(Don Kihot) fighting windmils

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dock

[SOLVED] (Don Kihot) fighting windmils

people!!!

i'm tyred of oposing each and every one of you.why don't someone, for a change, start to cooperate with me. you know that the way phisicist understand simple differential math just ain't right.i have hard evidence on this that no one can deny.just egknowledge my claims as rightfull.you have a junk-physics only to lose.

HERE IS WHAT I SAY IS WRONG IN GENERAL:

when you define v as dx/dt where x=x(t) then you cannot say that v is variable. you have to accept that v is constant.
same as if you assume that t=const then t=dx/dv where x=x(v).
if x=const then tdv=-vdt.
if none is constant then dx=vdt+tdv or x=vt.
it's not only dx=vdt but for every law with the form x=vt you do the same.for example dE=Fdx

Newton's dynamics:
dE(potential)=-FdR=-(wmMdR)/R^2 <=> E=(wmM)/R
dE(kinetic)=Fdx=madx=mdV(dx/dt)=mVdV <=> E=mVV/2
here you are talking about any (F and V) and yet F=const, V=const
FULL STOP.

PS: don't be afraid of the changes cause they are good.
 

chroot

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Exactly how long is the ignorant moron going to be allowed to continue to spam this forum?

- Warren
 

enigma

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The title of this thread is aptly named...
 
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Actually you are correct dock (the rest of you will love me in a second)

You say that you must assume v is a constant. Ok, check.

x = vt. Let's differentiate everything

dx = vdt + tdv Check.

NOW

You said that v is a constant. Well that means dv = 0 (since constants don't change).

Therefore:

dx/dt = v. Q.E.D.
 

Hurkyl

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when you define v as dx/dt where x=x(t) then you cannot say that v is variable.
Why is that true?

I would like to object to the abuse of notation x = x(t). Although it is a common abuse, the context of this discussion demands stricter adherence to mathematical grammar.


if you assume that t=const then t=dx/dv where x=x(v).
Why is that true?

Why would one assume t = constant?

Again I object to the abuse of notation x=x(v).


if x=const then tdv=-vdt.
I agree, since x = constant implies v = 0 implies dv/dt = 0.

I object to the ambiguity of your usage of dv and dt.


if none is constant then dx=vdt+tdv
Why?

Again I object to ambiguity.


... or x=vt.
Why?


Hurkyl
 
Last edited:

climbhi

Originally posted by Hurkyl
I would like to object to the abuse of notation x = x(t). Although it is a common abuse...
It certainly is a common abuse, my math professor uses this notation constantly.
 

dock

x = x(t) this only means that x is some function of t.

what is ambiguity?
 

Hurkyl

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The symbol 'x' is simultaneously being used as the symbol for coordinate position and as a symbol expressing the relationship between coordinate position and coordinate time.

Beyond that, the symbol 'x' isn't even standing for a unique function. In just this post alone you've used 'x' to refer to a function relating position and time and to refer to a completely different function relating position to velocity, and in the past you've even used the symbol 'x' as a (generally non-unique!) functional relation between x, v, and t.


This shorthand is used in practice because, if you know what's going on and apply everything correctly, they are all equivalent. I.E.

dx/dt = (d/dt)x(t) = (d/dt) (x(v(t))) = (d/dt) (x(v(t), t))

(and to be really technical, there's even more shorthand at work because I don't explicitly specify at what point the first expression is evaluated, and the latter three expressions should really be differentiated WRT a dummy variable and then evaluated at the same point as the first equation)


Given that dock denies the validity of even simple differential math, it would be incorrect for him to use abbreviations which rely on the results of differential math to remove the ambiguity. This abbreviation also obfuscates the underlying mathematics, making it entirely inappropriate to use when the underlying mathematics is being called into question and/or misunderstood.

Hurkyl
 
Last edited:

dock

OK!!
if you want that kind of precision, here it comes:
x=function1(t)
first of all you cannot write it this way cause x expresses in meters while t in seconds and function1(t) in function1(seconds) and in the end you end up with counterparts incompatibility.therefore you will have to write it like this:
x=function2(v,t)
if it is so then:
dx=(@x/@t)dt+(@x/@v)dv
according to the traditional physics
dx=vdt
follows
(dx=dx) and (@x/@t=v) and ((@x/@v=t=0) or (dv=0))
since t is any comes that v=const
to conclude:
dx=vdt <=> v=const

neither Hutkly nor any one can argue with me on this subject cause as you see i have hard evidence.
 
499
1
[q]according to the traditional physics
dx=vdt
follows
(dx=dx) and (@x/@t=v) and ((@x/@v=t=0) or (dv=0))
since t is any comes that v=const
to conclude:
dx=vdt <=> v=const[/q]

What are we looking for here? You have the same thing as the tradition physics.
 

FZ+

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first of all you cannot write it this way cause x expresses in meters while t in seconds and function1(t) in function1(seconds) and in the end you end up with counterparts incompatibility.therefore you will have to write it like this:
That is incorrect. Function 1 would be a function that convert time to distance. For another example, sin would be a function that converts degrees to fractions. And so on. Your objection is invalid. The velocity function is a function of time only.

To show this, let's give another example.
Say v = t^2 <=> v = f(t) where f: t -> t^2
v is function of t only, is it not?

Differentiating gives dv/dt = 2t

Now clearly as time changes, v changes. So v cannot be constant and is a function in t only. See?

If the differential of the velocity function with respect to time is non-zero, then velocity is variable.
 

Hurkyl

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FZ already discussed the invalidity of your argument that x cannot be written as a function of t alone.


I will discuss the flaw in your argument in situations where x can be written as a function of v and t.


x=function2(v,t)
if it is so then:
dx=(@x/@t)dt+(@x/@v)dv
This is complete gibberish. x is not a function, so partial differentiating it does not make sense.

Also, your usage of "dx" is, as previously mentioned, unclear as to which concept you are actually using. For the purposes of the rest of this post I will presume by "dx" you mean the differential form that shares that name.


The correct conclusion you could make from x = function2(v, t) is that over any curve &gamma satisfying

x(s) = function2(v(s), t(2))

you can conclude

&int&gamma dx = &int&gamma([pard]function2 / [pard]t) dt + ([pard]function2 / [pard]v) dv

which is abbreviated as

dx = ([pard]function2 / [pard]t) dt + ([pard]function2 / [pard]v) dv

Alternatively, if you don't like abbreviations, the above statement is true equality when you restrict your attention solely to a single 2-dimentional differential manifold where for any point (x, v, t) on that manifold the relation

x = function2(v, t)

holds.


dx=vdt
follows
(dx=dx) and (@x/@t=v) and ((@x/@v=t=0) or (dv=0))
Incorrect.

The equation dx = v dt is only valid when (x, v, t) is restricted to the trajectory of a particle (i.e. x is the position at time t and v is the velocity at time t for the particle). Trajectories are 1-dimentional (differential manifolds), meaning in particular that dx, dv, and dt are cannot linearly independant differential forms, because that would imply dimention greater than 1. Your scheme of equating coefficiencts on dx and dv is only a valid operation when dx and dv are linearly independant.

Besides that, [pard]x/[pard]v is, in general, inequal to t, so there's no reason for it to appear in your deductions.


Hurkyl
 

dock

Originally posted by Brad_Ad23
[q]according to the traditional physics
dx=vdt
follows
(dx=dx) and (@x/@t=v) and ((@x/@v=t=0) or (dv=0))
since t is any comes that v=const
to conclude:
dx=vdt <=> v=const[/q]

What are we looking for here? You have the same thing as the tradition physics.
no member i do not.i have more general one.according to mua(me) the instaneous velocity v equals to instaneous position x over instaneous time t;
x = vt
here v can be but doesn't have to be constant and this is a bit different then instaneous velocity v equals to the first derivate of some functional dependent position x from time t only.the difference is that the traditional (dx=vdt) aplies only on v=const because dv expression is missing in the equation.

feel free to ask me more about it.i'd be glad to answer.
 

dock

Originally posted by FZ+
That is incorrect. Function 1 would be a function that convert time to distance. For another example, sin would be a function that converts degrees to fractions. And so on. Your objection is invalid. The velocity function is a function of time only.

To show this, let's give another example.
Say v = t^2 <=> v = f(t) where f: t -> t^2
v is function of t only, is it not?

Differentiating gives dv/dt = 2t

Now clearly as time changes, v changes. So v cannot be constant and is a function in t only. See?

If the differential of the velocity function with respect to time is non-zero, then velocity is variable.
since dv<>0 (so you say) then why dx<>vdt+tdv but dx=vdt only?

you obviously don't know my favorite argument against that sort of objection.whatch this:
example 1:
let x = function1(t) = t;
dx = vdt so you say;
then v = dx/dt = dt/dt = 1 (now no counter parts);
we said that x = t then x = 1 * t but since v = 1 then
x = v * t remeber this as conclusion (1).
example 2:
let x = function2(t) = t^2;
v = dx/dt = d(t^2)/dt = 2tdt/dt = 2t (now seconds as counter parts);
it was x = t^2 then x = 2t(t/2) it was v = 2t then
x = v * t / 2 remeber this as conclusion (2)

first of all both coclusions (1) and (2) stongly suggest that x is function of both velocity v and time t and
second how come conclusions (1) and (2) yield different laws for different situation?this i can tell you:
the mightiest law of them all says that every change is simultaneous with some other.what you do with dx=vdt is making the change dx simultaneous with the change dt but the change dv simultaneous with the change of the law x=vt.

I'M THE ONE WHO MAY OBJECT!!!
 

dock

Originally posted by Hurkyl
This is complete gibberish. x is not a function, so partial differentiating it does not make sense.
Also, your usage of "dx" is, as previously mentioned, unclear as to which concept you are actually using. For the purposes of the rest of this post I will presume by "dx" you mean the differential form that shares that name.
is this an dekoy to destruct me from the main point.you are just making big fus about tiny little unimportant thigs.
dx=vdt
follows
(dx=dx) and (@x/@t=v) and ((@x/@v=t=0) or (dv=0))
Incorrect.
sorry hurkyl you missunderstood me about this.i wasd soling the system of this two equations:
(1st equation)dx=(@function2/@t)dt+(@function2/@v)dv
and
(2nd equation)dx=vdt+0
 

Hurkyl

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is this an dekoy to destruct me from the main point.you are just making big fus about tiny little unimportant thigs.
The devil is in the details! If you don't even know what the symbols you're using mean, how can you possibly expect to convince other people that establishment physics is wrong?

And how, pray tell, is it a distraction from your main point? I thought you were trying to explain how physics is wrong, wouldn't that necessitate proper usage of the physics?

Or is your main point just dogmatic self-aggrandizing?


sorry hurkyl you missunderstood me about this.i wasd soling the system of this two equations:
(1st equation)dx=(@function2/@t)dt+(@function2/@v)dv
and
(2nd equation)dx=vdt+0
I did not misunderstand. Your steps in solving this system of equations was flawed; equating coefficients in an equation of vectors is only legal when the vectors are independant.

Hurkyl
 

dock

Originally posted by Hurkyl
I did not misunderstand. Your steps in solving this system of equations was flawed; equating coefficients in an equation of vectors is only legal when the vectors are independant.
so what will be your solution to that system then?

by the way q:
if we come to an agreement with my point then what?
 

FZ+

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then v = dx/dt = dt/dt = 1
How the hell did you get that?

You seem not to understand the idea here at all.

say F(x) = k*x Where k is a constant.

This is a function in x, not a function in k and x.
In your equation v is a number which you use to convert t to x. But the v in physics does not represent a number, it is the expression itself. It is a function involving t, as the integral of acceleration.

Displacement is not a function of v and t. It is equal to a function of t on it's own. A function that happens to be the integral of velocity with respect to time.

x = v*t and x=f(t) are not equivalent. They can happen to be equivalent under certain conditions. When v is a constant, for example and contains no term in t.

By your argument, x = vt is not a function in v and t, but a function in a, b and t. As v = a * b. You can expand it forever like this, but it is not valid maths. Do you see?

let x = function2(t) = t^2;
v = dx/dt = d(t^2)/dt = 2tdt/dt = 2t (now seconds as counter parts);
it was x = t^2 then x = 2t(t/2) it was v = 2t then
x = v * t / 2 remeber this as conclusion (2)
This is incorrect maths. You can extract expressions out of thin air like this, but it reduces down to replacing v with a fudge function to make it fit your theory. v is dx/dt <=> x = &int v dt. In this case, you can separate this and make it a multiple of t, but x is a function of t only. You insert phantom functions in, but this is completely meaningless. Try:
x = e^t

All you will end up is a garbled function that has no relation to that little thing called the real world.
 
Last edited:

dock

Originally posted by FZ+
How the hell did you get that?

You seem not to understand the idea here at all.

say F(x) = k*x Where k is a constant.

This is a function in x, not a function in k and x.
In your equation v is a number which you use to convert t to x. But the v in physics does not represent a number, it is the expression itself. It is a function involving t, as the integral of acceleration.

Displacement is not a function of v and t. It is equal to a function of t on it's own. A function that happens to be the integral of velocity with respect to time.

x = v*t and x=f(t) are not equivalent. They can happen to be equivalent under certain conditions. When v is a constant, for example and contains no term in t.

By your argument, x = vt is not a function in v and t, but a function in a, b and t. As v = a * b. You can expand it forever like this, but it is not valid maths. Do you see?


This is incorrect maths. You can extract expressions out of thin air like this, but it reduces down to replacing v with a fudge function to make it fit your theory. v is dx/dt <=> x = &int v dt. In this case, you can separate this and make it a multiple of t, but x is a function of t only. You insert phantom functions in, but this is completely meaningless. Try:
x = e^t

All you will end up is a garbled function that has no relation to that little thing called the real world.
gibbrish A?
watch this:
(0)x=e^t right?
then
(1)t=ln(x) from (0)
(2)dx/dt=v your def
(3)dx/dt=e^t from (0)
(4)v=e^t from (2) and (3)
(5)t=ln(v) from (4)
(6)x=v=e^t and t=ln(v)=ln(x) from (1) and (5)
(7)x=1*v math says
(8)1=ln(x)/ln(v) from (6)
(9)x=v(ln(x)/ln(v)) from (7) and (8)
(a)x=vt/ln(v) from (1) and (9)
(b)x=function(v,t) from (a)

comment: in (a) there would not have been 1/ln(v) if you work as i say accordin to what dx=vdt+tdv.apearience of 1/ln(v) is due to your so called definition dx=vdt
 

Hurkyl

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so what will be your solution to that system then?
The solution would depend on knowledge of the trajectory at hand.

Hurkyl
 

dock

Originally posted by Hurkyl
The solution would depend on knowledge of the trajectory at hand.
i wanned to tell ya that considering only the derivates=d(something)
dv and dt are independent variables while dx is the dependent one in both equation.dx eliminates however you turn the system and what remains is:
vdt+0=(@f/@t)dt+(@f/dv)dv
therefore i'm sure that i can equalize the coeficients.

but any way, please pick one specific case and show us your approach.
 

FZ+

1,550
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Originally posted by dock
gibbrish A?
watch this:
(0)x=e^t right?
then
(1)t=ln(x) from (0)
(2)dx/dt=v your def
(3)dx/dt=e^t from (0)
(4)v=e^t from (2) and (3)
(5)t=ln(v) from (4)
(6)x=v=e^t and t=ln(v)=ln(x) from (1) and (5)
(7)x=1*v math says
(8)1=ln(x)/ln(v) from (6)
(9)x=v(ln(x)/ln(v)) from (7) and (8)
(a)x=vt/ln(v) from (1) and (9)
(b)x=function(v,t) from (a)

comment: in (a) there would not have been 1/ln(v) if you work as i say accordin to what dx=vdt+tdv.apearience of 1/ln(v) is due to your so called definition dx=vdt
And x = vt/lnv == x = v
So x is a function of v only.
And v is a function of t.

And what does all that mathematical contortion you just did mean? And your self cancelling out terms?

Bugger all.

Welcome to the real world.
 
Last edited:

dock

Originally posted by FZ+
And x = vt/lnv == x = v
So x is a function of v only.
And v is a function of t.

And what does all that mathematical contortion you just did mean? And your self cancelling out terms?

Bugger all.

Welcome to the real world.
Konfuche says:-"scratch my back and i'll scratch yours"

so x is function of v only.
but x is function of t only too cause x=exp(t)

if you prefer x=v instead of x=vt/ln(V) then prefer dx=dv instead of dx=vdt (the def)

by the way if dv=dx<>0 then why there is no dv in dx=vdt+0?
 

Hurkyl

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but any way, please pick one specific case and show us your approach.
Ok, how about the trajectory T defined by the system of differential equations:

x = e^t
v = dx/dt
for all (x, v, t) on T


This trajectory is equivalent to the functional system of equations:

x = e^t
v = e^t
for all (x, v, t) on T


If we take the total derivative of these equations with respect to t, we see that:

dx/dt = e^t
dv/dt = e^t


Now, I want to write the differential form dx in terms of the differential form dt. I can do this by applying dx to an arbitrary curve &gamma and applying the change of coordinates law of integration:

&int&gamma; dx = &int&gamma; (dx/dt) dt
= &int&gamma; e^t dt

Since &gamma was an arbitrary curve, we have the functional equality:

dx = e^t dt

similarly we have dv = e^t dt


Clearly dx, dv, and dt are not linearly independant because dx and dv are both simply dt multipled by a (function) coefficient.


So now we step back to the system of equations we want to solve:

(1st equation)dx=(@function2/@t)dt+(@function2/@v)dv
and
(2nd equation)dx=vdt+0

if we substitute in the functional equalities derived above, the first equation becomes:

e^t dt = [pard]function2/[pard]t dt + [pard]function2/[pard]v e^t dt

e^t dt = ([pard]function2/[pard]t + [pard]function2/[pard]v e^t) dt

Because dt is a nondegenerate functional, (a.k.a. the set { dt } is linearly independant), here we can equate coefficients yielding

e^t = [pard]function2/[pard]t + [pard]function2/[pard]v e^t

Let's now choose a function2. How about:

function2(a, b) = a * b / ln a

So that function2(v, t) = v * t / ln v, and via your derivation, x = function2(v, t) is a valid equation.


[pard]function2(v, t)/[pard]t = v / ln v
[pard]function2(v, t)/[pard]v = t * (ln v - 1) / (ln v)^2

So that your first equation in your system of differential equations is equivalent to:

e^t = v / ln v + e^t * t * (ln v - 1) / (ln v)^2

However, we know that v = e^t is a valid equation on the trajectory T, so if we substitute that equation into the above:

e^t = e^t / t + e^t * t * (t - 1) / (t)^2
= e^t / t + e^t * t^2 / t^2 - e^t * t / t^2
= e^t

Yay, it worked.


dv and dt are independent variables
That's your mistake; dv and dt are not independant variables. For the lifetime of any given particle, it has a unique velocity for any given time, which means that we can write a functional relationship between time and the velocity of that particle, which means that dv and dt are not independant functionals.

On a side note, dv and dt aren't even variables.



Hurkyl
 
Last edited by a moderator:

dock

Originally posted by Hurkyl
dx=([pard]function2(v, t)/[pard]t)dt+([pard]function2(v, t)/[pard]v)dv
[pard]function2(v, t)/[pard]t = v / ln v
[pard]function2(v, t)/[pard]v = t * (ln v - 1) / (ln v)^2
your mistake is that you say for every (x,v,t) on the curve T but write x=somefunction(t) and v=otherfunction(t).
insted it should be:

x=firstfunction(s)
v=secondfunction(s)
t=thirdfunction(s)
where s is independent

and then dx=(([pard]firstfunction(s)/[pard]s)/([pard]thirdfunction(s)/[pard]s))(ds/ds)dt

dx=(e^t / t)dt+(t * (t - 1) / t^2)dv
dx=(e^t / t)dt+(t * (t - 1) / t^2)e^tdt
dx=e^tdt

that's why inthe equation dx = vdt + 0 there is no dv?


if x is function2(v,t) then in no way

(dx is function of dt alone) and (dv is function of dt alone)

at the same time cause this way dt is the only indepedent tush x=function2(v,t) is 2-dimensional and we can eliminate dt and t to get dx as function of dv only and x as some function of v only (t is obsolite then).for convinience i actually got x=v under same conditions, didn't i?draw x=v in (x,v) enviroment you get strait line,do you?by the way x=v means that one (either x or v) is over suficient while describing the line if you introduce t.

instead it should be:
x=function2(v,t) is actually 3d surface (not 2d curve).
dx is function of (dt and dv) where dv and dt are both independent.
in this case

if (dx=Adt+Bdv) and (xd=Cdt+Ddv) then (Adt+Bdv=Cdt+Ddv)
and since dv and dt are independent => (A=C) and (B=D)


WHAT ABOUT THE COUNTERPARTS?
WHAT ARE THE COUNTERPARTS FOR X AND V HERE?
WHY THEY DON'T FIT?
 

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