Please explain what is wrong with my relativistic momentum problem

[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&lt;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.

 

[Dale]

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.

No, this is not always true.

Mathematica is correct in this case. You cannot generally simplify $\sqrt{x^2}=x$. That is an incorrect SYMBOLIC substitution for general x. Similarly with $\sqrt{(1+x^2)^2}=1+x^2$.

There are cases where Mathematica does not simplify as well as humans do, but this is not one of them. If you give correct assumptions for a and c then Mathematica will correctly simplify it, but if you do not give any assumptions then it assumes that they can have any value, including complex values, and then the simplification is not valid.

 

[Dale]

Virous, btw, I didn't follow this thread all the way, but I do have some simple Mathematica code for doing special relativity exercises. It is based on 4-vectors, so it is rather convenient to use for momentum and other similar problems.

 

[xox]

No, this is not always true.

Mathematica is correct in this case. You cannot generally simplify $\sqrt{x^2}=x$. That is an incorrect SYMBOLIC substitution for general x. Similarly with $\sqrt{(1+x^2)^2}=1+x^2$.

There are cases where Mathematica does not simplify as well as humans do, but this is not one of them. If you give correct assumptions for a and c then Mathematica will correctly simplify it, but if you do not give any assumptions then it assumes that they can have any value, including complex values, and then the simplification is not valid.

The point was that Mathematica did the correct job (replicating the one that I did) only AFTER Vitrous inserted the condition a<c, so it had nothing to do with the claim that Mathematica couldn't evaluate $\sqrt{x^2}=\pm x$. The real problem was that Mathematica got stuck on ANOTHER expression, \sqrt{1-a^2/c^2} and it needed a little "prodding" by being "told" that a<c.

 

[Dale]

It didn't get stuck, it gave a correct simplification given the (lack of) assumptions.

 

[xox]

It didn't get stuck, it gave a correct simplification given the (lack of) assumptions.

The correct result (produced by a human) is zero. Mathematica stopped way short of that, actually several steps short, leading Virous to the conclusion that the momentum was not conserved (turns out that it is). The reason could be traced to Mathematica's insistence in looking under the square root for reasons I cannot fathom. This is not a bug per se but it is definitely a shortcoming.

 

[Dale]

It is correct behavior. It is not a bug and it is not a shortcoming. If it were to not look under the square root, as you suggest, then it would be mathematically wrong.

In general, $\sqrt{x^2} \ne x$, and Mathematica correctly recognizes that fact by not making that substitution in general. If you place conditions on x such that the expression is true, then Mathematica will simplify it.

It is true that Simplify and FullSimplify are not perfect, but this particular thing that you are mentioning is not a flaw. One problem that it has is that it counts "leafs" to evaluate which expression is more complicated than another, and sometimes an expression with more "leafs" is judged by a human to be simpler.

 

[xox]

It is correct behavior. It is not a bug and it is not a shortcoming. If it were to not look under the square root, as you suggest, then it would be mathematically wrong.

In general, $\sqrt{x^2} \ne x$, and Mathematica correctly recognizes that fact by not making that substitution in general. If you place conditions on x such that the expression is true, then Mathematica will simplify it.

You need to go back and read the thread, Mathematica got stuck on a totally different type of expression, \sqrt{1-a^2/c^2}, \sqrt{1-b^2/c^2}. It got unstuck only after Virous "taught it" that a,b<c. $\sqrt{x^2} \ne x$ is not the issue, never was.

 

[Dale]

Sorry, I didn't read the thread in detail. Which post was this?

What was the problem with $\sqrt{1-a^2/c^2}$? That doesn't and shouldn't simplify further.

 

[xox]

Sorry, I didn't read the thread in detail. Which post was this?

Well, you need to go back and read the thread, I will provide you with the exact post.

What was the problem with $\sqrt{1-a^2/c^2}$? That doesn't and shouldn't simplify further.

Correct, it has nothing to do with any simplification, it has to do with Mathematica's inability to complete some symbolic computations when it encounters an expression that may be EITHER real OR imaginary.

 

[xox]

You need to g o back to post 90 and to start reading from there on. Virous has attached the offending Mathematica code as well as his "fix".

 

[Dale]

Correct, it has nothing to do with any simplification, it has to do with Mathematica's inability to complete some symbolic computations when it encounters an expression that may be EITHER real OR imaginary.

What is the specific symbolic computation that you are claiming that it does not complete and should?

 

[xox]

What is the specific symbolic computation that you are claiming that it does not complete and should?

It needs to bring the expression to a simpler form and to execute a subsequent reduction of like terms. The correct answer is "zero". Mathematica doesn't operate either the simplification, nor the reduction of like terms.

 

[Dale]

That is not a specific symbolic computation, that is a general complaint.

I put in the expression in post 100, and without the assumption you get terms under the square root that could be negative. So the behavior is correct.

 

[xox]

I put in the expression in post 100, and without the assumption you get terms under the square root that could be negative.

Yes, this is EXACTLY the problem, there is NO reason to check the terms under the square root.

 

[Dale]

Sure there is. You want to see if they simplify. If it didn't check the terms under the square root then you would complain that it didn't simplify things under square roots.