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I have an interesting question, one which I think may not work haha, but please bear with me.

Let's work with our sun, with mass of about $2 \times10^{30}$ kg. According to the following formula, the sun will have a Schwarzschild radius of

$$r_{s}=\dfrac{2GM}{c^{2}}$$ $$r_{s,\space \mathrm{sun}}=\dfrac{2 \big(6.67 \times 10^{-11}\big)\big(1.989 \times 10^{30}\big)}{\big (3 \times 10^{8}\big ) ^{2}}$$ $$r_{s,\space \mathrm{sun}}= 2948 \space \mathrm{m}$$

This means that if the sun were to collapse to a sphere with such a radius, it will become a black hole.

Let's assume now that our sun is travelling linearly at relativistic speeds. According to length contraction in relativity, observers at rest would see the sun shrink. The next question is, at what minimum speed would the sun have to travel such that it would appear to become a black hole to observers at rest? In order to answer this question, we must equate the observed length with the Schwarzschild diameter.

$$L = L_{0} \sqrt{1- \dfrac{v^{2}}{c^{2}}}$$ $$2r_{s,\space \mathrm{sun}}=2R_{☉}\sqrt{1- \dfrac{v^{2}}{c^{2}}}$$ $$r_{s,\space \mathrm{sun}}=R_{☉}\sqrt{1- \dfrac{v^{2}}{c^{2}}}$$ $$v=c\sqrt {1-\left(\dfrac {r_{s,\space \mathrm{sun}}}{R_{☉}}\right)^{2}}$$ $$v=c\sqrt {1-\left(\dfrac {2948}{6.957 \times 10^{8}}\right)^{2}}$$ $$v=0.9999999999911022c$$

But this, of course, would not happen! Why? Because by definition, stellar mass that collapses into a sphere with its radius being equal or less than its Schwarzschild radius would form a black hole. If the sun were to travel at relativistic speeds, it would only contract in the dimension that is parallel to its motion. Therefore, if the sun were to travel at $0.9999999999911022c$, its dimension parallel to its motion would appear to be contracted to a length equal to its calculated Schwarzschild radius. Its other dimensions would not appear to be contracted. A sphere that travels close to the speed of light would appear to become rather an ellipsoid.

Now, here is my actual question. If we calculate the sun's Schwarzschild radius, we can also calculate its Schwarzschild density. This would be

$$\rho_{s,\space \mathrm{sun}}=\dfrac {M_{☉}}{V}$$ $$\rho_{s,\space \mathrm{sun}}=\dfrac {M_{☉}}{\dfrac{4}{3}\pi r_{s,\space \mathrm{sun}}^{3}}$$ $$\rho_{s,\space \mathrm{sun}}=\dfrac {1.989 \times 10^{30}}{1.07 \times 10^{11}}$$ $$\rho_{s,\space \mathrm{sun}}=1.85 \times 10^{19} \space \mathrm {kg \space m}^{-3}$$

Would it be possible to observe a black ellipsoid if the sun were to travel at some relativistic speed such that its Schwarzschild density becomes equivalent to $1.85 \times 10^{19} \space \mathrm {kg \space m}^{-3}$?

Namely:

$$V_{\mathrm{ellipsoid}}= \dfrac {4}{3}\pi R_{1}R_{2}R_{3}$$

Let's assume $R_{1}=R_{2}=R_{☉}$ as they are not parallel to its motion, and $R_{3}$ is observed to be contracting since it is parallel to its motion.

$$\rho_{s,\space \mathrm{sun}}=1.85 \times 10^{19} \space \mathrm {kg \space m}^{-3}=\dfrac {4}{3}\pi R_{☉}^{2}R_{3}$$ $$\dfrac{3\rho_{s,\space \mathrm{sun}}}{4\pi R_{☉}^{2}}=R_{3}$$ $$L = L_{0} \sqrt{1- \dfrac{v^{2}}{c^{2}}}$$ $$R_{3}=R_{☉}\sqrt{1- \dfrac{v^{2}}{c^{2}}}$$ $$\dfrac{3\rho_{s,\space \mathrm{sun}}}{4\pi R_{☉}^{2}}=R_{☉}\sqrt{1- \dfrac{v^{2}}{c^{2}}}$$ $$v=c\sqrt {1-\left(\dfrac{3\rho_{s,\space \mathrm{sun}}}{4\pi R_{☉}^{3}}\right)^{2}}$$ $$v=0.99999999999999991398c$$

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    $\begingroup$ Effectively a duplicate of: physics.stackexchange.com/q/3436 $\endgroup$ – dmckee Feb 6 '18 at 21:43
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    $\begingroup$ The Schwarzschild radius is defined in the stationary frame of reference of the object. It has no validity in other frames. $\endgroup$ – Rob Jeffries Feb 6 '18 at 23:15

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