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Dale
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Yes, you said that above. What do you mean by that? Please clarify, not repeat. Do you mean the coordinates in the new basis, or are you referring to the underlying geometrical object?
DaleSpam said:Yes, you said that above. What do you mean by that? Please clarify, not repeat. Do you mean the coordinates in the new basis, or are you referring to the underlying geometrical object?
OK, thank you for the clarification, that is very helpful. Perhaps for clarity in this thread we can use ds² for the line element, a scalar, and g for the metric, a rank 2 tensor. If you will avoid calling ds² the metric then I will avoid interpreting it as a rank 2 tensor.Anamitra said:I am not considering ds^2 as a tensor.
ds^2 is defined by the relation
[tex]{ds}^{2}{=}{g}_{\mu\nu}{dx}^{\mu}{dx}^{\nu}[/tex]
This is incorrect. The coordinate transformations listed are valid diffeomorphisms. ds² is unchanged under any diffeomorphism.Anamitra said:Now dx(i) are being transformed by the usual rules but the metric chosen is a new one such that it cannot be obtained from (1) by transformations where ds^2 is preserved
It is not my interpretation, it is the standard interpretation, see the Carroll notes. However, I am willing to use your non-standard interpretation in this thread as long as you do so consistently and do not call ds² the metric.Anamitra said:You may have a different interpretation of ds^2 and a new formulation.That is a different issue----there is no reason to mix up the two issues.
As discussed above ds² is not a metric, it is a scalar, and I assume that ds'² is also a scalar and not a rank 2 tensor (and therefore not a metric). However, the scalar ds'² is not equal to ds² in any coordinate system, they are unrelated scalars that have nothing to do with each other. You do not get ds'² from ds² by a change in coordinates.Anamitra said:Now in the same r,theta,phi system we may consider the metric:
[tex]{ds'}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{sin}^{2}{\theta}{d}{\phi}^{2}{)}[/tex]
--------------- (3)
ds^ 2 has now changed for identical values of dt,dr,d(theta) and d(phi)
DaleSpam said:E.g. you cannot say, a hovering observer at infinity feels no proper acceleration therefore a hovering observer at the event horizon feels no proper acceleration. The reason you cannot say this is the same as above, a lack of symmetry, in the radial direction instead of in time or lambda.
Certainly. Do you understand the difference between a law of physics and the boundary conditions?Anamitra said:Actually the laws remain unchanged so long we are able to express them in tensor form.
Anamitra said:Suppose there was a body at a distance when we were in curved spacetime. We consider a process where the field is reduced. In flat spacetime we can evaluate the velocity and then we may think of scaling up the velocity in regard of curved spacetime to evaluate the velocity of a body at a distance in curved spacetime.
Anamitra said:So far as the metric in the last posting[#85] is concerned, proper speeds are defined terms--d(phi)/ds and d(theta)/ds.
Now for a null geodesic ds^2=0
How does the observer define the coordinate speed or physical speed of light unless one of the variables theta or phi is the time coordinate?This has not been clearly specified
The speed of light is independent of the motion of its source---What does he understand by the speed of light here unless one of the coordinates in the metric represents time.?
Ben Niehoff should clearly specify the time coordinate.
No, it should be clear that only coordinate transformations are allowed, not projections. Projections change the manifold, not just the coordinates. If you were talking about simple coordinate transformations then you would be able to provide those coordinate transformations explicitly, which you have never done.Anamitra said:In theoretical physics, general covariance (also known as diffeomorphism covariance or general invariance) is the invariance of the form of physical laws under arbitrary differentiable coordinate transformations.
It should be clear that the coordinate transformations involved may of projection type.
DaleSpam said:No, it should be clear that only coordinate transformations are allowed, not projections. Projections change the manifold, not just the coordinates.
This is true, but not relevant to the definition of diffeomorphism covariance. Read the definition you cited above carefully. It says that diffeomorphism covariance means that the form of the laws are invariant under coordinate transformations. It doesn't say that all transformations where the form of the laws are unchanged are diffeomorphisms.Anamitra said:1.The geodesic equations remain invariant in all manifolds
2. Maxwell's equations preserve their form in all manifolds--they remain invariant in form on transition from one manifold to another.
Diffeomorphisms, or coordinate transformations.DaleSpam said:Would you like to call them transformations where ds^2 is preserved?
Anamitra said:1.The geodesic equations remain invariant in all manifolds
2. Maxwell's equations preserve their form in all manifolds--they remain invariant in form on transition from one manifold to another.
When we use the term "invariant" we have in our view certain transformations.
What are these transformations in the above two cases?
Would you like to call them transformations where ds^2 is preserved?
Remember that if you have a diffeomorphism [itex]\phi :M \mapsto N[/itex] then [itex]\phi _{*}\mathbf{T}[/itex] can be expressed as [itex]T^{\mu ...}_{\nu ...} = \frac{\partial x^{\mu }}{\partial x^{\alpha }}...\frac{\partial x^{\beta }}{\partial x^{\nu }}...T^{\alpha ...}_{\beta ...}[/itex]. A map from from one manifold to another that doesn't preserve the structure of the original manifold doesn't need to keep [itex]ds^2[/itex] the same and arbitrary tensor components won't transform according to such a law.Anamitra said:Would you like to call them transformations where ds^2 is preserved?
DaleSpam said:No, it should be clear that only coordinate transformations are allowed, not projections. Projections change the manifold, not just the coordinates. If you were talking about simple coordinate transformations then you would be able to provide those coordinate transformations explicitly, which you have never done.
I think it is time for you to do some homework problems:
1) the ellipsoid problem
2) write the coordinate transformations to "flatten" the Schwarzschild metric
JDoolin said:Consider the Schwarzschild metric:
[tex]c^2 {d \tau}^{2} = \left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \left(1-\frac{r_s}{r}\right)^{-1} dr^2 - r^2 \left(d
\theta^2 + \sin^2\theta \, d\varphi^2\right)[/tex]
where
[tex]r_s= \frac{2 G M}{c^2}[/tex]
I don't understand your point. The standard definition of the Schwarzschild metric is not flat.JDoolin said:(2) The r, θ, and Φ coordinates of the Schwarzschild metric already indicate that those components are definable from an external reference frame comoving with the gravitational mass. Those coordinates can be easily translated into Cartesian coordinates in the external comoving reference frame.
Doesn't it stand to reason that the standard definition of the Schwarzschild metric describe exactly what you are saying, i.e., how to "flatten" the Schwarzschild metric?
DaleSpam said:I don't understand your point. The standard definition of the Schwarzschild metric is not flat.
WannabeNewton said:You can, for example, find a transformation that takes the [itex]t = t_{0}[/itex], [itex]\theta = \pi / 2[/itex] submanifold, of the 4 - manifold normally representing schwarzschild space - time, with the metric [itex]ds^{2} = (1 - \frac{2M}{r})dr^{2} + r^{2}d\phi ^{2}[/itex] that embeds this 2 - manifold in a flat 3 - space with metric [itex]ds^{2} = dr^{2} + r^{2}d\phi ^{2} + dz^{2}[/itex] but this is NOT a coordinate transformation. If you take the embedding map and evaluate the pushforward on, for example, a vector you will see that the vector's components will NOT transform according to the vector transformation law under coordinate transformations. You cannot find a COORDINATE transformation for the schwazrschild metric that makes [itex]R^{\alpha }_{\beta \mu \nu } = 0[/itex] identically.
JDoolin said:I'm not entirely sure what you mean either. I am referring to
[tex]c^2 {d \tau}^{2} = \left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \left(1-\frac{r_s}{r}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)[/tex]
[tex]r_s= \frac{2 G M}{c^2}[/tex]
as the standard definition of the Schwarzschild metric.
I would think that if I can take the coordinates (t, r, θ, Φ) and map them to a cartesian coordinate system plus time (t,x,y,z) then I have effectively mapped the events into a flat space-time. Actually, I guess there is a two-step process, because you have to go from [tex](\tau, r', \theta, \phi) \rightarrow (t,r,\theta,\phi)\rightarrow (t,x,y,z)[/tex]
by means of inverse Schwartzchild metric and then by use of spherical coordinates.
The point is that by the definition of the Schwarzschild metric, and its use of spherical coordinates, all of the events [itex](\tau, r', \theta, \phi)[/itex], (outside the schwarszchild radius) are embedded in a flat spacetime, and can be uniquely identified by values of (t,x,y,z).
Why would I want to set [itex]\theta = \pi / 2[/itex]? What is this requirement of [itex]R^{\alpha }_{\beta \mu \nu } = 0[/itex]? What does that mean, qualitatively, and why is it important?
Are you saying that I can't embed [itex](\tau, r', \theta, \phi)[/itex] into (t,x,y,z); that somehow there is no way to do it uniquely? I sincerely doubt that's true. You merely need to define a reference t=τ=0 event. Well, okay, I see a little bit of a problem since, for symmetry, you'd want to define that event at r=0; which is troublesome because that's under the event horizon--But that wouldn't matter to the person looking from the external frame.) Wouldn't you at least agree that every event outside the Schwarzschild radius can be mapped uniquely to an event (t,x,y,z) in cartesian coordinates?
Along the lines of the previous responses, you can certainly do a coordinate transform from (t,r,θ,Φ) to (t,x,y,z) but the metric expressed in the new coordinates will not be the usual flat spacetime metric regardless of the details of the transform.JDoolin said:I'm not entirely sure what you mean either. I am referring to
[tex]c^2 {d \tau}^{2} = \left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \left(1-\frac{r_s}{r}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)[/tex]
[tex]r_s= \frac{2 G M}{c^2}[/tex]
as the standard definition of the Schwarzschild metric.
I would think that if I can take the coordinates (t, r, θ, Φ) and map them to a cartesian coordinate system plus time (t,x,y,z) then I have effectively mapped the events into a flat space-time. Actually, I guess there is a two-step process, because you have to go from [tex](\tau, r', \theta, \phi) \rightarrow (t,r,\theta,\phi)\rightarrow (t,x,y,z)[/tex]
by means of inverse Schwartzchild metric and then by use of spherical coordinates.
Ben Niehoff said:Here is another variation of the Schwarzschild metric in "isotropic coordinates":
[tex]ds^2 = - \frac{\big(1 - \frac{m}{2R} \big)^2}{\big(1 + \frac{m}{2R} \big)^2} \; dt^2 + \big(1 + \frac{m}{2R} \big)^4 \; (dx^2 + dy^2 + dz^2)[/tex]
[tex]R = \sqrt{x^2 + y^2 + z^2}[/tex]
This is the closest you can get to "cartesian coordinates" for the Schwarzschild geometry. This metric is not flat. The relation between the original Schwarzschild coordinates [itex](t, r, \theta, \phi)[/itex] and the new coordinates [itex](t, x, y, z)[/itex] is given by
[tex]\begin{align} r &= R \big(1 + \frac{m}{2R} \big)^2 \\ x &= R \sin \theta \cos \phi \\ y &= R \sin \theta \sin \phi \\ z &= R \cos \theta \end{align}[/tex]
In general, however, you can't tell whether a metric is flat just by staring at it. In some cases it's obvious, such as Cartesian coordinates, but other cases might not be obvious. The only way to tell if a metric is flat is by computing its Riemann tensor. If (and only if) the Riemann tensor vanishes, then the metric is flat. The coordinates have no intrinsic meaning.
DaleSpam said:Along the lines of the previous responses, you can certainly do a coordinate transform from (t,r,θ,Φ) to (t,x,y,z) but the metric expressed in the new coordinates will not be the usual flat spacetime metric regardless of the details of the transform.
Similarly, you can take the usual flat metric (t,x,y,z) and do a coordinate transform to (t,r,θ,Φ) but the metric expressed in the new coordinates will still be flat. It may not be obvious that it is flat, but if you compute the curvature tensor it will be zero.
If there exists a diffeomorphism from coordinate system A to B and from B to C then there exists one directly from A to C. I wasn't suggesting skipping a step, but just using the direct transform for brevity.JDoolin said:I think you skipped a step. First do a coordinate transform from (t',r',θ,Φ) to (t,r,θ,Φ). Then do the transform from (t,r,θ,Φ) to (t,x,y,z).
No, there is nothing that you can do in two diffeomorphisms that cannot be done in one.JDoolin said:The two together create a mapping of the curved four-dimensional coordinates to a flat four-dimensional coordinates.
Coordinates do NOT have intrinsic meaning in general in GR (redundancy ftw). You can't define the r in the standard schwarzschild metric physically; it is not related to distance from the origin. How could you define the physical meaning of a coordinate when we don't have prescribed coordinate system when we solve the EFEs? I could just as easily transform to Eddington coordinates; how would you physically define the coordinates for those?JDoolin said:Please don't argue your case based on the claim that "whatever I'm saying has no intrinsic meaning." That's nonsense.