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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?

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If you like this question you may also enjoy reading this Phys.SE post. –  Qmechanic Jan 3 '13 at 1:51

2 Answers 2

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 $$g_{\mu \nu}a^{\mu}a^{\nu}= \frac{c^2r_S^2}{4r^4(1-\frac{r_S}{r})}=\frac{G^2M^2}{r^4(1-\frac{2GM}{c^2r})}$$ 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's not accelerating - the object held at rest at radius r is the one who's 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.

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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) –  Chris White Dec 23 '12 at 5:08
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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 5:40
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@LubošMotl that's because I only know how to calculate simple things! –  twistor59 Dec 23 '12 at 8:27

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.

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hi @Luboš Motl i am looking for mathematical equation –  user17093 Dec 22 '12 at 11:00
    
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
    
+1 The answer is brilliant in terms of the intuitions, though twistor gave a specific and explicit answer to a special case. –  New_new_newbie Jun 13 at 14:05

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