One dimensional collision of 2 unequal masses

[randombill]

Imagine two particles of uneven masses colliding at relativistic velocities. What is the final velocity of each particle after the collision (no they do not stick together).

This question pertains to the relativistic treatment of a one dimensional collision. Basically I tried to derive the equations of motion for the relativistic case but I'm stuck. I attached what I've done so far.

I thought that solving for the Newtonian conservation of momentum case would be easy simply by putting in the gammas and solving for a simultaneous equation except that I end up with a simultaneous equation for two unknown masses (these are the relativistic masses) while the final rest masses are known. In the attached picture the unknowns are the final relativistic masses while the initial rest masses and the initial relativistic masses are known (please see the last picture at the last line).


Thanks.

 

[starthaus]

Imagine two particles of uneven masses colliding at relativistic velocities. What is the final velocity of each particle after the collision (no they do not stick together).

This question pertains to the relativistic treatment of a one dimensional collision. Basically I tried to derive the equations of motion for the relativistic case but I'm stuck. I attached what I've done so far.

I thought that solving for the Newtonian conservation of momentum case would be easy simply by putting in the gammas and solving for a simultaneous equation except that I end up with a simultaneous equation for two unknown masses (these are the relativistic masses) while the final rest masses are known. In the attached picture the unknowns are the final relativistic masses while the initial rest masses and the initial relativistic masses are known (please see the last picture at the last line).


Thanks.

See https://www.physicsforums.com/blog.php?b=1857 , the attachment entitled "Collision"

 

[randombill]

I'm looking at the final equation (3.8) where:

γ (U '1 )m1 + γ (U '2 )m2 = γ (u '1 )m1 + γ (u '2 )m2 (3.8)

which I understand to be the velocity in the "other frame" but the equation just spits back the original problem. How do I find (u '1 ) and (u '2 )?

I think these proofs only show that the conservation of momentum is frame invariant but nothing else.

Say if you had a mass of 10 kg moving along the positive x direction at .8c and a mass of 5kg moving along the negative x direction at .7c. how would I solve that using these equations in the collision.doc?

thanks.

 

[starthaus]

I'm looking at the final equation (3.8) where:

γ (U '1 )m1 + γ (U '2 )m2 = γ (u '1 )m1 + γ (u '2 )m2 (3.8)

This is the conservation of relativistic energy . It reduces to the conservation of relativistic mass.

which I understand to be the velocity in the "other frame" but the equation just spits back the original problem. How do I find (u '1 ) and (u '2 )?

Add to (3.8) the equation of conservation of relativistic momentum:

γ (U '1 )m1 U'1+ γ (U '2 )m2 U'2= γ (u '1 )m1 u'1+ γ (u '2 )m2u'2

You now have a non-linear system of two equations with two unknowns : u'1 and u'2.

 

[jtbell]

This is the conservation of relativistic energy . [...].

Add to (3.8) the equation of conservation of relativistic momentum:

γ (U '1 )m1 U'1+ γ (U '2 )m2 U'2= γ (u '1 )m1 u'1+ γ (u '2 )m2u'2

You now have a linear system of two equations with two unknowns : u'1 and u'2, easy to solve.

Note that in non-relativistic mechanics you also have to start with both conservation of momentum and conservation of energy, in order to solve a one-dimensional collision. Relativistic mechanics is no different in this respect.

The main difference in the relativistic case is that momentum and energy depend on velocity in a more complicated way, so the algebra is messier.

 

[randombill]

This is the conservation of relativistic energy . It reduces to the conservation of relativistic mass.Add to (3.8) the equation of conservation of relativistic momentum:

γ (U '1 )m1 U'1+ γ (U '2 )m2 U'2= γ (u '1 )m1 u'1+ γ (u '2 )m2u'2

You now have a linear system of two equations with two unknowns : u'1 and u'2, easy to solve.

But what about the γ's for each velocity? Can those be factored out so:

(U '1 )m1 U'1+ (U '2 )m2 U'2= (u '1 )m1 u'1+ (u '2 )m2u'2I'm assuming that I cannot because the gammas are velocity dependent for each momentum and that's the part that makes it hard to solve. Basically that was the reason why I converted the relativistic momentum equation into the equation where:

(pc)^2 = - (m_0c^2)^2 + (mc^2)^2

(got it from here http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c1)

Basically how do I solve a simultaneous equation for this mess?

NOTE: I just noticed that I used lamba's instead of gamma's in my first post photos. Sorry about that.

Note that in non-relativistic mechanics you also have to start with both conservation of momentum and conservation of energy, in order to solve a one-dimensional collision. Relativistic mechanics is no different in this respect.

The main difference in the relativistic case is that momentum and energy depend on velocity in a more complicated way, so the algebra is messier.

Right, how do I solve for it is the problem. The algebra is the part that's messing me up because each γ's have an independent velocity under a square root and I am having trouble factoring it out so I can solve for each final velocity.

 

[starthaus]

But what about the γ's for each velocity? Can those be factored out so:

(U '1 )m1 U'1+ (U '2 )m2 U'2= (u '1 )m1 u'1+ (u '2 )m2u'2

No, they cannot. Different speeds produce different \gamma's.
Yet:

γ (U '1 )m1 + γ (U '2 )m2 = γ (u '1 )m1 + γ (u '2 )m2 (3.8)

gives you γ (u '2 ) (and, consequently, u '2) as a function of γ (u '1 ). Sibstitute in the equation of momentum conservation and you will get the formula for u '1.

 

[randombill]

No, they cannot. Different speeds produce different \gamma's.
Yet:

γ (U '1 )m1 + γ (U '2 )m2 = γ (u '1 )m1 + γ (u '2 )m2 (3.8)

gives you γ (u '2 ) (and, consequently, u '2) as a function of γ (u '1 ). Sibstitute in the equation of momentum conservation and you will get the formula for u '1.

I'm not entirely sure what you mean, could you show me that part by doing it... please, thanks.

 

[starthaus]

I'm not entirely sure what you mean, could you show me that part by doing it... please, thanks.

OK,

m_1v_1\gamma_1+m_2v_2\gamma_2=A
m_1\gamma_1+m_2\gamma_2=B

Multiplying the second equation by -v_1 and adding with the first one we get:

v_1=\frac{m_2v_2\gamma_2-A}{m_2\gamma_2-B}

From the above, you can calculate :

\gamma_1=1/\sqrt{1-(v_1/c)^2} as a function f(v_2)

Substitute in B=m_1\gamma_1+m_2\gamma_2=m_1f(v_2)+m_2\gamma_2 and you got yourself an algebraic equation in v_2

 

[randombill]

OK,m_1\gamma_1+m_2\gamma_2=B

Is B the final momentum equation. I'm lost as to what the initial and final velocities are.

From the above, you can calculate :

\gamma_1=1/\sqrt{1-(v_1/c)^2} as a function f(v_2)

Not sure how?

Substitute in B=m_1\gamma_1+m_2\gamma_2=m_1f(v_2)+m_2\gamma_2 and you got yourself an algebraic equation in v_2

Is v2 the velocity in the final momentum equation.

Sorry but I've never solved a simultaneous equation in such a way, plus my math isn't the greatest.

How would I solve this for example:

If you had a mass of 10 kg moving along the positive x direction at .8c and a mass of 5kg moving along the negative x direction at .7c, what are the final velocities of each mass?EDIT: I did find this page:
http://teachers.web.cern.ch/teacher...ch/mbitu/applications_of_special_relativi.htm

where the Lorentz transformation matrix is used to find the final momentum and energies but how do I find the velocities of the particles knowing the final energy for each particle?

 

[starthaus]

Is B the final momentum equation. I'm lost as to what the initial and final velocities are.

No, it is the initial momentum. You know it since you know the speeds before collision.

Not sure how?

Simple, substitute the formula for v_1 that I just derived for you, in the formula for \gamma(v_1)

Is v2 the velocity in the final momentum equation.

v_1 and v_2 are the final velocities of particle 1 and particle 2 (after the collision)

 

[randombill]

I tried substituting A and B into


[PLAIN]https://www.physicsforums.com/latex_images/28/2800728-3.png

and then substituted that into,

[PLAIN]https://www.physicsforums.com/latex_images/28/2800728-4.png

and all the terms cancel out and I got back the original equation for gamma?

But then I'm not sure what you mean at this part for,

[PLAIN]https://www.physicsforums.com/latex_images/28/2800728-5.png. I see that I can get v1 and v2 (the final velocities) by squaring gamma_1 and solving for v1.

The part I'm not getting is that B does not have any initial velocities so how would that find the final velocities?

 

[starthaus]

I tried substituting A and B into


[PLAIN]https://www.physicsforums.com/latex_images/28/2800728-3.png

and then substituted that into,

[PLAIN]https://www.physicsforums.com/latex_images/28/2800728-4.png

and all the terms cancel out and I got back the original equation for gamma?

I give up. This is basic algebra.

 

[randombill]

I give up. This is basic algebra.

C'mon you almost solved it. Algebra is never basic.

 

[starthaus]

OK,

m_1v_1\gamma_1+m_2v_2\gamma_2=A
m_1\gamma_1+m_2\gamma_2=B

Multiplying the second equation by -v_1 and adding with the first one we get:

v_1=\frac{m_2v_2\gamma_2-A}{m_2\gamma_2-B}

From the above, you can calculate :

\gamma_1=1/\sqrt{1-(v_1/c)^2} as a function f(v_2)

Substitute in B=m_1\gamma_1+m_2\gamma_2=m_1f(v_2)+m_2\gamma_2 and you got yourself an algebraic equation in v_2

\gamma_1=1/\sqrt{1-1/c^2*(\frac{m_2v_2\gamma_2-A}{m_2\gamma_2-B}})^2

Can you take it from here?

 

[randombill]

[PLAIN]https://www.physicsforums.com/latex_images/28/2801452-8.png

Can you take it from here?

Yes I substituted A and B into that equation and the terms reduced under the square root to:


[PLAIN]https://www.physicsforums.com/latex_images/28/2800728-4.png

Please let me know if there is something totally wrong with what I did, thanks.

 

[starthaus]

Yes I substituted A and B into that equation and the terms reduced under the square root to:


[PLAIN]https://www.physicsforums.com/latex_images/28/2800728-4.png

Please let me know if there is something totally wrong with what I did, thanks.

This is impossible since

A=γ (U '1 )m1 U'1+ γ (U '2 )m2 U'2

B=γ (U '1 )m1+ γ (U '2 )m2

where U'1 and U'2 are the speeds before the collision.

 

[randombill]

This is impossible since

A=γ (U '1 )m1 U'1+ γ (U '2 )m2 U'2

B=γ (U '1 )m1+ γ (U '2 )m2

where U'1 and U'2 are the speeds before the collision.

Oh, ok. So I retried doing that substitution for A and B and I attached two photos of the work. Let me know if the final result makes sense for v2 which is the final velocity. Do you know if there is a way to find v1 similar to Newtonian mechanics where:

v2f - v1f = v1i - v2i

These are the photos, I'm not good with tex either.

https://www.physicsforums.com/attachment.php?attachmentid=26983&stc=1&d=1279283932

https://www.physicsforums.com/attachment.php?attachmentid=26984&stc=1&d=1279283932

 

[starthaus]

Oh, ok. So I retried doing that substitution for A and B and I attached two photos of the work. Let me know if the final result makes sense for v2 which is the final velocity. Do you know if there is a way to find v1 similar to Newtonian mechanics where:

v2f - v1f = v1i - v2i

These are the photos, I'm not good with tex either.

https://www.physicsforums.com/attachment.php?attachmentid=26983&stc=1&d=1279283932

https://www.physicsforums.com/attachment.php?attachmentid=26984&stc=1&d=1279283932

No, it doesn't look right.

 

[starthaus]

OK,

m_1v_1\gamma_1+m_2v_2\gamma_2=A
m_1\gamma_1+m_2\gamma_2=B

Multiplying the second equation by -v_1 and adding with the first one we get:

v_1=\frac{m_2v_2\gamma_2-A}{m_2\gamma_2-B}

From the above, you can calculate :

\gamma_1=1/\sqrt{1-(v_1/c)^2} as a function f(v_2)

Substitute in B=m_1\gamma_1+m_2\gamma_2=m_1f(v_2)+m_2\gamma_2 and you got yourself an algebraic equation in v_2

I showed you how to get \gamma_1=1/\sqrt{1-(v_1/c)^2} as a function f(v_2)

Now, substitute \gamma_1 in the equation B=m_1\gamma_1+m_2\gamma_2 and you will get an algebraic equation in v_2. Solve it and you will find v_2. Substitute v_2 in the expression for \gamma_1 and you will find v_1. Try learning latex, or, at least, print your solutions.

 

[randombill]

I showed you how to get \gamma_1=1/\sqrt{1-(v_1/c)^2} as a function f(v_2)

Now, substitute \gamma_1 in the equation B=m_1\gamma_1+m_2\gamma_2 and you will get an algebraic equation in v_2. Solve it and you will find v_2. Substitute v_2 in the expression for \gamma_1 and you will find v_1. Try learning latex, or, at least, print your solutions.

I tried to do this:

Take A and B and substitute into:
[PLAIN]https://www.physicsforums.com/latex_images/28/2801452-8.png

and then I substituted that back into B and solved for v2. The solution was long and I had trouble with tex.

Let me know what your final solution looks like because at least then I could try to find that.

Thanks.

 

[starthaus]

I tried to do this:

Take A and B and substitute into:
[PLAIN]https://www.physicsforums.com/latex_images/28/2801452-8.png

and then I substituted that back into B and solved for v2. The solution was long and I had trouble with tex.

Let me know what your final solution looks like because at least then I could try to find that.

Thanks.

no, it isn't right, you forgot that \gamma_2 is a function of v_2

 

[randombill]

no, it isn't right, you forgot that \gamma_2 is a function of v_2

It would be impossible to factor out \gamma_2 out of the equation after doing this:

[PLAIN]https://www.physicsforums.com/latex_images/28/2801452-8.png

because \gamma_2 is mixed in with everything else. I read this book:

http://books.google.com/books?id=CC...&resnum=1&ved=0CCkQ6AEwAA#v=onepage&q&f=false

where he mentions toward the end of paragraph 2 on page 225 the use of numerical solutions for this problem. I guess this is why?

Basically I give up I can't solve it.

 

[starthaus]

It would be impossible to factor out \gamma_2 out of the equation after doing this:

[PLAIN]https://www.physicsforums.com/latex_images/28/2801452-8.png

because \gamma_2 is mixed in with everything else. I read this book:

http://books.google.com/books?id=CC...&resnum=1&ved=0CCkQ6AEwAA#v=onepage&q&f=false

where he mentions toward the end of paragraph 2 on page 225 the use of numerical solutions for this problem. I guess this is why?

Basically I give up I can't solve it.

You should get an equation degree 8 in v_2. This equation does not have symbolic solutions. Nevertheless, it is a good exercise for you to finish the calculations, never give up once you start something. There is satisfaction in forming the equation. How else would you solve it numerically if you don't have the final equation?

 

[randombill]

You should get an equation degree 8 in v_2.

Does that mean (x)^8 = v_2 where x is the final solution?

I don't need a numerical solution, unfortunately seeing how there is no symbolic solution makes me not need this result.

 

[starthaus]

Does that mean (x)^8 = v_2 where x is the final solution?

I don't need a numerical solution, unfortunately seeing how there is no symbolic solution makes me not need this result.

No, it desn't. It means:

v_2^8+a_1v_2^7+a_2v_2^6+...=0

 

[yuiop]

...
I don't need a numerical solution, unfortunately seeing how there is no symbolic solution makes me not need this result.

This is the solution for the relativistic one dimensional elastic collision according to Wikipedia.

http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_relativistic

It should give you an idea of what to aim for.

 

[randombill]

No, it desn't. It means:

v_2^8+a_1v_2^7+a_2v_2^6+...=0

I forgot, is that some kind of an expansion? I did learn something like that in class.

This is the solution for the relativistic one dimensional elastic collision according to Wikipedia.

http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_relativistic

It should give you an idea of what to aim for.

I've seen that, its junk.

 

[starthaus]

This is the solution for the relativistic one dimensional elastic collision according to Wikipedia.

http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_relativistic

It should give you an idea of what to aim for.

The wiki solution is valid only for speeds <<c. You can see that from the text, where they calculate p_T=m_1u_1+m_2u_2 valid only for u_1&lt;&lt;c and u_2&lt;&lt;c. The solution is invalid at relativistic speeds. Turns out that, as explained above, there are no symbolic solutions for this problem at relativistic speeds since the solution reduces to an equation degree 8. Numerical solutions exist, of course.

 

[starthaus]

I forgot, is that some kind of an expansion? I did learn something like that in class.

No, unfortunately it is the exact solution (a polinomial degree 8) for your problem. It is interesting indeed to see that such a simple problem has such a complicated solution.

I've seen that, its junk.

You are right, the solution works only at speeds <<C. So, the person who wrote the wiki page messed up.

 

[randombill]

No, unfortunately it is the exact solution (a polinomial degree 8) for your problem. It is interesting indeed to see that such a simple problem has such a complicated solution.

But how does someone go from solving for v2 to a solution for a polynomial. Or what steps would I take given what you've shown so far.

You are right, the solution works only at speeds <<C. So, the person who wrote the wiki page messed up.

And one of the references to a university page does not work.

 

[starthaus]

But how does someone go from solving for v2 to a solution for a polynomial. Or what steps would I take given what you've shown so far.

You need to form the equation, you haven't finished the calculations. Once you do that, there are sw packages that solve algebraic equations of higher order.

 

[randombill]

You need to form the equation, you haven't finished the calculations. Once you do that, there are sw packages that solve algebraic equations of higher order.

Well, if you're willing to type up the final solution that would be nice of you. I'm not sure where I went wrong the last time since I had trouble factoring out the gammas. What does the solution look like before using a computer algebra system?

 

[starthaus]

I showed you how to get \gamma_1=1/\sqrt{1-(v_1/c)^2} as a function f(v_2)

Now, substitute \gamma_1 in the equation B=m_1\gamma_1+m_2\gamma_2 and you will get an algebraic equation in v_2. Solve it and you will find v_2. Substitute v_2 in the expression for \gamma_1 and you will find v_1. Try learning latex, or, at least, print your solutions.

B=\frac{m_1}{\sqrt{1-(v_1/c)^2}}+\frac{m_2}{\sqrt{1-(v_2/c)^2}}

B-\frac{m_2}{\sqrt{1-(v_2/c)^2}}=\frac{m_1}{\sqrt{1-(v_1/c)^2}}

B^2+\frac{m_2^2}{1-(v_2/c)^2}-2B\frac{m_2}{\sqrt{1-(v_2/c)^2}}=\frac{m_1^2}{1-(v_1/c)^2}

B^2+\frac{m_2^2}{1-(v_2/c)^2}-\frac{m_1^2}{1-(v_1/c)^2}=2B\frac{m_2}{\sqrt{1-(v_2/c)^2}}

Square both sides one more time, replace v_1 by his expression in A,B and v_2 and you will get the algebraic equation in v_2 I told you about.

 

[yuiop]

This is the solution for the relativistic one dimensional elastic collision according to Wikipedia.

http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_relativistic

It should give you an idea of what to aim for.

I've seen that, its junk.

The wiki solution is valid only for speeds <<c. You can see that from the text, where they calculate p_T=m_1u_1+m_2u_2 valid only for u_1&lt;&lt;c and u_2&lt;&lt;c. The solution is invalid at relativistic speeds. Turns out that, as explained above, there are no symbolic solutions for this problem at relativistic speeds since the solution reduces to an equation degree 8. Numerical solutions exist, of course.

You are right, the solution works only at speeds <<C. So, the person who wrote the wiki page messed up.

You are both wrong. The wiki solution is a correct and exact symbolic solution for a 1D relativistic elastic collison of 2 unequal masses. I have checked it in a spreadsheet and it gives the correct answers for all masses and velocities. The <<C part comes after the exact solution where they compare the relativistic solution to the classical solution at low velocities where the two solutions should agree approximately.

The wiki solution is valid only for speeds <<c. You can see that from the text, where they calculate p_T=m_1u_1+m_2u_2 valid only for u_1&lt;&lt;c and u_2&lt;&lt;c.

They actually calculate:

P_T = \frac{m_1u_1}{\sqrt{1-u_1^2/c^2}} + \frac{m_2u_2}{\sqrt{1-u_2^2/c^2}} = \frac{m_1v_1}{\sqrt{1-v_1^2/c^2}} +\frac{m_2v_2}{\sqrt{1-v_2^2/c^2}}

Did you actually read the relativistic section of the Wiki article?

Using the information given in the Wiki article, the final velocities v1 and v2
can be calculated from the initial velocities, u1 and u2 using:

v_1 = \frac{c^2(2V_c-u_1)-u_1 V_c^2}{c^2+V_c(V_c -2u_1)}

v_2 = \frac{c^2(2V_c-u_2)-u_2 V_c^2}{c^2+V_c(V_c -2u_2)}

where the velocity of the centre of momentum frame is given by:

V_c = \frac{P_T c^2}{E_T}

and

E_T = \frac{m_1c^2}{\sqrt{1-u_1^2/c^2}} + \frac{m_2c^2}{\sqrt{1-u_2^2/c^2}} = \frac{m_1c^2}{\sqrt{1-v_1^2/c^2}} +\frac{m_2c^2}{\sqrt{1-v_2^2/c^2}}

These are exact symbolic solutions. No need for 8th degree polynomials.

For an example of a numerical solution (as in substituting numbers for the variables, rather than an iterative approximation)
to check your results against, consider the following initial conditions (using units of c=1):

m1 = 60
m2= 4
u1 = 0.8
u2 = -0.6

Initial total momentum P_T = 77
Initial total energy E_T = 105

then the velocity of the centre of mass frame is V_c = 0.733333 recurring
and the velocities after the elastic collison are:

v1 = 0.64878
v2 = 0.98824

Final total momentum P_T = 77
Final total energy E_T = 105

Note how the smaller mass has been accelerated to within 2% of the speed of light.

The total momentum and the total energy of the system, before and after the 1D elastic collision is conserved and everything is as it should be. When you or Starthaus get the correct solution, you should get the same results.

 

[starthaus]

You are both wrong. The wiki solution is a correct and exact symbolic solution for a 1D relativistic elastic collison of 2 unequal masses. I have checked it in a spreadsheet and it gives the correct answers for all masses and velocities. The <<C part comes after the exact solution where they compare the relativistic solution to the classical solution at low velocities where the two solutions should agree approximately.



They actually calculate:

P_T = \frac{m_1u_1}{\sqrt{1-u_1^2/c^2}} + \frac{m_2u_2}{\sqrt{1-u_2^2/c^2}} = \frac{m_1v_1}{\sqrt{1-v_1^2/c^2}} +\frac{m_2v_2}{\sqrt{1-v_2^2/c^2}}

Did you actually read the relativistic section of the Wiki article?

Yes, I did. Did you see where the author makes the approximation u_1&lt;&lt;c and u_2&lt;&lt;c? I bolded it for you. The author is also using :

v&#039;_1=-u&#039;_1
v&#039;_2=-u&#039;_2

with no proof whatsoever.
This means that he's calculating subrelativistic regimes, contrary to your claim.
By contrast, my solution is a simple algebraic derivation from base principles. So, it is correct by derivation.

 

[yuiop]

Yes, I did. Did you see where the authot makes the approximation u_1&lt;&lt;c and u_2&lt;&lt;c? I bolded it for you. The author is also using :

v&#039;_1=-u&#039;_1
v&#039;_2=-u&#039;_2

with no proof whatsoever.
This means that he's calculating subrelativistic regimes, contrary to your claim.

The author gives the final velocities as:

v_1 = \frac{v_1 &#039; _ V_c}{1+\frac{v_1 &#039; V_c}{c^2}}

v_2 = \frac{v_2 &#039; _ V_c}{1+\frac{v_2 &#039; V_c}{c^2}}

Since he has already given equations for v_1 &#039;, v_2 &#039; and V_c these are then the complete and exact symbolic solution to the problem of finding the final velocities. It is only after that, that he introduces "when u1 < < c and U2 << c," to discuss how the relativistic solutions reduce to the classical solution for low velocities.

The equations:

v&#039;_1=-u&#039;_1
v&#039;_2=-u&#039;_2

are justified by the assumption that momentum and energy are conserved in relativity just as in classical mechanins. The proof of that is in actual experiments conducted in particle accelerators.

The author arges it like this:

It is shown that u1 = − v1 (in the centre of momentum frame) remains true in relativistic calculation despite other differences. One of the postulates in Special Relativity states that the Laws of Physics should be invariant in all inertial frames of reference. That is, if total momentum is conserved in a particular inertial frame of reference, total momentum will also be conserved in any inertial frame of reference, although the amount of total momentum is frame-dependent. Therefore, by transforming from an inertial frame of reference to another, we will be able to get the desired results. In a particular frame of reference where the total momentum could be any,

My elaboration in bold parentheses. When you have gone through your lengthy derivations you will eventually, when you get it right, end up with the Wikipedia result.

You can not "prove" that momentum and energy must be conserved a priori, purely from mathematical considerations. At some point you have to actually observe the universe we live in and make some experimental measurements. Those experimental measurements suggest energy and momentum are conserved and with that knowledge you can deduce:

v&#039;_1=-u&#039;_1
v&#039;_2=-u&#039;_2

in the centre of momentum frame.

In your blog article on relativistic collisions you start the section on elastic collisions with equations 3.1 and 3.2 which clearly have the assumption (without proof) of conservation of relativistic momentum and conservation of relativistic energy already built into them. You are thertefore guilty of the same "crime" that you accuse the Wiki author of.

 

[starthaus]

The author gives the final velocities as:

v_1 = \frac{v_1 &#039; _ V_c}{1+\frac{v_1 &#039; V_c}{c^2}}

v_2 = \frac{v_2 &#039; _ V_c}{1+\frac{v_2 &#039; V_c}{c^2}}

No, he doesn't. He actually writes:

v_1 = \frac{v_1 &#039; + V_c}{1+\frac{v_1 &#039; V_c}{c^2}}

v_2 = \frac{v_2 &#039; + V_c}{1+\frac{v_2 &#039; V_c}{c^2}}

These are nothing but the inverse Lorentz transforms from COM frame to the original frame. These equations are devoid of any information per se.

Since he has already given equations for v_1 &#039;, v_2 &#039; and V_c these are then the complete and exact symbolic solution to the problem of finding the final velocities. It is only after that, that he introduces "when u1 < < c and U2 << c," to discuss how the relativistic solutions reduce to the classical solution for low velocities.

The equations:

v&#039;_1=-u&#039;_1
v&#039;_2=-u&#039;_2

are justified by the assumption that momentum and energy are conserved in relativity just as in classical mechanins. The proof of that is in actual experiments conducted in particle accelerators.

Nah, it is easy to show that both you and the author are wrong.

In COM:

v&#039;_1=\frac{v_1-v_c}{1-v_1v_c/c^2}

u&#039;_1=\frac{u_1-v_c}{1-u_1v_c/c^2}

When you claim :

v&#039;_1=-u&#039;_1

you get an equation in v_1

\frac{v_1-v_c}{1-v_1v_c/c^2}=-\frac{u_1-v_c}{1-u_1v_c/c^2}

The solution of the above equation contradicts the author's expression of v_1 as a function of u_1

This is what happens when one pulls equations out of ...thin air.

Those experimental measurements suggest energy and momentum are conserved and with that knowledge you can deduce:

v&#039;_1=-u&#039;_1
v&#039;_2=-u&#039;_2

in the centre of momentum frame.

Don't think so. See above.

 

[yuiop]

No, he doesn't. He actually writes:

v_1 = \frac{v_1 &#039; + V_c}{1+\frac{v_1 &#039; V_c}{c^2}}

v_2 = \frac{v_2 &#039; + V_c}{1+\frac{v_2 &#039; V_c}{c^2}}

These are nothing but the inverse Lorentz transforms from COM frame to the original frame. These equations are devoid of any information per se.

It is just that you do not not know how to extract the information. The overall algorithm is to transform to the COM frame where v1' = -u1' is valid, perform the substitution and then transform back to the original frame. Simple as that.

No
Nah, it is easy to show that both you and the author are wrong.

In COM:

v&#039;_1=\frac{v_1-v_c}{1-v_1v_c/c^2}

u&#039;_1=\frac{u_1-v_c}{1-u_1v_c/c^2}

When you claim :

v&#039;_1=-u&#039;_1

you get an equation in v_1

\frac{v_1-v_c}{1-v_1v_c/c^2}=-\frac{u_1-v_c}{1-u_1v_c/c^2}

The solution of the above equation contradicts the author's expression of v_1 as a function of u_1

This is what happens when one pulls equations out of ...thin air.

Again, I think you are having problems with simple algebra.

\frac{v_1-v_c}{1-v_1v_c/c^2}=-\frac{u_1-v_c}{1-u_1v_c/c^2}

Since the algebra is fairly lengthy, I will use v=v_1, u=u_1 and V = v_c to simplify the notation so that the above equation now reads:

\frac{v-V}{1-vV/c^2}=-\frac{u-V}{1-uV/c^2}

\Rightarrow \frac{v-V}{c^2-vV}=\frac{V-u}{c^2-uV}

\Rightarrow v-V = \frac{(V-u)(c^2-vV)}{(c^2-uV)}

\Rightarrow v-V = \frac{c^2(V-u)}{(c^2-uV)} - \frac{vV(V-u)}{(c^2-uV)}

\Rightarrow v + \frac{vV(V-u)}{(c^2-uV)}= V +\frac{c^2(V-u)}{(c^2-uV)}

\Rightarrow v(c^2-uV + V(V-u)) = V(c^2-uV) +c^2(V-u)

\Rightarrow v = \frac{V(c^2-uV) +c^2(V-u)}{c^2-uV+V(V-u)}

\Rightarrow v = \frac{2Vc^2-uV^2-uc^2}{c^2-2uV+V^2}

\Rightarrow v = \frac{c^2(2V-u)-uV^2}{c^2+V(V -2u)}

Which is the equation I gave earlier for v_x as a function of u_x:

...
Using the information given in the Wiki article, the final velocities v1 and v2
can be calculated from the initial velocities, u1 and u2 using:

v_1 = \frac{c^2(2V_c-u_1)-u_1 V_c^2}{c^2+V_c(V_c -2u_1)}

v_2 = \frac{c^2(2V_c-u_2)-u_2 V_c^2}{c^2+V_c(V_c -2u_2)}

where the velocity of the centre of momentum frame is given by:

V_c = \frac{P_T c^2}{E_T}

and

E_T = \frac{m_1c^2}{\sqrt{1-u_1^2/c^2}} + \frac{m_2c^2}{\sqrt{1-u_2^2/c^2}} = \frac{m_1c^2}{\sqrt{1-v_1^2/c^2}} +\frac{m_2c^2}{\sqrt{1-v_2^2/c^2}}

These are exact symbolic solutions. No need for 8th degree polynomials.

Note that the velocity V_c of the centre of momentum frame (COM) is a function of both u_1 and u_2.

If you have difficulties with basic algebra or simply to avoid errors, it is best to use symbolic algebraic software such as http://www.quickmath.com/webMathematica3/quickmath/page.jsp?s1=equations&s2=solve&s3=advanced

Simply copy and paste this expression:

(v-V)/(1-v*V/c^2)=-(u-V)/(1-u*V/c^2)

into the solver and enter v as the variable to solve for.

Just let me know when you need any more free algebra lessons

 

[starthaus]

It is just that you do not not know how to extract the information. The overall algorythm is to transform to the COM frame where v1' = -u1' is valid, perform the substitution and then then transform back to the original frame. Simple as that.

Yet, your claim is false. It is not clear why you insist on repeating such a false claim, especially in the face of you not being able to substantiate it.
So, once again, please prove that:

v&#039;_1=-u&#039;_1
v&#039;_2=-u&#039;_2

BTW, your entire solution is incorrect. I'll show you all the errors after you admit that the starting point of your derivation is incorrect (see above). Your claim is not only false in SR, it is also false in Newtonian mechanics.

 

[yuiop]

Yet, your claim is false. It is not clear why you insist on repeating such a false claim, especially in the face of you not being able to substantiate it.
So, once again, please prove that:

v&#039;_1=-u&#039;_1
v&#039;_2=-u&#039;_2

BTW, your entire solution is incorrect. I'll show you all the errors after we get afer you admit that the starting point of your derivation is incorrect (see above).

In the centre of momentum frame the total initial momentum is zero by definition. By the conservation of momentum principle the final momentum after the collision is also zero. The only solution that conserves the total momentum (and the total energy) after the collison is when v1 = -u1 and v2 = -u2. This is very well known from basic classical physics and is equally valid in relativistic physics. I do not know why you need me prove such a well known solution when you should already know it from a basic grounding in physics.

 

[starthaus]

In the centre of momentum frame the total initial momentum is zero by definition.

true

By the conservation of momentum principle the final momentum after the collision is also zero.

true

The only solution[/color] that conserves the total momentum (and the total energy) after the collison is when v1 = -u1 and v2 = -u2.

Prove it. Use math. Armwaving (even if repeated) does not count.
I'll make it easier for you, try solving the problem in the Newtonian approximation. Here:

m_1v_1+m_2v_2=m_1u_1+m_2u_2
m_1v_1^2+m_2v_2^2=m_1u_1^2+m_2u_2^2

 

[jtbell]

By definition the total momentum is zero in the COM frame before the collision. Therefore

<br /> P_{T(before)}^\prime = <br /> \frac {m_1 u_1^\prime} {\sqrt{ 1 - u_1^{\prime 2} / c^2} } +<br /> \frac {m_2 u_2^\prime} {\sqrt{ 1 - u_2^{\prime 2} / c^2} } = 0<br />

Now set up the total momentum in the COM frame after the collision, and make the substitutions v_1^\prime = -u_1^\prime and v_2^\prime = -u_2^\prime:

<br /> P_{T(after)}^\prime = <br /> \frac {m_1 v_1^\prime} {\sqrt{ 1 - v_1^{\prime 2} / c^2} } +<br /> \frac {m_2 v_2^\prime} {\sqrt{ 1 - v_2^{\prime 2} / c^2} } =<br /> <br /> \frac {m_1 (-u_1^\prime)} {\sqrt{ 1 - (-u_1^\prime)^2 / c^2} } +<br /> \frac {m_2 (-u_2^\prime)} {\sqrt{ 1 - (-u_2^\prime)^2 / c^2} }<br />

<br /> P_{T(after)}^\prime =<br /> - \frac {m_1 u_1^\prime} {\sqrt{ 1 - u_1^\prime^2 / c^2} }<br /> - \frac {m_2 u_2^\prime} {\sqrt{ 1 - u_2^\prime^2 / c^2} } =<br /> <br /> - \left( \frac {m_1 u_1^\prime} {\sqrt{ 1 - u_1^{\prime 2} / c^2} }<br /> + \frac {m_2 u_2^\prime} {\sqrt{ 1 - u_2^{\prime 2} / c^2} } \right) = 0<br />

Proceed similarly for the total energy.

 

[starthaus]

By definition the total momentum is zero in the COM frame before the collision. Therefore

<br /> P_{T(before)}^\prime = <br /> \frac {m_1 u_1^\prime} {\sqrt{ 1 - u_1^{\prime 2} / c^2} } +<br /> \frac {m_2 u_2^\prime} {\sqrt{ 1 - u_2^{\prime 2} / c^2} } = 0<br />

True

Now set up the total momentum in the COM frame after the collision, and make the substitutions v_1^\prime = -u_1^\prime and v_2^\prime = -u_2^\prime

Why would you do such a gratuitous thing? You are supposed to derive v&#039;_1 and v&#039;_2.
Nothing entitles you to set them by hand. Nor is v_1^\prime = -u_1^\prime and v_2^\prime = -u_2^\prime the only solution that conserves momentum after collision.

v&#039;_2=+u&#039;_2 and v&#039;_1=+u&#039;_1 are just as good.

In fact, there is a more general solution that does not guess the solution. The reason I am challenging so hard is that I have already derived a rigorous solution to the problem. I am interested in seeing how kev solves the simpler example for Newtonian mechanics. Let's see his solution to the simpler problem. I will post the correct results:

v_1=\frac{2m_2u_2+u_1(m_1-m_2)}{m_1+m_2}

v_2=\frac{2m_1u_1+u_2(m_2-m_1)}{m_1+m_2}

 

[jtbell]

We are assuming, hypothetically, that v_1^\prime = -u_1^\prime and v_2^\prime = -u_2^\prime give us the final velocities in terms of the initial velocities (in the COM frame). We verify that they are indeed the correct velocities by showing that they give us a final total momentum that equals the initial total momentum, and a final total energy that equals the initial total energy, thereby satisfying momentum and energy conservation. (Remember, all this is in the COM frame... to get the final solution, we still have to transform back to the lab frame.)

The solution must be unique, just as in classical mechanics, because with the masses and initial velocities given, we have two equations for two unknowns (the final velocities).

 

[starthaus]

We are assuming, hypothetically, that v_1^\prime = -u_1^\prime and v_2^\prime = -u_2^\prime give us the final velocities in terms of the initial velocities (in the COM frame).

This is the crux of my disagreement with the wiki solution (which kev has copied without any attempt to analyze for correctness).

We verify that they are indeed the correct velocities by showing that they give us a final total momentum that equals the initial total momentum, and a final total energy that equals the initial total energy, thereby satisfying momentum and energy conservation.

Why these values and not other values? In short, physics is not a guesswork, I can show you (in the end, after I see what kev does for a solution) how this problem is solved without such gratuitous assumptions. Let's wait for kev to solve the exercise above, agreed?

 

[yuiop]

Why would you do such a gratuitous thing? You are supposed to derive v&#039;_1 and v&#039;_2.
Nothing entitles you to set them by hand. Nor is v_1^\prime = -u_1^\prime and v_2^\prime = -u_2^\prime the only solution that conserves momentum after collision. In fact, there is a more general solution. The reason I am challenging so hard is that I have already derived a rigorous solution to the problem. I am interested in seeing how kev solves the simpler example for Newtonian mechanics. Let's see his solution to the simpler problem.

There is only one unique solution (with two roots) that conserves momentum AND energy at the same time. To satisfy both equations the techniques of simultaneous equations must be used.

Prove it. Use math. Armwaving (even if repeated) does not count.
I'll make it easier for you, try solving the problem in the Newtonian approximation. Here:

m_1v_1+m_2v_2=m_1u_1+m_2u_2
m_1v_1^2+m_2v_2^2=m_1u_1^2+m_2u_2^2

In the COM frame:

An obvious solution to the above equations is v_1 = u1 and v_2 = u_2 which is trivially true before the collision, or if the collision never happens because the particles are moving in the same direction with the slower one behind. For the less trivial solution do the following.

Using the conservation of momentum equation:

m_1v_1+m_2v_2= 0 \quad \Rightarrow m_1v_1 = -m_2v_2 \quad \Rightarrow (m_1v_1)^2 = (m_2v_2)^2 \qquad \qquad (1)

m_1u_1+m_2u_2= 0 \quad \Rightarrow m_1u_1 = -m_2u_2 \quad \Rightarrow (m_1u_1)^2 = (m_2u_2)^2 \qquad \qquad (2)

Using the conservation of energy equation:

m_1v_1^2+m_2v_2^2=m_1u_1^2+m_2u_2^2

\Rightarrow \frac{(m_1v_1)^2}{m_1}+\frac{(m_2v_2)^2}{m_2}=\frac{(m_1u_1)^2}{m_1}+\frac{(m_2u_2)^2}{m_2}

Now substitute in equations (1) and (2):

\Rightarrow \frac{(m_2v_2)^2}{m_1}+\frac{(m_2v_2)^2}{m_2}=\frac{(m_2u_2)^2}{m_1}+\frac{(m_2u_2)^2}{m_2}

\Rightarrow (m_2 +m_1)(m_2v_2)^2= (m_2+m_1)(m_2 u_2)^2

\Rightarrow (m_2v_2)^2= (m_2 u_2)^2

\Rightarrow v_2= \ \pm \ u_2

There are two roots, equal in magnitude but one positive and one negative. The positive root is the trivial solution before the collision. You can use the same method to determine that the there are only two possible roots to the solution for v_1 that satisfy both conservation equations simultaneously:

v_1= \ \pm \ u_1

However, we can do better than that.

Using the negative root v_2 = -u_2 and the conservation of momentum equation:

m_1v_1+m_2v_2= 0 \qquad \Rightarrow m_1v_1 -m_2u_2 = 0 \qquad \Rightarrow m_1v_1 = m_2u_2

m_1u_1+m_2u_2= 0 \qquad \Rightarrow m_1u_1 = -m_2u_2 \qquad \Rightarrow -m_1u_1 = m_2u_2

it is obvious that when v_2 = -u_2 there is a unique solution v_1 = - u_1

So, once again, please prove that:

v&#039;_1=-u&#039;_1
v&#039;_2=-u&#039;_2

BTW, your entire solution is incorrect. I'll show you all the errors after you admit that the starting point of your derivation is incorrect (see above). Your claim is not only false in SR, it is also false in Newtonian mechanics.

I have now proven mathematically that your assertion about the Newtonian solution is false.

If you can not handle the basic physics and algebra of classical mechanics, how can you hope to handle the more complicated relativistic calculations?

Also, because you have a habit of ignoring the context, you should be clear that:

v&#039;_1=-u&#039;_1
v&#039;_2=-u&#039;_2

is only generally true in the COM frame.

That is why we go to the trouble of transforming to the COM frame and back again.

 

[jtbell]

Why these values and not other values?

In the COM frame, the total momentum is zero before the collision. Therefore p_{1,before}^\prime = m_1 \gamma(u_1^\prime) u_1^\prime and p_{2,before}^\prime = m_2 \gamma(u_2^\prime) u_2^\prime have equal magnitudes and opposite signs (directions).

We require that the total momentum be conserved, so the total momentum must also be zero after the collision. Therefore p_{1,after}^\prime = m_1 \gamma(v_1^\prime) v_1^\prime and p_{2,after}^\prime = m_2 \gamma(v_2^\prime) v_2^\prime have equal magnitudes and opposite signs (directions).

There are three possibilities for |v_1^\prime|, according to whether it is equal to, greater than, or less than |u_1^\prime|. Let's examine each of these cases in turn:

Case 1: |v_1^\prime| = |u_1^\prime|.

This breaks down into two sub-cases:

Case 1A: v_1^\prime = u_1^\prime. This leads to v_2^\prime = u_2^\prime. It represents the trivial case in which the two objects "pass through" each other without affecting each other's energy or momentum.

Case 1B: v_1^\prime = -u_1^\prime. This leads to v_2^\prime = -u_2^\prime. I've already showed that this case conserves both total momentum and total energy.

Case 2: |v_1^\prime| &gt; |u_1^\prime|.

In this case |p_{1,after}^\prime| &gt; |p_{1,before}^\prime|, because p increases monotonically with v when m is held constant. Also, because

E = \sqrt{(pc)^2 + (mc^2)^2}

we must also have E_{1,after}^\prime > E_{1,before}^\prime (E increases monotonically with P when m is held constant).

In order to preserve conservation of momentum, we must also have, simllarly:

|v_2^\prime| &gt; |u_2^\prime|

|p_{2,after}^\prime| &gt; |p_{2,before}^\prime|

E_{2,after}^\prime[/itex] &gt; E_{2,before}^\prime<br /> <br /> But this would make<br /> <br /> E_{total,after}^\prime = E_{1,after}^\prime + E_{2,after}^\prime<br /> <br /> E_{total,after}^\prime &amp;gt; E_{1,before}^\prime + E_{2,before}^\prime<br /> <br /> E_{total,after}^\prime &amp;gt; E_{total,before}^\prime<br /> <br /> which violates conservation of energy. So case 2 is ruled out.<br /> <br /> <b>Case 3:</b> |v_1^\prime| &amp;lt; |u_1^\prime|.<br /> <br /> By a similar argument to Case 2, we can show that this leads to<br /> <br /> E_{total,after}^\prime &amp;lt; E_{total,before}^\prime<br /> <br /> which violates conservation of energy. So Case 3 is ruled out.<br /> <br /> The only non-trivial case that satisfies both conservation of momentum and conservation of energy is therefore Case 1B, in which<br /> <br /> v_1^\prime = -u_1^\prime<br /> <br /> v_2^\prime = -u_2^\prime

 

[starthaus]

There is only one unique solution (with two roots) that conserves momentum AND energy at the same time. To satisfy both equations the techniques of simultaneous equations must be used.
In the COM frame:

An obvious solution to the above equations is v_1 = u1 and v_2 = u_2 which is trivially true before the collision, or if the collision never happens because the particles are moving in the same direction with the slower one behind. For the less trivial solution do the following.

Using the conservation of momentum equation:

m_1v_1+m_2v_2= 0 \quad \Rightarrow m_1v_1 = -m_2v_2 \quad \Rightarrow (m_1v_1)^2 = (m_2v_2)^2 \qquad \qquad (1)

m_1u_1+m_2u_2= 0 \quad \Rightarrow m_1u_1 = -m_2u_2 \quad \Rightarrow (m_1u_1)^2 = (m_2u_2)^2 \qquad \qquad (2)

Using the conservation of energy equation:

m_1v_1^2+m_2v_2^2=m_1u_1^2+m_2u_2^2

\Rightarrow \frac{(m_1v_1)^2}{m_1}+\frac{(m_2v_2)^2}{m_2}=\frac{(m_1u_1)^2}{m_1}+\frac{(m_2u_2)^2}{m_2}

Now substitute in equations (1) and (2):

\Rightarrow \frac{(m_2v_2)^2}{m_1}+\frac{(m_2v_2)^2}{m_2}=\frac{(m_2u_2)^2}{m_1}+\frac{(m_2u_2)^2}{m_2}

\Rightarrow (m_2 +m_1)(m_2v_2)^2= (m_2+m_1)(m_2 u_2)^2

\Rightarrow (m_2v_2)^2= (m_2 u_2)^2

\Rightarrow v_2= \ \pm \ u_2

There are two roots, equal in magnitude but one positive and one negative. The positive root is the trivial solution before the collision. You can use the same method to determine that the there are only two possible roots to the solution for v_1 that satisfy both conservation equations simultaneously:

v_1= \ \pm \ u_1

However, we can do better than that.

Using the negative root v_2 = -u_2 and the conservation of momentum equation:

m_1v_1+m_2v_2= 0 \qquad \Rightarrow m_1v_1 -m_2u_2 = 0 \qquad \Rightarrow m_1v_1 = m_2u_2

m_1u_1+m_2u_2= 0 \qquad \Rightarrow m_1u_1 = -m_2u_2 \qquad \Rightarrow -m_1u_1 = m_2u_2

it is obvious that when v_2 = -u_2 there is a unique solution v_1 = - u_1I have now proven mathematically that your assertion about the Newtonian solution is false.

You don't have yet any solution to the problem. All you have done is that you have now established (after much prodding) that

v&#039;_1=-u&#039;_1

I can see two problems with all this work:

1. You aren't any closer to finding the solution to the Newtonian exercise (post 44). How will you use your trick to find the solutions?

2. You have blindly used the relationship v&#039;_1=-u&#039;_1 for the relativistic case. But, you have not established that the relationship is valid in SR as well. Can you prove it?

3. I have already solved both the Newtonian and the realativistic case without using the trick.

If you can not handle the basic physics and algebra of classical mechanics, how can you hope to handle the more complicated relativistic calculations?

Don't worry about my mathematical abilities, we have ample proof from prior encounters that they are much higher than yours. Worry about solving the problem.

 

[yuiop]

I can see two problems with all this work:

1. You aren't any closer to finding the solution to the Newtonian exercise (post 44). How will you use your trick to find the solutions?

This is a relativity forum. I am not going to spend of lot of time formulating the latex for the Newtonian solution for you when the result and derivation is readily available in any elementary classical physics book and many online texts. See for example http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_Newtonian

2. You have blindly used the relationship v&#039;_1=-u&#039;_1 for the relativistic case. But, you have not established that the relationship is valid in SR as well. Can you prove it?

Here is the intuitive way of proving that v&#039;_1=-u&#039;_1 and v&#039;_2=-u&#039;_2 holds in the COM frame in SR.

The conservation of energy equation in the COM frame is:

m_1 c^2\gamma(u&#039;_1)+m_2 c^2 \gamma(u&#039;_2) = m_1 c^2 \gamma(v&#039;_1) +m_2 c^2 \gamma(v&#039;_2)

Now we see if v&#039;_1=-u&#039;_1 and v&#039;_2=-u&#039;_2 is a correct solution to the above equation by performing the substitutions:

\Rightarrow m_1 c^2\gamma(u&#039;_1)+m_2 c^2\gamma(u&#039;_2) = m_1 c^2 \gamma(-u&#039;_1) +m_2 c^2 \gamma(-u&#039;_2)

Since \gamma(u&#039;_1) = \gamma(-u&#039;_1) and \gamma(u&#039;_2) = \gamma(-u&#039;_2) is always true:

\Rightarrow m_1 c^2\gamma(u&#039;_1)+m_2 c^2\gamma(u&#039;_2) = m_1 c^2 \gamma(u&#039;_1) +m_2 c^2 \gamma(u&#039;_2)

The above is obviously a true statement so v&#039;_1=-u&#039;_1 and v&#039;_2=-u&#039;_2 is a solution to the conservation of energy equation.

The conservation of momentum equations in the COM frame in SR are:

m_1 u&#039;_1 \gamma(u&#039;_1) +m_2 u&#039;_2 \gamma(u&#039;_2) = 0 \qquad \qquad (1)
m_1 v&#039;_1 \gamma(v&#039;_1) +m_2 v&#039;_2 \gamma(v&#039;_2) = 0 \qquad \qquad (2)

Now substituting v&#039;_1=-u&#039;_1 and v&#039;_2=-u&#039;_2 into (2) gives:

-m_1 u&#039;_1 \gamma(-u&#039;_1) -m_2 u&#039;_2 \gamma(-u&#039;_2) = 0

\Rightarrow -m_1 u&#039;_1 \gamma(u&#039;_1) -m_2 u&#039;_2 \gamma(u&#039;_2) = 0

\Rightarrow m_1 u&#039;_1 \gamma(u&#039;_1) +m_2 u&#039;_2 \gamma(u&#039;_2) = 0

which is the same as equation (1) proving that v&#039;_1=-u&#039;_1 and v&#039;_2=-u&#039;_2 is also a valid solution to the conservation of momentum equations in the COM frame. Now is this the unique solution to both the conservation of energy and momentum equations in the COM? (besides the trivial solution v&#039;_1=u&#039;_1 and v&#039;_2=u&#039;_2) The answer is yes. Intuition tells you that after the collison there is only one possible unique solution or the result of collisions in nature would be random and experimental evidence suggests this is not the case.

(QED)

The above proof seems fairly lengthy, but it is simple enough that you can do it in your head without writing anything down. In that sense it is intuitively obvious to myself and JTBell. He has also posted some proofs for you. Have you read any of them?

3. I have already solved both the Newtonian and the realativistic case without using the trick.

Well you have quoted the Newtonian solution (without proof) out of a standard textbook. 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.

Don't worry about my mathematical abilities, we have ample proof from prior encounters that they are much higher than yours. Worry about solving the problem.

We have ample proof that you have problems with basic physics and algebra and also made some major calculus blunders like the barbaric claim that dx/dt = 0 \Rightarrow d^2x/dt^2 =0 (that you still defend). We also have ample proof that you spend about about 200 posts trying to prove equations I have posted in other threads are wrong and ultimately failing. In this thread alone you have made a series of false assertions that have been proven wrong.

The wiki solution is valid only for speeds <<c. You can see that from the text, where they calculate p_T=m_1u_1+m_2u_2 valid only for u_1&lt;&lt;c and u_2&lt;&lt;c. The solution is invalid at relativistic speeds. Turns out that, as explained above, there are no symbolic solutions for this problem at relativistic speeds since the solution reduces to an equation degree 8. Numerical solutions exist, of course.

Proven wrong. You misread/did not understand the Wiki solution which has been proven to be correct. Your claim that "there are no symbolic solutions for this problem at relativistic speeds since the solution reduces to an equation degree 8" has also been proven wrong. An exact symbolic solution was given in URL="https://www.physicsforums.com/showpost.php?p=2805545&postcount=35"]post # 35 [/URL]and the elaboration in https://www.physicsforums.com/showpost.php?p=2806411&postcount=39"that fills in the details for you.

No, unfortunately it is the exact solution (a polinomial degree 8) for your problem. It is interesting indeed to see that such a simple problem has such a complicated solution.
...
You are right, the solution works only at speeds <<C. So, the person who wrote the wiki page messed up.

Wrong. See above.

\frac{v_1-v_c}{1-v_1v_c/c^2}=-\frac{u_1-v_c}{1-u_1v_c/c^2}

The solution of the above equation contradicts the author's expression of v_1 as a function of u_1

This is what happens when one pulls equations out of ...thin air.

Proven wrong. https://www.physicsforums.com/showpost.php?p=2806411&postcount=39".

Yet, your claim is false. It is not clear why you insist on repeating such a false claim, especially in the face of you not being able to substantiate it.
So, once again, please prove that:

v&#039;_1=-u&#039;_1
v&#039;_2=-u&#039;_2

BTW, your entire solution is incorrect. I'll show you all the errors after you admit that the starting point of your derivation is incorrect (see above). Your claim is not only false in SR, it is also false in Newtonian mechanics.

Your assertion that the above relationship does not hold in even in Newtonian mechanics has been proven wrong. https://www.physicsforums.com/showpost.php?p=2806645&postcount=47"