8
$\begingroup$

The universe described in the orthogonal series is one, in which the minus sign in the space time interval is replaced with a plus sign. So this universe has four fundamentally similar dimensions rather than three space like dimensions and one time like dimension.

In the universe described in the orthogonal series the electric force between two electric charges is attractive at some distances and repulsive at other distances. The electric potential energy between two electric charges in the universe described in the orthogonal series would be $$U_E=-\frac{cos(\omega_mr)Q_1Q_2}{4{\pi}r\varepsilon_0}$$ with $Q_1$ and $Q_2$ being the electric charge of each body, $r$ being the distances between the two bodies, $\omega_m$ being a constant that depends on the rest mass of the photon, $\varepsilon_0$ being the electric constant, and $U_E$ being the electric potential energy between the two electric charges.

In our universe the Schwarzschild Metric can be described by the equation $${\Delta}s^2=\frac{{\Delta}r^2}{1-\frac{2GM}{c^2r}}-\left(1-\frac{2GM}{c^2r}\right)c^2{\Delta}t^2+r^2(\Delta\theta^2+sin^2\theta\Delta\varphi^2)$$ with $s^2$ being the spacetime interval between two events, $G$ being the Gravitational Constant, $M$ being the rest mass of the massive body, ${\Delta}r$ being the distance in space between two events in spacetime relative to the massive body, $c$ being the speed of light, ${\Delta}t$ being the time passed between two events in spacetime relative to the massive body, $\theta$ being the colatitude, $\varphi$ being the longitude, and $r$ being the distance to the massive body. I noticed that $$\frac{2GM}{c^2r}$$ has the same relationship to distance as the electric potential energy between two electric charges, in our universe, as well as the gravitational potential for a massive body in newtonian physics.

In a universe with the same spacetime metric as the one described in the orthogonal series, but with massive gravitons, would gravity also be attractive at some distances, and repulsive at other distances?

I was thinking of a universe, with the same space time metric as the universe described in the orthogonal series, but with massive gravitons, and with gravity being repulsive at the closest distances.

In this type of universe, would this be the correct schwarzschild metric?

$${\Delta}s^2=\frac{{\Delta}r^2}{1-\frac{cos(\varrho_mr)2GM}{c^2r}}+\left(1-\frac{cos(\varrho_mr)2GM}{c^2r}\right)c^2{\Delta}t^2+r^2(\Delta\theta^2+sin^2\theta\Delta\varphi^2)$$

In this case $\varrho_m$ would be a constant, that would depend on the mass of the Graviton, and while all dimensions would be fundamentally the same, the world line of the massive body would be treated as the time axis.

Would this be correct for the Schwarzschild Metric for this type of universe?

$\endgroup$
13
  • 2
    $\begingroup$ I'm wondering how one would decide which of the four dimensions is the "time" and also how one would find a corresponding "r" in this universe. $\endgroup$
    – katatahito
    Jul 8, 2019 at 7:13
  • 1
    $\begingroup$ I'm concerned that the question would be moot if the whole space-time system you've looked at is prone to spontaneous dimensional collapse. How do you envisage that this could attain any stability and lead to worldbuilding? Have you tried asking on the Physics stack? $\endgroup$ Jul 8, 2019 at 9:01
  • $\begingroup$ Section 3.2.2 here seems to indicate that some black hole solutions in dRGT ghost-free massive gravity have a long-distance behavior similar to what you propose, but not exactly the same. The relevant function is $N(r)$; if I'm not mistaken, to pass from the Lorentzian to the Riemannian case you can change the sign of $C_1$ and replace $e^{-r\mu}$ by $\cos(r\mu)$. $\endgroup$
    – pregunton
    Jul 8, 2019 at 9:34
  • 5
    $\begingroup$ The question would be off topic on the Physics Stack Exchange. Please don't migrate it there or indeed repost it there. $\endgroup$ Jul 8, 2019 at 11:11
  • 3
    $\begingroup$ @StephenG not knowing the answer doesn't make a question off topic. Off topic is when the question is not about world building. Fromm the help center about what is on topic here: World building includes geography, culture and creatures for the world, not to mention magic and planetary physics, in short, everything from the physics underlying your reality to the entire universe you want to build. $\endgroup$
    – L.Dutch
    Jul 8, 2019 at 14:15

1 Answer 1

5
$\begingroup$

I would like to address several problems in your question.

  • If flat spacetime had a Kronecker delta metric instead of the usual Minkowski metric then the "invariant" spacetime interval wouldn't be invariant ( if you are still thinking of $t$ as time, otherwise it's an invariant space distance). Also while defining a spherically symmetric metric in spacetime one initially writes a general metric: $$ ds^2 = A(r) dt^2 + B(r)dtdr + C(r)dr^2 + r^2d\Omega^2$$ Then the terms can be rearranged in the form $$ds^2= A'(r)dt^2 + B'(r)dr^2 +r^2d\Omega^2$$ This can be done by considering (+,+,+,+) signature but time loses its uniqueness. There wouldn't be any difference between $r$ and $t$. In your universe there is not any difference if one changes $t \leftrightarrow r$. Basically by introducing (+,+,+,+) signature you have removed the concept of time from your universe.

  • A massive electromagnetic field has many problems. First of all any massive force field will have to be short ranged because of the Yukawa Potential term. This force will be limited to very short distances. Massive electromagnetic field will also break the internal gauge symmetry. Though this can be avoided by coupling it with a massive scalar field. The similar problems will be faced when working with massive gravity (see this).

  • When you introduce mass to a spin 2 field, it no longer follows the Einstein field equations anymore. It will follow a new set of equations given by the action for the ghost-free de Rham-Gabadadze-Tolley massive gravity. Solving this will give you a completely different answer. So you can't say that the new Schwartchild metric is given by the metric in the question, it will be totally different.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .