Transformation of the Line-Element

[Dale]

First let us consider the Right –hand sides of the relations (1) and (3) of Post #55.
The metric coefficients are different—they represent different manifolds.

I thought you had given up the idea of projecting onto different manifolds a couple of pages ago. You certainly never addressed any of the objections earlier.

If you want to work on projections you probably should learn about projective geometry. One of the most important aspects of projective geometry is that it is non-metric. So the very concept of a metric doesn't make sense in when you are projecting from one manifold to another. Because you lose the metric you also lose distance, time, angles, velocity, mass, etc. After losing all of that you don't have much physics left.

Although you have not said so explicitly, I suspect that the primary reason that you want to work in a projected flat manifold is to regain the uniqueness of parallel transport. Is that correct?

Between (3) and (4) of Post #55
The transformation (dt,dr)-->(dt,dR) follows the transformation rule of the contravariant tensors:

Oh, I missed that. What is the coordinate transformation between r and R?

 

[Anamitra]

Let’s consider a metric given by:

{ds}^{2}{=}{ax1}{dt}^{2}{-}{bx3}{dx1}^{2}{-}{c}{x1}^{2}{dx2}^{2}{-}{e}{x2}{dx3}^{2} ----------------- (1)
Each term on the RHS should have the unit corresponding to length square. I have used constants[a,b,c and e] for maintaining dimensional consistency. The constant”c” on the RHS does not represent the speed of light.
If one uses the transformation:
T’=t/17
x1’=x1/5
x2’=x2/7
x3’=x3/6
But now we consider a new metric lfor instance:
{ds'}^{2}{=}{dt'}^{2}{-}{dx1'}^{2}{dx2'}^{2}{-}{dx3}^{2}
the new metric[generally speaking] will have a different value for ds^2.
Let us go into a more general case:
{ds}^{2}{=}{a}{f}_{0}{(}{x1}{,}{x2}{,}{x3}{)}{dt}^{2}{-}{b}{f}_{1}{(}{x1}{,}{x2}{,}{x3}{)}{dx1}^{2}{-}{c}{f}_{2}{(}{x1}{,}{x2}{,}{x3}{)}{dx2}^{2}{-}{e}{f}_{3}{(}{x1}{,}{x2}{,}{x3}{)}{dx3}^{2} ----------------- (2)
Again, each term on the RHS should have the unit corresponding to length square. I have used constants for maintaining dimensional consistency. The constant”c” on the RHS does not represent the speed of light

We choose transformations of the following type:
{t’}{=}{k}_{0}{(}{x1}{,}{x2}{,}{x3}{)}
{x1’}{=}{k}_{1}{(}{x1}{,}{x2}{,}{x3}{)}

{x2’}{=}{k}_{2}{(}{x1}{,}{x2}{,}{x3}{)}
{x3’}{=}{k}_{3}{(}{x1}{,}{x2}{,}{x3}{)}
K0,k1,k2 and k3are well behaved functions in terms of continuity differentiability etc.
As before,we now consider a new metric lfor instance:
{ds}^{2}{=}{dt'}^{2}{-}{dx1'}^{2}{dx2'}^{2}{-}{dx3}^{2}
For the above relations we have:

{dx’}{=}\frac{dx’}{{dx}^{\alpha}}{dx}^{\alpha}
The rule for tensor transformation[contravariant] holds. It should also hold for the inverse transformations.

So the invariance of ds^2 is not associated with the laws in relation to the transformation of the tensors.
It is an additional condition.
The important consequence that follows is that mutual relationships are preserved in the transformed. The important thing that we should do is to make sure that the dimension/unit of each of the quantities do not change during the transformation. This is the condition that we shall impose on the transformations [at the cost of compromising with the invariance of ds^2]

Consequences: The physical nature of each quantity does not change.
The laws relating to mutual relationships between the physical quantities do not change.[ The individual values of the physical quantities represented by tensors may change: the form of the laws should not change]

[The non-metric nature of projection transformations imply: norms are not preserved,angles are not preserved etc---to state in short:the value of ds^2 changes on transformation. The transformations shown here have similar properties in so far as the non-preservation of ds^2 is concerned]

 

[Dale]

Let’s consider a metric given by:

{ds}^{2}{=}{ax1}{dt}^{2}{-}{bx3}{dx1}^{2}{-}{c}{x1}^{2}{dx2}^{2}{-}{d}{x2}{dx3}^{2} ----------------- (1)
Each term on the RHS should have the unit corresponding to length square. I have used constants for maintaining dimensional consistency. The constant”c” on the RHS does not represent the speed of light.
If one uses the transformation:
T’=t/17
x1’=x1/5
x2’=x2/7
x3’=x3/6
the new metric will have a different value for ds^2.

You need to be careful here to distinguish between the geometric object and its representation in a particular coordinate basis. The geometric object ds² has not changed. What has changed is the coordinate basis and therefore the components used to express the same underlying geometric object in terms of the new coordinate basis.

So, if by "different value" you mean that ds² has different components in the primed basis than in the unprimed basis, then you are correct. But if you mean that ds² itself has changed, then you are incorrect. Can you clarify your meaning?

 

[JDoolin]

When you are finding the length of a space-time interval between two events in special relativity, you can find it is either positive or negative; (c dt)^2 - dx^2 - dy^2 - dz^2

If that turns out to be positive, you know that its a time-like interval, but if it turns out to be negative, you know it's a space-like interval.

What is it that distinguishes between a time-like interval from a space-like interval in the Schwarzschild metric?

Let me go ahead and post what I was going to post before I realized I had that question, keeping in mind that I probably made a BLUNDER in confusing space-like and time-like intervals: Edit: I've struck out the parts of my argument that are flawed because I was thinking of the metric for Δτ instead of Δs.

I have been wondering a great deal lately about the question of physical interpretation of line elements ds and or dτ vs.

dr, in the Schwarzschild metric. I notice, you are mostly focusing on the length of ds. However, I was able to make a

bit of progress thinking about the length of dτ, i.e. the timelike component between two events. Perhaps some insight

into the nature of the time element could help resolve some confusion about the line element.

Consider the Schwarzschild metric:

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<br /> <br /> \theta^2 + \sin^2\theta \, d\varphi^2\right)

where

r_s= \frac{2 G M}{c^2}

There is a fairly straightforward but lengthy process for calculating the coefficient of g_{00} = \left ( \frac<br /> <br /> {\partial \tau}{\partial t} \right )^2 = \left ( 1 - {\frac{2 G M}{c^2 r}} \right ) which I explored in some detail

https://www.physicsforums.com/showpost.php?p=3415913&postcount=18".

we were comparing frequencies, which is essentially the same as comparing Δτ to Δt. Or more specifically, the length of

the four-vector from (0,r,θ,Φ) to (Δt,r,θ,Φ). Δt is the time as measured from the external coordinate system. But from

the point-of-view of an internal coordinate system, that four-vector appears to go from (0,r,θ,Φ) to (Δτ,r,θ,Φ). In

other words, the clock that is located at that point is going to go slower, but that is the only difference.

Since then, I've been trying to figure out how to get the second coefficient,g_{11}=\left(1-\frac{2 G M}{r c^2}\right)^{-1}
, which has been difficult, because I have not even really been able to establish a common-sense interpretation to what that coefficient even means. However, this morning, I think I was able to grasp something.

The key is that instead of considering a time-like interval in (t,r,θ,Φ), we will be considering a space-like interval in the r-component alone. Specifically, where dt=dθ=dΦ, but r is allowed to change. Hence, two events which appear to happen simultaneously, but at different locations outside the gravitational field, [STRIKE]but will appear to happen at different times from within the gravitational field.
This is a significant qualitative (big) difference from the Rindler coordinates, so [/STRIKE]I think it would be enlightening to try to make a distinctive comparison of the Schwarzschild metric to the Rindler metric at this point.[STRIKE]


Why do I say this is a big difference?

Because [/STRIKE] in the Rindler coordinates, events that happen at t=0 also happen at τ=0.

If events seem to happen simultaneously in the "rocket" frame, then they also seem to happen simultaneously in the momentarily comoving rest-frame. Here is a diagram:

In this diagram, the future events, C and D, are on a line of constant τ. However, using a line of constant t, it appears that event C will happen before event D. As time passes, the line of constant τ will rotate (technically a Lorentz skew Transform, with eigenvectors of slope ±c) down and coincide with the line of constant t. At the point in time where event C and D actually HAPPEN, they are simultaneous.

Similarly, though events A and B appear to HAVE happened at different times, at the time that they actually HAPPENED, (when they crossed the t=0 line) they were simultaneous.

In the Rindler coordinates, events that appear to be simultaneous in the rocket also appear to be simultaneous in the momentarily comoving reference frame. But this is the point that surprised me: In the schwarzschild coordinates, events which appear to be simultaneous from the external reference frame DO NOT appear to be simultaneous locally.

[STRIKE]That means that the t=0 line and the τ=0 line will only intersect at one point, and do not represent the same set of events. I'm not sure what the implications of this are. [/STRIKE]

But now that this has occurred to me, I wonder whether anyone has a graph of Schwarzschild coordinates just showing curves of constant τ and curves of constant r' in the Schwarzschild metric, analogous to the image I posted above for the Rindler coordinates, but perhaps on a space-time scale near a body of high gravity?
​

 

[JDoolin]

There still is a big difference from the Rindler coordinates, in that using the Rindler coordinates, you will find that the length of a meter-stick is the same on the rocket as it is in the momentarily comoving rest frame.

Whether I understand why its this way or not, let me see if I understand the meaning of what's going on with the ds element:

For instance, let's calculate the length of a ruler as observed at the surface of the earth, seen from a stationary position in outer-space.

We use two simultaneous events at t=0 at the top of the ruler and the bottom of the ruler, so dt=0, dθ=0, dΦ=0, and since the scale factor doesn't change much over the length of the ruler, we can replace ds and dr with Δs and Δr:

\Delta s = \sqrt{\frac{1}{1 - \frac{2 G M}{r c^2}}}\Delta r

Setting \begin{matrix} G=6.673 \times 10^{-11} m^3/kg \cdot s^2\\ M=5.9742 \times 10^{24} kg\\ r=6.3781 \times 10^6 m\\ c=299792458 m/s \end{matrix}

Then \Delta s = \sqrt{\frac{1}{1 - 1.39 \times 10^{-9}}}\Delta r

which means that locally, the ruler will be just a tiny tiny tiny bit longer than it appears to be from space.

I think I may be saying that wrong, though. If I take a ruler from space, and put it on earth, it will look shorter from space, but measure the same length on earth? If I take a clock from space, and put it on earth, the clock really runs slower, but the scale of time on the Earth is determined by the clock; which is really slower deeper in the gravitational well. But if I take a ruler from space, and put it on the Earth vertically, then the ruler looks shorter from space, but it's the real scale of the vertical space, and you'd need to have a longer ruler to look like it was the same length from space.

Maybe there are better ways to express that.

But basically, as observed from space, an object carried (not falling) down into a gravity well, and held vertically will appear to get shorter and shorter as it goes down, according to the Schwarzschild metric, right?

 

[Anamitra]

I have done a simple modification on Post #62. I have repeated the relevant portion here:

Let us consider a general[stationary] case:
{ds}^{2}{=}{a}{f}_{0}{(}{x1}{,}{x2}{,}{x3}{)}{dt}^{2}{-}{b}{f}_{1}{(}{x1}{,}{x2}{,}{x3}{)}{dx1}^{2}{-}{c}{f}_{2}{(}{x1}{,}{x2}{,}{x3}{)}{dx2}^{2}{-}{e}{f}_{3}{(}{x1}{,}{x2}{,}{x3}{)}{dx3}^{2} ----------------- (2)
Each term on the RHS should have the unit corresponding to length square. I have used constants for maintaining dimensional consistency. The constant”c” on the RHS does not represent the speed of light

We choose transformations of the following type:
{t’}{=}{k}_{0}{(}{x1}{,}{x2}{,}{x3}{)}
{x1’}{=}{k}_{1}{(}{x1}{,}{x2}{,}{x3}{)}

{x2’}{=}{k}_{2}{(}{x1}{,}{x2}{,}{x3}{)}
{x3’}{=}{k}_{3}{(}{x1}{,}{x2}{,}{x3}{)}
K0,k1,k2 and k3are well behaved functions in terms of continuity differentiability etc.
But we now consider a new metric for instance:
{ds&#039;}^{2}{=}{dt&#039;}^{2}{-}{dx1&#039;}^{2}{dx2&#039;}^{2}{-}{dx3}^{2}
For the above relations we have:

{dx’}{=}\frac{dx’}{{dx}^{\alpha}}{dx}^{\alpha}
The rule for tensor transformation[contravariant] holds. It should also hold for the inverse transformations.

So the invariance of ds^2 is not associated with the laws in relation to the transformation of the tensors.
It is an additional condition.
The important consequence that follows is that mutual relationships are preserved in the transformed. The important thing that we should do is to make sure that the dimension/unit of each of the quantities do not change during the transformation. This is the condition that we shall impose on the transformations [at the cost of compromising with the invariance of ds^2]

Consequences: The physical nature of each quantity does not change.
The laws relating to mutual relationships between the physical quantities do not change.[ The individual values of the physical quantities represented by tensors may change: the form of the laws should not change]

It is important to note that in a transformation where ds^2 changes g(mu,nu) does not behave like a second rank tensor.

For transformations where ds^2 remains unchanged :
{g&#039;}_{\mu\nu}{dx&#039;}^{\mu}{dx&#039;}^{\nu}{=}{g}_{\alpha\beta}{dx}^{\alpha}{dx}^{\beta}
Therefore we may show that:
{g&#039;}_{\mu\nu}{=}\frac{{dx}^{\alpha}}{{dx}^{\mu}}{\frac}{{dx}^{\beta}}{{dx^\nu}}{g}_{\alpha\beta}

If ds^2 is not preserved in a transformation then g(mu,nu) will not behave as a covariant tensor.

But so far our transformations are concerned the infinitesimals[dt,dx1,dx2,dx3] will continue to behave as contravariant tensors.

[The non-metric nature of projection transformations imply: norms are not preserved,angles are not preserved etc---to state in short:the value of ds^2 changes on transformation. The transformations shown here have similar properties in so far as the non-preservation of ds^2 is concerned]

 

[Anamitra]

In view of the previous posting,relationships which do not contain g(mu,nu) explicitly will remain unmodified.
The relation:

{A}^{\mu}{B}^{\nu}{=}{C}^{\mu}{D}^{\nu}

Should be preserved in the new frame for vectors that satisfy.

{A&#039;}^{\alpha}{=}{k}\times\frac{{dx&#039;}^{\alpha}}{{dx}^{\mu}}{{A}^{\mu}}

K is a scale factor which may be point[spacetime] dependent. The scale factor has been taken since the norm/mod of a vector changes in our transformation [where ds is not preserved]
[If each component equation is preserved [due to the cancellation of the scale factor at the point concerned], the overall equation is also preserved]

[For the transformation of the coordinate values of tensor components the scale factor is not needed. But for the physical values of these components the scale factor is needed:you may see post # 73 for this]

 

[Anamitra]

Transformation of g(mu,nu) when ds^2 changes.

Point-wise transformation of g(mu,nu) may also be included in the scale factor mentioned in the previous post.

Let

{ds&#039;}^{2}{=}{f}{(}{x}{,}{y}{,}{z}{)}{ds}^{2}
Or,
{g&#039;}_{\mu\nu}{dx}^{\mu}{dx}^{\nu}{=}{f}{(}{x}{,}{y}{,}{z}{)}{g}_{\alpha\beta}{dx}^{\alpha}{dx}^{\beta}

Or,
{g&#039;}_{\mu\nu}{=}{Scale}{\,}{Factor}\times\frac{{dx}^{\alpha}}{{dx}^{\mu}}\frac{{dx}^{\beta}}{{dx}^{\nu}}{g}_{\alpha\beta}

Relations in general should be preserved

In Post #55 we had the relation

E=Constant ,for geodetic motion [E: Energy per unit rest mass]

The relation holds true for both the manifolds , Schwarzschild's Geometry and Flat space-time.

 

[Dale]

Anamitra, when someone asks for clarification it is both useless and rude to proceed without answering the question.

 

[Anamitra]

You need to be careful here to distinguish between the geometric object and its representation in a particular coordinate basis. The geometric object ds² has not changed. What has changed is the coordinate basis and therefore the components used to express the same underlying geometric object in terms of the new coordinate basis.

So, if by "different value" you mean that ds² has different components in the primed basis than in the unprimed basis, then you are correct. But if you mean that ds² itself has changed, then you are incorrect. Can you clarify your meaning?

The value of ds^2 itself is changing in view of the modification.

 

[Dale]

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?

 

[Anamitra]

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?

I am not considering ds^2 as a tensor.
ds^2 is defined by the relation

{ds}^{2}{=}{g}_{\mu\nu}{dx}^{\mu}{dx}^{\nu}
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

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.

 

[Anamitra]

We may consider the following metric in the rectangular x,y,x,t system:

{ds}^{2}{=}{dt}^{2}{-}{dx}^{2}{-}{dy}^{2}{-}{dz}^{2} ---------------- (1)

The same metric in the r,theta ,phi,t reads

{ds}^{2}{=}{dt}^{2}{-}{dr}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{sin}^{2}{\theta}{d}{\phi}^{2}{)} --------------- (2)
ds^2 has not changed

Now in the same r,theta,phi system we may consider the metric:
{ds&#039;}^{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}{)}
--------------- (3)
ds^ 2 has now changed for identical values of dt,dr,d(theta) and d(phi)
The relationship between x,y,z,t and r,theta,phi,t is not changing when we are considering (1) and (2) or (1) and (3)

Each term on the RHS of (3) has a scale factor[=1 in some cases] wrt to each corresponding terms in (2)

 

[Dale]

I am not considering ds^2 as a tensor.
ds^2 is defined by the relation

{ds}^{2}{=}{g}_{\mu\nu}{dx}^{\mu}{dx}^{\nu}

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.

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

This is incorrect. The coordinate transformations listed are valid diffeomorphisms. ds² is unchanged under any diffeomorphism.

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.

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.

 

[Dale]

Now in the same r,theta,phi system we may consider the metric:
{ds&#039;}^{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}{)}
--------------- (3)
ds^ 2 has now changed for identical values of dt,dr,d(theta) and d(phi)

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.

Simply calling one quantity A and another quantity A' does not mean that they are in any way related to each other.

 

[Anamitra]

Suppose we have spherical body with its associated Schwarzschild's Geometry.Equation (3) in post #73 is valid and ds^2 is given by it.

Now due to some catastrophic effect the entire mass gets annihilated and the energy flies off to infinity. The present situation is described by the second or the first equation of the same posting. We have a new value of ds^2 on the same coordinate grid[t,r,theta,phi, system]

Thus a physical process can cause a change in the metric coefficients in the same coordinate grid leading to a change of the value ds^2.

In the present context our transformation relates to the change of the metric coefficients wrt the same coordinate grid. This may be due to the redistribution of mass or due to annihilation.

 

[Dale]

I have no problem with that, but note that is a physical change, not merely a coordinate transformation. You have a new metric, a new manifold, and as I said above ds² is completely unrelated to ds'².

You can do physics in either manifold, but there is no way to transform the results of physics in one to get answers in the other since they are not related by a coordinate transformation. In particular, ds'² may tell you how fast a clock is ticking in the new manifold but it cannot tell you about gravitational time dilation in the previous manifold.

 

[Anamitra]

to understand let us consider a time dependent metric :

{ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2} ----------- (1)

The coefficients, g(mu,nu) are time dependent ones.We assume that the metric coefficients remain time independent for one hour. Then they change for the next 10 minutes and again they become constant for the next one hour.[The next change may be of a different type]The process goes on repeating.
The same metric is taking care of different stationary manifolds --each manifold corresponding to the period when the metric coefficients are not changing. Several different manifolds are coming under the fold of the same metric
[The ds^2 values of the different stationary manifolds are related to each other by the general nature of the time dependent metric ,that is the overall metric given by (1)]

 

[Anamitra]

We may replace the variable time in the metric coefficients[of equation (1) in the last posting] by some variable lambda which is not time.This would give us an ensemble picture.[Of course dt^2 has to be maintained as before].
Different values of lambda would give us different stationary manifolds---having different values of ds^2

 

[Dale]

Sure. But since your metric is asymmetric wrt t or lambda you cannot take results from one t or lambda and use them for doing physics at another t or lambda. It is just not the same physical situation.

Similarly you cannot use results at one r to make predictions about another r in the Schwarzschild metric. 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.

 

[Anamitra]

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.

The dependence of proper acceleration on the radial distance [in general on the spacetime coordinates]will be our law.

 

[Dale]

Yes, and in these other non-static spacetimes you describe it will depend on time as well.

 

[Anamitra]

A Thought Experiment to Perform
We are on initially flat spacetime and we undertake to perform experiments on geodesic motion, Maxwell's equations and other equations that may be expressed in tensor form.Then gravity is turned on and the field is increased in stages. Well the equations we were considering do remain unchanged so far as they are expressed in tensor form.
We also perform some other experiments simultaneously.
While in flat space-time we consider the length of an arrow lying on the ground. This arrow may represent a vector.We stick labels on its tip and tail.When the field is turned on the physical distance between the labels should change due to change in the values of g(mu,nu) .
One may consider a cuboidal box[on the ground] with the main diagonal representing a vector.
When the field[gravitational] is turned on the physical lengths of the components change.The norm as well as the orientation of the vector changes[in short, ds^2 changes].This is something similar to what is supposed to happen in projection geometry.

Actually the laws remain unchanged so long we are able to express them in tensor form.

Now let us think in the reverse direction.We are in curved space-time and the field is turned off gradually in stages. Finally we reach flat spacetime. The laws[in tensor form] remain unchanged at each and every stage of the process/experiment

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.

 

[Dale]

Actually the laws remain unchanged so long we are able to express them in tensor form.

Certainly. Do you understand the difference between a law of physics and the boundary conditions?

Changing coordinates does not change the boundary conditions, changing the curvature does change the boundary conditions.

 

[Ben Niehoff]

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.

Here you've given a vague description of some kind of procedure. To understand your procedure better, you should do an explicit calculation on a specific example. I want you to do the following example, because it will reveal the error in your reasoning:

The metric of an ellipsoid is

\begin{align*}ds^2 &amp;= \Big( (a^2 \cos^2 \phi + b^2 \sin^2 \phi) \; \cos^2 \theta + c^2 \sin^2 \theta \Big) \; d\theta^2 + (a^2 \sin^2 \phi + b^2 \cos^2 \phi) \; \sin^2 \theta \; d\phi^2 \\ &amp;\qquad \phantom{a} + 2 (b - a) \cos \theta \sin \theta \cos \phi \sin \phi \; d\theta \; d\phi \end{align*}

where a, b, c are some constants. Now, Alice is sitting at \theta = \pi/4, \; \phi = 5\pi/6 and has local velocity vector 2 \partial_\theta - \partial_\phi at that point. Bob is sitting at \theta = \pi/2, \; \phi = \pi/4 and has local velocity vector 2 \partial_\phi at that point.

What is the difference between their vectors? Remember that the vectors are at different points on the ellipsoid! Do the calculation explicitly, using the method you have described. If you do not attempt to do your calculation explicitly, I will assume you are not interested in learning why your reasoning is wrong.

Alternatively, if you do realize what is wrong with your reasoning, then explain.

 

[Anamitra]

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.

 

[Anamitra]

General Covariance/Diffeomprphism covariance:
[Link: http://en.wikipedia.org/wiki/General_covariance ]
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. One has to include transformations wherer ds^2 change, where this quantity is no longer an invariant[this has been indicated in post #83: The fundamental laws should have the same form in all manifolds---flat spacetime and the different varieties of distinct curved spacetimes]. Interestingly for such transformations, as we pass between different manifolds, g(mu,nu) is not a tensor of rank two[since ds^2 is not preserved].Norms ,angles and dot products need not be preserved in these transformations[where ds^ changes].

 

[Ben Niehoff]

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.

There is no time coordinate. Erase the word "velocity" if it makes you feel better.

You claimed that you could define a canonical way to compare vectors at different points of a general curved manifold. So I am asking you to demonstrate your method on this manifold. Call Alice's vector A, and Bob's vector B. Is there any sensible way to define the vector C = A - B? If so, what is it? If not, why not? Explain, show work, etc.

I chose an ellipsoid because it has no continuous symmetries. Hopefully that will get you to think about it properly.

 

[Dale]

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.

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

 

[Anamitra]

No, it should be clear that only coordinate transformations are allowed, not projections. Projections change the manifold, not just the coordinates.

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?