What is the gravitational force acting on a massless body?

Question

It's a well known fact that acceleration due to gravity is independent of the mass of the accelerating body, and only depends on the mass of the body it is accelerating towards and the distance from it. One can prove this mathematically very easily.

$$F = \frac{GMm}{r^2}\tag1,$$

$$F = ma\tag2.$$

So, $ma = \frac{GMm}{r^2}$ and $m$ cancels out giving

$$a = \frac{GM}{r^2}.\tag3$$

But what if we are to consider the acceleration acting on a massless object (like a photon)? From equation $(3)$, there would still be an acceleration due to gravity, but from equation $(1)$, the product of the masses is zero, and therefore the force would be zero.

This means that the massless particle will experience acceleration with zero net force.

What is the contradiction here? Is it because we cannot divide by $m$ when $m$ is zero?

Answer 1

This question is definitely an interesting one: indeed, trying to work within Newtonian mechanics one finds themselves in an impasse. Should we take the result computed with $m = 0$ exactly, or the limit as $m \to 0$ as our prediction?

This was uncertain in 1919 when Dyson, Eddington and Davidson, during a solar eclipse, measured the displacement due to the Sun's gravity of the light coming from a star which appeared close to the Sun's surface.

They proposed three scenarios:

1. light could be completely unaffected by gravity (zero $a$);
2. it could be affected like an object with negligible but nonzero mass in Newtonian gravity;
3. it could obey the then-new theory of General Relativity, which turned out to predict twice the displacement of case 2.

As you can read in the freely-available paper, interpretation 2 was indeed considered quite a reasonable possibility, especially in light of the Equivalence Principle.

Still, the experiment agreed with case 3: the discussion is purely academic, since in the end light is curved by gravity according to General Relativity.

Answer 2

There are no mass-less particles in Newtonian mechanics and generally classical mechanics.

A photon belongs to the realm of quantum mechanics and special relativity. It cannot be accelerated because by mathematical construction of special relativity it always moves with speed c, the speed of light (as for all mass-less particles ).

At the quantum level force is represented by dp/dt in the interactions between particles, and a photon interacting with an effective quantum gravitational field has a momentum and it can change , but its speed will always be c.

Answer 3

Is it because we cannot divide by $m$ when $m$ is zero?

Yes. You have

$ma = \frac {GMm}{r^2}$

but you cannot cancel $m$ from both sides of this equation if $m=0$, since both sides then equal $0$ and $0/0$ is undefined.

In Newtonian mechanics we can conclude from $F = \frac {GMm}{r^2}$ that the gravitational force on a massless body is zero. So if gravity is the only force acting on a massless body then its acceleration is zero.

Answer 4

F=ma is a simplification of Newton's second law. Newton did not actually assume that mass is a constant. Newton's second law can be more correctly quoted "the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it." And photons do have momentum.

Answer 5

I'm going to have to start with your premise:

It's a well known fact that acceleration due to gravity is independent of the mass of the accelerating body, and only depends on the mass of the body it is accelerating towards and the distance from it.

I hope this is not a well known fact, because it is not correct.

Let's get some basics out of the way:

0. You're using Newtonian Physics Equations, so my answers are in that context. There are already good answers above dealing with relativistic and quantum contexts.

1. The force due to gravity is a mutual attraction between masses. There is no "accelerating body" and "body accelerating towards". They're both accelerating towards each other.

Since we're used to M='The Earth', and m='an apple', it's easy (but wrong) to assume the apple is not significant. If one mass is so much greater than the other, the the second mass can be approximated as "1", but never zero. This is why I always use (F = G m1 m2 / r^2) with my students.

2. The problem you're having with your mathematical proof is as follows:

F = ma <-- this equation applies for one rigid body, where an external acceleration is applied.

F = GMm/r2 <-- this equation is for an attractive force between two rigid bodies, and is a property of the two bodies.

By trying to remove "m", what you're really doing is equating two very different definitions of "F", and getting nonsensical results.

This is another problem - just because two variables have the same units (Newtons), does not mean the phenomenon they measure are interchangeable.

In short, the Newtonian Force of Gravity equation makes no sense with a single body, or two bodies where one has no mass.

Answer 6

Even in Newtonian physics it is possible to see that a massless particle may undergo an acceleration. The problem is ill-defined because whilst the force between a massive and a massless particle is zero, it is also that case that a particle with no inertial mass would have infinite acceleration with the application of any force. As you say, the acceleration is $0/0$ and undefined. It is better to use Newton's second law as force being a rate of chnge of momentum. Since light does have a momentum, then applying any force to it will result in a change in momentum and an acceleration.

Prior to quantum mechanics and General Relativity it was commonly assumed (and calculated) that a gravitational field would bend the trajectory of light - and therefore accelerate it, since the velocity was changed, even if the speed was not. Notably, in 1801 - J. G. von Soldner, (Astronomisches Jahrbuch für das Jahr 1804 nebst einer Sammlung der neuesten astronomischen Wissenschaften, einschlagenden Abhandlungen, Beobachtungen und Nachrichten 1801, 29, 161) calculated a value for the deflection of a light beam by the Sun's gravitational field.

This is a very simple calculation (or Treder & Jackisch 1981) that uses the finite speed of light as its input but assumes that light has a "gravitational mass" that equals its inertial mass. The exact mass of the particle does not feature and cancels when working out the deviation of a light beam. Of course the result is incorrect by a factor of two compared with the predictions of General Relativity, but it is not zero.

It is also possible to do the calculation assuming a photon has a mass given by $E/c^2$ (e.g., Deines 2016).

Answer 7

Ol' Isaac Newton goofed. He wrote:

$F = \frac {G M_1 M_2}{d^2}$

when he should have written:

$F = \frac {G E_1 E_2}{d^2}$

(with a different value and units for G) where it is to be understood that $E$ is the total energy of something.

For light, $E=hf$. Any electromagnetic wave has energy and momentum. For a simple plane wave, this is it. (We'll ignore momentum for now.)

For a bowling ball or planet, $E=\frac {c^2 m} {\sqrt{1-(v/c)^2}}$.

I like to put the $c$ before the $m$ because it's just a coefficient, a constant, and that it the convention throughout most of physics. Everyone is brainwashed to write $mc^2$ but let me get off that soapbox and instead preach about this basic formula for energy of a moving massive particle (meaning things as big as stars, galaxies, etc).

If the particle isn't moving, it gets simpler: $E = c^2 m$. For the situations Newton was pondering, planets and apples, $v \ll c$, and we can have a pretty good approximation by just setting $v=0$.

This way I rewrite Newton's law of gravity resembles Einstein's equation for general relativity.

We're only half done. There's also $F=ma$. Again, Newton goofed. He should have written $F = \frac{dp}{dt} $ where $p$ is momentum. Again, light has momentum, and so do bowling balls and planets. Photons being massless, you can't use the old formula $p=mv$ but waves do have momentum, $p= \frac{hf}{c}$, and this can change with time due to "force". The old formula is fine for bowling balls, planets and expensive vases.

The truly modern way to describe gravity and the motions of things flying around, massive or massless, it to describe curved space-time and the geodesics that follow the curvature, which are the paths of things that aren't being acted on by any forces other than gravity. But this is a different viewpoint, kind of like the difference between saying Earth has gravity which caused expensive vases to fall and break, versus being an astronaut in orbit enjoying "zero gee".

All questions about light, or photons, not fitting the familiar classical formulas as taught in high school level physics, is due to not using the more general formulas we've learned from relativity and quantum mechanics.

Modern theoretical physics is all done with Lagrangians and Hamiltonians. There are ways to write these, which are formulas dealing with energy, for waves and hard objects, massless or massive, in all kinds of plain or exotic situations.

From the Lagrangian or Hamiltonian we can derive the familiar relativistic formulas for massive or massless quanta, or the simpler classical formulas for massive objects. The trouble comes from confusing these different approximations.

Silly of Newton to not know about 20th Century relativity!

Answer 8

In classical physics you have an indeterminacy of the type

$$a=\frac{GM}{r^2} \frac{0}{0} \tag{1}$$

This means that unless you have some new information to find that limit you cannot proceed. However, with some basic knowledge of special relativity an approximate solution can be found. First rewrite equation $(1)$ as $$\frac{dp}{dt}=\frac{GMm}{r^2}\frac{c^2}{c^2} \approx \frac{GM}{r^2c^2} \epsilon \tag{2}$$ Where $\epsilon=c\sqrt{p^2+m^2c^2}$ is particle’s energy (I'm assuming $p^2<<m^2c^2$). Case $m=0$ equation $(2)$ becomes $$\frac{dp}{p} \approx \frac{GM}{c}\frac{1}{r^2}dt \tag{3}$$ Or $$p \approx p_0 e^{\frac{GM}{c}\int{\frac{dt}{r^2}}} \tag{4}$$ As can be seen, a point mass $M$ causes a change of momentum even for a massless particle. To answer your question, the force $F$ would be

$$F \approx \frac{p_0GM}{cr^2} e^{\frac{GM}{c} \int{\frac{dt}{r^2}}} \tag{5}$$

However, the concept of force without mass is senseless and useless to say the least. That’s why it is much better to study the dynamics of particles using Lagrangian or Hamiltonian approaches as those formulations only make reference to energies $E$ and momentum $p$ which are universal for all particles.

Its importance to notice that gravity actually affects objects through the energy-momentum tensor a sixteen element tensor and the “mass” term used here just correspond to its $00^{th}$ component ($T^{00}$). So there is much more to say than equation $(4)$ may suggest.

Answer 9

The answer in unified field dynamics, an underdeveloped science.

There is no such thing as a massless body. Energy indicates mass. Electricity indicates magnetism. Gravity indicates mass. Black holes suck in light. Attraction is due to indicated attracting of opposites. To top it off it's all related to the centrifugal force. A unified field, but it's all in Ivy League physics class, these snippets.

While we're at it, f doesn't = mass * acceleration. f = mass * velocity. Please don't confuse the two like Ivy League does!

The very fact light falls into a black hole shows there's gravity and mass, even of light (lightweight) photons and other masses falling in. Real mass. Black hole.

In fact, very little integration of physics, biology, chemistry, and math sciences is going on in college stuff. This indicates they're not capable of telling any of it exactly the way it is. I know because I've experienced the lies.

It's in first year physics, but most people won't integrate it all. The math confusion is due to relativistic zero term where the (relativistic) particle actually shifted out of time dimension. No mass one place; it went somewhere else. If you measure it where it went you'd get a nonzero figure.

The master solution to your problem is that light is not massless; it's very light-weight, very light-massed. there, no zero figure. Problem solved. All you need to do is integrate

1. energy indicates mass and vice versa
2. gravity indicates mass and vice-versa
3. when you apply relativity, don't measure the particle mass in the place it disappeared; measure it where it went.
4. do the complex math when you run into zeroes, and realize that one universe is, when multiplied, only one, not zero, because there's only one universe. That's one all-in-all universe, not zero.