Scenario of Crossing the speed limit of our Universe [duplicate]

One of Important conclusions of Einstein's Theory of Relativity is that we cannot cross or break the speed limit set by the laws of physics that is the speed of causality

consider this scenario:

suppose I somehow built a spaceship and is able to reach 99.99.......% the speed of light. After achieving this much speed i somehow turned on the boosters on my shuttle.

will I be able to cross the limit by turning on the boosters if no what will actually happen to spaceship and where will the work-done by the boosters go. can expect to be in another universe having different laws and speed limits.

• if this has already an answer please refer link. i couldn't find exactly this question..Though, i may have missed. Jan 10, 2017 at 19:24
• Due to Lorentz factor the closer you are to the speed of light the less effect you will get out of your boosters. So nothing special will happen, your speed will increase by some minuscule fraction and still stay under the speed of light, that is where the work of the boosters will go. The reason you expect something else is that you are thinking of the classical formulas for kinetic energy, in relativity they get multiplied by that same factor. Jan 10, 2017 at 19:36
• Jan 10, 2017 at 20:27

You would get one very little bit closer to the speed of light.

One of the important conclusions of the Theory of Special Relativity is, that 'The faster you go, the more inertial mass you get' (but your invariant mass doesn't change.) If something (other than massless particles ) would reach c, it's mass would be infinite, so the energy needed to accelerate this mass to c would be infinite too. So the work of your thrusters would go 'in the shuttle's mass, instead off it's velocity.'

With different approach, on such extreme velocities, the velocities don't add like in our everyday world. The sum of 0.9c and 0.7c, is not 1.6c, but slightly below c.

https://en.wikipedia.org/wiki/Mass_in_special_relativity

The Lorentz transformations are $$dt'~=~\gamma(dt~+~vdx/c^2)$$ $$dx'~=~\gamma(dx~+~vdt)$$ where $\gamma~=~1/\sqrt{1~-~(v/c)^2}$ and the velocity $u~=~dx/dt$. We now compute $dx'/dt'$ that results from the addition of the velocity $u$ with $v$, $$\frac{du'}{dt'}~=~\frac{\gamma(dx~+~vdt)}{\gamma(dt~-~vdx/c^2)}$$ which leads to $$\frac{du'}{dt'}~=~\frac{u~+~v}{1~-~uv/c^2}.$$ If is clear that no additional $u$ on the velocity $v$ is going to make $du'/dt'$ equal to $c$ the speed of light.