Mechanical Symmetry?

[Sensei]

Thank you Sensai for the link & (excellent) explaination.

But I don't understand the need for the redundant complexity

with the hatted i,j,k syntax.

I thought the direction was already implied

in x,y,z.

Why is that redone with hats (on or off, pun; & primes ' )?

 

E.g.

(I guess (the unit vector is an intermediate bridge (=tool))

for converting to circular coordinates,

using Pythagorus's rule r^2=(x^2)+(y^2), + etc.)

Defining (direction) twice, seems backwards to me. (E.g. not progress.)

I isolate bidirection, with polarity + or -, e.g. 2D x,y.

Velocity is vector.

It has three components in 3D (x,y,z), or two components in 2D (x,y).

 

So in 3D we have

vx,vy,vz

 

Calculate velocity scalar from it:

[math]v=\sqrt{v_x^2+v_y^2+v_z^2}[/math]

 

Then such velocity scalar can be used in classic E.K.

[math]E.K.=\frac{1}{2}*m*v^2[/math]

 

so you can substitute and receive:

[math]E.K.=\frac{1}{2}*m*(v_x^2+v_y^2+v_z^2)[/math]

 

Velocity is change in position of object in some period of time.

You can go even further and expand vx,vy,vz like this:

 

[math]v_x=\frac{x1-x0}{t1-t0}[/math]

 

[math]v_y=\frac{y1-y0}{t1-t0}[/math]

 

[math]v_z=\frac{z1-z0}{t1-t0}[/math]

 

It just works for small non-relativistic velocities.

 

3D is definitely a challenge, to get the twist right; which Lorentz did 1904 (GR).

Maybe you can say some more to clear my confusion?

Would somebody please clarify that for me?

Special Relativity SR, is earlier than GR, and they are Einstein work, not Lorentz.

Lorentz factor is used in SR, for relativistic velocities.

Lorentz created LET Lorentz ether theory

https://en.wikipedia.org/wiki/Lorentz_ether_theory

Edited January 4, 2017 by Sensei

[Capiert]

Can we agree that

KE3~(m1*KE1+m2*KE2+mom1*mom2)/(m1+m2)

?

Edited April 30, 2017 by Capiert

[Capiert]

(..instead of #19's).

Edited May 1, 2017 by Capiert

[swansont]

Can we agree that

KE3~(m1*KE1+m2*KE2+mom1*mom2)/(m1+m2)

?

 

 

Two object of equal mass and KE inelastically colliding will come to rest. KE3 = 0

 

So no.

[studiot]

Thank you.

Maybe I've labled (=named) things wrong?

What is it called when the final masses barely touches,

continuing to travel both at the same speed,

but did NOT collide or bang together?

Neither elastic nor inelastic?

 

This is a fair question that does not seem to have been answered.

 

It is called kissing contact.

[Capiert]

Thanks, both of you.

Two object of equal mass and KE inelastically colliding will come to rest. KE3 = 0

 

So no.

Excuse me please,

this time

I mean your non_polar KE

(not my vectorized KE).

Let

m1=1 kg

m2=1 kg

v1=1 m/s

v2=-1 m/s

 

mom1=m1*v1=1 kg*1 m/s=1 kg*m/s

mom2=m2*v2=1 kg*(-1 m/s)=-1 kg*m/s

KE1~m1*(v1^2)/2=1 kg*((1 m/s)^2)/2=0.5 J

KE2~m2*(v2^2)/2=1 kg*((-1 m/s)^2)/2=0.5 J (not my negative value)

 

KE3~(m1*KE1+m2*KE2+mom1*mom2)/(m1+m2)

KE3~(1 kg*0.5 J+1 kg*0.5 J+1 [kg*m/s]*(-1 [kg*m/s]))/(1 kg + 1 kg)

KE3~(1 kg*J - 1 kg*kg*m*m/(s*s))/(2 kg), J=kg*m*m/(s*s)

KE3~0 kg*J/(2 kg)

KE3~(0/2) J

KE3~0 J

 

Sorry, your arguement does not hold.

Can we agree?

Edited May 2, 2017 by Capiert

[swansont]

But if KE is a vector, you don't get zero. The KE1 and KE2 terms should cancel. But you still have the mom1*mom2 term.

[Capiert]

But if KE is a vector, you don't get zero.

Yes. (But it's no problem to devectorize: simply square, then root.

At present I don't see a need to maintain that superfluous clutter, yet.

So I didn't bother with it (=my vectorized energy).)

The KE1 and KE2 terms should cancel.

Yes.

But you still have the mom1*mom2 term.

I'm not sure what you mean there. ?

Please clarify. What do you mean, still have mom1*mom2?

That (KE3) equation does not work for energy vectors, (as stated above),

it works for (non_polar) energy.

As you've seen they (=mom1*mom2) cancel the mass*energy terms, if needed;

& they (all, together) become the remaining energy term (KE3)

when divided by the (total) mass.

As far as I am concerned,

(of course) they (=mom1*mom2)

must be in the equation (now)

otherwise you will not get the correct energy (value) out.

 

(All that is only algebra.)

 

If you want a formula

that uses my KE vectors (instead)

it is (something like)

 

KE3~((m1*((KE1^2)^0.5)+m2*((KE2^2)^0.5)+mom1*mom2)/(m1+m2))*(mom1+mom2)/(((mom1+mom2)^2)^0.5)

 

where to produce polarity (or vectorize)

KE1~m1*((v1^2)/2)*v1/((v1^2)^0.5)

KE2~m2*((v2^2)/2)*v2/((v2^2)^0.5).

 

But (divide by) zero (e.g. speed) does not work in excel,

so an if statement

has to be used instead

for the zero case.

 

The last set of terms (as a factor)

(mom1+mom2)/(((mom1+mom2)^2)^0.5)

is only used

to derive the (total) KE3's

"polarity" from

the total momentum.

I.e. to vectorize

the KE3 answer.

Edited May 2, 2017 by Capiert

[Capiert]

You've marked me with a -1.

Would you please help show me

what I've done wrong?

Maybe I've made a typo?

 

If I'm not mistaken,

your 3 sentences (#82)

were discussing KEs as vectors:

meaning if KE1=0.5 J & KE2=-0.5 J cancel (to zero),

then you would expect mom1*mom2=-1 kg*J

would still remain.

But I tried to tell you

energy vectors don't work

(the way I want (them to), KE2 needs to be positive, instead)

so I don't persist

in maintaining them (KE as vectors).

That means (instead),

that the (negative) mom1*mom2 term

will cancel some of the (non_vector, =positive) KE instead.

I was only being curtious

(for your benefit)

in presenting

a redundant vector format

(that doesn't really work),

because you persisted

in maintaining (commenting on) vector energy syntax.

 

If you want me to develop

a vector energy formula

that works (=fits),

then I'm sorry

but it's going to take me a lot of time

(if it's possible, at all?).

 

Your -1 evaluation

was moot

because you've mixed up formulas

& thus were not on track.

But if KE is a vector, you don't get zero.

Yes, (but why do you want to do such a wrong thing with that formula?)

 

The KE1 and KE2 terms should cancel.

I said, "yes" (but with what? Answer: mom1*mom2.)

 

That's why I could not follow you

when you said:

 

But you still have the mom1*mom2 term.

Because mom1*mom2 had already been cancelled with the KEs.

 

Are we getting any (bit) closer to an agreement?

Edited May 3, 2017 by Capiert

[swansont]

Yes, (but why do you want to do such a wrong thing with that formula?)

You're the one insisting that KE is a vector, not me.

[Capiert]

That's what I originally intended to do;

but with your help

I changed my mind a bit

trying to clean things up

(& figure things out).

(I thought #81 made it clear,

we're not dealing with energy vectors, anymore.)

I would still like to get things to vector energy

but the math tells me

that's not completely possible

(at least not yet).

I'm still open for the idea,

but I've given up a bit

& have settled with

that normal standard KE

will do the job,

so why should I make things more difficult

than that,

when I can't reap

any advantage

(that I intended).

The math is set up

so I can research

the hows & whys

it fails or works.

 

It's not that I didn't warn you

occational errors had happened

& things I couldn't explain.

 

I came here for help

to straighten things out

& try to figure

these things out.

 

I must start with some sort of assumption (as the truth,

even if it turns out not to be true),

to put it to the test;

& I use it (intensely) until it fails,

& convinces me otherwise.

That has always been my basic strategy.

 

Some hypothesises are obvious,

while others are much more subtle.

Edited May 3, 2017 by Capiert

[Capiert]

You're the one insisting that KE is a vector, not me.

If we state

KE3/v3=(KE1/v1)+((KE2/v2)

as

KE3=((KE1/v1)+(KE2/v2))*v3

we can see vector_like attributes (=character)

in the use of the KE(s in brackets).

E.g. Let

m1=1 kg

m1=4 kg

v1=1 m/s

v2=-1 m/s

 

(

where

KE1~m1*(v1^2)/2=1 kg*((1 m/s)^2)/2=0.5 J

KE2~m2*(v2^2)/2=4 kg*((-1 m/s)^2)/2=2 J

 

m3=m1+m2=1 kg+4 kg=5 kg

 

v3=(mom1+mom2)/m3=((1 kg*1 m/s)+(4 kg*(-1 m/s))/5 kg=-0.6 m/s

 

KE3~m3*(v3^2)/2=5 kg*((-0.6 m/s)^2)/2=0.9 J

 

so

)

 

0.9 J=((0.5 J/(1 m/s))+(2 J/(-1 m/s)))*(-0.6 m/s)

particularly so as

0.9 J=(0.5 J*(s/m) - 2 J*(s/m))*(-0.6 m/s)

i.e. the 0.5 J & -2 J part (of the equation);

continued, that gives

0.9 J=-1.5 J*(s/m)*(-0.6 m/s).

The -1.5 J (part)

is the vector_like part

we are searching for;

but must be multiplied

by 0.6 to give the correct value.

However, polarity is lost

because minus is squared.

 

I have done nothing exceptional

in that demonstration,

only restating com

conservation of momentum.

 

But by now it should be obvious,

that energy

does NOT add correctly.

That is how

2 J plus 0.5 J gives 0.9 J.

 

If I am not mistaken:

The person that declared

the conservation of energy

must have been out of his mind!

 

Or do I see things wrong?

If so please explain.

Edited May 4, 2017 by Capiert

[swansont]

If we state

KE3/v3=(KE1/v1)+((KE2/v2)

 

...

 

Or do I see things wrong?

If so please explain.

 

 

The statement is not, in general, true. So why go on from there? KE is not a conserved quantity. Consider the case of 1 and 2 are traveling toward each other, colliding inelastically.

 

Conservation of momentum says that m1v1+m2v2 = m3v3, where the v's are vectors

 

Lets make the masses equal, so we can cancel them. v1+v2 = 2v3

 

If v1= 1 m/s and v2 = -0.9 m/s (since it's moving in the opposite direction) we find that v3 = 0.05 m/s

 

But with your equation, which is all scalars, we have (simplifying) 2v3 = v1 + v2 = 1.9 m/s so v3 = .95 m/s

 

 

 

 

If you're saying that this is a scalar equation

 

Can we agree that

KE3~(m1*KE1+m2*KE2+mom1*mom2)/(m1+m2)

?

 

 

Then consider particle 2 at rest, and all masses are equal. You have KE3 = m1KE1/(m1+m2), or KE3 = 1/2 KE1

 

But that's wrong. They should be equal.

 

 

And your equations don't cover the possibility that two particles are moving after a collision

[Capiert]

So why go on from there?

..

Consider .. inelastically.

 

..the possibility that two particles are moving after a collision

Elastically:

 

There I would propose (introducing) a 4th term KE4(/v4)

because the (formula's) structure

is already established.

 

For the syntax,

we may let

even postscript numbers

be for the same mass;

& the same for odd (postscript numbers).

E.g. before vs after (collision),

let

m1=m3

m2=m4.

At that point (in time, instance=circumstance, (=in that) case)

I would then propose

that the common

mutual speed

be called u

for united (speed vector)

of a non_elastic collision,

to reduce confusion;

instead of v3.

 

(& perhaps (something like)

mt=m1+m2

for the (common=) total mass (syntax,

instead of m3.)

 

In that way

I could avoid prime (') notation

for (other things; instead of for) after collision;

(but (unfortunately) my notes & files are cluttered with it (=primes)).

 

COM (before & after):

mom1+mom2=mom3+mom4.

 

Let

m4*KE4=-mom1*mom2.

 

For me an inelastic collision

is the continuation

of (what happened after)

a non_elastic collision.

 

Said differently,

an elastic collision

has an intermediate (state, actually several (states, or stages); each=all)

called a non_elastic collision.

 

I hope you understand what I mean. ?

 

I see a continuum.

Edited May 5, 2017 by Capiert

[Capiert]

From your 1st Example, you forgot something (made a false assumption, & over_simplified (if I'm not mistake). Sorry.).

Conservation of momentum says that m1v1+m2v2 = m3v3, where the v's are vectors

 

Lets make the masses equal, so we can cancel them. v1+v2 = 2v3

 

If v1= 1 m/s and v2 = -0.9 m/s (since it's moving in the opposite direction) we find that v3 = 0.05 m/s

 

But with your equation, which is all scalars, we have (simplifying) 2v3 = v1 + v2 = 1.9 m/s so v3 = .95 m/s

 

If you're saying that this is a scalar equation

 

The equation

KE3=(m1*KE1+m2*KE2+mom1*mom2)/(m1+m2)

is NOT purely scalars,

(but instead)

it is a mixture

of KE (scalars)

& momentum (vectors).

 

You did NOT include the momentum term (mom1*mom2=-0.9 kg*J)

in your simplification

which should read (before simplifing)

m3*(v3^2)/2=(m1*m1*(v1^2)/2+m2*m2*(v2^2)/2+m1*m2*v1*v2)/(m1+m2)

simplifing m=m1=m2=1 kg, (m3=m1+m2), & cancelling mass, we get

v3^2=((v1^2)+(v2^2)+2*v1*v2)/(2^2)

v3^2=(((1 m/s)^2)+((0.9 m/s)^2)+2*(1 m/s)*(-0.9 m/s))/4

v3^2=(1 ((m/s)^2)+0.81 ((m/s)^2)-1.8 ((m/s)^2))/4

v3^2=(0.01/4) ((m/s)^2)

v3^2=0.0025 ((m/s)^2

v3=(+/-) 0.05 m/s.

 

That is what you should have gotten

from my formula

even by simplifying,

but (it looks like) you forgot

the speeds were squared terms

(if I'm not mistaken).

 

(Too much simplification is not healthy,

too much complexity neither.)

 

The equation

KE3/v3=KE1/v1+KE2/v2

is also NOT only scalars

but (instead, it is)

a mixture

of KE (scalars)

& v speed (vectors).

 

Using

KE1~m1*((v1^2)/2=1 kg*((1 m/s)^2)/2=0.5 J

KE2~m2*((v2^2)/2=1 kg*((-0.9 m/s)^2)/2=0.405 J

KE3~m3*((v3^2)/2=2 kg*((0.05 m/s)^2)/2=0.0025 J

 

we get

 

0.0025 J/(0.05 m/s)=0.5 J/(1 m/s)+0.405 J/(-0.9 m/s)

0.05 J*s/m=0.5 J*s/m-0.45 J*s/m.

 

A scalar*vector is a vector (which you know).

 

The division operation

can be seen

as multiplying the inverse.

 

Your 2nd example is more interesting,

& thanks for pointing it out to me.

 

m3*KE3=m1*KE1+m2*KE2+mom1*mom2

0.5 kg*J=0.5 kg*J+0 kg*J+0 kg*J

 

But I think you are confusing things (a bit?)

because I said (to you)

energy does NOT (always) add correctly.

That is my complaint

& the classic (=best) example

to show that it (=energy) doesn't add correctly!

How can you expect

to get the same energy output

as input?

The KE before a collision

will NOT give

the same energy

as after a non_elastic collision.

That is exactly what I am complaining about.

You are (unintentionally)

turning the table around

& trying to (help) tell me

that I have made a mistake there;

but I am trying to (help) tell you

the error lies in energy:

when you try to add it (=energy)

it (=energy) will not do that (addition) correctly.

At least not always.

I have given you a formula

which will equate energy;

& it is NOT simple addition

as you can see.

I have known your 2nd example

for many years (in my excel sheet

as m1=4 kg, & m2=4 kg

v1=1 m/s, & v2=0 m/s;

the total KE is 1 J;

NOT 2 J, like the input (energy) of the 1st mass, m1 !)

Your 2nd example

is not new to me, (instead)

it is my major proof_complaint (arguement, evidence, clue, hint).

Knowing that (fault in energy)

I have tried to make the numbers balance.

 

An equation (now) exists.

& you have seen it.

Edited May 6, 2017 by Capiert

[swansont]

From your 1st Example, you forgot something (made a false assumption, & over_simplified (if I'm not mistake). Sorry.).

The equation

KE3=(m1*KE1+m2*KE2+mom1*mom2)/(m1+m2)

is NOT purely scalars,

(but instead)

it is a mixture

of KE (scalars)

& momentum (vectors).

 

You did NOT include the momentum term (mom1*mom2=-0.9 kg*J)

in your simplification

 

 

I was analyzing KE3/v3=(KE1/v1)+((KE2/v2)

 

which I had quoted, and should have been obvious.

[Capiert]

I was analyzing KE3/v3=(KE1/v1)+(KE2/v2)

 

which I had quoted, and should have been obvious.

Yes, I noticed that (it was that formula) later (again),

& also commented (about that formula) in #90,

(due to the ambiguous uncertainty on my side, just to be sure).

 

That formula is also a mixture

of scalars & vectors;

instead of only scalars

(as you said).

But thank you,

for clearing my uncertainty.

 

Both of your derived (simplified) formulas

2*v3=v1+v2

thus receives vectors;

NOT scalars.

 

That's what perhaps confused me into ambiguity

because I did not expect

you (might) make a wrong assumption.

 

(I thought maybe you meant another formula

because your assumption

did NOT make sense to me.)

 

The speeds v1 & v2

are vectors

(NOT scalars).

 

Thus, in both cases

(mine & yours)

we have the same value

for v3 (=0.05 m/s),

so there is no difference between them.

 

So against what you wrote,

0.95 m/s does NOT happen for v3

in #88 (example 1).

 

Or have I missed something,

& made a mistake?

(If so) Please explain.

Edited May 6, 2017 by Capiert

[swansont]

Yes, I noticed that (it was that formula) later (again),

& also commented (about that formula) in #90,

(due to the ambiguous uncertainty on my side, just to be sure).

 

That formula is also a mixture

of scalars & vectors;

 

 

 

 

Which you posted right after agreeing that KE is a scalar, and that you had given up on making it a vector.

 

(The equation can't be a vector equation; dividing by a vector is not a valid formulation)

 

Or have I missed something,

& made a mistake?

(If so) Please explain.

 

 

Your mistake is coming up with equations that apply to only one situation, and based on dubious math, rather than applying actual physics to the problem.

 

Learn the physics and drop the nonsense.

[imatfaal]

!

Moderator Note

 

Capiert

 

Sorry but this is getting nowhere and, I firmly believe, causing more harm than good. Off the wall speculations can be fun and harmless - but are fairly fruitless; but when you start playing fast and loose with mathematics then you are more than likely to come up with ideas that are not even wrong - and the danger is that you and others will learn bad and incorrect methods/techniques.

 

I am afraid your mangling and flip-flopping between Vectors and Scalars is of the latter sort described above - it is not wholesome and not helpful. I am locking this thread and at present you do not have permission to reopen a new thread on the same topic.

 

I would heartily recommend you take your undoubted enthusiasm for this subject to the next level and spend some time at the online schools/universities/moocs - the resources available at present for free are extraordinary (Gross on Multivariable Calculus / Strang's lectures on Linear Algebra are two recent courses I have looked at). A fair dose of study will equip you to answer your own questions and perhaps provide more rigorous speculative ideas.

 

This thread has been locked