3
$\begingroup$

Imagine an alternate universe where FTL do happens, let's just say the inhabitants discovered a hypothetical particles that allows ordinary matter to temporarily be in superposition of any amount of mass including negative value... [I'm drunk don't mind me]

Okay straight to the point, there should be luminous boom, a flash when the spaceship goes faster than light so I'm wondering if we can measure the blue shift or any properties of this cherenkov effect to determine how much faster than the speed of light the spaceship moved? Please show your working and round off to the third sig fig.

$\endgroup$
  • $\begingroup$ lets rethink this, first we need to know the starting frequency $\endgroup$ – Topcode Apr 4 at 1:56
  • $\begingroup$ also you mean redshift because when it moves away it stretches the light $\endgroup$ – Topcode Apr 4 at 1:59
  • $\begingroup$ and cherenkov radiation is already blue talking about reactors at least $\endgroup$ – Topcode Apr 4 at 2:03
  • $\begingroup$ but since we are talking about a spaceship i assume it would have different effects $\endgroup$ – Topcode Apr 4 at 2:05
  • 2
    $\begingroup$ being drunk doesn't justify lack of research $\endgroup$ – L.Dutch - Reinstate Monica Apr 4 at 3:04
4
$\begingroup$

If you had just googled you'd have found your answer here...

The frequency spectrum of Cherenkov radiation by a particle is given by the Frank–Tamm formula:

$d^2E \over dxd\omega$$=$$q^2 \over 4\pi$$\mu(\omega)\omega$$(1-{c^2 \over v^2n^2(\omega)})$

The Frank–Tamm formula describes the amount of energy E emitted from Cherenkov radiation, per unit length traveled x and per frequency $\mu$. $\mu(\omega)$ is the permeability and $n(\omega)$ is the index of refraction of the material the charge particle moves through. q is the electric charge of the particle, v is the speed of the particle, and c is the speed of light in vacuum.

Unlike fluorescence or emission spectra that have characteristic spectral peaks, Cherenkov radiation is continuous. Around the visible spectrum, the relative intensity per unit frequency is approximately proportional to the frequency. That is, higher frequencies (shorter wavelengths) are more intense in Cherenkov radiation. This is why visible Cherenkov radiation is observed to be brilliant blue. In fact, most Cherenkov radiation is in the ultraviolet spectrum—it is only with sufficiently accelerated charges that it even becomes visible; the sensitivity of the human eye peaks at green, and is very low in the violet portion of the spectrum.

As in sonic booms and bow shocks, the angle of the shock cone is directly related to the velocity of the disruption. The Cherenkov angle is zero at the threshold velocity for the emission of Cherenkov radiation. The angle takes on a maximum as the particle speed approaches the speed of light. Hence, observed angles of incidence can be used to compute the direction and speed of a Cherenkov radiation-producing charge.

But mind that all of the above applies to particles moving in a medium at a velocity higher than the speed of light in that medium.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.