Asked by: Joe Thomas

If the photon, the 'particle' of light, is thought of as behaving like a massive object, it would indeed be accelerated to higher speeds as it falls toward a black hole. However, the photon has no mass and so behaves in a manner that is not intuitively obvious.

The reason for the qualification 'properly defined' above is that the speed of light depends upon the vantage point (frame of reference) of the observer. When we say that the speed of light is decreased, we mean from the perspective of an observer fixed relative to the black hole and at an essentially infinite distance. On the contrary, to an observer free falling into the black hole, the speed of light, measured locally, would be unaltered from the standard value of c.

Most of us have heard of the result from

Local inertial frames in general relativity are just those frames of reference in which the observer is in gravitational free fall. A fancy way of looking at it is that the _local_ frame of reference of a free falling observer corresponds to a small patch of _flat_ spacetime tangent to the globally curved spacetime. As long as the observer confines measurements to a small enough local region, the approximation provided by the small tangent patch of flat spacetime can be made to be an arbitrarily good approximation to the true spacetime, which is actually curved in the main. The speed of light in flat spacetime is, of course, the usual value of c.

For example, if one had a closed laboratory in orbit (i.e., in free fall) around the earth and one did an experiment inside that laboratory to measure the speed of light, one would get the usual published value of c. All such observers would get one and the same value for c.

If, however, the distance through which the light travelled in the course of measuring its speed was too great, the deviation of the reference frame from being 'flat' would become apparent. That is, gravitational effects would begin to become apparent.

So, it is absolutely true that the speed of light is _not_ constant in a gravitational field [which, by the equivalence principle, applies as well to accelerating (non-inertial) frames of reference]. If this were not so, there would be no bending of light by the gravitational field of stars. One can do a simple Huyghens reconstruction of a wave front, taking into account the different speed of advance of the wavefront at different distances from the star (variation of speed of light), to derive the deflection of the light by the star.

Indeed, this is exactly how Einstein did the calculation in:

which predated the full formal development of general relativity by about four years. This paper is widely available in English. You can find a copy beginning on page 99 of the Dover book 'The Principle of Relativity.' You will find in section 3 of that paper, Einstein's derivation of the (variable) speed of light in a gravitational potential, eqn (3). The result is,

c' = c

where V is the gravitational potential relative to the point where the speed of light c

You can find a more sophisticated result derived later by Einstein from the full general theory in the weak field approximation in the book:

This book is widely available, and should be in your university library.

A non-mathematical discussion of this can be found in:

See, in particular, pages 65-66 and, especially, the first full paragraph on page 66. Here, Bergmann takes the deflection of light by the gravitational field of a star as evidence of the decreased speed of light in a gravitational field.

The speed of light is _not_ constant in a gravitational field, but depends upon the reference frame of the observer. An observer anywhere in free fall will measure (locally) the traditional value of c. An observer sufficiently far away from the source of the field will conclude likewise that the speed of light is c (locally).

Answered by: Warren Davis, Ph.D., President, Davis Associates, Inc., Newton, MA USA

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