# How can travelling at light speed affect light itself? [duplicate]

~Hypothetical Scenario:

Your driving in a car (with the headlights on high beam) and you gradually get faster until you reach the speed of light.

Raised Assumptions:

• You can see in and out of the car whilst moving at the speed of light.
• Ignore air resistance.
• Ignore all biological affects.
• The headlights on the car still work at the speed of light.
• Bystanders can see objects move at the speed of light.
• The car can physically make it to the speed of light.

Questions

1. Would the driver be able to see the high beam on in front of the car and be travelling twice as fast as the speed of light? Or will the photons produced pool up within the light bulb itself? Would there be no light produced even though the light bulb is on?

2. If a bystander saw the car drive by would they see the light from the headlights?

Note: For those who think this is a duplicate question, I cannot find sufficient information from the previous not-so-related questions to get a full understanding.

## marked as duplicate by Carl Witthoft, Kyle Kanos, user36790, ACuriousMind♦, David HammenOct 9 '15 at 14:08

• Nobody can answer your question because in our current theories a massiv object (e.g. car) cannot reach the speed of light. Your question can be answered of the car is even a tiny bit slower than the speed of light. Then in the viewpoint of the driver the light will still travel at the speed of light away from him. Nothing is changed to the situation where he is standing still. – Jannick Oct 9 '15 at 11:22
• Mix of this, this, this, this and this, really. As Jannick says, it cannot happen so asking about it really is a moot point. – Kyle Kanos Oct 9 '15 at 12:07

Einstein stated that the speed of light is constant and that space and time will change to keep it that way.

EDIT: my mistake, this section is more relevant than length contraction based on the OP's original question (see also here).

Einstein's Theory of Relativity showed that when two objects are traveling in the same direction, their speeds don't just add up: walking at velocity $v_1$ in a bus at traveling at $v_2$ does NOT mean that your velocity is now $v_1+v_2$ relative to the ground, but rather $$\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$$ Let's plug in the numbers of the arbitrarily fast car. Let's say that the car is traveling at $0.99c$. The light is, obviously, traveling at $c$. $$v_{light}=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}=\frac{0.99c+c}{1+\frac{0.99c^2}{c^2}}=\frac{1.99c}{1+0.99}=\frac{1.99c}{1.99}=c$$ This shows that the light travels at the speed of light even when being emitted from a fast-moving source.

As mentioned by @Jannick in his comment, it is impossible for any massive object ($m>0$) to reach the speed of light. However, if your car were to travel at a velocity arbitrarily close to $c$, Lorentz length contraction is what solves the problem of others seeing the light traveling faster than it should. Objects traveling at high speeds will appear shorter than they really are to observers who are stationary (i.e. not moving). Length contraction is defined by

$$L=\frac{L_o}{\gamma(v)}=L_o\sqrt{1-\frac{v^2}{c^2}}$$ where $L$ is the length of the fast moving object (in this case the car) as measured by a stationary observer (assuming they were fast enough to measure the length of such a fast car), $L_o$ is the "proper" length of the object (i.e. the length of the car when it's stationary), $v$ is the speed of the car, and $c$ is the speed of light (299,792,458m/s).

Would the driver be able to see the high beam on in front of the car and be travelling twice as fast as the speed of light? Or will the photons produced pool up within the light bulb itself? Would there be no light produced even though the light bulb is on?

No, no, and no.
(EDITED: interestingly, nobody noticed my math error here)
In order to let all observers perceive the light as traveling at $c$, the object emitting the light gets shorter. For example, for a car of length 4m traveling at 99.9% of $c$:

$$L=L_o\sqrt{1-\frac{0.999^2c}{c}}=4m\sqrt{1-0.998001}=4m\sqrt{0.001999}=0.1788407112m$$ The car will appear to contract to a tenth of its original length. The world surrounding the car will also contract to a tenth of its original length but only in the view of the driver. It is important here to understand what is called a "frame of reference". In the frame of reference of a stationary observer, the world is still and the car is traveling at $0.999c$ relative to the observer. In the frame of reference of the driver, the interior of the car and the driver himself are stationary, whereas the world is travelling at $0.999c$ backwards.

The driver will see everything in the world around him become shorter: people will get thinner, trains will get shorter, etc. In fact, if the car were to travel at the speed of light (again, this is impossible, but hypothetically) it would have no length (and the driver's perception of time and space will be completely distorted).

$\gamma(v)$ is known as the Lorentz Factor and is given by: $$\gamma(v)=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $v$ is the velocity of the object and $c$ is the speed of light.

In case you're interested, the reason massive objects can't reach $c$ is shown by Einstein's redefinition of Isaac Newton's Second Law of Motion (originally $F=ma$): $$F=\frac{ma}{\left(1-\frac{v^2}{c^2}\right)^{\frac32}}$$ where $m$ is mass, $a$ is acceleration, $v$ is the object's velocity, and, well, you already know what $c$ is. As the velocity approaches the speed of light, more and more force is required to accelerate the object even slightly. Eventually it will require a practically infinite amount of force to accelerate the object further.