I was thinking about this recently and I think this is a reasonable question to ask. SR can handle forces just like Newtonian physics can; the difference is that the four-force is defined a bit differently than the classical force.

So this made me wonder; what's the difference between that a body is experiencing a force, of let's say 5 newtons, due to electrostatic forces and that the same body is experiencing the same force due to gravity?

Can't we just treat the force due to gravity like any other four-force in SR when it comes to gravity? Why does the origin of the force matter? A force is still a force, so what exactly does SR fail at? Can't we just treat gravity as a four-force acting on a body, like any other force in SR?

  • $\begingroup$ This is almost a duplicate, but when questions similar to this were marked as duplicates the "originals" to which they refer tend to be about the particle theory level description of gravity, which is somewhat orthogonal to what's being asked here. $\endgroup$ – DanielSank Jul 8 '15 at 23:22
  • $\begingroup$ "Can't we just treat the force due to gravity like any other four-force in SR when it comes to gravity?" Briefly, (and I believe I've answered this similarly before here), if gravity were described by a four-force, gravitational waves would transport negative energy. Courtesy of Google Books and "Gravitation": [1]: i.stack.imgur.com/eyWKU.png $\endgroup$ – Alfred Centauri Jul 9 '15 at 0:29
  • $\begingroup$ Well, general relativity doesn't treat gravity as a real force - it's an inertial/apparent force. That's a bit of a paradigm shift :) One of the key observations is that inertial mass is the same as gravitational mass - very suspicious. So GR redefined universe in a way where gravitation is an inertial force - and it works marvellously. Indeed, SR handles gravity-as-a-real-force just fine - GR throws this explanation out of the window, and in doing so, explains a few suspicious things, as well as some pretty annoying things - like "why would light bend in a gravitational field" :) $\endgroup$ – Luaan Jul 9 '15 at 10:16
  • $\begingroup$ @AlfredCentauri If we described gravity as just an other four-force, then there would be no such thing as "gravitational waves" since GR would not even exist to begin with. $\endgroup$ – Madde Anerson Jul 9 '15 at 12:14
  • $\begingroup$ @MaddeAnerson, you're assuming that, by gravitational waves, one means propagating disturbances in spacetime curvature. But that isn't the case. Since the context is SR, there is no instantaneous action at a distance and thus, gravitational 'influence' must propagate with finite speed regardless. $\endgroup$ – Alfred Centauri Jul 9 '15 at 13:34

While DanielSank's answer is correct, I don't think it is the complete story. Sure the equivalence principle naturally leads to a geometric description of gravity but it does not necessitate such a description in and of itself.

For example, Newtonian gravity can be described geometrically in a manner completely analogous to GR but at the same time this description is entirely equivalent to the usual Newtonian description of gravity-both respect the equivalence principle but the geometric description comes from a reinterpretation of inertial frames.

So there is more that needs to be said regarding your question. For example, coming back to SR, what if I posit a scalar field $\varphi$ for the gravitational field that obeys a Lorentz covariant linear field equation? It is easy to convince oneself that such a field equation would have to take the form $\square \varphi = -4\pi T$ where $T$ is the trace of the stress-energy tensor of the matter field coupled to gravity.

There are immediately two problems with this. The less problematic of the two is that this equation implies the gravitational field does not couple to itself since $\square$ is a flat space-time operator and $T$ knows nothing about gravity (remember, in SR the metric is non-dynamical). This can however be fixed by making the gravitational field equations non-linear. More importantly is the issue that $T =0$ for the electromagnetic field. This means light would not be deflected by gravity, which immediately contradicts experiment and throws a scalar theory of gravity out of the picture.

Let us now consider a 4-vector $A^{\mu}$ for the gravitational field. It is actually very easy to show that this route leads to a dead end because any relativistic classical field theory of a vector field will yield repulsive like charges (see reference below), and we know gravity is attractive.

We thus finally come to a dimensionless symmetric tensor field $h_{\mu\nu}$ on flat space-time (see reference below for why a symmetric tensor is sufficient). Here the story is much more subtle and rather detailed. If we assume the gravitational interaction is linear then a rather detailed and involved set of arguments (again see reference below) lead to the field equations $-\square \bar{h}^{\mu\nu}- \eta^{\mu\nu}\partial_{\alpha}\partial_{\beta}\bar{h}^{\alpha\beta} + 2\partial_{\alpha}\partial^{(\nu}\bar{h}^{\mu)\alpha} \propto T^{\mu\nu}$, where $\bar{h}_{\mu\nu}$ is the trace-reverse.

However the gauge freedom $h_{\mu\nu} \rightarrow h_{\mu\nu} - \partial_{\mu}\xi_{\nu} - \partial_{\nu}\xi_{\mu}$ of the equations of motion demands that $\partial_{\mu}T^{\mu\nu} = 0$. This can only hold approximately since one cannot have energy conservation in the matter fields when they interact dynamically with the gravitational field.

This can be fixed by instead considering a non-linear gravitational interaction obtained from the above equations of motion through reduction of order. However such an order reduction process requires additional input in order to get a unique, gauge-invariant result; one can in fact show that with the appropriate inputs this iteration can yield the Einstein equations but this requires us knowing GR beforehand which defeats the purpose.

Thus it is much easier and far more elegant to simply adopt Einstein's conceptual leap through the equivalence principle and consider a dynamical metric theory of gravity (metric due to its elegant coherence with the equivalence principle and dynamical due to background independence/diffeomorphism invariance).


"Gravitation: Foundations and Frontiers"-T.Padmanabhan, sections 2.8, 3.3.

  • $\begingroup$ This is an interesting answer but OP really just asked why gravity can't be treated as a usual 4-force. I think the blatant contradictions between $F=ma$ and how gravity works given in my answer definitely answer that question. Again, I still think this answer is interesting and well written. $\endgroup$ – DanielSank Jul 9 '15 at 1:46
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    $\begingroup$ This seems to be the only answer so far that addresses the original question of why SR is needed. I wonder if the scalar field theory you mentioned would be a reasonable “low field” approximation, and give more than Newtonian gravity (gravitational waves, precession of Mercury...). Or maybe this should be a separate question? $\endgroup$ – Edgar Bonet Jul 9 '15 at 7:35
  • $\begingroup$ @DanielSank, fair point; I don't disagree with what you've said. $\endgroup$ – FenderLesPaul Jul 9 '15 at 17:40
  • $\begingroup$ @Edgar Bonet, That might be good to ask separately. I would love to try and answer it if you do. $\endgroup$ – FenderLesPaul Jul 9 '15 at 17:41

As Einstein himself put it, "If a person falls freely he will not feel his own weight."

Consider yoruself sitting in your chair right now. You don't see to be accelerating, yet you feel a force on your butt. This is very strange because $F=ma$$^{[a]}$ would suggest that if you are under the influence of a force (the one on your butt) you should be accelerating. Something is strange.

Now suppose you jump of a cliff and are in free fall. Now you most certainly seem to be accelerating, yet you feel no forces at all. This, again, runs against what you'd expect based on $F=ma$.

The resolution of this is that what we normally think of as accelerating under the influence of gravity is not really acceleration. An object in free fall is moving along its most natural unperturbed trajectory. It's actually you, sitting on your chair, under the influence of gravity but not moving, that are being deflected from your natural trajectory. You are, in a real sense, being accelerated away from your geodesic path through space-time.

$[a]$: And the special relativistic extension.

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    $\begingroup$ This argument proves too much. If you were sitting on a chair in space, and were being pulled down by electromagnetic forces rather than by gravity, you would "feel a force on your butt" despite the fact that you weren't accelerating. Similarly, if you took the chair away, you would start accelerating despite the fact that you don't feel any force. (I'm assuming here that the charge in your body is uniformly distributed.) Since that argument doesn't prove that the electromagnetic force is geometric in nature, it can't be used to prove that the gravitational force is either. $\endgroup$ – Harry Johnston Jul 9 '15 at 2:37
  • $\begingroup$ (The observation that gravitation always behaves this way but electromagnetic forces only behave that way under specific circumstances suggests GR as an elegant solution - but does not prove that it is necessary. And at the time in question, I don't believe we had the experimental evidence necessary to rule out the possibility that the equivalence principle is only approximately true, or only true under certain circumstances.) $\endgroup$ – Harry Johnston Jul 9 '15 at 2:44
  • $\begingroup$ @HarryJohnston isn't part of what makes gravity special that it does in fact act on all matter (and light) whereas things like electrical forces do not? Therefore doesn't my answer indeed address OP's question of why gravity cannot be treated like a usual force? I'm not claiming this answer is a complete motivation of geometrical space-time, just that it's sufficient to answer the original question. $\endgroup$ – DanielSank Jul 9 '15 at 17:41
  • $\begingroup$ I don't read the question quite the same way. But my objection isn't that, it's that I think it misleading to suggest that there's something inherently mysterious about feeling a "force on your butt" when you're not accelerating - IMO, this description misses the point of the equivalence principle, and could cause a novice to become confused when dealing with classical mechanics. $\endgroup$ – Harry Johnston Jul 9 '15 at 21:59
  • $\begingroup$ @HarryJohnston definitely see your point. $\endgroup$ – DanielSank Jul 10 '15 at 5:14

General relativity is not a replacement of special relativity because the latter fails with gravity, rather it is the extension of its kinematic and dynamical quantities to accelerated systems, which, in turn, correspond to writing down the same equations of motion but with a different (non-Euclidean) metric.

Why then accelerated systems are equivalent to gravitation fields is nothing that concerns general relativity but it is instead a consequence of the fact that inertial mass and gravitational mass are (so far) equivalent. This undergoes the name of, in fact, equivalence principle and there is a lot of literature explaining why it is so: see for example here or any other similar question asked in the database.

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    $\begingroup$ You can describe accelerated systems perfectly well from within special relativity. You just have a global notion of an unaccelerated frame. Also, there are many theories that are not general relativity that admit the equivalence principle. Namely, any metric theory will. $\endgroup$ – Jerry Schirmer Jul 8 '15 at 23:46
  • $\begingroup$ How do you describe accelerated systems in special relativity? $\endgroup$ – gented Jul 8 '15 at 23:54
  • $\begingroup$ You do the coordinate transformations to the accelerated frames. There will be effects analogous to Coriolis forces, and it will be clear that you are not in an inertial frame, but you can still make predictions and know things like how much proper time has been elapsed in the accelerated frame. $\endgroup$ – Jerry Schirmer Jul 8 '15 at 23:56
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    $\begingroup$ Yes, and that's exactly how you pass to general relativity once you realise that the same additional non-inertial forces can be described by having a non-Euclidean metric. This is exactly the step taking you from SR to GR, isn't it? $\endgroup$ – gented Jul 9 '15 at 0:00
  • $\begingroup$ In principle, yes. You still need an equation of motion for the metric. $\endgroup$ – Jerry Schirmer Jul 9 '15 at 0:14

protected by Qmechanic Jul 12 '15 at 13:10

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