# Question about light speed (is velocity cumulative?)

1. Dec 12, 2009

### blockstyl

If a spaceship were to fly at 1/10 the speed of light and project a laser onto a distant moon, would the light cast from the ship reach the moon at 1.1 times the speed of light, relative to a stationary mounted laser fired at the same time/distance?

2. Dec 13, 2009

### hamster143

No, it would reach the target at exactly the same time. Velocity addition law in special relativity is such that c + any speed is still c.

3. Dec 13, 2009

### blockstyl

How is this reconciled with observations from within the moving ship? Wouldn't they find the recorded time of impact to indicate that light had travelled more slowly than expected?

4. Dec 13, 2009

### hamster143

No, why would they?

5. Dec 13, 2009

### HallsofIvy

Not if they were knowlegable not to "expect" that their speed would be added to light speed. And the whole point of relativity is that experiment shows that speeds do NOT add like that.

6. Dec 13, 2009

### blockstyl

Still going to need some help with this. If there was another ship matching speed and trajectory, who was at the moon at the time the laser was fired, how could the laser directed at the moon travel at .9C, while another laser directed at the other ship travel at C? It shouldn't make a difference between the two ships how fast they are going, but it wouldn't make sense for the beam to impact the moon at the same time as the other ship.

7. Dec 13, 2009

### JesseM

Keep in mind that each observer uses rulers and synchronized clocks at rest relative to themselves to measure speed = distance/time, and because of length contraction, time dilation, and the relativity of simultaneity, each observer will judge that the rulers and clocks of other observers are distorted--their rulers will appear shrunk, their clocks will appear slowed down and out-of-sync. Here's a numerical example of how this works out to ensure everyone measures light to move at c, which I wrote up a while ago for another thread:

To measure the speed of anything, you need to measure its position at one time and its position at another time, with the time of each measurement defined in terms of a local reading on a synchronized clock in the same local region as the measurement; then "speed" is just (change in position)/(change in time). So, time dilation, length contraction, and the relativity of simultaneity all come into play. From the stationary observer's perspective, the ruler which the moving observer used to measure the distance was shrunk by a factor of $$\sqrt{1 - v^2/c^2}$$, the time between ticks on the moving clocks is expanded by $$1 / \sqrt{1 - v^2/c^2}$$, and the two clocks are out-of-sync by $$vx/c^2$$ (where x is the distance between the clocks in their own rest frame, where they are synchronized).

So, say there's a ruler that's 50 light-seconds long in its own rest frame, moving at 0.6c in my frame. In this case the relativistic gamma-factor (which determines the amount of length contraction and time dilation) is 1.25, so in my frame its length is 50/1.25 = 40 light seconds long. At the front and back of the ruler are clocks which are synchronized in the ruler's rest frame; because of the relativity of simultaneity, this means that in my frame they are out-of-sync, with the front clock's time being behind the back clock's time by vx/c^2 = (0.6c)(50 light-seconds)/c^2 = 30 seconds.

Now, when the back end of the moving ruler is lined up with the 0-light-seconds mark of my own ruler (with my own ruler at rest relative to me), I set up a light flash at that position. Let's say at this moment the clock at the back of the moving ruler reads a time of 0 seconds, and since the clock at the front is always behind it by 30 seconds in my frame, then in my frame the clock at the front must read -30 seconds at that moment. 100 seconds later in my frame, the back end will have moved (100 seconds)*(0.6c) = 60 light-seconds along my ruler, and since the ruler is 40 light-seconds long in my frame, this means the front end will be lined up with the 100-light-seconds mark on my ruler. Since 100 seconds have passed, if the light beam is moving at c in my frame it must have moved 100 light-seconds in that time, so it will also be at the 100-light-seconds mark on my ruler, just having caught up with the front end of the moving ruler.

Since 100 seconds passed in my frame, this means 100/1.25 = 80 seconds have passed on the clocks at the front and back of the moving ruler. Since the clock at the back read 0 seconds when the flash was set off, it now reads 80 seconds; and since the clock at the front read -30 seconds, it now reads 50 seconds. And remember, the ruler was 50 light-seconds long in its own rest frame! So in its frame, where the clock at the front is synchronized with the clock at the back, the light flash was set off at the back when the clock there read 0 seconds, and the light beam passed the clock at the front when its time read 50 seconds, so since the ruler is 50-light-seconds long, the beam must have been moving at 50 light-seconds/50 seconds = c as well! So you can see that everything works out--if I measure distances and times with rulers and clocks at rest in my frame, I conclude the light beam moved at 1 c, and if a moving observer measures distance and times with rulers and clocks at rest in his frame, he also concludes the same light beam moved at 1 c.

If you want to also consider what happens if, after reaching the front end of the moving ruler at 100 seconds in my frame, the light then bounces back towards the back in the opposite direction towards the back end, then at 125 seconds in my frame the light will be at a position of 75 light-seconds on my ruler, and the back end of the moving ruler will be at that position as well. Since 125 seconds have passed in my frame, 125/1.25 = 100 seconds will have passed on the clock at the back of the moving ruler. Now remember that on the clock at the front read 50 seconds when the light reached it, and the ruler is 50 light-seconds long in its own rest frame, so an observer on the moving ruler will have measured the light to take an additional 50 seconds to travel the 50 light-seconds from front end to back end.

8. Dec 15, 2009

### blockstyl

If I ever have time, I will try to model all that. I don't suppose anyone's made it into a visual format that can be played with?

9. Dec 15, 2009

### lalbatros

Last edited by a moderator: Apr 24, 2017
10. Dec 16, 2009

### deadcat

I believe it would just start from 1/10 SOL then reach SOL and continue at that speed. SOL is confusing though sometimes.

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