One dimensional collision of 2 unequal masses

[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 $$u_1<<c$$ and $$u_2<<c$$? I bolded it for you. The author is also using :

$$v'_1=-u'_1$$
$$v'_2=-u'_2$$

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 $$u_1<<c$$ and $$u_2<<c$$? I bolded it for you. The author is also using :

$$v'_1=-u'_1$$
$$v'_2=-u'_2$$

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

The author gives the final velocities as:

$$v_1 = \frac{v_1 ' _ V_c}{1+\frac{v_1 ' V_c}{c^2}}$$

$$v_2 = \frac{v_2 ' _ V_c}{1+\frac{v_2 ' V_c}{c^2}}$$

Since he has already given equations for $v_1 ', v_2 '$ 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:

$$v'_1=-u'_1$$
$$v'_2=-u'_2$$

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:

$$v'_1=-u'_1$$
$$v'_2=-u'_2$$

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:

$$v_1 = \frac{v_1 ' _ V_c}{1+\frac{v_1 ' V_c}{c^2}}$$

$$v_2 = \frac{v_2 ' _ V_c}{1+\frac{v_2 ' V_c}{c^2}}$$

No, he doesn't. He actually writes:

$$v_1 = \frac{v_1 ' + V_c}{1+\frac{v_1 ' V_c}{c^2}}$$

$$v_2 = \frac{v_2 ' + V_c}{1+\frac{v_2 ' V_c}{c^2}}$$

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 $v_1 ', v_2 '$ 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:

$$v'_1=-u'_1$$
$$v'_2=-u'_2$$

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:

$$v'_1=\frac{v_1-v_c}{1-v_1v_c/c^2}$$

$$u'_1=\frac{u_1-v_c}{1-u_1v_c/c^2}$$

When you claim :

$$v'_1=-u'_1$$

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:

$$v'_1=-u'_1$$
$$v'_2=-u'_2$$

in the centre of momentum frame.

Don't think so. See above.

 

[yuiop]

No, he doesn't. He actually writes:

$$v_1 = \frac{v_1 ' + V_c}{1+\frac{v_1 ' V_c}{c^2}}$$

$$v_2 = \frac{v_2 ' + V_c}{1+\frac{v_2 ' V_c}{c^2}}$$

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:

$$v'_1=\frac{v_1-v_c}{1-v_1v_c/c^2}$$

$$u'_1=\frac{u_1-v_c}{1-u_1v_c/c^2}$$

When you claim :

$$v'_1=-u'_1$$

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:

$$v'_1=-u'_1$$
$$v'_2=-u'_2$$

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:

$$v'_1=-u'_1$$
$$v'_2=-u'_2$$

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 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

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

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

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

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

Proceed similarly for the total energy.

 

[starthaus]

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

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

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 $$v'_1$$ and $$v'_2$$.
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.

$$v'_2=+u'_2$$ and $$v'_1=+u'_1$$ 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 $$v'_1$$ and $$v'_2$$.
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:

$$v'_1=-u'_1$$
$$v'_2=-u'_2$$

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:

$$v'_1=-u'_1$$
$$v'_2=-u'_2$$

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: $|v_1^\prime| > |u_1^\prime|$.

In this case $|p_{1,after}^\prime| > |p_{1,before}^\prime|$, 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:

$$|v_2^\prime| > |u_2^\prime|$$

$$|p_{2,after}^\prime| > |p_{2,before}^\prime|$$

[tex]E_{2,after}^\prime[/itex] > [itex]E_{2,before}^\prime[/tex]

But this would make

$$E_{total,after}^\prime = E_{1,after}^\prime + E_{2,after}^\prime$$

$$E_{total,after}^\prime > E_{1,before}^\prime + E_{2,before}^\prime$$

$$E_{total,after}^\prime > E_{total,before}^\prime$$

which violates conservation of energy. So case 2 is ruled out.

Case 3: $|v_1^\prime| < |u_1^\prime|$.

By a similar argument to Case 2, we can show that this leads to

$$E_{total,after}^\prime < E_{total,before}^\prime$$

which violates conservation of energy. So Case 3 is ruled out.

The only non-trivial case that satisfies both conservation of momentum and conservation of energy is therefore Case 1B, in which

$$v_1^\prime = -u_1^\prime$$

$$v_2^\prime = -u_2^\prime$$

 

[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 $v_1 = - u_1$I 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

$$v'_1=-u'_1$$

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 $$v'_1=-u'_1$$ 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 $$v'_1=-u'_1$$ 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 $v'_1=-u'_1$ and $v'_2=-u'_2$ holds in the COM frame in SR.

The conservation of energy equation in the COM frame is:

$$m_1 c^2\gamma(u'_1)+m_2 c^2 \gamma(u'_2) = m_1 c^2 \gamma(v'_1) +m_2 c^2 \gamma(v'_2)$$

Now we see if $v'_1=-u'_1$ and $v'_2=-u'_2$ is a correct solution to the above equation by performing the substitutions:

$$\Rightarrow m_1 c^2\gamma(u'_1)+m_2 c^2\gamma(u'_2) = m_1 c^2 \gamma(-u'_1) +m_2 c^2 \gamma(-u'_2)$$

Since $\gamma(u'_1) = \gamma(-u'_1)$ and $\gamma(u'_2) = \gamma(-u'_2)$ is always true:

$$\Rightarrow m_1 c^2\gamma(u'_1)+m_2 c^2\gamma(u'_2) = m_1 c^2 \gamma(u'_1) +m_2 c^2 \gamma(u'_2)$$

The above is obviously a true statement so $v'_1=-u'_1$ and $v'_2=-u'_2$ is a solution to the conservation of energy equation.

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

$$m_1 u'_1 \gamma(u'_1) +m_2 u'_2 \gamma(u'_2) = 0 \qquad \qquad (1)$$
$$m_1 v'_1 \gamma(v'_1) +m_2 v'_2 \gamma(v'_2) = 0 \qquad \qquad (2)$$

Now substituting $v'_1=-u'_1$ and $v'_2=-u'_2$ into (2) gives:

$$-m_1 u'_1 \gamma(-u'_1) -m_2 u'_2 \gamma(-u'_2) = 0$$

$$\Rightarrow -m_1 u'_1 \gamma(u'_1) -m_2 u'_2 \gamma(u'_2) = 0$$

$$\Rightarrow m_1 u'_1 \gamma(u'_1) +m_2 u'_2 \gamma(u'_2) = 0$$

which is the same as equation (1) proving that $v'_1=-u'_1$ and $v'_2=-u'_2$ 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 $v'_1=u'_1$ and $v'_2=u'_2$) 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 $$u_1<<c$$ 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.

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:

$$v'_1=-u'_1$$
$$v'_2=-u'_2$$

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"

 

[starthaus]

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

In other words, you are unable to solve this simple problem. (the wiki page doesn't show it so you don't have a readily available place for "cut and paste").

Here is the intuitive way of proving that $v'_1=-u'_1$ and $v'_2=-u'_2$ holds in the COM frame in SR.

The conservation of energy equation in the COM frame is:

$$m_1 c^2\gamma(u'_1)+m_2 c^2 \gamma(u'_2) = m_1 c^2 \gamma(v'_1) +m_2 c^2 \gamma(v'_2)$$

Now we see if $v'_1=-u'_1$ and $v'_2=-u'_2$ is a correct solution to the above equation by performing the substitutions:

$$\Rightarrow m_1 c^2\gamma(u'_1)+m_2 c^2\gamma(u'_2) = m_1 c^2 \gamma(-u'_1) +m_2 c^2 \gamma(-u'_2)$$

Since $\gamma(u'_1) = \gamma(-u'_1)$ and $\gamma(u'_2) = \gamma(-u'_2)$ is always true:

$$\Rightarrow m_1 c^2\gamma(u'_1)+m_2 c^2\gamma(u'_2) = m_1 c^2 \gamma(u'_1) +m_2 c^2 \gamma(u'_2)$$

The above is obviously a true statement so $v'_1=-u'_1$ and $v'_2=-u'_2$ is a solution to the conservation of energy equation.

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

$$m_1 u'_1 \gamma(u'_1) +m_2 u'_2 \gamma(u'_2) = 0 \qquad \qquad (1)$$
$$m_1 v'_1 \gamma(v'_1) +m_2 v'_2 \gamma(v'_2) = 0 \qquad \qquad (2)$$

Now substituting $v'_1=-u'_1$ and $v'_2=-u'_2$ into (2) gives:

$$-m_1 u'_1 \gamma(-u'_1) -m_2 u'_2 \gamma(-u'_2) = 0$$

$$\Rightarrow -m_1 u'_1 \gamma(u'_1) -m_2 u'_2 \gamma(u'_2) = 0$$

$$\Rightarrow m_1 u'_1 \gamma(u'_1) +m_2 u'_2 \gamma(u'_2) = 0$$

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?
Can you derive the value for $$V_{COM}$$? Without it , the solution on the wiki page is useless.

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?

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.

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

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. 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?

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.

LOL.

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).

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.

 

[yuiop]

In other words, you are unable to solve this simple problem. (the wiki page doesn't show it so you don't have a readily available place for "cut and paste").

The Newtonian derivation is hinted at in the Wikipedia page, but you seem unable to comprehend it, so I will fill in the gaps and spell it out for you step by step.

Step 1: Rearrange the conservation of energy equation:

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

$$\Rightarrow m_1(v_1-u_1)(v_1+u_1) = m_2(u_2-v_2)(u_2+v_2) \qquad \qquad (eq1)$$

Step 2: Rearrange the conservation of momentum equation:

$$m_1v_1+m_2v_2=m_1u_1 + m_2u_2$$

$$\Rightarrow m_1(v_1-u_1)=m_2(u_2-v_2) \qquad \qquad (eq2)$$

Step 3: Divide (eq1) by (eq2). This is basic simultaneous equations. Remember them or were you playing hooky that day?

$$(v_1+u_1)=(u_2+v_2)$$

$$\Rightarrow v_1=v_2+u_2-u_1 \qquad \qquad (eq3)$$

$$\Rightarrow v_2=v_1+u_1-u_2) \qquad \qquad (eq4)$$

Step 4: Now substitute (eq4) back into (eq2) and solve for v1:

$$m_1(v_1-u_1)=m_2(u_2-v_1-u_1+u_2)$$

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

Similarly substitute (eq3) back into (eq2) and solve for v2:

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

Clear now? 4 easy steps compared to the 12 convoluted steps in your blog solution.

 

[starthaus]

The Newtonian derivation is hinted at in the Wikipedia page, but you seem unable to comprehend it, so I will fill in the gaps and spell it out for you step by step.

Step 1: Rearrange the conservation of energy equation:

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

$$\Rightarrow m_1(v_1-u_1)(v_1+u_1) = m_2(u_2-v_2)(u_2+v_2) \qquad \qquad (eq1)$$

Step 2: Rearrange the conservation of momentum equation:

$$m_1v_1+m_2v_2=m_1u_1 + m_2u_2$$

$$\Rightarrow m_1(v_1-u_1)=m_2(u_2-v_2) \qquad \qquad (eq2)$$

Step 3: Divide (eq1) by (eq2). This is basic simultaneous equations. Remember them or were you playing hooky that day?

Err, you can't do that because in certain frames, like COM $$v_1=u_1$$, remember? So, your hack doesn't work when you blindly try to do the division. This is thought in the beginner algebra classes. You may have to go all the way back and retake them.

$$(v_1+u_1)=(u_2+v_2)$$

$$\Rightarrow v_1=v_2+u_2-u_1 \qquad \qquad (eq3)$$

$$\Rightarrow v_2=v_1+u_1-u_2) \qquad \qquad (eq4)$$

Step 4: Now substitute (eq4) back into (eq2) and solve for v1:

$$m_1(v_1-u_1)=m_2(u_2-v_1-u_1+u_2)$$

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

Similarly substitute (eq3) back into (eq2) and solve for v2:

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

Clear now? 4 easy steps compared to the 12 convoluted steps in you blog solution.

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?

 

[yuiop]

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.

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:

$$\frac{P}{E}c^2 = \frac{M\gamma(V_{COM})V_{COM}}{M\gamma(V_{COM})c^2}c^2 = V_{COM}$$

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.

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.

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.

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.

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.

LOL.

YOU have still not posted your derivation or solution for v1 and v2 in the relativistic case.

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.

 

[yuiop]

Err, you can't do that because in certain frames, like COM $$v_1=u_1$$, remember? So, your hack doesn't work when you blindly try to do the division. This is (sic)thought in the beginner algebra classes. You may have to go all the way back and retake them.

If you happen to be in the COM frame, then yes the result is 0/0 and v1 and v2 can not be determined using this method, but then we already know that in the COM, v1= -u1 and v2 = -u2 and so the methoid is not required. For any other reference frame other than the COM the method gives the correct results. The method is pretty standard in any basic classical physics textbook. Perhaps you should buy one.

In the COM frame use:

$$v_1 = -u_1$$

$$v_2 = -u_2$$

In any other frame use:

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

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

 

[starthaus]

Wrong. See below.
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.

Err, this is not a valid proof, it s another one of your hacks. Remember, the COM is defined as the frame in which the total momentum is null. It is from this information you needed to determine $$V_{COM}$$.
I gave you a hint how this is done rigorously in the attachment dealing with the Newtonian case. You have read the attachment already, right? :-)

Once you have obtained $V_{COM} = Pc^2/E$ then it is trivial to obtain:

$$V_{COM} = \frac{m_1u_1\gamma(u_1)+m_2 u_2 \gamma(u_2)}{m_1c^2\gamma(u_1) + m_2c^2\gamma(u_2)}$$

Err, you got this one wrong, do you want another crack at it? Another "chancie"?

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.

This is not a proof, this is just aniother hack, exactly as in the case of the SR case above. You need to use the COM definition, which you did not.

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.

I don't use the property $$|v'_i|=|u'_i|$$ in my proof <shrug>. Nevertheless, I can produce a proof that is different from jtbell's and it is, unlike yours, rigorous.

There is no point in my duplicating jtbell's work. We have different styles and that makes the world a more interesting place.

Yes, he has rigor, you are just hacking.

Assuming something and then verifying it is correct is good enough for me.

Precisely: the difference between a scientist and a hacker.

How sad it is that despite being shown simple counterproofs to your assertion by various people,

...who are as bad as you at calculus.

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 $f(t)=x^2$.

$f(t)=x^2$? LOL

When x=0, $dx/dt = 2x = 0$

This is getting worse. You do not understand basic function theory, let alone differentiation of functions.

and $d^2x/dt^2 = 2$

...and worse.

so your assertion that $dx/dt = 0 \Rightarrow d^2x/dt^2 =0$ is proven false.

I think that you truly need to take an introductory calculus class (high school would be the appropriate level), there is no other way around it.

 

[yuiop]

Err, this is not a valid proof, it s another one of your hacks. Remember, the COM is defined as the frame in which the total momentum is null. It is from this information you needed to determine $$V_{COM}$$.
I gave you a hint how this is done rigorously in the attachment dealing with the Newtonian case. You have read the attachment already, right? :-)

You have yet to demonstrate that you can derive $V_{COM}$ or the final velocities v1 or v2 in the relativistic case.

Once you have obtained $V_{COM} = Pc^2/E$ then it is trivial to obtain:

$$V_{COM} = \frac{m_1u_1\gamma(u_1)+m_2 u_2 \gamma(u_2)}{m_1c^2\gamma(u_1) + m_2c^2\gamma(u_2)}$$

Err, you got this one wrong, do you want another crack at it? Another "chancie"?

I think it obvious that I accidently left out the c^2 factor and effectively gave $V_{COM} = P/E$ rather than the $V_{COM} = Pc^2/E$ that I obviously intended. The final equation when corrected for the typo should have read:

$$V_{COM} = \frac{m_1u_1\gamma(u_1)+m_2 u_2 \gamma(u_2)}{m_1\gamma(u_1) + m_2\gamma(u_2)}$$

Other than the typo, the derivation I gave of $V_{COM} = Pc^2/E$ is rigorous enough, while your derivation does not exist.

In the Newtonian case the total mass is simply the sum of the individual masses so the total momentum is given by:

$$(m_1 + m_2) V_{COM} = m_1u_1 + m_2u_2$$

$$V_{COM} = \frac{ m_1u_1 + m_2u_2}{ m_1 + m_2}$$

This is not a proof, this is just aniother hack, exactly as in the case of the SR case above. You need to use the COM definition, which you did not.

You do not realize it, but the above equations ARE a definition of the velocity of the COM frame. The equation above essentially says the velocity of the COM frame is the total mometum of the system, divided by the total mass of the system, which is a definition of $V_{COM}$. You do NOT need the definition that the total momentum is zero in the COM frame, in order to derive the velocity of the COM frame (which is also zero in the COM frame by defintion). Let's see your SR derivation and see how it is better. You call me a hacker, but I always produce correct results long before you do. Like I said, I am sure you will find a more complicated way to do it.

I don't use the property $$|v'_i|=|u'_i|$$ in my proof <shrug>. Nevertheless, I can produce a proof that is different from jtbell's and it is, unlike yours, rigorous.

So let's see you relativistic solution of $$|v'_i|=|u'_i|$$ and your relativistic solution for the final velocities v1 and v2 without using $$|v'_i|=|u'_i|$$ and quit bluffing.

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.

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.

$f(t)=x^2$? LOL

I think it is obvious that I meant to say if $f(t)=t^2$ then when t=0,

$$f(t) = t^2 = 0,$$

$$\frac{d}{dt}f(t) = 2t = 0$$

and

$$\frac{d^2}{dt^2}f(t) = 2$$

so your assertion that $df/dt = 0 \Rightarrow d^2f/dt^2 =0$ is not generally true.

...who are as bad as you at calculus.

I hope you do not mean George Jones who gave essentially the same counterproof here:

No. For example, the parabola y = x^2 has first derivative y' = 2x and second derivative y'' = 2. At x = 0, the first derivative is zero and the second derivative equals two.

The Wikipedia derivation and solutions remain correct despite your attempts to invalidate them and you are now reduced to nit picking my typos, that are not contained in the Wikipedia article, (which is word perfect and mathematically sound).

 

[yuiop]

Err, you can't do that because in certain frames, like COM $$v_1=u_1$$, remember? So, your hack doesn't work when you blindly try to do the division. This is thought in the beginner algebra classes. You may have to go all the way back and retake them.

This is pathetic. You are saying that the two equations can not be divided by each other, because when $v_1=u_1$, then $(v_1-u_1)/(v_1-u1) = 0/0$ so that invalidates the whole operation. Let us say we have an equation like:

$$xb = b^2$$

where b is an unknown variable and we want to solve for the variable x, so we divide both sides by b and obtain:

$$x = \frac{b^2}{b} = b$$

This is perfectly vaild, but wait! Starthaus points out that it is possible that b could conceivably have the value zero and the right hand side becomes 0/0 which means you are not allowed to divide both sides by b, implying there is no possible solution for x. If the "Starthaus objection" was true, all elementary algebra that involved division by unknown variables would be impossible. Hmmm.. I wonder who needs to retake their beginner algebra classes?

 

[starthaus]

Other than the typo, the derivation I gave of $V_{COM} = Pc^2/E$ is rigorous enough,

Err, no, it is a hack.

while your derivation does not exist.

It does, I am simply waiting for you to admit that what you produced is a hack for both the Newtonian case and the SR case. I have already given you https://www.physicsforums.com/blog.php?b=1887 , try copying and pasting and you'll get the correct proof. After that, see if you can apply it for the SR case. It is really trivial.

You call me a hacker, but I always produce correct results long before you do.

"Correct results" obtained via using the known result in a fake derivation don't count.

I hope you do not mean George Jones who gave essentially the same counterproof here:

As opposed to you, George understood why his example does not apply as I explained to him in the very next post. You, on the other hand still don't get it after all this time.

I think it is obvious that I meant to say if $f(t)=t^2$ then when t=0,

After all these months, you still miss the point:

$$f(t)=0$$ for all $$t$$

$$f(t) = t^2 = 0,$$

$$\frac{d}{dt}f(t) = 2t = 0$$

and

$$\frac{d^2}{dt^2}f(t) = 2$$

Just as bad. I don't understand why you keep embarassing yourself on this subject in a futile attempt to discredit my math. It is really basic calculus. Will you ever learn? I doubt it.

 

[starthaus]

Starthaus points out that it is possible that b could conceivably have the value zero and the right hand side becomes 0/0 which means you are not allowed to divide both sides by b,

...provided that $$b$$ is not zero. In the case of your "proof", $$b=0$$ so you can't divide by it. I simply pointed out that your "solution" for the Newtonian case is invalid. You have been given the correct solution already, no point in posting hacks.

 

[yuiop]

... I simply pointed out that your "solution" for the Newtonian case is invalid. You have been given the correct solution already, no point in posting hacks.

You say the my Newtonian solution is wrong, because at step 3 in the derivation quoted below:

$$\Rightarrow m_1(v_1-u_1)(v_1+u_1) = m_2(u_2-v_2)(u_2+v_2) \qquad \qquad (eq1)$$
...
$$\Rightarrow m_1(v_1-u_1)=m_2(u_2-v_2) \qquad \qquad (eq2)$$
...
Step 3: Divide (eq1) by (eq2) ...

$$\Rightarrow (v_1+u_1)=(u_2+v_2)$$

... I divided both sides of (eq1) by (eq2) that contains the expression $v_1-u_1$ and that because $v_1-u_1$ might conceivably have the value zero, it is an invalid operaton. I pointed out that if your objection is true then all algebraic operation that involve division by an unknown variable are invalid. I then gave this simple example:

... Let us say we have an equation like:

$$xb = b^2$$

where b is an unknown variable and we want to solve for the variable x, so we divide both sides by b and obtain:

$$x = \frac{b^2}{b} = b$$

This is perfectly vaild, but wait! Starthaus points out that it is possible that b could conceivably have the value zero and the right hand side becomes 0/0 which means you are not allowed to divide both sides by b,...

Your response:

...provided that $$b$$ is not zero. In the case of your "proof", $$b=0$$ so you can't divide by it.

is wrong. If b is a variable, then it can have a range of values including the value zero, and it is perfectly valid to divide both sides by b even when there is a possibility that b is zero. When we take the final result x=b, we see that when b=0, then x=0 which is correct. It is only when b is a constant equal to zero that the operation is invalid. Do you see the difference between the behavior of variables and constants? If not then you need to revist your basic understanding of algebra. In the Newtonian drivation that I gave, $v_1 - u_1$ is a varible so the operation is not invalid.

On a further point of correctness:

Err, you can't do that because in certain frames, like COM $$v_1=u_1$$, remember? So, your hack doesn't work when you blindly try to do the division. This is thought in the beginner algebra classes. You may have to go all the way back and retake them.

the relationship is actually [tex]v_1=-u_1[/itex] in the COM frame, so division by $v_1-u_1$ in the COM frame is actually division by $2v_1$ and not division by zero as you are suggesting. $v_1=u_1$ is only true before the collision (the trivial case) when there is no need to carry out any calculations.

$$V_{COM} = \frac{m_1u_1\gamma(u_1)+m_2 u_2 \gamma(u_2)}{m_1\gamma(u_1) + m_2\gamma(u_2)}$$

... if $f(t)=t^2$ then when t=0,

$$f(t) = t^2 = 0,$$

$$\frac{d}{dt}f(t) = 2t = 0$$

and

$$\frac{d^2}{dt^2}f(t) = 2$$

so your assertion that $df/dt = 0 \Rightarrow d^2f/dt^2 =0$ is not generally true.

After all these months, you still miss the point:

$$f(t)=0$$ for all $$t$$

You are missing the point.

If f(t)=0 for all t, then f is not a function of t, but a constant wrt t.

Your original assertion in the other thread was that $df/dt = 0 \Rightarrow d^2f/dt^2 =0$ is always true, was in the context of the trajectory of a particle at its apogee. At the apogee, the velocity is momentarily zero at that point in time. Along the entire trajectory, the velocity is not always zero, so the function of position versus time is obviously not always zero, or the particle would be permanently stationary. Your assertion that $df/dt = 0 \Rightarrow d^2f/dt^2 =0$ is only true if you specify that f(t)=0 for all time, but that was not the case in the context of a free falling particle following a geodesic as in the other thread.

Without qualification, your assertion that $df/dt = 0 \Rightarrow d^2f/dt^2 =0$ is not generally true .

f(t) implies a function of the variable t. Your declaration that f(t)=0 for all t implies that f is a constant and not a function of t. In both the counter proofs above, it is clear that the root of all your confusion, is that you are not clear on the differences between variables and constants.

 

[yuiop]

... I am simply waiting for you to admit that what you produced is a hack for both the Newtonian case and the SR case. I have already given you https://www.physicsforums.com/blog.php?b=1887 , try copying and pasting and you'll get the correct proof. After that, see if you can apply it for the SR case. It is really trivial.

Why does it follow that if stand by my derivation, you are unable to post your derivation? Just stalling for time, while I explain to you how it all works and then claim you had the solution all along? LOL.

In your blog, your derivation includes:

$$m_1v'_1 + m_2v'_2 = 0 \qquad \qquad (6)$$

The above is satisfied if and only if

$$V = \frac{m_1u_1 + m_2u_2}{m_1 + m_2} \qquad \qquad (7)$$

You have not derived (7) from (6) but simply assumed (7) and state it is the only solution that satisfies (6) without showing any of the maths or logic of the "verification". It is obvious you have not derived (7) from (6) as the two equations do not even contain the same variables. Remember, you have stated that assuming an answer and verifying it, is not a correct derivation:

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.

This is the correct way to derive the velocity of the COM frame in both Newtonian and Relativistic physics.

Wikipedia gives $V_{COM}$ as:

$$V_{COM} = \frac{\textrm{Total Momentum}}{\textrm{Total Energy}}c^2 = \frac{P_T}{E_T}c^2$$

which while it is technically correct, it is not intuitive, because there is no corresponding relationship between momentum and energy in Newtonian physics.

I hinted at a more intuitive solution in post #57:

...

$$V_{COM} = \frac{m_1u_1\gamma(u_1)+m_2 u_2 \gamma(u_2)}{m_1\gamma(u_1) + m_2\gamma(u_2)}$$
...
The equation above essentially says the velocity of the COM frame is the total mometum of the system, divided by the total mass of the system, which is a definition of $V_{COM}$. You do NOT need the definition that the total momentum is zero in the COM frame, in order to derive the velocity of the COM...

The much more intuitive statement is:

$$V_{COM} = \frac{\textrm{Total Momentum}}{\textrm{Total Mass}} = \frac{P_T}{M_T}$$

which is true in Newtonian physics and also true in SR (if the relativistic form of mass is used.)

Newtonian solution:

$$V_{COM} = \frac{\textrm{Total Momentum}}{\textrm{Total Mass}} = \frac{ m_1u_1 + m_2u_2}{ m_1 + m_2}$$

SR solution:

$$V_{COM} = \frac{\textrm{Total Momentum}}{\textrm{Total Mass}} = \frac{(m_1\gamma_{u_1})u_1+(m_2 \gamma_{u_2})u_2}{(m_1\gamma_{u_1}) + (m_2\gamma_{u_2})}$$

This is the simple, clear, consistent and intuitive way to obtain $V_{COM}$ in either case. If you are not satisfied with the relativistic case, you can always refer to the lengthier method derived from the relativistic energy-momentum equation that I gave in post #54, which simply verifies the solution I give here.

 

[starthaus]

On a further point of correctness:the relationship is actually [tex]v_1=-u_1[/itex] in the COM frame, so division by $v_1-u_1$ in the COM frame is actually division by $2v_1$ and not division by zero as you are suggesting. $v_1=u_1$ is only true before the collision (the trivial case) when there is no need to carry out any calculations.



You are missing the point.

If f(t)=0 for all t, then f is not a function of t, but a constant wrt t.

Your original assertion in the other thread was that $df/dt = 0 \Rightarrow d^2f/dt^2 =0$ is always true, was in the context of the trajectory of a particle at its apogee. At the apogee, the velocity is momentarily zero at that point in time. Along the entire trajectory, the velocity is not always zero, so the function of position versus time is obviously not always zero, or the particle would be permanently stationary. Your assertion that $df/dt = 0 \Rightarrow d^2f/dt^2 =0$ is only true if you specify that f(t)=0 for all time, but that was not the case in the context of a free falling particle following a geodesic as in the other thread.

Without qualification, your assertion that $df/dt = 0 \Rightarrow d^2f/dt^2 =0$ is not generally true .

f(t) implies a function of the variable t. Your declaration that f(t)=0 for all t implies that f is a constant and not a function of t. In both the counter proofs above, it is clear that the root of all your confusion, is that you are not clear on the differences between variables and constants.

No, I am not missing anything. espen180 was trying to hack in $$\frac{df}{dt}=0$$ into his geodesic equation. I pointed out to him that if he did that hack, then the second order terms disappear. espen180 understood his error and he stopped using that hack. You did not and you continue wasting your time and posting nonsense.

 

[starthaus]

Why does it follow that if stand by my derivation, you are unable to post your derivation? Just stalling for time, while I explain to you how it all works and then claim you had the solution all along? LOL.

You don't have a derivation, you just produced another hack.

In your blog, your derivation includes:

You have not derived (7) from (6)

Insert (6) into (5) and you get an equation dgree 1 in $$V$$. The solution is (7).
If you follw the same pattern, you will be able to derive $$V$$ for the SR case as well.

 

[yuiop]

Insert (6) into (5) and you get an equation dgree 1 in $$V$$. The solution is (7).
If you follw the same pattern, you will be able to derive $$V$$ for the SR case as well.

Your method works in the Newtonian case, but when you go the relativistic solution for a 1D collision of 2 unequal masses, your methods become intractable. I guess that is why you still not derived the relativistic case.

 

[starthaus]

Your method works in the Newtonian case, but when you go the relativistic solution for a 1D collision of 2 unequal masses, your methods become intractable.

LOL. The methiod works just the same in finding out $$V$$ in the relativistic case. You should try it, it is a simple exercise.

I guess that is why you still not derived the relativistic case.

Will you ever stop making unsubstantiated claims about things that you don't understand?

 

[yuiop]

LOL. The methiod works just the same in finding out $$V$$ in the relativistic case. You should try it, it is a simple exercise.

That part is easy enough. But once you have the velocity of the centre of momentum frame, can you obtain the relativistic solution for the final velocities without using v1=-u1 as you claim to be able to do? I am sure it is possible, but it is a lot harder and that is why you have not done it yet.

 

[starthaus]

That part is easy enough.

Yet, so far, you have been unable to derive it. Can you do it?

But once you have the velocity of the centre of momentum frame, can you obtain the relativistic solution for the final velocities without using v1=-u1 as you claim to be able to do? I am sure it is possible, but it is a lot harder

Of course I did it and it isn't much harder. I am waiting to see you derive $$V_{COM}$$, a much easier task indeed that, despite my hints, you haven't done yet.

and that is why you have not done it yet.

Why do you keep making claims that you can't substantiate?