\begingroup$

If I understand your question properly, this is not generally possible. Take Schwarzschild, for example, which is spherically symmetric. You have one "special" point at the singularity, but you have nothing else physical to which you could make reference for specifying the metric. (Depending on your view of your goal, you might say the event horizon is also a "physical" surface, but even with that you cannot specify angles relative to anything special in the spacetime.) Your only choice is to introduce some structure of your own for writing it down, which is typically a coordinate system.

If you have a spacetime with more structure, it might or might not be possible depending on the structure. There's no way to answer exactly how to do it at this level of generality.

$\endgroup$
  • $\begingroup$ Well why would I have to make up some structure that in the end is not just irrelevant but tends to produce even more non physical structure one has to discern to see what the actual dynamics is. It seems weird to write down a physical quantity in such a way. It's like writing down a field of a charged particle using coordinates.. $\endgroup$ – Leo Kovacic Aug 1 at 14:14
  • $\begingroup$ Part of the metric is not physical. Free choice of coordinates is the gauge freedom inherent in the theory. So, taken the other way, it's not clear why you think you should be able to write something that, inherently, has non-physical coordinate choices worked into it in a way that does not reference any coordinates. @LeoKovacic $\endgroup$ – Brick Aug 1 at 14:17
  • $\begingroup$ This differs from the EM field, by the way, in that the gauge freedom in GR is the coordinate choice whereas the gauge freedom in EM is the EM field potential. Coordinates are wrapped into the GR theory in a fundamentally different way. $\endgroup$ – Brick Aug 1 at 14:21
  • $\begingroup$ If coordinates are a fundamental part of GR than the theory has no hope of being any part of fundamental physics theory $\endgroup$ – Leo Kovacic Aug 1 at 16:35
  • 1
    $\begingroup$ You're mixing up different things. Relativity has at its core the assumption that no reference frame is special, which manifests as freedom to choose coordinates. In a field theory (generally) this appears as unconstrained degrees of freedom in the fields (i.e. gauge freedom). For GR the primary field is the metric and you have gauge fields (e.g. "lapse" and "shift") that have no direct physical meaning but are part of the metric. In EM you have the vector potential $A^\mu$ playing the same role - Not completely arbitrary but not unique and not directly physical. $\endgroup$ – Brick Aug 1 at 17:40
0
$\begingroup$

An interesting approach that achieves what OP is after in geometries of physical importance is detailed in a series of papers by Vasiliev and collaborators [1–3]. There a series of asymptotically AdS black hole solutions (Kerr–AdS${}_4$ and its various generalizations) are given a coordinate-free descriptions as a deformation of a background anti-de Sitter geometry with a distinguished Killing vector (with the parameter of deformation in the simplest case being the mass of the black hole, and the spin parameter determined by the kinematics of the Killing vector).

An important feature of the construction is that metrics belong to the Kerr–Schild class, which means that the Kerr–Schild vector is null and geodesic vector of both the background and deformed geometries.

  1. Didenko, V. E., Matveev, A. S., & Vasiliev, M. A. (2008). Unfolded description of AdS4 Kerr black hole. Physics Letters B, 665(4), 284-293, DOI:10.1016/j.physletb.2008.05.067, arXiv:0801.2213.

  2. Didenko, V. E., Matveev, A. S., & Vasiliev, M. A. (2009). Unfolded dynamics and parameter flow of generic AdS(4) black hole, arXiv:0901.2172.

  3. Didenko, V. E. (2011). Coordinate independent approach to 5D black holes. Classical and Quantum Gravity, 29(2), 025009, doi:10.1088/0264-9381/29/2/025009, arXiv:1108.4321.

$\endgroup$
  • $\begingroup$ Sorry but this is just the opposite of what I am looking for. I thought that maybe coordinates are there only to simplify some calculations. For me a metric is just a product of absolute values and a cosine of the angle, for euclidian space that is. Why can't you just use a similar formula for other geometry? $\endgroup$ – Leo Kovacic Aug 1 at 20:41
  • $\begingroup$ Yes of course I want geometry has physical importance it's just that coordinate geometry a priori has no physical value. $\endgroup$ – Leo Kovacic Aug 1 at 20:46
  • $\begingroup$ I guess it wasn't a very good question and nothing can be done without coordinates . Well that sucks.. $\endgroup$ – Leo Kovacic Aug 1 at 20:47
  • $\begingroup$ @LeoKovacic: Let us step back a little: can you write the metric of AdS space without coordinates. Since this is maximally symmetric space, this should not be a problem (in the first paper cited this is eqs. 1.4-1.6). $\endgroup$ – A.V.S. Aug 2 at 5:17
  • $\begingroup$ Ok I must admit that I just scrolled quick and closed the PDF file as soon as I saw a matrix. Eqs. 1.4,1.6 are what I'm looking for. I am not a big fan of index calculations in any context. But this is nice cause pde s are dealt with manifestly free of coordinates. However this goes very far, I just wanted to see how would a simple solution ( metric) look without coordinates. I am not familiar with all of the formalism in the articles. $\endgroup$ – Leo Kovacic Aug 2 at 8:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.