# Since quantum mechanics give you that photons have (relativistic) mass $m=\frac{hf}{c^2}$, why gravity does not accelerate it?

Since quantum mechanics give you that photons have (relativistic) mass $m=\frac{hf}{c^2}$ why gravity does not accelerate it?? I know it changes its energy hence its frequency hence its wave length-colour.But why it does not speed up? If you consider photons massless then it is obvious but then you would not take in consideration that Energy equals mass*$c^2$ and since photons have energy they can't have zero mass. (I'm a mathematics undergraduate took a course on an introduction to quantum physics so try to give a more intuitive answer than or if you use math please be rigorous to the interpretation of the quantities.) (h=planck's constant ,c=speed of light constant , f=frequency).I partly found the answer .I can accelerate without changing speed just direction hence i have bending of light in directions thus acceleration.But what if i send light Straight to the centre of the mass.Nor its speed will change nor its direction.How will i explain acceleration then?

• Not all energy is in the form of mass. Oct 22, 2016 at 15:14
• Have a read, e.g., through en.wikipedia.org/wiki/Gravitational_lens Oct 22, 2016 at 15:15
• @WillO then the equation for the mass of a photon i wrote is wrong?
– Jam
Oct 22, 2016 at 15:15
• @ManolisLyviakis: The rest mass of a photon is $0$. The relativistic mass of a photon is infinite. Your expression is equal to neither $0$ nor $\infty$. Oct 22, 2016 at 15:16
• Possible duplicates: physics.stackexchange.com/q/98750/2451 , physics.stackexchange.com/q/218051/2451 and links therein. Oct 22, 2016 at 15:16

The concept of relativistic mass is considered, basically, out-dated. In this case, gravity does exert a force on photons, but because the relationship between energy and momentum for photons is $$E=pc$$ force alters the momentum of the photon and the energy of the photon, but not the speed of the photon. See, force is not equal to $m\mathbf{a}$, that's a derived relationship for particles that have mass. The fundamental definition of force that applies for all particles is $$\mathbf{F} = \frac{\operatorname{d} \mathbf{p}}{\operatorname{d} t},$$ in words, force is the time rate of change of momentum. For conservative forces, we also have $\mathbf{F} = -\nabla U$, with $U$ the potential energy. Combining these we get the equation: $$-\nabla U = \frac{\operatorname{d} \mathbf{p}}{\operatorname{d} t},$$ in words, the relationship that defines forces is about the rate at which potential energy changes over space gets transferred into the time rate of change of momentum.

• Exaclty as i said, but yours is a complete answer. Oct 22, 2016 at 15:55
• So if i use formulas like Potential energy=relativisticmass* gravitationalforce *Height is wrong? For photons?Or use Kinetic energy=1/2mc^2 ??
– Jam
Oct 23, 2016 at 16:43
• Off the top of my head, I don't know. See if doing that allows you to replicate the formula of gravitational redshift verified in the Pound-Rebka experiment: $$f_r = \sqrt{\frac{1 - \frac{2GM}{(R+h)c^2}}{1 - \frac{2GM}{Rc^2}}} f_e.$$ Oct 24, 2016 at 4:06

Quantum mechanics or any other mechanics that I know of does not give an equation for the mass of a photon.
Saying that a photon has mass is an invalid concept.

The bending of light travelling from a source to an observer because it has passed some matter which is between the source and the observer is a prediction of the general theory of relativity and was proved correct when Eddington observed the solar eclipse of May 1919.

It comes from the idea that matter can bend space(time).
Here is a demonstration which tries to illustrate the point.

Another important example of the bending of a beam of light is gravitational lensing using which astronomers have been able to detect exoplanets.

• So the equation for the mass of a photon i wrote is wrong?
– Jam
Oct 22, 2016 at 15:38
• Think of photons in terms of energy and momentum. Mass by itself is not a useful feature of photons. Oct 22, 2016 at 15:47
• But when you enclose two photons for eg inside a perfectly insulated box, you can measure the increase in box's mass. Again to reiterate, this does not mean photon has mass , its 'mass' is the kinetic energy which we are actually measuring Oct 22, 2016 at 15:59

"Accelerate" is not the right word since the velocity of a photon has to remain constant as a result of its being massless. The confusion I think comes from thinking of gravity as an instantaneous force that attracts things but this idea does not survive special relativity. What we call gravity is a result of curvature of spacetime. So I would rephrase your question as do photons "feel" the bending of spacetime? The answer is yes. For example for the schwartzchild solution of GR, photons can have a circular orbit about a massive body at $r =\frac{3GM}{c^2}$.

The answers already answered the question of acceleration and what really happens. Light bends into the well of a gravitational field, or in general relativistic terms, towards a high curvature region of spacetime.

If it is going directly in it clearly cannot bend any more. But in all cases when it goes towards a higher gravitational field, i.e., towards a higher curvature spacetime, it will experience a blue shift. Ie, the energy of each photon will increase. As the OP mentioned, the energy of a photon E = hf. So as it goes into a gravitational well it increases its energy, and thus its frequent, it blue shifts. When it tries to get out it red shifts.

No need to bring mass in for a photon, it means nothing except really it's equivalent energy.

For a photon travelling toward a gravitational source, MC Physics would suggest that gravity would increase the kinetic energy of the rotation of a photon (ie., increase its frequency) and not its linear KE and velocity, which is limited by relativity.

For a photon travelling away from a gravitational source, MC Physics would suggest that gravity would decrease KE rotation (ie., cause a red-shift in frequency). That is an clear and simpler explanation of Hubble's red-shift of distant stars and galaxies than of an 'ever expanding and accelerating universe'.

"But what if i send light Straight to the centre of the mass.Nor its speed will change nor its direction."

Its speed will change. General relativity predicts that the speed of light falling towards the source of gravity decreases (in the gravitational field of the Earth the acceleration is negative, -2g:

http://www.speed-light.info/speed_of_light_variable.htm "Einstein wrote this paper in 1911 in German. [...] ...you will find in section 3 of that paper Einstein's derivation of the variable speed of light in a gravitational potential, eqn (3). The result is: c'=c0(1+φ/c^2) where φ is the gravitational potential relative to the point where the speed of light c0 is measured. Simply put: Light appears to travel slower in stronger gravitational fields (near bigger mass). [...] You can find a more sophisticated derivation later by Einstein (1955) from the full theory of general relativity in the weak field approximation. [...] Namely the 1955 approximation shows a variation in km/sec twice as much as first predicted in 1911."

Newton's emission theory of light predicts that the speed of light falling towards the source of gravity varies like the speed of ordinary falling objects (in the gravitational field of the Earth the acceleration is g):

http://www.einstein-online.info/spotlights/redshift_white_dwarfs Albert Einstein Institute: "One of the three classical tests for general relativity is the gravitational redshift of light or other forms of electromagnetic radiation. However, in contrast to the other two tests - the gravitational deflection of light and the relativistic perihelion shift -, you do not need general relativity to derive the correct prediction for the gravitational redshift. A combination of Newtonian gravity, a particle theory of light, and the weak equivalence principle (gravitating mass equals inertial mass) suffices. [...] The gravitational redshift was first measured on earth in 1960-65 by Pound, Rebka, and Snider at Harvard University..."