Let an extended mass M moving at four-velocity v have a constant four-force F applied to some point so that M doesn't rotate. Up to now, I've assumed that a suitable equation of motion is:

F = d/dtau ( Mv)

The problem IMO is that during acceleration, internal forces will drive the Lorentz contraction to maintain rigidness in its proper frame so that there's a reduction in the momentum at the leading edge relative to the trailing edge of the extended body:
there's a conversion of mechanical to internal energy-momentum; hence maintaining the mass of M. From classical electrodynamics we have the LAD equation; so using that as a template, I'd guess a possible modification would be:

F = d/dtau ( Mv - Ca)

where C is a constant and a the four acceleration.

Thanks in advance for any helpful comments and criticisms

Larry

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 2023-04-20 23:01:33 +0000, larry harson said:

Let an extended mass M moving at four-velocity v have a constant
four-force F applied to some point so that M doesn't rotate. Up to now,
I've assumed that a suitable equation of motion is:

F = d/dtau ( Mv)

The problem IMO is that during acceleration, internal forces will drive
the Lorentz contraction to maintain rigidness in its proper frame so
that there's a reduction in the momentum at the leading edge relative
to the trailing edge of the extended body

Internal stresses don't affect the motion of the body, only its shape. Therefore the same equation is valid anyway.

However, Special Relativity says that there are no rigid bodies. The
body does change its shape unless the external force is distributed
so that there are no internal stresses.

Mikko

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 4/20/23 6:01 PM, larry harson wrote:

Let an extended mass M moving at four-velocity v have a constant
four-force F applied to some point so that M doesn't rotate. Up to
now, I've assumed that a suitable equation of motion is: F = d/dtau
(Mv)

OK. I'll come back to this, but first must dispel you misunderstanding.

The problem IMO is that during acceleration, internal forces will
drive the Lorentz contraction

This is nonsense. The "Lorentz contraction" of a moving object is just
the way an inertial fame measures a moving object. It simply does not
apply here.

to maintain rigidness in its proper frame

You are trying to consider Born rigid motion, in which the object
continuously maintains its dimensions in each of its successive
instantaneously co-moving inertial frames. For typical objects
considered at timescales of a second or longer, the internal vibrations
from the strain induced by the force will damp out -- they can be
considered to execute Born rigid motion for applied forces small enough
to not distort or damage the object.

[...]

Consider a simpler problem in classical mechanics: two objects located
along the X axis connected by a massless spring that is initially in its equilibrium state. Apply a force f along X to one mass and what happens?
This is simple enough that it can be worked out analytically. While both individual masses oscillate back-and-forth, the center of mass
accelerates smoothly with value f/M (M = mass of system = total mass of
both objects).

The same occurs in your example, except the oscillations damp out.

Tom Roberts

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Friday, April 21, 2023 at 7:53:55 PM UTC+1, Tom Roberts wrote:

On 4/20/23 6:01 PM, larry harson wrote:

Let an extended mass M moving at four-velocity v have a constant four-force F applied to some point so that M doesn't rotate. Up to
now, I've assumed that a suitable equation of motion is: F = d/dtau
(Mv)

OK. I'll come back to this, but first must dispel you misunderstanding.

The problem IMO is that during acceleration, internal forces will
drive the Lorentz contraction

This is nonsense. The "Lorentz contraction" of a moving object is just
the way an inertial fame measures a moving object. It simply does not
apply here.

The Lorentz contraction of a moving object, as found in mainstream texts authored by professional physicists, is demonstrated using the Lorentz transformations; comparing the spatial distance between two events simultaneous in one frame with their
spatial distance measured in another frame for the same two events where they won't be simultaneous. Typically, this is done using a rod of length L in its proper frame as an example.

My claim above is that if I take a rod of length L and move it rigidly to another frame so every part is now moving at velocity v: then there is a consistent physical model for this physical transfer, consistent with SR, that models its length L' to be
measured smaller than L using the Lorentz transformations. The fact that the parts that make up the rod are accelerated differently to one another, to maintain rigidness in its proper frame, is a physical model for this: yes?

[snipped]

Larry

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 4/21/23 6:47 PM, larry harson wrote:

On Friday, April 21, 2023 at 7:53:55 PM UTC+1, Tom Roberts wrote:

On 4/20/23 6:01 PM, larry harson wrote:

Let an extended mass M moving at four-velocity v have a constant
four-force F applied to some point so that M doesn't rotate. Up
to now, I've assumed that a suitable equation of motion is: F =
d/dtau (Mv)

OK. I'll come back to this, but first must dispel you
misunderstanding.

The problem IMO is that during acceleration, internal forces
will drive the Lorentz contraction

This is nonsense. The "Lorentz contraction" of a moving object is
just the way an inertial fame measures a moving object. It simply
does not apply here.

The Lorentz contraction of a moving object, as found in mainstream
texts authored by professional physicists, is demonstrated using the
Lorentz transformations; [...]

Yes. But then it simply does not apply here, because the entire
description is in the rod's rest frame.

My claim above is that if I take a rod of length L and move it
rigidly to another frame so every part is now moving at velocity v:
then there is a consistent physical model for this physical transfer, consistent with SR, that models its length L' to be measured smaller
than L using the Lorentz transformations.

You made the basic mistake of not mentioning IN WHICH FRAME L' is
measured. Here L' is measured in the first frame. From your description,
the proper length of the rod is unchanged, and remains L in its new rest
frame. That is, the "Lorentz contraction" applies ONLY in the inertial
frame used to measure the moving rod, and does not affect the rod itself
(its proper length remains L).

The fact that the parts that make up the rod are accelerated
differently to one another, to maintain rigidness in its proper
frame, is a physical model for this: yes?

Yes, to maintain the proper length of the rod you must accelerate
different parts of it differently. Or you can apply the accelerating
force to a single point and let the internal forces that maintain the
rod adjust accordingly -- after all, those forces will always act to
maintain the rod's proper length at L -- this is just a way of
accelerating different parts of the rod differently.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)