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Assume that a man is travelling in a space ship at a certain relativistic speed with respect to a man at rest at some point in space, such that 3 minutes in the ship is equal to 5 minutes for the person at rest.

Also assume that the man in the ship has a lighter which contains gas of a certain amount such that the lighter can be lit for 5 minutes .

Now if the man in the space ship lights the lighter for 3 minutes, then he would have 2 minutes' worth of gas left, but the stationary observer would have seen light emitted for about 5 minutes (since 3 in that space ship = 5 minutes for the stationary observer)

How is it possible for the stationary observer to see light for 5 minutes? And in this case, how is energy conserved?

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    $\begingroup$ Apart from various practical confusions (the light need not be equally bright for both observers, and in any case represents only a fraction of the energy freed by burning the gas), there is a fundamental misunderstanding here of what conservation of energy means: it means that for each observer individually all energy is exactly accounted for; it does not mean that the emitted light represents the same amount of energy to both observers (and indeed it need not, due to Doppler shift). $\endgroup$ – Marc van Leeuwen Aug 16 '17 at 11:18
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From the perspective of the stationary observer, the light is dimmer. Why? Because the chemical reaction is happening more slowly, so the fire emits fewer photons per second. The flame burns for longer, but it emits less energy per second. Both observers will agree on the total energy emitted by the flame (once they have accounted for possible red-shift) and thus total energy is conserved.

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    $\begingroup$ Although energy is conserved, it is not invariant. It is part of a 4-vector. How does this fit with your answer? $\endgroup$ – lalala Aug 16 '17 at 1:30
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    $\begingroup$ @lalala "Once they have accounted for possible red-shift." Obviously the amount of energy they actually measure will be different. $\endgroup$ – Jahan Claes Aug 16 '17 at 2:02
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Following on from @Jahan Claes' answer we can do some maths!

Say the man on the spaceship is using an ethanol burner. Ethanol has an enthalpy change of $1058 \text{ kJ}/\text{mol}$ but for simplicity let's say that it's equal to $1050$. If he burns $1$ mole of ethanol then it gives out approximately $1050 \text{ kJ}$. For the moving observer this is equal to $350\,000$ joules/min. For the stationary observer this will be equal to $210\,000$ joules/min.

Let's also say that the ethanol burner is giving out orange light, of a wavelength of $600 \text{ nm}$. Using the equation $$E = \frac{hc}{\lambda}$$ we can work out the energy of each photon to be roughly $3.315\times 10^{-19}$ (luckily for us the speed of light is the same for all observers). This means that the moving observer sees approximately $1.056\times 10^{24}$ photons every minute, whilst the stationary observer sees only approximately $6.335\times 10^{23}$ photons every minute.

From this we can see that $1.056(...)\times 10^{24}\times3 = 6.335(...)\times 10^{23}\times 5$ and everything works out just fine :)

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  • $\begingroup$ Putting an approximately sign in a defense of conservation seems wrong. Since the terms are already approximates perhaps you can round them differently to be equal in the last step or something. Or identify were the discrepancy happens. $\endgroup$ – user118047 Aug 15 '17 at 21:44
  • $\begingroup$ That's a good point. I just didn't want to start spewing significant figures everywhere though! I'll edit in a few "...'s" and set a proper equals sign. $\endgroup$ – CooperCape Aug 15 '17 at 21:46
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    $\begingroup$ Well, it is not correct. The photons get redshifted. Your wavelength is at emission, the person who sees it burning for 5 minutes sees lower freq, sees less total energy. Energy as measured by the two is not the same, energy in special relativity is a component of a vector, it is less for that observer, and so is the momentum to give a photon rest mass of zero, which is the only invariant, along with c. The other answer is wrong in the same way $\endgroup$ – Bob Bee Aug 15 '17 at 22:50
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    $\begingroup$ @BobBee You're assuming the ship is travelling AWAY from the observer. If the ship is travelling towards the stationary observer it would be blue shifted, and if it was travelling perpendicular to the observer it wouldn't be red or blue shifted. The point is that once you've adjusted for red/blue shift, you'll agree on the total energy emitted. You know this MUST be true, since in at least one possible setup--the perpendicular setup--there IS no red or blue shift. $\endgroup$ – Jahan Claes Aug 16 '17 at 0:08
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    $\begingroup$ @bobbee You're welcome to write your own, but I think the crux of the question centers around why the flame can last longer in one reference frame, not the details of the Lorentz boost for a particular setup. $\endgroup$ – Jahan Claes Aug 17 '17 at 13:56
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You say the gas in the lighter is sufficient to be lit for 5 minutes, but whose 5 minutes? I guess you mean 5 mins as reckoned by the spaceship observer. In that case, for the observer at rest, the amount of gas in lighter is sufficient to be lit for $5\times (5/3)=8.3$ mins. So when spaceship observer lights the gas only for 3 mins and has 2 mins worth of gas left over, observer at rest sees that the gas has been lit for 5 mins and 3.3 mins worth of gas is left over. There is no contradiction or violation of any law.

P.S. If you say that the amount of gas in the lighter can burn for 5 mins as per observer at rest, then from spaceship observer's point of view the lighter contains only 3 mins worth of gas. Then if he lights it for 3 mins there cannot be any gas left inside the lighter.

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protected by ACuriousMind Aug 16 '17 at 9:41

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