Thread: approaching with the light-speed

hi to all this is my first msg on this board but i can't find a satisfiying answer on the net so maybe someone here can tell me how that happens thx for spending your time to read and for your answers. Here is the question.

"x" is a human static in space and "y" is a mass approacing to "x" with a speed of light. can "x" see (or recognise) "y" before the hit.

Originally Posted by ckcem

hi to all this is my first msg on this board but i can't find a satisfiying answer on the net so maybe someone here can tell me how that happens thx for spending your time to read and for your answers. Here is the question.

"x" is a human static in space and "y" is a mass approacing to "x" with a speed of light. can "x" see (or recognise) "y" before the hit.

It is impossible for a massive body Y to be traveling at the speed of light.

But no matter what the speed of Y might be, if a photon is emitted from Y it will approach X at c, the speed of light, in the reference frame of X. So as the speed of Y approaches c, X will have less and less time to react to a photon emitted from Y before Y arrives a the location of X.

thank you very much for you answer.

but as you know the science just formulate all this data up to speed "c".

i mean "travelling at c for a mass is impossible".

but do you know the outcome happens before the beginning when you cross the light-speed. there is no time factor. x=v.t for calculating the speed but if you loose the time there can't be a referance point. the formule changes...!

so the e=mc2 loose its validity.

i still can't find the answer or understand what's happening.

so i repeat the question with a little change for you to understand properly.

we have "x" static. and "y" approaching to "x" with 2X speed of c. can "x" see the "y" before the hit. i know the answer is no but (even a human can't see a bullet approaching) just asking to formulate this.

thx...

Originally Posted by ckcem

thank you very much for you answer.

but as you know the science just formulate all this data up to speed "c".

i mean "travelling at c for a mass is impossible".

but do you know the outcome happens before the beginning when you cross the light-speed. there is no time factor. x=v.t for calculating the speed but if you loose the time there can't be a referance point. the formule changes...!

so the e=mc2 loose its validity.

i still can't find the answer or understand what's happening.

so i repeat the question with a little change for you to understand properly.

we have "x" static. and "y" approaching to "x" with 2X speed of c. can "x" see the "y" before the hit. i know the answer is no but (even a human can't see a bullet approaching) just asking to formulate this.

thx...

You are misunderstanding the physics. No massive body can travel at c or higher in any reference frame. e=mc^2 does not loose its validity, but it has no bearing on the question at hand. Nothing can travel at twice the speed of light -- not light, not information, not gravitational waves, and certainly not any body with non-zero rest mass.

I gave you the formulation. Light travels at c in all inertial reference frames. So the light signal traveling from Y to X travels at c in the reference frame of X, and it does not matter how fast Y is traveling. If Y is traveling at nearly the speed of light then Y will arrive at X at nearly the same time as any light signal that is emitted by Y. But light will get there a little earlier, since the speed of Y must be slightly slower than c.

It is not that special relativity just formulates the physics for speeds less than c. It is that the speed of light is c in all inertial reference frames, and the laws of physics take the same form in all inertial reference frames. From those two principles one mathematically and logically derives the Lorentz transformation. And from that one conlcudes as a matter of logic that it is impossible for any mass, energy or information to travel faster than light.

The disturbing aspect of this little mind experiment is that x sees only little difference in the velocities of y and light, while y sitting in its own reference frame also measures light with the velocity of c, i.e. a huge difference. It is the essence of Special Relativity that the solution in this apparent contradiction is time dilation (observed by x) and length contraction (observed by y).

ok i'll tell you what i found during my search.

they can cross the speed of light. and photons have a calculative mess but at this day knowledge of physics can't even recognise it.

they sent a photon through a glass tube over the speed of c and the photon pass the tube before going inside. outcome happens before the beginning.

the answer is "x" cannot see before the hit because as you said -observed by "x" it will appear like "y" hit before and then approach. but from "y" just normal approach and hit.

A,B travelling in the speed of light. and observing each other travelling in the speed of light in all 3 conditions below.

1.
A ----> <-----B

2.
<---- A B------>

3.
A ------>
B ------>

but when it comes to crossing c, everything changes.

Hi, doesn't it depend on the distance between x and y ?

BARCUD

there is no distance...

i understand your point but you can't think in traditional ways. today's physics acceptance and the way of understanding events belongs to same formule which loose it's validity over speed of c. on the ground maybe you can talk about the mechanics cause of gravity. but in space i'll tell you what. Everything is in the same coordinate it's just one point. no distance no dimensions...!

that's because when you cross the speed of light, the "c" outcome happens before the beginning. but i need some one who have knowlegde to discuss without prejudice.