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Suppose that you are driving in a car with 100m/s and shoot a bullet with initial velocity 500m/s. Relative to the stationary observer, the bullet would be traveling at (500+100=600m/s).
Lets say you are driving at night with 100 m/s and put on the headlight, will the light beam be traveling (3x10^8 + 100) m/s relative to the stationary observer? It is apparently greater than the speed of light. Clear my confusion please..

2007-08-16 01:39:59 · 4 answers · asked by Twist of Fate 1 in Science & Mathematics Physics

4 answers

No because of time dilation. This is Einstein's theory of special relativity. The velocity of light (in a vacuum) is always c. As your velocity through space increases your passage through time decreases by the Lorentz factor:
1/(sqrt(1-v^2/c^2) ) keeping the speed of light constant.

Yes the stationary observer will measure your velocity to be 100 m/s. However if you had a giant clock on your car he would also notice that its ticking slower than his. So when he calculates the speed of light shining from your headlights using the formula v=d/t he will always come up with c.

You can also look at from the perspective of length contraction that Black wolf explains below.

2007-08-16 01:46:32 · answer #1 · answered by kennyk 4 · 0 0

both the travelling and stationary observer see light travel at the speed c
The fundamental concept required to understand the derivation of the equation's of the special theory of relativity is that

you cannot throw light

so where does the 100m/s of initial energy go ?
If you try and throw light (and sound also) you essentially squash it, that is you shorten it's wavelength (search doppler effect), so the stationary observer will see a shorter wavelength (bluer) light than the traveler, but they will both observe the same speed c.
Energy is conserved since the energy of light increases as the wavelength gets shorter, so the stationary observer observes higher energy light than the traveler, just as he observes a higher energy bullet than the traveler. it's just that rather than seeing a higher speed light, in his reference frame, the initial energy shortens the wavelength of the light instead.

the end
.

2007-08-16 09:23:21 · answer #2 · answered by The Wolf 6 · 1 1

I believe Einstein asked this same question, so good thinking on your part.

Lazily answering this, I will copy and past from the Faster-than-light Wikipedia article, explains why this is not possible.

An observer may incorrectly conclude that two objects are moving faster than the speed of light relative to each other, if that observer incorrectly adds velocities according to Galilean relativity.

For example, fast-moving particles on opposite sides of a circular particle accelerator will appear to be moving at slightly less than twice the speed of light, relative to each other, from the point of view of an observer standing at rest relative to the accelerator, if the observer incorrectly adds velocities according to Galilean relativity. However, if the observer understands special relativity, and makes a correct calculation, and the two particles are moving at velocities v and -v, or expressed in units of c, β and − β, where

β=v/c

then from the observer's point of view, the correct relative velocity (again in units of the speed of light c) is

βrel = [β-(-β)] / (1+β^2) = 2β / (1+β^2)

which is less than the speed of light.


Basically when things are going that fast, you can't use General Relativity anymore.

2007-08-16 08:47:37 · answer #3 · answered by Jon G 4 · 0 2

Relativity between moving masses can be a dilemma to many people.

The confusion comes from Eisntein 1st Relativity postulate,which says: The velocity of light is unchanging irespective of the velocity of the source.

The velocity of light relative to the source is a relative velocity .
So relativity of two moving bodies follows the law of Galilean relativity,which is the difference between velocities.

Assuming that a source is moving at 1X10^8 m/sec relative to the zero point that its motion started(called frame of reference).
Assuming that the micromass of light particle is moving at the Speed of light = C ,relative to the same frame of reference as the the moving source.

The relativity between the source moving mass and the light particle moving mass would exist as follows;

C- Vs = Vr

Vr = the relative velocity betwen the source and the motion of the light particle.

C = the velocity of the light particle mass relative to the same zero point of frame reference that the source motion started.

Vs= The velocity of the moving mass source of light relative to the zero point of reference frame.


Notice the relative velocity of light is Vr.
Vr is less than the velocity of light. However the speed of light relative to the zero point of reference does not change.

To Avoid confusion ;Einteins relativity 1st postulate should have read as follows.
The average Speed of light relative ot a point in a local frame of reference is a constant ,if a source is moving relative to the same frame of reference ,the Speed of light becomes relative to the moving source.
in the case you have stated= a moving car is moving at 100 m/sec and than light are turned on whould the speed of light =C+100?
In reality the folowing actually occurs;
The relative velocity of light = C minus 100m/Sec,
( C-100 =Vr))
The velocity of light relative to the local frame of reference which is the road of the moving car,remains the same as C=3 x10^8 m/sec
The same rule applies to any mass whether its light mass particle or a bullet. The reason is because the Power causing a particular velocity to a mass in motion is invariant in local space.
Note ;Speed of light that was measured on Earth (local space)is an average speed. The reason is that we cannot measure instantaneous velocity .(see also Heinsenberg's Underterminency principle)

2007-08-16 09:34:07 · answer #4 · answered by goring 6 · 0 1

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