# Tidal forces and geodesic deviation: a principle question

1. Sep 6, 2008

### tsahi

hi all,

i kind of have a black hole regarding my understanding of the consistensy of
tidal forces and geodesic deviation analysis. one can determine some coefficients
of the riemman tensor from the tidal forces equations, by getting to a form
that is like the form of the geodesic deviation equation.
however, in the tidal forces analysis we assume that spacetime is flat,
and in geodesic deviation analysis we assume spacetime is curved. how come
we can than compare the two equations? that is, in flat space time
we have a bilocal meaning to vectors, where in curved spacetime we have a local
meaning?

Last edited: Sep 6, 2008
2. Sep 6, 2008

### tsahi

i will sharpen the question:

how can one even have a notion of a seperation vector in curved spacetime?
if we are only looking locally where the spacetime is flat, then there should
be no geodesic deviation.

3. Sep 6, 2008

### atyy

A first order derivative is more non-local than a zeroth order, because it compares two positions. Similarly, a second order derivative is more non-local than a first order derivative. The tidal forces equation is not local in the sense that it has a second order derivative. Also, the tidal forces equation does not occur in flat special relativistic spacetime - it occurs in flat Newtonian spacetime with gravity - so it is curved. Curved spacetime is locally flat by the equivalence principle, but only to first order. The Riemann tensor is also a second order derivative.

4. Sep 6, 2008

### tsahi

yes, i understand,
but what is it that enables us to regard spacetime as flat (newton, tidal forces analysis) and curved (geodesic analysis) at the same time?

5. Sep 6, 2008

### MeJennifer

Two objects of mass change their relative speed as if they accelerate towards each other, however their proper acceleration is zero, hence spacetime cannot possible be flat. Newton apparently did not think of that at the time he developed his theory.

Space is simply an arbitrary slice of spacetime so whether that slice is curved simply depends on how you wish to slice it. :)

Last edited: Sep 6, 2008
6. Sep 6, 2008

### tsahi

i did not understand the last explanation. proper acceleration?

7. Sep 6, 2008

### atyy

Magic. Your suspicion is well justified! Newton's theory and Einstein's theory are 2 completely different conceptions of gravity. In a mathematically consistent theory, which statements we regard as assumptions and which are derived is a matter of taste. However, Newton's theory and Einstein's theory are not mathematically consistent. Einstein's theory cannot be derived from Newton's theory. Newton's theory can be derived only approximately from Einstein's theory. Even in the weak field situation, Einstein's theory predicts perihelion precession, but Newton's doesn't. The relationship between tidal forces and geodesic deviation should be seen as a way of transitioning between two extremely useful but inconsistent theories.

There's a useful analogy I learnt from the Feynman lectures. Newtonian mechanics has 3 equivalent viewpoints: Newtonian, Lagrangian and Hamiltonian. You can pick any viewpoint you like. But in generalizing to quantum mechanics, (trying to guess the more correct theory to which Newton's theory is only an approximation), only the Hamiltonian and Lagrangian viewpoints work. Note that till this day we do not know why classical reality emerges from our supposedly more correct quantum theory!

Around 1910, Einstein knew Newton's theory was right, but not completely right. So he was in a similar situation as to where we are now with respect to his equations (he had some excellent competitors, and the choice between them was based on experiment, not Einstein's genius). Which aspects of his theory should we use to transition to the new more correct theory?

As summarized by Gu and Wen: There are many different approaches to quantum gravity based on different principles. Some approaches, such as loop quantum gravity, stress the gauge structure from the diffeomorphism of the space-time. Other approaches, such as superstring theory stress the renormalizability of the theory. In this paper, we follow a different rule of game by stressing finiteness and locality.

Or Padmanabhan: The quantum description of spacetime is likely to be as different from the classical description, as the atomic description of a solid is from the macroscopic continuum description of a solid. The latter uses concepts like density, stress and strain, bulk flow velocity etc., none of which has much relevance in the microscopic scale; the quantum description of molecules in a solid cannot be obtained by quantising the classical macroscopic variables. ... Quantisation of the metric has as much relevance — in this paradigm — as quantising, say, the density and bulk flow velocity of a solid with the hope of obtaining a quantum theory of molecules.

http://arxiv.org/abs/gr-qc/0606100
http://arxiv.org/abs/gr-qc/0408051

Last edited: Sep 6, 2008
8. Sep 7, 2008

### tsahi

all that is good and well. but when you look at the 2 equations,
the geodesic deviation has a proper time of the fiducial geodesic as parameter,
the tidal force has
regular time as parameter. these parameters are not the same thing.
so how can we simply compare the coefficients there to get riemann?

9. Sep 7, 2008

### tsahi

also, the comparison does not seem to take under account the fact the there are spatial contractions when moving to the reference frame of the fiducial particle...

10. Sep 7, 2008

### atyy

The assumption is made that the speed is small, so proper time and coordinate time are very nearly the same.

11. Sep 7, 2008

### atyy

We are only using one coordinate system here. Length contraction only occurs when comparing lengths measured in two different coordinate systems.

12. Sep 7, 2008

### tsahi

but we have two - lorenz of the fiducial particle and newtonian, and i meant generally
the not trivial lorentz transformation between coordiate systems.
my problem lies in the very ability of newton to think space is flat and time as seperate.
how come the measures came out right? what were his rods and clocks?

13. Sep 8, 2008

### atyy

Yes, there seems to be something strange going on here. We cannot interpret our mathematics unless we have a scalar product or metric which encodes our operational definition of distance. In Newtonian physics, the metric is fixed, and so we always have an interpretation of our coordinates. In GR, the metric is not fixed, and we cannot interpret our coordinates until we have a metric. Yet here it seems that we are interpreting our coordinates to arrive at the metric.

I suppose a more rigourous way would be to take a well-interpreted static solution to GR such as the Schwarzschild solution, take the weak field and low velocity limit, and see that we approximately recover Newtonian tidal forces.

14. Sep 9, 2008

### atyy

I am now recalling that Rindler does give a procedure for measuring the metric locally (Relativity: Special, General, and Cosmological, 2006). I don't remember what it is though.

15. Sep 9, 2008