Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why a constant determinant of the metric?

  1. Jan 18, 2014 #1
    Hi,

    In his original paper, Schwarzschild set the "'equation of the determinant" to be: |g|=-1. In other words, he imposed the determinant of the metric to be equal to minus one when solving the Einstein's equations. Must we impose this equality systematically in general relativity and why?


    (see english translation: "On the Gravitational Field of a Mass Point according to Einstein’s Theory", K. Schwarzschild, in General Relativity and Gravitation, vol.35, No.5, 2003, pp.951-959; http://link.springer.com/article/10.1023/A:1022971926521?LI=true
    or
    http://www.phys.huji.ac.il/Phys_Hug/Lectures/77632/SchwarzschildTranslated.pdf [Broken])

    Rob
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 18, 2014 #2
    The determinant of metric is not invariant under coordinate transformation. In GR , the sign of g is important which is invariant.In Schwarzchild coordinate , as a case , this quantity is -1 but actually if you change the coordinate it may change.
     
  4. Jan 18, 2014 #3
    Invariance

    Hello,

    I am agree, the signature of the metric $g$ is an invariant and changes of coordinates can change the value of the determinant $|g|$ of $g$. From the Sylvester’s law, the signature is an invariant under changes of coordinates.

    Besides, invariance is always defined with respect to something.

    Now, if the determinant of the metric is not an invariant, it is not an invariant with respect to what? Necessarily, it is not an invariant with respect to the determinant of another metric.

    In addition, this means also that the scalar product defined by $g$ of any two 4-vectors, say $v$ and $u$, is no more an invariant. Only, the quantity $g(u,v)/\sqrt{|g(u,u)\,g(v,v)|}$ may be considered as an invariant since this quantity does not depend on the value of the determinant of the metric. Thus, only ‘’angles’’ are the invariants of the geometry, i.e., we have a conformal geometry. Then, only the ‘’normalized metric’’ $-g/|g|$ may be invariant. And, we could say: $g$ is not invariant with respect to the ‘’invariant’’ metric $-g/|g|$.

    Generalizing this to covariant 2-tensors with components $T_{i,j}$, for instance, means that only the components $T_{i,j}/\sqrt{|g_{ii}\,g_{jj}|}$ are invariant.

    Hence, if we consider that not only the angles are invariant, then, this involves that the scalar products $g(u,v)$ are invariant with respect to $g$ which must not change and, in particular, its determinant. Thus, we can take a constant determinant and, in particular, a determinant equals to minus one. As a consequence, any change of coordinates must be compatible with this claim, and not any change is admissible. Also, we have no more a conformal geometry but an isometric geometry. I consider that general relativity is based on an isometric geometry and thus $|g|$ must be constant.

    Then, the question is: must we impose an isometric geometry in general relativity?

    Rob
     
    Last edited: Jan 18, 2014
  5. Jan 18, 2014 #4

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    For invariants, you have to match covariant and contraviariant. Thus, for dot product of two (contravariant) vectors, you use the covariant metric. This results in determinant factors cancelling, so it is invariant even under transforms that change the determinant.
     
  6. Jan 18, 2014 #5

    atyy

    User Avatar
    Science Advisor

    There is no constraint on the determinant of the metric in general relativity. (So I don't think Schwarzschild was right on that point.)

    't Hooft, http://www.staff.science.uu.nl/~hooft101/lectures/gr.html (p49, footnote 9):
    "In his original paper, using a slightly different notation, Karl Schwarzschild replaced ... by a new coordinate r that vanishes at the horizon, since he insisted that what he saw as a singularity should be at the origin, claiming that only this way the solution becomes "eindeutig" (unique), so that you can calculate phenomena such as the perihelion movement (see Chapter 12) unambiguously. The substitution had to be of this form as he was using the equation that only holds if g = 1 . He did not know that one may choose the coordinates freely, nor that the singularity is not a true singularity at all. This was 1916. The fact that he was the first to get the analytic form, justifies the name Schwarzschild solution."

    There is something called unimodular gravity in which the determinant of the metric is 1.
    http://www.phy.olemiss.edu/~luca/Topics/u/unimodular.html
     
  7. Jan 18, 2014 #6

    WannabeNewton

    User Avatar
    Science Advisor

    It can't vanish.
     
  8. Jan 18, 2014 #7

    atyy

    User Avatar
    Science Advisor

    Yes, thanks!
     
  9. Jan 18, 2014 #8
    Volume 4-form

    Schwarzschild was wrong? Well!

    Anyway, in GR the volume 4-form is $\omega_S\equiv\sqrt{-|g|}\,d^4x$.

    But, on the one hand, under any change of frame expressed by the Jacobian matrix $J$, the volume 4-form $\omega_G\equiv\tfrac{1}{\sqrt{-|g|}}\,d^4x$ is invariant.
    Indeed, $d^4x$ is transformed into $|J|\,d^4x$ and $|g|$ into $|J|^2\,|g|$.
    So, the ratio is preserved, and thus, $\omega_G$ is invariant under any $J$ being an element of the general linear group $GL(4,R)$.

    Now, on the other hand, $\omega_S$ is transformed into $|J|^2\,\omega_S$.

    Hence, if we assume that $\omega_S$ is the volume 4-form of the general relativity, as I can read in a lot of papers, then, necessarily, it is considered as the invariant volume 4-form. But, it can be invariant only if $J$ is an element of the special linear group $SL(4,R)$, i.e., $|J|=1$.
    And thus, implicitly, GR is an isometric theory preserving the determinant of the metric which can be, therefore, set to -1.
    If the GR is not an isometric theory, then, people ought to use the volume 4-form $\omega_G$ which is the only $GL(4,R)$-invariant measure.

    About, 't Hooft, you know his polemic with Martinus Veltman about Schwarzschild...

    Rob
     
  10. Jan 18, 2014 #9

    atyy

    User Avatar
    Science Advisor

    No, tell me about it?
     
  11. Jan 18, 2014 #10
    The question

    Well, I think that we move away from my initial question.
    No clear or convincing replies. I would like to avoid things such that: "someone said that..." such as Mr. 't Hooft or others. I would prefer scientific answers.... Thank you.
     
  12. Jan 18, 2014 #11

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Sadraj gave you one in #2. To amplify on what s/he said, GR is agnostic about coordinates. Any smooth change of coordinates (diffeomorphism) is allowed. Under such a change of coordinates, the determinant of the metric changes. Schwarzschild was writing in the very early days of GR when this kind of thing was not yet well understood.

    Amplifying on WannabeNewton's #6, the standard formulation of GR doesn't allow the determinant of the metric to vanish, because it depends on the ability to raise and lower indices, which you can't do if the metric isn't an invertible matrix.
     
  13. Jan 18, 2014 #12

    atyy

    User Avatar
    Science Advisor

    Thanks for letting me know you were being rude!
     
  14. Jan 19, 2014 #13
    The "physical tensors"

    My apologies to « atyy ». I felt a slipping slope and a spreading of the initial problem. Sorry. I have also to improve my english to find all the correct civil sentences.

    Now, the things remains not clear for me. « bcrowell » after « Sadraj » said that GR is « agnostic » to changes of coordinates.

    But, there are two approaches about the changes of coordinates.

    First, coordinates are associated with charts on a manifold, say M, and changes of charts are associated with changes of coordinates. These changes of charts can be smooth and the Jacobian matrices can have nonvanishing determinants different from unity. Such changes of charts are associated with the notion of covariance (indices, etc…).

    Second, the diffeomorphisms on M are independent on the charts and some consider that change of coordinates are also associated with such diffeomorphisms. Clearly, it is wrong. There are defined independently on the charts and thus independently on the coordinates.These diffeomorphisms can be conformal or not, i.e., the metric is scaled passing from one point m in M to another point m’ in M. Also, there exist manifolds on which conformal diffeomorphisms do not exist. As a consequence, conformal changes of charts do not exist as well. On contrary, an example of conformal theory is the H. Weyl theory among more recent others.

    But, problems appear in conformal theories. If you admit scaling changes passing from a point m in M to another point m’ in M, then, you need to have a physical observable to indicate the change of scale passing from m to m’; and, in particular, the change of scale between the metric at m and the metric at m’; and also, passing from a chart at m to another chart at m’.

    If you admit that such scaling changes on M are not related, in any way, to physical observables at disposal (it was one of the criticisms historically made against the conformal Weyl theory), then, necessarily, you admit to be unable to reach physically the information on scalings. If you consider that you have all the informations and that any information on scaling is then void of meaning then you consider that scalings do not exist; And therefore, that scaling changes on the determinants of the various changes of « … » are forbidden because not physical. And, in particular, that changes of coordinates with scalings on the Jacobian matrices are also forbidden. And thus,... the determinant of the metric in any charts, coordinates, must be set to -1.
    Equivalently, this means that you are able to compare physically different scales at different points in M; in other words, to compare the lengths of the rulers at different points in M; but more, to be able to effectively use rulers during the time (since making a length measurement is a process in time considering that the lengths of the rulers you use do not change in time). If you consider that you are unable to make physically these sorts of comparisons or to have these potential uses, then you admit that the value of the nonvanishing determinant of the metric can vary as you want; as well as the determinants of the Jacobian matrices associated with the different changes of charts (coordinates) you consider.

    Questions about covariance is not the problem since covariance is related to changes of frames at a given point m in M, that is to changes of charts. But physical admissible changes of charts related to scalings, physical or not, is also the problem translated in the question: must the determinant of the metric be equal to -1 or not?

    Of course, you can make changes of coordinates with determinants of the Jacobian matrices differing from unity, but then, that means that you abandon the possibilities to make comparisons between those geometrical objects expressed in the different systems of coordinates which are not « agnostic » with respect to changes of scale. Thus, you loose the means of scaling comparison between the two systems of coordinates. And then, it is related to this well-known notion of « physical tensors » of the tensorial calculus such as the cotensor $T_{i,j}/\sqrt{|g_{ii}\,g_{jj}|}$ associated with the cotensor $T_{i,j}$. You will be able to make comparisons and interpretations only between these « physical tensors » which are « agnostic » with respect to scaling changes.

    Hence, the question rises to known if setting $|g|=-1$ is simpler or not and if this « equation of the determinant » must be or not included in practice in any computation in general relativity.

    Bye

    Rob
     
    Last edited: Jan 19, 2014
  15. Jan 19, 2014 #14

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Try to simply state your concern. I answered already that a properly formed invariant, including a volume integral, does not change under a coordinate transform that changes the determinant, because the balancing of contravariant and covariant components introduces exactly cancelling factors of the determinant. This is true for a volume integral as well because the coordinate differentials have contravariant character.

    All observables (including geometric comparisons) are supposed to phrased in terms of invariants. Thus I see no possible advantage to restricting the metric to a determinant of -1.
     
  16. Jan 19, 2014 #15
    Equivalence principle?

    Hello PAllen,

    I am completely in agreement with you modulo the definition of the volume element or volume integral we consider. I am always very surprised that in GR the definition of the volume 4-form is $\sqrt{-|g|}\,d^4x$ and not $\tfrac{1}{\sqrt{-|g|}}\,d^4x$ because the former is not invariant under coordinate transforms (in the general linear group), as I indicated and shown in a precedent post above, whereas the latter is. The volume element $\sqrt{-|g|}\,d^4x$ is only invariant with respect to the special linear group so prohibiting the most general coordinate transforms. Why such choice? An explanation could be, precisely, that we restrict the set of coordinate transforms to those for which the Jacobian matrices $J$ have determinant equal to 1 (and thus preserving the orientation); and, additionally, that this could be the result of a kind of abuse of notations. I mean that we would write rather $\sqrt{|J|}\,d^4x$ and substituting $-|g|$ by $|J|$. But, because of habits, somehow, I don’t know, this is the notation used, implying that, indeed, we restrict the coordinate transforms to those in the special linear group.

    Moreover, what could be the advantage of restricting the metric to a determinant -1? This is my question. But more generally, what could be the advantage of restricting the metric to a constant determinant? Is it a mathematical advantage or a constraint imposed by the physics? Schwarzschild imposed this constraint. He found only one solution which is continuous as not very well-known (you can have a look on his solution in the formula (14) in the english translation; see url in the first post). Is it because we obtain continuous solutions of the Einstein’s equations when setting this constraint? If this constraint is not imposed we may have vanishing limits of the metric at certain points; this could be impossible if the determinant of the metric is constant; but there are certainly exceptions.

    Also, is it related to the equivalence principle in GR? In other words, must we consider that the correspondence between the manifold M in the vicinity of a given point m and the tangent space at this point is isometric? Is this constraint a sort of safety belt to ensure the equivalence principle works everywhere?
    Question…

    Rob
     
    Last edited: Jan 19, 2014
  17. Jan 19, 2014 #16

    atyy

    User Avatar
    Science Advisor

    The volume element ##\sqrt{-g}d^{4}x## is generally covariant, as explained by Carroll in http://preposterousuniverse.com/grnotes/grtinypdf.pdf [Broken] (Eq 27-29).

    You can also find more information on unimodular gravity, including forms which are equivalent to GR in Smolin's http://arxiv.org/abs/0904.4841 (just read the classical bits, ignore the parts about quantization). There are many more references in http://www.phy.olemiss.edu/~luca/Topics/u/unimodular.html.
     
    Last edited by a moderator: May 6, 2017
  18. Jan 19, 2014 #17

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    There is no advantage I see to restricting the metric determinant. All it does is make a funny gauge group rather than all diffeomorphisms.
     
  19. Jan 19, 2014 #18
    Many thanks to atyy.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why a constant determinant of the metric?
Loading...