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 $$u_1<<c$$ and $$u_2<<c$$. 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 $$u_1<<c$$ and $$u_2<<c$$. 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 $$u_1<<c$$ and $$u_2<<c$$.

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.