Image 1

Imagine that there is a car and it is not moving but its headlights are on. There is a wall in front of the car but is very far away. Right now energy is being used only in switching on the headlights. Now the car starts moving at a very high speed.

image 2

As I have shown in the picture, there is a blueshift of light and so the energy of light emitted per unit time has increased. Now my question is that from where does this extra energy come from.

Some arguments that prove that extra energy is generated. If the car was moving without the headlights off but at the same speed, the energy would have been used in the movement of the car. Now if the car was not moving and only the headlight was on, the energy would have been used in powering the headlight. But when we take both the cases simultaneously, then we see that there is an increase in the net energy. For further explanation I will give some equations.

Case 1 when the car is moving but the headlights are off

$Q_1 = \frac{dE}{dt} = \frac{d\sqrt{p^2c^2 + m^2c^4}}{dt} = 0;$

Case 2 when the car is not moving but the headlights are on

$Q_2 = \frac{dE}{dt} = \frac{d(mc^2)}{dt} < 0$ since energy is being radiated by the lights on;

Case 3 when the car is moving and the headlights are on

$Q_3 = \frac{dE}{dt} = \frac{d\sqrt{p^2c^2 + m^2c^4}}{dt} < Q_2$, because the power of radiation is higher than in Case 2. This is because the light is blue-shifted and its quanta have higher energy.

So from where does this extra energy of light come from?

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    $\begingroup$ The fact that the light is blue (together with the quantum idea of light, where blue light quantum has higher energy than a red light quantum) does not by itself guarrantee that the power of radiation will be higher for blue light. This is because the power of radiation depends also on the number of quanta emitted per unit time. This will be lower for the blue light, so it is not clear what the resulting power of radiation will be. $\endgroup$ – Ján Lalinský Feb 8 '14 at 16:51
  • $\begingroup$ Much easier is to analyze this within classical EM theory, where it turns out that the radiation of the light in the direction of car's velocity will have greater power (the frequency does not play any role). See mpv's answer. $\endgroup$ – Ján Lalinský Feb 8 '14 at 16:52
  • $\begingroup$ Aren't photons massless? So shouldn't it be $E=pc$ in all three cases? $\endgroup$ – Kyle Kanos Feb 8 '14 at 19:46
  • $\begingroup$ My impression was that $E$ is energy of the car. $\endgroup$ – Ján Lalinský Feb 8 '14 at 21:38
  • $\begingroup$ @KyleKanos $E = hv$. Though I am just 15 and do not know much but I think that is right. $\endgroup$ – rahulgarg12342 Feb 9 '14 at 2:28

The extra energy comes from the kinetic energy of the moving car. The radiated light is carrying away some momentum and is decreasing the speed (and therefore kinetic energy) of the car.

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  • $\begingroup$ That is not true as I have already mentioned the kinetic energy part in case 1 equations. Read my explanation once again. $\endgroup$ – rahulgarg12342 Feb 8 '14 at 17:00
  • $\begingroup$ rahulgarg12342, I've edited your question to make it more readable and corrected some errors. Mpv's answer is right - when the car moves and radiates light, some kinetic energy of the car is being transformed into light energy. Or we misunderstood your question (check the question and edit again if necessary). $\endgroup$ – Ján Lalinský Feb 8 '14 at 18:32
  • $\begingroup$ @rahulgarg12342 The extra energy comes from the kinetic energy of the car exactly like mpv says. $\endgroup$ – Brandon Enright Feb 8 '14 at 18:37
  • $\begingroup$ @JánLalinský I just did not understand how is $Q_2 < 0$ $\endgroup$ – rahulgarg12342 Feb 9 '14 at 2:30
  • $\begingroup$ When car radiates, its energy decreases, $Q_2 = dE/dt$ is positive if energy of the car increases in time, negative if it decreases in time. $\endgroup$ – Ján Lalinský Feb 9 '14 at 9:23

There is no extra energy - the blueshift is due to an increase in the observed frequency, not the actual frequency. The light is still being emitted at the same frequency so the same amount of energy is used.

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