# Transformation of the Line-Element

by Anamitra
Tags: lineelement, transformation
Mentor
P: 16,489
 Quote by Anamitra Let us assume that the law changes on passing from the old frame to the new frame[flat spacetime].
You are not changing from an old frame to a new frame, that is just a coordinate transformation. You are changing to a new Riemannian manifold. You cannot do what you are talking about simply by changing frames.
Mentor
P: 16,489
 Quote by Anamitra Calculations are much simpler in flat spacetime context
That is highly unlikely. I would have to see a clear example to believe that.
 P: 621 Lets come to a basic issue: Consider two infinitesimally close spacetime points A ans B[on the spacetime surface]. The separation between them is given by: $${ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}-{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}$$ -------------------- (1) Now I connect A and B by several paths[long ones or infinitesimally small ones] lying on the spacetime surface and calculate integral ds for different paths. In all likelihood I would be getting different values. In fact if you connect A and B by one million infinitesimal paths [on the spacetime surface] we may get different values of pathlength. Is there any guarantee that a physical path should exist between A and B whose value is given by (1) To what extent ds itself is a perfect differential is an issue to rekcon with! The General Relativity metric at most is a valid concept only in a path dependent situation.
 P: 621 Radial Motion in Schwarzschild's Geometry Metric for Radial Motion $${ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}$$ --- (1) $$\frac{{ds}^{2}}{{1}{-}\frac{2m}{r}}{=}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-2}{dr}^{2}$$ --- (2) $${ds’}^{2}{=}{dt}^{2}{-}{dR}^{2}$$ -------------- (3) LHS of the above corresponds to the fact that ds itself is a path dependent quantity. dR is defined by: $${dR}{=}{(}{1}{-}\frac{2m}{r}{)}^{-2}{dr}$$ $${R-R0}{=}{\int}{(}{1}{-}\frac{2m}{r}{)}^{-2}{dr}$$ The integral on RHS is a definite integral and so the constant which may contain time gets eliminated. [It should be noted that $$\frac{{\partial}{R}}{{\partial}{t}}{=}{0}$$ ] Equation (3) is a flat spacetime metric which allows non local velocities, relative velocities etc [One may take the definition of radial speed as V=dR/dt] ds and ds’ are not exact differentials[they are path dependent quantities] and one must not use the the formula for exact differentials to expand them. Whenever we apply a General Relativity metric we have a path in our mind. The metric itself expresses a relationship between quantities which are not perfect differentials[though we use "d" for ds]
 Mentor P: 16,489 ds' is no longer invariant, i.e. it is not a metric
 P: 621 We are supposed to choose transformations wrt which ds' is invariant. Suppose: t=F(k) R=F1(m) $${ds'}^{2}{=}{d}{(}{F}{(}{k}{)}{)}^{2}{-}{d}{(}{F1}{(}{m}{)}{)}^{2}$$ ---- (1) [(k,m) are the new coordinates.] $${ds'}^{2}$$ is not changing. Arbitrary transformations like orthogonal projections must be excluded! [This has already been suggested by you]
 Mentor P: 16,489 Yes, it is changing. ds' is now a function of r. In other words, according to ds' two identically constructed clocks at different r will measure different amounts of proper time per tick. ds' is not the line element, it is not the metric. The expression on the rhs is not a flat spacetime, and you cannot do parallel transport or covariant differentiation with it. Furthermore, your laws of physics now change from place to place.
 P: 621 s was a function of r ant t ds was a function of r and t ds'=ds/F(r) s' is a function of r and t. Alternatively ds' is a function of t and R From equation (3) of post #22 ds' is a line element in flat spacetime.
 Mentor P: 16,489 No, the line element is not a function of r and t because the connection is metric-compatible. This is, in fact, the whole point of having a metric-compatible connection.
P: 621
 Quote by DaleSpam Yes, it is changing. ds' is now a function of r. In other words, according to ds' two identically constructed clocks at different r will measure different amounts of proper time per tick.
Lets take Schwarzschild's Equation for radial motion:

$${ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}$$
We take two points along the radial line for which
1.The values of r are different[say r1 and r2]
2. Same values dt and dr are considered.

ds will be different for the two points.

The only thing is that ds has to be invariant--that too wrt to a certain class of transformation-----for which ds is invariant

[For any pair of events dt and dr should be the same for all observers]
Mentor
P: 16,489
 Quote by Anamitra Lets take Schwarzschild's Equation for radial motion: $${ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}$$ We take two points along the radial line for which 1.The values of r are different[say r1 and r2] 2. Same values dt and dr are considered. ds will be different for the two points.
That shows that dt and dr functions of r, not that ds is.

If you want to show whether or not something is constant wrt some variable you take a derivative, in this case a covariant derivative. The Levi-Civta connection is metric compatible, meaning that the covariant derivative of the metric is 0, but not your modified metric.
 P: 621 The parallel Transport issue: Suoppse you are parallel transporting (dt,dr) in your system and I am parallel transporting (dt,dR) in my rectangular system. In my case the components dt and dR are not supposed to change in magnitude individually at any point of the transport[parallel-transport] For my parallel transport: $${dR}{=}{[}{{1}{-}\frac{2m}{r}}{]}^{-1}{dr}$$ Should not change at any point of the journey. “dt” should not change also.[Norms of the component vectors are preserved in a parallel transport for each and every point of the movement] For your case: $${[}{1}{-}\frac{2m}{r}{]}^{-1/2}{dr}$$ should not change And $${(}{1}{-}\frac{2m}{r}{)}^{1/2}dt$$ should not change. The norm of each vector dt and dr should not change Now let me go back to my parallel transport for a moment again: $${dR}{=}{[}{{1}{-}\frac{2m}{r}}{]}^{-1}{dr}$$ $${dR}{=}{[}{{1}{-}\frac{2m}{r}}{]}^{-1/2}{dr}{*}{[}{{1}{-}\frac{2m}{r}}{]}^{-1/2}{=}{Const}$$ Which means: $${[}{1}{-}\frac{2m}{r}{]}^{-1/2}{dr}$$ is not constant One may apply similar logic to the time component So far as my motion[parallel transport] is concerned ds is not constant . So far as your parallel transport is concerned ds is constant! [For parallel transport the magnitude of the component vectors do not change-the individual magnitudes/norms are preserved at each point of the transport]
 P: 621 Let us have a look at the metric: $${ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}$$ The left side is dot product between the identical vectors each being given by(dt,dx1,dx2,dx3) Now the dot product is a scalar. The LHS ie ds^2 is a scalar since it is the result of a dot product. One should not apply covariant differentiation to it--ordinary differentiation should be applied to the LHS and hence to the RHS
 P: 621 Points to Observe: 1. ds^2 is a scalar . It is invariant wrt to coordinate transformations[ a class of transformations] 2. For parallel transport of the vector(dt,dr), ds^2 does not change. The same points hold for the metric that I have introduced [in Relation 3 in Post #22]: 1. ds'^2 is a scalar 2. ds'^2 does not change for the parallel transport of(dt,dR) [see Posts #30 and #33]
 P: 621 Some Details in Relation to Post #30 Schwarzschild’s Metric for Radial Motion: $${ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}$$ Reduced metric: $${ds’}^{2}{=}{dt}^{2}{-}{dR}^{2}$$ Where, $${dR}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}^{-2}{dr}^{2}$$ $${ds’}^{2}{=}\frac{{ds}^{2}}{{1}{-}\frac{2m}{r}}$$ For the parallel transport of my vector (dt,dR) 1. dt=k1 => $${(}{1}{-}\frac{2m}{r}{)}^{-1/2}{(}{1}{-}\frac{2m}{r}{)}^{+1/2}{dt}{=}{k1}$$ =>$${(}{1}{-}\frac{2m}{r}{)}^{+1/2}{dt}{=}{k1}{(}{1}{-}\frac{2m}{r}{)}^{+1/2}$$------------------------------(1) 2. dR=k2 =>$${(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}{=}{k2}$$ $${(}{1}{-}\frac{2m}{r}{)}^{-1/2}{(}{1}{-}\frac{2m}{r}{)}^{-1/2}{dr}{=}{k2}$$ Therefore, $${(}{1}{-}\frac{2m}{r}{)}^{-1/2}{dr}{=}{k2}{(}{1}{-}\frac{2m}{r}{)}^{1/2}$$------------------------------(2) From (1) and (2) ds^2 for the parallel transport of my vector is given by: $${ds}^{2}{=}{k1}^{2}{(}{1}{-}\frac{2m}{r}{)}{-}{k2}^{2}{(}{1}{-}\frac{2m}{r}{)}$$ Since k1 and k2 are constants for my parallel transport, ds^2 is not a constant for this transport Rather, $$\frac{{ds}^{2}}{{1}{-}\frac{2m}{r}}{=}{k1}^{2}{-}{k2}^{2}{=}{Const}$$ That is, ds’^2 is constant for parallel transport of (dt,dR) [It is to be noted that k1 and k2 are constants]
Mentor
P: 16,489
 Quote by Anamitra [For parallel transport the magnitude of the component vectors do not change-the individual magnitudes/norms are preserved at each point of the transport]
That is part of the definition of parallel transport. Although it is true, it is uninformative because it is true for all vectors.

What we are interested in is not the parallel transport of some vector or scalar, but the covariant derivative of the metric ds²=g and your ds'²=g', which are rank 2 tensors. If you compute the covariant derivatives you get:
$$\nabla_{\eta}g_{\mu \nu} = 0$$
and
$$\nabla_{\eta}g'_{\mu \nu} \ne 0$$
specifically, there are 4 non-zero components
$$\nabla_{r}g'_{tt} = -\frac{R}{r(r-R)}$$
$$\nabla_{r}g'_{rr} =\frac{rR}{(r-R)^3}$$
$$\nabla_{r}g'_{\theta \theta} =\frac{r^2 R}{(r-R)^2}$$
$$\nabla_{r}g'_{\phi \phi} =\frac{r^2 R \sin^2(\theta)}{(r-R)^2}$$

So the ds² is constant everywhere, and the dependence that you would expect by naively looking at the expression is simply the coordinates. But ds'² is not constant in a covariang sense; instead, ds'² changes wrt changes in r.

 Quote by Anamitra Points to Observe: 1. ds^2 is a scalar . ... 1. ds'^2 is a scalar
 Quote by Anamitra Let us have a look at the metric: $${ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}$$ The left side is dot product between the identical vectors each being given by(dt,dx1,dx2,dx3) Now the dot product is a scalar. The LHS ie ds^2 is a scalar since it is the result of a dot product. One should not apply covariant differentiation to it--ordinary differentiation should be applied to the LHS and hence to the RHS
This is not correct, but the notation is indeed sloppy and can be confusing. For reference please see Sean Carrol's Lecture Notes on General Relativity (http://lanl.arxiv.org/abs/gr-qc/9712019v1) on page 48 in the first full paragraph following equation 2.32.

In the expression for the line element the dt and similar terms are not infinitesimal displacements, but rather they are basis dual vectors (one-forms) and the dt² refers to a tensor product, not a dot product. So the line element is a rank 2 tensor, not a scalar. Perhaps this is the source of some of your confusion.
 C. Spirit Sci Advisor Thanks P: 4,941 But at each point p on a manifold $g_{p}: T_{p}(M) \times T_{p}(M)\mapsto \mathbf{R}$ and $ds^{2} = g_{\mu \nu }dx^{\mu }\otimes dx^{\nu }$ so how could $ds^{2}$ be a 2 - tensor when it is the result of the metric tensor mapping the members of the tangent space at p (or across the entire manifold by mapping all members of the tangent bundle if the distinction matters) to the reals?
 Mentor P: 16,489 Because the notation is sloppy and inconsistent. See the link I posted above. Sean Carroll explains it far better than I can. But in the form that Anamitra has been discussing (${ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}-{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}$) it is clear that ds² is a rank 2 tensor, not a scalar.

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