Please explain what is wrong with my relativistic momentum problem

[Virous]

I`m very sorry for taking your time. You said, that I`m omitting x components. I included them in the way you showed me. If I did it incorrectly, please, tell me where.

 

[Virous]

Without x components of momenta:

 

[Virous]

After:

 

[xox]

I`m very sorry for taking your time. You said, that I`m omitting x components. I included them in the way you showed me. If I did it incorrectly, please, tell me where.

Yes, you corrected that error. You also corrected the fact that you were missing one term in your first post. Now, stop being so stubborn and separate the equation of conservation into the x and y components, as I showed you. You have a sign error someplace but it is very difficult to debug since you insist on adding the x and y components of the momentum.

 

[Virous]

Just forget about x-components. They are fine. Y components are just the two posts above your last one.

 

[xox]

Just forget about x-components. They are fine. Y components are just the two posts above your last one.

This is better, now please put the "before" and "after" in one post, so I can compare the results.

 

[Virous]

Sure!

Before:

After:

 

[PeterDonis]

Virious, your $u_{by}$ values are wrong; the denominator should have $\left( 1 - (a / c )^2 \right)$, not $\left( 1 + (a / c)^2 \right)$.

Also, you really need to learn to use the forum LaTeX feature to write equations, instead of posting images. Check out this guide:

https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[Virous]

PeterDonis, Ball B moves in the negative direction in frame S and frame S' moves in the positive. Hence we have (-a)*(a) which gives -a^2. And - times - gives +

 

[Virous]

Yes, yes, sorry about LaTeX. I`m using it whenever I can. But because I have all of these equations in Wolfram Mathematica, I just screenshot them to save time. They are quite huge and it will take a lot to copy them. Sorry again!

 

[Virous]

To derive these equations (in the table) I used velocity addition formula:

$u'=\frac{u}{\gamma (1-\frac{u_{x}v}{c^2})}$

v here is a velocity of S' in S. It is a.
ux is -a, since Ball B moves to the left.
uy is -b.

If you`ll substitute, you`ll get the denominator with "+"

 

[Virous]

Wow, your previous post just disappeared :)

 

[xox]

Sure!

Before:

After:

OK, it is getting close, we are down to an apparent sign inversion between "before" and "after" total momentum.

Now, I will give you a hint: look at the first term in the "before" and work the expression under the radical

$$(1+\frac{a^2}{c^2})^2-4\frac{a^2}{c^2}-\frac{b^2}{c^2}(1-\frac{a^2}{c^2})=(1-\frac{a^2}{c^2})(1-\frac{a^2}{c^2}-\frac{b^2}{c^2})$$

 

[PeterDonis]

PeterDonis, Ball B moves in the negative direction in frame S and frame S' moves in the positive. Hence we have (-a)*(a) which gives -a^2. And - times - gives +

Actually, looking back at my post #37, I was wrong. The correct values for $u_{ay}$ and $u_{by}$ before (the only change in the after values is that the signs are swapped) are, as I noted in an earlier post, the ratios $u^y / u^t$ of the values from my post. Those are (again, I'm using units where $c = 1$):

$$
u_{ay} = \frac{b}{\sqrt{1 - a^2}}
$$

$$
u_{by} = \frac{- b \sqrt{1 - a^2}}{1 + a^2}
$$

So the corresponding momentum values are (note that I'm using your values for $u_{ax}$ and $u_{bx}$, which are correct):

$$
p_{ay} = \frac{m u_{ay}}{\sqrt{1 - u_{ax}^2 - u_{ay}^2}} = \frac{m b}{\sqrt{1 - a^2} \sqrt{1 - b^2 / \left( 1 - a^2 \right)}} = \frac{m b}{\sqrt{1 - a^2 - b^2}}
$$

$$
p_{by} = \frac{m u_{by}}{\sqrt{1 - u_{bx}^2 - u_{by}^2}} = \frac{- m b \sqrt{1 - a^2}}{\left( 1 + a^2 \right) \sqrt{1 - \left( 2 a \right)^2 / \left( 1 + a^2 \right)^2 - b^2 \left( 1 - a^2 \right) / \left( 1 + a^2 \right)^2}} = \frac{- m b}{\sqrt{\left[ \left( 1 + a^2 \right)^2 - 4 a^2 \right] / \left( 1 - a^2 \right) - b^2}} = \frac{- m b}{\sqrt{1 - a^2 - b^2}}
$$

So the $y$ momentum does cancel, as desired.

 

[Virous]

It was like this even before all the corrections :) The point is, that if we use the classical momentum formula (p=mu), we get this error. Introduction of gamma according to the book must solve this problem, but it doesn`t.

 

[PeterDonis]

Introduction of gamma according to the book must solve this problem, but it doesn`t.

Check my post #84, your $y$ velocity components were indeed wrong, I just gave a mistaken correction previously.

 

[xox]

It was like this even before all the corrections :) The point is, that if we use the classical momentum formula (p=mu), we get this error. Introduction of gamma according to the book must solve this problem, but it doesn`t.

This is not the issue, you haven't finished your calculations, use the hint I gave you, some very nice simplification will happen.

 

[Virous]

Your hint didn`t exist, when I saw your post the first time. Sorry :(

 

[xox]

Your hint didn`t exist, when I saw your post the first time. Sorry :(

post 83, you are very close, if you do the calculations correctly you should get the total momentum to be null.

 

[Virous]

Oh, yeeeeees! I found it! It simplifies to zero. My problem was, that I fully relied on my Math software, and it for some reason didn`t simplified it completely, because of possiblity of imaginary roots in case of overcoming the speed of light!

Thanks to everyone! So, summarizing everything, my initial mistake was in the fact, that I had to add velocity components geometrically via pyphagora`s theorem!

Thanks to all!

 

[xox]

Oh, yeeeeees! I found it! It simplifies to zero. My problem was, that I fully relied on my Math software, and it for some reason didn`t simplified it completely, because of possiblity of imaginary roots in case of overcoming the speed of light!

Thanks to everyone! So, summarizing everything, my initial mistake was in the fact, that I had to add velocity components geometrically via pyphagora`s theorem!

Thanks to all!

Yes, you had the "gammas" computed incorrectly, this and the fact that you did not persevere enough, were your only flaws.
On an amusing note, humans are still better than software, your tool was incapable of detecting the expression simplification. Out of curiosity, what software are you using?

 

[Virous]

It was. The problem was, that if the speed is bigger than the speed of light, simplification won`t work (because of imaginary numbers). Since the software had to consider all the cases, it stopped there. It worked well after I added a<c and b<c into assumptions list.

I`m using Wolfram Mathematica.

 

[xox]

It was. The problem was, that if the speed is bigger than the speed of light, simplification won`t work (because of imaginary numbers). Since the software had to consider all the cases, it stopped there. It worked well after I added a<c and b<c into assumptions list.

I`m using Wolfram Mathematica.

interesting...this is a rather spectacular failure to complete a simple symbolic computation, the result is 0 , you shouldn't be forced to add the conditions a<c and b<c. I think those fabled East Europeans coding Mathematica are not that hot after all :-)

 

[PAllen]

One thing that I see caused me confusion was terminology (either from the book or Virous, not sure). The formulas given for deriving image 4 in the OP are just Lorentz transform of coordinate velocities. They are not velocity addition formulas in the sense I was thinking. In the case of all x motion, the Lorentz transform for Ux is, of course, the simplest velocity addition formula. In all other cases, they don't give you the resultant speed or gamma (as the formula I gave does). They just give you components in the frame boosted by v in the x direction. You then have to compute resultant speed and gamma.

[Virous: is that a British virus? ]

 

[xox]

Virious, your $u_{by}$ values are wrong; the denominator should have $\left( 1 - (a / c )^2 \right)$, not $\left( 1 + (a / c)^2 \right)$.

Also, you really need to learn to use the forum LaTeX feature to write equations, instead of posting images. Check out this guide:

https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

No, his speeds are transformed correctly, his only errors were:

-incorrect formula for some of the "gammas"

-relying on Mathematica to do the final simplification of the expressions (Mathematica failed in a spectacular way)

I have encountered some failures in Matlab but never before in Mathematica. This was an interesting experience.

 

[PeterDonis]

No, his speeds are transformed correctly

Yes, you're right, he expressed them differently, but looking again at his expressions they are equivalent to mine.

 

[xox]

[Virous: is that a British virus? ]

A Russian one.

 

[Virous]

A Russian one.

How do you know that? :)

(my articles, probably) :D

 

[xox]

How do you know that? :)

(my articles, probably) :D

Yep.

 

[Virous]

I have encountered some failures in Matlab but never before in Mathematica. This was an interesting experience.

This is not a failure. It should be like this, since the conservation law does not work for speeds >c.

Final results (everything works, assuming, that a+b<c geometrically):

 

[xox]

This is not a failure. It should be like this, since the conservation law does not work for speeds >c.

Except that Mathematica treats the problem as an algebra problem, it doesn't "know" anything about "speeds>c". My contention was that the factorizations and reduction of like terms SHOULD have worked WITHOUT you having to add the conditions $a,b<c$.

 

[Virous]

No, because during the simplification you assumed, that $\sqrt{a}^2=a$ and only $a$, while in fact there are other interpretations. If you take some random data, like c=15, a=20, b=25 and substitute it you`ll see, that it does not make momenta zero.

 

[xox]

No, because during the simplification you assumed, that $\sqrt{a}^2=a$ and only $a$, while in fact there are other interpretations. If you take some random data, like c=15, a=20, b=25 and substitute it you`ll see, that it does not make momenta zero.

I ONLY used $\sqrt{(1+a^2/c^2)^2}=1+a^2/c^2$. This is ALWAYS true, you don't need to add the condition a<c.

 

[Virous]

Did that for you :)

You see

 

[xox]

Did that for you :)

You see

That is the problem when you start EVALUATING expressions, I operated exclusively in SYMBOLIC algebra, never evaluated any of the expressions. The Mathematica coders need to do some work, their program is not quite at the level of human beings. This is not quite a Mathematica bug but it is a limitation.