# If a car was traveling at the speed of light and it turned it's headlights on, what would happen?

## Question:

If a car was traveling at the speed of light and it turned it's headlights on, what would happen?

## The Speed Of Light

The special theory of relativity is based on two postulates:

1) The laws of physics have the same mathematical form in all inertial frames of reference. The correct transformations from one inertial frame to another are the Lorentz Transformations.

2) The speed of light is the same in all inertial frames.

Together these postulates imply that the classical notions of invariant lengths, time intervals, and invariant masses need to be abandoned. The classical velocity addition formula is also in need of modification.

## Answer and Explanation:

Once the postulates of relativity are accepted then there is no escape from the fact that the speed of light is the same in all inertial frames. Thus if the headlights are switched on while the car is moving then in the driver's frame of reference the speed of light will be measured to be {eq}\displaystyle {c} {/eq}. A ground-based observer would also see the same speed {eq}\displaystyle {c} {/eq}. The mathematical machinery of relativity is geared to generate the same vacuum speed for light in all inertial frames.

Now consider the pre-relativistic situation:

If a train is travelling with a speed {eq}\displaystyle {v_1} {/eq} to the right and a person inside the train throws a ball to the right with a speed {eq}\displaystyle {v_2} {/eq} relative to the train, then a ground-based stationary observer would measure the speed of the ball to be,

{eq}\displaystyle { v_g=v_1+v_2} {/eq}.

However, relativity insists that the above addition formula is a mere approximation to a more general relativistic formula. The relativistic formula is,

{eq}\displaystyle { v_g=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}}----------(1) {/eq}

For small velocities, the second term in the denominator would be mighty close to zero. Then the pre-relativistic formula is a good approximation. But at large velocities, the exact formula is a must.

Suppose {eq}\displaystyle {v_2=c} {/eq} then we have the situation given in the problem. The car has the speed {eq}\displaystyle {v_1} {/eq} and the light has the speed {eq}\displaystyle {c} {/eq}.

Then in the ground frame,

{eq}\displaystyle { v_g=\frac{v_1+c}{1+\frac{v_1c}{c^2}}=c} {/eq}.

So clearly in both the car frame and the ground frame the same speed obtains for light.