# (Don Kihot) fighting windmils

[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
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.

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

- Warren

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The title of this thread is aptly named...

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.

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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

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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.

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

what is ambiguity?

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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

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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
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 anyone can argue with me on this subject cause as you see i have hard evidence.

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+
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.

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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

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.

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!

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

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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?

(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

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+
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.

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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)
(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

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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

Originally posted by Hurkyl
The solution would depend on knowledge of the trajectory at hand.
i wanned to tell you 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+
Originally posted by dock
gibbrish A?
watch this:
(0)x=e^t right?
then
(1)t=ln(x) from (0)
(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:
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?

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Gold Member
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:
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) environment 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.

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

and since dv and dt are independent => (A=C) and (B=D)

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

FZ+
Originally posted by dock
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?

Let's repeat: x is the integral of v with respect to t.
x is NOT a function of v
Ths only way you involved two variables in that function was to create one that canceled itself up. That, like saying x is a function in a b and c where x = a/a * b/b * c/c is not valid maths. And your values still bear no relevance to the real world.

v is an expression. It is not a variable. If it varies, what does it vary with respect to? Time. So ultimately x is based on t, not v.

You still refuse to get it.

x = &int v dt in all cases. This is the general definition.
In this case, it is also true that &int v dt = v but this is not the general rule.

Let's try another one in case you actually learn something.

x = cos t.

Originally posted by FZ+
dx=vdt this is the general definition.
Let's try another one in case you actually learn something.

x = cos t.
at first place saying x=cos(t) means that there is no need ov v. v becomes involved only because of your def dx=vdt which by the way i find particular instead of general because there is no dv in it or you are working with v=const only.
you can throw any function at me and i think i can handle it.
x=cos(t)
am introducing the following substitute:
tg(t/2)=s
2acrtg(s)=t
cos(t)=(1-ss)/(1+ss)
sin(t)=2s/(1+ss)
dt=2ds/(1+ss)
now

x=vt*(tg^2(t)-1)/(arcsin(-v)tg(t/2))

FZ+
And do you now see why this is completely pointless? By your method, you can have an infinite number of variables by simply adding more complexity into the equation. You have shown nothing, but how you can introduce a self cancelling out term. In each you have show precisely the relation between v and t. And hence shown that x is still based on t, not v.

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Gold Member
your mistake is that you say for every (x,v,t) on the curve T but write x=somefunction(t) and v=otherfunction(t).

Why is it a mistake? Any rank-1 differential manifold embedded in 3-space (such as a trajectory in (x, v, t) space) can be written as the solution set of two independant functional equations.

We can do better. Suppose we do parametrize the curve T in s, as you suggest. Then the curve (x, v, t) is described by the three equations:

x = f(s)
v = g(s)
t = h(s)

for some functions f, g, and h.

If dh(s)/ds is never zero, then by the implicit function theorem there exists a function p such that for every point on the curve:

s = p(t)

And if we define the functions:

q(t) = f(p(t))
r(t) = g(p(t))

Then substitution yields the following system of equations

x = q(t)
v = r(t)
t = h(p(t))

Notice, though, that s = p(t) = p(h(t)), so p and h are inverses, so the above system of equations is identical to:

x = q(t)
v = r(t)
t = t

Since the last equation is redundant, it can be eliminated to yield the equivalent system of equations:

x = q(t)
v = r(t)

So the solution set to the above two equations is identical with the trace of the original curve T.

To sum up the implications of this:

if (x, v, t) is a point on the curve (f(s), g(s), h(s))
and dh(s) / ds is never equal to 0,
then there exist functions q and r such that x = q(t) and v = r(t)

I would like to point out that these are equations, not definitions.

The implicit function theorem is compatable with the arithmetic on differential forms. i.e. the following equalities hold on the curve T:

dx = (df(s) / ds) ds = (dq(t) / dt) dt
dv = (dg(s) / ds) ds = (dr(t) / dt) dt

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

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

On the contrary, if (x, v, t) is constrained by two independant functional equations (such as restricting (x, v, t) to lie on some fixed curve), then {dx, dt} is linearly dependent, as well as {dv, dt} and {dx, dv}

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)

Partially correct. The solution set to the equation x = function2(v, t) is generally a 2-dimensional surface. However, this equation is not the only constraint on (x, v, t). In particular, the fact (x, v, t) lies on a curve generally forces there to be two independant functional constraints on (x, v, t).

for convinience i actually got x=v under same conditions, didn't i?draw x=v in (x,v) environment 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.

Yep. Over regions where the right derivatives never vanish, you can choose any of the three coordinates (x, v, t) to parametrize the curve. The fact {x, v, t} are functionally dependant mean there are lots of choices of functions with f(x, v, t) = 0, including functions where any of those 3 parameters are absent from the function.

x=function2(v,t) is actually 3d surface (not 2d curve).

If you meant to say:

The solution set to x = function2(v, t) is generally a 2-dimensional surface embedded in 3-dimensional space, you'd be correct.

dx is function of (dt and dv) where dv and dt are both independent.
in this case

If (x, v, t) are allowed to range over the entire solution set of

x = function2(v, t)

then in general you'd be correct. However, (x, v, t) as physical variables have further constraints.

Hurkyl

Staff Emeritus
Gold Member

I have been using differential geometry as the basis of my previous posts, because you seemed to want to go that way.

In the interest of demonstrating one of my previous points that the notation you are using is an abbreviation for lots of things, I will do the same for a more traditional approach, which FZ is using.

For a given particle, we define the function:

x(t) := the position of the particle at time t

This means we have some way of coordinatizing position and time, and at time coordinate t, the position coordinate of the particle is x(t).

Notice t is NOT DEFINED; t is merely a dummy parameter. The above definition is exactly equivalent to:

x(s) := the position of the particle at time s

or

x(&xi) := the position of the particle at time &xi

So we have defined the function x. Let us define another function v as:

v := dx/dt

Is that a good equation? Yes... but it's an abbreviated equation. If you wanted to do it without abbreviation, it would look like one of the following (all are equivalent):

v := x'

v(t) := x'(t)

v(t) := dx(s)/ds |s=t

v(t) := lims->t (x(t) - x(s)) / (t - s)

Again, t and s in the above equation are dummy variables. They only exist to indicate which parameters in the expressions must be equal.

Let us again work with the same example:

x(t) = et

However, this time is different. In the differential geometry approach, x was defined as the dummy parameter that was always representing the first coordinate of our 3 dimensional space R3. Here, x is a function that maps any real number to the position coordinate of the given particle at time coordinate equal to that real number (which just happens to be e to the power of that real number). Again, t is a dummy variable and we can use any symbol in its place to get an equivalent expression:

x(&psi) = e&psi

we have defined v above, so now that we have an explicit formula for x, we can apply our definition to get an explicit formula for v:

v(t) = des/ds |s = t
= es |s = t
= et

You derived the formula:
x=vt/ln(v)

properly notated, this becomes

x(t) = v(t) * t / ln v(t)

t is, again a dummy parameter.

Now, per your program, we would like to differentiate both sides of this equation. Applying the laws of differentiation yield (and defining a := v'):

x'(t) = v'(t) * t / ln v(t) + v(t) * 1 / ln v(t) - v(t) * t / (ln v(t))2 * (1 / v(t)) * v'(t)

v(t) = a(t) t / ln v(t) + v(t) / ln v(t) - a(t) t / (ln v(t))2

This equation will still be valid if we substitute in our above equations for v and the one we can derive for a:

et = et t / t + et / t - et t / t2
= et + et / t - et / t
= et

You like to ask the question differently, you ask what dx equals.

Well, dx is an overloaded expression. One of its meanings, as I mentioned before, is that of a differential form. The relevant definition is:

dx(t) := x'(t) dt

Let us rewind a moment. You like to write the equation x = f(v, t) (NOTE: this is an equation, not a definition). This is an abbreviated equation, the full form is:

x(t) = f(v(t), t)

where

f(&alpha, &beta) = &alpha * &beta / ln &alpha

We can define a new function g(t) := f(v(t), t), and then x(t) = g(t) and there is no problem, but it is instructive to see how calculus works when we don't do that!

Let us again take the derivative of your equation.

x(t) = f(v(t), t)
implies
x'(t) = f1(v(t), t) * v'(t) + f2(v(t), t)

I've used a standard subscript notation for partial differentiation operators. Another (unabbreviated) way to write this same equation is:

x'(t) = [pard]f(&nu, &tau)/[pard]&nu |(&nu, &tau) = (v(t), t) * v'(t) + [pard]f(&nu, &tau)/[pard]&tau |(&nu, &tau) = (v(t), t)

This deduction, of course, only holds whenever f is a function such that:

x(t) = f(v(t), t)

is true for all t.

You like to write differential forms, so let's apply the relevant theorems on differential forms to your equation:

dx(t) = df(v(t), t) = f1(v(t), t) dv(t) + f2(v(t), t) dt

We also know that:

dx(t) = x'(t) dt = v(t) dt

So now we can substitute above:

v(t) dt = f1(v(t), t) dv(t) + f2(v(t), t) dt

However, recall that dv(t) is defined as:

dv(t) := v'(t) dt = a(t) dt

So we substitute in above:

v(t) dt = f1(v(t), t) a(t) dt + f2(v(t), t) dt

And now, since there is a theorem that let's us do so, we can equate coefficients on dt:

v(t) = f1(v(t), t) a(t) + f2(v(t), t)
(except possibly on a set of measure 0)

In physics we would typically presume that the functions we use are all continuous. When we do so, then the above equation holds for every t.

Hurkyl

Last edited:
ObsessiveMathsFreak
What the hell is going on here.

I read someone arguing that velocity was a constant and now there is talk of making t a dependant variable.

Can someone explain what points are trying to be made here?

actually the problem was:
whenever one defines V as dx/dt V has to be constant.
if V is variable then dx=Vdt+tdV instead of dx=Vdt.

what do you think of it as a mathematician, ObsessiveMathsFreak?

Staff Emeritus
Gold Member
And to me it seems that Dock is going through a phase I once went through... except that Dock is promoting his misunderstandings to the level of belief while I spent effort trying to figure out the underlying structures that rigorously support these types of manipulations.

I thought it might help dock to see the explicit pieces that go into some of these manipulations; I know it would've helped me back then.

Hurkyl

Actually, dx&nbsp;=&nbsp;vdt does not imply that v is a constant, it merely implies that x&nbsp;=&nbsp;vt&nbsp;&minus;&nbsp;&int;tdv. I have no problem with this.

ObsessiveMathsFreak
v = dx/dt

I think this fact is agreed upon by all at least.

What does it mean. It means that velocity is the instantaneous rate of change of distance. In other words, at any given instant, v is a measure of how fast you are going.

but what happens in the case of

x = Sin(t)