I have a metric formula that does some interesting things for me. It's excellent at predicting the luminosity of Sne 1a. I'd like to see what the EFE solutions are, but I need to convert it from displacements to infinitesimals. The model of expansion is:

enter image description here

Where $\Delta\sigma$ is the proper space displacement and $\Delta\tau$ is the proper time displacement and $\Delta x^r$ is the coordinate space displacement and $\Delta x^t$ is the coordinate time displacement. The metric formula is:$$\Delta s^2=-\frac{1}{4}\left(\tau_0-\tau_1 \right)^2\left(-2V + A\left(\tau_0+\tau_1 \right) \right)^2+\frac{\left(-2V\left(t_0+t_1 \right) + A\left(\tau_0^2+\tau_1^2 \right)\right)^2}{4 \tau_1^2\left(-2V+A\tau_1\right)}\left(\Delta x^r \right)^2$$ Where $A$ and $V$ are constants.

Δs2 == (-(1/4))*(τ0 - τ1)^2*(-2*V + A*(τ0 + τ1))^2 + (Δxr^2*(-2*V*(τ0 + τ1) +
  A*(τ0^2 + τ1^2))^2)/(4*τ1^2*(-2*V + A*τ1)^2)

Given $\Delta s$, how would one find the metric tensor for $ds$? I understand the basic concept of taking the derivative with respect to each of the dimensions, but I'm having trouble coming up with the exact steps. The main problem for me seems to be that the spatial term has two displacements in it, $\Delta x^r$ and $\Delta \tau=\tau_1-\tau_0$ that I want to turn into infinitessimals.

  • 1
    $\begingroup$ You diagram shows that $\Delta \tau = \tau_2 -\tau_1$ so your first term is something * $(\Delta \tau)^2$. Your second term is something else * $(\Delta x')^2$. Just set the two finite deltas to infinitesimal $d\tau$ and $dx'$. It then has the generic form of a metric line element after you replace $(\Delta s)^2$ with $ds^2$. $\endgroup$
    – user87932
    Oct 4, 2023 at 2:38

1 Answer 1


You can replace (τ0 - τ1) by dt and Δxr by dx using the rule:

(τ0 - τ1) -> dt, Δxr -> dx

Then the remaining τ0 and τ1 can be replaced by t:

τ0 -> t, τ1 -> t

If we apply this to expr:

expr = (-(1/4))*(τ0 - τ1)^2*(-2*V + A*(τ0 + τ1))^2 + (Δxr^2*(-2* V*(τ0 + τ1) + A*(τ0^2 + τ1^2))^2)/(4*τ1^2*(-2*V +  A*τ1)^2);

expr/. {(τ0 - τ1) -> dt, Δxr -> dx, τ0 -> t, τ1 -> t}

enter image description here

  • $\begingroup$ This is definitely on the right track, but it doesn't yield the same answer as the version using displacements in the OP. if you set $ds=0$ and solve for $dx$, and then integrate that formula over $\tau_0$ to $\tau_1$, you get an answer that is 1/4 of the $\Delta x$ from the original formula. The correct solution is $$ds^2=-(V-A\tau)^2dt^2+\frac{1}{4}dx^2$$I'm having trouble coming up with a justification to set $t_0=0$, but that seems to be what makes the infinitesimal integration match the displacement formula. $\endgroup$ Oct 4, 2023 at 14:19
  • $\begingroup$ You can actually prove that this doesn't work by plugging in some numbers. Use A->1, V->1 and then create a function $f[t0,t1]=dx/dt$ then integrate that function over t0 to t1. You don't get the same values as solving the original equation for $\Delta x^r$ and plugging in t0 and t1. $\endgroup$ Oct 4, 2023 at 20:11
  • $\begingroup$ I think one potential problem is that $\tau$ is proper time, and you need to convert all the variables into coordinates in order for this to work. I.e. $d\tau^2 = dt^2 - dx^2$. The supplied expression contains both $\tau$ and t, which are different. $\endgroup$
    – user87932
    Oct 6, 2023 at 1:56

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