# Physical interpretation of describing mass in units of length

I'm working in Taylor and Wheeler's "Exploring Black Holes" and on p.2-14 they use two honorary constants: Newton's constant divided by the speed of light squared e.g. $G/c^2$ as a term to convert mass measured in $kg$ to distance.

Without doing the arithmetic here, the "length" of the Earth is 0.444 cm; and of the sun is 1.477 km. To what do these distances correspond? What is their physical significance, generally?

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What do you mean by "in the metric"? (I know what a metric is, but I'm just not seeing why you use that phrase.) – David Z Aug 25 '11 at 19:03
Nasty edit. I took it out b/c it added nothing but confusion, as evidenced by your question and comment... – user756 Aug 25 '11 at 19:06

They represent the scale on which general relativisic effects dominate physics related to bodies of that mass.

For instance if you were to create a (un-rotating, uncharged) black hole of 1 Earth mass it's event horizon would have a radius of about $9\text{ mm} = 2 * M_\text{Earth}$ in those units.

For scales much, much larger than the "length" of the mass, general relativity may be neglected. For intermediate scale in comes in as corrections on order of $\frac{l}{L}$ where $l$ is the mass in the scaled units and $L$ is the length scale of the problem.

This is similar to what particle physicists do by setting $c = \hbar = 1\text{ (dimensionless)}$ energy scales and length scales become inter-changeable.

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This answer gets my vote! – user756 Aug 25 '11 at 20:11

I'm not sure it's terribly helpful, but it seems like the following analysis helps explain dmckee's response.

The force of gravity is

$F = G \frac{m M}{r^2}$.

Rearranging and dividing by $c^2$ gives

$\frac{G}{c^2} = \frac{F r^2}{M (m c^2)}$

where the $mc^2$ is the rest mass energy $(E_0)$ of the object experiencing the force caused by mass $M$. When you multiply through by the mass of the "large" object you get

$M \frac{G}{c^2} = l = \frac{F_g r^2}{E_0}$

Since we are interested in the length $l$, at that distance we have

$M\frac{G}{c^2} = r = \frac{F_g r^2}{E_0}$

or simply

$E_0 = F_g r$.

In words, this is the distance at which the energy of the system due to the rest mass of an object in a gravitational field is the same as the potential energy due to gravitation.

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Earth is 0.444 cm; and of the sun is 1.477 km

It corresponds to half of the respective Schwarzschild radius.

The $\frac{G}{c^2}$ is covered there and also in Adam’s answer.

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