Calculating mass and speed from kinetic energy and linear momentum

1. Aug 21, 2004

niko2000

Hi,
I have tied to solve an exercise where I have kinetic energy and linear momentum and I have to calculate the speed and mass (not relativistic mass)
I have these specifications:
Wk = 5713.13 MeV
P = 6584.92 Mev/c0

I have used these formulas:
Wk = mc^2*((gamma)-1)
P = (gamma)*m*v
I have tried to calculate v and m, but I haven't got sensible results.
I have probably done some mistake, but I can't find it. If anyone had time to take a look at this exercise it would be really helpful.
Regards,
Niko

2. Aug 21, 2004

pmb_phy

Usually when someone posts what appears to be a homework question we ask to see what work they have done so far so that we don't do all the work for them. Can you please show us what you did and why you don't think your results were sensible? Thanks.

Until then this is what I got. Let K = kinetic energy, m = proper mass, E = inertial energy = K + mc2. Then

$$E^2 - p^2c^2 = m^2c^4 = (K + mc^2)^2 - p^2c^2$$

Multiply the term in the parenthese out to obtain

$$K^2 + 2kmc^2 + m^2c^4 - p^2c^2 = m^2c^4$$

Cancel like terms and solve for m to obtain

$$m = \frac {p^2c^2 - K^2}{2Kc^2}$$

Now that you have m you can use p = gamma*m*v and solve for v. I'll let you work that part out since I did the first part of the work to obtain m. Let us know if you need more help.

Pete

Last edited: Aug 21, 2004
3. Aug 21, 2004

niko2000

I don't know if the first equation is right:
E = mr*c^2 = m0*c^2 + K,
where mr = relativistic mass and m0 = proper mass, so I thing this can't lead us to the right soluton.
When I was doing this exercise I have got the speed greater then speed of light, so it's not right. I have probably made some calculation mistakes. I think the formulas I have written guide us to the right answer, but I simply couldn't get it because I have made some mistakes I can't find.

4. Aug 21, 2004

DW

Using a relativistic mass substitution somewhere you shouldn't have is probably what led to your mistake. Do not use relativistic mass and try again.
There is no place for relativistic mass in modern relativity.

5. Aug 21, 2004

krab

Here's a simpler way: Square the momentum. That gives you $\beta^2\gamma^2m^2c^2$ and use the identity $\beta^2\gamma^2=\gamma^2-1$. Divide this by K.E. and note $\gamma^2-1/(\gamma-1)=\gamma+1$. This gives you gamma, and the rest follows easily.

6. Aug 21, 2004

pmb_phy

Yes. That equation is correct, but not useful. I posted the formula for m as a function of kinetic energy and momentum above.
Not so fast. His data does not correspond to a particle which moves at speeds less than the speed of light (i.e. tardyons). This can be seen by noticing that for the proper mass to be positive, (pc)2- k2 must be positive. But with his data this comes out negative. His data there therefore does not describe a tardyon. It describes a tachyon.

Pete

7. Aug 21, 2004

pmb_phy

Sorry but I don't see how you got that. Momentum is given by

$$p = mv = \gamma m_0 v$$

Therefore

$$p^2 = \gamma^2 m_0^2 v^2$$

You claim this is

$$p^2 = \beta^2 \gamma^2 m_0^2 c^2$$

Why?

Kinetic energy is given by

$$K = (\gamma - 1)m_0c^2$$

so when you divide p2 by K2 to obtain (p/K)2 you will obtain

$$(p/K)^2 = \frac{\gamma^2 m_0^2 v^2}{ (\gamma - 1)m_0c^2}$$

Canceling a factor of m0 gives

$$(p/K)^2 = \frac{\gamma^2 m_0 v^2}{ (\gamma - 1)c^2}$$

Simplify to obtain

$$(p/K)^2 = (\gamma+1)m_0\beta^2}$$

I don't see that this is useful since you haven't eliminated either v or m0.

Pete

Last edited: Aug 21, 2004
8. Aug 21, 2004

robphy

Try a rapidity calculation.

$$K= m_0 c^2(\cosh\theta - 1)$$
$$p= m_0 c\sinh\theta$$

So,
$$\cosh\theta = \frac{K}{m_0c^2}+1$$
$$\sinh\theta = \frac{p}{m_0c}$$

Now, given values for K and p,
$$1 =\cosh^2\theta - \sinh^2\theta = (\frac{K}{m_0c^2}+1 )^2- ( \frac{p}{m_0c} )^2$$
solve for $m_0$.
Armed additionally with $m_0$,
$$v = c\tanh\theta = c\frac{\sinh\theta }{\cosh\theta } = c\frac{\frac{p}{m_0c} }{ \frac{K}{m_0c^2}+1 }$$

9. Aug 22, 2004

pmb_phy

That doesn't solve his problem since this last expression is still a function of m0. He needs m0(p,K) and v(p,K). You gave v(m0,p,K).

Pete

10. Aug 22, 2004

DW

Perhaps you should do the algebra right. You could also stop innappropriately subscipting the mass with a zero as he did not do that and don't claim that p = mv as it was not. It was clearly stated for this thread that m was not relativistic mass.

Last edited by a moderator: Aug 23, 2004
11. Aug 22, 2004

robphy

Let me fill in some steps in algebra.
$$\textcolor{red}{m_0=\frac{ (pc)^2-K^2 }{ 2Kc^2} }$$

\begin{align*} v &=c\frac{\frac{p}{m_0c} }{ \frac{K}{m_0c^2}+1 }\\ &=\frac{p}{ \frac{K}{c^2}+m_0 }\\ &=\frac{p}{ \frac{K}{c^2}+\textcolor{red}{\frac{ (pc)^2-K^2 }{ 2Kc^2} } } = \frac{2K pc^2}{K^2+p^2c^2} \end{align*}

12. Aug 22, 2004

krab

Like DW, I truly do not understand some of pete's comments. But then since the dynamics of relativistic particles is my field of daily work, things that are obvious to me might not be obvious to others...

Anyway, I showed the trick of taking p^2/K to get something proportional to (gamma+1)m. (Sorry, I erred in saying this gave you gamma; I lost sight that it was m that is not known.) Since you already have K, which is prop.to (gamma-1)m, just subtract the two to get mass. In detail,

$${1\over 2}\left({p^2c^2\over K}-K\right)=mc^2$$

This agrees with the red formula given by robphy.

13. Aug 23, 2004

pmb_phy

Which ones? If you're refering to the "Why?" question then it was simply a question and nothing else. For some dumb reason I didn't recognize beta2c2 as v2. I don't know WHAT I was thinking. I reserve the right to have a bad day and miss simple things like this.

I also wrote down K2 on the left side instead of K. But I did divide by K on the right. I can't seem to edit that post to correct it. Why is that?

That all seems like a lot of work to get gamma. Gamma can easily be found from the equation in my first post. That gives the proper mass m0 from the given values of p and K. Once you have m0 then use K = (gamma - 1)m0c2 and solve for gamma. Once you have gamma solve for v.

Note: I have to read posts and respond in a hurry for the next several months. Due to a back injury I can't sit in front of the computer for very long otherwise there are severe consequences. The pain is bit distracting too. So please understand that I will make more errors, due to necessary haste (and distraction) in reading and posting, until this gets better. Thanks for your understanding.

Pete

Last edited: Aug 23, 2004