| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
16 May 2007 02:23:23 PM |
| Object: |
@T/@qj = 0 in Cartesian co-ordinates? |
On page 20 of Goldstein he says that the partial derivative of the
kinetic energy with respect to qj vanishes when Cartesian Coordinates
are used. For a single particle of mass m, T = 1/2mv^2, qj = x, so
this means @/@x(x_dot^2) = 0, correct?
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| User: "John C. Polasek" |
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| Title: Re: @T/@qj = 0 in Cartesian co-ordinates? |
16 May 2007 09:42:47 PM |
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On 16 May 2007 12:23:23 -0700, wrote:
On page 20 of Goldstein he says that the partial derivative of the
kinetic energy with respect to qj vanishes when Cartesian Coordinates
are used. For a single particle of mass m, T = 1/2mv^2, qj = x, so
this means @/@x(x_dot^2) = 0, correct?
Yes. xdot has nothing to do with x.
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| User: "blackhead" |
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| Title: Re: @T/@qj = 0 in Cartesian co-ordinates? |
21 May 2007 11:02:08 AM |
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On May 17, 3:42 am, John C. Polasek <jpola...@cfl.rr.com> wrote:
On 16 May 2007 12:23:23 -0700, wrote:
On page 20 of Goldstein he says that the partial derivative of the
kinetic energy with respect to qj vanishes when Cartesian Coordinates
are used. For a single particle of mass m, T = 1/2mv^2, qj = x, so
this means @/@x(x_dot^2) = 0, correct?
Yes. xdot has nothing to do with x.
It's strange that I know you are right, yet I have comprehension
problems with not really understanding what an independent variable
is! Even though Tx may vary over x and t, that does not mean it's
dependent on these variables any more than it's dependent upon the
length of a continuously increasing line, L(t), I might decide to
draw. Mathematically, T can be written as T(L(t)), but physically T is
still independent of L(t). Is this reasoning correct?
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| User: "John C. Polasek" |
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| Title: Re: @T/@qj = 0 in Cartesian co-ordinates? |
21 May 2007 03:05:26 PM |
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On 21 May 2007 09:02:08 -0700, blackhead <larryharson@softhome.net>
wrote:
On May 17, 3:42 am, John C. Polasek <jpola...@cfl.rr.com> wrote:
On 16 May 2007 12:23:23 -0700, wrote:
On page 20 of Goldstein he says that the partial derivative of the
kinetic energy with respect to qj vanishes when Cartesian Coordinates
are used. For a single particle of mass m, T = 1/2mv^2, qj = x, so
this means @/@x(x_dot^2) = 0, correct?
Yes. xdot has nothing to do with x.
It's strange that I know you are right, yet I have comprehension
problems with not really understanding what an independent variable
is! Even though Tx may vary over x and t, that does not mean it's
dependent on these variables any more than it's dependent upon the
length of a continuously increasing line, L(t), I might decide to
draw. Mathematically, T can be written as T(L(t)), but physically T is
still independent of L(t). Is this reasoning correct?
No, T(L(t)) makes T dependent on position, but kinetic energy is
purely dependent on velocity. Just for the practice, try to write the
function of L.
Your expression @/@x(x_dot^2) tries to express how xdot is affected by
a change in x, and it isn't. x and xdot can be considered orthogonal
to each other so there's no partial coefficient of sensitivity between
them. Yes, xdotdot is affected by x in gravity but then you'd be using
something that wasn't given in the Lagrangian.
In polar coordinates T could be mr^2*w^2/2, therefore the derivative
with r exists as Goldstein said.
John Polasek
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| User: "blackhead" |
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| Title: Re: @T/@qj = 0 in Cartesian co-ordinates? |
22 May 2007 08:03:39 AM |
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On 21 May, 21:05, John C. Polasek <jpola...@cfl.rr.com> wrote:
On 21 May 2007 09:02:08 -0700, blackhead < >
wrote:
On May 17, 3:42 am, John C. Polasek <jpola...@cfl.rr.com> wrote:
On 16 May 2007 12:23:23 -0700, wrote:
On page 20 of Goldstein he says that the partial derivative of the
kinetic energy with respect to qj vanishes when Cartesian Coordinates
are used. For a single particle of mass m, T = 1/2mv^2, qj = x, so
this means @/@x(x_dot^2) = 0, correct?
Yes. xdot has nothing to do with x.
It's strange that I know you are right, yet I have comprehension
problems with not really understanding what an independent variable
is! Even though Tx may vary over x and t, that does not mean it's
dependent on these variables any more than it's dependent upon the
length of a continuously increasing line, L(t), I might decide to
draw. Mathematically, T can be written as T(L(t)), but physically T is
still independent of L(t). Is this reasoning correct?
No, T(L(t)) makes T dependent on position, but kinetic energy is
purely dependent on velocity. Just for the practice, try to write the
function of L.
Your expression @/@x(x_dot^2) tries to express how xdot is affected by
a change in x, and it isn't. x and xdot can be considered orthogonal
to each other so there's no partial coefficient of sensitivity between
them. Yes, xdotdot is affected by x in gravity but then you'd be using
something that wasn't given in the Lagrangian.
In polar coordinates T could be mr^2*w^2/2, therefore the derivative
with r exists as Goldstein said.
John Polasek- Hide quoted text -
- Show quoted text -
I just find it confusing that xdot will in general vary with x, even
though it isn't directly dependent on it, so that xdot(x) and @xdot/@x
can be formed. For example, a constant force Fx applied to a particle
gives rise to an xdot that varies with x. Even though x does not
directly affect xdot, @xdot/@x can still be formed. Correct?
.
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| User: "John C. Polasek" |
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| Title: Re: @T/@qj = 0 in Cartesian co-ordinates? |
22 May 2007 09:14:18 AM |
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On 22 May 2007 06:03:39 -0700, blackhead <larryharson@softhome.net>
wrote:
On 21 May, 21:05, John C. Polasek <jpola...@cfl.rr.com> wrote:
On 21 May 2007 09:02:08 -0700, blackhead < >
wrote:
On May 17, 3:42 am, John C. Polasek <jpola...@cfl.rr.com> wrote:
On 16 May 2007 12:23:23 -0700, wrote:
On page 20 of Goldstein he says that the partial derivative of the
kinetic energy with respect to qj vanishes when Cartesian Coordinates
are used. For a single particle of mass m, T = 1/2mv^2, qj = x, so
this means @/@x(x_dot^2) = 0, correct?
Yes. xdot has nothing to do with x.
It's strange that I know you are right, yet I have comprehension
problems with not really understanding what an independent variable
is! Even though Tx may vary over x and t, that does not mean it's
dependent on these variables any more than it's dependent upon the
length of a continuously increasing line, L(t), I might decide to
draw. Mathematically, T can be written as T(L(t)), but physically T is
still independent of L(t). Is this reasoning correct?
No, T(L(t)) makes T dependent on position, but kinetic energy is
purely dependent on velocity. Just for the practice, try to write the
function of L.
Your expression @/@x(x_dot^2) tries to express how xdot is affected by
a change in x, and it isn't. x and xdot can be considered orthogonal
to each other so there's no partial coefficient of sensitivity between
them. Yes, xdotdot is affected by x in gravity but then you'd be using
something that wasn't given in the Lagrangian.
In polar coordinates T could be mr^2*w^2/2, therefore the derivative
with r exists as Goldstein said.
John Polasek- Hide quoted text -
- Show quoted text -
I just find it confusing that xdot will in general vary with x, even
though it isn't directly dependent on it, so that xdot(x) and @xdot/@x
can be formed. For example, a constant force Fx applied to a particle
gives rise to an xdot that varies with x. Even though x does not
directly affect xdot, @xdot/@x can still be formed. Correct?
You can make an artificial case. Ex: v = sqrt(2gx) constant g, but v
comes first from time, and x from time and with algebra you can make
v(x). But then it comes down to
dv/dx = g/v
but I don't believe it is valid to write
@v/@x
but you can write
@(v^2/2)/@x = g
but v^2 isn't v.
I think you are wasting your time; what are you going to do with it?
John Polasek
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| User: "blackhead" |
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| Title: Re: @T/@qj = 0 in Cartesian co-ordinates? |
22 May 2007 12:45:13 PM |
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On May 22, 3:14 pm, John C. Polasek <jpola...@cfl.rr.com> wrote:
On 22 May 2007 06:03:39 -0700, blackhead < >
wrote:
On 21 May, 21:05, John C. Polasek <jpola...@cfl.rr.com> wrote:
On 21 May 2007 09:02:08 -0700, blackhead < >
wrote:
On May 17, 3:42 am, John C. Polasek <jpola...@cfl.rr.com> wrote:
On 16 May 2007 12:23:23 -0700, wrote:
On page 20 of Goldstein he says that the partial derivative of the
kinetic energy with respect to qj vanishes when Cartesian Coordinates
are used. For a single particle of mass m, T = 1/2mv^2, qj = x, so
this means @/@x(x_dot^2) = 0, correct?
Yes. xdot has nothing to do with x.
It's strange that I know you are right, yet I have comprehension
problems with not really understanding what an independent variable
is! Even though Tx may vary over x and t, that does not mean it's
dependent on these variables any more than it's dependent upon the
length of a continuously increasing line, L(t), I might decide to
draw. Mathematically, T can be written as T(L(t)), but physically T is
still independent of L(t). Is this reasoning correct?
No, T(L(t)) makes T dependent on position, but kinetic energy is
purely dependent on velocity. Just for the practice, try to write the
function of L.
Your expression @/@x(x_dot^2) tries to express how xdot is affected by
a change in x, and it isn't. x and xdot can be considered orthogonal
to each other so there's no partial coefficient of sensitivity between
them. Yes, xdotdot is affected by x in gravity but then you'd be using
something that wasn't given in the Lagrangian.
In polar coordinates T could be mr^2*w^2/2, therefore the derivative
with r exists as Goldstein said.
John Polasek- Hide quoted text -
- Show quoted text -
I just find it confusing that xdot will in general vary with x, even
though it isn't directly dependent on it, so that xdot(x) and @xdot/@x
can be formed. For example, a constant force Fx applied to a particle
gives rise to an xdot that varies with x. Even though x does not
directly affect xdot, @xdot/@x can still be formed. Correct?
You can make an artificial case. Ex: v = sqrt(2gx) constant g, but v
comes first from time, and x from time and with algebra you can make
v(x). But then it comes down to
dv/dx = g/v
but I don't believe it is valid to write
@v/@x
but you can write
@(v^2/2)/@x = g
but v^2 isn't v.
I think you are wasting your time; what are you going to do with it?
John Polasek- Hide quoted text -
- Show quoted text -
I'm just looking at how Goldstein extends D'Alembert's principle to
generalised forces and coordinates so that he ends up with the
expression:-
Sigma_j { [ d/dt ( @T/@qdot) - @T/@qj] - Qj } delta_qj = 0
Upto this point, nothing has been said about the nature of the forces,
and he makes the statement that @T/@qj vanishes for Cartesian
coordinates.
.
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| User: "John C. Polasek" |
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| Title: Re: @T/@qj = 0 in Cartesian co-ordinates? |
22 May 2007 02:00:25 PM |
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On 22 May 2007 10:45:13 -0700, blackhead <larryharson@softhome.net>
wrote:
On May 22, 3:14 pm, John C. Polasek <jpola...@cfl.rr.com> wrote:
On 22 May 2007 06:03:39 -0700, blackhead < >
wrote:
On 21 May, 21:05, John C. Polasek <jpola...@cfl.rr.com> wrote:
On 21 May 2007 09:02:08 -0700, blackhead < >
wrote:
On May 17, 3:42 am, John C. Polasek <jpola...@cfl.rr.com> wrote:
On 16 May 2007 12:23:23 -0700, wrote:
On page 20 of Goldstein he says that the partial derivative of the
kinetic energy with respect to qj vanishes when Cartesian Coordinates
are used. For a single particle of mass m, T = 1/2mv^2, qj = x, so
this means @/@x(x_dot^2) = 0, correct?
Yes. xdot has nothing to do with x.
It's strange that I know you are right, yet I have comprehension
problems with not really understanding what an independent variable
is! Even though Tx may vary over x and t, that does not mean it's
dependent on these variables any more than it's dependent upon the
length of a continuously increasing line, L(t), I might decide to
draw. Mathematically, T can be written as T(L(t)), but physically T is
still independent of L(t). Is this reasoning correct?
No, T(L(t)) makes T dependent on position, but kinetic energy is
purely dependent on velocity. Just for the practice, try to write the
function of L.
Your expression @/@x(x_dot^2) tries to express how xdot is affected by
a change in x, and it isn't. x and xdot can be considered orthogonal
to each other so there's no partial coefficient of sensitivity between
them. Yes, xdotdot is affected by x in gravity but then you'd be using
something that wasn't given in the Lagrangian.
In polar coordinates T could be mr^2*w^2/2, therefore the derivative
with r exists as Goldstein said.
John Polasek- Hide quoted text -
- Show quoted text -
I just find it confusing that xdot will in general vary with x, even
though it isn't directly dependent on it, so that xdot(x) and @xdot/@x
can be formed. For example, a constant force Fx applied to a particle
gives rise to an xdot that varies with x. Even though x does not
directly affect xdot, @xdot/@x can still be formed. Correct?
You can make an artificial case. Ex: v = sqrt(2gx) constant g, but v
comes first from time, and x from time and with algebra you can make
v(x). But then it comes down to
dv/dx = g/v
but I don't believe it is valid to write
@v/@x
but you can write
@(v^2/2)/@x = g
but v^2 isn't v.
I think you are wasting your time; what are you going to do with it?
John Polasek- Hide quoted text -
- Show quoted text -
I'm just looking at how Goldstein extends D'Alembert's principle to
generalised forces and coordinates so that he ends up with the
expression:-
Sigma_j { [ d/dt ( @T/@qdot) - @T/@qj] - Qj } delta_qj = 0
Upto this point, nothing has been said about the nature of the forces,
and he makes the statement that @T/@qj vanishes for Cartesian
coordinates.
Your formula says the sum of the generalized forces in brackets dotted
with the conjugate displacement Dq = 0, so Q is the annulling external
force. (I didn't find this in my Goldstein).
T is only the kinetic energy part of the Lagrangian and T is no
function of position. The Lagrangian is very sparse in what it can
include so my sqrt(2gx) would be disallowed I think. In any case even
to introduce viscous friction would require you to use the Rayleigh
dissipation function (I don't know how it works).
John Polasek
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| User: "Eric Gisse" |
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| Title: Re: @T/@qj = 0 in Cartesian co-ordinates? |
21 May 2007 03:22:22 PM |
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On May 16, 12:23 pm, wrote:
On page 20 of Goldstein he says that the partial derivative of the
kinetic energy with respect to qj vanishes when Cartesian Coordinates
are used. For a single particle of mass m, T = 1/2mv^2, qj = x, so
this means @/@x(x_dot^2) = 0, correct?
Looks like! Though it isn't clear to me why from what you have stated.
1) [good] Generalized coordinates are orthogonal. The dot product of a
q_dot and q is zero, for example.
2) Cartesian kinetic energy is only a function of q_dot.
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