Photon Energy

[Proteus]

I had thought that the energy of a photon is entirely kinetic, since it has no rest mass. Is this true?

 

At the edge of the Schwarzschild radius of a neutral non-rotating black hole, a photon can only just escape the black hole if it goes in the opposite direction of the black hole. If the kinetic energy equals the potential gravitational energy, then

 

mc^2 = GMm/r

<=>

r =GM/c^2

 

Yet, this is HALF the actual distance to the black hole, since the Schwarzschild radius equals 2GM/c^2 . What's wrong?

Edited November 15, 2009 by Proteus

[ajb]

You said it yourself, the photon is massless.

 

For a massive object you can equate the kinetic energy the object with gravitational potential energy.

 

It does not work for massless object as in Newtonian gravity massless objects do not feel gravity.

[timo]

There's quite a few things that went wrong. The short version is: The equation you started off with is neither explained (what do the letters stand for?) nor justified (you seem to think in terms of Newtonian gravity when working on the prime example for gravity beyond Newtonian physics).

[swansont]

There's quite a few things that went wrong. The short version is: The equation you started off with is neither explained (what do the letters stand for?)

 

I think that in this context, G, M, m, r and c should all be familiar to anyone with a little bit of physics background. Put another way, anyone who doesn't know what the letters stand for is probably not in a position to contribute to the discussion. (You must be this tall to ride the ride)

[timo]

I think that in this context, G, M, m, r and c should all be familiar to anyone with a little bit of physics background. Put another way, anyone who doesn't know what the letters stand for is probably not in a position to contribute to the discussion. (You must be this tall to ride the ride)

I think excluding the OP from the discussion is not really an option, big boy.

[swansont]

I think excluding the OP from the discussion is not really an option, big boy.

 

Excuse me? The terms appear in the original post — that's the only place the equations are present. How does that exclude the poster?

[timo]

Proteus starts with an equation probably involving a relativistic expression with m being the relativistic mass on the left side (the kinetic energy of a photon) and an expression with m being the rest mass on the right side (the potential energy in Newtonian gravity). At least that what it seems like to me. For me that seems like a case of not knowing what the letters stand for which would become more apparent if it was properly specified - it could well be that he wanted to couple the classical gravity field to energy rather than mass, or <something else>.

[Proteus]

You're right, wrong formula. Sorry for the stupid question.