# Why would light go backwards.

1. Oct 16, 2013

### aleemudasir

Why would light go to the back of spacecraft in the earth reference frame as shown in figure b? Didn't get that.

#### Attached Files:

• ###### Relativity.jpg
File size:
27.2 KB
Views:
76
2. Oct 16, 2013

### Simon Bridge

The pic is for the situation that the spacecraft could travel faster than light, while keeping the speed of light the same for all observers.

3. Oct 16, 2013

### HallsofIvy

Staff Emeritus
You may be thinking of the situation where I, riding in the back of a bus at 40 mph, throw a ball forward at, relative to me and the bus, 50 mph. Relative to you, standing at the side or the road, the ball is traveling at 50+ 40= 90 mph (approximately).

But the basic concept to relativity is that the speed of light is the same for all observers. If, instead of throwing a ball, I pointed a flashlight forward, you would see the light still moving at speed c, not at c+ 40. If I were on a spaceship moving with speed v, you would see light moving at c, not c+ v. If v were greater than c you would see the light moving backward relative to the rocket because the rocket is faster. (0f course, relativity says it is impossible for an object to move faster than light so this whole scenrio is fishy!)

(I said, above, that the ball's speed, relative to you on the side of the road, would be approximately "50+ 40= 90 mph". The reason for the "approximately" is that the Gallilean "addition of velocities", $v_1+ v_2$, does not exactly apply even for slower moving objects. The relativistic formula is
$$\frac{u+ v}{1+ \frac{uv}{c^2}}$$
Since the denominator is larger than 1 for any non-zero u and v, the resultant speed is slightly less than "u+ v". Of course, c is so much larger than "40 mph" or "50 mph" (about 186000 miles per second which is 669600000 miles per second, $uv/c^2= (40)(50)/(669600000)^2$ would be about 0.00000000000000446) that would be indistinguishable from 90 mph (89.999999999999598540614843079346 mph). But with v equal to c, that would become (u+ c)/(1+ u/c)= (uc+ c^2)/(c+ u)= c(u+ c)/(c+ u)= c.)

4. Oct 16, 2013

### Staff: Mentor

Not only that, but rocket observer also has to observe the speed of light to be c, meaning that if it started in front of him moving away, it will stay in front of him and move further away. Try reconciling that with your view of the light falling behind the rocket, and you'll conclude that the scenario isn't just "fishy", we're talking serious hakarl here.

(HallsOfIvy knows this perfectly well, so I'm not correcting him here, just adding something for anyone else who happens to come upon this thread)

Last edited: Oct 16, 2013
5. Oct 16, 2013

### HallsofIvy

Staff Emeritus
And would do great damage to our karma.

For those who are wondering, I looked up "hararl". It is a food consumed in iceland consisting of fermented Greenland shark. The Greenland shark's meat is poisonous, containing large quantities of trimethalamine oxide, which is removed through the fermentation process. However, the result is considered an "aquired taste" both smelling and tasting really, really bad!
http://en.wikipedia.org/wiki/Hákarl

(HallsOfIvy knows this perfectly well, so I'm not correcting him here, just adding something for anyone else who happens to come upon this thread)[/QUOTE]

6. Oct 16, 2013

### dauto

It wouldn't. That's the point of the picture. It shows that FTL (faster than light) travel is inconsistent with a constant light speed which is one of the pillars of the relativity theory.

7. Oct 16, 2013

### A.T.

In my opinion the argument against FTL shown in the picture is flawed, because it omits length contraction. If you want to play the "what if the rocket was going FTL"-game, then you also have to extrapolate length contraction beyond the c limit. And depending on how you do this, you can create consistency with the two postulates of SR. For example, if the extrapolated length contraction would mirror the rocket along its movement direction (negative contraction factor) the light could move away from the astronaut at the same speed in both frames.

DISCLAIMER FOR MODS: I'm not trying to speculate what happens at v>c. I'm merely pointing out, that the presented argument assumes v>c and pretends to know how the rocket would like at v>c. The way the rocket is drawn in the Earth's frame contains unjustified hidden assumptions.

8. Oct 16, 2013

### aleemudasir

got that.

9. Oct 16, 2013

### Simon Bridge

@AT: the argument is demonstrating the need for some sort of transformation to make the results consistent rather than demonstrates that the postulate of invariance for some speed means that nothing faster will be observed. Yah. So there are a bunch of unspoken assumptions based on the assumed education level of the audience perhaps?