B Is gravity a force?

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Gravity can be described not as a force but a curvature of spacetime. I assume this can’t be done to the other 3 fundamental forces. If so, then we cannot treat gravity in a way similar to the other forces. Why then does QFT postulate the existence of gravitons? Why does it attempt to treat gravity in a way similar to the other forces?

It is also postulated that the 4 fundamental forces were once a single fundamental force. But if gravity is not a force, why do people still treat it as one of the fundamental forces? And worse, how could people still think that gravity were once the same force as the other forces?

I say gravity is not a force because a force model of gravity cannot explain the bending of light near a massive object. So shouldn’t we give up trying to grand-unify gravity, a non-force, with the “other” forces?
 

Vanadium 50

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Orodruin

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Why then does QFT postulate the existence of gravitons?
It doesn’t. It is a general expectation that a quantum theory of gravitation will contain gravitons.

Also, at the quantum level we usually talk about ”interactions”, not ”forces” as force is a rather classical concept.

I say gravity is not a force because a force model of gravity cannot explain the bending of light near a massive object.
This is incorrect. Even Newtonian gravity suggests that light bends in the particle model of light.

So shouldn’t we give up trying to grand-unify gravity, a non-force, with the “other” forces?
You cannot base opinion such as this on popularised descriptions of science. An informed opinion on this can only stem from looking at similarities and differences in the actual theory.
 

vanhees71

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I would abandon the word "force" already at the classical level of relativistic physics and keep "force" strictly in the Newtonian realm, where you have instantaneous interactions at a distance.

The great insight by Faraday was the field concept, which paved the way for the field-theoretical view and finally to relativity. What appears as "force" within classical point-particle mechanics is the local interaction of the point particle with the field, which itself is caused by the presence of other particles due to their corresponding "charges".

The gravitational interaction is special due to the equivalence principle, which admits us to reinterpret the corresponding spin-2 field as the fundamental form of a pseudo-Riemannian spacetime, but the principle is the same: The presence of energy, momentum, and stress of some entity (matter or fields, which in the quantum field theoretical view are anyway not much different) cause a curvature of spacetime which determines the motion of test particles to look for an observer as if there is some "force" acting on it, though it's just moving along a straight line ("geodesic") in the pseudo-Riemannian spacetime, and this geodesic is again governed by local laws.
 
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Even Newtonian gravity suggests that light bends in the particle model of light.
Could you explain further? If light is massless, then its motion should be unaffected by Newtonian gravity.
 
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Newtonian gravity says the gravitational force on massive particles is F=km (with some prefactor k) and links this to an acceleration via F=ma. Combine both: am=km. For every non-zero mass a=k, the acceleration does not depend on the mass. It is plausible to take the limit for massless particles, in that case they have the same acceleration.
 

Nugatory

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Gravity can be described not as a force but a curvature of spacetime. I assume this can’t be done to the other 3 fundamental forces.
That's right (at least for purposes of a B-level thread). Curvature works as an explanation when everything is affected the same way, as with gravity; when a road curves to the left all the traffic on that road is going to follow that curve. It doesn't work when positive-charged particle have their paths deflected in one direction, negative-charged ones in another, and uncharged particles aren't deflected at all - that can't be explained by saying that they're all following the same curved road.
If so, then we cannot treat gravity in a way similar to the other forces. Why then does QFT postulate the existence of gravitons? Why does it attempt to treat gravity in a way similar to the other forces?
Current quantum field theories work well when gravitational effects are negligible and general relativity works well when quantum effects are negligible. But neither theory works well when both quantum effects and gravitational effects are too strong to ignore so the challenge is to develop a theory of quantum gravity that does work under those conditions (and agrees with GR and QM everywhere else). Gravitons will naturally appear in such a theory, for reasons that are analogous to the way that photons naturally appear in quantum electrodynamics.
It is also postulated that the 4 fundamental forces were once a single fundamental force. But if gravity is not a force, why do people still treat it as one of the fundamental forces? And worse, how could people still think that gravity were once the same force as the other forces?
You are being misled by the way that the popular press misuses the word “force” in describing quantum theories. The more accurate word “interaction” would reduce the confusion... but the truth is in the math not the words, and there is no confusion in the math.
 
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It is plausible to take the limit for massless particles, in that case they have the same acceleration.
So a photon on the Earth surface experiences an acceleration of 9.81 ##ms^{-2}## too? So its radius of curvature ##r=\frac{c^2}{9.81}\approx10^{16}## m?

But what if it travels straight down towards the Earth? It cannot move faster than ##c##, so its acceleration would have to be zero.

So given the angle ##\theta## light makes with the horizontal, the radius of curvature ##r=\frac{(c\cos\theta)^2}{9.81}\approx10^{16}\cos^2\theta## m. Is this the result based on Newtonian gravity?
 

Orodruin

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So a photon on the Earth surface experiences an acceleration of 9.81 ##ms^{-2}## too? So its radius of curvature ##r=\frac{c^2}{9.81}\approx10^{16}## m?

But what if it travels straight down towards the Earth? It cannot move faster than ##c##, so its acceleration would have to be zero.

So given the angle ##\theta## light makes with the horizontal, the radius of curvature ##r=\frac{(c\cos\theta)^2}{9.81}\approx10^{16}\cos^2\theta## m. Is this the result based on Newtonian gravity?
We said Newtonian gravity a priori has acceleration of massless objects, not that it is accurate. It makes predictions that are incompatible with observations, such as the amount of bending.
 

PeroK

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So a photon on the Earth surface experiences an acceleration of 9.81 ##ms^{-2}## too? So its radius of curvature ##r=\frac{c^2}{9.81}\approx10^{16}## m?

But what if it travels straight down towards the Earth? It cannot move faster than ##c##, so its acceleration would have to be zero.

So given the angle ##\theta## light makes with the horizontal, the radius of curvature ##r=\frac{(c\cos\theta)^2}{9.81}\approx10^{16}\cos^2\theta## m. Is this the result based on Newtonian gravity?
If you start with Newton's Law ##F = ma##, then that has nothing to say about a massless electromagnetic wave.

If light did have some residual mass, then it would, indeed, accelerate under gravity as any other particle.

You could even, as has been suggested, extend Newton's laws to postulate a gravitational acceleration of light as massless radiation or particle.

All this, however, would be a (perhaps unnecessary and unjustifiable) extension to classical physics. Note that in classical physics there is, in principle, no invariant speed of light.

None of this, however, has any bearing on Relativity. Except that, once GR had predicted light bending round the Sun, it was important to establish that the curvature was different from the hypothetical bending of light under Newtonian gravity.

In any case, you cannot appeal to classical physics on one hand (light accelerating in a plunge orbit) and special relativity on the other (invariant speed of light) at the same time. That simply leads to nonsense.
 
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So a photon on the Earth surface experiences an acceleration of 9.81 ##ms^{-2}## too?
Relative to an observer standing at rest on the Earth's surface, yes, a photon accelerates downward (in the sense of coordinate acceleration) just like any other free-falling object.

Note that this "local" acceleration is not the same as the global bending of light when it passes close to a massive object. The global light bending depends on the object's speed relative to local observers at rest in the gravitational field; heuristically, it is ##1 + v## times the global bending you would infer from the local Newtonian acceleration. For a photon, ##v = 1## (we are using units where ##c = 1##), so the actual global bending of light is twice the bending you would infer from the local Newtonian acceleration.
 

A.T.

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But what if it travels straight down towards the Earth? It cannot move faster than ##c##, ...
It cannot move faster than ##c## in inertial frames. The rest frame of the Earth's surface is not an inertial frame in General Relativity.
 

pervect

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A photon falling down a gravity well does wind up gaining energy, but this is due to a frequency shift (a blueshift, as it's commonly called for gaining energy), and the releation ##E = h \nu##, ##\nu## being the frequency.

The speed of the photon as it falls, measured by a co-located observer, is always c. This is the sort of speed that's physically meaningful, the relative speed of the photon to an observer at the same position as the photon is.

Coordinate speeds are a bit different, but the details of the coordinate speeds depend on the coordinates chosen. Because the choice of coordinates is a convention, "coordinate speeds" don't have any physical significance directly, though physically significant things can be calculated from them if one knows what the coordinate choice is.
 

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