2
$\begingroup$

I was reading Energy-momentum, and I came across this simplified equation: $$E^2 = (mc^2)^2 + (pc)^2$$

where $m$ is the mass and $p$ is momentum of the object. That said, the equation is pretty fundamental and nothing is wrong when looked upon, I similarly also believed this but I came across a "special" cases where this does not apply:

  • If the body's speed $v$ is much less than $c$, then the equation reduces to $E = (mv^2/2) + mc^2$.

I find this really crazy, because first Einstein, always wanted to create a theory\equation that applied to every aspect of physics and has no "fudge" factors, that said irony is present from Einstein. Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given.

Most importantly, why does this not work, if velocity is "much" slower than light? What do they mean by "much slower", what is the boundary for "much slower"?

Regards,

$\endgroup$
6
  • $\begingroup$ So now that you have corrected your mistake, we have to ask what makes you believe that the simplification of a formula for a special case is not useful to those who know how to apply it correctly? $\endgroup$
    – CuriousOne
    Sep 20, 2014 at 13:20
  • $\begingroup$ Well, All I am saying is why does this equation not apply universally and rather has very "blurry" boundary of when this applies, not to mention why is this? Surely a equation worthy in physics should be able to work in any situation rather than a "blurry" boundary that is not defined. Next, why does this not apply to things that is going super-subliminal velocities? $\endgroup$
    – user43495
    Sep 20, 2014 at 13:25
  • $\begingroup$ What's blurry about it? You can easily put v=1m/s into both expressions for E and calculate the difference. What you will get is a very small number that is negligible for most applications. If you put v=10km/s in, it stops being negligible for e.g. the people who design and operate the GPS system. What's "good enough" depends solely on the application and physicists learn how to determine PRECISELY, if an approximation is good enough, or if they have to use the exact formula. $\endgroup$
    – CuriousOne
    Sep 20, 2014 at 13:34
  • 2
    $\begingroup$ The limit of "much slower" is whatever fits your application. Some people need 12 digits of precision, others can do with a relative error of 10%. Neither case is wrong, it's just not the same for every application. $\endgroup$
    – CuriousOne
    Sep 20, 2014 at 14:03
  • 2
    $\begingroup$ The phrase "the equation reduces to" means it is basically the same equation, except that some very small terms deemed insignificant have been dropped, and some simplification has been made (here taking square roots on both sides). So (apart from being a very slight approximation) the new equation does not contradict, but just restates the old equation. $\endgroup$ Sep 20, 2014 at 15:01

10 Answers 10

14
$\begingroup$

First, the non-relativistic equation $$ E= mc^2 + \frac{mv^2}{2} $$ is equivalent to its second power, $$ E^2 = (mc^2)^2 + m^2 c^2 v^2+ \frac{m^2v^4}{4} $$ If $v/c\ll 1$, then the last term is much smaller than the previous two, and the first two terms on the right hand side are equivalent to the correct relativistic $$ E^2 = (mc^2)^2+ (pc)^2 $$ which completes the proof that the two formulae are the same in the $v/c\ll 1$ limit.

The last, relativistic formula is always right. The first one, if we want to consider "only absolutely correct and exact" formulae, is never correct – except for the case $v=0$. However, the non-relativistic equation may be written in a completely rigorous way (to describe that it is approximate) as $$ E = mc^2 + \frac{mv^2}{2} + O(mv^4/c^2) $$ The symbol $O$ represents "a function that in the relevant limit, here $v/c\to 0$, has a finite limiting ratio with the function in the parentheses after $O$", and this concept may be and is defined 100% rigorously.

The boundary of the values of $v$ where the non-relativistic formula applies is indeed "fuzzy" – one can't quote any exact value of $v$ (except for $v=0$, in the useless sense described above) where the non-relativistic formula ceases to hold. But for $v/c\lt 0.1$ or so, the error is smaller than one percent. For greater speed than $v=c/2$, the non-relativistic formula becomes so bad that it can't be use in any quantitative context.

The error of the non-relativistic energy formula – or, more democratically, the difference between the two formulae – simply gradually increases from $0$ at $v=0$ to something comparable to 100% at $v=c/2$ and a huge error for $v\to c$.

Physics is fundamentally based on continuous numbers which means that pretty much all of its quantities are gradually changing and their differences and errors are gradually changing, too. Also, errors smaller than a certain threshold are experimentally undetectable which allows one to say, in a very specific empirically rooted sense, that the error is basically zero.

Because of the omnipresence of limits and limiting claims about formulae, expressions, and theories in physics, one may say that if you won't comprehend and embrace these important concepts about limits and expressions' being equivalent in limits, you have virtually no chance to understand anything in physics.

$\endgroup$
10
$\begingroup$

Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given.

Yes, it would be nice if we could find an equation that was "universal". In the case of mass m, energy E, and momentum p, there is such a universal equation:

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

Einstein, always wanted to create a theory\equation that applied to every aspect of physics and has no "fudge" factors

Yes, it is a great triumph that this equation always applies to every aspect of physics, with no fudge factors: $$E^2 = (mc^2)^2 + (pc)^2$$

I came across a "special" cases where this does not apply

Really? My understanding is that the above equation does always apply. Have you actually calculated or measured what the alleged difference is?

Most importantly, why does this not work,

I don't understand. The equation $$E^2 = (mc^2)^2 + (pc)^2$$ always works, no matter what the velocity.

If the body's speed $v$ is much less than $c$, then the equation reduces to $E = (mv^2/2) + mc^2$.

What do they mean by "much slower", what is the boundary for "much slower"?

What they mean by "reduces to" and "much slower" is that people are often willing to tolerate some small amount of error, and given any particular amount of tolerable error, there is some range of speeds (including 0) where the equation

$E = (mv^2/2) + mc^2$

, while not mathematically exactly correct, is within that tolerable error of the correct equation.

So, if we're willing to tolerate an error of 0.01%, then this technically incorrect formula is adequate for speeds v slower than 0.1 c =~= 108,000,000 km/h.

Sometimes we do much higher-precision measurements -- in that case, there is some smaller range of slower speeds that gives adequate precision with the Newtonian formula.

Historically this "reduces to" was important, because over a century of research showed that Newtonian equations always matched up with the actual observed results, to within experimental error. People were reluctant to switch to Einstein's equations -- that, as you pointed out, are not the same as the Newton's equations. Why would they switch from something that works to a different equation? The point is that all those experiments were at velocities so slow that the difference between the "correct" equation and the "incorrect" equation was too small to measure. And so to decide which equation was right, we needed to do new experiments where the difference between these equations was big enough to measure.

$\endgroup$
7
$\begingroup$

If you put $p = \gamma mv $ where $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$ and $\beta = \frac{v}{c}$ in Einstein's equation, you get

$E^2 = (pc)^2 + (mc^2)^2 = (\gamma mvc)^2+ (mc^2)^2 $

$= (\frac{c^2}{c^2 - v^2})v^2(mc)^2 + (mc^2)^2= (\frac{v^2}{c^2 - v^2})(mc^2)^2 + (mc^2)^2$

$ = (\frac{c^2}{c^2 - v^2})(mc^2)^2 = (\gamma mc^2)^2$

$ \therefore E = \gamma m c^2\ .$

Now apply a binomial expansion, assuming $ v \ll c $ i.e. $\beta \ll 1$:

$ E = (1-\beta^2)^{-\frac{1}{2}}mc^2$

$\implies$$ E = (1 -\frac{1}{2}(-\beta^2) + O(\beta^4))mc^2$

Now you get the answer of everything. If the accuracy you want in your calculations is less than what is provided by the $O(\beta^4)$ terms, i.e. if v is so "small" that in the energy calculation, the fourth power of $\frac{v}{c}$ won't create much difference to you, then you can use the formula by taking only the first two terms and ignoring the rest, in which case you get the result you have stated in your question:

$E = (1 -\frac{1}{2}(-\beta^2))mc^2 = mc^2 + \frac{1}{2}m\frac{v^2}{c^2}c^2 = mc^2 + \frac{1}{2}mv^2$

$\endgroup$
6
$\begingroup$

If the body's speed $v$ is much less than $c$, then the equation reduces to

That is an imprecise wording, and the cause of your confusion, I believe.

The correct description is that the second equation is an approximation of the first, and it will be closer to accurate the lower $v$ is.

The approximation is important for understand the connection between Newtonian and relativistic physics, but I don't think there is much practical purpose to the approximation beyond that.

Either you do your calculations using Newtonian physics where you disregard the mass-energy relation, or you do them using relativistic rules where you use the full equations.

$\endgroup$
4
$\begingroup$

First, your findings are correct and are found by Einstein. However the fact that a new term appears is only showing that our previous knowledge was incomplete. The new term expresses the Energy related to the rest mass of the body which was not considered before. And it makes sense because before, in equating the Energy of a body only were included those terms related to forms of Energy into which we knew some interchange could happen. Since mass was not known to interchange with Energy, the law for mass conservation was known but separated.

In conclusion, the new term is not incompatible with the reproducibility of previous physics, it is part of the new knowledge brought.

As for what is understood for "much slower", it depends on the sensitivity of your experiment or phenomena. But a good rule of thumb is: when the kinetic energy is of the order of the rest mass. Remember that the total energy of a particle relates to the rest mass $m_0$ as $$E = \frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}$$ Expanding for $\frac{v}{c} << 1$ you get $$E = m_0c^2 + m_0\frac{v^2}{2} + \frac{3}{8}m_0 \frac{v^4}{c^2} + ...$$ so you should account for relativity at least qhe the 3rd term is important for you.

$\endgroup$
3
$\begingroup$

Special equations of the kind you mention are useful as they elucidate limiting behaviors.

Your example is no different from a statement like: the hypotenuse $c$ of a right triangle with legs $a$ and $b$ with $b \ll a$ is given by $c = a + \frac12 b^2/a$. The Pythagorean relation $c^2 = a^2 + b^2$ is valid for any right triangle, but the special case $b \ll a$ is useful to elucidate how much the hypotenuse extends beyond the larger leg in the limit of ultra-sharp right triangles.

Similarly, the equation $E = mc^2 + \frac12 mv^2$ describes how much larger is the total energy $E$ of a moving object compared to its rest energy $mc^2$, in the limit of the ratio of the object speed over the speed of light dropping to zero. This energy difference is known as the kinetic energy, and the limiting equation makes it clear that in the limit of slow movements, kinetic energies in Einstein's theory are consistent with those in Newton's theory.

$\endgroup$
3
$\begingroup$

I didn't see anyone mention the practical reason to use an approximation for energy. It is that in most problems you will be computing differences in energy. In that case, for small velocities, you can not only go the approximation ${m v^2\over 2}+m c^2$, but if you are also not converting mass to energy or vice-versa, you can drop the $m c^2$ as well since it will subtract out. Now you have Newton's ${m v^2\over 2}$. This is much more appropriate for most Earthly problems, and you can calculate it more readily without the $m c^2$. If you wanted to compute the energy change from accelerating an object by 20 m/s, and you tried to use $E^2=(m c^2)^2+(p c)^2$, then the change would be out in the 14th decimal place of $E$. That is where approximations and dropping constant terms becomes not only useful but essential.

$\endgroup$
2
$\begingroup$

The above answers are all good, but I want to add something else. You can derive the energy-momentum relation from at least principle, the action is $A=-m\int \sqrt{1-v^2}{\rm d}t$ . Lagrangian is $\cal L$$=-m\sqrt{1-v^2}$, then you can get momentum $p$ from derivative with respect to $v$ (this is the real momentum, not $mv$). Energy $H=E=pv-\cal L$ , just as $E^2=p^2+m^2$ you get.

$\endgroup$
2
$\begingroup$

A direct answer:

Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given.

The equation $E^2 = (mc^2)^2 + (pc)^2$ does work for all values.

As other answers have explained, at low speeds (or equivalently, low momenta), $E = \frac{1}{2}mv^2 + mc^2$ gives approximately the same result, and is slightly easier to calculate. That's the reason we use it.

$\endgroup$
8
  • $\begingroup$ But will it work? Because if the momentum is 2 because v = 2 and m = 1 then will the overall mass-energy suddenly be $E^2 = (mc^2)^2 + (pc)^2$? therefore be $E^2 = (2 * c^2)^2 + (2c)^2$? $\endgroup$
    – user43495
    Sep 21, 2014 at 18:30
  • $\begingroup$ (1) Momentum can't be 2; it has units. It could be $2\text{ kg m/s}$ if you use SI units, for example. (2) Suppose you use SI units, $v=2\text{ m/s}$ and $m=1\text{ kg}$. In these units, $c=299792458\text{ m/s}$. Plug those values into both formulas and check that you get basically the same result no matter which one you use. $\endgroup$
    – David Z
    Sep 21, 2014 at 18:53
  • $\begingroup$ So If we accelerate an object by $1$ meter we get $599584 916 KG/s$ more joules in the energy-mass? Sorry I am a middle schooler $\endgroup$
    – user43495
    Sep 21, 2014 at 18:58
  • $\begingroup$ @RohanVijjhalwar I'm not sure what you mean by "accelerate an object by 1 meter." A meter is a distance, not a change in speed or an acceleration. $\endgroup$
    – David Z
    Sep 21, 2014 at 19:06
  • $\begingroup$ I mean 1 m/s sorry, I missed the per second. $\endgroup$
    – user43495
    Sep 21, 2014 at 19:25
0
$\begingroup$

The first two nonzero terms of the Taylor Series expansion of $E$ in

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

are

$$E = (mv^2/2) + mc^2.$$

The remaining nonzero terms will include $v^4$, $v^6$, etc, and are practically irrelevant for everyday speeds. The purpose of this approximation is to show how relativistic kinetics reduces to Newtonian kinetics at low speeds.

It's important to note that there is no boundary, but rather the difference becomes smaller and smaller.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.