Is the sun a black hole for low frequency radiation?

[MikeGomez]

Consider a photon of a certain frequency emitted from a distance slightly greater than the event horizon of a black hole. It seems possible that if a certain low frequency photon is not able to escape (due to becoming infinitely red-shifted), then a photon of a higher frequency emitted from the same location would able to escape (not become infinitely red-shifted).

If this is true, then is seems all mass has an event horizon for low frequency radiation. Even though that may be an extremely low frequency, it is non-zero.

 

[D H]

Consider a photon of a certain frequency emitted from a distance slightly greater than the event horizon of a black hole. It seems possible that if a certain low frequency photon is not able to escape (due to becoming infinitely red-shifted), then a photon of a higher frequency emitted from the same location would able to escape (not become infinitely red-shifted).

If this is true, then is seems all mass has an event horizon for low frequency radiation. Even though that may be an extremely low frequency, it is non-zero.

Your mistake here is that you appear to think redshift is a subtractive operation. It isn't. It is a multiplicative operation:

\lambda_{\mbox{observed}} = \frac 1 {\sqrt{1 - \frac {r_s} r}}\,\lambda_{\mbox{source}}

where

• r_s is the Schwarzschild radius of the radiating object,
• r is the distance between the center of mass of the source and the source,
• λ_source is the wavelength at the surface of the radiating object, and
• λ_observed is the wavelength as measured by a remote observer.

 

[Nabeshin]

Consider a photon of a certain frequency emitted from a distance slightly greater than the event horizon of a black hole. It seems possible that if a certain low frequency photon is not able to escape (due to becoming infinitely red-shifted), then a photon of a higher frequency emitted from the same location would able to escape (not become infinitely red-shifted).

It's buried (not very deeply) in what D H says, but just to make explicit: A photon emitted from a distance slightly greater than the event horizon will have a finite redshift. Basically, a photon's motion through spacetime is independent of its energy (read: wavelength).

 

[MikeGomez]

DH. I.m confused about the distances/locations. Is something is missing? Don't we need to know the distance from the observer to the center of mass of the source, as well as the distance from where the photon is emitted?

I'm assumming the 'center of mass of the source' refers to the BH or star or whatever, but we also need to consider the location of the photon emission.

 

[D H]

Sorry. I typed that post rather badly. I corrected it. The equation I gave is for the redshift of some light as observed by an observer well-removed from the gravity well. For an observer at some distance R from the center of the massive object, the redshift is\lambda_{\mbox{observed}}<br /> = \sqrt{\frac {1 - \frac {r_s} R} {1 - \frac {r_s} r}}<br /> \,\lambda_{\mbox{source}}

where

• r_s is the Schwarzschild radius of the radiating object,
• r is the distance between the center of mass of the source and the source,
• R is the distance between the center of mass of the source and the observer,
• λ_source is the wavelength at the surface of the radiating object, and
• λ_observed is the wavelength as measured by a remote observer.

In the limit R→∞, this reduces to the form given in post #2. Note however that r is the distance from the center of mass to the source rather than to the observer.

In any case, the bottom line is that redshift is proportional to wavelength.

 

[MikeGomez]

Thanks DH and Nabeshin. My original thinking was that the gravitational effect on photons is due to the photon’s equivalent mass. Normal mass loses velocity when traveling in a direction opposite the gravity source, and will be able to escape the gravitations influence if it reaches escape velocity. Photons behave differently in that they maintain a constant speed of c, but the gravitational influence in seen in a change in frequency. This lead me to think that the frequency of a photon can be thought of as the equivalent of rotational velocity, and photons would have an escape frequency instead of an escape velocity.