# How much Gravity is required to stop time?

Clocks free of gravitational influence run faster than those experiencing gravity. Is it possible for gravitational influence to bring time to a stop? Additionally can acceleration affect clocks in the same manner as gravity with time running fastest with no acceleration. In a hypothetical case where a clock in deep space is not subjected to gravitational influence and is therefore not moving, does the rate of time reach maximum?

• Clocks are always running at the same rate. Observers that are observing a non-local clock may be seeing the remote clock run at a different rate than their own. It's also not clear what you mean by "feeling of gravitational influence". One can't feel gravity. The only thing that one can feel are the forces that are preventing one from falling freely. – CuriousOne May 30 '16 at 21:40
• The fact that you're asking those specific questions suggests that you're thinking about relativity in the right way, you just need to explore the maths a little more. – The Geoff May 31 '16 at 0:13
• Curious one..clocks do not always run at the same rate.Period – RaSullivan Jan 5 '17 at 11:48
• You need to distinguish between local and global notions. Locally, a clock is just a clock, and there is no way of defining how fast the clock runs except to verify that it runs at the same rate as other nearby clocks. Globally, simultaneity is not defined, so we can't in general compare clocks, although there is a meaningful notion of comparison of distant clocks in a stationary spacetime. – user4552 Mar 9 '18 at 20:52
• You talk about acceleration, but gravitational time dilation depends on the potential, not the field. – user4552 Mar 9 '18 at 20:52

More formally this can be see with the Schwarzschild metric for a nonrotating black hole $$ds^2 = \left(1 - \frac{2GM}{rc^2}\right)dt^2 - \left(1 - \frac{2GM}{rc^2}\right)^{-1}dr^2 - r^2d\Omega^2.$$ We look at the situation for light rays leaving a body close to the black hole hole with the horizon at $r = 2GM/c^2$. The interval for null rays is zero $ds = 0$ and we integrate the time $$\int^Tdt = \int^R\left(1 - \frac{2GM}{rc^2}\right)^{-1}dr = R - \frac{2GM}{c^2}ln\left(R - \frac{2GM}{c^2}\right).$$ This diverges as $R \rightarrow \frac{2GM}{c^2}$, which means you have to wait an infinite time to see the clock tick off the time it crossed the horizon.