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I recently got a more complete proof of photons having no mass. (I knew it before, but now I really know it.) But now, I'm curious how gravitational lensing can occur without a mass to act on.

I have heard that space is like a sheet and gravity works because the more massive an object is, the more it bends space. I heard that when I was five years old. When I got older I questioned how that would work, seeing as space is 3-dimensional. The answer I eventually cobbled together from a plethora of excellent resources was this:

Gravity is like a point light source. At the center, you have the most intense light. As you move outward the intensity decreases with the square of the distance. Like light, gravity radiates in all directions simultaneously.

This works well for me, and I still believe it to be accurate. However, when I was thinking about photons, I realized that you cannot apply a force to an object without mass. At least, you can't by standard Newtonian thinking. This is because $F=ma$. With no mass, you can have no force. Alternately, you could rearrange to $\frac{F}{m}=a$. With no mass, and no force, you can have no acceleration.

Yet gravity is able to refract light.

How is this possible? Like $E=mc^2$, does this only apply to a specific set of conditions?

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  • $\begingroup$ Photons feel gravity. Read about the gravitational red-shift and the Pound-Rebka experiment. $\endgroup$
    – suresh
    Commented Aug 29, 2014 at 0:22
  • $\begingroup$ I know something about redshift. I was asking if someone could explain how gravity affects photon in general relativity $\endgroup$
    – CoilKid
    Commented Aug 29, 2014 at 0:25
  • $\begingroup$ @suresh David Hammen mentioned in his answer that in GR, gravity is based on geometry, not force. Could someone please expand on that? $\endgroup$
    – CoilKid
    Commented Aug 29, 2014 at 0:27
  • $\begingroup$ @suresh also, I meant gravitational lensing as in black holes. We can't see them, but you could see where one was, in theory, by watching for a distortion of the normal background starlight. Like the edges of a glass lens. $\endgroup$
    – CoilKid
    Commented Aug 29, 2014 at 0:30
  • $\begingroup$ Since you are discovering things for yourself, I did/will not answer your question! The word redshift appears a lot but not all redshifts are the same. So read about the (cool) Pound-Rebka experiment and understand it. $\endgroup$
    – suresh
    Commented Aug 29, 2014 at 0:34

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$F=ma$

Don't do that! You can't mix Newtonian mechanics and special relativity, let alone Newtonian mechanics and general relativity.

Gravitation is fundamentally very different between Newtonian mechanics and general relativity. In general relativity, gravitation is a result of geometry. It is not quite a force. Mass-energy tells space-time how to curve. Curved space-time tells mass-energy how to move.

$E=mc^2$

Don't do that, either! A better expression is $E^2 = (mc^2)^2 + p^2c^2$. Note how this allows objects with zero mass to have energy and momentum, objects with zero momentum to have mass and energy. Another way to look at this expression: Energy, mass, and momentum are just different aspects of one fundamental concept. It is this common concept that results in gravitation and interacts with gravitation.


Objects with non-zero mass have energy thanks to that intrinsic mass, making massive objects subject to gravitation. Light has energy thanks to its momentum, so light too is subject to gravitation.

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  • $\begingroup$ $F=ma$ Yes, I asked if it has special rules like $E=mc^2$. I know that it should be $E^2=(mc^2)^2 + p^2c^2$ In the top of my OP, I said I got a better understanding of why photons have no mass. I meant, I learned that E=mc^2 does not apply to them, and you should instead apply $E^2=(mc^2)^2 + p^2c^2$. $\endgroup$
    – CoilKid
    Commented Aug 29, 2014 at 0:19
  • $\begingroup$ And Also, I know not to mix Relative and Newtonian physics. I was asking if $F=ma$ had a more applicable version of itself, much like $E=mc^2$ has the Relative physics version of $E^2=(mc^2)^2 + p^2c^2$. That was much the point of my question. $\endgroup$
    – CoilKid
    Commented Aug 29, 2014 at 0:21
  • $\begingroup$ Thank you for the knowledge that in General Relativity, gravity is based on geometry, not force. Could anyone please expand on that? $\endgroup$
    – CoilKid
    Commented Aug 29, 2014 at 0:23
  • $\begingroup$ No, for three reasons. Reason #1: You have to learn a lot (a whole lot!) of mathematics to understand general relativity. If you don't have that mathematical understanding you're just fooling yourself. Reason #2: I'm nowhere near as smart as Einstein; I can't reduce GR to a "explain it like I'm five" explanation. Reason #3: While general relativity says what happens with amazing precision, it says nothing about why that stuff happens. General relativity is essentially a kinematic theory. $\endgroup$
    – David Hammen
    Commented Aug 29, 2014 at 0:39
  • $\begingroup$ Then how the heck can I calculate how much a massive object changes the trajectory of a photon moving at a tangent to the gravity well? I don't need it explained like I'm five. Give me the formula, or a link to somewhere with the formula, and let me figure it out from there! $\endgroup$
    – CoilKid
    Commented Aug 29, 2014 at 0:42

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