Using Solve returns unnecessary Root, overcomplicated formula, and erroneous negative value

Question

I want to find positive solution to an equation $\text{Solve}\left[t==\frac{\sqrt{\left(\frac{L}{\gamma }+t v\right)^2+x^2}}{c}\land \text{assume},t\right]$.

Although using many assumptions, Mathematica provides a negative solution as explained in what follows.

A similar but simpler equation works as expected.

How can I obtain the positive solution to the equation

First a little context:

I'm using Solve to solve for t:

t11s1 = Solve[t == Sqrt[(L + v*t)^2 + (x)^2]/(c), t]

The solution are two roots:

$$ \left\{\left\{t\to \frac{L v-\sqrt{c^2 L^2+c^2 x^2-v^2 x^2}}{c^2-v^2}\right\},\left\{t\to \frac{\sqrt{c^2 L^2+c^2 x^2-v^2 x^2}+L v}{c^2-v^2}\right\}\right\} $$

This I Refine with some assumptions, and get a positive solution:

$$ \left\{\frac{\sqrt{c^2 L^2+c^2 x^2-v^2 x^2}}{c^2-v^2}+\frac{L v}{c^2-v^2}\right\} $$

However, if instead of $L$ I use $L\rightarrow\frac{L}{\sqrt{1-\frac{v^2}{c^2}}}$ which is a positive, real multiplicative factor on $L$, the solution changes to

$$ \left\{\left\{t\to \text{ConditionalExpression}\left[\text{Root}\left[\text{$\#$1}^4 \left(c^8-2 c^6 v^2+c^4 v^4\right)+\text{$\#$1}^2 \left(-2 c^6 L^2-2 c^6 x^2+2 c^4 v^2 x^2+2 c^2 L^2 v^4\right)+c^4 L^4+2 c^4 L^2 x^2+c^4 x^4-2 c^2 L^4 v^2-2 c^2 L^2 v^2 x^2+L^4 v^4\&,4\right],v>0\land c>v\land x>0\land L>x\land 0<a<x\land 0<n<\frac{c}{v}\right]\right\}\right\} $$

which if I force to be expressed as radicals (to eliminate the Root[,4]) and again I refine using the same assumptions I get

$$ \left\{-\sqrt{\frac{c^2 \left(L^2+x^2\right)+2 c L v \sqrt{L^2+x^2}+L^2 v^2}{c^4-c^2 v^2}}\right\} $$

which is a negative value because all variables are Reals !!

My assumptions are:

$$ \text{assume}=v>0\land v\in \mathbb{R}\land L>0\land L\in \mathbb{R}\land \text{L2}>0\land \text{L2}\in \mathbb{R}\land a>0\land a\in \mathbb{R}\land t>0\land t\in \mathbb{R}\land c>0\land c\in \mathbb{R}\land n>0\land n\in \mathbb{R}\land \gamma >0\land \gamma \in \mathbb{R}\land \beta \geq 0\land \beta <1\land \beta \in \mathbb{R}\land c>v\land x\in \mathbb{R}\land x>0\land n<\frac{c}{v}\land L>x\land L>a\land x>a $$

and I use them as:

$$ \text{t11s2}=\text{Solve}\left[t==\frac{\sqrt{\left(\frac{L}{\gamma }+t v\right)^2+x^2}}{c}\land \text{assume},t\right] $$

$$ \text{t2}=\text{Simplify}[\text{ToRadicals}[\text{Refine}[t\text{/.}\, \text{t11s2},\text{assume}]],\text{assume}] $$

to get

$$ t2=\left\{-\sqrt{\frac{c^2 \left(L^2+x^2\right)+2 c L v \sqrt{L^2+x^2}+L^2 v^2}{c^4-c^2 v^2}}\right\} $$

What can I do to recover a positive real values for t? Why is Mathematica behaving like this?

Answer 1

I think it would be simpler to use Solve without assumptions, and then simplify after:

res = t /. Solve[t == Sqrt[(L/Sqrt[1-v^2/c^2]+v*t)^2 + x^2]/c, t];
Simplify[res, 0 < v < c] //TeXForm

$\left\{\frac{c L v-\sqrt{c^4 \left(L^2+x^2\right)-2 c^2 v^2 x^2+v^4 x^2}}{\left(c^2-v^2\right)^{3/2}},\frac{\sqrt{c^4 \left(L^2+x^2\right)-2 c^2 v^2 x^2+v^4 x^2}+c L v}{\left(c^2-v^2\right)^{3/2}}\right\}$

On the other hand, to fix your approach, you can use the undocumented option Assumptions for ToRadicals:

assum = 0<v<c && L>0 && t>0 && x>0;

res = t /. First @ Solve[
    t==Sqrt[(L/Sqrt[1-v^2/c^2]+v*t)^2 + x^2]/c && assum,
    t
];

FullSimplify[
    ToRadicals[res, Assumptions->assum],
    assum
] //TeXForm

$\sqrt{\frac{c^4 L^2+c^2 L^2 v^2+x^2 \left(c^2-v^2\right)^2+2 c L v \sqrt{c^4 L^2+x^2 \left(c^2-v^2\right)^2}}{\left(c^2-v^2\right)^3}}$