# New paradox in special relativity?

1. Mar 6, 2013

### djsourabh

we can see from length contraction of inclined bodies to the velocity,
consider situation in figure.A spaceship is moving with respect to observer 1 at velocity V.
as shown in the diagram,the observer inside observs length equal to PQ.
the observer outside observes length to be PS .
if this is true that is the inclination of say rod PQ changes to PS then,

consider figure 2.

the same situation is there in figure 2 except MN or MD is a reflecting surface instead of rods.
a ray of light comes from the outside of spaceship.(shown in orange colour)
the angles of incidence = angle of reflection x=x' and y=y'.... (law of nature)
the observer inside will see the light reflect as OAC. (say horizontal surfaces absorb light).
while the observer outside will see light reflect as OBO.
now if the vertical surface inside the spaceship (shown in yellow colour) is a light sensitive surface,
which on incident of light triggers 'SELF DISTRUCTION CIRCUIT' of the spaceship.
so observer outside should see ship blown, but observer inside light never hits yellow vertical surface.
this cannot happen according to special relativity.
so does the law of nature 'angle of incidence = angle of reflection' is violated/not applied?
or we have another explaination?

2. Mar 6, 2013

### phinds

You have failed to shorten everything equally. Your representation of the contracted mirror is correct but your representation of the reflection off of it is not shown contracted.

The light hits point C, which the external observer sees as C'. If you had a stiff rod from A to C, the same rod would go from B to C' according to the external observer and the light would go there as well.

Last edited: Mar 6, 2013
3. Mar 6, 2013

### Staff: Mentor

I have not looked into this in detail. The law of nature is Maxwell's equations. I am not certain that "angle of incidence = angle of reflection" is applicable in all cases. It may assume a stationary mirror.

4. Mar 6, 2013

### jartsa

Let's say the speed is very close to c.

The outside observer says the mirrors are almost vertical and the reflected beam is almost horizontal.

Then he continues: "the light does not hit the wall in the front, because the light is approaching the front wall extremely slowly, and has enough time to reach the side wall instead"

5. Mar 6, 2013

### phinds

Which is exactly what I illustrated in my redrawn image. You don't have to go near-vertical to get the picture.

6. Mar 6, 2013

### djsourabh

angle of incidence = angle of reflection' is violated/not applied?

7. Mar 6, 2013

### djsourabh

it seems that 'angle of incidence = angle of reflection' applies here but as light hits mirror from the frame outside, ship moves ahead some distance before light strikes somewhere again.and it always will strike horizontal surface.

have i understood it correctly?

8. Mar 6, 2013

### phinds

Exactly. But not "somewhere again", rather "exactly the same place as if the ship were not moving"

9. Mar 6, 2013

### jartsa

Ok, but my answer was intuitive.

Although, thinking more carefully, my observer was speaking carelessly.

He said: "when the spaceship accelerates to speed very close to c, the inclined mirror becomes almost vertical, while the inclined light beam becomes almost horizontal"

That sounds like a contradiction.

He should have said something like: "To me it seems like the light beam goes almost straight up, inside the very short space ship, but the spaceship moves to the right very fast, so the light beam can also be said to go almost straight from left to right"

Last edited: Mar 7, 2013