starthaus said:
So...you don't kow how to find the speed of the center of mass in Newtonian physics.
Why would I ask you to find it for the SR case?
Wrong. See below.
starthaus said:
Can you derive the value for V_{COM}? Without it , the solution on the wiki page is useless.
Yes, I can.
Derivation of speed of COM in Special Relativity.
The total (invariant) rest mass (M) of a system in SR can be found from the momentum-energy eqaution:
E = \sqrt{Mc^2 + (Pc)^2}
\Rightarrow M = \frac{\sqrt{E^2 - (Pc)^2}}{c^2}
where E and P are the total energy and total momentum of the system respectively.
The total momentum of the system in SR is then given by:
\Rightarrow P = \frac{M}{\sqrt{1-(V_{COM}/c)^2}} \ V_{COM}
\Rightarrow P = \frac{\sqrt{E^2-(Pc)^2}}{c^2 \sqrt{1- (V_{COM}/c)^2} } \ V_{COM}
\Rightarrow \frac{V_{COM}}{c} = \frac{P c}{\sqrt{E^2-(Pc)^2}} \sqrt{1- (V_{COM}/c)^2}
\Rightarrow \left(\frac{V_{COM}}{c}\right)^2 = \frac{(Pc)^2}{E^2-(Pc)^2}} (1-(V_{COM}/c)^2)
\Rightarrow \left(\frac{V_{COM}}{c}\right)^2 \left(1+ \frac{(Pc)^2}{E^2-(Pc)^2}} \right) = \frac{(Pc)^2}{E^2-(Pc)^2}}
\Rightarrow \left(\frac{V_{COM}}{c}\right)^2} \left(\frac{E^2}{E^2-(Pc)^2}} \right) = \frac{(Pc)^2}{E^2-(Pc)^2}}
\Rightarrow \left(\frac{V_{COM}}{c}\right)^2 = \frac{(Pc)^2}{E^2-(Pc)^2}}\left(\frac{E^2-(Pc)^2}{E^2} \right)
\Rightarrow \left(\frac{V_{COM}}{c}\right)^2 = \frac{(Pc)^2}{E^2}
\Rightarrow V_{COM} = \frac{P}{E}c^2
QED.
Once you have obtained V_{COM} = Pc^2/E then it is trivial to obtain:
V_{COM} = \frac{m_1u_1\gamma(u_1)+m_2 u_2 \gamma(u_2)}{m_1\gamma(u_1) + m_2\gamma(u_2)}
from the expressions for total momentum and total energy. See
http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_relativistic
[EDIT]This final equation has been edited to fix a minor typo that was noticed by Starthaus who has kindly proof read the above.
Note that:
\frac{P}{E}c^2 = \frac{M\gamma(V_{COM})V_{COM}}{M\gamma(V_{COM})c^2}c^2 = V_{COM}
starthaus said:
You did not provide a proof, you just posted a hack. By contrast, jtbell provided a proof. It would behoove on you to learn what a valid proof looks like.
There is no point in my duplicating jtbell's work. We have different styles and that makes the world a more interesting place.
starthaus said:
Err, false again. I knew that you'll start with this nonsense type of claims and I uploaded https://www.physicsforums.com/blog.php?b=1887 yesterday. You can learn how to produce valid proofs.
See my simpler proof for the Newtonian solution in my previous post.
starthaus said:
The relativistic proof simply uses a different formula for the speed of COM wrt the original frame. Do you even know how to derive the speed of COM? Either in Newtonian or in SR mechanics?
Yes. See above for the derivation in SR. See below for the classical derivation.
Derivation of the speed of COM in the Newtonian case.
In the Newtonian case the total mass is simply the sum of the individual masses so the total momentum is given by:
(m_1 + m_2) V_{COM} = m_1u_1 + m_2u_2
V_{COM} = \frac{ m_1u_1 + m_2u_2}{ m_1 + m_2}
Simple as that, but I am sure you can find a much harder way to derive it.
starthaus said:
This is not a proof, this is another one of your hacks. You are assuming the solution from the beginning and you are only verifying that it fulfills the conditions.
I asked you if you could produce a legitimate proof. Can you?
Assuming something and then verifying it is correct is good enough for me. You can always prove to yourself that if v'_1 \ne \ \pm u_1 in the COM frame then momentum and energy are not conserved, (proof by contradiction), if that makes you happier. I leave that as an exercise for you
. If you want to prove that v'_1 = u_1 is true without assuming it in the first place, then go ahead and knock yourself out.
I will be impressed if you can do it, but somehow I don't think you are able to.
kev said:
You have yet to post your relativistic solution. When you do, it will be the same as
https://www.physicsforums.com/showpost.php?p=2805545&postcount=35"or it will be wrong.
starthaus said:
YOU have still not posted your derivation or solution for v1 and v2 in the relativistic case.
starthaus said:
Firstly, try to stay on topic. Secondly, it is not my problem that your calculus knowledge is limited to the use of software packages for calculating basic differentials. I'll give you a hint: f(t)=0 \Rightarrow \frac{df}{dt}=0. Thus, \frac{df}{dt}=0 \Rightarrow \frac{d^2f}{dt^2}=0. Feel free to consult your high school calculus textbook for confirmation.
How sad it is that despite being shown simple counterproofs to your assertion by various people, that you still cling to your misconception. Your example above is only true for the case f(t)=0 and in that particular case if the value of f(t) is always zero, then it is not a function of t but a constant. Do you see that? Your assertion that dx/dt = 0 \Rightarrow d^2x/dt^2 =0 is not generally true. The simple counterproof is this. Let us say f(t)=x^2. When x=0, dx/dt = 2x = 0 and d^2x/dt^2 = 2 so your assertion that dx/dt = 0 \Rightarrow d^2x/dt^2 =0 is proven false.
Back on topic, all your objections to the Wikpedia derivation of the final velocities in the relativistic case have been proven groundless. The Wikipedia derivation is far superior to your derivation, because the Wikipedia derivation for the relativistic case actually exists,
unlike your derivation, which we have yet to see.
