# An apparent contradiction to $m = m_0/\sqrt{1-v^2/c^2}$ [duplicate]

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Using theoretical framework of the special relativity, we can show that the quantity that we classically regard as energy does have a property of inertia. And particularly, if the total energy of a box is $L$ then its inertia is $L/c^2$. Now suppose I have an infinitely large parallel plate capacitor which is at rest in an inertial frame $O$. Uniformly charged plates of the capacitors have such a charge density so that the uniform electric field between the plates is $E$.

Now in a frame of reference $O'$ which is moving uniformly along the area vector of the plate plane with respect to $O$, the electric field will be $E$ and magnetic field would still remain zero.

In $O$ frame, the total energy of capacitor is $L_1 = l_0 \int\frac{1}{2}\epsilon_0E^2 dA$ and thus its mass is $m_1 = L_1/c^2$.Where $l_0$ is the perpendicular separation between the plates and $dA$ is the infinitesimal small area element of one of the plates in $O$ frame.

In $O'$ frame, the total energy of the capacitor is $L_2 = l_0\sqrt{1-v^2/c^2} \int\frac{1}{2}\epsilon_0E^2 dA$ . The infinitesimal $dA$ and limits of integration remain exactly the same because no length contraction happen in the direction perpendicular to the relative motion. Here also the mass is thus $m_2 = L_2/c^2$.

So, $m_2$ = $m_1$ $\sqrt{1-v^2/c^2}$. But in general, in relativity we prove that $m = m_0/\sqrt{1-v^2/c^2}$. So why does this apparent contradiction is happening? Does it have to do anything with the fact that actually both the $m_1$ and $m_2$ are infinities? Because I doubt that it might be the case that we cannot directly write $m_2$ = $m_1$ $\sqrt{1-v^2/c^2}$ because $m_2$ and $m_1$ are infinities. But on the other hand that relation seems to hold pretty well from their corresponding expressions in the integral form.

## marked as duplicate by Community♦Nov 1 '15 at 20:16

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• Treating $m[1 - \beta^2]^{-1/2}$ as inertia has problems from the very beginning. It is the correct inertia for forces acting transverse to the direction of relative motion, but is not correct for longitudinal forces. – dmckee Nov 1 '15 at 18:53
• @dmckee But while proving the $m = m_0/\sqrt{1-v^2/c^2}$ relation, we analyse a collision which has longitudinal forces. Can you suggest the reason why the expression shouldn't hold for some cases? – Dvij Mankad Nov 1 '15 at 19:02
• @dmckee Can you provide some link where I can read the relevant literature on the problems with $m_0/\sqrt{1-v^2/c^2}$ expression? – Dvij Mankad Nov 1 '15 at 19:06
• mass is velocity-independent. $m=\gamma m_0$ is a historical mistake, and shouldnt be taken seriously in any context. – AccidentalFourierTransform Nov 1 '15 at 19:26
• @qftishard I know in recent literature we consider rest mass to be the mass. And rest mass is obviously velocity independent. But are you suggesting that the inertia of a body doesn't change with the speed? I think then it creates a lot of problems with momentum conservation in the relativistic context. Can you provide some web link to read about what you are suggesting? – Dvij Mankad Nov 1 '15 at 19:35

## 1 Answer

Using theoretical framework of the special relativity, we can show that the quantity that we classically regard as energy does have a property of inertia.

It's isn't as straightforward as having a scalar $m$ like in $\vec F=m\vec a.$ And if you have mechanical momentum like $\vec p=\vec vE/c^2$ then mass doesn't appear at all and $d\vec p/dt$ has to definitely adjust the energy and maybe the velocity or the speed too. But since the energy depends on the full 3d motion you don't get the simple separation of changes in velocity components for each of the three components like you do in Newtonian mechanics. But you can get a nice change in momentum.

No word games can change that momentum can steadily increase while energy and/or velocity can be a more complicated change.

And particularly, if the total energy of a box is $L$ then its inertia is $L/c^2$.

That's a pretty big claim. If it weren't so vague I'd say its provably wrong.

In $O$ frame, the total energy of capacitor is $L_1 = l_0 \int\frac{1}{2}\epsilon_0E^2 dA$ and thus its mass is $m_1 = L_1/c^2$.

That's not true. A capacitor has massive plates and charges and such and some stress holding all the charges together and so there is more energy than the fields, and there is stress so there is more than just energy.

In $O'$ frame, the total energy of the capacitor is $L_2 = l_0\sqrt{1-v^2/c^2} \int\frac{1}{2}\epsilon_0E^2 dA$ .

Now there is kinetic energy of the massive charges too, and the stress is involved as well.

Here also the mass is thus $m_2 = L_2/c^2$.

No. Mass isn't just energy divided by $c^2$ you need the total energy (not just some of it) and you need the momentum (and in this frame there is momentum for the charges).

But in general, in relativity we prove that $m = m_0/\sqrt{1-v^2/c^2}$.

That's not true either. Sure if the total energy momentum vector exists and is timelike then there is a center of momentum frame and in that frame there is an energy and if you divide that by $c^2$ you can call that the rest mass of the system. And then you can compare the total energy in other frames to that energy and it will be larger in the other frames.

But the spatial integral of the field energy is not the total energy and it isn't the time component of the total energy momentum of a system. And a good way you can tell physically is ...

There is no rest frame or center of momentum frame for a bare electric field with no charges.

So clearly the field energy isn't a tine component of a timelike vector. Which you can tell by the fact that there isn't a fixed frame where it is smaller than any other frame.

So why does this apparent contradiction is happening?

There is no contradiction. The energy of a bare electric field simply doesn't transform the way you claimed it does. No one said it did.

Were you expecting field energy to have a mass all by itself?

You did completely ignore whatever force is keeping those charged plates from slamming into each other (you ignored it in both frames), and that force has some energy too.

Does it have to do anything with the fact that actually both the $m_1$ and $m_2$ are infinities?

No. You could talk about finite sized capacitors. Or even do a calculation in a finite universe, such as a universe that is infinite in z the direction but if you go in the x or y directions it repeats Pac Man style.

The key is to avoid the bad physics. And just like two frames can disagree about which objects have kinetic energy, you must include absolutely every bit of energy, the stress of the plates and wires, the energy of the fields, the kinetic energy and rest energy of the parts of the plates, everything.

• Special relativity suggests that mass indicates the energy content of the body. The 'useful' mass is rest mass which is by definition invariant. But still there exists a quantity called relativistic mass which is not a lorentz scalar. Also, if relativistic mass of a body is $m$ then it essentially means that the total energy of the body is $mc^2$. Total energy of a body transforms as $E = E_0/\sqrt{1-v^2/c^2}$ and so does the relativistic mass. – Dvij Mankad Nov 2 '15 at 12:47
• @Dvij Be intellectually honest. Don't pretend that using two words (energy and relativistic mass) for the same thing is an accomplishment or any kind of insight. Call it energy and move on with your life. Special relativity didn't "indicate" that you should make up two words for the same thing and call it an accomplishment. You say there is a thing called relativistic mass and I say there is a thing called energy. But the point is you can't just take equations and replace mass with energy and expect it to now be correct, that's wrong. Momentum conservation works if you use correct equations. – Timaeus Nov 2 '15 at 12:59
• @Dvij Just a bit of advice: If you consider a system of two objects, and increase that system's rest mass by accelerating the two objects to opposite directions, then you are effectively using relativistic mass but nobody can complain! ;) – stuffu Nov 2 '15 at 13:23
• @user7027 In the specific question the OP asked, we can and do complain. Ignoring some of the stress or momentum or energy does give wrong answers. Just as two objects with relative velocity can have different parts have kinetic energy depending on frame so can different energies of different parts be important to the total energy of a system. The OP is definitely making mistakes and playing semantic games won't fix it, the OP needs to actually learn relativity and track all the energy, stress, and momentum instead of picking and choosing. – Timaeus Nov 2 '15 at 13:35
• I am not telling total energy of the entire "capacitor+assembly holding capacitor plates" to be the integration of $\frac{1}{2}\epsilon_0E^2$. But I am particularly choosing the system consisting of only the fields between the plates. – Dvij Mankad Nov 2 '15 at 14:56