Let there be a hypothetical vehicle/spaceship traveling at the speed of light(c). The headlights of this spaceship are turned on which emit light. What will be the speed of light from the headlights (a) from person sitting inside spaceship's frame of reference (b) a nearby observer at rest. c =3× 10⁸m/s (approx.)
Answers
Explanation:
No. As others have explained, you cannot exceed the speed of light. In one dimension this prohibits your light speed car from shining at all, because the light cannot move faster from the car, and so can never leave the headlights. However, we live in multiple dimensions and so not all light shines in the same direction.
Let us set up a 2-dimensional, massless (i.e. moving-at-lightspeed) car that fires off two photons, one going up and one going down
Answer:
Explanation:Sadly this question and all others about experiences at the speed of light do not have a definitive answer. You cannot go at the speed of light so the question is hypothetical. Hypothetical questions do not have definitive answers. Only massless particles such as photons can go at the speed of light. As a massive object approaches the speed of light the amount of energy needed to accelerate it further increases so that an infinite amount would be needed to reach the speed of light.
Sometimes people persist: What would the world look like in the reference frame of a photon? What does a photon experience? Does space contract to two dimensions at the speed of light? Does time stop for a photon?. . . It is really not possible to make sense of such questions and any attempt to do so is bound to lead to paradoxes. There are no inertial reference frames in which the photon is at rest so it is hopeless to try to imagine what it would be like in one. Photons do not have experiences. There is no sense in saying that time stops when you go at the speed of light. This is not a failing of the theory of relativity. There are no inconsistencies revealed by these questions. They just don't make sense.
Despite these empty answers, nobody should feel too put down for asking such questions. They are exactly the kind of question that Einstein often asked himself from the age of 16 until he discovered special relativity ten years later. Einstein reported that in 1896 he thought,
``If I pursue a beam of light with the velocity c (velocity of light in a vacuum), I should observe such a beam of light as a spatially oscillatory electromagnetic field at rest. But there seems to be no such thing, whether on the basis of experience or according to Maxwell's equations. From the very beginning it appeared to me intuitively clear that, judged from the standpoint of such an observer, everything would have to happen according to the same laws as for an observer who, relative to the earth, was at rest. For how, otherwise, should the first observer know, i.e., be able to determine, that he is in a state of fast uniform motion? One sees that in this paradox the germ of the special relativity theory is already contained. Today everyone knows, of course, that all attempts to clarify this paradox satisfactorily were condemned to failure as long as the axiom of the absolute character of time, viz., of a simultaneous, unrecognizedly was anchored in the unconscious. Clearly to recognize this axiom and its arbitrary character really implies already the solution to the problem.''
In 1905 he realised how it could be that light always goes at the same speed no matter how fast you go. Events that are simultaneous in one reference frame will happen at different times in another that has a velocity relative to the first. Space and time cannot be taken as absolute. On this basis Einstein constructed the theory of special relativity, which has since been well confirmed by experiment.
Questions of relative velocity in relativity can be answered using the velocity subtraction formula v = (w − u)/(1 − wu/c2) (see relativity FAQ: velocity addition). If you are driving at a speed u relative to me and you measure the speed of light in the same direction (w = c in my frame), the formula gives v the speed of light in your reference frame as, v = (c − u)/(1 − u/c). For any speed u less than c this gives v = c so the speed of light is the same for you. But if u = c the formula degenerates to zero divided by zero; a meaningless answer.