According to the Schwarzchild metric an interval of proper time [itex]d\tau[/itex] at a fixed distance [itex]r[/itex] from an object of mass [itex]M[/itex] is related to an interval of co-ordinate time [itex]dt[/itex] measured by a distant observer:(adsbygoogle = window.adsbygoogle || []).push({});

[tex] d\tau = dt \sqrt{1 - \frac{2GM}{rc^2}} [/tex]

This is just the gravitational time dilation formula. Rearraging this expression we get

[tex] \frac{1}{dt} = \frac{1}{d\tau} \sqrt{1 - \frac{2GM}{rc^2}} [/tex]

Now according to quantum mechanics the total energy of any system is proportional to its quantum frequency.

Thus the rest mass of a quantum system, [itex]m_0[/itex], is inversely proportional to the proper period of its quantum oscillation [itex]d\tau[/itex]

[tex] m_0 \propto \frac{1}{d\tau} [/tex]

The mass of the system from the point of view of the distant observer who uses co-ordinate time [itex]t[/itex] is

[tex] m \propto \frac{1}{dt} [/tex]

Substituting into the rearranged time dilation formula we get

[tex] m = m_0 \sqrt{1 - \frac{2GM}{rc^2}} [/tex]

Thus as a mass [itex]m_0[/itex] is lowered towards a gravitating mass [itex]M[/itex], the mass as seen by a distant observer at [itex]r=\infty[/itex], [itex]m[/itex], is reduced.

If the gravitating body is a black hole then as the mass approaches the Schwarzchild radius it disappears from the point of view of the distant observer.

If we imagine the whole mass of the black hole as having been lowered into its own metric then does this imply that the mass/energy of the black hole is zero from the point of view of a distant observer?

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# Is a black hole's mass zero according to a distant observer?

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