# Why is the Newtonian expression for kinetic energy called the “first order” approximation of the relativistic expression?

In many texts, the non-relativistic (Newtonian) kinetic energy formula $$\text{KE}_\text{Newton} =\frac{1}{2}mv^2$$ is referred to as a first order approximation of the relativistic kinetic energy $$\text{KE}_\text{relativistic} = \gamma mc^2 - mc^2$$ The same is also said of the classical momentum formula in relation to its relativistic counterpart.

However, comparing the Newtonian approximations to their respective relativistic formulas, the Newtonian KE formula appears to be a second order approximation while the momentum formula appears to be of first order.

Let's begin with momentum. The relativistic formula for momentum is $$p=\gamma mv=\frac{mv}{\sqrt{1-\left(\frac{v}{c}\right)^2}} \, .$$ For non-relativistic velocities ($v \ll c$), we use the Taylor series $$\frac{x}{\sqrt{1-x^2}} \approx x\left(1 + \frac{x^2}{2}\right) \, ,$$ giving $$p/c \approx mv/c \left[ 1 + \frac{1}{2}\left( \frac{v}{c} \right)^2 \right] \approx m (v/c)$$ which is first order in $v/c$. In other words, $p\approx mv$ which is the usual Newtonian expression.

On the other hand, the relativistic kinetic energy is \begin{align} \text{KE}_\text{relativisitic} = \gamma mc^2 - mc^2 = \frac{mc^2}{\sqrt{1-\left( \frac{v}{c}\right)^2}} - mc^2 \end{align} which for $v \ll c$ is $$\text{KE}_\text{relativistic} \approx mc^2 \left[ 1 + \frac{1}{2}\left( \frac{v}{c} \right)^2\right] - mc^2 = mc^2 \frac{1}{2} \left( \frac{v}{c} \right)^2 = \frac{1}{2} m v^2$$ which is obviously second order in $v$.

If we compare plots of the Newtonian forms for kinetic energy and linear momentum against their respective relativistic formulas, there appears to be a closer agreement for the approximation of kinetic energy than can be seen for linear momentum.

And hence my question: why is the Newtonian formula for kinetic energy referred to as a first order approximation when it appears to be of a second order?

• A few tips: 1) "Classical" means "not quantum mechanics" and has nothing to do with the distinction between relativistic and Newtonian physics. When you want to say that something is in the limit $v/c \ll 1$, say "Newtonian" or "non-relativistic". 2) You really don't have to go through all that extra rigamarole with showing the Taylor series. I simplified the post a lot to make it easier to get through the math. If you don't like it, you can always roll back the edit. 3) I don't understand what the plots are showing because you didn't define "Relativistic linear" and other such terms. – DanielSank May 11 '17 at 15:25
• @DanielSank: when the context is clear and QM is nowhere in sight, it's common to say "classical" when you mean "non-relativistic". – Javier May 11 '17 at 15:27
• @Javier I disagree. Relativity is unquestionably a "classical" theory. If modern usage is changing such that "classical" means "not relativistic and not quantum mechanical", then we're all going to be very, very confused when we try to make sense of historical literature. We can talk about this in chat if you want. – DanielSank May 11 '17 at 15:28
• @DanielSank like it or not, he is not incorrect. The use of "classical" in this sense is quite common. – Rococo May 11 '17 at 20:50
• @DanielSank In all fields of physics Iv'e known, "classical" can be either non-relativistic, non-quantum, or both; so if the context is not clear, you specify which it is. With all the other physicists Iv'e talked to (from many countries and many fields), we always said things like "classical in the relativistic sense" or "classical in the quantum sense". – user1803551 May 12 '17 at 12:57

The way I see it, there are four possible answers. You can pick the one you like the most, because in the end it doesn't matter very​ much.

1. You're right, it's a second order approximation and those who say it's first order are making a terminology mistake.

2. When we say first order, we really mean first non-null order, since the linear term vanishes.

3. It's actually first order in $v^2$.

4. It doesn't really matter. We all know what the non relativistic approximation is, its properties are not going to change if we call it by a different name.

Personally I support answer 4, and I suggest you get used to it because physics is not known for its rigor and formality.

• I agree very much with 1 through 3, particularly 2 and 3, but I strongly disagree with 4. The sloppy language used in physics is the source of a huge amount of unnecessary confusion. – DanielSank May 11 '17 at 15:07
• +1 I would also like to mention that a second order expansion in a first integral of motion (e.g., energy) corresponds to a first order approximation to the equations of motion. – AccidentalFourierTransform May 11 '17 at 15:15
• Thank you for your response. I find answers 1 and 3 to be particularly satisfactory. Answer 2 flies in the face of most literature, so I don't think it is appropriate. Answer 4 is a perspective I do not share. However the question may be approached, I think there is something to be said about the quantitatively better approximating power of the non-relativistic formula for kinetic energy than the non-relativistic formula for linear momentum, which has yet to be addressed in the above response. – Master Drifter May 11 '17 at 16:52
• I always read it as "the first that is useful" – PlasmaHH May 12 '17 at 13:43

The issue of the terminology used for expansions has been pretty well addressed in Javier's response. I wanted to address another part of your question that you expressed concern about in a comment, namely your assertion that

we can see a much closer agreement for the approximation of kinetic energy than can be seen for linear momentum.

But in fact, the level of agreement between classical vs. relativistic values is actually worse for kinetic energy than it is for momentum at a given speed $v$. The only reason that the gap looks larger for the momentum in your graph is that the quantity itself is larger.

To see this, define $r$ to be the ratio of the relativistic value of some quantity to its classical value. For the momentum, this would be $$r_p = \frac{mv \gamma}{mv} = \frac{1}{\sqrt{1 - v^2}}$$ (using units where $c = 1$), while for the kinetic energy it would be $$r_\text{KE} = \frac{ m \gamma - m}{\frac{1}{2} mv^2} = 2 \frac{1/\sqrt{1 - v^2} - 1}{v^2}.$$ Plotting both of these, we get the following graph:

We can also plug in numbers to the above expressions. For example, at $v = \frac{1}{2}$, the relativistic momentum is $\frac{2}{\sqrt{3}} \approx 1.15$ times larger than the classical momentum, while the relativistic KE is $8(\frac{2}{\sqrt{3}} - 1) \approx 1.24$ times larger than the classical KE. The "ratio of the ratios", which we can think of as measuring the "relative badness" of the classical expressions for kinetic energy and momentum, is $$\frac{r_\text{KE}}{r_p} = 2 \frac{1/\sqrt{1 - v^2} - 1}{v^2/\sqrt{1 - v^2}} = 2 \frac{ 1 - \sqrt{ 1 - v^2}}{v^2}.$$ This ratio approaches 1 in the limit $v \to 0$, which means that the classical expressions are "equally good" in this limit; and this ratio approaches 2 in the limit $v \to 1$, which implies that the classical expression for the kinetic energy is "twice as bad" in this limit.

• Thank you for your response and catching my error. It's interesting nevertheless that some Newtonian formulas are more accurate than others. – Master Drifter May 12 '17 at 6:41
• Would you mind elaborating on your final result r_KE -> 2r_p as v -> 1? – Master Drifter May 15 '17 at 9:01
• @MasterDrifter: See my edited answer. – Michael Seifert May 15 '17 at 13:34