# I Relativistic Kinetic Energy Derivation

1. Mar 28, 2016

### PCB

Hello,
I tried a different route to derive relativistic kinetic energy and I cannot see why it doesn't work. Here is my work:
8.00000000000000E+01 RM, Rest mass of object
7.50000000000000E+05 v, velocity of object
6.00001877636573E+07 Momentum, p,= RM/Sqrt(1-(v^2/c^2))*v
4.50001408227429E+13 Energy= p*v = RM/Sqrt(1-(v^2/c^2))*v^2 equation 1

4.50002112348160E+13 kE= 2*((RM/Sqrt(1-(v^2/c^2))*c^2)-(RM*c^2)) equation 2
My question is, why doesn't equation 1 yield the same quantity as equation 2?

2. Mar 28, 2016

### Staff: Mentor

What units are you using? When you say $v=7.5\times{10}^5$ I'm assuming that is not in nanofurlongs per megafortnight.... But it would be nice to know what units you are using if only so we know what value to use for $c$.

Does your equation 2 have an extraneous factor of two? It looks like $E_k=2(m_0c^2/\sqrt{1-v^2/c^2}-m_0c^2)$.... Posts are much easier to read, and you'll get better answers, if you use Latex. To see how I did that, reply to this post and look at the quoted text.

And finally, did you actually calculate those two values or did you just key them into a calculator? The latter technique can easily introduce and magnify rounding errors.

3. Mar 28, 2016

### Staff: Mentor

That doesn't look like any formula for relativistic energy that I know of, either for total energy or kinetic energy. Where did you get it?

4. Mar 29, 2016

### PCB

Nugatory,
The units I am using are kg, meters, seconds. Specifically, v = 750,000 m/s and 299,790,000 m/s for c.
I will look into writing in Latex. I write in "Excel" presently. Because I do not trust my algebra, I calculate with real numbers so I can check the calculations.

jtbell,
I didn't get equation 1 from anywhere specifically, I am trying to derive relativistic kE by this particular route.

5. Mar 29, 2016

### robphy

Try
$K=\int{v\ dp}$

6. Mar 29, 2016

### PCB

Thanks robphy. I am aware of several different derivations of relativistic kE, but I am trying to find out why my derivation does not work

7. Mar 29, 2016

### Staff: Mentor

My point was, more explicitly, what made you think E = pv might work? robphy's $K = \int {vdp}$ does give $K=\frac{1}{2}mv^2$ in classical (non-relativistic) mechanics, but that equals $K=\frac{1}{2}pv$ not $K=pv$. And then there's the difference between [total] energy (your equation 1) and kinetic energy (your equation 2)... unless that was just a typo on your part.

8. Mar 29, 2016

### robphy

If p is the relativistic momentum,
you get the relativistic kinetic energy.

9. Mar 29, 2016

### PCB

jtbell,
First of all, thank you for your response(s).
I think E=pv might work because I am using relativistic p, meaning m/gamma*v. (Sorry, I haven't figured out LaTex yet).
Concerning the extraneous 2, I am assuming that the kE imparted on an object is 1/2E when the reference frame is independent of the Object and the object or whatever that imparted the energy onto the Object, and thus half the energy is imparted on the object and half is imparted equally and oppositely on the "whatever object". I am assuming the reference frame of the "whatever object" which means 100% of the energy is imparted onto the Object.

10. Mar 29, 2016

### PCB

robphy,
Yes, p is relativistic momentum, as can be seen in the equation 1. Thus my problem--I cannot figure out why my equation 1 doesn't equal equation 2.

11. Mar 29, 2016

### robphy

This is integrates to $(\gamma-1)m$.

12. Mar 29, 2016

### PCB

robphy,
Yes, it integrates into equation 2.

13. Mar 29, 2016

### robphy

$K=\int{v\ dp}$, which is not equal to your Eq.1,
does integrate to twice your Eq. 2.

Added in edit: Note that in the integrand, v is not the "final" velocity.

Added in edit: I didn't see that extra factor of 2. I just went on what you said.

Last edited: Mar 29, 2016
14. Mar 29, 2016

### PCB

Yes I know. Your equation is equivalent to my equation 2.
The p in my equation 1 is relativistic momentum, thus (I think) equation 1 should equal equation 2.

15. Mar 29, 2016

### robphy

$K=\int_0^{p_f} v(p)\ dp=m\left(\frac{1}{\sqrt{1-v_f^2}}-1\right)\quad \neq\quad v_f p_f = v_f \left(mv_f\frac{1}{\sqrt{1-v_f^2}}\right)$

16. Mar 29, 2016

### Staff: Mentor

Even if $p$ is relativistic momentum, $p=\gamma{m_0}$, neither total nor kinetic energy is given by $pv$. However, when $p$ is the classical momentum, $pv$ is twice the classical kinetic energy.

Meanwhile, your equation 2 is the correct expression for twice (because of that extraneous factor of two) the relativistic kinetic energy.

Finally, the speed value you're using, $7.5\times{10}^5$ m/sec, is only about .2% of the speed of light so relativistic effects are going to be very small. Thus equation 1 is wrong for the classical kinetic energy but not by much... All you've discovered is that when the speed is small compared with the speed of light, the correct relativistic value of the kinetic energy (equation 2) doesn't differ much from the classical kinetic energy, whether computed correctly ($E_k=pv/2=mv^2/2$) or using your incorrect (if $p$ is relativistic momentum) equation one. And that's why you have to look out to six decimal places to see a difference between the two.

Last edited: Mar 29, 2016
17. Mar 29, 2016

### vanhees71

The most simple derivation of the expression for relativistic kinetic energy is through Hamilton's principle. The only scalar action you can built from $\dot{\vec{x}}$ alone is given by
$$S[\vec{x}]=-m c^2 \int \mathrm{d} t \sqrt{1-\dot{\vec{x}^2/c^2}}.$$
Energy, by definition is the conserved quantity due to time-translation invariance according to Noether's theorem, and that's the Hamilton function
$$H=\vec{p} \cdot \dot{\vec{x}}-L$$
with the canonical momentum
$$\vec{p}=\frac{\partial L}{\partial \dot{\vec{x}}}=m \frac{\dot{\vec{x}}}{\sqrt{1-\dot{\vec{x}}^2/c^2}},$$
$$E=H=\frac{m c^2}{\sqrt{1-\dot{\vec{x}}^2/c^2}}.$$
The momentum due to Noether's theorem is the canonical momentum and thus four-momentum
$$(p^{\mu})=(E/c,\vec{p})=\frac{m}{\sqrt{1-\dot{\vec{x}}^2/c^2}}(c,\dot{\vec{x}}).$$
$$p_{\mu} p^{\mu}=\frac{m^2}{1-\dot{\vec{x}}^2/c^2}(c^2-\dot{\vec{x}}^2)=m^2 c^2,$$
identifying $m$ with the invariant mass.

18. Mar 29, 2016

### PCB

Thanks vanhees71 but you have missed my point in that I am trying to fin why my derivation is wrong.
And thanks Nugatory. I know there is a difference between classical and relativistic P and kE etc... What I am trying to do is relativise the classic formula which is why I am using gamma. You say my equation 1 is wrong, and I agree, I agreed in my original post. Thus I come back to asking why it is wrong. I am asking why equation 1 does not equal equation 2. It cannot be pure chance they are very close.

19. Mar 29, 2016

### robphy

So, we agree that your expression 1 ( $m \gamma v^2$ ) is not the expression for relativistic kinetic energy.
So, we agree that your expression 2 ( $2m(\gamma -1)$ ) is not the expression for relativistic kinetic energy [it is actually twice the relativistic kinetic energy].

Note that your velocity is small compared to c (that is, $v/c\ll 1$)
The Taylor expansion of the "difference between your equation 2 and equation 1" has a leading term of order $(v/c)^4$, which is very tiny
$$\frac{v^4}{4}+\frac{v^6}{4}+\frac{15}{64} v^8+O(v^{10})$$
http://www.wolframalpha.com/input/?i=series(2*(1/sqrt(1-v^2)-1)-v^2/sqrt(1-v^2)))

So, in conclusion, your two expressions (neither of which is the relativistic energy) agree to order $(v/c)^4$ for small $(v/c)$.

Last edited: Mar 29, 2016
20. Mar 29, 2016

### Staff: Mentor

You can't "relativise" a classical formula just by plugging in $\gamma$ in various places. That's not a valid method, so it's not surprising that it gives you a wrong answer. The reason the answer is still close for $v \ll c$ is simply that $2 \left( \gamma - 1 \right)$ is very close to $\gamma v^2$, so your incorrect formula just happens to be very close to the correct formula.