Special Relativity: Kinetic Energy Expansion and Contradictions

[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
is there any contradiction?

 

[Herman Trivilino]

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

 

[PeroK]

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
is there any contradiction?

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

 

[Das apashanka]

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

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

 

[Herman Trivilino]

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

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

 

[Das apashanka]

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

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

 

[Das apashanka]

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

yes

 

[PeroK]

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

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

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

 

[Herman Trivilino]

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

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 $m_o v$.

 

[Das apashanka]

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 $m_o v$.

ok thanks, I have mislooked that thing

 

[Das apashanka]

ok thanks, I have mislooked that thing

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

 

[PeroK]

ok thanks, I have mislooked that thing

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

 

[Das apashanka]

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

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

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

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

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

 

[PeroK]

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

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.

 

[Herman Trivilino]

When expanding Kinetic energy as
T=mc²-m₀c²
where m₀=rest mass and
m=m₀/sqrt(1-β²)

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

the first term correction is coming out to be -p⁴/8m₀³c²

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

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

but taking T=E-m₀c²
where E=sqrt(c^2p^2+m₀²c⁴) the correction is coming to be 3p^4/8m₀³c²
is there any contradiction?

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

By the way, you may find this thread interesting:
https://www.physicsforums.com/threads/kinetic-energy-and-momentum-of-a-relativistic-particle.895897/
There the usual notation is used, where $m$ is the ordinary mass, what you are calling the rest mass.