Preferred Frame Theory indistinguishable from SR
in [Relativity]
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From: mpc755 on 20 Jul 2010 00:07 On Jul 19, 6:34 pm, colp <c...(a)solder.ath.cx> wrote: > On Jul 20, 4:40 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Jul 18, 12:49 am, colp <c...(a)solder.ath.cx> wrote: > > > > On Jul 18, 3:53 pm, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > > On Jul 17, 5:54 pm, colp <c...(a)solder.ath.cx> wrote: > > > > > > On Jul 18, 4:29 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > On Jul 16, 11:39 pm, colp <c...(a)solder.ath.cx> wrote: > > > > > > > > On Jul 17, 2:07 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > > > Just look at the aspects of kinetic theory which underpins ALL! fluid > > > > > > > > theory. There is no concept of absolutes there. > > > > > > > > Nope. For example, take two molecules, each with mass m, and separated > > > > > > > by distance d. Give the molecules velocities such that they will > > > > > > > approach each others center of mass and collide at time t. The > > > > > > > calculated energy of the collision will depend on how you measure your > > > > > > > velocities. > > > > > > > > For either molecule, the relative velocity of the other molecule will > > > > > > > be d/t or -d/t. > > > > > > > The calculated energy of collision is 1/2 mv^2 = 1/2 m.d^2/t^2, since > > > > > > > one molecule is deemed to be stationary. > > > > > > > > If, instead, we say that each molecule is moving toward a stationary > > > > > > > collision point, then v = d/2t or -d/2t > > > > > > > The calculated energy of collision is now 2 * 1/2 mv^2 = m.d^2/4t^2 = > > > > > > > 1/2 m.d^2/2t^2 > > > > > > > > The two calculations disagree by a factor of two. > > > > > > > So? What's your point. > > > > > > That the concept of absolute motion is inherent to kinetic theory.. > > > > > > > Instead of two molecules use you and the > > > > > > Earth... From your perspective it's the Earth is hurling towards you > > > > > > with a momentum of Mv and you're motionless. OTOH, from the Earth > > > > > > perspective it is motionless and you are hurling towards it at mv... > > > > > > You have addressed the issue of momentum, not kinetic energy. > > > > > You have GOT to be kidding? > > > > Not at all. You said: > > > > "Just look at the aspects of kinetic theory which underpins ALL! fluid > > > theory. There is no concept of absolutes there." > > > > The gas laws are part of fluid theory, and one of the gas laws > > > describes pressure in terms of average kinetic energy. The calculation > > > of kinetic energy depends on how you measure velocity; i.e. the > > > velocity relative to a co-ordinate system which is deemed to be > > > absolute. > > > Nonsense. It is velocity relative to the center of mass of the fluid. > > You are contracting Paul, who said: "Just look at the aspects of > kinetic theory which underpins ALL! fluid theory. There is no concept > of absolutes there." If everything were truly relative two synchronized clocks would always be synchronized when in the same state.
From: PD on 20 Jul 2010 10:37 On Jul 19, 5:34 pm, colp <c...(a)solder.ath.cx> wrote: > On Jul 20, 4:40 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Jul 18, 12:49 am, colp <c...(a)solder.ath.cx> wrote: > > > > On Jul 18, 3:53 pm, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > > On Jul 17, 5:54 pm, colp <c...(a)solder.ath.cx> wrote: > > > > > > On Jul 18, 4:29 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > On Jul 16, 11:39 pm, colp <c...(a)solder.ath.cx> wrote: > > > > > > > > On Jul 17, 2:07 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > > > Just look at the aspects of kinetic theory which underpins ALL! fluid > > > > > > > > theory. There is no concept of absolutes there. > > > > > > > > Nope. For example, take two molecules, each with mass m, and separated > > > > > > > by distance d. Give the molecules velocities such that they will > > > > > > > approach each others center of mass and collide at time t. The > > > > > > > calculated energy of the collision will depend on how you measure your > > > > > > > velocities. > > > > > > > > For either molecule, the relative velocity of the other molecule will > > > > > > > be d/t or -d/t. > > > > > > > The calculated energy of collision is 1/2 mv^2 = 1/2 m.d^2/t^2, since > > > > > > > one molecule is deemed to be stationary. > > > > > > > > If, instead, we say that each molecule is moving toward a stationary > > > > > > > collision point, then v = d/2t or -d/2t > > > > > > > The calculated energy of collision is now 2 * 1/2 mv^2 = m.d^2/4t^2 = > > > > > > > 1/2 m.d^2/2t^2 > > > > > > > > The two calculations disagree by a factor of two. > > > > > > > So? What's your point. > > > > > > That the concept of absolute motion is inherent to kinetic theory.. > > > > > > > Instead of two molecules use you and the > > > > > > Earth... From your perspective it's the Earth is hurling towards you > > > > > > with a momentum of Mv and you're motionless. OTOH, from the Earth > > > > > > perspective it is motionless and you are hurling towards it at mv... > > > > > > You have addressed the issue of momentum, not kinetic energy. > > > > > You have GOT to be kidding? > > > > Not at all. You said: > > > > "Just look at the aspects of kinetic theory which underpins ALL! fluid > > > theory. There is no concept of absolutes there." > > > > The gas laws are part of fluid theory, and one of the gas laws > > > describes pressure in terms of average kinetic energy. The calculation > > > of kinetic energy depends on how you measure velocity; i.e. the > > > velocity relative to a co-ordinate system which is deemed to be > > > absolute. > > > Nonsense. It is velocity relative to the center of mass of the fluid. > > You are contracting Paul, who said: "Just look at the aspects of > kinetic theory which underpins ALL! fluid theory. There is no concept > of absolutes there." Read what I just said. The center of mass of the fluid is in no way an absolute frame.
From: PD on 20 Jul 2010 10:38 On Jul 19, 11:07 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Jul 19, 6:34 pm, colp <c...(a)solder.ath.cx> wrote: > > > > > > > On Jul 20, 4:40 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Jul 18, 12:49 am, colp <c...(a)solder.ath.cx> wrote: > > > > > On Jul 18, 3:53 pm, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > > > On Jul 17, 5:54 pm, colp <c...(a)solder.ath.cx> wrote: > > > > > > > On Jul 18, 4:29 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > > On Jul 16, 11:39 pm, colp <c...(a)solder.ath.cx> wrote: > > > > > > > > > On Jul 17, 2:07 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > > > > Just look at the aspects of kinetic theory which underpins ALL! fluid > > > > > > > > > theory. There is no concept of absolutes there. > > > > > > > > > Nope. For example, take two molecules, each with mass m, and separated > > > > > > > > by distance d. Give the molecules velocities such that they will > > > > > > > > approach each others center of mass and collide at time t. The > > > > > > > > calculated energy of the collision will depend on how you measure your > > > > > > > > velocities. > > > > > > > > > For either molecule, the relative velocity of the other molecule will > > > > > > > > be d/t or -d/t. > > > > > > > > The calculated energy of collision is 1/2 mv^2 = 1/2 m.d^2/t^2, since > > > > > > > > one molecule is deemed to be stationary. > > > > > > > > > If, instead, we say that each molecule is moving toward a stationary > > > > > > > > collision point, then v = d/2t or -d/2t > > > > > > > > The calculated energy of collision is now 2 * 1/2 mv^2 = m.d^2/4t^2 = > > > > > > > > 1/2 m.d^2/2t^2 > > > > > > > > > The two calculations disagree by a factor of two. > > > > > > > > So? What's your point. > > > > > > > That the concept of absolute motion is inherent to kinetic theory. > > > > > > > > Instead of two molecules use you and the > > > > > > > Earth... From your perspective it's the Earth is hurling towards you > > > > > > > with a momentum of Mv and you're motionless. OTOH, from the Earth > > > > > > > perspective it is motionless and you are hurling towards it at mv... > > > > > > > You have addressed the issue of momentum, not kinetic energy. > > > > > > You have GOT to be kidding? > > > > > Not at all. You said: > > > > > "Just look at the aspects of kinetic theory which underpins ALL! fluid > > > > theory. There is no concept of absolutes there." > > > > > The gas laws are part of fluid theory, and one of the gas laws > > > > describes pressure in terms of average kinetic energy. The calculation > > > > of kinetic energy depends on how you measure velocity; i.e. the > > > > velocity relative to a co-ordinate system which is deemed to be > > > > absolute. > > > > Nonsense. It is velocity relative to the center of mass of the fluid. > > > You are contracting Paul, who said: "Just look at the aspects of > > kinetic theory which underpins ALL! fluid theory. There is no concept > > of absolutes there." > > If everything were truly relative two synchronized clocks would always > be synchronized when in the same state. What? What on earth makes you say that?
From: Paul Stowe on 20 Jul 2010 20:21 On Jul 20, 7:37 am, PD <thedraperfam...(a)gmail.com> wrote: > On Jul 19, 5:34 pm, colp <c...(a)solder.ath.cx> wrote: > > > > > > > On Jul 20, 4:40 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > On Jul 18, 12:49 am, colp <c...(a)solder.ath.cx> wrote: > > > > > On Jul 18, 3:53 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > On Jul 17, 5:54 pm, colp <c...(a)solder.ath.cx> wrote: > > > > > > > On Jul 18, 4:29 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > > On Jul 16, 11:39 pm, colp <c...(a)solder.ath.cx> wrote: > > > > > > > > > On Jul 17, 2:07 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > > > > Just look at the aspects of kinetic theory which underpins ALL! fluid > > > > > > > > > theory. There is no concept of absolutes there. > > > > > > > > > Nope. For example, take two molecules, each with mass m, and separated > > > > > > > > by distance d. Give the molecules velocities such that they will > > > > > > > > approach each others center of mass and collide at time t. The > > > > > > > > calculated energy of the collision will depend on how you measure your > > > > > > > > velocities. > > > > > > > > > For either molecule, the relative velocity of the other molecule will > > > > > > > > be d/t or -d/t. > > > > > > > > The calculated energy of collision is 1/2 mv^2 = 1/2 m.d^2/t^2, since > > > > > > > > one molecule is deemed to be stationary. > > > > > > > > > If, instead, we say that each molecule is moving toward a stationary > > > > > > > > collision point, then v = d/2t or -d/2t > > > > > > > > The calculated energy of collision is now 2 * 1/2 mv^2 = m.d^2/4t^2 = > > > > > > > > 1/2 m.d^2/2t^2 > > > > > > > > > The two calculations disagree by a factor of two. > > > > > > > > So? What's your point. > > > > > > > That the concept of absolute motion is inherent to kinetic theory. > > > > > > > > Instead of two molecules use you and the > > > > > > > Earth... From your perspective it's the Earth is hurling towards you > > > > > > > with a momentum of Mv and you're motionless. OTOH, from the Earth > > > > > > > perspective it is motionless and you are hurling towards it at mv... > > > > > > > You have addressed the issue of momentum, not kinetic energy. > > > > > > You have GOT to be kidding? > > > > > Not at all. You said: > > > > > "Just look at the aspects of kinetic theory which underpins ALL! fluid > > > > theory. There is no concept of absolutes there." > > > > > The gas laws are part of fluid theory, and one of the gas laws > > > > describes pressure in terms of average kinetic energy. The calculation > > > > of kinetic energy depends on how you measure velocity; i.e. the > > > > velocity relative to a co-ordinate system which is deemed to be > > > > absolute. > > > > Nonsense. It is velocity relative to the center of mass of the fluid. > > > You are contracting Paul, who said: "Just look at the aspects of > > kinetic theory which underpins ALL! fluid theory. There is no concept > > of absolutes there." > > Read what I just said. The center of mass of the fluid is in no way an > absolute frame. Where is the 'center of mass' for the pacific ocean??? I think what you really wanted to say is the frame where the local sum of the population vector momenta is zero. But, note what I said kinetic theory 'which underpins' fluid mechanics. At the scales small enough where kinetic theory is necessary... But, either way, there is no physical preference for any imagined coordinate system, aka, reference frame. Paul Stowe
From: PD on 21 Jul 2010 09:40
On Jul 20, 7:21 pm, Paul Stowe <theaether...(a)gmail.com> wrote: > On Jul 20, 7:37 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Jul 19, 5:34 pm, colp <c...(a)solder.ath.cx> wrote: > > > > On Jul 20, 4:40 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Jul 18, 12:49 am, colp <c...(a)solder.ath.cx> wrote: > > > > > > On Jul 18, 3:53 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > On Jul 17, 5:54 pm, colp <c...(a)solder.ath.cx> wrote: > > > > > > > > On Jul 18, 4:29 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > > > On Jul 16, 11:39 pm, colp <c...(a)solder.ath.cx> wrote: > > > > > > > > > > On Jul 17, 2:07 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > > > > > Just look at the aspects of kinetic theory which underpins ALL! fluid > > > > > > > > > > theory. There is no concept of absolutes there. > > > > > > > > > > Nope. For example, take two molecules, each with mass m, and separated > > > > > > > > > by distance d. Give the molecules velocities such that they will > > > > > > > > > approach each others center of mass and collide at time t.. The > > > > > > > > > calculated energy of the collision will depend on how you measure your > > > > > > > > > velocities. > > > > > > > > > > For either molecule, the relative velocity of the other molecule will > > > > > > > > > be d/t or -d/t. > > > > > > > > > The calculated energy of collision is 1/2 mv^2 = 1/2 m.d^2/t^2, since > > > > > > > > > one molecule is deemed to be stationary. > > > > > > > > > > If, instead, we say that each molecule is moving toward a stationary > > > > > > > > > collision point, then v = d/2t or -d/2t > > > > > > > > > The calculated energy of collision is now 2 * 1/2 mv^2 = m.d^2/4t^2 = > > > > > > > > > 1/2 m.d^2/2t^2 > > > > > > > > > > The two calculations disagree by a factor of two. > > > > > > > > > So? What's your point. > > > > > > > > That the concept of absolute motion is inherent to kinetic theory. > > > > > > > > > Instead of two molecules use you and the > > > > > > > > Earth... From your perspective it's the Earth is hurling towards you > > > > > > > > with a momentum of Mv and you're motionless. OTOH, from the Earth > > > > > > > > perspective it is motionless and you are hurling towards it at mv... > > > > > > > > You have addressed the issue of momentum, not kinetic energy. > > > > > > > You have GOT to be kidding? > > > > > > Not at all. You said: > > > > > > "Just look at the aspects of kinetic theory which underpins ALL! fluid > > > > > theory. There is no concept of absolutes there." > > > > > > The gas laws are part of fluid theory, and one of the gas laws > > > > > describes pressure in terms of average kinetic energy. The calculation > > > > > of kinetic energy depends on how you measure velocity; i.e. the > > > > > velocity relative to a co-ordinate system which is deemed to be > > > > > absolute. > > > > > Nonsense. It is velocity relative to the center of mass of the fluid. > > > > You are contracting Paul, who said: "Just look at the aspects of > > > kinetic theory which underpins ALL! fluid theory. There is no concept > > > of absolutes there." > > > Read what I just said. The center of mass of the fluid is in no way an > > absolute frame. > > Where is the 'center of mass' for the pacific ocean??? This is relatively straightforward to calculate. Why would you think it's difficult? In general R_cms = int[R*rho*dv]/int[rho*dv], where R_cms and R are vectors, R is the location of a volume element dv weighted by mass density rho (rho(R)). The momentum of any of the elements doesn't enter into it. Can you see why? > I think what > you really wanted to say is the frame where the local sum of the > population vector momenta is zero. I think that's equivalent. > But, note what I said kinetic > theory 'which underpins' fluid mechanics. At the scales small enough > where kinetic theory is necessary... But, either way, there is no > physical preference for any imagined coordinate system, aka, reference > frame. That's not quite right. There's the presumption that (1/2)m(v_rms)_x^2 = 1/2)m(v_rms)_y^2 = 1/2)m(v_rms)_z^2. This won't be true where there is a general drift from left to right of the molecules. > > Paul Stowe- Hide quoted text - > > - Show quoted text - |