If an observer at the origin O of an inertial frame were to fire an object (an object that emits light continuously) at the speed of light, would the light from the object still reach the observer? I know that this requires infinite work (for a particle with mass) and this is therefore impossible, but if this factor is neglected, the photons that this object emits at the speed of light (in the direction of the observer) still reach at the observer?
On the one hand I would think so, because the speed of light is invariant, and since the object will one day be at point P (on the orbit), photons will still leave from point P to O on the orbit. speed of light (and will they reach O anyway)?
But on the other hand, if you look at the formulas for the time dilation, then the denominator becomes zero and time stands still (the time interval lasts infinitely long) for the observer in the inertial frame (relative to the object). In addition, length contraction will make the length of the object appear zero to the observer. Does the object become point-shaped?
Thanks for your reply
Answer
Dear Jonas,
As you know, in the theory of relativity, velocities are not simply added up. Assume that a rocket moves away from Earth at speed v and the astronaut fires a bullet toward Earth at speed u (relative to the rocket). Then, according to the theory of relativity, the velocity u’ of the bullet with respect to the earth will be as follows:
u’=(uv)/(1-uv/c2).
If the speeds are small compared to the speed of light, you can neglect the term uv/c² in the denominator and you get the daily formula u’=uv. In other words, you see the bullet approaching with the speed it has in the rocket reduced by the speed of the rocket.
But what you are proposing is a rocket with speed of light (v=c) that instead of a bullet emits light radiation, so also u=c. Then the formula becomes an indefinite fraction 0/0. However, try a rocket with v=c-ε (ε< The atoms in the rocket that, due to their vibrations, emit light of wavelength λ0 vibrate with the frequency c/λ0. You can think of them as clocks that tick at that frequency. As they move away from the earth at speed v, successive taps take place at increasing distances, and the light of a tap has a greater path to travel to the earth than that of the previous tap; it will therefore also arrive on earth correspondingly later. To the terrestrial observer of the light, the ticks are thus further apart, the frequency appears smaller and the wavelength larger. This is the well-known Doppler effect, also known for sound waves. But relativity does something more: thanks to the time dilation, the clocks (the atoms) themselves start to tick more slowly. The combination of the two effects gives the following relationship between the observed wavelength λ on Earth and the wavelength in the rocket: =λ0 [(1+v/c):(1-v/c)]1/2 For v≅c the λ becomes very large: we perceive light in the far infrared. In the limit v=c the wavelength becomes infinite: you only observe a non-vibrating electromagnetic field. Your last question about the apparent deformation of a luminous moving object I cannot address here. It took 50 years to discover it. After the discoverers it is called the Penrose-Terrell rotation. You can read it (only English) on e.g. http://www.math.ubc.ca/∼cass/courses/m309-01a/cook/terrell1.html I hope this helped you further. Relativity is an exciting subject, but if you want to delve into it, in-depth mathematics is involved.
Answered by
Prof. dr. French Cerulus
physics, especially classical theoretical mechanics, electromagnetism, quantum mechanics, history of physics .
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