Minkowski metric and proper time interpretation

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TL;DR
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.
 
on Phys.org
Your basic assumption is wrong: ##a < 1## for the Schwarzschild metric.
 
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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?
 
msumm21 said:
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.
 
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Dale said:
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!
 
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