Education Forums
| View previous topic :: View next topic |
| Author |
Message |
Dono
Guest
|
 Posted: Sun Jul 13, 2008 4:39 am Post subject: Relativistic Dynamics |
 |
|
Sinnce m_0 *d^2x/dt^2=k (constant) is Galilei invariant but not
Lorentz invariant, we have to redefine the impulse in SR as p=
\gamma(v)*m_0*v
We can prove that in the proper frame of the object F :
d(\gamma(v)*v)/dt=\gamma(v)^3*dv/dt
We can also prove (after some computations) that , in a frame F'
moving with constant speed V wrt the frame F:
d(\gamma(v')*v')/dt'=\gamma(v')^3*dv'/dt'
We can further prove that
\gamma(v')^3*dv'/dt'=\gamma(v)^3*dv/dt=k/m_0
This means that by redefining the relativistic impulse as
\gamma(v)*m_0*v , the equations of motion under constant force (F=k)
are Lorentz invariant. So far, so good.
I need some help with the situation when F is NOT constant.
Here are several examples:
1. F= - q*x (common spring)
2. F= -q *sin(theta) (common pendulum)
3. An even nastier case is the case of the torsion pendulum where we
need the relativistic equivalent for the Newtonian p=I*d(theta)/dt
where I is the momentum of inertia . It is not obvious what that
formula would be.
In these particular cases, the fact that the left term of the equation
is invariant (\gamma(v')^3*dv'/dt'=\gamma(v)^3*dv/dt) is of no good,
since the right term is obviously not Lorentz invariant since neither
x, nor theta are Lorentz invariants.
Of course, Hooke law and the pendulum law are laws derived
empirically, so the obvious approach would be to redefine them in
such a fashion that they become Lorentz invariant. Did you see any
literature on this? |
|
| Back to top |
|
 |
| |
Ads |
Advertising
Sponsor
|
|
Eric Gisse
Guest
|
| Posted: Sun Jul 13, 2008 8:49 am Post subject: Re: Relativistic Dynamics |
|
|
On Jul 12, 8:39pm, Dono <sa...@comcast.net> wrote:
[...]
There are three ways, and yours is the option that nobody should pick.
a) Covariantly: Use the geodesic equation and a four-vector
formulation of the force.
b) Covariantly: Construct the Lagrangian from four-vectors and use a
four-force derivable from a four-potential. Then build a new four-mine
to get new fours, as the world is running out.
c) Not-covariantly: Write the regular Lagrangian -mc^2sqrt[1-v^2/c^2]
and go as normally.
Going with c, I basically get d/dt [mv*gamma] = F. Not _terrifically_
surprising but it is good to know it behaves as expected.
| Quote: |
I need some help with the situation when F is NOT constant.
Here are several examples:
1. F= - q*x (common spring)
|
Maple reduces this to a quadrature that appears to be solvable exactly
if given initial conditions.
| Quote: |
2. F= -q *sin(theta) (common pendulum)
|
Also reduces to quadratures, but appears to be only solvable in terms
of elliptic functions. That shouldn't come as a surprise, if you think
about it.
| Quote: |
3. An even nastier case is the case of the torsion pendulum where we
need the relativistic equivalent for the Newtonian p=I*d(theta)/dt
where I is the momentum of inertia . It is not obvious what that
formula would be.
|
Well there is a covariant way to form angular momentum, and thus
inertia, so I expect it exists. It isn't interesting enough for me to
tackle though.
[...] |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Dono
Guest
|
| Posted: Mon Jul 14, 2008 1:02 am Post subject: Re: Relativistic Dynamics |
|
|
On Jul 13, 1:49am, Eric Gisse <jowr...@gmail.com> wrote:
| Quote: |
On Jul 12, 8:39pm, Dono <sa...@comcast.net> wrote:
[...]
There are three ways, and yours is the option that nobody should pick.
|
...except it worked perfectly :-)
| Quote: |
a) Covariantly: Use the geodesic equation and a four-vector
formulation of the force.
b) Covariantly: Construct the Lagrangian from four-vectors and use a
four-force derivable from a four-potential. Then build a new four-mine
to get new fours, as the world is running out.
c) Not-covariantly: Write the regular Lagrangian -mc^2sqrt[1-v^2/c^2]
and go as normally.
Going with c, I basically get d/dt [mv*gamma] = F. Not _terrifically_
surprising but it is good to know it behaves as expected.
I need some help with the situation when F is NOT constant.
Here are several examples:
1. F= - q*x (common spring)
Maple reduces this to a quadrature that appears to be solvable exactly
if given initial conditions.
|
I think you missed the point, please read my post again. The problem
is that one cannot get a covariant formulation of the Hooke law
because of the right hand term. What I asked is : how do we need to
modify Hooke's law in order to have a covariant formulation. |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Juan R." Gonzlez-lvarez
Guest
|
| Posted: Mon Jul 14, 2008 10:50 am Post subject: Re: Relativistic Dynamics |
|
|
Dono wrote on Sat, 12 Jul 2008 21:39:42 -0700:
(...)
| Quote: |
Of course, Hooke law and the pendulum law are laws derived empirically,
so the obvious approach would be to redefine them in such a fashion
that they become Lorentz invariant. Did you see any literature on this?
|
Here one 'good' reference on the Lagrangian you may use
http://sci.tech-archive.net/Archive/sci.physics.relativity/2008-07/
msg00824.html
ha ha ha ha ha ha
--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org
--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Juan R." Gonzlez-lvarez
Guest
|
| Posted: Mon Jul 14, 2008 10:53 am Post subject: Re: Relativistic Dynamics |
|
|
Dono wrote on Sun, 13 Jul 2008 18:02:13 -0700:
| Quote: |
On Jul 13, 1:49 am, Eric Gisse <jowr...@gmail.com> wrote:
On Jul 12, 8:39 pm, Dono <sa...@comcast.net> wrote:
[...]
There are three ways, and yours is the option that nobody should pick.
...except it worked perfectly :-)
|
(...)
| Quote: |
I think you missed the point, please read my post again. The problem is
that one cannot get a covariant formulation of the Hooke law because of
the right hand term. What I asked is : how do we need to modify Hooke's
law in order to have a covariant formulation.
|
Ha ha ha ha
Round one for crackpots Eric and Dono
--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Eric Gisse
Guest
|
| Posted: Mon Jul 14, 2008 11:26 am Post subject: Re: Relativistic Dynamics |
|
|
On Jul 13, 5:02pm, Dono <sa...@comcast.net> wrote:
| Quote: |
On Jul 13, 1:49am, Eric Gisse <jowr...@gmail.com> wrote:
On Jul 12, 8:39pm, Dono <sa...@comcast.net> wrote:
[...]
There are three ways, and yours is the option that nobody should pick.
...except it worked perfectly 
|
No denying that - I just believe it is more effort than it is worth
since there are easier methods.
| Quote: |
a) Covariantly: Use the geodesic equation and a four-vector
formulation of the force.
b) Covariantly: Construct the Lagrangian from four-vectors and use a
four-force derivable from a four-potential. Then build a new four-mine
to get new fours, as the world is running out.
c) Not-covariantly: Write the regular Lagrangian -mc^2sqrt[1-v^2/c^2]
and go as normally.
Going with c, I basically get d/dt [mv*gamma] = F. Not _terrifically_
surprising but it is good to know it behaves as expected.
I need some help with the situation when F is NOT constant.
Here are several examples:
1. F= - q*x (common spring)
Maple reduces this to a quadrature that appears to be solvable exactly
if given initial conditions.
I think you missed the point, please read my post again. The problem
is that one cannot get a covariant formulation of the Hooke law
because of the right hand term. What I asked is : how do we need to
modify Hooke's law in order to have a covariant formulation.
|
My bad - couldn't see the forest for the trees.
I think you are overthinking what you want to do unless I'm still
missing the point. Why not go with the simplest answer: Write the
force as a four-vector like F = (0, qx, 0, 0) ?
Writing the force as a four-vectors is sufficient to ensure
covariance, imho. Then all you'd have to do is shove it in the right
hand side of the geodesic equation! This would then be true in both SR
and GR in any coordinate system. |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Eric Gisse
Guest
|
| Posted: Mon Jul 14, 2008 11:26 am Post subject: Re: Relativistic Dynamics |
|
|
On Jul 14, 2:53am, "Juan R." Gonzlez-lvarez
<juanREM...@canonicalscience.com> wrote:
[...]
Quiet. Adults are talking. |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Juan R." Gonzlez-lvarez
Guest
|
| Posted: Mon Jul 14, 2008 12:17 pm Post subject: Re: Relativistic Dynamics |
|
|
Eric Gisse wrote on Mon, 14 Jul 2008 04:26:33 -0700:
Why cannot your mistakes be noticed 'adult'?
--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Dono
Guest
|
| Posted: Mon Jul 14, 2008 3:58 pm Post subject: Re: Relativistic Dynamics |
|
|
On Jul 14, 4:26am, Eric Gisse <jowr...@gmail.com> wrote:
| Quote: |
On Jul 13, 5:02pm, Dono <sa...@comcast.net> wrote:
On Jul 13, 1:49am, Eric Gisse <jowr...@gmail.com> wrote:
On Jul 12, 8:39pm, Dono <sa...@comcast.net> wrote:
[...]
There are three ways, and yours is the option that nobody should pick..
...except it worked perfectly :-)
No denying that - I just believe it is more effort than it is worth
since there are easier methods.
a) Covariantly: Use the geodesic equation and a four-vector
formulation of the force.
b) Covariantly: Construct the Lagrangian from four-vectors and use a
four-force derivable from a four-potential. Then build a new four-mine
to get new fours, as the world is running out.
c) Not-covariantly: Write the regular Lagrangian -mc^2sqrt[1-v^2/c^2]
and go as normally.
Going with c, I basically get d/dt [mv*gamma] = F. Not _terrifically_
surprising but it is good to know it behaves as expected.
I need some help with the situation when F is NOT constant.
Here are several examples:
1. F= - q*x (common spring)
Maple reduces this to a quadrature that appears to be solvable exactly
if given initial conditions.
I think you missed the point, please read my post again. The problem
is that one cannot get a covariant formulation of the Hooke law
because of the right hand term. What I asked is : how do we need to
modify Hooke's law in order to have a covariant formulation.
My bad - couldn't see the forest for the trees.
|
No problem, let's move on.
| Quote: |
I think you are overthinking what you want to do unless I'm still
missing the point. Why not go with the simplest answer: Write the
force as a four-vector like F = (0, qx, 0, 0) ?
Writing the force as a four-vectors is sufficient to ensure
covariance, imho. Then all you'd have to do is shove it in the right
hand side of the geodesic equation!
|
I am not sure how this would solve the problem, can you do the
calculations (at least the few starting steps)? so we have a few
equations to look at? I would greatly appreciate this. |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Dono
Guest
|
| Posted: Mon Jul 14, 2008 6:01 pm Post subject: Re: Relativistic Dynamics |
|
|
On Jul 14, 10:52am, stevendaryl3...@yahoo.com (Daryl McCullough)
wrote:
| Quote: |
In article <272d6ef2-e464-4110-9a45-30ab41fa4...@b1g2000hsg.googlegroups.com>,
Dono says...
I think you missed the point, please read my post again. The problem
is that one cannot get a covariant formulation of the Hooke law
because of the right hand term. What I asked is : how do we need to
modify Hooke's law in order to have a covariant formulation.
If you have access to a physics library, or are willing to pay $19,
you can read an article on this topic in the American Journal of Physics:
Oyvind Gron,
"Covariant formulation of Hooke's Law",
Am. J. Phys. 49 (1981)
http://scitation.aip.org/vsearch/servlet/VerityServlet?KEY=AJPIAS&ONL....
--
Daryl McCullough
Ithaca, NY
|
Thank you,
This is precisely what I was after |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Daryl McCullough
Guest
|
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Eric Gisse
Guest
|
| Posted: Wed Jul 16, 2008 12:38 pm Post subject: Re: Relativistic Dynamics |
|
|
On Jul 14, 7:58am, Dono <sa...@comcast.net> wrote:
| Quote: |
On Jul 14, 4:26am, Eric Gisse <jowr...@gmail.com> wrote:
On Jul 13, 5:02pm, Dono <sa...@comcast.net> wrote:
On Jul 13, 1:49am, Eric Gisse <jowr...@gmail.com> wrote:
On Jul 12, 8:39pm, Dono <sa...@comcast.net> wrote:
[...]
There are three ways, and yours is the option that nobody should pick.
...except it worked perfectly :-)
No denying that - I just believe it is more effort than it is worth
since there are easier methods.
a) Covariantly: Use the geodesic equation and a four-vector
formulation of the force.
b) Covariantly: Construct the Lagrangian from four-vectors and use a
four-force derivable from a four-potential. Then build a new four-mine
to get new fours, as the world is running out.
c) Not-covariantly: Write the regular Lagrangian -mc^2sqrt[1-v^2/c^2]
and go as normally.
Going with c, I basically get d/dt [mv*gamma] = F. Not _terrifically_
surprising but it is good to know it behaves as expected.
I need some help with the situation when F is NOT constant.
Here are several examples:
1. F= - q*x (common spring)
Maple reduces this to a quadrature that appears to be solvable exactly
if given initial conditions.
I think you missed the point, please read my post again. The problem
is that one cannot get a covariant formulation of the Hooke law
because of the right hand term. What I asked is : how do we need to
modify Hooke's law in order to have a covariant formulation.
My bad - couldn't see the forest for the trees.
No problem, let's move on.
I think you are overthinking what you want to do unless I'm still
missing the point. Why not go with the simplest answer: Write the
force as a four-vector like F = (0, qx, 0, 0) ?
Writing the force as a four-vectors is sufficient to ensure
covariance, imho. Then all you'd have to do is shove it in the right
hand side of the geodesic equation!
I am not sure how this would solve the problem, can you do the
calculations (at least the few starting steps)? so we have a few
equations to look at? I would greatly appreciate this.
|
Ok, but it changes nothing. It just makes the formulation covariant.
Write the Lagrangian L as -m sqrt[eta_uv U^u U^v] where eta is the
metric and U is the four velocity. The generalized coordinates are
velocity and position, as usual.
The force f^u is -@V/@x^u so V is, as expected, -1/2 q x^2.
The geodesic equation reduces to d^2x/dTau^2 = qx, as expected.
Is that helpful? |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Albertito
Guest
|
| Posted: Wed Jul 16, 2008 12:57 pm Post subject: Re: Relativistic Dynamics |
|
|
On Jul 16, 1:38 pm, Eric Gisse <jowr...@gmail.com> wrote:
| Quote: |
On Jul 14, 7:58 am, Dono <sa...@comcast.net> wrote:
On Jul 14, 4:26 am, Eric Gisse <jowr...@gmail.com> wrote:
On Jul 13, 5:02 pm, Dono <sa...@comcast.net> wrote:
On Jul 13, 1:49 am, Eric Gisse <jowr...@gmail.com> wrote:
On Jul 12, 8:39 pm, Dono <sa...@comcast.net> wrote:
[...]
There are three ways, and yours is the option that nobody should pick.
...except it worked perfectly :-)
No denying that - I just believe it is more effort than it is worth
since there are easier methods.
a) Covariantly: Use the geodesic equation and a four-vector
formulation of the force.
b) Covariantly: Construct the Lagrangian from four-vectors and use a
four-force derivable from a four-potential. Then build a new four-mine
to get new fours, as the world is running out.
c) Not-covariantly: Write the regular Lagrangian -mc^2sqrt[1-v^2/c^2]
and go as normally.
Going with c, I basically get d/dt [mv*gamma] = F. Not _terrifically_
surprising but it is good to know it behaves as expected.
I need some help with the situation when F is NOT constant.
Here are several examples:
1. F= - q*x (common spring)
Maple reduces this to a quadrature that appears to be solvable exactly
if given initial conditions.
I think you missed the point, please read my post again. The problem
is that one cannot get a covariant formulation of the Hooke law
because of the right hand term. What I asked is : how do we need to
modify Hooke's law in order to have a covariant formulation.
My bad - couldn't see the forest for the trees.
No problem, let's move on.
I think you are overthinking what you want to do unless I'm still
missing the point. Why not go with the simplest answer: Write the
force as a four-vector like F = (0, qx, 0, 0) ?
Writing the force as a four-vectors is sufficient to ensure
covariance, imho. Then all you'd have to do is shove it in the right
hand side of the geodesic equation!
I am not sure how this would solve the problem, can you do the
calculations (at least the few starting steps)? so we have a few
equations to look at? I would greatly appreciate this.
Ok, but it changes nothing. It just makes the formulation covariant.
Write the Lagrangian L as -m sqrt[eta_uv U^u U^v] where eta is the
metric and U is the four velocity. The generalized coordinates are
velocity and position, as usual.
The force f^u is -@V/@x^u so V is, as expected, -1/2 q x^2.
The geodesic equation reduces to d^2x/dTau^2 = qx, as expected.
Is that helpful?
|
Yes, thanks, very helpful. That is precisely what
I was after. I'm in the toilet, but there aren't
ass wipes |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Eric Gisse
Guest
|
| Posted: Wed Jul 16, 2008 1:18 pm Post subject: Re: Relativistic Dynamics |
|
|
On Jul 16, 4:57am, Albertito <albertito1...@gmail.com> wrote:
| Quote: |
On Jul 16, 1:38 pm, Eric Gisse <jowr...@gmail.com> wrote:
On Jul 14, 7:58 am, Dono <sa...@comcast.net> wrote:
On Jul 14, 4:26 am, Eric Gisse <jowr...@gmail.com> wrote:
On Jul 13, 5:02 pm, Dono <sa...@comcast.net> wrote:
On Jul 13, 1:49 am, Eric Gisse <jowr...@gmail.com> wrote:
On Jul 12, 8:39 pm, Dono <sa...@comcast.net> wrote:
[...]
There are three ways, and yours is the option that nobody should pick.
...except it worked perfectly :-)
No denying that - I just believe it is more effort than it is worth
since there are easier methods.
a) Covariantly: Use the geodesic equation and a four-vector
formulation of the force.
b) Covariantly: Construct the Lagrangian from four-vectors and use a
four-force derivable from a four-potential. Then build a new four-mine
to get new fours, as the world is running out.
c) Not-covariantly: Write the regular Lagrangian -mc^2sqrt[1-v^2/c^2]
and go as normally.
Going with c, I basically get d/dt [mv*gamma] = F. Not _terrifically_
surprising but it is good to know it behaves as expected.
I need some help with the situation when F is NOT constant.
Here are several examples:
1. F= - q*x (common spring)
Maple reduces this to a quadrature that appears to be solvable exactly
if given initial conditions.
I think you missed the point, please read my post again. The problem
is that one cannot get a covariant formulation of the Hooke law
because of the right hand term. What I asked is : how do we need to
modify Hooke's law in order to have a covariant formulation.
My bad - couldn't see the forest for the trees.
No problem, let's move on.
I think you are overthinking what you want to do unless I'm still
missing the point. Why not go with the simplest answer: Write the
force as a four-vector like F = (0, qx, 0, 0) ?
Writing the force as a four-vectors is sufficient to ensure
covariance, imho. Then all you'd have to do is shove it in the right
hand side of the geodesic equation!
I am not sure how this would solve the problem, can you do the
calculations (at least the few starting steps)? so we have a few
equations to look at? I would greatly appreciate this.
Ok, but it changes nothing. It just makes the formulation covariant.
Write the Lagrangian L as -m sqrt[eta_uv U^u U^v] where eta is the
metric and U is the four velocity. The generalized coordinates are
velocity and position, as usual.
The force f^u is -@V/@x^u so V is, as expected, -1/2 q x^2.
The geodesic equation reduces to d^2x/dTau^2 = qx, as expected.
Is that helpful?
Yes, thanks, very helpful. That is precisely what
I was after. I'm in the toilet, but there aren't
ass wipes
|
Contribute or shut the fuck up. |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
Dono
Guest
|
| Posted: Wed Jul 16, 2008 1:42 pm Post subject: Re: Relativistic Dynamics |
|
|
On Jul 16, 5:38am, Eric Gisse <jowr...@gmail.com> wrote:
| Quote: |
On Jul 14, 7:58am, Dono <sa...@comcast.net> wrote:
On Jul 14, 4:26am, Eric Gisse <jowr...@gmail.com> wrote:
On Jul 13, 5:02pm, Dono <sa...@comcast.net> wrote:
On Jul 13, 1:49am, Eric Gisse <jowr...@gmail.com> wrote:
On Jul 12, 8:39pm, Dono <sa...@comcast.net> wrote:
[...]
There are three ways, and yours is the option that nobody should pick.
...except it worked perfectly :-)
No denying that - I just believe it is more effort than it is worth
since there are easier methods.
a) Covariantly: Use the geodesic equation and a four-vector
formulation of the force.
b) Covariantly: Construct the Lagrangian from four-vectors and use a
four-force derivable from a four-potential. Then build a new four-mine
to get new fours, as the world is running out.
c) Not-covariantly: Write the regular Lagrangian -mc^2sqrt[1-v^2/c^2]
and go as normally.
Going with c, I basically get d/dt [mv*gamma] = F. Not _terrifically_
surprising but it is good to know it behaves as expected.
I need some help with the situation when F is NOT constant.
Here are several examples:
1. F= - q*x (common spring)
Maple reduces this to a quadrature that appears to be solvable exactly
if given initial conditions.
I think you missed the point, please read my post again. The problem
is that one cannot get a covariant formulation of the Hooke law
because of the right hand term. What I asked is : how do we need to
modify Hooke's law in order to have a covariant formulation.
My bad - couldn't see the forest for the trees.
No problem, let's move on.
I think you are overthinking what you want to do unless I'm still
missing the point. Why not go with the simplest answer: Write the
force as a four-vector like F = (0, qx, 0, 0) ?
Writing the force as a four-vectors is sufficient to ensure
covariance, imho. Then all you'd have to do is shove it in the right
hand side of the geodesic equation!
I am not sure how this would solve the problem, can you do the
calculations (at least the few starting steps)? so we have a few
equations to look at? I would greatly appreciate this.
Ok, but it changes nothing. It just makes the formulation covariant.
Write the Lagrangian L as -m sqrt[eta_uv U^u U^v] where eta is the
metric and U is the four velocity. The generalized coordinates are
velocity and position, as usual.
The force f^u is -@V/@x^u so V is, as expected, -1/2 q x^2.
The geodesic equation reduces to d^2x/dTau^2 = qx, as expected.
Is that helpful?- Hide quoted text -
- Show quoted text -
|
Yes, it is |
|
| Back to top |
|
|
| |
Ads |
Advertising
Sponsor
|
|
|
|
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
|
93 Attacks blocked
Powered by phpBB © 2001, 2005 phpBB Group
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
