# What is dt/dg?

1. May 28, 2014

### ABunyip

I want to consider the rate at which time slows as gravity increases near a massive object such as a sun or even a black hole. Obviously there is a distance component here but I am after a generalisation that simply shows the relationship between time and gravity (ought to be possible).

Surely someone has done this derivation - Hawking, Susskind, Cox? Why cant I find it anywhere?

2. May 28, 2014

### pervect

Staff Emeritus
Time doesn't slow in any physical sense. However, coordinate time slows relative to proper time. The details depend on the coordinate system you choose - for instance Schwarzschild, or isotropic coordinates.

If you're not familiar with the distinction between coordinate time and proper time, try looking up the later on wiki. Coordinate time is simple in principle , it's just an assignment of a number that gives the "time coordinate" of every event. The important part to realize about coordinate time is that it isn't anything that a physical clock measures, it's just a convention as to how to assign the time coordinate to specify events.

Coordinate time implies a synchronzation convetion (all clocks with the same coordinate time are synchronized). Proper time is measured with a single clock present at both events and doesn't need or imply any synchronization convention.

The difference in clock rate between coordinate time and proper time is due to the curvature of space-time. It is rather similar how degrees of longitude (coordinate changes) correspond to smaller distances (physical distances which can be measured) as one approaches the poles on the curved surface of the Earth. The important point is that the coordinate time is a convention, while the time that clocks actually keep (proper time) has a direct physical interpretation

3. May 29, 2014

### Staff: Mentor

Can you describe an actual experiment that would measure this "rate at which time slows"? Giving a concrete example would help.

4. May 29, 2014

### Staff: Mentor

How do you quantify "gravity" as a single independent variable? What does the variable measure? The term "gravity" does not have a single unambiguous meaning in GR.

5. May 29, 2014

### DeIdeal

It has no meaning because "gravity" cannot be fully characterised by a single scalar in GR(*). There's no well-defined g in the sense you're using it.

(*) In more than two dimensions.

Last edited: May 29, 2014
6. May 29, 2014

### Staff: Mentor

Is $h$ supposed to be height here? If so, why do you think "gravity" is the square root of the reciprocal of the height? What is this supposed to mean, physically? It doesn't correspond to anything with physical meaning that I'm aware of.

7. May 29, 2014

### Jonathan Scott

Time experienced by an object does not slow in any subjective sense. Some sort of external time reference is needed to compare time rates, such as an observer using a suitable coordinate system to describe the whole region.

In a simple situation involving gravitational sources which are at rest relative to the coordinate system, the time rate only depends on the gravitational potential, not the field (or acceleration).

In the weak field (semi-Newtonian) approximation, the relative rate at which a standard clock runs at the location of an object in Newtonian gravitational potential $\Phi$ (equal to the sum of $-Gm/r$ for all relevant sources) compared with the rate at a distant point is approximately $(1 + \Phi/c^2)$. For stronger fields the solution in GR for a single central mass is that given by xox.

If the object is moving, then there is an additional time dilation effect from Special Relativity due to the speed, giving another factor of $\sqrt{1 - v^2/c^2}$. These factors can be multiplied together to give the total effect.

8. May 29, 2014

### grav-universe

Well, in GR, we have
$$a = - m c^2 L_t^2 / (r^2 z)$$
where L_t is the tangent coordinate length contraction, z is the local time dilation, m is r_s / 2, and a is the local acceleration of gravity, whereas in Schwarzschild L_t = 1 and z = sqrt(1 - 2 m / r), so
$$z^2 = 1 - 2 m / r$$
$$2 m / r = 1 - z^2$$
$$r = 2 m / (1 - z^2)$$
which gives us
$$a = - m c^2 / ((2 m / (1 - z^2))^2 z)$$
$$a = - c^2 (1 - z^2)^2 / (4 m z)$$

Since a and z are invariant for a particular shell and m and c are constant, these values will remain the same regardless of the coordinate system used, so the equation in this form should be coordinate system independent and it doesn't matter that we applied Schwarzschild coordinates originally.

9. May 29, 2014

### grav-universe

If you want da / dz, then the derivative of that works out to
$$da / dz = c^2 (1 - z^2) (1 + 3 z^2) / (4 m z^2)$$

10. May 29, 2014

### ABunyip

Sorry. It should have read: g(h) ≈ 1/h2

The consequence of supposing I can do anything useful at 11pm after a very long day.

This is the "intensity" of the gravitational "radiation" over distance h.

Last edited by a moderator: May 30, 2014