Comparing predictions and reality for the gravitational attraction due to light beams

Question

While doing some on-the-side reading, I stumbled across this question: Do two beams of light attract each other in general theory of relativity?. Great question and a great, easily understandable answer. In short, it explains that General Relativity allows photons/beams of light to gravitationally attract other things. It mentions a very important and well known phenomenon; that the deflection of light passing through the gravitational field of a massive particle is twice that predicted by Newtonian Gravitation. It also mentions and links to a wonderful article that shows that light beams travelling anti-parallel are deflected by each other's gravitation by four times what is predicted by Newtonian methods.

The Newtonian predictions were able to be made because of the commonly accepted gravitational mass for a photon, which effectively uses Einstein's $E=mc^2$ and Planck's $E=h f$ to get $m=h f/c^2$. Not a bad strategy.

My question is why we choose to equate the photon's gravitational mass with a hypothetical particle's rest mass? Given that the total energy of a photon (if you rescale the potential energy to 0) can be written as: $$Total~Energy=Kinetic~Energy+Rest~Energy$$ And given that it is nice to set the rest energy of our photon to $0$. Why then should we choose the mass on which to base the predictions using Newtonian Gravity to be a rest mass? Especially when Newtonian physics provides an adequate way of obtaining mass from kinetic energy (as long as that mass is used only in other Newtonian physics). I mean, why can we not say the following for purely Newtonian calculations: $$E=hf,~~K=\frac{1}{2}mv^2,~~Rest~Energy=E_o=0$$ $$\therefore hf=\frac{1}{2}mv^2\rightarrow m=2hf/v^2=2hf/c^2$$ This effectively doubles the gravitational mass of a light beam without altering the actual momentum of it. When predicting the deflection of a beam due to a massive particle, this would make the force of Newtonian gravitation twice as large and the fact that momentum didn't change means the deflection prediction would be twice as large. For the deflection of two antiparallel beams, since the gravitational masses of both are doubled, this would quadruple the force of attraction again without modifying each beam's momentum, making the Newtonian prediction four times that compared to using mass from the rest energy equation. Both of these new predictions are what is shown to actually happen.

Understandably, if this were a good and valid idea, it would have been used or realized a long time ago. My question is centred around understanding why the rest mass equation must be used; I am not trying to say what has been done is wrong, just trying to understand why it makes sense.

Answer 1

When you say "without altering the actual momentum of it" is that really true?

$$ E^2 = p^2c^2 + m^2c^4 $$

so for a photon $E = pc$, since rest mass is zero.

Now according to your first "traditional" calculation of m, we would have $E = pc = m_1c^2$, and therefore $p=m_1c$, where $m_1$ is mass according to the first "traditional" calculation.

For your second "non-traditional" calculation from $p=mv$ we would have $p = m_2c$, where $m_2$ is mass according to the second "non-traditional" calculation.

But $m_2 = 2m_1$, so $p_2 = 2p_1$

Answer 2

For a particle of fixed mass $m$ moving in a fixed gravitational potential $\phi(\vec{r})$ the motion is independent of the mass of the particle. The equations are $$ \vec{F}=-m\nabla\phi $$ and $$ \vec{F} = \frac{d\vec{p}}{dt} = m \frac{d\vec{v}}{dt} $$ It's clear that the $m$'s cancel when combining these equations. So from this point of view it doesn't matter what (non-zero) mass is taken for the photon in the calculation. It sounds like you've read a derivation of photon deflection which assumes the photon mass is $m=E/c^2$, but this assumption isn't necessary.

In your argument you calculate the gravitational force using a mass derived from $E=\frac{1}{2}m v^2$ but then take the momentum to be $p=mv=E/c$. This implies taking the gravitational mass of the photon (the $m$ in the first equation I wrote) to be twice the inertial mass of the photon (the $m$ in the second equation).

Of course, there's nothing to stop you modifying Newton's gravity by assuming this, and it does correct the deflections you mention (at least to first order in $\phi/c^2$, in the higher order terms your results would still disagree with those of general relativity). However, such a choice would violate the equivalence principle, for which there is a lot of experimental evidence (albeit mostly with massive particles). For me, this seems like the biggest reason not to consider your modification.

So, in summary, there is no requirement for the photon rest mass to be taken as $E/c^2$. As long as the inertial and gravitational masses are assumed to be equal the usual Newtonian deflection is found (i.e. half the predition of GR for weak fields). Your argument assumes that the gravitational mass is twice the inertial mass, which doubles the deflection, but violates the equivalence principle.

Update:

What I wrote above is true for the case of deflection in a constant field. For the interaction of two photons travelling in opposite directions the choice of mass is important in the Newtonian model. If both photons are modelled as having mass $m$ then the force between them is proportional to $m^2$ and so their acceleration is proportional to $m$. Therefore a larger choice of $m$ would result in a greater deflection. However, the comment I made about your idea being equivalent to choosing different inertial and gravitational masses applies in this case too.

If the paper you cite shows that the choice of $m=E/c^2$ gives a quarter of the GR prediction for two interacting photons then this is the only choice of mass for which this is true (I can't actually access that paper atm, so I can't check). This choice of Newtonian mass is reasonable since GR implies that the curvature of space depends upon the energy density, rather than the mass density.

Regarding your comment about Newtonian physics already violating the equivalence principle, I think it's important to be precise. If it is assumed that coordinates change between observers with different constant velocities by a Galilean transformation then Newtonian predictions do satisfy the equivalence principle. However, under the Lorentz transformation (which special relativity tells us is the correct one since it conserves the speed of light) Newtonian predictions violate the equivalence principle. This was one of the observations Einstein used to develop the theory of General Relativity.

Answer 3

Usually the Newtonian limit is described as taking $v << c$ but a much better way to express it is saying that the kinetic energy is much less than the rest energy $$ \frac{1}{2}m v^2 << m c^2 $$ of course this runs into trouble when we talk about photons since we don't have a well defined concept of velocity, in the Newtonian sense.

This is because in special relativity four-velocity is only defined for massive particles as the derivative with respect to the proper time $\tau$ (defined from $ds^2=-d\tau^2$) as $$ v^\mu = \frac{dx^\mu}{d\tau} $$ Photons follow null trajectories which means $ds^2=0$ and so you can't define a proper time.

Anyway, the point is that in the newtonian limit the leading order contribution to the energy is the rest energy. The fact that it is always ignored in calculations is because it is a constant factor. Nevertheless if you want to make the kind of calculations you are trying to make you must take $mc^2$ as the leading term for the energy.