Stopping time, by speeding it up inside a bubble

Well, let's see how such a field would work. The Hamiltonian equations are $$\frac{\mathrm{d}{x}_k}{\mathrm{d}t}=\frac{\partial H}{\partial p_k}\frac{\mathrm{d}{p}_k}{\mathrm{d}t}=-\frac{\partial H}{\partial x_k}$$ Now we want the field to affect the speed of things happening, so a natural assumption would be that the field just acts as factor of the Hamiltonian: $$H(x,p)=\exp (f(x,t))H_0(x,p)$$ Here the exponential function mainly is there to make time go normally when the field is zero. It however also makes sure time cannot go reverse.

Now for a constant non-zero field, you'd just get a rate of temporal change proportional to $\exp (f)$, so if the field is positive, things are going faster, as intended.

However what happens in the "bubble wall", assuming a static field? Well, here we have to use the product rule: $$\begin{array}{rl}\frac{\mathrm{d}{x}_k}{\mathrm{d}t}& =\exp f(x)\frac{\partial H_0}{\partial p_k}\\ \frac{\mathrm{d}{p}_k}{\mathrm{d}t}& =-\exp (f(x))\frac{\partial H_0}{\partial x_k}-f^{\prime }(x)\exp (f(x))H_0(x,p)\end{array}$$ Note the extra term on the second equation. This is an extra force proportional to the derivative of the field and the total Hamiltonian (which more or less gives the total energy). Assuming that energy is positive, this means that this acts like an extra repulsive force. Note that this repulsive force is in addition to the slowing down due to the local factor, and its strength depends on how rapidly the field grows. Note that since the kinetic energy is proportional to the mass, this leads also to a force term proportional to mass, similar to gravitation. Any potential energy would, however, give rise to an extra force that is not mass dependent.

Let's look at the energy, now also with time-dependent field: $$\frac{\mathrm{d}E}{dt}=\frac{\partial H}{\partial t}=\exp (f(x,t))\frac{\partial H}{\partial t}+\frac{\partial f}{\partial t}\exp (f(x,t))H_0(x,p)$$ So energy change only happens with field change, and proportional to it; that can well be explained as energy going into the field (of which I only included the effect). As long as the field remains constant, energy is conserved.

In effect, the outside world would look colder than would be expected from the slowdown, as the additional force would draw away more energy from incoming particles;on the other hand, outgoing particles would get an extra boost, so to the outside world, you'd be hotter than expected from the speedup.

Indeed, with a sufficiently small border zone (and corresponding rapid onset of the field) the border might even act like a wall for all normal-speed particles coming from outside.

posted over 8 years ago

celtschk‭

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