What is the weight equation through general relativity?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

Little late but I'll try anyway.

I thought I'd put this here because it doesn't require world-lines or Christoffel symbols or spacetime curvature to get to.

The equivalence principle tells us that while falling in a gravitational field, you are not being accelerated towards the ground, but in a mathematical sense the ground is accelerating towards you.

It's interesting because if you study accelerated vs inertial reference frames there is indeed a tangible difference, this differs from Newtonian physics where absolute acceleration can seem relative.

Because of this it doesn't really make sense to think of the weight equation as "the force pulling you down", but instead the force you need to supply to remain stationary in the field.

By thinking of each stationary observer in a gravitational field as being like a rocket ship accelerating upwards, we find that time should appear to tick slower below them and faster above them. This is because as an object becomes faster --relative to some inertial frame of reference-- its spatial coordinate in the direction of motion appears to "look" further in to the future and backwards in time in the direction opposite. These effects balance out in such a way that time appears to run faster along the direction of acceleration and slower in the opposite.

Imagining I'm in one of these "stationary yet accelerating" rockets, I can associate a constant $T$ as a ratio of how fast clocks below and above me are ticking relative to my clock.

So if some clock above me is ticking at twice the rate, $T=2$.

We can, by doing some offscreen math, find a formula to calculate of rate of change of $T$ per meter moved up or down.

Being: $\frac{dT}{dx}=\frac{1}{c^{2}}a\cdot T$

Where $a$ is the acceleration needed to stay stationary at that point.

But lets say instead that we already had $T$ and instead wanted to find $a$, the "weight equation".

$a=\frac{c^{2}}{T}\cdot\frac{dT}{dx}$

Now comes in the Schwarzschild Metric, which gives the relevant coefficients for a massive point particle.

$ds^{2}=\left(1-\frac{r_{S}}{r}\right)c^{2}dt^{2}-\frac{1}{1-\frac{r_{S}}{r}}dr^{2}\ ...$

Where $r_{S}$ is the Schwarzschild Radius, defined as $r_{S}=\frac{2GM}{c^{2}}$

("$...$" Because I'm omitting the other terms since we only care about the radial and time directions.)

Anyways, we can see that $T$ in this case is just the coefficient attached to the $dt$ term: $\left(1-\frac{r_{S}}{r}\right)$. Well, really its square-root because all of the terms are themselves squared, so $\sqrt{1-\frac{r_{S}}{r}}$, (the $c^{2}$ isn't part of it, it's a unit conversion factor).

Another thing we have to account for is that our formula differentiates with $dx$, not $dr$, but as can be seen by the Schwarzschild Metric, $dx^{2}=\frac{1}{1-\frac{r_{S}}{r}}dr^{2}\ \ \ \to\ \ \ dx=\frac{1}{\sqrt{1-\frac{r_{S}}{r}}}dr$

And with this we can finally find a, the answer for our "weight equation".

Substituting in for $T$ and $dx$:

$a=\frac{c^{2}}{T}\cdot\frac{dT}{dx}\ \ \ \to\ \ \ a=\frac{c^{2}}{\sqrt{1-\frac{r_{S}}{r}}}\cdot\frac{dT}{\frac{1}{\sqrt{1-\frac{r_{S}}{x}}}dr}$

$a=c^{2}\frac{dT}{dr}$

Finally differentiating $T$ with respect to $r$:

$\frac{dT}{dr}=\frac{1}{2\sqrt{1-\frac{r_{S}}{r}}}\cdot\frac{r_{S}}{r^{2}}$

$r_{S}=\frac{2GM}{c^{2}}$

$\frac{dT}{dr}=\frac{GM}{c^{2}r^{2}}\cdot\frac{1}{\sqrt{1-\frac{r_{S}}{r}}}$

$a=\frac{GM}{r^{2}}\cdot\frac{1}{\sqrt{1-\frac{r_{S}}{r}}}$

Of course this is the accleration to stay stationary, if you want the force/weight, by $F=ma$, $F=\frac{GMm}{r^{2}}\cdot\frac{1}{\sqrt{1-\frac{r_{S}}{r}}}$.

There you go.