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The gravitational force on your body, called your weight, pushes you down onto the floor. $$W=mg$$ So, what is the weight equation through general relativity?

Waffle's Crazy Peanut
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2 Answers2

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Start with the Schwarzschild metric $$ds^2 = (1-\frac{r_S}{r})c^2dt^2-(1-\frac{r_S}{r})^{-1}dr^2-r^2d\Omega^2 $$ where $$r_S=\frac{2GM}{c^2} $$ A particle at rest at radius $r$ and angular parameters zero from the centre of mass has worldline $$ x^{\mu}=(t, r, 0, 0)$$ Its four velocity is thus $$ u^{\mu}=\frac{dx^{\mu}}{d\tau}=((1-\frac{r_S}{r})^{-\frac{1}{2}}, 0, 0, 0)$$ Its four-acceleration is $$a^{\mu}= \frac{du^{\mu}}{d\tau}+\Gamma^{\mu}_{\alpha \beta}u^{\alpha}u^{\beta} $$ After looking up the Christoffel symbols, because I'm lazy, I get $$ a^{\mu} = (0, \frac{c^2r_S}{2r^2}, 0, 0)$$ So the Lorentz norm squared of the four-acceleration is $$g_{\mu \nu}a^{\mu}a^{\nu}= \frac{c^4r_S^2}{4r^4(1-\frac{r_S}{r})}=\frac{G^2M^2}{r^4(1-\frac{2GM}{c^2r})}$$ Now the proper acceleration of an object at time t is the acceleration relative to an observer in free fall, who is momentarily at rest w.r.to the object at time t. The free fall guy is the one who is not accelerating - the object held at rest at radius r is the one who is accelerating. As we've shown, his acceleration is $$\frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} $$ So if you want to define a force, it would be $$F=ma=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} $$ As $c\rightarrow \infty$ we recover the Newtonian definition, but nobody bothers phrasing it in these terms.

user2299067
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twistor59
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    @CrazyBuddy Hah, it's only awful when you try to shoehorn concepts like forces into GR. That's why textbooks avoid them - all you need is the Einstein equation and the geodesic equations as Lubos mentioned. Then it becomes an object of beauty! – twistor59 Dec 22 '12 at 13:57
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    "How do you guys learn such things..?" - very slowly and painfully! :-) – John Rennie Dec 22 '12 at 14:43
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    @CrazyBuddy Or at a decent rate, and very much not painfully, if you happen to get the one-in-a-million combination of a great instructor and a great text ;) (GR is the only branch of physics where this happened for me) –  Dec 23 '12 at 05:08
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    Twistor, a fun answer but bizarre one, too. You at most tried to incorporate one particular GR correction and it is not the largest one. In your metric, you neglected the spin of the Earth and the associated interesting GR phenomena including frame-dragging and, more importantly, the already Newtonian centrifugal and Coriolis force that get mixed with the relativistic corrections. All those things are irrelevant in practice because they combine to some $W=mg$ at the end with some $g$ when the impact on classical mechanics is calculated. – Luboš Motl Dec 23 '12 at 05:40
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    @LubošMotl that's because I only know how to calculate simple things! – twistor59 Dec 23 '12 at 08:27
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    I have almost no training in Physics. Just curious: what if $1-\frac{2GM}{c^2r}<0$? Does your result imply that the mass $M$ of an object cannot exceed $\frac{c^2r}{2G}$?! Or if $r$ is sufficiently small, the weight is not defined? – user1551 Oct 12 '13 at 14:59
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    @user1551 the condition $1-\frac{2GM}{c^2r}<0$ would mean that you're at a point inside the Schwarzschild radius i.e. inside the horizon of a black hole. So, $M$ can exceed $\frac{c^2r}{2G}$, but it just means that a black hole is present. – twistor59 Oct 14 '13 at 06:38
  • @twistor59 You don't need the Einstein equation and the geodesic equations, just Einstein's equation is sufficient. – Timaeus Sep 05 '14 at 23:29
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    Einstein didn't know the Schwarzschild solution when he published, yet he must have established this equivalence in the limit with Newton. So how did he do it? – Mark Adler Dec 24 '14 at 19:15
  • So to get the acceleration of someone standing on Earth, you need to add about 0.7 parts per billion to Newton's result. For a non-spinning Earth anyway. – Mark Adler Dec 24 '14 at 19:22
  • @MarkAdler see e.g. here for an explanation of the Newtonian limit in GR. As you can see, there is no Schwarzschild solution involved. – Danu Mar 18 '16 at 16:34
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    A step by step derivation is given here: http://www.physicspages.com/2013/12/31/schwarzschild-metric-acceleration/ See also Hartle, pages 434-435. – Peter4075 Mar 22 '17 at 14:31
  • @Timaeus What do you mean? – user76284 Feb 11 '20 at 19:34
  • If weight is to be talked about in relativity, wouldn't it make more sense to say that W = Eg, where E is the total energy including the rest mass? Isn't it the total energy that determines the amount of curvature of spacetime? Or are you using m in the pre-1970's sense (in the context of professional relativists) of the mass equivalent of the total energy of the object? I found out recently that since the 1970's pro relativists don't say that mass increases with speed, but they do say that energy (meaning total energy including m0) and therefore curvature of spacetime increases with speed. – Matthew Christopher Bartsh Jul 14 '22 at 21:02
  • The physicspages article linked by Peter4075 is now at https://physicspages.com/pdf/Relativity/Schwarzschild%20metric%20-%20acceleration.pdf – PM 2Ring Aug 29 '23 at 11:26
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Morally speaking, the formula is still obeyed in general relativity. However, both $W$ and $g$ become sort of obsolete quantities in general relativity, so we would never describe the behavior of the physical system in this way.

In general relativity, the gravitational acceleration $g$ has to be carefully replaced by a quantity that encodes the Christoffel connection for the metric. The connection isn't a proper tensor so the numerical values strongly depend on the choice of the coordinates. In various coordinates, one could write an equation that would resemble $W=mg$. At the end, however, we would be interested in the motion of an object, so we would be forced to rewrite $W$ as $ma$ in general relativity as well as $a$ would have to be calculated from the world line of the moving object etc.

In the same, as the previous sentences already hinted, $W$ is kind of obsolete. In general relativity, it's easiest to study the free fall of the objects – gravity is the only force that acts in this case. The free fall is described by the condition that the world line is a geodesic, $\delta \tau_{\rm proper} = 0$: the proper time along the world line is maximized (yes, maximized, it's not a mistake, the sign is unusual due to the Minkowski signature). In this form, the law is independent of the mass of the object, which isn't surprising given the equivalence principle underlying GR (all objects are affected by the gravitational field to the same extent).

If the object weren't moving in a free fall, one would have to describe the other forces that act on the object using the language of mechanics of the continuum – essentially field theory. General relativity doesn't describe macroscopic objects by "several numbers" such as positions and velocities. It requires us to describe the pressure etc. at every point of the objects and study how the pressure evolves and how it is affected by the spacetime curvature. So mechanical formulae of the type $W=mg$ are only good to describe point masses in GR: they're just inappropriate for extended objects. And for point masses, the only "long distance forces" that could act in a controllable way are the electromagnetic forces. Forces caused by the mutual contact of bodies require the bodies to be macroscopic, and then the usage of the field theory formalism of GR is necessary.

To summarize, $W=mg$ is an example of the obsolete language of Newton's mechanics and general relativity doesn't modify just some precise functional dependence in similar equations – which is what special relativity does. It forces us to describe the same phenomena using different, much more general concepts, too.

Luboš Motl
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    hi @Luboš Motl i am looking for mathematical equation –  Dec 22 '12 at 11:00
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    The mathematical equations governing everything in GR are either variations of $G_{\mu\nu}=8\pi G T_{\mu\nu}$ for the background geometry or variations of $\delta \tau=0$ for the world lines of objects moving on a spacetime background, or some hybrid equations with additional fields. At any rate, they're partial differential equations for tensors and you would have to be more specific about what the desired equation should describe. There is no GR counterpart of $W=mg$. – Luboš Motl Dec 22 '12 at 12:03
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    +1 The answer is brilliant in terms of the intuitions, though twistor gave a specific and explicit answer to a special case. – 299792458 Jun 13 '14 at 14:05
  • If weight is to be talked about in relativity, wouldn't it make more sense to say that W = Eg, where E is the total energy including the rest mass? Isn't it the total energy that determines the amount of curvature of spacetime? Or are you using m in the pre-1970's sense (in the context of professional relativists) of the mass equivalent of the total energy of the object? I found out recently that since the 1970's pro relativists don't say that mass increases with speed, but they do say that energy (meaning total energy including m0) and therefore curvature of spacetime increases with speed. – Matthew Christopher Bartsh Jul 14 '22 at 20:58
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    Matthew, science doesn't say what one "should" do and which concepts should be fashionable. Science decides, using empirical data and reasoning, which precise statements are correct and which are wrong. You didn't list any precise statements so I cannot give you precise answers. Yes, the modern recent terminology is such that the mass tends to be reserved for the rest mass (an invariant, a scalar) while energy is reserved for the time component of the energy-momentum vector (not Lorentz-invariant) and in this sense, the words mass and energy are not the same. But context is needed. – Luboš Motl Jul 16 '22 at 04:07
  • When a weight sits on a scale, we know its velocity, it should be zero, relatively to the scale, otherwise it's flying somewhere, it is not being measured. Because v is zero, we shouldn't pretend that we are writing some very general formulae that hold for all speeds and all components. OK, now, when the speed is zero, there may still be some internal energy inside the weight, atoms bouncing back and forth (heat) and balls bouncing back and forth, and indeed, this internal motion or energy does increase the total gravitational mass as well as energy (E=mc^2 holds) of the weight & the force. – Luboš Motl Jul 16 '22 at 04:09
  • In short, even the simplest phenomena become inextricably complicated in GR. – my2cts Jan 16 '24 at 09:42
  • It is just somewhat harder than in Newton's physics. And more accurate, and in some fancy sense even simpler and requiring less independent ideas. – Luboš Motl Jan 17 '24 at 13:08