# Thread: Can we discuss this?

1. ## Can we discuss this?

The speed of Light is frequency times wavelength.

Will let ν = frequency
Will let λ = wavelength

Where c = νλ

Where plank's constant is h
Where photon energy = νh

Gravitational potential energy is given as GMm/r
And a photon mass is νh/c2
We will write it as νh/c^2
It is not a rest mass.

So the potential energy of a photon to leave a gravitational well GM(νh/c^2)/r

So the photon will after leaving the gravitational well have the energy of ν'h

Where νh - GM(νh/c^2)/r = ν'h

And the new wavelength λ'

Where λ' = c / ν'

I may have left out some steps. But they can be inserted were needed.  Reply With Quote

2. You lost me after that part about "the speed of light".   Reply With Quote

3. You are developing some approximate gravitational redshift calculations; the best framework for this calculation is relativity.

It gets bit awkward to speak of the "mass" of a photon; it actually has energy, for which an equivalent mass value can be calculated; but this isn't mass in the normal sense and you can't always just apply the same equations that you'd apply for massive objects. I appreciate you have already noted this point in observing that the mass you calculate is not a rest mass. Personally I avoid the term mass entirely for photons.

The exact solution for the simple case of gravitational redshift for a non-rotating uncharged spherically symmetric mass is

v/v' = λ'/λ = 1/sqrt(1-2GM/rc^2)

r is the distance of the photon from the center of mass when it is emitted, M is the mass, G is gravitational constant, and c is speed of light. λ' is the observed frequency of the photon outside the gravitational well (at distance infinity) and λ is the frequency at the point of emission, inside the well. (To get formally precise; r is distance in Schwartzchild co-ordinates. You can simply think of it as distance as we normally understand it; but if we are in an exceptionally strong field -- that is, if r is close to the Schwartzchild radius of the mass -- then distance can get a bit ambiguous. Relativity does that.)

Now, take your formula: νh - GM(νh/c^2)/r = ν'h

A bit of algebra (divide through by vh), and this becomes v'/v = 1 - GM/rc^2

This is actually the standard Newtonian approximation in mathematics for the true formula, which is v'/v = sqrt(1 - 2GM/rc^2)

The approximation is sqrt(1-ε) ~= 1-ε/2 for small values of ε

You need to use relativity, not the approximations of Newtonian physics, to get the exact answers, but the approximations get very close as long as the r is significantly bigger than the Schwartzchild radius: that is, as long as we are not talking about photons being emitted from close to a black hole.

Cheers -- sylas

PS. The Schwartzchild radius for a mass M is 2GM/c^2. If r is equal to this, we get a divide by zero in the calculations; and if r is less than this we have the square root of a negative number. Photons can't escape out from inside the Schwartzchild radius; this is the case with a black hole.  Reply With Quote

4. Originally Posted by sylas You are developing some approximate gravitational redshift calculations; the best framework for this calculation is relativity.

It gets bit awkward to speak of the "mass" of a photon; it actually has energy, for which an equivalent mass value can be calculated; but this isn't mass in the normal sense and you can't always just apply the same equations that you'd apply for massive objects. I appreciate you have already noted this point in observing that the mass you calculate is not a rest mass. Personally I avoid the term mass entirely for photons.

The exact solution for the simple case of gravitational redshift for a non-rotating uncharged spherically symmetric mass is

v/v' = λ'/λ = 1/sqrt(1-2GM/rc^2)

r is the distance of the photon from the center of mass when it is emitted, M is the mass, G is gravitational constant, and c is speed of light. λ' is the observed frequency of the photon outside the gravitational well (at distance infinity) and λ is the frequency at the point of emission, inside the well. (To get formally precise; r is distance in Schwartzchild co-ordinates. You can simply think of it as distance as we normally understand it; but if we are in an exceptionally strong field -- that is, if r is close to the Schwartzchild radius of the mass -- then distance can get a bit ambiguous. Relativity does that.)

And that solution has 1-2GM/(rc^2) in it.

So if we were to write: 1/sqrt(1 - v^2/c^2) = 1/sqrt(1-2GM/rc^2)

We get v = sqrt(2GM/r)

If we take Newton's mv^2/2 = GMm/r
We also get v = sqrt(2GM/r)

Now Einstein's kinetic energy = mc^2 (1/sqrt(1 - v^2/c^2) - 1)
And the first term of the Taylor series for it happens to be kinetic energy = mv^2/2 + . . .

Now where do you want to go from here? Or do you want to take a step back, we talk there?

Now, take your formula: νh - GM(νh/c^2)/r = ν'h

A bit of algebra (divide through by vh), and this becomes v'/v = 1 - GM/rc^2

This is actually the standard Newtonian approximation in mathematics for the true formula, which is v'/v = sqrt(1 - 2GM/rc^2)

The approximation is sqrt(1-ε) ~= 1-ε/2 for small values of ε

You need to use relativity, not the approximations of Newtonian physics, to get the exact answers, but the approximations get very close as long as the r is significantly bigger than the Schwartzchild radius: that is, as long as we are not talking about photons being emitted from close to a black hole.
So I think we need to talk about velocity and kinetic energy. Using Einstein's equation.

Cheers -- sylas

PS. The Schwartzchild radius for a mass M is 2GM/c^2. If r is equal to this, we get a divide by zero in the calculations; and if r is less than this we have the square root of a negative number. Photons can't escape out from inside the Schwartzchild radius; this is the case with a black hole.
That is the accepted view.  Reply With Quote

5. Originally Posted by 37818 And that solution has 1-2GM/(rc^2) in it.
Of course. The gravitational redshift of this case can be written ν'/ν = sqrt(1-R/r), where R is the Schwartzchild radius and r is the radius at which the photon was emitted with frequency ν Originally Posted by 37818 So if we were to write: 1/sqrt(1 - v^2/c^2) = 1/sqrt(1-2GM/rc^2)
This invites all kinds of horrible confusion. The LHS of that is the time dilation for a moving observer, using v as a VELOCITY. This is bound to get mixed up with ν that we have been using as a frequency. You are going to get awfully confused if you don't stick to a specific problem and consistent naming of terms. Nothing previously involved the time dilation of a moving observer; this is unrelated to your original post. If you are going to talk about moving observers AS WELL as observers in gravitational wells, then use f for frequency, so you don't confuse v as velocity with ν as frequency.

37818, I am not really joining in to "talk about" this. There's nothing difficult or contentious here, and I can explain some points if you are confused or would like to know more about anything. Ask a question if you like, and if you think it would help. But the formula v = sqrt(2GM/r) isn't what you appear to think. It's a VELOCITY calculation; and specifically it happens also to be the escape velocity for a particle in a gravitational field. It has nothing at all to do with photon frequencies.

Cheers -- sylas  Reply With Quote

6. unsubscribing --- it makes my head hurt just to read this! I need to stick with "on" and "off" utilizing a light switch. Oh, and dimmers! I CAN operate dimmers!  Reply With Quote

7. It's a great calculation if you're falling into a Black Hole.

Relevance to age of Earth/History?

K54  Reply With Quote

8. Yeah, I'm just wondering if there might be some point coming here. If there is, I can't see it.  Reply With Quote

9. I'm not sure what the point is either, or what it is you want us to discuss.

It looks you're using Newton's laws of mechanics, combined with a fictiotious mass for a photon, to give a red shift as a photon leaves a gravitational well. Of course photons don't have mass, and they don't ever change velocity in vacuum, but you'd not be the first to try to give the photon a mass like that (unfortuntely you'll get the wrong gravitional bending of the light by a factor of 2).

Are you asking whether your formulas are correctly derived from your assumptions, or how they compare against the results derived properly from the General Theory of Relativity?

You get the correct first-order approximation using Newton's laws for the redshifting, though you also get an erroneous results which gets folded away, namely you predict a new velocity for the photon. Which simple isn't part of the reality: the speed of light in vacuum is constant in all inertial frames of rest.  Reply With Quote

10. FYI: There's nothing in Decorum that says you can't just discuss science for the heck of it - or that every discussion must lead to throwing pencils at each other. Of course, this being Nat Sci, we might wanna change that last rule...  Reply With Quote

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