# Does the inertia of an object changes with its temperature? [closed]

$$F=ma$$ is not accurate enough at relativistic speeds.

Does the inertia of a rigid body related to its temperature?

## closed as unclear what you're asking by AccidentalFourierTransform, Mozibur Ullah, ZeroTheHero, M. Enns, BuzzDec 10 '18 at 2:30

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At relativistic speeds it's easier to work with the more fundamental version of Newton's second law, namely $$\textbf{F}=\frac{d(\gamma m \textbf{u})}{dt}.$$ $$m$$ is the body's Lorentz-invariant mass or 'rest mass'. This includes (multiplied by $$\frac{1}{c^2})$$ the body's internal energy' and therefore raising the temperature of a body increases its mass. $$\textbf{u}$$ is the body's velocity and $$\gamma=(1-\frac{u^2}{c^2})^{-\frac{1}{2}}.$$ [$$\gamma m \textbf{u}$$ is the body's momentum.]

The equation has been tested experimentally by observing charged particle trajectories in electric and magnetic fields, and comparing with the predictions of the equation $$q(\textbf{E}+\textbf{v} \times \textbf{B})=\frac{d(\gamma m \textbf{u})}{dt}.$$ You'll recognise the left hand side as the Lorentz force.

A body of larger mass but the same charge will, for example, go in a wider circle in a given magnetic field, just as in Newtonian Physics, but it would be very difficult to detect this happening by raising the temperature of a macroscopic charged body, as the effect on its mass would be so small.

You can derive relativistic forms of $$\textbf{F} =m\textbf{a}$$ from $$\textbf{F}=\frac{d(\gamma m \textbf{u})}{dt},$$ but they are rather messy, involving different values for 'mass' (that is the coefficient of $$a$$) according to whether the force is parallel to the direction of motion or transverse to it.

• Right, I got confused, and placed the γ in the wrong place in my head. Of course there should be no $\gamma$s in tensorial expressions, or IOW your time ought to not be an invariant. My apologies. I removed my comment to avoid confusing people. Thanks for clarifying which quantities you mean. – tobi_s Dec 9 '18 at 12:23
• Not a problem. As you see, I've removed my comments, too. – Philip Wood Dec 9 '18 at 12:31

You are right to be suspicious, and indeed (relativistically), the answer is no. The higher-temperature object will experience less acceleration (even - and this is important, in the "relativistically correct" sense of "proper acceleration"), and can thus be considered to have more mass.

The reason for this is that the mass of a system - here a collection of vibrating atoms or molecules - is not necessarily equal to the sum of its parts. In particular the actual total mass is equal to the masses of all parts plus the mass-equivalent of all forms of energy (i.e. excluding the masses of parts just counted) contained within the system, thanks to $$E = mc^2$$. And one of these is thermal energy - kinetic energy of random vibrations of atoms relative to each other. It is important to point out, though, that this is a distinct phenomenon from what is called "relativistic mass", which is the notion that the energy of translation of the system as a whole, or perhaps, of its center of mass, should be counted as a form of mass, which makes mass frame dependent. This added mass from thermal energy is not frame dependent, since no reference frame shift will change the patterns of relative motions of system components with respect to each other, and thus it is a real increase in mass. It is also not best thought of as the result of relativistic mass increases of the individual particles because if we treat any particle individually we have the same concern, rather it's better thought of as inhering within the system as a whole, as an additional term in the total accounting required to compute its mass.

In practice, of course, the contributions are very small, and typically too small to measure. For example, if I have a 1 kg jug of water at room temperature, say 295 K, versus one with the water boiling hot at 373 K, then the difference in mass $$\Delta m$$ can be found by

$$\Delta m = \frac{m_0 c_\mathrm{th} \Delta T}{c^2}$$

where $$m_0$$ is the mass at the "baseline" temperature (here 295 K), thus here $$m_0 = 1\ \mathrm{kg}$$, $$\Delta T$$ is the temperature difference, here $$373 - 295 = 78\ \mathrm{K}$$, and $$c_\mathrm{th}$$ (subscripted to disambiguate against the Einstein constant, $$c$$) is the specific heat capacity for water, $$4184\ \frac{\mathrm{J}}{\mathrm{kg \cdot K}}$$. With these parameters the $$\Delta m$$ is about $$3.6 \times 10^{-12}\ \mathrm{kg}$$, or $$3.6\ \mathrm{ng}$$. This is about the same mass change that would result as from dropping a single human cell (~3 ng) into the original water. I do not believe we have instruments sensitive enough to measure such tiny mass changes on the order of one part in $$10^{12}$$ as of this writing, but I could be wrong. (Moreover, for a realistic jug, evaporative losses would, of course, dwarf this, and so we'd be best off using a piece of solid like a metal being heated over a wider range to compensate for lower heat capacities, say up to orange heat at 1200 K instead, giving almost 900 K of temperature difference and this compensating for roughly 10-fold lower heat cap for metals like iron.)

• @Elio Fabri : Yes, you'd be right. I removed that claim just now. – The_Sympathizer Dec 9 '18 at 14:03
• OK, I've removed my comment. But what is a "non-mass energy"? I'm afraid of misunderstanding by naive readers. Those who are accustomed to think of "matter converting into energy" and the like. – Elio Fabri Dec 9 '18 at 14:11
• @Elio Fabri : Modified it some more. – The_Sympathizer Dec 9 '18 at 14:47