The interference paradox This is a thought experiment I created myself.
Imagine two radio sources fixed at the ends of a  rocket and they produce signal of same frequency such that they arrive at the center simultaneously and at the center there is a device that explodes if the signals arriving at it interfere constructively, so obviously the device explodes in rocket’s frame. Now from the point of view of the observer on ground what will happen? If the signals don’t arrive simultaneously the device must not explode.
I personally don't find any difficulty in creating such a device as you can have antennas at the centre pointing in opposite directions.
 A: 
Now from the point of view of the observer on ground what will happen? If the signals don’t arrive simultaneously the device must not explode.

Constructive interference occurs at a given event when the phase, $\phi$, of the two waves differs by an even multiple of $\pi$ at that event. Destructive interference occurs when they differ by an odd multiple of $\pi$. So, the question is whether or not the phase is a relativistic invariant. If it is an invariant then all frames will agree whether or not device is triggered because all frames will agree if there is constructive interference. 
Now, we introduce the four-position $R^{\mu}=(ct,x,y,z)$ and the four-wavevector $K^{\mu}=(\omega/c,k_x,k_y,k_z)$ (see Four-vector in Wikipedia for details). These are both four-vectors so their product is an invariant. Expanding their product in an inertial coordinate system we get $g_{\mu\nu}R^{\mu}K^{\nu}=\omega t-k_x x-k_y y-k_z z = \phi$. We immediately recognize this quantity as the phase of a wave. 
Therefore, because the product of two four-vectors is manifestly invariant and because the phase can be written as the product of two four-vectors then we conclude that the phase is an invariant. This in turn implies that any device whose function is based on the phase, such as the detector in this scenario, will be predicted to behave identically in all frames. 
A: Unfortunately, Dale's answer lacks sufficient intuition! Assume that, instead of explosive material, there is a thin screen at the midpoint of the rocket. From the viewpoint of the observer in the lab frame of reference, contrary to the observer in the rocket's rest frame, the wavelengths meeting each other on the screen are no longer equal, and thus it is expected that no simple interference pattern is produced on the screen. (Paradox)
However, If you name each part of the left wave (crests, troughs, etc.) as $A$, $B$, $C$, ..., and  $A^\prime$, $B^\prime$, $C^\prime$, ... for the right wave, it can be easily calculated that the arrival of each counterpart pair of points (say $B$ and $B^\prime$) is simultaneous on the screen as seen by the observer in the lab frame. [See the attached figure for a destructive interference.] 
That is to say, although $B$ is closer to the screen than $B^\prime$, it needs to reach a receding screen during a time $t$; and although $B^\prime$ is farther from the screen than $B$, it needs to reach an approaching screen during a time $t^\prime$ so that we have: $t=t^\prime$ 
Therefore, we can deduce that both of the observers measure the same interference pattern on the screen and there is no paradox in this example. I just calculate the real wavelengths measured by the observer in the lab frame of reference for when they interfere with each other on the screen:
$$\lambda_1+vt_D=ct_D\rightarrow t_D=\frac{\lambda_1}{c-v},$$
and:
$$\lambda_2-vt_{D^\prime}=ct_{D^\prime}\rightarrow t_{D^\prime}=\frac{\lambda_2}{c+v}.$$
Using $\lambda_1=\sqrt{\frac{c-v}{c+v}}{\lambda_0}$ and $\lambda_2=\sqrt{\frac{c+v}{c-v}}{\lambda_0}$, we get:
$$t_D=t_{D^\prime}=\frac{\lambda_0}{\sqrt{c^2-v^2}}.$$
The wavelength of both waves that make an interference patern on the screen are calculated to be:
$$\lambda=ct_D=ct_{D^\prime}=\frac{c\lambda_0}{\sqrt{c^2-v^2}},$$
where ${\lambda_0}$ is the wavelength of the radio sources measured by the observer in the rocket's rest frame. The above calculations show that not only the phase of the waves are similar as hitting the screen, but also the wavelengths measured by the observer in the lab frame are equal and Lorentz expanded during the absorption by the screen. Remember that although ${\lambda_1}$ and ${\lambda_2}$ are not equal from the viewpoint of the observer in the lab frame of reference, he obtains similar values for the wavelengths as they meet the screen.

A: There is a very simple answer. The bomb will explode if and only if a detector on the ship detects constructive interference.
Thus, the interference needs to be observed on the ship (by the detector), to explode the bomb.
The explosion is based on that observation on the ship. It is a local observation on the ship that triggers the bomb.
Now you are asking maybe two questions:


*

*if the observer on ground (stationary I assume), does not see the two EM waves interfere constructively, why does the bomb explode in hos view

*if the observer on the ground does not see the EM waves interfere constructively, and the bomb explodes, how will the observer cope with this contradiction
The answer to 1. is very simple, the observer has nothing to do with the (explosion of the) bomb, the trigger is a local observer (detector) on the ship.
The answer to 2. is more complicated, the observer will definitely see a contradiction, but only if the observer does not understand relativity and SR. If the observer on the ground understands SR, then it will be clear, that in his own frame the EM waves did create constructive interference (but the frequency will be different based on the correct comment), like in the frame of the ship they did.
It is the same phenomenon why one observer might see the same EM field as electric and the other observer as magnetic.
