anonymous
  • anonymous
subject=Special Relativity The StartTrek Enterprise is following a Klingoon ship to destroy it with its phasers (laser, I suppose). Both ships move at exactly the same relativistic speed (>99% of light) on a straight line at a distance of 1Km. Which is the maximum speed allowed for the Enterprise to hit the enemy ?
Physics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
I don't understand your question.But if both the ships have exactly same speed then they are in the same frame of reference, Hence for each of them other ship appears stationary, i.e. not moving.(We can experience this when the two trains move parallel to each other in the same direction with the same speed, Then while looking from the window towards other train, passengers of the both the trains experience that both the trains are stationary. Now answer to your question, If Star trek ship increases the current speed, while the enemy ship remains at the current speed then they can catch the enemy.But still I don't get your exact question.
anonymous
  • anonymous
Thanks madhura30, this comes from my not understanding the implications of speed closed to light. Let me try to say it in other words: the ships travel at 299.999 km/s. The laser beam comes out from gun at light speed, assuming that light speed is exactly 300.000 km/sec. Respect to the ships, the laser travels at 1km/sec, correct? (the speeds can not be summed with this values, right?) If it correct, then the laser will hit the Klingoon after 1 second at 1km of separation; should ships be separed further, let's say by 1000km - a more realist distance, the 1000 secs will be too much to await for an effect. ...or, I'm totally confused by relativity.
anonymous
  • anonymous
Both the ships have the same speed 299,999 km/s.Hence they are in the same frame of reference so for each of them the other ship appears stationary. Alright. Now light possess special property that, the speed of light(300000km/s) is constant in all frame of reference. So for the frame of reference in which both the ships are(frame of reference of speed 299,999 km/s) , the speed of light remains the same 300000km/s. This implies that, for the ships, the speed of the laser beam i.e. light is also 300000km/s and not 1km/s. This is confusing but its true.

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anonymous
  • anonymous
Uhmm, interesting this. This means that for and external observer (in a different frame?) the light speed will be 599,999 km/s ? Confusing, confusing...
anonymous
  • anonymous
nope, it will remain 300000km/s only, Because according to Einstein's second postulates of special theory of relativity, The speed of light in free space is same in all directions and in all inertial frames. Hence for an external observer, as he is also in some frame of reference, the speed of light remains the same 300000km/s. As for that frame of reference also the speed of light is same, according to second postulate. I know this is confusing, you should talk to your physics teacher or go for any relativity book Introduction to Relativity by Robert Resnick .
anonymous
  • anonymous
Thanks madhura, I will ask to Captain Kirk if he agrees :-) since I did not understand at all this matter. As per this other suggestion "you should talk to your physics teacher", I could try, but I'm now 60 and is not sure if he is still alive :-) :-)
anonymous
  • anonymous
Oh sorry sir, I thought you must be a student as you asked a question about referring to Star Trek.I just told you the things which I read in my physics book(graduation level). Ya, It is written there, According to Einstein's second postulates of special theory of relativity, The speed of light in free space is same in all directions and in all inertial frames. Hence for the rest frame also, the speed of light is 300000km/s.
anonymous
  • anonymous
If the ship is moving in x-direction, Then according to Lorentz transformation of velocity,\[u' _{x}=\frac{ u _{x}-V }{ 1-\frac{ u _{x}V }{ c ^{2} } }\] V=velocity of the frame of reference(velocity of the ships) \[u _{x}=\]velocity of laser beam Solve it and you will get the answer.

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