Determinism, Stokes' Theorem and Relativity

Phrak
Messages
4,266
Reaction score
7
I don't think these three: {Determinism, Stokes' Theorem, Relativity Theory}, are compatible.

The notion of determinism, as applied to spacetime physics, means that if we know everything on an R3 spacelike hypersurface at time ta, we can predict what will be will be the state of things on an R3 spacelike hypersurface at some future time, tc.

The Kelvin-Stokes theorem,

\int_{t_a}^{t_c}f(t')dt' = F(t_c)-F(t_a)​

seems to say something else:-

It tells me that what I see at time tb, where ta<tb<tc, is defined by the events at both ta and tc up to a constant of integration.
 
Last edited:
Physics news on Phys.org
There are different kinds of boundary value problems. Typically you specify one value everywhere on the boundary, and the field inside is determined. For a second order field equation like the wave equation, the "value" can either be the field Φ itself or its normal derivative ∇Φ. With an appropriate choice of the surface of integration, the surface integral boils down to an integration over all space at t = ta and t=tb. As you point out this is unphysical, since it makes Φ appear to depend on its value at some future time. So you use instead Cauchy conditions, which specify both Φ and ∇Φ on just the initial surface t = ta.

The situation is clearer for first order equations, which is why Dirac sought a set of first order equations to describe electrons. For a first-order matrix equation Φt = M Φ, you just need to specify Φ at t = ta.
 
Bill_K said:
As you point out this is unphysical, since it makes Φ appear to depend on its value at some future time. So you use instead Cauchy conditions, which specify both Φ and ∇Φ on just the initial surface t = ta.

I didn't point out that it was nonphysical but take serious consideration. Consider ∇Φ to be the measurable quantity and Φ to be a gauge field, Φ <-- Φ' = Φ + dφ.
 
Phrak said:
I don't think these three: {Determinism, Stokes' Theorem, Relativity Theory}, are compatible.

The notion of determinism, as applied to spacetime physics, means that if we know everything on an R3 spacelike hypersurface at time ta, we can predict what will be will be the state of things on an R3 spacelike hypersurface at some future time, tc.

Actually, this isn't quite the case. As the other posters mentioned, you need to know not only everything on the space-like hypersurface at some time, you also need the boundary conditions, even in classical mechanics.

For example, a photon or "light packet" could come in "from infinity" and upset things, unless one specifies the boundry conditions so this doesn't happen.

There is extensive discussion in Wald, "General Relativity", of the initial value formulation of GR, if you have the textbook. Chapter 10. So if you're trying to argue that GR doesn't have a well posed initial value formulation, you'll have to show where Wald went wrong...
 
pervect said:
Actually, this isn't quite the case. As the other posters mentioned, you need to know not only everything on the space-like hypersurface at some time, you also need the boundary conditions, even in classical mechanics.

You can't get away with anything around here. I was attempting to state the issue in the simplest possible manner, but not too simple. I guess I didn't.

Of course, we need all the entire R^3 hypersurface encompassing a 4-volume, to be precise. This shouldn't distract from the question.
 
In attempting to decern which fields are measurable and which are not I've asked, "Can charge density be measured without electric or magnetic fields?" Also, "Can electric and magnetic fields be measured without charge?" The answer seems to be no.

Given this (and please correct me if I am wrong) only the products of second and third derivatives of the 4-vector electromagnetic potential are at our disposal to determine the current, past, and future values of electromagnetic phenomena.

Bill_K, you've given much food for thought. In the above, I mean we can set up many textbook problems with bounary conditions in ∇Φ and Φ, where Φ is some tensor, but not all of these field values are really accessible by experimentation, and so these textbook execises are somewhat constrived and nonphysical.
 
Last edited:
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top