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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.

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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.

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  • $\begingroup$ definitely my favourite answer so far. But given that the EP states that one cannot distinguish between a constant gravitational field and a frame undergoing constant acceleration, and given that the current Newtonian predictions predict that a photon deflects by half (a quarter for photon-photon deflection) as much as observed when considering its gravity but are accurate in predicting momentum exchange from collisions, etc. Wouldn't this seem to indicate that the current Newtonian predictions already violate the EP? $\endgroup$ – Jim Feb 28 '14 at 15:14
  • $\begingroup$ Which, by the way, isn't problematic since we already know Newtonian physics doesn't really hold any water in a relativistic limit $\endgroup$ – Jim Feb 28 '14 at 15:14
  • $\begingroup$ @Jim - I have responded to your question in the answer. $\endgroup$ – Mark A Mar 1 '14 at 22:14
  • $\begingroup$ So, if I may oversimplify your response into one short statement, you are saying that the mass of $m=hf/c^2$ was used in the Newtonian predictions because it is the same mass used in GR where it gives in the correct results. Essentially, consistency is the key? $\endgroup$ – Jim Mar 2 '14 at 15:06
  • $\begingroup$ @Jim... well, I would put it like this: In Newtonian physics the gravitational force is coupled to the mass whereas in GR it is coupled to the energy, and an energy of $hf$ in GR corresponds to a mass of $hf/c^2$ in the Newtonian case. At least that's how I would justify it, as I said I haven't actually read the paper in question (yet). Thanks for the bounty! :-) $\endgroup$ – Mark A Mar 2 '14 at 23:23
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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 $p_2 = 2p_1$

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  • $\begingroup$ I definitely see your point, and I considered it when I wrote my question. However, the statement that it wouldn't change the actual momentum is, I believe, still true. The momentum of a photon can actually be measured because we know that $E=pc$ and we can easily measure its energy. However, gravitational mass is not measurable except through the effect of an object on the local curvature of space (i.e. how it affects nearby objects). Add to that the fact that we can always compute $E=E\rightarrow\frac{1}{2}mv^2=pc$ and recover the original momentum. $\endgroup$ – Jim Feb 28 '14 at 14:16
  • $\begingroup$ Still a great point though, I'm just hesitant to accept this as the complete explanation; I feel there must be some additional and possible deeper reasoning. $\endgroup$ – Jim Feb 28 '14 at 14:18
  • $\begingroup$ But doesn't $\frac{1}{2}mv^2=pc$ contradict the premise that purely Newtonian physics is being used? You are arbitrarily ignoring Newton when it comes to momentum. $\endgroup$ – DavePhD Feb 28 '14 at 15:41
  • $\begingroup$ True, but the current version of the Newtonian predictions doesn't follow the premise of using purely Newtonian physics anyway, they use $E=mc^2$. The idea is not to go purely Newtonian, but as far as possible $\endgroup$ – Jim Feb 28 '14 at 16:13
  • $\begingroup$ I wouldn't agree that the current version of the Newtonian prediction relies on special relativity to calculate mass. It was already known that light had momentum prior to knowledge of special relativity. Light pressure was already measured. I would say the mass value comes from $p = mv = mc$, which is purely Newtonian. So the current version goes further in being purely Newtonian than your proposed version. $\endgroup$ – DavePhD Feb 28 '14 at 17:09
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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.

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  • $\begingroup$ but if the rest energy of a photon (at least a theoretical photon) is exactly and identically zero, how can it be the leading order contribution? Also, I'm well aware of the restrictions and the pointlessness of using Newtonian physics in the limit where $v\rightarrow c$, but that is what was used so that is what I'm focused on. Most of Newtonian physics runs into trouble in the relativistic limit, so if you must use it, why not use its kinetic energy? It's the one energy you know a photon has. $\endgroup$ – Jim Feb 28 '14 at 14:24
  • $\begingroup$ @Jim if you want to give a mass to the photon then its rest energy won't be zero. That is the problem you can't first assume is zero and then select the second term (KE) and say it isn't. $\endgroup$ – jpm Mar 1 '14 at 23:05
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    $\begingroup$ I've come to realize that a question asking for the reasoning of using a certain value when analyzing a physical system with physics that don't apply well to in the limit of that system isn't exactly a fair question. You do make a valid point. So +1 and thanks for the discussion! $\endgroup$ – Jim Mar 2 '14 at 15:20
  • $\begingroup$ That's what I was trying to explain. You're welcome and thank you too for making me thing about this. I believe this is very constructive because only when you analyse the limits and correspondences of physical theories you can come to a complete knowledge of them. And that the first step in trying to extend them. $\endgroup$ – jpm Mar 2 '14 at 16:10

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