A question about time in Special and General Relativity. [closed]

Question

I apologize if this question is considered too mathematically imprecise. My understanding of Special and General Relativity comes from reading books which attempt to explain them to non-expertsin these fields. Let C denote a clock (such as an atomic clock) fixed at the earth's surface. It seems that the rate at which another clock B runs can be "slowed down" to almost any extent one chooses (when compared with A's rate) if either (1) B moves at a speed sufficiently close to that of light-with respect to A-or (2) B is at rest with respect to A but is located in a sufficiently strong gravitational field. I have never found mention of any "scenario" in which B's rate could be "speeded up" to any desired extent (when compared with A's rate). Are there any such "scenarios"? If not, is there some fact or law which implies that their existence is impossible?

Answer 1

It is possible, but not in special relativity (because of the answer given by Youloush). Use a 2-dimensional space-time diffeomorphic to $\mathbb{R}^2$ and define a pseudo-riemannian metric (using coordinates for the tangent space from the flat case):

$g_{t,x}((a,b), (c,d))=(1+x^4)\cdot a\cdot c-b\cdot d$

The coordinate $t$ is identical to the proper time of an observer resting at point $0$. Now consider the proper time of an observer moving with constant velocity (with respect to our coordinates) from $(0,0)$ to $(2,2)$ and back to $(4,2)$ (he starts at the speed of light, but we do not have to worry, we would get the same result when we would start a little bit slower). The proper time elapsed is:

$2\cdot\int_0^2 \sqrt{(1+t^4)-1}\mathrm{d}t=2\cdot\int_0^2 t^2\mathrm{d}t=\frac{16}{3}$

This number is larger than $4$ (the proper time elapsed for the resting observer).

However, you can generalize the result from special relativity: Geodesics are those paths (locally) extremising the elapsed proper time. In the most typical cases you actually maximise the elapsed proper time (for example in flat space-time, where the geodesics are given by movements at constant velocity, look at the example above: you move far away (to 2), where the metric is very different, to get the effect, if you would stay within $(-1,1)$, the effect would not happen and it would be like in the flat case: dilatation). Then on every non-geodesic path you get a smaller proper time. However, in general it is not the case.

Answer 2

This is really basic special relativity. Whichever "way" an observer moves relative to another, his perception of the time in the other frame will always be slowed down. This is called "time dilation". Each observer will experience it, as absurd as it may sound.

Basically in the case of a constant relative speed $v$, applying a Lorentz transformation shows that the elapsed time is multiplied by $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ Which is obviously greater than 1. The case of any other kind of movement requires more work; if you know what an inertial frame is then it is obvious, otherwise just believe it works the same.

The case of a gravitational field is just the same, under the equivalence principle.

Either way I don't think this belongs on mathoverload.