Question

eliminate \(v\)

\(p = \dfrac{mv}{\sqrt{1-(v/c)^2}}\)

\(K=mc^2\left(\dfrac{1}{\sqrt{1-(v/c)^2}}-1\right)\)

Answer 1

this is similar to eliminating v from the momentum, KE equations and getting the relation p^2/2m = ke

Answer 2

eliminate \(v\)

\(p = \dfrac{mv}{\sqrt{1-(v/c)^2}}\)

\(K=mc^2\left(\dfrac{1}{\sqrt{1-(v/c)^2}}-1\right)\)

Answer 3

\[\left(\frac{K}{mc^2} +1\right)^{-2} = 1 - (v/c)^2\]So on to solve for \(v\) and substituting in the first equation.

Answer 4

I don't see a good form for this though :|

Answer 5

Nice!

\[1-\left(\frac{K}{mc^2} +1\right)^{-2} = (v/c)^2\]

Answer 6

im thinking of squaring the first equation and substituting above thingy

Answer 7

Oh, yeah, that'd work I guess.

Answer 8

\[p^2 = \dfrac{(mv)^2}{1-(v/c)^2} \implies (p/c)^2 = \dfrac{m^2(v/c)^2}{1-(v/c)^2} \]

Answer 9

that still looks nasty... somehow i need to arrive at below relation :
\[(pc)^2=K^2+2Kmc^2\]

Answer 10

can we say something like\[\left(1 + \frac{K}{mc^2} \right)^{-2} \approx 1 - 2\frac{K}{mc^2}\]this only works if \(K \ll mc^2\) and I'm not really sure if that's true in relativistic mechanics haha.

Answer 11

approximating using taylor series ?

Answer 12

yes, but again that only works if \(K \ll mc^2\).

Answer 13

kinetic energy can be a million times greater than the rest energy mc^2, so that approximation is not so useful here..

Answer 14

for example
rest energy (mc^2) of an electron is just 0.511 MeV, but we can accelerate the electrons to huge kinetic energies as 50 GeV

Answer 15

that's weird. as long as you're not doing anything wrong, you should be able to get that expression. and yeah, I just looked at the original expression for kinetic energy and it makes sense how for low values of velocity it's way bigger than \(mc^2\)

Answer 16

yeah this looks messy... il just substitute K and p in the final equation and verify if it balances..

Answer 17

I just looked at your expression and seeing some patterns\[(pc)^2 = K^2 + 2K mc^2+ \color{#C00}{(mc^2)^2 - (mc^2)^2} \]

Answer 18

as you said the kinetic energy can be a million times greater than the rest energy. it's possible that the actual expression is\[(pc)^2 = K^2 + 2K mc^2 + (mc^2)^2 = (K + mc^2)^2\]and that they neglected the final term. wait, let me check if this is true. :P

Answer 19

nah im dumb

Answer 20

that looks very interesting! we can't give away K or mc^2 though... their relative values depend on the mass of the particles...

Answer 21

for an electron, rest energy may be negligible compared to the kinetic energy
but for an uranium atom, rest energy may not be negligibel..

Answer 22

hey did you plug in the values of K and p in the equation you're given in the end?
i think plugging in can have two benefits: it can tell us if an equation is exactly true, and if not, we can simplify it and end up with something in the form \(\alpha = \beta\) which would mean that we've approximated \(\alpha\) to be equal to \(\beta\) if you know what i mean.

basically we're reverse-solving it.

Answer 23

i haven't plugged in p and K yet, but i plan to do it as soon as i find a paper and pen...
here is a screenshot from textbook :

Answer 24

looks the textbook is not referring to any approximation...