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The question stated: By what percentage does your rest mass increase when you climb 30m to the top of a ten-story building?

New to the concept of relativistic energy, I was a bit confused with figuring out how to approach this question.

The relativistic energy of a particle (its normal total energy but incorporating relativity effects due to its high speeds) is equal to:

$E= K+mc^2=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma m c^2 $

Since the equation is only the sum of the particles kinetic energy and rest mass, I am uncertain in how to find the rest mass increase knowing only the change in height (potential energy increase). Since I am not moving at relativistic speeds I can just use the normal conservation of energy formula:

$E_{initial}=E_{final} $

(assuming I started at height h=0m, and all energy was conserved while climbing)

$\frac{1}{2} m v^2 = mgh $

which when simplified and rearranged for speed v equals to:

$v=\sqrt{2gh}$

The formula for relativistic mass is:

$m_{rel}=\frac{m_{rest}}{\sqrt{1-\frac{v^2}{c^2}}}$

Since the question asked me "by what percentage does your rest mass increase", I am assuming I have to find:

$\frac{m_{rel}}{m_{rest}} \times 100$

Rearranging the formula for relativistic mass and substituting in the equation for speed v from above:

$\frac{m_{rel}}{m_{rest}} = \frac{1}{\sqrt{1-\frac{2gh}{c^2}}} $

However this gives me an answer of 1, which is clearly incorrect. The answer stated in the text is $3.3 \times 10^{-13} $ %.

I have checked over my work, but I don't know whether I have made an error in my reasoning or my calculations. All help is greatly appreciated.

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This is a poor question because your rest mass does not increase when you climb 30m. However I can see how the examiner expects you to answer.

If your mass is m then your energy when static at the bottom of the building is $mc^2$. To climb a distance h you have to hav work $mgh$ done on you, so your energy is now $mc^2 + mgh$. So the percentage change in your energy is:

$$ \%E = \frac{mgh}{mc^2} = \frac{gh}{c^2} $$

and if you set $h = 30$m you find this comes to $3.27 \times 10^{-15}$ or $3.27 \times 10^{-13}$%.

But let me emphasise that to say your mass has increased when you climb is meaningless nonsense. If you take a bound system (e.g. you and the Earth) and add some energy $E$ to it then it's true that the mass of the bound system will increase by $E/c^2$. For example this is the origin of the notorious mass deficit in nuclei. However to claim that the rest mass of any particular part of the bound system changes is meaningless.

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  • $\begingroup$ It's bad enough that "mass" can mean different things in relativity, and now we have people explicitly saying "rest mass" when they mean the exact opposite! $\endgroup$ – Javier Aug 13 '15 at 18:04
  • $\begingroup$ There is a nice Wikipedia article on mass in general relatvity:en.wikipedia.org/wiki/Mass_in_general_relativity . What John has described is known as the Newtonian limit for the mass. $\endgroup$ – Virgo Aug 14 '15 at 1:39
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I believe John's answer is sufficient to guide you through the question. My answer will try to pinpoint your mistake in your working. Your working is mostly correct except the ratio $\frac{m_{rel}}{m_{rest}}$.

To get the correct value for $\frac{m_{rel}}{m_{rest}}$, use Taylor expansion:

$\frac{1}{\sqrt{1-\frac{2gh}{c^2}}} = (1-\frac{2gh}{c^2})^{-\frac{1}{2}} = 1 + (-\frac{1}{2})(-\frac{2gh}{c^2}) + \ldots$

Terms with power higher than 1 is omitted since they are very small compared to the first and second term. The increase in mass is the second term, which you can verify is $3.27\times10^{-15}$. The reason you get 1 is because the number differs from 1 by such a small quantity (after 15 figures) that a scientific or graphical calculator cannot display.

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