# Production of light in a gravitational field

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1. Feb 7, 2015

### Gary Boothe

If light was being produced in an intense gravity field, would the light be of higher or lower frequency/energy to a distant observer? I understand that the light escaping the gravitational field would be red shifted, but that is not what I'm asking. I'm asking about the relativistic effects on light production.

2. Feb 7, 2015

### Staff: Mentor

I
I don't know what you mean by this.

Last edited: Feb 7, 2015
3. Feb 8, 2015

### Gary Boothe

Sorry to be unclear. Consider the mechanical vibrations of a tuning fork. In one reference frame, say the Earth's surface, the frequency of the vibration is a certain value, depending on the properties of the tuning fork. To an observer far from the Earth, is the frequency (measured to, let's say, 23 significant figures) of the tuning fork the same? I ask this question because I know that time, distance, and mass are different in different reference frames (surface of Earth versus off Earth). Consider also a pendulum. For any given pendulum with constant physical parameters, the period, or frequency of the oscillations, is determined by the gravitational field (the gravitational constant). So, a pendulum at sea level oscillates slightly faster than a pendulum on Mt. Everest. Light, in a classical view, is also an oscillation, but of course it is not like a tuning fork or a pendulum, and gravity has essentially no effect on the atomic level. But still, I believe asking if the light frequency varies in a gravitational field is a legitimate question. I know that the speed of light with respect to an observer, is always measured as c. I also know that the frequency of light with respect to an observer is not measured the same (e.g, red shift). So, is the energy of light dependent on reference frame? Perhaps the best way to ask my question is: It is well known that the half-life of a radioactive atom, say a gamma emitter, is longer on the Earth's surface than in outer space, due to time running slower on the Earth's surface. That is, the decay constant is changed. But is the energy or frequency (Energy = Planck's constant times frequency) of the gamma ray changed, due to relativistic effects? I can't make this any clearer, but if you still don't understand the question, or if you think it is nonsense, please ignore. I will not be offended. I frequently frustrate other physicists.

4. Feb 8, 2015

### Staff: Mentor

It depends on the way you define frequency. If you consider how the observer far away sees it (visually, for example), the frequency will be lower. You can use a light clock (or anything else clock-like) and the answer will be the same.
True, but that has nothing to do with the effects of time dilation. The pendulum clock will also run slower if you heat it (because the pendulum gets longer).

A local observer will always measure the same energy. A different observer can observe something different - but that is not surprising, as the gamma ray has to go from the surface of earth to our observer in space, and gets redshifted on the way.

5. Feb 8, 2015

### Staff: Mentor

The confusing part about the question is that you say that you are not asking about redshift, but then everything else that you write seems to be asking about gravitational redshift. What I don't understand is the distinction that you are trying to make. I will answer your questions below, but there is no attempt to avoid gravitational redshift since to me it seems to be precisely what you are asking about.
Assuming a tuning fork which was stable/reproducible enough then to an observer far from earth it would be vibrating at slower than the "reference" frequency.

Sorry if I gave the impression that I felt your question was not legitimate. I just don't think that I understand what you are actually asking. I suspect that my answers are not what you are looking for because of that.

In any local inertial frame this is true. However, the coordinate speed of light can differ from c in non-inertial coordinates.

Yes, energy is frame variant. It is the timelike component of the four-momentum, meaning that energy has the same relationship to momentum as time has to space. http://en.wikipedia.org/wiki/Four-momentum

Locally, it is not changed, but to the observer at infinity it is reduced. For example, suppose that you had a radioactive emitter which emitted a very precise frequency and an absorber which also absorbs that frequency very precisely. If placed next to each other, the emitter would emit and the absorber would absorb, regardless of how deep into a gravitational potential they are. But an absorber held by an observer at infinity would not absorb the emitted radiation.

6. Feb 8, 2015

### Staff: Mentor

As measured locally, the decay constant is not changed. Even at a distance, the change in observed half-life is not usually attributed to a change in the decay constant; it is attributed to the curvature of spacetime affecting relative clock rates. That is, the "decay constant" is defined to be the locally measured value, so that it can be viewed as an intrinsic property of the radioactive nucleus.