# Correction to non relativistic KE

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1. Jul 4, 2015

### ognik

1. The problem statement, all variables and given/known data
find series expansion of relativistic energy formula, At what speed is there a 10% correlation to the non relativistic KE.
2. Relevant equations
$$E=\frac{mc^2}{\sqrt(1- \frac{v^2}{c^2})} = mc^2 + \frac{1}{2}mv^2 + ...$$
3. The attempt at a solution
I found the series above, confirmed in the text.
This is one where I don't understand the (2nd part) question. Isn't the non relativistic KE just the 2nd term of the above equation? I would just get a velocity in the opposite direction to c which is meaningless .....

2. Jul 4, 2015

### andrewkirk

Are you sure they asked about a 'correlation'? That would be an odd thing to ask for, as correlation is a statistical concept and there is no stochasticity in this problem. Correlations crop up all the time in quantum mechanics, but never in pure relativity.

3. Jul 4, 2015

### SammyS

Staff Emeritus
By any change do you mean to say:
At what speed is there a 10% correction to the non relativistic KE ?​

4. Jul 5, 2015

### ognik

You're quite right, it should be 'correction' - which doesn't help me unfortunately.

5. Jul 5, 2015

### andrewkirk

The question is very poorly posed, because it is so vague that it is hard to know what they want. But here is my attempt to read the lecturer's mind:

They are thinking of the 'non-relativistic KE' as $\frac{1}{2}mv^2$ and the 'relativistic KE' (ie non-relativistic KE plus corrections for the effect of relativity, but not for rest-mass-energy) as the whole expansion without the rest-mass-energy part $mc^2$. So if we write the expansion as

$$\frac{mc^2}{\sqrt(1- \frac{v^2}{c^2})} = mc^2 + \frac{1}{2}mv^2 + Corrections$$

Then (my guess is that) the lecturer wants you to find what value of $v$ makes the 'correction ratio' $\frac{Corrections}{\frac{1}{2}mv^2}$ equal to 0.1

If you divide both sides of the equation by $\frac{1}{2}mv^2$ you will get a formula that gives that ratio in terms of $\frac{v}{c}$. Setting the ratio to 0.1 and solving for $\frac{v}{c}$ will give you the answer.

6. Jul 5, 2015

### vela

Staff Emeritus
Just out of curiosity, what do you mean by that last sentence?

Anyway, the problem is asking you to find when the relativistic kinetic energy, $(\gamma-1)mc^2$, differs from the classical kinetic energy, $\frac 12 mv^2$, by 10%.

7. Jul 5, 2015

### ognik

Slightly relieved I am not the only one who thinks it is vague, thanks for helping under those circs.
Did you mean dividing the 'relevant equation' above? I get a very messy expression which would be unusual for a text book problem:
$$\frac{2c^2}{v^2} (\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1) = 1.1$$
Maybe there is a trick I can't see to simplifying that? I'll battle through it though - if you think it is right.....

An alternative your insight triggered was that the 'correction' maybe of the order of the next term in the series expansion, which I got as ...+
$$\frac{3mv^4}{8c^3}$$
If I let that term =0.1, then its still a bit messy to get v/c out ....
------------
The reply from Vela came while I was still replying, looks good, so I will work with that for a while now ... Thanks Vela
$$( \frac{1}{10} \frac{mv^2}{2} = mc^2 + \frac{mv^2}{2} ... )$$

Last edited: Jul 5, 2015
8. Jul 5, 2015

### rude man

Hint: (1-x)n = 1 - nx + n(n-1)x2/2! - n(n-1)(n-2)x3/3! + ...
where n can be any real number.

Last edited: Jul 9, 2015
9. Jul 9, 2015

### ognik

Hi - could you please check that my 'messy expression' above is correct? When I solve that, I get v=1.096c which - while nice to think of - cant be right ...Thanks.

10. Jul 9, 2015

### andrewkirk

How did you solve it? Quite apart from the physical impossibility, $\frac{v}{c}=1.096$ can't give a solution because it makes the inside of the square root negative. So re-substituting the candidate solution in the equation shows that it's not right.

I solved it by just running the following R code and observing where the sign changed. The answer was approximately 0.3475, and re-substituting that in the LHS of the formula gives approx 1.1, which is what we want.

y=(1:100)/100
gamma=(1-y^2)^(-0.5)
2*(gamma-1)/(y*y)-1.1

11. Jul 9, 2015

### SammyS

Staff Emeritus
Did you try the expansion given by rude man ?

12. Jul 9, 2015

### ognik

Hi, I expanded
$$\left(1 - \frac{{v}^{2}}{{c}^{2}}\right)^{-\frac{1}{2}} - 1 = \left(\frac{v}{c}\right)^{\!{2}} - \frac{3}{8}\left(\frac{v}{c}\right)^{\!{4}} + \frac{5}{16}\left(\frac{v}{c}\right)^{\!{6}} + ....$$
Multiply that by $$2\left(\frac{c}{v}\right)^{\!{2}} => 2-\frac{3}{4} \left(\frac{v}{c}\right)^{\!{2}} + \frac{5}{8}\left(\frac{v}{c}\right)^{\!{4}} = 1.1$$
But there I'm stuck (as andrewkirk said, the resultant quadratic leaves a - sign in the sqrt) - and I haven't encountered R code before.

13. Jul 9, 2015

### rude man

Probably didn't help that I got a sign wrong in my post #8 (it's correct now).

So, expand mc2{1 - nx + n(n-1)x2/2! - n(n-1)(n-2)x3/3! + ... }

with n = -1/2 as you had it, and x = v2 /c2.
The 1st term is the rest energy, the 2nd term is the Galilean kinetic energy, the third term is the correction you're looking for.

14. Jul 11, 2015

### ognik

Hi rude man (you're quite polite) - I hadn't noticed the sign (I prefer to do suggestions myself once I understand the approach) so no problem. I had looked at that previously (my post #7), where I got the third term =
$$\frac{3m{v}^{4}}{8{c}^{3}} = \frac{1}{10}.\frac{1}{2}m{v}^{2}$$
$$\therefore v=c.\sqrt{\frac{2}{15}} = 0.365c$$
Although I feel comfortable with the method, I wonder how well I understand it: As andrewkirk expressed it in post #5, using a relativistic formula when v is around 1/3 c, gives an error of 10% in the KE, is that right?

15. Jul 11, 2015

### vela

Staff Emeritus
This hardly qualifies as a messy expression, much less a very messy expression! It helps to rewrite the equation in terms of $\beta = v/c$.
$$\frac{1}{\sqrt{1-\beta^2}}-1 = \left(1+\frac{1}{10}\right)\left(\frac 1 2 \beta^2\right).$$ Hopefully, you see that the lefthand side essentially corresponds to the relativistic kinetic energy and the righthand side to the classical kinetic energy plus 10%.

There are two approaches to solving for $\beta$: (1) It's just basic algebra – crank it out. Isolate $\frac{1}{\sqrt{1-\beta^2}}$ and then square both sides of the equation. If you do this, you'll recover andrewkirk's numerical solution. (2) Approximate the lefthand side by expanding in a series and discarding high-order terms. This will get you a simpler equation to solve, but the solution will only be approximate. This is essentially what you've already done in finding $\beta \cong 0.365$.

This is wrong. This equation says 10% of the classical kinetic energy is equal to the total relativistic energy.

16. Jul 11, 2015

### rude man

Unfortunately, that equation has inconsistent units ...
The ratio you are looking for is the third term divided by the second term. That ratio is of course dimensionless ... and it happens to be 1/10.

Last edited: Jul 11, 2015
17. Jul 12, 2015

### ognik

Hi rude man, that was a typo, should be 8c^2 on the bottom ...
Then what I have is the equivalent of the 3rd term / 2nd.
-------------
Hi Vela, you're right about the algebra, but this is in a section on series .....but out of interest I worked through the algebra (with a little help from Mathematica)
It simplified to
$$0.1 -0.7975 b^2-0.3025 b^4$$
which has 4 roots, 2 are < 0 and we can ignore. The others are 0.346317 and 0.3025 . What have I got wrong this time please? (andrewkirk got 0.3475)

18. Jul 12, 2015

### vela

Staff Emeritus
It's impossible to say what you did wrong because you didn't show your work. We can't read your mind or see your paper from here! Keep in mind andrewkirk's answer is only approximate. He didn't solve the equation either.

19. Jul 12, 2015

### andrewkirk

The 0.3475 was an interpolation based on a grid of granularity 0.01, so there's no discrepancy with your calc. On a grid of granularity 0.0001 the solution is between 0.3463 and 0.3464. Your 0.346317 is consistent with that. I think you can regard this problem as solved

20. Jul 12, 2015

### ognik

Thanks Andrew & Vela, I didn't appreciate the R code figure was approximate. This problem wasn't homework BTW, and as I wanted to - because of everyone's patience -I have learned more than just how to solve this particular type of problem, thanks again :-)
(sorry, but I can't find 'thread tools' to mark this as solved, pls let me know how?)

Last edited: Jul 12, 2015