by Henning Makholm » Sun Jul 20, 2008 3:50 pm
henk21cm wrote:In the Newtonian approximation where the tides are the derivative of the "acceleration" it is hard to find a function which is infinite at some value z0, whereas the derivative of that function with respect to z at z0 is finite. tan(z) at z=π/2 does not qualify, just like any function in the family ln(z) and 1/z^n.
Correct. So what I was trying to say is that according to GR the tides are real, but the Newtonian gravitational field whose gradient the tides are supposed to be is not.
henk21cm wrote:Henning Makholm wrote:The tidal effect is a quantifiable physical reality at every point in GR spacetime, it's described by the "Weyl tensor", which can be derived from the curvature of spacetime.
Correct me if i'm wrong, first picture a perfectly flat space, no curvature. Now introduce a mass, which curves space. In a 1D analogon: a flat straight line and some curve, e.g. a Gaussian bell, upside down. The vertical distance between a point on the flat line and the corresponding point on the curve is a measure for the gravitational force between the introduced mass and a very small test mass. The slope of the curve would be a measure for the tide. Is this what you mean?
I don't think I follow you here. Tides in GR are just a fact of
how space is curved; they are not related to
why space is curved. The standard gedankenexperiment (IIUC) is to deposit a small ball of test particles (with negligible gravity of their own) somewhere in space, initially at rest with respect to each other, and watch how they start to drift under gravity. In general it may start to shrink or enlarge, and/or to deform into an ellipsoid. The rate at which the ball deforms into an ellipsoid is a measure of the strength of the tides.
According to
this nice explanation by John Baez, Einstein says that the rate at which our test ball changes its
volume is proportional to the energy density at its location, measured from the rest frame of the ball. If the ball floats in a vacuum, its volume will stay constant. Thus, a tendency to become ellipsoidal will look as if something is tugging it apart in one direction and squeezing in another -- presto, tides!
When I speak of "rate" here, I mean the second-degree coefficient in a Taylor expansion of the behavior of the ball as a function of time. The first-degree coefficient is zero because we stipulate that the test particles initially have no mutual velocity. Accordingly, the natural dimension of tides becomes 1/s², consistent with the Newtonian idea of acceleration per meter.
henk21cm wrote:Henning Makholm wrote:In GR that is done by requiring that B has constant spatial coordinates in a "stationary" coordinate system. Being "stationary" is a rather special condition on a coordinate system; essentially it requires that the curvature of spacetime is the same at all events that share the same spatial coordinates.
First idea was: that should not be too difficult. Any point on a sphere that surrounds a mass has a constant space-time curvature. But then: that means trouble. Since the original mass is not alone in this universe, there more spheres, intersecting the original sphere. That leaves us with a deformed sphere, which in a mathematical sense has entirely lost its spherical symmetry (although within a few percent may be a sphere).
It sounds like you're misunderstanding me here. You don't need spherical symmetry to have stationarity. Different points of space can have different curvatures, but each point has to
keep its curvature as time passes, forever. So stationarity in this sense is possible only in a changeless universe.
In a general two-body system the distance between the bodies varies periodically, so it is not changeless, and perfect stationarity is not possible. (We can have approximate stationarity. An ordinary coordinate system for the solar system, including the interior of the sun and the planets, is close enough to stationarity for the Newtonian approximation to be useful. Inside a black hole, however, stationarity cannot even be approximated, because any timelike path in it ends up at the singularity, and thus passes through regions with arbitrarily large curvature).
henk21cm wrote:There must be a reason why the acceleration of B with respect to A (=free falling) reaches infinity at the event horizon and i fail to see it. The only reason i can think of is "by definition". At the event horizon there exists no escape to the outside world. No force or energy is sufficient to increment the distance between A and the center of mass. That means that the force must be infinte. Is that the background for "infinity"?
Yes, I think this is good, but it is perhaps more convincing to think in terms of "no particle world-line with finite proper acceleration can touch the horizon and escape". Eliminating appeals force and energy gives a cleaner and more geometric picture.
If point B is just outside the horizon, it could be the worldline of a rocket hovering just above the horizon. Another rocket with more thrust (i.e. a greater acceleration) could be descending towards the black hole, decelerating to come to a standstill just beside B, and going off into space. In fact, since the other rocket is stronger, it can afford to come to its standstill a bit further down than B. If the acceleration a rocket a B needs to exert to stand still were to approach a finite limit as B comes closer to the horizon, the lowest point from which a rocket with twice as good an acceleration as this limit can escape ought to be
below the horizon, contradicting the definition of the horizon.
[quote="henk21cm"]In the Newtonian approximation where the tides are the derivative of the "acceleration" it is hard to find a function which is infinite at some value z0, whereas the derivative of that function with respect to z at z0 is finite. tan(z) at z=π/2 does not qualify, just like any function in the family ln(z) and 1/z^n.[/quote]
Correct. So what I was trying to say is that according to GR the tides are real, but the Newtonian gravitational field whose gradient the tides are supposed to be is not.
[quote="henk21cm"][quote="Henning Makholm"]The tidal effect is a quantifiable physical reality at every point in GR spacetime, it's described by the "Weyl tensor", which can be derived from the curvature of spacetime. [/quote]
Correct me if i'm wrong, first picture a perfectly flat space, no curvature. Now introduce a mass, which curves space. In a 1D analogon: a flat straight line and some curve, e.g. a Gaussian bell, upside down. The vertical distance between a point on the flat line and the corresponding point on the curve is a measure for the gravitational force between the introduced mass and a very small test mass. The slope of the curve would be a measure for the tide. Is this what you mean?[/quote]
I don't think I follow you here. Tides in GR are just a fact of [i]how[/i] space is curved; they are not related to [i]why[/i] space is curved. The standard gedankenexperiment (IIUC) is to deposit a small ball of test particles (with negligible gravity of their own) somewhere in space, initially at rest with respect to each other, and watch how they start to drift under gravity. In general it may start to shrink or enlarge, and/or to deform into an ellipsoid. The rate at which the ball deforms into an ellipsoid is a measure of the strength of the tides.
According to [url=http://math.ucr.edu/home/baez/einstein/]this nice explanation by John Baez[/url], Einstein says that the rate at which our test ball changes its [i]volume[/i] is proportional to the energy density at its location, measured from the rest frame of the ball. If the ball floats in a vacuum, its volume will stay constant. Thus, a tendency to become ellipsoidal will look as if something is tugging it apart in one direction and squeezing in another -- presto, tides!
When I speak of "rate" here, I mean the second-degree coefficient in a Taylor expansion of the behavior of the ball as a function of time. The first-degree coefficient is zero because we stipulate that the test particles initially have no mutual velocity. Accordingly, the natural dimension of tides becomes 1/s², consistent with the Newtonian idea of acceleration per meter.
[quote="henk21cm"][quote="Henning Makholm"]In GR that is done by requiring that B has constant spatial coordinates in a "stationary" coordinate system. Being "stationary" is a rather special condition on a coordinate system; essentially it requires that the curvature of spacetime is the same at all events that share the same spatial coordinates. [/quote]
First idea was: that should not be too difficult. Any point on a sphere that surrounds a mass has a constant space-time curvature. But then: that means trouble. Since the original mass is not alone in this universe, there more spheres, intersecting the original sphere. That leaves us with a deformed sphere, which in a mathematical sense has entirely lost its spherical symmetry (although within a few percent may be a sphere).[/quote]
It sounds like you're misunderstanding me here. You don't need spherical symmetry to have stationarity. Different points of space can have different curvatures, but each point has to [i]keep[/i] its curvature as time passes, forever. So stationarity in this sense is possible only in a changeless universe.
In a general two-body system the distance between the bodies varies periodically, so it is not changeless, and perfect stationarity is not possible. (We can have approximate stationarity. An ordinary coordinate system for the solar system, including the interior of the sun and the planets, is close enough to stationarity for the Newtonian approximation to be useful. Inside a black hole, however, stationarity cannot even be approximated, because any timelike path in it ends up at the singularity, and thus passes through regions with arbitrarily large curvature).
[quote="henk21cm"]There must be a reason why the acceleration of B with respect to A (=free falling) reaches infinity at the event horizon and i fail to see it. The only reason i can think of is "by definition". At the event horizon there exists no escape to the outside world. No force or energy is sufficient to increment the distance between A and the center of mass. That means that the force must be infinte. Is that the background for "infinity"?[/quote]
Yes, I think this is good, but it is perhaps more convincing to think in terms of "no particle world-line with finite proper acceleration can touch the horizon and escape". Eliminating appeals force and energy gives a cleaner and more geometric picture.
If point B is just outside the horizon, it could be the worldline of a rocket hovering just above the horizon. Another rocket with more thrust (i.e. a greater acceleration) could be descending towards the black hole, decelerating to come to a standstill just beside B, and going off into space. In fact, since the other rocket is stronger, it can afford to come to its standstill a bit further down than B. If the acceleration a rocket a B needs to exert to stand still were to approach a finite limit as B comes closer to the horizon, the lowest point from which a rocket with twice as good an acceleration as this limit can escape ought to be [i]below[/i] the horizon, contradicting the definition of the horizon.