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Constant speed of light

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Robb Phoenix | 22:12 Sat 29th Sep 2007 | Science
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If a man in a car is going at say 200km/h fires a bullet which travels at a speed of 400km/h: 1. the bullet will hit a stationary vehicle ahead it at 600km/h
2. " " " behind it at 200km\h
Q1.But if a person is going at half the speed of light why does the speed of light remain constant? What actually does happen? Why doesnt initial speed add? as the speed of light and the distance is constant in this case, does time itself expand?
Q2.Can ya guys explain(with equations) why mass and energy to obtain that speed becomes infinity at light speed.
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Time and distance are constant only within a given inertial reference frame. Time and distance conform to the velocity of light when comparing two reference frames in motion relative to each other.

The 'rules' apply consistently in a given reference frame. Disagreement arises from different reference frames.
An object having a constant velocity and direction is 'at rest' within its own reference frame, (it is not moving toward or away from itself). For an object not undergoing acceleration, time, distance, and the velocity of light all remain fixed.

Observers moving relative to each other occupy different reference frames with both measuring constant passage of time and velocity of light within their own reference frames. If one of these observers takes a trip upon returning their clock will exhibit a reduced passage of time due to the contraction of time that is a consequence of approaching the velocity of light. This discrepancy is not apparent to the individual observers within their own reference frame but comes about in conjunction with changes in their respective reference frames resulting from their relative motion.

The illusion of increases in matter and energy of an object approaching the elusive velocity of light are only �observed� in external reference frames.

some equations
Question Author
The photons fired ahead experience a dilation and photons behind a contraction in time.Correct?
Question Author
in other words
the photons fired ahead experience time slower while the photons behind experience time faster to keep the keep c constant.
Correct?
To illuminate (pun only slightly intended) what has already been stated concerning your speed of light query: Start with understanding that the entire series of equations to answer your question depends on understanding that E = mc2 is derived directly from Special Relativity and a comprehension of that relationship. Simply put, Special Relativity (more correctly The Theory of...) rests on the foundation of only two requisites: 1. the laws of physics are the same in all inertial references and 2. The speed of light (c) (in a vacuum) has the same constant value in all inertial references (frames).
So:
m = relativistic mass, i.e. mass at the speed it is traveling.
m^0 = "rest mass", i.e. mass of object when stationary.
v = speed of object.
c = speed of light.
Therefore m= m^0 divided by the square root of 1- v^2/c^2 (I don't have the ability to frame this for you. The results can also be expressed on an x/y graph where: the x value is mass increase while the y value is percentage of the speed of light. The resultant indicators are that as the percentage of c increases the mass increases exponentially. However, the mass increase isn't felt by the object itself, just as the time dilation of special relativity isn't felt by the object. It is only apparent to an external observer; hence it is "relative" and depends on the frame of reference used. (With thanks to Jim Doyle: E=mc^2; Deriving the Equation)...

Contd.
Contd.
Next, we have to take into account (Kinetic energy). Newtonian math tells us kinetic energy is equal to half the mass multiplied by the velocity squared... it deals only with slower speeds, however. Therefore, at the higher, theoretical speeds, our original formula, must be corrected for rest mass, i.e. the mass the body has when it is at rest as well as understanding that E=mc^2 applies to energy and energy is somehow inherently within the mass, (which caused Einstein to conclude that mass and energy are really the same thing) and thus we are returned to our original formula. The mathematics of special relativity explains the way in which we take into account the kinetic energy of E = mc^2 and any body not at rest with a mass of a value greater than zero will take into account the total energy (E), the mass of the body (m), and the speed of the body (v). As such it accounts for both the relativistic mass increase and the relativistic kinetic energy�. Obviously this is greatly simplified or Einstein would not have to have been so, well� Einsteinian, would he?
The passage of time is a variable when compared in separate reference frames in motion relative to each other with the velocity of light being the "common denominator". With the velocity of light applied to the equation one can predict the rate at which time passes in an alternate reference frame for which its apparent velocity has been determined.

It is not the direction, it is the velocity of relative motion which determines "dilation/contraction" and the velocity is equally valid for an object orbiting around another as for one moving away or toward the object remaining at rest and vice versa. Starting from a condition of rest, (co-moving non-accelerating objects), it is the object that undergoes a change in velocity for which time dilation (the slowing of elapsed time) takes place.

The reason behind the infinite requirement of energy to achieve the velocity of light is that time dilation corresponds to acceleration and approaches infinity as the velocity of light is approached so doubling your velocity would get you to your destination in half the time as you measure its passage but relative to an observer at your original starting point you would only be going a fraction of the velocity of light faster and would arrive at your destination only a small fraction of time sooner. Redoubling your velocity dilates time exponentially the closer you get to C.
This part was supposed to come first so, read this first, then the above: lol

Photons by virtue of traveling at C do not experience any passage of time. Time is essentially frozen at C. The velocity of light is "absolute zero" in reference to time. The passage of time is observed only by objects traveling at less than the velocity of C.
Question Author
How do we derive the equation between relative mass and rest mass which gives mass becomes infinity at light speed in the first place?
btw should'nt we become black holes at near light speed as our mass tends to infinity? this question is countered by answering that only relativistic mass is infinity and that inertial mass is same so the body doesnt collapse into a black hole.I say so what shouldnt the body still turn into a black hole? Can anyone please elaborate.
I always thought of it more as a big bang myself but to each their own . . .

The term "relativistic mass" is used to lump an objects kenetic energy due to motion together with its rest mass. An accelerated body only behaves as though it has an increased mass due to the substantial energy it acquires from momentum as it approaches the velocity of light. Neither an increase of mass nor energy is observed within the co-moving reference frame of a moving object.

mr = E/c^2 or m0 = sqrt(E^2/c^4 - p^2/c^2)

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