How can energy have inertia?

To my intuition, inertia is so closely associated with mass that my intuition says "Huh?"

Indirectly by mass energy equivalence it works fine, for example:

I have a closed system, and add energy. Now it has more mass according to $E=mc^2$, and the inertia associated with that mass increased.

I do not doubt it's right, but does it mean anything associated with physics, beyond being true?

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    $\begingroup$ Relativistic rocket works by pushing on the energy it carries along, which is mostly kinetic energy. But the question seems quite unanswerable. $\endgroup$
    – stuffu
    Jan 18, 2020 at 8:24
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    $\begingroup$ @jinawee maybe you can try to reach prof. Einstein to explain him that the title he gave to the paper where he derived $E=mc^2$ was wrong. $\endgroup$ Jan 18, 2020 at 11:07
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    $\begingroup$ @GiorgioP $E=mc^2$ does not imply the other claim. Unless you also accept claims like frequency has energy $\nu=E/h$. $\endgroup$
    – jinawee
    Jan 18, 2020 at 11:14
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    $\begingroup$ @GiorgioP Of course, "inertia depends on energy content" and "energy has inertia" are two different statements. $\endgroup$ Jan 18, 2020 at 14:06
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    $\begingroup$ I don't understand what beyond "being true" you want from that statement. Physics is a collection of models, it does not, in and of itself, possess a unique ontology (cf. e.g. the many different quantum interpretations). Being true is all a model can ever hope. (And if you want, we can even argue about what "truth" is science, cf. e.g. Popper vs. Kuhn vs. Feyerabend vs. ...) $\endgroup$
    – ACuriousMind
    Jan 18, 2020 at 22:30

6 Answers 6


Here's an intuitive way, I believe, to understand it.

First off, we need to have some definitions, to understand exactly what we're talking about - in particular, we need to know what exactly we mean by a "closed system" or even a "system" in general, here. This is an important and crucial part of any deductive reasoning (and missing it is the source of many errors therein) - for it is, in effect, part of what supplies the premises, beyond also, plain facts regarding the situation.

You see, relativity, in its core insights, is really "just" a theory of principles describing space, time, and information. What you get in textbooks about "relativity" is actually a hybrid of the "true" core thrust of the theory, together with a Newtonian-style (in that it involves particles with infinite position and momentum information guaranteed at all times, forces, acceleration, and other such things) system of mechanics placed on top of it. This is important, because we need to distinguish that this statement, $E = mc^2$, more properly belongs to the "mechanics" part, and not the more fundamental "space-time-information" part.

Moreover, even more advanced theories of physics - in particular, relativstic quantum field theory - ditch many parts of the Newtonian mechanical framework apparatus, yet are still talked about as "combining special relativity with quantum mechanics", which further implies that these are not the core essence o the theory.

So how do we deal with this, then? Well, the archetypal "system" is a swarm of particles interacting by forces, just as in fully-Newtonian mechanics. Indeed, this should make some sense because "real" matter is kind of like this, though a precise microscopic description also requires us to take account of quantum mechanics (its gist is limitation of the information content) - so we can take as our imagined, intuitive scenario being a block of material which we will then be made to undergo heating. We will imagine, of course, an idealized material that can heat to arbitrary temperatures without vaporizing or other such things just to keep the amount of thought required down, though one should ultimately be able to show rigorously that the same result holds in all situations.

Now, you should know from studies of relativistic mechanics that one of the basic results that must obtain is that an elementary particle with nonzero mass is confined to move at speeds below the speed of light, once you fix a suitable reference frame with respect to which to talk about speeds.

So, consider the acceleration, to the observer in the ground frame, of such a particle undergoing steady force. At first, the acceleration will be steady - but then as it approaches the speed of light, it seems to tail off: for some reason, the force involved is becoming less and less effective at accelerating the mass, even though nothing about it has changed. This is because we're seeing the acceleration process, in effect, distorted by the geometry of space and time. For someone moving by it at a speed close to the speeds in this regime, they would see, at least for a time, a more normal acceleration profile.

Moreover, the process applies in reverse: once a particle is near the speed of light, it also follows that it is very hard - but quite crucially, not in a symmetric fashion - to deflect it left or right, or up or down, as well - harder than we'd expect from fully-Newtonian mechanics to make it curve its trajectory, even if by doing so, such curving would not cause its speed to exceed the speed of light. ("Not in a symmetric fashion" means that deflecting it left or right, or other kinds of deflections, has a different difficulty than speeding or slowing it.)

So, now, let us return to our magic block of material. Think about such a magic block of material that can be heated up to any temperature. As it does so, its particles jiggle around faster. We are adding energy to the system. Initially, their jiggling will be well below the speed of light, so we should not expect any noticeable difference from the Newtonian situation. But as the speed of light is approached, the particles' speeds with converge onto it.

Suppose now that, you were to try and grab (assuming you, also, are protected by a magic spell of invincibility) hold of the object and to shake it around. What would you notice? Well, "shaking it around" implies that every particle in it must be undergoing more-or-less synchronized deflections from their normal trajectories. Given that they are virtually all moving near the speed of light, and it's much harder to deflect such particles, it then becomes likewise harder to deflect the block as a whole, even though as a whole, the block is not moving initially! In effect, the "stickiness" I just mentioned similarly makes the particles "sticky" to the points in space at which they oscillate about in their thermal vibrations, and so the object as a whole gets similarly "stuck" - particle by particle - more firmly in one spot in space.

Since mass, perhaps by definition, is the physical parameter which characterizes the response curve of an object when subjected to a force, and it's now responding differently to the force from our hand than it would were it cold, we find that it seems like the mass of the whole object has changed. And, indeed, if you try to calculate this via rigorous mathematical derivation, you will find that its "effective mass" rises exactly in proportion to the added energy:

$$\Delta m_\mathrm{sys} = \frac{1}{c^2} (\Delta E)$$

or, in a more familiar but less directly-connected rearrangement,

$$\Delta E = (\Delta m_\mathrm{sys}) c^2$$

where $m_\mathrm{sys}$ is the system mass. :) And even more, that this doesn't depend on the distribution of speeds, either - so there's nothing particular about assuming a thermal (Maxwell-Boltzmann, or even better, Maxwell-Jüttner) distribution other than as a guide for setting intuition.

And of course, the factor $\frac{1}{c^2}$ explains why we don't notice this in real life, everyday objects, being equal to about $1.11 \times 10^{-14} \mathrm{\frac{kg}{kJ}}$. Hence if, say, I heat up a pot with 1 kg of water on the stove to boil, maybe a rise of 80 degrees Celsius (assuming room temp as 20 C and at standard pressure so boils at 100 C), then it should take roughly 320 kJ (since the specific heat capacity of water is roughly 4 $\mathrm{\frac{kJ}{kg \cdot K}}$), and gain a mass of $3 \times 10^{-12}\ \mathrm{kg}$ - utterly negligible and unmeasurable.

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    $\begingroup$ This nicely explains how kinetic energy gives rise to inertia. Is there a similarly intuitive explanation for potential energy? $\endgroup$
    – Ruslan
    Jan 21, 2020 at 7:17

How can energy have inertia?

The title of Einstein's paper where he introduced what we call nowadays the equivalence mass-energy was "Does inertia of a body depend upon its energy content?" ( Annalen der Physik, 18(13), 639-41 (1905) ). The main conclusion of the paper was (adapting the notation)

The mass of a body is a measure of its energy-content; if the energy changes by $\Delta E$, the mass changes in the same sense by $\Delta E/c^2$.

However, as far as I understand, the present question takes for granted the equivalence, but it is asking for a better physical insight.

I feel that for that purpose, analogies are dangerous. Moreover, a possible additional source of confusion is the survival of the old concept of relativistic mass which is not directly related to the variation of mass without change of velocity which is the main content of Einstein's result. I think that by following the key point of Einstein's reasoning and making clear what is the meaning of inertia in the present context could be the best strategy. I have also to say that, since the whole derivation heavily hinges on Special Relativity (SR) results, it is not obvious how a really intuitive explanation can be found, since our intuition is built on a non-relativistic experience.

Let me start with a couple of almost trivial observation.

  1. Any meaningful statement on energy should be intended as a statement on variations of energy. This is due to the fact that energy is defined within an arbitrary constant, and physical effects only depend on variations of energy. Therefore, a relation between inertia (whatever it is its meaning) and energy, should be based on variation of inertia and variation of energy.
  2. whatever is the meaning of inertia, classical mechanics does not allow to derive a relation between inertia and energy. So, discussion on such an issue requires and it is meaningful only within the SR conceptual frame.

What is the meaning of inertia in the present context? Reading Einstein's paper, one can see that he used the term inertia only twice, in the title and in the conclusions. In between, he worked with the mass of a system, and conclusions were based on results for the mass. Even if in classical mechanics inertia is not always equivalent to mass, I think that in the present discussion one should consider the two concepts equivalent. However, notice that the mass we are speaking about is what nowadays is also called invariant mass and previously was indicatd as the rest mass, i.e. the mass in the rest frame of the system.

Then, how can we understand why changes of energy should imply changes of mass?

The way Einstein arrived to his famous result is a simple but quite subtle analysis. Nevertheless, I think that there is no best way to understand origin and meaning of the inertia-energy relation. Basically, Einstein's few lines derivation uses the analysis of an event of simultaneous emission of radiation carrying the same amount of energy $\Delta E/2$ in opposite directions. Then, the energy of the emitting body before and after emission are related by $$ E_{\mathrm{before~emission}}=E_{\mathrm{after~emission}} + \Delta E.~~~~~~~~~~[1] $$ The same event, described in another inertial frame moving relatively to the first with velocity $v$, using relativistic formulae, is $$ E^{\prime}_{\mathrm{before~emission}}=E^{\prime}_{\mathrm{after~emission}} + \Delta E^{\prime}.~~~~~~~~~~[2] $$

SR allows to relate radiated energy in the two refernce frames: $$ \Delta E^{\prime} = \frac{\Delta E}{\sqrt{1-\frac{v^2}{c^2}}}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[3] $$ Taking into account that $E^{\prime}-E$ is the energy difference of the same system observed in two reference frames, one of them being the rest frame, it is the kinetic energy of that system within a possible additive constant.

Subtracting eq.1 from eq.2, and using relation [3], we can get the difference of kinetic energy $$ E^{\prime}_{\mathrm{before~emission}} - E_{\mathrm{before~emission}} - (E^{\prime}_{\mathrm{after~emission}} - E_{\mathrm{after~emission}} ) = K_{\mathrm{before~emission}}-K_{\mathrm{after~emission}} = \Delta E \left( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1 \right) $$ This allows to find that the kinetic energy of the emitting body must be different, before and after emission, in the primed reference frame where the emitting body is moving. Since the relative velocity of the two inertial frames is arbitrary, one can study the limit of vanishing relative velocity. From such analysis, one obtains that the mass of the body should undergo a variation $\Delta m = \Delta E/c^2$.

Notice that the limit of vanishing relative velocity of the two frames is important to deduce the main novelty of this formula:

the mass variation consequence of an energy variation is present even in the reference frame where the emitting body is at rest.

Such an observation should be clear to everybody who has used the mass-energy relation to deduce the binding energy of nuclear forces from the so-called mass-defect. Still it is quite common to see statements which make confusion between Einstein's result and the more trivial change of energy when changing reference frame.

Notice that since the relation $\Delta m = \Delta E/c^2$ holds in the frame where $\sum_i {\bf p_i}=0$, it is valid also for a fixed box containing photons. Even if a single photon is massless, a gas of photons under such condition does have a non-zero mass and that mass increases with the energy. That is a nice example of the so-called non-additivity of masses in SR (see for example Okun, Lev B. 1989. “The Concept of Mass (Mass, Energy, Relativity).” Soviet Physics Uspekhi 32: 629-638 ).

Note added two days after.

What about your conclusive question:

I do not doubt it's right, but does it mean anything associated with physics, beyond being true?

After understanding the meaning of the formula, it should be clear that it is actually saying something on the physics. A sudden increase or decrease of energy is reflected in a proportional change of mass


Short answer: Due to the properties of Lorentz covariance.

Explanation: Acceleration is not invariant under Lorentz transformations. Hence the acceleration of an object subjected to a given force depends on the frame of reference. Since acceleration is a measure of the object’s inertia, this implies that the object’s 'inertial mass' depends on the frame of reference.

Notice that the kinetic energy of an object also depends on the frame of reference. If you consider two frames of references with different accelerations, the difference in kinetic energy of the same particle turns out to be exactly $c^2$ times the variation in 'inertial mass', where $c$ is the speed of light. This exact proportionality between the extra inertia and the extra energy of a moving particle naturally suggests that the energy itself has contributed the inertia, and this in turn suggests that all of the particle’s inertia corresponds to some form of energy.


Consider such situation. You stand in the street and some bad guy rushes with his bicycle (along street) near you which you want to push with hands so that he would acquire momentum along direction perpendicular to street. However your time window to push that guy is $$ \Delta t = \frac{L}{v_x}$$, where $L$ is bicycle length and $v_x$ - bicycle speed.

According to Newton second law, momentum acquired by that guy with your help is : $$ p_y = F_y \Delta t = F_y \frac{L}{v_x} $$, where $p_y$ is bicycle momentum along direction perpendicular to street, $F_y$ - your pushing force perpendicularly to street.

Now consider that next time near you this bad guy doubles his speed (quadruples his kinetic energy), so you have two times smaller time window for same moment induction to driver. And this means when he doubles speed, you need to double your pushing force $F_y$ for setting same moment to him. And if you raised your pushing force for same output, then this means that bicycle inertia has increased only because bicycle now has more kinetic energy !


Now imagine sub-luminal alien spaceship which flies near Earth at some time moment and our army forces wants to shoot a rocket projectile into that spaceship when it will be at point closest to Earth. In this case you will need to account for a spaceship length contraction, thus including Lorentz factor into the formula :

$$ p_y = F_y \frac{L_{0}{\sqrt {1-v_x^{2}/c^{2}}}}{v_x} $$

If you take the limits, you will see that when $v_x \to c $, then $p_y \to 0$. Meaning that if spaceship speed is almost the speed of light, then you will have almost zero size time window for projectile impact onto ship, thus inducing zero momentum to space ship with any value of projectile force. So this means that you simply can't affect ship's trajectory, because it has infinite inertia.


Well, if energy were to have inertia then it implies that it's going to be affected by gravity. We all know that a matter having a mass gets affected by gravity (I mean it accelerates in presence of gravity) so by induction and converse we can say that anything that gets affected by gravity has a mass.

Light gets bent in presence of high gravitational field, one way to explain this bent is : Light is an electromagnetic wave and thus a form of energy, since it gets bent by gravity (i.e. gets affected) therefore we can conclude that energy (which is light in this case) behaves like a matter with a mass. So, energy have inertia.

I have a closed system, and add energy. Now it has more mass according to $𝐸=𝑚𝑐^2$, and the inertia associated with that mass increased.

That may happen (hypothetically), the energy you give may produce electron and positron hence increase the mass of the closed system.

Hope it helps!

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    $\begingroup$ Look I'm no relativity (SR) expert but as I know the use of $E = mc^2$ is wrong in the context of light aslight does not have a rest mass. The true equation which works for all cases is $E^2 = m^2c^4+ p^2c^2$. For an object with zero rest the equation turns out to be $E = pc$ which is the case for light light. $\endgroup$
    – user249968
    Jan 18, 2020 at 9:15
  • $\begingroup$ @JohanLiebert I never said light have mass. $\endgroup$
    – user240696
    Jan 18, 2020 at 10:29
  • $\begingroup$ Neither do i. I am saying that use of $E = mc^2$ in the context of light is wrong. $\endgroup$
    – user249968
    Jan 18, 2020 at 10:30
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    $\begingroup$ Make sure you are careful with how you interpret your equations. For example, does the mathematical statement of Newton's second law $F=ma$ mean that the more acceleration a mass has then the more force it can produce? $\endgroup$ Jan 18, 2020 at 13:09
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    $\begingroup$ The OP is about special relativity. Bringing here the equivalence principle (which, in the framework of the SR, is irrelevant) is quite unneeded, and, in the way it's done here, I don't think this answers the question. $\endgroup$
    – Ruslan
    Jan 20, 2020 at 18:40

If a potato has mass and inertia, that's because it has energy.

If a potato decides to use its energy to accelerate itself to high speed, it loses mass, as kinetic energy has no mass.

If a potato is carefully lowered very close to a black hole and then dropped, we notice that neither the mass or the inertia of the black hole increases.

Potato hanging next to a black hole has no energy. Potato without energy has no inertia and no mass.

Energy has mass and inertia, except kinetic energy, which has no mass.

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    $\begingroup$ Kinetic energy does have mass: a hot gas has more mass than a cold one. $\endgroup$
    – Ruslan
    Jan 20, 2020 at 18:46
  • $\begingroup$ @Ruslan No no, Object's mass is the rest masses of its particles added together using relativistic addition of rest masses with different speeds. It would be much simpler if kinetic energy had mass, but relativistic mass has been banished. $\endgroup$
    – stuffu
    Jan 21, 2020 at 11:57
  • $\begingroup$ Regardless of how you describe it, the effect is there: the acceleration due to the same force is different for an object that has different speeds. $\endgroup$
    – Ruslan
    Jan 21, 2020 at 12:50

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