Why are massless photons affected by gravity?

jtbell
Staff Emeritus
Science Advisor
Homework Helper
Messages
16,023
Reaction score
7,585
In general relativity, gravitation is a manifestation of the curvature of spacetime. The motion of all objects is affected by this curvature, regardless of whether they have mass or not. Light follows geodesic paths in spacetime, which are straight lines in flat spacetime, and curved paths in curved spacetime.

Note that by "mass" above I mean "invariant mass" as discussed in the following FAQ:

https://www.physicsforums.com/showthread.php?t=511175

because it is the invariant mass that is zero for a photon. If you prefer to think in terms of "relativistic mass" (which is related to energy via E = m_{rel} c^2, note that all photons (as far as we know) follow the same geodesics, regardless of their energy. This has been verified, for example, by comparing the deflection of visible light as it passes close to the sun, with the deflection of radio waves from distant sources.

The following forum members have contributed to this FAQ:
jtbell
 
Last edited by a moderator:
  • Like
Likes Manpriyam
Physics news on Phys.org
jtbell said:
In general relativity, gravitation is a manifestation of the curvature of spacetime. The motion of all objects is affected by this curvature, regardless of whether they have mass or not.
So, in GR there are no attractive forces between two masses?
 
Mass produces spacetime curvature (as do energy and momentum), so two masses affect each other gravitationally. Whether to call this "force" is a matter of semantics. If you are falling freely under only the influence of gravity, you do not "feel" it, unless the curvature is so strong as to produce tidal stresses in your body. Therefore many people do say that "gravity is not a force."
 
  • Like
Likes afcsimoes
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
Thread 'Dirac's integral for the energy-momentum of the gravitational field'
See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by ##T_1 \le x^0 \le T_2## and ##\mathbf{|x|} \le R##, where ##R## is sufficiently large to include all the matter-energy fields in the system. Then \begin{align} 0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\ &= \left(...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top