15
$\begingroup$

Nordström's 1912 proposed theory of relativistic gravity posited that the gravitational potential field $\phi$ and the matter density field $\rho$ are both scalar fields, simply related by the wave equation $\partial^2 \phi = 4 \pi G \rho$.

The Wikipedia article linked above says that Abraham proposed the force law $a_\mu = -\partial_\mu \phi$, where $a_\mu := d(u_\mu)/d\tau$ is the four-acceleration of a test particle, $u_\mu$ is its four-velocity, and $\tau$ is its proper time. But the article just says that

Nordström knew that it wouldn't work. Instead he proposed $a_\mu = -\partial_\mu \phi - \frac{d\phi}{d\tau} u_\mu$.

Why wouldn't Abraham's proposed force law work?

A second question, which I'll combine with the previous one because I suspect that the answers might be the same: the Wikipedia article says that Nordström's first theory of gravity (with his replacement force law) "disagrees drastically with observation". How so? It seems clear that (with the possible exception of the new term on the RHS of Nordström's force law) this theory reduces to Newtonian gravity in the nonrelativistic limit, and Newtonian gravity is an excellent description of all gravitational phenomena experimentally known in 1912 except for the slight discrepancy with the perihelion of Mercury, so it's hard for me to see how any theory that reduces to Newtonian gravitation in the nonrelativistic limit could possibly have "disagree[d] drastically with observation" at the time.

(Note that these are physics questions, not history questions. I'm not asking "Who understood what when?"; I'm asking about the physical problems with the theories themselves.)

$\endgroup$
1
  • 5
    $\begingroup$ According to arxiv.org/abs/1205.5966, Nördstrom's theory doesn't predict deflection of light near the Sun and precession of Mercury's perihelion $\endgroup$
    – atarasenko
    Mar 26, 2022 at 20:46

1 Answer 1

19
$\begingroup$

Because it doesn't conserve the magnitude of the four velocity. From the definition of proper time, in special relativity we must always have $u_\mu u^\mu = -1$, which implies $a^\mu u_\mu = 0$: the acceleration has to be orthogonal to the velocity, so that its magnitude is constant.

If we use the originally proposed formula $a_\mu = -\partial_\mu \phi$, then $a_\mu u^\mu = - u^\mu \partial_\mu \phi = - d\phi/d\tau$, which is not zero. Adding the extra term fixes this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.