Mechanical Symmetry?

[swansont]

KE3~(((2^0.5) kg*((1 m/s)^2)^0.5)+(2^0.5) kg*(((-1 m/s)^2)^0.5)^2)/4 kg

KE3~(((2^0.5) kg*(1 m/s)+(2^0.5) kg*(-1 m/s))^2)/4 kg

KE3~((((2^0.5) kg*m/s)-(2^0.5) kg*m/s)^2)/4 kg

 

 

When you square a negative number, the result is a positive number.

[Capiert]

When "I" (do a) square (operation) & then reverse that operation with rooting,

then I expect (to get) the original (as NOT modified).

That is reversable math. (Otherwise that is NOT reversible math, by definition.)

For each operation, an anti_operation should (exist &) be available.

E.g.

Addition vs subtraction

multiplication vs division

squaring vs rooting.

 

Am I demanding too much (rigor)?

 

"You" (are perhaps a little more careless in that respect, if I may say, &) generalize & make the approximation:

that a squared minus is positive.

 

I however distinguish

between subtraction

vs negative polarity.

For me they are NOT identical.

Negative polarity is equivalent

to 180 degrees rotated, into the opposite direction.

The root of negative polarity is 90 degrees.

Polarities are multiplied to add angles.

(-1)*(-1)=(--1)

180 degrees+180 degrees=360 degrees.

360 degrees is (to) the (angle) position as 0 degrees (starting place),

but they do NOT mean (exactly) the same thing.

 

Does that help explain that technique?

 

The polarity (symbolic for angle) is treated like a unit

so it is multiplied by the quantity (=amount)

like you do with length, for meter e.g. 1 m.

I could have written it (=negative polarity) so

(1-)*(1-)=1*1*(-)*(-).

It is simply another syntax (unit).

[swansont]

KE is always a positive value. There is no direction information in it.

 

Basically you're making stuff up and wondering why it's not working.

[Bignose]

"You" (are perhaps a little more careless in that respect, if I may say, &) generalize & make the approximation:

that a squared minus is positive.

 

I however distinguish

between subtraction

vs negative polarity.

Like swansont said, KE is an always positive quantity. Defining it as such has been supremely successful. If you think it isn't, the onus is on your to provide copious examples where it doesn't. And no, the examples you've given don't count because they are well described by the existing definitions even if you don't understand them.

[Capiert]

KE is always a positive value.

Yes, but you do NOT use that "+" symbol (for 360 degrees),

as a compromise indicating 2 crossed negative symbols i.e.

1 horizontal & the other vertical.

Instead, you drop it completely.

 

You've seen how the rooting operation was cancelled out

with the squared operation

in #45.

 

Can you tell me why that (double operation, as a "do & undo" should not work, or) is not allowed?

 

At present what I've shown you in #49 works.

 

Please show me a(n algebra) math law broken.

 

There is no direction information in it.

That's not true.

The speed has polarity.

 

I've only made polarity observations

& reported them.

 

No one will prevent you from closing your eyes, if you want.

No one will force you to observe.

 

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

Have I forgot anything in that rigor?

--1 kg*((m/s)^2).

Negative speeds have 2 negatives in their KE.

"You" can keep them or get rid of them.

I'm only telling you they are there,

something you already know.

If you get rid of them

then you turn the equation

into an ambiguous thing

& can no longer extract the original polarity.

It's thus your choice,

what you want

& want to do.

Not mine.

You can "assume" they are the same,

but I don't.

Thus I can not (completely) agree with you,

when you say KE is always positive.

 

Basically you're making stuff up and wondering why it's not working.

Newton made things up too. I.e.

They were "new" equations meaning nobody else knew about them.

The rest is an attempt to explain how the things work. (Self defining.)

But I think you are guessing when you say that "wondering why"

because with your help

I have narrowed in on the problem.

(But I still don't know why, the algebra rule is not allowed.

It's in a different place not yet mentioned.)

When you square a negative number, the result is a positive number.

When you root a squared number what do yo get?

When you square a rooted number,

you get a positive.

The other (way around) is NOT guaranteed.

There is a sequence priority.

 

(E.g. The egg must come before the chicken,

otherwise no chicken, e.g. minus or error.)

Edited January 2, 2017 by Capiert

[swansont]

An object has 8 Joules of kinetic energy. What direction is it moving? Where is the "polarity" information?

[Sriman Dutta]

Capiert you must first understand what's scalar and what's vector.

Force and momentum are vectors because they involve scalar multiplication of vectors ( in force we have the multiplication of scalar mass and vector acceleration, and in momentum we have multiplication of scalar mass and vector velocity).

But energy is not a vector.

Also you are not clear about squaring and rooting. Any integer squared gives a positive integer. But, on extracting the square root we put a plus-minus sign before the number whose square we have rooted. And if we extract the square root of a negative integer we call that an imaginary number. [math]\sqrt{-1}=i[/math]

[Capiert]

An object has 8 Joules of kinetic energy. What direction is it moving? Where is the "polarity" information?

If I gave you that task (instead)

(with my syntax technique)

then I (& you) could say (for a 1D number line)

"to the right".

But since you are giving me the task

it can NOT be said

whether it is going

to the right or the left,

it's ambiguous

because you are not observing polarity so,

& have thrown away that (option, possiblity) ability.

 

Is that a satisfactory answer?

Edited January 2, 2017 by Capiert

[Sriman Dutta]

Energy does not have polarity (I don't know what you mean by polarity---I know polarity of magnets, but not this).

[Capiert]

Capiert you must first understand what's scalar and what's vector.

Force and momentum are vectors because they involve scalar multiplication of vectors (in force we have the multiplication of scalar mass and vector acceleration, and in momentum we have multiplication of scalar mass and vector velocity).

Thank you very much for explaining vectors vs scalars.

That (simple) is how I saw it before I came to this forum,

but some of their crew has made me insecure

with further details

about dot & cross products.

Perhaps you would like to continue.

 

energy is not a vector.

Is not a scalar, multiplied by a scalar, a scalar?

& is not a scalar, multiplied by a vector, a vector?

With that I have every indication that KE is a vector,

unless you can show me otherwise.

 

Also you are not clear about squaring and rooting. Any integer squared "gives" a positive integer.

We don't have to accept every gift given. E.g. Trojan horse.

I distinguish between what things (really) are

& what they "can" become (virtual, equivalence), but do NOT have to be.

 

But, on extracting the square root we "put"

Oh! We make an exception, to try & fix the damage done.?

a plus-minus sign before the number whose square we have rooted. And if we extract the square root of a negative integer we "call" that an "imaginary" number. [math]\sqrt{-1}=i[/math]

That sounds like a lot of exceptions to me.

I guess nobody complains if it (still) works.

 

Energy does not have polarity (I don't know what you mean by polarity---I know polarity of magnets, but not this).

I think some are forgetting my goal

is to put this (formula) into an excel sheet

& that I am dealing with algebra.

A number line can be (for (all) practical purposes)

treated as a vector.

It has (both) minus & positive (sign) polarity,

you say "directions".

E.g. a battery is also labled positive & negative

with the same symbols "+" & "-"

for the A, B, (baby) C, AA, cells & 9 V block cells (etc).

"Swap poles" or "swap polarity" is a common expresion.

Please make a suggestion,

I'd like to explain things to kids too.

 

Bi_mention (etymology), you say "dimension",

& mention (=talk about) that it is a pair.

 

But maybe I have selected the wrong word "polarity", e.g.

the ends of a pole (is 3D thick, as a real physical example,

for common speach

& explaining)

instead of (a virtual zero thickness) number line.

Edited January 2, 2017 by Capiert

[swansont]

 

Is not a scalar, multiplied by a scalar, a scalar?

& is not a scalar, multiplied by a vector, a vector?

With that I have every indication that KE is a vector,

unless you can show me otherwise.

 

We don't have to accept every gift given. E.g. Trojan horse.

I distinguish between what things (really) are

& what they "can" become (virtual, equivalence), but do NOT have to be.

 

Oh! We make an exception, to try & fix the damage done.?

That sounds like a lot of exceptions to me.

I guess nobody complains if it (still) works.

 

 

 

 

So you are proposing something new. Including a unit vector with KE. You could write the term as sqrt(mE). And then sqrt(m₁E₁) + sqrt(m₂E₂) = sqrt(m₃E₃), as a vector equation. That would work, because it's conservation of momentum. p = sqrt(2mE), and the sqrt(2) is divided out, being common to all terms. This is nothing new. IOW, it adds nothing, and is not a new insight.

 

What you had was

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

 

[sqrt(m₁E₁) + sqrt(m₂E₂₎]²/ (m₁ + m₂)

 

With a vector, that's still just applying conservation of momentum. (It's wrong by a factor of two, however. KE = p²/2m)

 

What you've succeeded at is a more complicated way of doing something that physics already does.

[Capiert]

 

Correction for #47

Note: vf & vi must be swapped

to eliminate (Sensai's #46, relativity) gammas,

Thus

 

final speed vf=c (NOT vi).

 

The initial speed vi

is NOT zero

& is the isotopes' speed.

The kinetic energy is KE=m*v*va

for mass m

difference speed v=vf-vi

& average speed va=(vi+vf)/2.

 

(I still need more time to prepare.)

So you are proposing something new. Including a unit vector with KE. You could write the term as sqrt(mE). And then sqrt(m₁E₁) + sqrt(m₂E₂) = sqrt(m₃E₃), as a vector equation. That would work, because it's conservation of momentum. p = sqrt(2mE), and the sqrt(2) is divided out, being common to all terms. This is nothing new. IOW, it adds nothing, and is not a new insight.

It's missing the factor 2. I don't know why you are presenting something wrong 1st

(unless it's to show something near (=very similar) which is very encouraging & helpful);

but I am happy you are finally recognizing what I am presenting.

I had to finally state it as new,

because you rejected it previously

(wrongly?) stating mine was wrong.

It's very difficult to get you guys

to recognize what you already know.

I have been (wrongly?) accused of not understanding

the material.

 

What you had was

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

That's also missing the factor 2.

 

[sqrt(m₁E₁) + sqrt(m₂E₂₎]²/ (m₁ + m₂)

dito

 

With a vector, that's still just applying conservation of momentum.

What was the thesis #1?

(It's wrong by a factor of two, however. KE = p²/2m)

Finally recognized. I'm very happy.

 

What you've succeeded at is a more complicated way of doing something that physics already does.

Thanks for recognizing. Edited January 3, 2017 by Capiert

[swansont]

I had to finally state it as new,

because you rejected it previously

(wrongly?) stating mine was wrong.

It's very difficult to get you guys

to recognize what you already know.

I have been (wrongly?) accused of not understanding

the material.

It was wrong. You can't just arbitrarily change what standard terminology means. It would be easier to recognize if you didn't misuse the language.

[Capiert]

It (=?) was wrong. You can't just arbitrarily change what standard terminology means. It would be easier to recognize if you didn't misuse the language.

Please be more specific & show me the formula errors.

Your "It" (pronoun) does not pinpoint the individual errors, for me enough.

I'm still a little foggy (=vague) which formula (mine or yours) you were referring

in your previous post #58, #59

"It's wrong by a factor of 2, however."

Didn't mine have a factor 2, but yours that you said were mine did not?

Won't com also work with the factor 2?

Or doesn't it (= factor 2) belong there, at all?

I don't understand the distortion. (=I'm confused, please clarify.)

E.g.

Quoting my formula wrong, makes it wrong

(unless typos, mistakes, etc caused that).

But then you can say it (=the formula that is supposed to be mine, but isn't) is wrong?

I don't follow.

 

Surely there must be a misunderstanding.

 

It looks like #19 could be wrong?

My error, sorry.

& Thanks.

Edited January 3, 2017 by Capiert

[swansont]

Your math was wrong; every insistence that you were doing it differently is an example of misusing established terminology (such as KE being negative, or a vector). The explanations are located in the previous posts.

[Capiert]

Including a unit vector with KE.

You could write .. sqrt(m₁E₁) + sqrt(m₂E₂) = sqrt(m₃E₃), as a vector equation.

That would work, because it's conservation of momentum.

p = sqrt(2mE),

and the sqrt(2) is divided out,

being common to all terms.

 

What you had was

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

 

[sqrt(m₁E₁) + sqrt(m₂E₂₎]²/ (m₁ + m₂)

 

With a vector, that's still just applying conservation of momentum. (It's wrong by a factor of two, however. KE = p²/2m)

Please clarify.

Above did you say (or mean) the formula above would work with a unit vector, because common 2 was divided out?

If it works, that nears saying it's right, it's not (completely) wrong, if it's working.?

 

What you've succeeded at is a more complicated way of doing something that physics already does.

If physics is already doing what I've done,

then it (=what I've "done" (in the formuls)) can't be wrong,

e.g. extracting KE from momentum.

(Because physics is suppose to be right.)

 

So we (=physics & me) are both either doing something right, or else something wrong.

(Because we get the "same" results.)

 

But then later you say "It's wrong by a factor of 2, however",

although the factor 2 seemed correctly cancelled.

 

So I'm still confused, what's right & wrong, & about that 2 (if it's ok=allowed to do cancelling so?).

 

Facit: What (or where) are the problems (=errors) if it is working?

 

Perhaps a few words (=sentences) about "unit" vector would help too.

I don't (really) know what is meant by "unit" (without an s for plural) other than e.g. kg, m, s units; or 1 (a single thing, e.g. radio).

 

Thanks in advance.

Edited January 3, 2017 by Capiert

[swansont]

Please clarify.

Above did you say (or mean) the formula above would work with a unit vector, because common 2 was divided out?

If it works, that nears saying it's right, it's not (completely) wrong, if it's working.?

The unit vector and the factor of 2 are separate issues. One is direction, the other is magnitude.

 

If physics is already doing what I've done, then it can't be wrong. (Because physics is suppose to be right.)

No, physics is doing what you intended to do. What you actually did was wrong.

 

Perhaps a few words (=sentences) about "unit" vector would help too.

Thanks in advance.

I don't (really) know what is meant by "unit" (without an s for plural) other than e.g. kg, m, s units; or 1 (a single thing, e.g. radio).

A unit vector has a length of 1 so it doesn't change the magnitude.

[Capiert]

The unit vector and the factor of 2 are separate issues. One is direction, the other is magnitude.

 

No, physics is doing what you intended to do. What you actually did was wrong.

Ok (=good, accepted). What is the actual mistake? The missing factor 2?

Meaning I'm not allowed to cancel the factor(s) 2?

I'm still confused there, needing orientation to help me.

Or do you mean other errors too? (Naturally, yes?)

Please show me to help me pinpoint them.

 

Thanks again.

Edited January 3, 2017 by Capiert

[swansont]

Ok (=good, accepted). What is the actual mistake? The missing factor 2?

Meaning I'm not allowed to cancel the factor(s) 2?

I'm still confused there, needing orientation to help me.

Or do you mean other errors too? (Naturally, yes?)

Please show me to help me pinpoint them.

 

Thanks again.

 

 

One error was treating KE as a vector when it isn't, as I explained.

 

The other is that KE = p²/2m, as I also explained, and which you acknowledged a few posts back ("Finally recognized. I'm very happy.")

 

Do you not recognize this mistake anymore?

[Capiert]

One error was treating KE as a vector when it isn't, as I explained.]

What is (the formal name for) KE called after you use the unit vector with it. "KE vector"? Please give it an appropriate name.

& what is the operation used? (Dot product? or multiply?)

 

The other is that KE = p²/2m, as I also explained, and which you acknowledged a few posts back ("Finally recognized. I'm very happy.")

 

Do you not recognize this mistake anymore?

My happyness was generally the summary for the whole post (icing on the cake, to top it off)

but with the detailed quotes (inserted later, that statement) seemed to refer only to the last quote,

instead of reinforced restatement (=getting better).

Generally you were recognizing the steps,

& the steps were going in a common direction,

climaxing finally with a correct formula,

instead of doubtful ones.

A day after I posted the formula #19

I thought I had (really) goofed!

But as the 2nd day rolled by

it occurred to me

that it (=that posted formula) still might work,

& so I then forgot about it along during the week.

That formula had a very complicated history,

& I had too separate myself from at least

half of its original development (stumbling on it)

after finding the approach with coe

shockingly failed on me,

& let me down.

I wasn't able to work on a recovery strategy

for more than a month, due to another job,

so I was in a terrible state (of) not knowing

what the score is.

So (later) I retried based on com instead

& got good results

except a few times

when coe seemed valid again

(but not com). Very puzzling.

Over the years I've noticed

about 5 errors

I can NOT explain

on other projects.

My indecision has gone on too long

so I thought I should post the formulas here

to get some help (with the math algebra (intended for the excel sheet))

& (try to) bring it (=the project) to an end, as finished.

(=Decided, & without errors.)

 

To answer your question:

 

I immediately recognized your #59 formulas

were missing the factor 2

because in #19 I had told our india friend (Sriman Dutta)

mom^2=2*m*E is pronounced "to me"

for quick identification (strategy).

That formula is only correct with the factor 2;

not without.

But that's not my problem.

I'm confused with the #19 "equation"

whether that formula (which now has the common factor 2 cancelled)

is valid

because (according to standard algebra rules)

if you asked me I would say yes;

but coming to this website forum

everybody has said no it (=that (algebra) formula) is wrong.

 

Can you understand how I feel?

I'm getting mixed up & confused.

Edited January 4, 2017 by Capiert

[Sensei]

Perhaps a few words (=sentences) about "unit" vector would help too.

I don't (really) know what is meant by "unit" (without an s for plural) other than e.g. kg, m, s units; or 1 (a single thing, e.g. radio).

 

"Unit vector" is other name for "normalized vector".

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

 

If you have 3D vector x,y,z

to calculate its length there is used length=sqrt(x^2+y^2+z^2)

then divide each dimension by length (it's called normalization):

x'=x/length

y'=y/length

z'=z/length

 

And x',y',z' will be normalized vector.

 

"unit" in its name has nothing to do with "unit" in physics.

 

You can reverse normalization by multiplication of normalized vector by previously calculated length:

 

x=x' * length

y=y' * length

z=z' * length

Edited January 4, 2017 by Sensei

[Capiert]

"Unit vector" is another name for "normalized vector".

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.

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?

But that's not my problem.

I'm confused with the #19 "equation"

whether that formula (which now has the common factor 2 cancelled)

is valid

because (according to standard algebra rules)

if you asked me I would say yes;

but coming to this website forum

everybody has said no it (=that (algebra) formula) is wrong.

Can you understand how I feel?

I'm getting mixed up & confused.

Would somebody please clarify that for me?

Either confirm; or reject it (=the formula in question)

& show me why it's wrong

& will fail,

so I can clean up things on my side.

Edited January 4, 2017 by Capiert

[swansont]

 

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.

 

 

 

Because KE does not have a direction. Neither does speed. They just give magnitudes. To make speed into a vector (velocity), you need direction information, too. It's not redundant.

[Capiert]

So you are proposing something new. Including a unit vector with KE. You could write the term as sqrt(mE). And then sqrt(m₁E₁) + sqrt(m₂E₂) = sqrt(m₃E₃), as a vector equation. That would work, because it's conservation of momentum. p = sqrt(2mE), and the sqrt(2) is divided out, being common to all terms. This is nothing new. IOW, it adds nothing, and is not a new insight.

 

What you had was

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

 

[sqrt(m₁E₁) + sqrt(m₂E₂₎]²/ (m₁ + m₂)

 

With a vector, that's still just applying conservation of momentum. (It's wrong by a factor of two, however. KE = p²/2m)

 

What you've succeeded at is a more complicated way of doing something that physics already does.

Is this following formula correct? (Yes?)

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

Is it not conservation of momentum? (No it is com?)

 

Aren't these 2 following equations equal?

(((2*m1*KE1)^0.5)+((2*m2*KE2)^0.5)^2)/(2*(m1+m2))=(((m1*KE1)^0.5)+((m2*KE2)^0.5)^2)/(m1+m2)

(Yes?)

 

Why then should that #19 formula be wrong?

Edited January 4, 2017 by Capiert

[swansont]

Is this following formula correct? (Yes?)

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

Is it not conservation of momentum? (No it is com?)

It is not correct. There are only scalars in the equation. There are many instances where this will yield the wrong answer, if you use the accepted mathematical and physics definitions.

 

 

Aren't these 2 following equations equal?

(((2*m1*KE1)^0.5)+((2*m2*KE2)^0.5)^2)/(2*(m1+m2))=(((m1*KE1)^0.5)+((m2*KE2)^0.5)^2)/(m1+m2)

(Yes?)

 

Why then should that #19 formula be wrong?

I've realized that the factor of 2 is not strictly the issue. (It's harder to decipher this since you hadn't derived anything in post 19). The equation is just flat-out wrong.

 

How do you apply it to a situation where both objects are moving both before and after a collision?

 

What of object 2 is at rest, and the masses are equal, in an elastic collision? The answer should be KE1 = KE3, but that's not what you get. Even in an inelastic collision, what you get will depend on how inelastic it is. The only situation where you can apply this is in a completely inelastic collision. And then, of course, it's possible to get KE3 = 0 since KE is not a vector.

 

Your derivation in post 21 fails when you fail to account for momentum being a vector.