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We agree that energy is higher in liquids than solids and is higher in gases than liquids. How can we fit this into $E=mc^2$. As an example; 1 kg of ice has less energy than 1 L of water, right? But masses are the same specially in the case of water with density of 1,000 kg/m³. Could you please help me out with this paradox?! All answers are appreciated.

Kevin Kwok
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Ali
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    It's not exactly clear what you're trying to do here: Think about what type of energy the $E$ in $E=mc^2$ is and what type of energy you are talking about when you say it is "higher in liquids". – ACuriousMind Oct 09 '16 at 11:51
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    @Ali "1 kg of ice has less energy than 1 L of water": it looks like you're referring to the specific latent heats here. – BLAZE Oct 09 '16 at 13:47
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    So, why the downvotes? Sure, the question is naive, but how is it a bad question? It seems a question is voted up or down based on whether the question is obvious to the voters or not... – anon01 Oct 09 '16 at 14:14
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    I too think the downvoting is unfortunate. This is a perfectly good question and the same number of water molecules will indeed have a different mass in different states. – John Rennie Oct 09 '16 at 14:16

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When one gram of ice becomes one gram of water, it gains 80 cal of energy, sometimes called latent heat. 80 cal is 335 J. Einstein tells us that $$m=\frac{E}{c^2}$$ so the increase in the mass of the sample is $3.7 \times 10^{-12}$ grams.

Hardly measurable.

garyp
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  • Well said, that was exactly what I was trying to say, but you explained it better so (+1). – BLAZE Oct 09 '16 at 14:01
  • Is this true? This is using older convention for relativistic mass that has fallen out of favor, no? – anon01 Oct 09 '16 at 14:31
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    @ConfusinglyCuriousTheThird No, this is a calculation of the increase of the rest mass of an object due to internal energy. The internal energy in this case is largely potential energy of inter-molecular bonds. What you are referring to is an increase in inertial mass due to increased velocity. – garyp Oct 09 '16 at 14:36
  • This is not accurate. The phase transition is a thermodynamic property. It effects the kinetic energy of the molecules and not the mass. Mass changes happen in chemical and nuclear reactions not thermodynamic ones. – Michael Oct 09 '16 at 21:32
  • @Michael The temperature of the substance remains constant during the phase transition, so the kinetic energy, and any other contribution quadratic in position or momentum, remains constant. The latent heat can be thought of as a kind of binding energy, analogous to the nuclear binding energy responsible for the mass defect in nuclei. – garyp Oct 09 '16 at 22:03
  • @garyp I think that we may need to agree to disagree on this point. The transitional energy for the phase transition is to remove it from the negative potential well that is inside the lower energy state. I can see a parallel to a general relativistic effect, but I am skeptical about the analogy actually holds on the physical state. If this is correct, the difference should be measurable. I am going to look to see if anyone has done any work on this type of experiment. It would make an interesting paper. – Michael Oct 09 '16 at 23:00
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$E=mc^2$ is probably the most known and misunderstood equations in physics by the general population. Lets look where it comes from. The relativistic energy equation is $E=\gamma mc^2$ where $\gamma$ is the Lorentz factor $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. We set $\beta = \frac{v}{c}$ giving $\gamma = \frac{1}{\sqrt{1-\beta^2}}$. Doing a Maclaurin expansion of the Lorentz factor around $\beta$ gives you get $\gamma\approx 1+\frac{1}{2}\beta^2+\frac{3}{8}\beta^4+...$

When we combine the expansion back into the energy equation we get:

$E\approx mc^2+\frac{1}{2}mc^2\frac{v^2}{c^2}+\frac{3}{8}mc^2\frac{v^4}{c^4}+...$

So velocity is $0$ this becomes the equation $E=mc^2$. This is the mass energy conversion equation that everyone knows and loves. It indicates how much energy you have if you convert mass into energy. As velocity increases the other terms are important. The first term simplified to $\frac{1}{2}mv^2$ is the kinetic energy term giving the connection to Newtonian physics. This is also the term for things like temperature as temperature is the average random molecular kinetic energy in a system. The third term $\frac{3}{8}m\frac{v^4}{c^2}$ is the first relativistic correction in special relativity. Note that velocity must be very high ($\approx 0.75c$) before this term comes into noticeable effect.

Michael
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To compare apples with apples, let's consider a particular substabnce that can exist in the three well-known phases: solid, liquid and gas. The thermal energy per unit volume (thermal energy density) in the gas phase would in general be larger than that of the liquid phase, which in turn would have a larger thermal energy density than the solid. However, the thermal energy is not the only form of energy that would be present in a particular volume of a articular substance. There can also be other forms of energy such as chemical energy.

If we would consider the total mass of a particular volume of some substance in terms of energy through Einstein's relation $E=m c^2$ then we end up with an extremely large amount of energy that exceeds to amount of thermal energy by so many orders of magnitude that the contribution of thermal energy to the total amount of energy is completely insignificant. Therefore, within the accuracy, intrinsic in the statement that water has a density of 1000 kg/m$^3$, it would not be possible to notice the contribution of thermal energy.

flippiefanus
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  • So in fact the water is heavier than the ice? – Mike Dunlavey Oct 09 '16 at 13:36
  • Yes, ice floats on water, but I don't see how this follows from the argument provided here. – flippiefanus Oct 09 '16 at 13:39
  • Ice doesn't float on water because it's lighter. It floats because it's larger. What I mean is, take the same number of water molecules. If they are warmer, does that make them heavier? (If you say yes, I'm inclined to believe you.) – Mike Dunlavey Oct 09 '16 at 13:45
  • Ice floats of water because it has a lower mass density. Yes, in principle one can say the warmer molecules would be heavier, but one would be hard pressed to see that, because the difference is miniscule. – flippiefanus Oct 09 '16 at 13:51
  • So yes, it is heavier, though miniscule, as you say. – Mike Dunlavey Oct 09 '16 at 16:04
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In short: this is not the correct equation to describe the physical scenarios you have in mind.

$E=mc^2$ is probably the most misunderstood and misapplied equation in all of physics. This equation gives some conversion factor ($c^2$) if some mass is converted to energy - think atomic bombs, and is not useful to describe the scenarios you're considering.

A more complete version would contain the kinetic energy term:

$E^2 = (mc^2)^2 + (pc)^2$.

From this, you could calculate the average kinetic energy of your solid/liquid/gas, and would indeed find the total energy higher. Alas, this is still incomplete, because it doesn't account for the potential energy of the system, which is important in this case.

anon01
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