Can Special Relativity Solve the Twin Paradox in a Gravitational Field?

[lightarrow]

It is possible to draw world-lines of the two twins in the non-accelerating one (A) frame of ref. The aging of the other (B), with respect to the first, is evaluated integrating its proper time differential (and comparing it with A's proper time).

My question is (certainly already asked): is there a difference, about our ability to draw those diagrams and make that computation, if B accelerates because of a star's gravitational field and not because of his rocket's engines?

Could we avoid knowing GR and using SR only, in this case? If the answer is yes, in which other cases (where B is accelerating) would be possible to use SR only?

 

[pervect]

It is possible to draw world-lines of the two twins in the non-accelerating one (A) frame of ref. The aging of the other (B), with respect to the first, is evaluated integrating its proper time differential (and comparing it with A's proper time).

My question is (certainly already asked): is there a difference, about our ability to draw those diagrams and make that computation, if B accelerates because of a star's gravitational field and not because of his rocket's engines?

Could we avoid knowing GR and using SR only, in this case? If the answer is yes, in which other cases (where B is accelerating) would be possible to use SR only?

While one does have to use GR when one is in the gravitational field of a massive body, not very much of GR is needed.

Basically, one has to replace the line element used in SR

$$d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$$

by the more general line element for GR

$$d\tau^2 = \sum_{i,j} g_{ij} dx^i dx^j$$

Specifically, for a single massive body, one would use the Schwarzschild metric, and one would typically write out the sum as

$$d\tau^2 = (1 - r/r_s) dt^2 - \frac{dr^2}{1-r/r_s} - r^2 d\theta^2 - r^2 \sin^2 \theta d\phi^2$$

which equates $dx^0$=dt, $dx^1$=dr, $dx^2$=$d\theta$ $dx^3$ = $d\phi$

and replaces the general metric with the Schwarzschild metric, so that for instance $g_{00}$ = (1-r/$r_s$) and $g_{11}$= $$\frac{1}{1-r/r_s}$$

$r_s$ is the Schwarzschild radius

One solves for $d\tau$ by taking $\sqrt{d\tau^2}$ which is given by the metric, and then one integrates $d\tau$ in the usual manner you describe, just as one does for SR, except that one uses the general metric to compute $d\tau$ rather than the flat Minkowski metric.

 

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So, if twin B (the one who accelerates) doesn't look out of the window, how can he establish that he accelerates because of a grav. field and not because of his rockets and so, that he have to use another metric instead of the Minkowsky one?

 

[pervect]

So, if twin B (the one who accelerates) doesn't look out of the window, how can he establish that he accelerates because of a grav. field and not because of his rockets and so, that he have to use another metric instead of the Minkowsky one?

An observer in a flat Minkowski space-time does not HAVE to use a Minkowski metric, it's just the easiest and simplest choice. An accelerating observer in a Minkowski space-time might chose to use Rindler coordinates, for instance.

Take a look at http://www.eftaylor.com/pub/chapter2.pdf

This has some material from one of Taylor's book I'll quote a brief excerpt:

Nothing is more distressing on first contact with the idea of curved spacetime
than the fear that every simple means of measurement has lost its
power in this unfamiliar context. One thinks of oneself as confronted with
the task of measuring the shape of a gigantic and fantastically sculptured
iceberg as one stands with a meterstick in a tossing rowboat on the surface
of a heaving ocean.

Were it the rowboat itself whose shape were to be measured, the procedure
would be simple enough (Figure 1). Draw it up on shore, turn it
upside down, and lightly drive in nails at strategic points here and there
on the surface. The measurement of distances from nail to nail would
record and reveal the shape of the surface. Using only the table of these
distances between each nail and other nearby nails, someone else can
reconstruct the shape of the rowboat. The precision of reproduction can be
made arbitrarily great by making the number of nails arbitrarily large.
It takes more daring to think of driving into the towering iceberg a large
number of pitons, the spikes used for rope climbing on ice. Yet here too the
geometry of the iceberg is described—and its shape made reproducible—
by measuring the distance between each piton and its neighbors.But with all the daring in the world, how is one to drive a nail into spacetime
to mark a point? Happily, Nature provides its own way to localize a
point in spacetime, as Einstein was the first to emphasize. Characterize the
point by what happens there: firecracker, spark, or collision! Give a point
in spacetime the name event

So a metric is defined by setting up a number of events in space-time, and assigning them coordinates.

If the events are close enough to each other, the distance (in GR, the Lorentz interval) will be determined by some quadratic form written in terms of the coordinates. The coefficients of this quadratic form are the metric.

It might or might not be obvious, but one can recover the notion of distances and times given the Lorentz intervals between all pairs of points. I don't want to digress on how this is done right at the moment, but if you want more info, ask.

So you can have many metrics that describe the same physical situation, by chosing different coordinates, i.e. assigning different numbers for the coordinates of some physically defined set of events.

If you know the metric coefficients in some particular neighborhood, there is a way to determine if the underlying space-time is flat, but it's a bit technical. One basically computes the curvature tensor - if the space-time is flat, there will be no curvature, i.e. all the components of the curvature tensor will vanish.

By this procedure, an accelerating observer using Rindler coordinates and an inertial observer using some Minkowski coordinates will have different coordinates for the same points, and also will have different metrics.

Both will, however, compute a zero curvature tensor.

I hope this helps some.

 

[lightarrow]

Sorry if I keep asking, even if you have already answered; I have understood something but not everything.
My question is: having all the necessary events for the twins, can I avoid using GR and use SR only if I find an inertial frame of ref. with respect to which I draw the twin's world lines?

 

[MeJennifer]

An observer in a flat Minkowski space-time does not HAVE to use a Minkowski metric, it's just the easiest and simplest choice.

An observer in Minkowski space has no choice but to use the Minkowski metric, but he is free to choose any suitable coordinate chart.

So you can have many metrics that describe the same physical situation, by chosing different coordinates, i.e. assigning different numbers for the coordinates of some physically defined set of events.

You are really describing coordinate charts here not metrics.

The difference is not very important in flat spaces but in curved spaces the difference is essential.

 

[pervect]

An observer in Minkowski space has no choice but to use the Minkowski metric, but he is free to choose any suitable coordinate chart.


You are really describing coordinate charts here not metrics.

The difference is not very important in flat spaces but in curved spaces the difference is essential.

I tend to use "metric" and "coordinate chart" interchangeably.

Most people who make a big deal of the distinction are mathematicians rather than physicists.

Looking at http://mathworld.wolfram.com/MinkowskiMetric.html

it appears to me that the Minkowski metric is defined in terms of a particular coordinate chart, and that metrics with a vanishing Riemann may be equivalent to the Minkowski metric but are not identical to it.

This URL may only prove that other people are as sloppy as I am about the distinction between the two.

 

[pervect]

Sorry if I keep asking, even if you have already answered; I have understood something but not everything.
My question is: having all the necessary events for the twins, can I avoid using GR and use SR only if I find an inertial frame of ref. with respect to which I draw the twin's world lines?

If you can find some inertial frame of reference, then you can use SR to draw the twins worldlines on a flat sheet of paper, using the coordinates of that inertial frame.

If you can't find some inertial frame of reference (for instance if you have gravity), then you'd have to draw your SR diagram on some non-flat sheet of paper to get the right answer.

Often times the gravitational effects are small, and the error involved in using a flat sheet of paper rather than the GR approach is small. For instance, the gravitational time dilation causes a difference of .7 parts per billion between the surface of the Earth and a point infinitely far away.

(see for instance)
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html

So if you don't mind a part per billion error, you can ignore the gravity of the Earth in a "twin paradox" type example.