Breaking The Light Speed Barrier

[Photon Guy]

     Apparently the light speed barrier can't be broken by conventional means. Well how about in a case such as this? Lets say I've got a spaceship that goes at over half the speed of light. Now lets say I've got a gun that shoots a bullet at over half the speed of light. Now, Im in my spaceship and its going forward at maximum velocity. I fire my gun forward in the same direction the ship is going, would my bullet break the light speed barrier?

[dimreepr]

9 minutes ago, Photon Guy said:

would my bullet break the light speed barrier?

Nope, it would stretch to fit.

[swansont]

11 minutes ago, Photon Guy said:

     Apparently the light speed barrier can't be broken by conventional means. Well how about in a case such as this? Lets say I've got a spaceship that goes at over half the speed of light. Now lets say I've got a gun that shoots a bullet at over half the speed of light. Now, Im in my spaceship and its going forward at maximum velocity. I fire my gun forward in the same direction the ship is going, would my bullet break the light speed barrier?

Nope. Speeds don't add linearly, but you don't notice this until you are moving a reasonable fraction of c. Your result would be a bullet moving at 0.8c (relative to the observer)

https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

 

 

[HallsofIvy]

  The "Gallilean theory" says that velocities add as $v_1+ v_2$.  If you see a person driving by at 40 mph, relative to yourself, on the back of a flatbed truck and, just as they pass you, that person throws a ball forward at 30 mph, relative to the truck, Gallilean theor y says that you would see that ball having speed 40+ 30= 70 mph relative to you.

Relativity theory says that the velocities add as [tex]\frac{v_1+ v_2}{1+ \frac{v_1v_2}{c^2}}[/tex].    If the velocities are small then [tex]\frac{v_1v_2}{c^2}[/tex] will be negligible and that the two formulas give the same thing to within measurement error.  If [tex]v_1[/tex] and [tex]v_2[/tex] are close to c, as [tex]v_1= v_2= 0.6c[/tex] then [tex]\frac{v_1v_2}{c^2}= (0.6)^2= 0.36[/tex] so the velocity of the bullet relative to you is [tex]\frac{0.6c+ 0.6c}{1+ 0.36}= \frac{1.2}{1.36}c= 0.88c[/tex].

Edited October 20, 2020 by HallsofIvy

[Curious layman]

11 minutes ago, HallsofIvy said:

  The "Gallilean theory" says that velocities add as $v_1+ v_2$... Relativity theory says that the velocities add as [tex]\frac{v_1+ v_2}{1+ \frac{v_1v_2}{c^2}}[/tex].    If the velocities are small then [tex]\frac{v_1v_2}{c^2}[/tex] will be negligible and that the two formulas give the same thing to within measurement error.  If [tex]v_1[/tex] and [tex]v_2[/tex] are close to c, as [tex]v_1= v_2= 0.6c[/tex] then [tex]\frac{v_1v_2}{c^2}= (0.6)^2= 0.36[/tex] so the velocity of the bullet relative to you is [tex]\frac{0.6c+ 0.6c}{1+ 0.36}= \frac{1.2}{1.36}c= 0.88c[/tex].

Good answer, next time someone asks me this question, I'll know what to say. 😁