# Car travelling at 0.8 c

1. Apr 19, 2015

### RisingSun361

The following is my understanding of how light travels:

I’m driving my car at 0.8 c, and turn on the headlights. The light comes out at 0.2 c, producing a net speed of exactly 1.0 c. I see it as 1.0 c, and an observer on the side of the road sees it as 1.0 c, although the light itself is actually moving at much less than c, just 0.2 c. The faster the car goes, the slower the light, always producing a net speed of 1.0 c.

Is my understanding correct?

2. Apr 19, 2015

### VantagePoint72

No. The light is traveling at c in both the car's and the roadside observer's reference frames, it does not travel at 0.2c according to anyone. c is the speed that light is "actually" moving at in all inertial reference frames (in a vacuum), it is never "actually" moving at less than that.

3. Apr 19, 2015

### Ibix

Always ask yourself "with respect to what?" when you hear yourself (or anyone else) talking about velocity. If the answer isn't explicitly stated, and isn't obvious from context, the problem isn't specified properly. That is part of what is going wrong with your picture.

Sitting in your car, you can consider yourself to be at rest. From your viewpoint, the light is travelling away from you at c - not 0.2c, or anything else. From the viewpoint of an observer at rest on the roadside, you are doing 0.8c and the light is doing c.

In fact, everyone agrees that the light is doing c. They do not (in general) agree what speed you are doing, so will not agree on the rate at which the light is opening the gap between you. But they will always note that your clocks are time-dilated and your rulers length-contracted, so will be unsurprised that you consider light to be opening the gap at c.

4. Apr 19, 2015

### Staff: Mentor

Google for "relativistic velocity addition". The relevant formula will be $w=(u+v)/(1+uv)$, where $u$ is the velocity of the car relative to you, $v$ is the velocity relative to the car of something thrown from the car, and $w$ will be the velocity of the something relative to you. Here we have $u=.8c$ and $v=c$.

Without relativity, you would expect $w=(u+v)$, but experiments (including some done long before the discovery of relativity, so the results were somewhat baffling at the time) have confirmed that the relativistic formula is correct. It's very hard to see any difference if $u$ and $v$ are both small compared to the speed of light, which is why we don't often notice.

5. Apr 19, 2015

### VantagePoint72

Don't forget relativity of simultaneity! If we use the standard Einstein clock synchronization scheme, all three are needed to "conspire together" to ensure that the roadside observer ("Alice") correctly notes that according to the car observer ("Bob"), the gap between car and light beam is widening at c (even though she herself marks it down as c-v). Bob's frame marks the light beam's progress at two events: when it's originally emitted and when it's propagated some distance from the car (a distance that, according to Alice will be exaggerated by Bob since the latter is observed to use contracted rulers). The two clocks that Bob has to use are at different places in his reference frame, and so Alice would say that Bob's account of the light propagating away from him at c is because he measures its recession with contracted rulers and with clocks that are both running too slow and aren't synchronized with each other.

Of course, when it comes to this sort of thing it's much easier to just forget about length contraction, time dilation, and relativity of simultaneity and just get the velocity composition formula straight from the Lorentz transformations. The synchronization scheme is just a convention anyway. But it's nice to be reminded now and then of the completely internally consistent "stories" observers can tell to explain why relatively moving observers observe the things they do.

Last edited: Apr 19, 2015