Question

\[\huge m_pc^2\sqrt{1+{v^2_{m_p}\over c^2}\ \over 1-{v^2_{m_p}\over c^2}}+m_pc^2\over {\sqrt{1+{v^2_M \over 1-{v^2_m \over c^2}}}}\]

@LastDayWork
think there's a way to simplify that?

Answer 1

@LastDayWork i didn't know if you knew what gamma was or not so I wrote out the whole thing

Answer 2

gamma the Lorentz factor

Answer 3

That's how far I could go without using Binomial approximation -
\[mc^{2} \frac{ \sqrt{c^2 + v^2} +\sqrt{c^2 - v^2}}{ \sqrt{c^2 + c^2 v^2 - v^2} }\]

& I know about gamma, but never remembered its formula..So you did me a favor by substituting it ;)

Answer 4

BTW the v^2 in the denominator is also V(mp)...right ??

Answer 5

no, there were two m's
there were two gammas, the gamma of a particle of mass mp and the gamma of a particle of mass M

so the denominator was for the gamma of mass M and therefore velocity M and the other was for the velocity of a mass of mp

Answer 6

basically it was a relativistic conservation of linear momentum of two particles in in inelastic collision. but one particle was stationary so it went bye bye

the mass M was the result and mass mp was the initial moving particle

Answer 7

there was also conservation of energy; forgot to mention that

Answer 8

In that case, you should directly apply binomial..as I can't see any valid simplification..

Answer 9

the only question really then is WHERE to begin the binomial...

Answer 10

\[\huge \sqrt{{(m_pc^2)^2}+{(\gamma_m m_p v_mc)^2}}+m_pc^2 = \huge \sqrt{{(Mc^2)^2}+{(\gamma_M M v_Mc)^2}}\]

Answer 11

aww, huge is too big

Answer 12

\[\huge \sqrt{{(Mc^2)^2}+{(\gamma_M M v_Mc)^2}}\]

Answer 13

where gamma is 1-v^2/c^2
and I'm solving for Mc^2

Answer 14

This question looks too scary to me :(
@douglaswinslowcooper is online...maybe he can help here..

Answer 15

lol

Answer 16

give me a sec to post the original question then

Answer 17

@roadjester Where did this algebraic expression come from?

Answer 18

@agent0smith - Can you help here ??

Answer 19

oh from conservation of energy
\[E^2={(mc)}^2+{(pc)}^2\]
each particle has it's own energy
the letter p is momentum which is
\[p=\gamma mv\]

Answer 20

Okay, so how did you manage to screw it up?

Answer 21

well that was rude, and I didn't "screw it up" as you so nicely put it

Answer 22

@ybarrap

Answer 23

There are no egos here. We all all buddies in math.

Answer 24

this is the physics group bro

Answer 25

73. The creation and study of new and mery massive elementary particles is an important part of contemporary physics. To create a particle of mass M requires an energy Mc^2. With enough energy, en exotic particle can be created by allowing a fast-moving proton to collide with a similar target particle. Consider a perfectly inelastic collision between two protons: an incident proton with mass m_p, kinetic energy K, and momentum magnitude p joins with an originally stationary target proton to form a single product particle of mass M. Not all the kinetic energy of the incoming proton is available to create the product particle because conservaton of momentum requires that the system as a whole still must have some kintetic energy after the collision. Therefore, only a fraction of the energy of the incident particle is available to create a new particle. (a) Show that the energy available to create a product particle is given by [SEE ATTACHMENT] This result shows that when the kinetic energy K of the incident proton is large compared with its rest energy m_pc^2, then M approaches (2m_pK)^(1/2)/c. Therefore, if the energy of the incoming proton is increased by a factor of 9, the mass you can create increases only by a factor of 3, not by a factor of 9 as would be expected. (b) This problem can be alleviated by using colliding beams as is the case in most modern accelerators. Here the total momentum of a pair of interacting particles can be zero. The center of mass can be at rest after the collision, so, in principle, all the initial kinetic energy can be used for particle creation. Show that [SEE ATTACHMENT] where K is the kinetic energy of each of the two identical colliding particles. Here, if K>>mc^2, we have M directly proportional to K as we would desire.

Answer 26

You're right. It is physics. I hope you lift.

Answer 27

Still, at what point did you decide you needed to expand \(\gamma\)?

Answer 28

@roadjester
you interpreted @wio wrong..it was meant as a humor..

@ybarrap
Can you explain how to simplify the expression in the question...using binomial and stuff..

Answer 29

Are you sure this,
$$
\huge{ m_pc^2\sqrt{1+{v^2_{m_p}\over c^2}\ \over 1-{v^2_{m_p}\over c^2}}+m_pc^2\over {\sqrt{1+{v^2_M \over 1-{v^2_m \over c^2}}}}}\\
$$
Isn't supposed to be this,
$$
\huge{ m_pc^2\sqrt{1+{v^2_{m_p}\over c^2}\ \over 1-{v^2_{m_p}\over c^2}}+m_pc^2\over\sqrt{1+{v^2_{m_p}\over c^2}\ \over 1-{v^2_{m_p}\over c^2}}}\\
$$
?

Answer 30

actually there is a mistake in the square roots but it isn't the one you suggested.
the 1 in the numerator of the top square root should also be pulled out like the denominator