Minkowski metric and proper time interpretation

[msumm21]

TL;DR Summary

I'm trying to learn general relativity, but misunderstanding how the metric implies that time appears to pass slower for something near a heavy mass, as viewed from something far away

Using an example of 1 space dimension and 1 time dimension, consider the metric $d\tau^2 = a dt^2 - dx^2$ near a heavy mass ($a>1$).

From what I've read a clock ticks slower near a heavy mass, as viewed from an observer far away. A clock tick would be representative of $d\tau$ right (not $dt$)? If so, then my confused understanding is below.

If $a$ is large, then small $dt$ results in large $d\tau$. If the far away observer's $d\tau$ is approximately $dt$, then his clock tick, say $dt=1$ corresponds to $d\tau >> 1$ near the mass. My interpretation of this is that the clock near the mass ticks $d\tau >> dt$ ticks (it ticks more than the clock far from the mass), and hence the clock near the mass moves faster. I realize this is wrong, but not clear what part is wrong.

 

[Orodruin]

Your basic assumption is wrong: $a < 1$ for the Schwarzschild metric.

 

[msumm21]

The book I'm reading (General Relativity: The Theoretical Minimum by Susskind) says the metric is approximately $d\tau^2 = (1+2gy)dt^2 - dy^2$ where the grav potential is $gy$ but yes I see this doesn't jive with stuff I see on Wikipedia. I must have misunderstood what this metric was supposed to be in the first place. Does anyone know what this metric is?

 

[Dale]

I must have misunderstood what this metric was supposed to be in the first place. Does anyone know what this metric is?

This is a local metric, only valid in a small region. The reference is not a clock at infinity, but a clock at $y=0$. Clocks at higher $y$ will be faster and clocks at lower $y$ will be slower compared to the reference clock.

 

[msumm21]

This is a local metric, only valid in a small region. The reference is not a clock at infinity, but a clock at y=0. Clocks at higher y will be faster and clocks at lower y will be slower compared to the reference clock.

Oh yes I think I'm getting it now, thanks!