One dimensional collision of 2 unequal masses

[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 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.<br /> <br /> <blockquote data-attributes="" data-quote="kev" data-source="post: 2809763" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> kev said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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)} </div> </div> </blockquote><br /> <blockquote data-attributes="" data-quote="kev" data-source="post: 2809763" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> kev said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> ... if f(t)=t^2 then when t=0,<br /> <br /> f(t) = t^2 = 0,<br /> <br /> \frac{d}{dt}f(t) = 2t = 0 <br /> <br /> and<br /> <br /> \frac{d^2}{dt^2}f(t) = 2<br /> <br /> <b>so your assertion that df/dt = 0 \Rightarrow d^2f/dt^2 =0 is not generally true.</b> </div> </div> </blockquote> <blockquote data-attributes="" data-quote="starthaus" data-source="post: 2810000" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> starthaus said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> After all these months, you <b>still</b> miss the point: <br /> <br /> f(t)=0 for <b>all</b> t </div> </div> </blockquote> You are missing the point.<br /> <br /> <b>If f(t)=0 for all t, then f is not a function of t, but a constant wrt t.</b> <br /> <br /> 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 <b>only</b> true if you specify that f(t)=0 <b>for all time</b>, but that was not the case in the context of a free falling particle following a geodesic as in the other thread.<br /> <br /> <b>Without qualification, your assertion that df/dt = 0 \Rightarrow d^2f/dt^2 =0 is not generally true </b>.<br /> <br /> 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&#039;_1 + m_2v&#039;_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 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.<br /> <br /> <br /> <br /> You are missing the point.<br /> <br /> <b>If f(t)=0 for all t, then f is not a function of t, but a constant wrt t.</b> <br /> <br /> 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 <b>only</b> true if you specify that f(t)=0 <b>for all time</b>, but that was not the case in the context of a free falling particle following a geodesic as in the other thread.<br /> <br /> <b>Without qualification, your assertion that df/dt = 0 \Rightarrow d^2f/dt^2 =0 is not generally true </b>.<br /> <br /> 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.

<br /> <br /> 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?