# I Special relativity

1. Dec 12, 2017

### Das apashanka

When expanding Kinetic energy as
T=mc2-m0c2
where m0=rest mass and
m=m0/sqrt(1-β2)
the first term correction is coming out to be -p4/8m03c2
but taking T=E-m0c2
where E=sqrt(c^2p^2+m02c4) the correction is coming to be 3p^4/8m03c2

2. Dec 12, 2017

### Mister T

You would need to show your work. Are you using the relativistic expression for $p$?

Last edited: Dec 12, 2017
3. Dec 12, 2017

### PeroK

Are you trying to get $\frac12 m_0v^2$?

4. Dec 12, 2017

### Das apashanka

this term is coming from both but the second term is coming different

5. Dec 12, 2017

### Mister T

I think he's looking at the next term in the series.

6. Dec 12, 2017

### Das apashanka

for the second case p remains as it is but for the first case I have taken p=m0v

7. Dec 12, 2017

### Das apashanka

yes

8. Dec 12, 2017

### PeroK

You must have made a mistake with the binomial expansion in the first case.

I don't see how you can take $p = m_0 v$.

9. Dec 12, 2017

### Mister T

Well, that's likely the reason for the discrepancy in your two results!

In the relation $E=\sqrt{(pc)^2+(m_oc^2)^2}$, $p$ is the relativistic momentum given by $\frac{m_o v}{\sqrt{1-\beta^2}}$, not the newtonian $m_o v$.

10. Dec 12, 2017

### Das apashanka

ok thanks, I have mislooked that thing

11. Dec 12, 2017

### Das apashanka

but that doesn't match the coefficient of -1/8 and 3/8

12. Dec 12, 2017

### PeroK

I'm not sure how you apply the binomial expansion to that, given that neither term need be larger than the other.

13. Dec 12, 2017

### Das apashanka

the term is coming as
m0c2(-.5C2(-v2/c2)2)

14. Dec 12, 2017

### PeroK

It's a mess in any case.

If you assume $p$ is small, then you can get an expansion in terms $p/c$. But, $p$ has a gamma factor, so you cannot compare this expansion directly with the expansion involving $v/c$.

The first terms are $\frac12 m_0v^2$ and $\frac{p^2}{2m_0}$ respectively. But, these are not equal; and neither are the second terms, which in fact have different signs.

15. Dec 12, 2017

### Mister T

You should get $T \approx m_oc^2\left(\frac{1}{2}\beta^2 + \frac{3}{8}\beta^4\right)$

Mine is $\frac{3m_ov^4}{8c^2}$.

Replace $m_o v$ with $p$ and you get $\frac{3p^4}{8m_o^3c^2}$.

I'm not sure how you did that, but it matches what I got above!

By the way, you may find this thread interesting:
There the usual notation is used, where $m$ is the ordinary mass, what you are calling the rest mass.