A Newtonian response to Special Relativity

[RLH]

The changes required by Lorentz, Einstein and others to Newton revolve around the observations, experiments, etc. using applications of the Second Law

F = ma

The other 2 are left unchanged.

The change is to

F = gamma ma

where gamma is derived from equations by Lorentz involving velocity, v and the speed of light, c.

All that we know indicates that this relationship best describes what we can see/measure. So that should be the end, all is well and Special Relativity is proven to be correct. Not quite.

We can rearrange the terms so that

F / gamma = ma

All the known work then continues as correct, as the relationship remains the same, only the explanations differ.

We can stop talking about Relativistic Mass and start talking about Effective Forces.

We can restate the above as

The Effective Force, as applied between 2 frames of reference, decreases to zero as the relative velocity, v, between them approaches the speed of light, c.

As before, the other 2 laws do not need to change.

There is no way to determine which of the above two approaches is 'correct' as both will give identical outcomes in all cases.

E = mc^2

is left unchanged by the above as it does not change the actual equations and the relationship between left and right, only rearranges the terms.

[anne]

On Saturday, September 30, 2017 at 12:14:26 PM UTC+1, RLH wrote:
> The changes required by Lorentz, Einstein and others to Newton revolve around the observations, experiments, etc. using applications of the Second Law
>
> F = ma
>
> The other 2 are left unchanged.
>
> The change is to
>
> F = gamma ma

well, ok, but this equation is not the whole story if the acceleration and force are not parallel to the momentum.
see for example: http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

[RLH]

Mass, Force and thence Acceleration operate as regards the Second Law to the First and can be derived by setting F, and thus a, to 0. The m doesn't disappear.

Momentum can only be derived by bringing another inertial frame in to refer to the first frame to or set it to 0 if this is the reference frame. So momentum of 0 is the 'rest mass' or 'Newtonian Mass'.

Please note that Mass is scalar, Distance is scalar, Force Magnitude is Scalar and Force Direction, fd, is a vector and that fd = - fd' from the Third Law.

[Dirk Van de moortel]

Op 30-sep-2017 om 13:14 schreef RLH:

> The changes required by Lorentz, Einstein and others to Newton
> revolve around the observations, experiments, etc. using applications
> of the Second Law
>
> F = ma
>
> The other 2 are left unchanged.
>
> The change is to
>
> F = gamma ma
>
> where gamma is derived from equations by Lorentz involving velocity,
> v and the speed of light, c.
>
> All that we know indicates that this relationship best describes what
> we can see/measure. So that should be the end, all is well and
> Special Relativity is proven to be correct. Not quite.
>
> We can rearrange the terms so that
>
> F / gamma = ma
>
> All the known work then continues as correct, as the relationship
> remains the same, only the explanations differ.

Certainly not.
The equation
F = m a
results, when F is known, in a differential equation
involving the second derivative of location with respect
to time:
a = d^2x/dt^2.

The equation
F / gamma = m a
which can also be written as
f sqrt(1-v^2/c^2) = m a
results, when F is known, in a differential equation
involving ALSO the *first¨* derivative of location w.r.t.
time:
v = dx / dt

So none of the work remains correct.

Dirk Vdm

[Tom Roberts]

On 9/30/17 9/30/17 6:14 AM, RLH wrote:
> The changes required by Lorentz, Einstein and others to Newton revolve around the observations, experiments, etc. using applications of the Second Law
> F = ma
> The other 2 are left unchanged.
> The change is to
> F = gamma ma
> where gamma is derived from equations by Lorentz involving velocity, v and the speed of light, c.

Not really. That applies ONLY when the force is perpendicular to the 3-velocity.

And your "revolve around" is completely wrong: that is not at all the underlying
structure of SR, which is Lorentz invariance. Indeed a much better way to state
the correspondence between Newtonian mechanics and SR mechanics, which expresses
the symmetry of SR, is that 3-vectors become 4-vectors. Then

F = dP/d\tau = m dV/d\tau = m A

(F is the 4-force on an object, P is its 4-momentum, \tau is its proper time, m
is its mass, V is its 4-velocity, and A is its 4-acceleration). The last two
equalities hold only when m is constant.

Note that when projected onto the object's instantaneously
co-moving inertial frame, the components of F and A are equal
to their Newtonian counterparts, with a zero time component.

> [... excessively naive claims that miss the whole point of SR]

Tom Roberts

[Ned Latham]

RLH wrote:
>
> The changes required by Lorentz, Einstein and others to Newton
> revolve around the observations, experiments, etc. using
> applications of the Second Law
>
> F = ma
>
> The other 2 are left unchanged.
>
> The change is to
>
> F = gamma ma
>
> where gamma is derived from equations by Lorentz involving velocity,
> v and the speed of light, c.

See http://www.users.on.net/~nedlatham/Science/Physics/index.html
and consider the equations there for the gravitational and
electro-mitive force between two bodies:

F = G v[e] m[1] m[2] / d^2

F = EM v[e] c[1] c[2] / d^2

where G is the gravitational factor, a positive constant, EM is the
electro-motive factor, a negative constant, v[e] is the impact energy
factor, m[1] and m[2] are the bodies' masses, c[1] and c[2] are their
charges (positive or negative), and d is the distance between them.

The inpact eberfy factor fulfills there the function of your "gamma".
It is defined as "a number >= 1 which varies with the velocity
components of the potential impact energy between the two bodies".

Definition of exactly *how* it varies awaits research; at present,
we have only two data points.

Those two formulae are the only alterations Newtonian physics
requires to render SR and GR and all of the nonsense built on
them obsolete.

[RLH]

[... excessively naive (sic) claims that miss the whole point of SR]

Umm. Perhaps you would like to say how SR changes the 3 laws then. Which laws change and to what and why?

[RLH]

"So none of the work remains correct."

Are you saying that moving terms from one side of an equation to the other is somehow incorrect (provided that the 'correct' rules are applied). In this case multiply on one side can be replaced by divide on the other.

[RLH]

"Those two formulae are the only alterations Newtonian physics
requires to render SR and GR and all of the nonsense built on
them obsolete."

Changing just the one formula seems simpler somehow. And that just by rearranging the terms.

[Odd Bodkin]

He’s saying that you’ll have a harder time solving the equations of motion
since you’ve moved some of the derivatives to the other side of the
equation.

--
Odd Bodkin -- maker of fine toys, tools, tables

[RLH]

"He’s saying that you’ll have a harder time solving the equations of motion
since you’ve moved some of the derivatives to the other side of the
equation."

That's not correct. Terms can be moved from side to side without affecting the overall relationship. That's the whole point.

The derived equations for motion work out the same either way.

[Dirk Van de moortel]

Op 30-sep-2017 om 22:08 schreef Odd Bodkin:

> RLH <richardli...@gmail.com> wrote:
>> "So none of the work remains correct."
>>
>> Are you saying that moving terms from one side of an equation to the
>> other is somehow incorrect (provided that the 'correct' rules are
>> applied). In this case multiply on one side can be replaced by divide on the other.
>>
>
> He’s saying that you’ll have a harder time solving the equations of motion
> since you’ve moved some of the derivatives to the other side of the
> equation.

I'm not going to bother trying to educate this conehead.
Be my guest ;-)

Dirk Vdm

[RLH]

"I'm not going to bother trying to educate this conehead.
Be my guest ;-"

Well thank you.

I notice that you failed to come up with anything that refutes my conjecture.

The plain fact is that nothing changes at all in the maths, only the explanations for why the maths is correct.

Thus there is no way to actually decide which is better,

F = gamma ma

or

[RLH]

Please note that it is only the 'effective force' that is changed, not the force itself. Exactly in the same way as 'relativistic mass' is different to 'rest mass'.

There is no way to determine if the 'effective force' goes down or the 'relativistic mass' goes up. Either way it all still balances.

[Odd Bodkin]

Note that the force itself is customarily velocity-independent. For
example, the force of gravity acting on a falling ball doesn’t depend on
the speed of the ball. It is therefore also Independent of inertial frame,
because while the speed will change with change of frame, the force itself
does not. This leads to the classical observation that acceleration is the
same regardless of frame.

Note that your “effective force” does not share these traits, because gamma
isn’t a constant but is in fact a function of speed. This immediately shows
the additional complexity.

[RLH]

On Sunday, October 1, 2017 at 2:09:36 PM UTC+1, Odd Bodkin wrote:

> Note that the force itself is customarily velocity-independent.

Indeed. It is that assumption that I am questioning.

I am stating that Einstein, Loretnz, etc. can be satisfied in an alternative fashion.

The same maths. An alternative explanation.

[Tom Roberts]

On 10/1/17 10/1/17 8:09 AM, Odd Bodkin wrote:
> Note that the force itself is customarily velocity-independent. For
> example, the force of gravity acting on a falling ball doesn’t depend on
> the speed of the ball. It is therefore also Independent of inertial frame,
> because while the speed will change with change of frame, the force itself
> does not. This leads to the classical observation that acceleration is the
> same regardless of frame.

This is true for gravity (and some other forces) in Newtonian mechanics. It is
not true in SR or GR for any force, unless you mean a force that is a 4-vector
(in which case it is manifestly invariant).

For instance, the electromagnetic force on a charged particle cannot be
represented as a 4-vector, and is manifestly velocity dependent.

Tom Roberts

[Tom Roberts]

On 10/1/17 10/1/17 8:15 AM, RLH wrote:
> The same maths. An alternative explanation.

Your original post in this thread did NOT use "the same maths" as SR, and is not
at all a valid "alternative explanation".

Before you can reasonably propose alternatives, you must LEARN what SR actually
says.

Hint: it is NOT "F = gamma ma" -- that is NOT the generalization
of Newton's second law in SR. I gave the correct equation in an
earlier post in this thread.

Tom Roberts

[Daphne Charney]

Tom Roberts wrote:

> On 10/1/17 10/1/17 8:09 AM, Odd Bodkin wrote:
>> Note that the force itself is customarily velocity-independent. For
>> example, the force of gravity acting on a falling ball doesn’t depend
>> on the speed of the ball. It is therefore also Independent of inertial
>> frame,
>> because while the speed will change with change of frame, the force
>> itself does not. This leads to the classical observation that
>> acceleration is the same regardless of frame.
>
> This is true for gravity (and some other forces) in Newtonian mechanics.
> It is not true in SR or GR for any force, unless you mean a force that
> is a 4-vector (in which case it is manifestly invariant).

Unbelievable. That I need to say. The ball is in freefall, no force is
presented, sensed, detected or applied. You guys are unbelievable
unemployable. Do your work.

[Tom Roberts]

On 10/1/17 10/1/17 3:55 AM, RLH wrote:
> Thus there is no way to actually decide which is better,
> F = gamma ma
> or
> F / gamma = ma

There's no need to "decide" -- NEITHER is the correct generalization of Newton's
second law. See my earlier post for the correct equation.

Tom Roberts

[RLH]

On Sunday, October 1, 2017 at 4:12:05 PM UTC+1, tjrob137 wrote:

> There's no need to "decide" -- NEITHER is the correct generalization of Newton's
> second law

It is correct for one dimensional maths which is what we are talking here.

[danco...@gmail.com]

On Saturday, September 30, 2017 at 4:14:26 AM UTC-7, RLH wrote:
> The changes required by Lorentz, Einstein and others to Newton revolve
> around the observations, experiments, etc. using applications of the

> Second Law, F = ma, The other 2 are left unchanged. The change is to
> F = gamma ma.

No, that isn't the "change". (Let me use "g" for gamma.) If you are just talking about force in the direction of motion, the relevant equation of special relativity has gamma cubed, i.e., the equation is F = g^3 ma, and the expression g^3 m was sometimes called the "longitudinal mass". However, if the force is in the transverse direction to the speed of the particle, the relation is F = gma, and the expression gm was sometimes called the "transverse mass".

Now, it was soon realized (by Planck, Tolman and Lewis, etc.) that this is not the best way to express things, and in fact Newton's second law as stated in Principia was not really F=ma, it was F=dp/dt where p is the momentum, p=mv. So, if we define the "relativistic mass" m = g m0, where m0 is the rest mass, then F = dp/dt corresponds to F = d(mv)/dt = d(g m0 v)/dt, which gives the correct forces in the longitudinal, transverse, and every other direction.

Feynman famously said (somewhat loosely) that if you just want to learn enough special relativity to compute the right answers, all you really need to learn is to replace m with gm, but this assumes you use F=dp/dt and not F=ma. Anyway, you should be able to see the problem with your idea, because the gamma factor is not just an algebraic factor on the right side, it is inside the differentiation, so it can't just be divided. Also, even if you could just divide it to give a different definition of force, it would be a directionally dependent force, which would upset the Third Law. So, for multiple reasons, your idea doesn't work.

[Tom Roberts]

No, it isn't. It is correct ONLY for a force PERPENDICULAR to the velocity of
the object, and your F and a are the MAGNITUDES of the underlying 3-vectors
(they point in different directions).

For one-dimensional motion the correct simplification is:
F = gamma(v)^3 m a
(F and a are parallel or anti-parallel 3-vectors, m is constant, v is not)

Note that for non-colinear force and velocity the 3-acceleration is not parallel
to the 3-force, so no such simplification is possible and the real formula must
be used.

One thing I keep repeating: you MUST understand what the symbols
mean in the equations you use. Here you failed miserably in that.

Tom Roberts

[RLH]

On Sunday, October 1, 2017 at 6:00:11 PM UTC+1, danco...@gmail.com wrote:
> On Saturday, September 30, 2017 at 4:14:26 AM UTC-7, RLH wrote:
> > The changes required by Lorentz, Einstein and others to Newton revolve
> > around the observations, experiments, etc. using applications of the
> > Second Law, F = ma, The other 2 are left unchanged. The change is to
> > F = gamma ma.
>
> No, that isn't the "change". (Let me use "g" for gamma.) If you are just talking about force in the direction of motion, the relevant equation of special relativity has gamma cubed, i.e., the equation is F = g^3 ma, and the expression g^3 m was sometimes called the "longitudinal mass". However, if the force is in the transverse direction to the speed of the particle, the relation is F = gma, and the expression gm was sometimes called the "transverse mass".

Then Longitudinal Effective Force is F / g^3 = ma

and

Transverse Effective Force is F / g = ma

See what I mean about the maths being the same. Whatever you come up with as a valid scaling for mass, time, length in Lorentz/Einstein, the inverse could apply to Force to derive Effective Force instead.


A logical Newtonian response might be:

If the result of forces, propagated at the speed of light, decreases in their effectiveness as any relative velocity approaches the speed of light then look to your equations, not mine.

It only seems reasonable to place terms that deal with forces propagated at c should contain terms that refer to c and v as required.

[danco...@gmail.com]

On Sunday, October 1, 2017 at 10:17:41 AM UTC-7, RLH wrote:
> Then Longitudinal Effective Force is F / g^3 = ma
> and
> Transverse Effective Force is F / g = ma
> See what I mean about the maths being the same. Whatever you come up with as a valid scaling for mass, time, length in Lorentz/Einstein, the inverse could apply to Force to derive Effective Force instead.

You overlooked my comment that if you define a direction-dependent force, then the Third Law is no longer satisfied, meaning we no longer have equal action and reaction.

Overall your approach doesn't make sense, because you agree that the correct answer is given by special relativity, and you're just trying to say you can get exactly those same answers by re-defining symbols so that the equations formally resemble those of Newtonian physics, but this is just silly. The plain fact is that, according to Newtonian physics, an object subjected to a constant force will undergo constant acceleration, whereas in special relativity this is not the case. This is because energy has inertia, which was absent from Newtonian physics. No amount of re-defining symbols will change this crucial underlying difference between Newtonian mechanics and special relativity.

[RLH]

On Sunday, October 1, 2017 at 7:48:34 PM UTC+1, danco...@gmail.com wrote:
> On Sunday, October 1, 2017 at 10:17:41 AM UTC-7, RLH wrote:
> > Then Longitudinal Effective Force is F / g^3 = ma
> > and
> > Transverse Effective Force is F / g = ma
> > See what I mean about the maths being the same. Whatever you come up with as a valid scaling for mass, time, length in Lorentz/Einstein, the inverse could apply to Force to derive Effective Force instead.
>
> You overlooked my comment that if you define a direction-dependent force, then the Third Law is no longer satisfied, meaning we no longer have equal action and reaction.

As action and reaction are equal and opposite then that also follows with Effective Force.

The Effective Force as experienced in one frame must be equal and opposite to that experienced in another frame. That tells us nothing about the magnitude of that force. 3 dimensions gives the direction of the vector. The magnitude of effective force varies with only with d, v and c. The Third Law stands.

> Overall your approach doesn't make sense, because you agree that the correct answer is given by special relativity, and you're just trying to say you can get exactly those same answers by re-defining symbols so that the equations formally resemble those of Newtonian physics, but this is just silly. The plain fact is that, according to Newtonian physics, an object subjected to a constant force will undergo constant acceleration, whereas in special relativity this is not the case. This is because energy has inertia, which was absent from Newtonian physics. No amount of re-defining symbols will change this crucial underlying difference between Newtonian mechanics and special relativity.

I understand what Relativity says. I am pointing out that there is a Newtonian, mathematical and logical alternative which gives the same observed behaviour.

[RLH]

As I pointed out elsewhere if you wish to choose the Longitudinal Effective Force it is indeed F / g^3 = ma, which corresponds to the "Longitudinal Relativistic Mass" of F = g^3 ma.

The Transverse Effective Force is F / g = ma which corresponds to the "Transverse Relativistic Mass" of F = g ma.

[danco...@gmail.com]

On Sunday, October 1, 2017 at 1:33:18 PM UTC-7, RLH wrote:
> As action and reaction are equal and opposite then that also follows
> with Effective Force.

No it doesn't, because your "effective force" is directionally and velocity dependent, and hence the formerly equal and opposite forces are no longer equal.

> I understand what Relativity says.

Your messages indicate clearly that you don't.

> I am pointing out that there is a Newtonian, mathematical and
> logical alternative which gives the same observed behaviour.

This is what you were *claiming*, but it's been disproven. First you claimed that for longitudinal forces we have F = gma, and then your mistake was pointed out, and you immediately (with no apparent embarrassment at all) switched to claiming a directionally and velocity dependent factor (necessitated by the fact that the "gamma" factor is really inside the differentiation), and then when it's pointed out that this violates the Third Law you begin (obviously improvising) spouting verbiage that is devoid of rational content.

You really should focus on the fact that, overall, your approach doesn't make sense, because you agree that the correct answers are given by special relativity, and you're just trying to say we can get exactly those same answers by re-defining symbols so that the equations *formally* resemble those of Newtonian physics, but this is just silly. You can re-define "cat" to mean "dog", but what's the point?

[Odd Bodkin]

You keep saying is a Newtonian alternative. In a Newtonian model, the force
and acceleration are independent of inertial reference frame. Your
effective force is not frame-independent and so does not follow the
Newtonian concept set.

[Tom Roberts]

The world is 3 dimensional. Consider a 45-degree angle, a 30-degree angle, ....
You'll run out of special "mass" names before I run out of angles.

Moreover, mass is a property of AN OBJECT, not a "property" of a combination of
the object, its velocity, a force applied to it, and which inertial frame one uses.

And also remember that the 3-force and the 3-acceleration ARE NOT PARALLEL in
general. That completely confounds Newtonian mechanics. Your quest is hopeless
-- the world we inhabit does not behave that way.

Tom Roberts

[RLH]

On Sunday, October 1, 2017 at 11:08:14 PM UTC+1, danco...@gmail.com wrote:
> On Sunday, October 1, 2017 at 1:33:18 PM UTC-7, RLH wrote:
> > As action and reaction are equal and opposite then that also follows
> > with Effective Force.
>
> No it doesn't, because your "effective force" is directionally and velocity dependent, and hence the formerly equal and opposite forces are no longer equal.

Effective force only has magnitude. No direction is stated or implied. It's value is related to d, v and c.

> > I understand what Relativity says.
>
> Your messages indicate clearly that you don't.

Hmm.

> > I am pointing out that there is a Newtonian, mathematical and
> > logical alternative which gives the same observed behaviour.
>
> This is what you were *claiming*, but it's been disproven. First you claimed that for longitudinal forces we have F = gma, and then your mistake was pointed out, and you immediately (with no apparent embarrassment at all) switched to claiming a directionally and velocity dependent factor (necessitated by the fact that the "gamma" factor is really inside the differentiation), and then when it's pointed out that this violates the Third Law you begin (obviously improvising) spouting verbiage that is devoid of rational content.

The Third Law is not violated. EF on one frame will still be -EF in the other regardless of its magnitude/direction.

You offered another equation with g^3 in it. I observed that the g^3 could still be on either side of the equation.

> You really should focus on the fact that, overall, your approach doesn't make sense, because you agree that the correct answers are given by special relativity, and you're just trying to say we can get exactly those same answers by re-defining symbols so that the equations *formally* resemble those of Newtonian physics, but this is just silly. You can re-define "cat" to mean "dog", but what's the point?

None at all, but that is not what has happened.

> The plain fact is that, according to Newtonian physics, an object subjected to a constant force will undergo constant acceleration, whereas in special relativity this is not the case. This is because energy has inertia, which was absent from Newtonian physics. No amount of re-defining symbols will change this crucial underlying difference between Newtonian mechanics and special relativity.

I am not redefining symbols.

The plain fact is that the overall relationship between mass and force has been to be determined to be

F = g ma

or

F = g^3 ma

depending on which case of F you are looking at.

That is all we know experimentally. SR has decided that g is on the right hand side of the equation.

[RLH]

You what? Of course it is frame independent. It still follows the Third Law.

Why should a force that propagates at the speed of light not be affected by a velocity between the two frames?

[RLH]

On Monday, October 2, 2017 at 12:48:21 AM UTC+1, tjrob137 wrote:

> The world is 3 dimensional.

Really.

>
> Moreover, mass is a property of AN OBJECT, not a "property" of a combination of
> the object, its velocity, a force applied to it, and which inertial frame one uses.

I never said it did.

> And also remember that the 3-force and the 3-acceleration ARE NOT PARALLEL in
> general. That completely confounds Newtonian mechanics. Your quest is hopeless

IYHO of course.

[Odd Bodkin]

It cannot possibly be frame independent. Your effective force is F/gamma.
Gamma is a function of speed. Speed is frame dependent, so gamma is frame
dependent, so your effective force is frame dependent. That is not
Newtonian, which treats forces as frame independent. Read this over twice,
slowly. What part of this do you not understand?

>
> Why should a force that propagates at the speed of light not be affected
> by a velocity between the two frames?
>

Because NEWTONIAN physics treats forces as frame INdependent.

[mlwo...@wp.pl]

W dniu poniedziałek, 2 października 2017 01:05:38 UTC+2 użytkownik Odd Bodkin napisał:

> You keep saying is a Newtonian alternative. In a Newtonian model, the force
> and acceleration are independent of inertial reference frame. Your
> effective force is not frame-independent and so does not follow the
> Newtonian concept set.

Imagination, poor idiot. BTW, it's not him redefining,
it's your gurus.

[mlwo...@wp.pl]

And planes can not fly, because they are heavier than air.

[Tom Roberts]

On 10/1/17 10/1/17 8:21 PM, RLH wrote:
> The Third Law is not violated. EF on one frame will still be -EF in the other
> regardless of its magnitude/direction.

[By "EF" you presumably mean "effective force", which seems to
be a concept and name of your own devising: "Effective force
only has magnitude. No direction is stated or implied."]

Newton's third law refers to 3-vector forces, not your "EF".

Moreover, Newton's third law is not valid in SR and classical electrodynamics.
In some sense it is replaced by conservation of 4-momentum.

Consider the force between charged particle A and charged
particle B, which are both moving arbitrarily. The net
3-force on A is not equal and opposite to the net 3-force
on B, because radiation is generated and it carries
momentum (and energy) away. Note, however, that when one
totals the energy and momentum of the particles and fields
(including radiation), both energy and 3-momentum are
conserved (or better, total 4-momentum is conserved).

Exercise for experts: In general the 3rd law is not valid
for this, but there are certain special situations in which
it happens to hold. Describe them. Hint: the very first
case you think of is probably one of the special ones.

> You offered another equation with g^3 in it. I observed that the g^3 could
> still be on either side of the equation.

The g^3 is for colinear 3-force and 3-velocity, and g^1 is for perpendicular
3-force and 3-velocity. For other angles between them there is no simple power
of g that makes the equation correct. Any general "law" must hold for all
possible values; yours doesn't even hold for these two (you need two separate
"laws").

> The plain fact is that the overall relationship between mass and force has
> been to be determined to be
> F = g ma
> or
> F = g^3 ma
> depending on which case of F you are looking at.

No. Not even close. Those are two very special cases out of an infinity of
similar ones with different angles between 3-force and 3-velocity.

I repeat: LEARN what SR actually says. It is considerably more complicated than
your guesses.

Tom Roberts

[RLH]

On Monday, October 2, 2017 at 3:13:29 AM UTC+1, Odd Bodkin wrote:

> It cannot possibly be frame independent. Your effective force is F/gamma.
> Gamma is a function of speed. Speed is frame dependent, so gamma is frame
> dependent, so your effective force is frame dependent. That is not
> Newtonian, which treats forces as frame independent. Read this over twice,
> slowly. What part of this do you not understand?

Velocity magnitude is frame independent, velocity direction is equal and opposite between 2 frames from the Third Law i.e. Frame dependant.

Please do stop trying to say that the 2 somehow alter if you rearrange the algebra.

What part of it is only the magnitude and the effects of that magnitude that is being altered am I not conveying to you and that it is still distributed in exactly the same way as before between frames between the frames?

If I were to say the magnitude has halved you would not complain. Yet you say that moving to a situation where the magnitude varies as the speed approaches c is impossible, reaching 0 at c = v.

To give a concrete example lets take a coil and a particle. The coil accelerates the particle with force F. The propagation of F is limited to c in any particular frame.

If a particle were to pass a coil at v = c (let's leave aside the problem of actually getting to c in the first place for now) it is irrelevant to the magnitude of the change that occurs when it is switched on. That change only propagates at c so it will never catch up (in the frame of the coil) with the particle if it is already moving away at c. At no time will the particle and the change in force be coincident in space, regardless of the frame it is viewed from.

Thus the effective force must be zero when v = c. Of course SR requires an infinite mass at that point instead. Dealing with zeros and what happens if you move beyond it I much easier to deal with than what happens beyond infinity.

> >
> > Why should a force that propagates at the speed of light not be affected
> > by a velocity between the two frames?
> >
>
> Because NEWTONIAN physics treats forces as frame INdependent.

Relativistic physics treats mass as a variable against velocity.
Newtonian physics treats effective force as a variable against velocity instead.

The equation F = gamma ma is the algebraic equivalent of f / gamma = ma.

Running around claiming that it is somehow incorrect kinda says that algebra is observer dependant.

[RLH]

On Monday, October 2, 2017 at 4:47:55 AM UTC+1, tjrob137 wrote:
> On 10/1/17 10/1/17 8:21 PM, RLH wrote:
> > The Third Law is not violated. EF on one frame will still be -EF in the other
> > regardless of its magnitude/direction.
>
> [By "EF" you presumably mean "effective force", which seems to
> be a concept and name of your own devising: "Effective force
> only has magnitude. No direction is stated or implied."]
>
> Newton's third law refers to 3-vector forces, not your "EF".

Well I have to use words to describe the effect of a force that is propagated at c and how it impacts m as v reaches c. Relativistic Mass and Effective Force seem to be reasonable choices.

Newton's Laws as they stand are in a single vector. It's realisation in 3d space gives a 3 vector force.

> Moreover, Newton's third law is not valid in SR and classical electrodynamics.

The 3 laws exist and apply across all physics. Suspending one in order to get the maths/algebra to get the equations to work is a definite sign that all is not well.

> In some sense it is replaced by conservation of 4-momentum.
>
> Consider the force between charged particle A and charged
> particle B, which are both moving arbitrarily. The net
> 3-force on A is not equal and opposite to the net 3-force
> on B, because radiation is generated and it carries
> momentum (and energy) away. Note, however, that when one
> totals the energy and momentum of the particles and fields
> (including radiation), both energy and 3-momentum are
> conserved (or better, total 4-momentum is conserved).

3d space requires magnitude and direction. F = gamma ma only gives magnitude. Relative position gives direction. That is unchanged by using F /gamma = ma.

> Exercise for experts: In general the 3rd law is not valid
> for this, but there are certain special situations in which
> it happens to hold. Describe them. Hint: the very first
> case you think of is probably one of the special ones.
>
> > You offered another equation with g^3 in it. I observed that the g^3 could
> > still be on either side of the equation.
>
> The g^3 is for colinear 3-force and 3-velocity, and g^1 is for perpendicular
> 3-force and 3-velocity. For other angles between them there is no simple power
> of g that makes the equation correct. Any general "law" must hold for all
> possible values; yours doesn't even hold for these two (you need two separate
> "laws").

The change from a a 1d space to a 3d space is what creates the 3 vector.

> > The plain fact is that the overall relationship between mass and force has
> > been to be determined to be
> > F = g ma
> > or
> > F = g^3 ma
> > depending on which case of F you are looking at.
>
> No. Not even close. Those are two very special cases out of an infinity of
> similar ones with different angles between 3-force and 3-velocity.
>
> I repeat: LEARN what SR actually says. It is considerably more complicated than
> your guesses.

These are not guesses as you put it.

It is an observation that you can only determine the overall relationship described by an equal sign. You cannot decide which side of an equation a term sits, only that if it swaps sides it must follow the rules as regards add/subtract and multiply/divide.

I know what SR says, I am just observing that there are 2 cases that fit the overall behaviour, not one.

[Poutnik]

Dne 2.10.2017 v 10:15 RLH napsal(a):

>
>
> Relativistic physics treats mass as a variable against velocity.

No, it does not.
The term relativistic mass is obsolete and abandoned.
Current SR considers mass as velocity invariant.

In SR, 2 original formulations of the 2nd Newton law

F = m . a
F = dp / dt

are not considered equivalent any more
and only the latter is taken as valid.

The equivalence is taken just as an approximation for low speeds.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.

[RLH]

I'll try and restate that to improve clarification.

As the relative velocity between frames approaches the speed of light then the apparent mass increases towards infinity. At c, m is infinite.

We see this in cloud chambers, particle accelerators, etc.

This presumes that the actual forces applied are constant, i.e. velocity independent.

If, however, the mass is constant, then the apparent forces will have to decrease to zero instead.

This is the equivalent of saying, use Lorentz on Maxwell, not on Newton.

[Poutnik]

Dne 2.10.2017 v 11:52 RLH napsal(a):

> On Monday, October 2, 2017 at 10:45:41 AM UTC+1, Poutnik wrote:
>>
>> No, it does not.
>> The term relativistic mass is obsolete and abandoned.
>> Current SR considers mass as velocity invariant.
>>
>> In SR, 2 original formulations of the 2nd Newton law
>>
>> F = m . a
>> F = dp / dt
>>
>> are not considered equivalent any more
>> and only the latter is taken as valid.
>>
>> The equivalence is taken just as an approximation for low speeds.
>>

>

> I'll try and restate that to improve clarification.
>
> As the relative velocity between frames approaches the speed of light then the apparent mass increases towards infinity. At c, m is infinite.

Invariant mass m is still dp/dt.

Apparent mass is nonsense, as it is direction dependent.
See radial and longitudal mass
For high speed, it may differ by many orders,
as for various directions it varies
from using gamma to gamma^3.

https://en.wikipedia.org/wiki/Mass_in_special_relativity#The_relativistic_mass_concept

[Poutnik]

Dne 1.10.2017 v 10:55 RLH napsal(a):

>
> Thus there is no way to actually decide which is better,
>
> F = gamma ma
>
> or
>
> F / gamma = ma
>

The way is easy. They are equivalent, so equally valid.

Which is better is a subjective criteria
with meaning which of the forms
better fits your intentions with the equation in wider context.

[RLH]

On Monday, October 2, 2017 at 11:03:23 AM UTC+1, Poutnik wrote:
> Dne 2.10.2017 v 11:52 RLH napsal(a):
> > On Monday, October 2, 2017 at 10:45:41 AM UTC+1, Poutnik wrote:
> >>
> >> No, it does not.
> >> The term relativistic mass is obsolete and abandoned.
> >> Current SR considers mass as velocity invariant.
> >>
> >> In SR, 2 original formulations of the 2nd Newton law
> >>
> >> F = m . a
> >> F = dp / dt
> >>
> >> are not considered equivalent any more
> >> and only the latter is taken as valid.
> >>
> >> The equivalence is taken just as an approximation for low speeds.
> >>
>
> >
> > I'll try and restate that to improve clarification.
> >
> > As the relative velocity between frames approaches the speed of light then the apparent mass increases towards infinity. At c, m is infinite.
>
> Invariant mass m is still dp/dt.
>
> Apparent mass is nonsense, as it is direction dependent.
> See radial and longitudal mass
> For high speed, it may differ by many orders,
> as for various directions it varies
> from using gamma to gamma^3.

Apparent mass is the mass as observed from a different frame. It is just an alternative name for Relativistic Mass, including both Mass and Momentum as required.

[RLH]

On Monday, October 2, 2017 at 11:09:48 AM UTC+1, Poutnik wrote:
> Dne 1.10.2017 v 10:55 RLH napsal(a):
> >
> > Thus there is no way to actually decide which is better,
> >
> > F = gamma ma
> >
> > or
> >
> > F / gamma = ma
> >
>
> The way is easy. They are equivalent, so equally valid.

You don't say.

> Which is better is a subjective criteria
> with meaning which of the forms
> better fits your intentions with the equation in wider context.

SR has one viewpoint. Another, logical, alternative exists. That gamma has been incorrectly placed with the equation.

If changes in a force propagates at c, how can an object travelling with respect to that change of force be anything other than 0 at v = c?

In all frames the change will never be in the same location as the perticle.

[Poutnik]

Dne 2.10.2017 v 12:54 RLH napsal(a):

> On Monday, October 2, 2017 at 11:03:23 AM UTC+1, Poutnik wrote:

>
> Apparent mass is the mass as observed from a different frame. It is just an alternative name for Relativistic Mass, including both Mass and Momentum as required.

Direction dependent, therefore useless parameter.

Its value depends on direction, in which it is measured.
If we consider electron at 0.9999 .c,
gamma is cca 70.7
gamma^3 is cca 353 000

So moving electron would have "apparent mass"
70.7 . m_e in one direction
353 000 . m_e in other direction,
where m_e is electron invariant mass.

Or something between in directions something between.
------------------------------------------------

It is not good to introduce the concept of the mass M = m *gamma
(*) of a moving body for which no clear definition can be given. It is
better to introduce no other mass concept than the ’rest mass’ m.
Instead of introducing M it is better to mention the expression for the
momentum and energy of a body in motion.

(*) edited for use in plain text

— Albert Einstein in letter to Lincoln Barnett, 19 June 1948 (quote
from L. B. Okun (1989), p. 42[2])


Many contemporary authors such as Taylor and Wheeler avoid using the
concept of relativistic mass altogether:

"The concept of "relativistic mass" is subject to misunderstanding.
That's why we don't use it. First, it applies the name mass - belonging
to the magnitude of a 4-vector - to a very different concept, the time
component of a 4-vector. Second, it makes increase of energy of an
object with velocity or momentum appear to be connected with some change
in internal structure of the object. In reality, the increase of energy
with velocity originates not in the object but in the geometric
properties of spacetime itself."[7]

[Poutnik]

Dne 2.10.2017 v 12:59 RLH napsal(a):

> On Monday, October 2, 2017 at 11:09:48 AM UTC+1, Poutnik wrote:
>> Dne 1.10.2017 v 10:55 RLH napsal(a):
>>>
>>> Thus there is no way to actually decide which is better,
>>>
>>> F = gamma ma
>>>
>>> or
>>>
>>> F / gamma = ma
>>>
>>
>> The way is easy. They are equivalent, so equally valid.
>
> You don't say.

Have you heard about equivalent operation
of dividing both equation sides by the same nonzero number ?

F = gamma . ma
F / gamma = gamma . ma / gamma

F / gamma = ma

>

>> Which is better is a subjective criteria
>> with meaning which of the forms
>> better fits your intentions with the equation in wider context.
>
>
> SR has one viewpoint. Another, logical, alternative exists. That gamma has been incorrectly placed with the equation.
>
> If changes in a force propagates at c, how can an object travelling with respect to that change of force be anything other than 0 at v = c?
>
> In all frames the change will never be in the same location as the perticle.
>

It has nothing to do with any theory nor physics,
but with mathematical formalism.

[RLH]

As distance is only a magnitude there is no direction in f = Gamma ma or f / gamma = ma.

[RLH]

On Monday, October 2, 2017 at 1:24:09 PM UTC+1, Poutnik wrote:
> Dne 2.10.2017 v 12:59 RLH napsal(a):
> > On Monday, October 2, 2017 at 11:09:48 AM UTC+1, Poutnik wrote:
> >> Dne 1.10.2017 v 10:55 RLH napsal(a):
> >>>
> >>> Thus there is no way to actually decide which is better,
> >>>
> >>> F = gamma ma
> >>>
> >>> or
> >>>
> >>> F / gamma = ma
> >>>
> >>
> >> The way is easy. They are equivalent, so equally valid.
> >
> > You don't say.
>
> Have you heard about equivalent operation
> of dividing both equation sides by the same nonzero number ?

Sure. If I were to plot the results what change would I see.

> F = gamma . ma
> F / gamma = gamma . ma / gamma
> F / gamma = ma

Why include the irrelevant step? You can move a term from left to right without that.

So you agree that the two are equally valid choices.

> >
> >> Which is better is a subjective criteria
> >> with meaning which of the forms
> >> better fits your intentions with the equation in wider context.
> >
> >
> > SR has one viewpoint. Another, logical, alternative exists. That gamma has been incorrectly placed with the equation.
> >
> > If changes in a force propagates at c, how can an object travelling with respect to that change of force be anything other than 0 at v = c?
> >
> > In all frames the change will never be in the same location as the perticle.
> >
> It has nothing to do with any theory nor physics,
> but with mathematical formalism.

As that is what I pointed out in the first place I am not sure what you want me to say.

[Poutnik]

Dne 2.10.2017 v 14:45 RLH napsal(a):

> On Monday, October 2, 2017 at 1:17:27 PM UTC+1, Poutnik wrote:
>> Dne 2.10.2017 v 12:54 RLH napsal(a):
>>> On Monday, October 2, 2017 at 11:03:23 AM UTC+1, Poutnik wrote:
>>
>>>
>>> Apparent mass is the mass as observed from a different frame. It is just an alternative name for Relativistic Mass, including both Mass and Momentum as required.
>>
>> Direction dependent, therefore useless parameter.
>>
>> Its value depends on direction, in which it is measured.
>> If we consider electron at 0.9999 .c,
>> gamma is cca 70.7
>> gamma^3 is cca 353 000
>>
>> So moving electron would have "apparent mass"
>> 70.7 . m_e in one direction
>> 353 000 . m_e in other direction,
>> where m_e is electron invariant mass.
>>
>> Or something between in directions something between.
>> ------------------------------------------------

[.....]

>
> As distance is only a magnitude there is no direction in f = Gamma ma or f / gamma = ma.
>

Direction of apparent mass measurement
relatively to direction of moving body velocity.

Apparent inertial mass in direction of motion is gamma^3 . m
Apparent inertial mass perpendicularly to motion is gamma . m

[Odd Bodkin]

RLH <richardli...@gmail.com> wrote:
> On Monday, October 2, 2017 at 3:13:29 AM UTC+1, Odd Bodkin wrote:
>
>> It cannot possibly be frame independent. Your effective force is F/gamma.
>> Gamma is a function of speed. Speed is frame dependent, so gamma is frame
>> dependent, so your effective force is frame dependent. That is not
>> Newtonian, which treats forces as frame independent. Read this over twice,
>> slowly. What part of this do you not understand?
>
> Velocity magnitude is frame independent,

I’m sorry? What drugs are you on? I didn’t read the rest, after I saw this.

[Poutnik]

Dne 2.10.2017 v 14:49 RLH napsal(a):

> On Monday, October 2, 2017 at 1:24:09 PM UTC+1, Poutnik wrote:
>>
>> Have you heard about equivalent operation
>> of dividing both equation sides by the same nonzero number ?
>
> Sure. If I were to plot the results what change would I see.

...that it is the equivalent operation.
If equation was true before it, it is true after.
If equation was false before it, it is false after.

>
>> F = gamma . ma
>> F / gamma = gamma . ma / gamma
>> F / gamma = ma
>
> Why include the irrelevant step? You can move a term from left to right without that.

You looked like you needed that. Sory if you did not.

>
> So you agree that the two are equally valid choices.

I was trying to persuade you they are.
It is not matter of physics, but of elementary algebra,
whatever F, gamma<>0, "m" and "a" are.

>> It has nothing to do with any theory nor physics,
>> but with mathematical formalism.
>
> As that is what I pointed out in the first place I am not sure what you want me to say.

Why did you then ask in the first place ?

The fact the equations are equivalent is clear
after late basic school algebra lessons.

[RLH]

Distance has no direction.

[Poutnik]

Dne 2.10.2017 v 15:28 RLH napsal(a):

> On Monday, October 2, 2017 at 2:00:02 PM UTC+1, Poutnik wrote:
>>
>> Direction of apparent mass measurement
>> relatively to direction of moving body velocity.
>>
>> Apparent inertial mass in direction of motion is gamma^3 . m
>> Apparent inertial mass perpendicularly to motion is gamma . m

> Distance has no direction.

Velocity has.
So do force and acceleration.

[RLH]

Magnitudes, such as Distance, do not have a direction. I am 10 miles away from something. I may add a direction, but I am not required to do so.

Changes in Magnitude are signed/directional, but not Magnitude itself.

Mass has no direction but Momentum does (see the First Law).

[RLH]

Others were suggesting that they are not. I apologise if you were not amongst them.

As we agree they are equivalent, then the choice of which is 'correct' must come from other things.

SR believes the gamma term(s) MUST reside to the RHS.
I believe that it could be on the LHS.

[Poutnik]

Dne 2.10.2017 v 15:35 RLH napsal(a):

>
> Magnitudes, such as Distance, do not have a direction. I am 10 miles away from something. I may add a direction, but I am not required to do so.
>

It depends..
Polar coordinates are based on oriented distances.

[RLH]

Velocity has a Magnitude AND a direction.
Acceleration has a Magnitude AND a direction.

Force only has a Magnitude, it does not say in which direction it is applied. Thus a force in a particular direction will cause an acceleration in that direction.

[Poutnik]

Dne 2.10.2017 v 15:39 RLH napsal(a):

> Others were suggesting that they are not. I apologise if you were not amongst them.
>
> As we agree they are equivalent, then the choice of which is 'correct' must come from other things.

If they are equivalent, both are correct,
F = gamma .m.a is used more often.

>
> SR believes the gamma term(s) MUST reside to the RHS.
> I believe that it could be on the LHS.

SR believes nothing.
It is just a custom to write it that way.

[RLH]

Polar or Cartesian I am still 10 miles away. The method of describing the directions do not affect that.

[Poutnik]

Dne 2.10.2017 v 15:45 RLH napsal(a):

> On Monday, October 2, 2017 at 2:32:30 PM UTC+1, Poutnik wrote:
>>
>> Velocity has.
>> So do force and acceleration.
>>

>

> Velocity has a Magnitude AND a direction.
> Acceleration has a Magnitude AND a direction.
>
> Force only has a Magnitude, it does not say in which direction it is applied. Thus a force in a particular direction will cause an acceleration in that direction.

You contradict yourself.

"....it does not say in which direction it is applied.
Thus a force in a particular direction ...."

In Newtonian mechanics,

vector_F = m . vector_a

Both F and A are vectors with the same direction
and the mass is the proportionality constant.

[RLH]

SR states that mass, more accurately perceived or apparent mass, rises towards infinity as v approaches c.

That is why gamma is to the RHS.

[Poutnik]

Dne 2.10.2017 v 16:11 RLH napsal(a):

>
> SR states that mass, more accurately perceived or apparent mass, rises towards infinity as v approaches c.
>
> That is why gamma is to the RHS.

This is an obsolete information.

SR uses now just invariant mass.

That is why gamma is to the RHS.

Otherwise it would be F=ma regardless of speed
and not F=gamma.m.a

[danco...@gmail.com]

On Sunday, October 1, 2017 at 8:47:55 PM UTC-7, tjrob137 wrote:
> Moreover, Newton's third law is not valid in SR and classical
> electrodynamics.

Not true. People who say this are overlooking the momentum of the interaction fields. They think that, in special relativity, although Newton's third law is valid for contact interactions, it is ambiguous or false for separate particles interacting via electromagnetic forces (for example), but this is because they are focusing only on the particles, overlooking the momentum of the electromagnetic field itself. It's true that the third law is valid only for local interactions, but in relativity all interactions are local, i.e., there is no action at a distance. This was all hashed out in the early days of relativity, by Poincare, Lorentz, Einstein, and Laue. The fundamental thing missing from Newtonian physics was the realization that energy has momentum.

[danco...@gmail.com]

On Sunday, October 1, 2017 at 6:21:56 PM UTC-7, RLH wrote:
> Effective force only has magnitude. No direction is stated or
> implied. It's value is related to d, v and c.

That makes no sense at all. If your "effective force" is just a magnitude, with no direction, then it isn't even possible to state (for example) Newton's Third law, because that law refers to equal and *opposite* forces, which entails opposite *directions*.

> The Third Law is not violated. EF on one frame will still be -EF in
> the other regardless of its magnitude/direction.

Again, that makes no sense at all, for multiple reasons. First, the third law has nothing to do with different frames, it applies in any single frame. Second, you already said your "effective force" is just a magnitude, with no direction, so you cannot have "opposite" effective forces, let alone perpendicular effective forces, or effective forces at any other angle. Everything you're saying is nonsense.

> > > I understand what Relativity says.
> > Your messages indicate clearly that you don't.
> Hmm.

Hmm? For example, your original post said relativity says F=gma for longitudinal forces, whereas in fact is it F=g^3 ma. So there's no "hmm" about it. You've clearly shown that you don't understand special relativity.

> You offered another equation with g^3 in it. I observed that the
> g^3 could still be on either side of the equation.

More accurately, I pointed out that you didn't know what you were talking about, because the real relativistic equation for longitudinal force involves gamma^3, and that you didn't even know Newton's law is F=dp/dt rather than F=ma, and that gamma actually appears inside a differentiation, which explodes your entire premise (based on the belief that gamma was a simple algebraic factor.)

> You can re-define "cat" to mean "dog", but what's the point?
> None at all, but that is not what has happened.

But it *is* what happened, or rather, it is what you are trying to do. You accept that special relativity gives the right answers, and you are simply trying to redefine terms so that the "cat" of Newtonian physics is the "dog" of special relativistic physics. But the two kinds of physics are still different. You can't change this simply by re-defining terms. (I emphasize that this is what you are trying to do, not what you have done, because your attempt to re-define terms has been gibberish.)

> I am not redefining symbols.

Sure you are. You are re-defining F from force to effective force. If you begin with the special relativistic formula and you don't re-define any terms, then you are left with the special relativistic formulas. Your original intent was to say "F=ma" applies in relativistic mechanics, provided we understand "F" to be the effective force, right? But this (failed) attempt to re-define F is just silly.

> The plain fact is that the overall relationship between mass and force has been to be determined to be F = g ma or F = g^3 ma depending on which case of F you are looking at.

No, there are infinitely many different expression for force when taken outside the differential form, because they are infinitely many directions, not just parallel and perpendicular. Newton's law is a vector equation, it does not deal just with magnitudes, and the same is true in special relativity. Overall, your idea is just nonsense.

[Odd Bodkin]

Being frame dependent has nothing to do with whether the quantity has a
direction or not. Kinetic energy is a scalar and has no direction, and it
is also frame dependent. If none of this makes sense at all, then you
should be learning basics. Not here, but from a book.

>
> Changes in Magnitude are signed/directional, but not Magnitude itself.
>
> Mass has no direction but Momentum does (see the First Law).
>

[Odd Bodkin]

This is flat wrong. See a freshman textbook.

[RLH]

A force 'n' times as big in one direction will produce 'n' times as much acceleration in that same direction. 'n' times is only a magnitude.

[Dono,]

My only question is : are you an old idiot using a new nick or are you just a new idiot?

[Dono,]

On Sunday, October 1, 2017 at 6:24:22 PM UTC-7, RLH wrote:
>
> Why should a force that propagates at the speed of light not be affected by a velocity between the two frames?

New imbecile (or old imbecile with new nick)

Force is a vector, as such, you need to start with its VECTORIAL defuinition:

\vec{F}=m\frac{d \gamma(v) \vec{v}}{dt}

If you ever manage to do the differentiation of the above expression, you will understand your errors. But I doubt you can.

[Stného Esovány]

RLH wrote:

>> A wise man guards words he says,
>> as they say about him more, than he says about the subject.
>
> A force 'n' times as big in one direction will produce 'n' times as much

Why not less?

[RLH]

On Monday, October 2, 2017 at 4:24:51 PM UTC+1, danco...@gmail.com wrote:
> On Sunday, October 1, 2017 at 6:21:56 PM UTC-7, RLH wrote:
> > Effective force only has magnitude. No direction is stated or
> > implied. It's value is related to d, v and c.
>
> That makes no sense at all. If your "effective force" is just a magnitude, with no direction, then it isn't even possible to state (for example) Newton's Third law, because that law refers to equal and *opposite* forces, which entails opposite *directions*.

Equal in magnitude, opposite in direction between objects or frames.

> > The Third Law is not violated. EF on one frame will still be -EF in
> > the other regardless of its magnitude/direction.
>
> Again, that makes no sense at all, for multiple reasons. First, the third law has nothing to do with different frames, it applies in any single frame. Second, you already said your "effective force" is just a magnitude, with no direction, so you cannot have "opposite" effective forces, let alone perpendicular effective forces, or effective forces at any other angle. Everything you're saying is nonsense.

The Third Law applies in the same way regardless if you deal with two objects in a single frame or two frames interacting together. That is rather what it is all about.

> > > > I understand what Relativity says.
> > > Your messages indicate clearly that you don't.
> > Hmm.
>
> Hmm? For example, your original post said relativity says F=gma for longitudinal forces, whereas in fact is it F=g^3 ma. So there's no "hmm" about it. You've clearly shown that you don't understand special relativity.
>
> > You offered another equation with g^3 in it. I observed that the
> > g^3 could still be on either side of the equation.
>
> More accurately, I pointed out that you didn't know what you were talking about, because the real relativistic equation for longitudinal force involves gamma^3, and that you didn't even know Newton's law is F=dp/dt rather than F=ma, and that gamma actually appears inside a differentiation, which explodes your entire premise (based on the belief that gamma was a simple algebraic factor.)

F = dp/dt

and

F = gamma dp/dt

or

F/gamma = dp/dt

doesn't change much does it?

> > You can re-define "cat" to mean "dog", but what's the point?
> > None at all, but that is not what has happened.
>
> But it *is* what happened, or rather, it is what you are trying to do. You accept that special relativity gives the right answers, and you are simply trying to redefine terms so that the "cat" of Newtonian physics is the "dog" of special relativistic physics. But the two kinds of physics are still different. You can't change this simply by re-defining terms. (I emphasize that this is what you are trying to do, not what you have done, because your attempt to re-define terms has been gibberish.)

I accept that the calculations that SR gave rise to can be satisfied by

F = gamma ma.

> > I am not redefining symbols.
>
> Sure you are. You are re-defining F from force to effective force. If you begin with the special relativistic formula and you don't re-define any terms, then you are left with the special relativistic formulas. Your original intent was to say "F=ma" applies in relativistic mechanics, provided we understand "F" to be the effective force, right? But this (failed) attempt to re-define F is just silly.

I am not redefining it. I am creating at alternative to it in the same way as Relativistic mass is an alternative to Newtonian mass.

> > The plain fact is that the overall relationship between mass and force has been to be determined to be F = g ma or F = g^3 ma depending on which case of F you are looking at.
>
> No, there are infinitely many different expression for force when taken outside the differential form, because they are infinitely many directions, not just parallel and perpendicular. Newton's law is a vector equation, it does not deal just with magnitudes, and the same is true in special relativity. Overall, your idea is just nonsense.

In the quoted form a single vector equation.

N times as much Force results in N times as much acceleration. What part of N times is directional?

[Poutnik]

Dne 2.10.2017 v 18:16 RLH napsal(a):

> On Monday, October 2, 2017 at 2:50:58 PM UTC+1, Poutnik wrote:

>> You contradict yourself.
>>
>> "....it does not say in which direction it is applied.
>> Thus a force in a particular direction ...."
>>
>> In Newtonian mechanics,
>>
>> vector_F = m . vector_a
>>
>> Both F and A are vectors with the same direction
>> and the mass is the proportionality constant.
>>
>

> A force 'n' times as big in one direction will produce 'n' times as much acceleration in that same direction. 'n' times is only a magnitude.
>

But force is a vector.
A vector multiplied by a scalar is still a vector.

[Dono,]

On Monday, October 2, 2017 at 9:35:52 AM UTC-7, RLH wrote:
>
> F = dp/dt
>
> and
>
> F = gamma dp/dt
>
> or
>
> F/gamma = dp/dt
>

I knew you were a cretin, gamma is variabile in time, imbecile.

[danco...@gmail.com]

On Monday, October 2, 2017 at 9:35:52 AM UTC-7, RLH wrote:

> Equal in magnitude, opposite in direction...

But you said they have no direction. You seem to blatantly contradict yourself.

> The Third Law applies in the same way regardless if you deal with
> two objects in a single frame or two frames interacting together.

That makes no sense at all. Frame are abstract systems of reference, they don't "interact together". Also, according to your proposed equations, the third law is violated, so your proposed equations are wrong. Also, you must not just consider "objects", you must consider fields as well, i.e., anything that carries energy.

> F = dp/dt and F = gamma dp/dt

What?? That's nonsense. Gamma is not a constant, so you can't take it outside the differentiation. It depends on the angle between the force and the velocity, both of which are vectors.

> doesn't change much does it?

It merely shows (again) that you have no idea what you're talking about.

> I accept that the calculations that SR gave rise to can be satisfied by
> F = gamma ma.

But we've been over this before. It depends on the angle between the force and the velocity. For longitudinal cases (when force is parallel to velocity) the factor is gamma^3. So you've gone back to your original misconception. Have you forgotten this already?

> I am not redefining [F]. I am creating at alternative to it in the

> same way as Relativistic mass is an alternative to Newtonian mass.

But going from Newton's conception of mass to the concept of relativistic mass (which is really just energy) is what yields special relativity. This represented a real physical difference, not just a different definition or different way of looking at things. It has real measurable consequences, because of the new principle that all energy has inertia. Now, your goal is to "create an alternative to" the concept of force, but you aren't proposing any new physical principle nor any new measurable effects, so at best you are just re-defining terminology (even though you deny this). But you haven't even proposed a coherent re-definition of terms, you've just typed some gibberish based on misunderstandings of both Newtonian physics and special relativity.

> In the quoted form a single vector equation.

Is that supposed to be a sentence in English? It doesn't convey any rational content, and doesn't even parse as a sentence.

> N times as much Force results in N times as much acceleration.
> What part of N times is directional?

The force is directional. Force is a vector, both in Newtonian physics and in special relativity.

[RLH]

On Monday, October 2, 2017 at 6:10:22 PM UTC+1, danco...@gmail.com wrote:
> On Monday, October 2, 2017 at 9:35:52 AM UTC-7, RLH wrote:
> > Equal in magnitude, opposite in direction...
>
> But you said they have no direction. You seem to blatantly contradict yourself.

A magnitude has no direction. A vector has a magnitude and direction.

> > The Third Law applies in the same way regardless if you deal with
> > two objects in a single frame or two frames interacting together.
>
> That makes no sense at all. Frame are abstract systems of reference, they don't "interact together". Also, according to your proposed equations, the third law is violated, so your proposed equations are wrong. Also, you must not just consider "objects", you must consider fields as well, i.e., anything that carries energy.
>
> > F = dp/dt and F = gamma dp/dt
>
> What?? That's nonsense. Gamma is not a constant, so you can't take it outside the differentiation. It depends on the angle between the force and the velocity, both of which are vectors.

A differential is applied over a range of point values in a field to produce a sum.

I am taking about the point values, not the range.

> > doesn't change much does it?
>
> It merely shows (again) that you have no idea what you're talking about.

No it just appears that I am failing to convey my meaning appropriately.

> > I accept that the calculations that SR gave rise to can be satisfied by
> > F = gamma ma.

Thank you.

> But we've been over this before. It depends on the angle between the force and the velocity. For longitudinal cases (when force is parallel to velocity) the factor is gamma^3. So you've gone back to your original misconception. Have you forgotten this already?

Newton's 3 laws can be applied in 1 or 4 dimensions. I am discussing the 1 dimensional case.

> > I am not redefining [F]. I am creating at alternative to it in the
> > same way as Relativistic mass is an alternative to Newtonian mass.
>
> But going from Newton's conception of mass to the concept of relativistic mass (which is really just energy) is what yields special relativity.

SR came about because the relationship between mass and force failed to follow Newton's Laws as v approached c.

>
> > In the quoted form a single vector equation.
>
> Is that supposed to be a sentence in English? It doesn't convey any rational content, and doesn't even parse as a sentence.

A vector in a single direction, towards the attractor.

> > N times as much Force results in N times as much acceleration.
> > What part of N times is directional?
>
> The force is directional. Force is a vector, both in Newtonian physics and in special relativity.

What part of 'N times' is directional?

Twice the force = twice the acceleration.
Twice contains no directional information.

[RLH]

I know that. It was a very poor example.

I hadn't noticed the subtle difference of F as point values and F as a range of point values and using the same word 'Force'.

Language is a tricky thing :-)

[Dono,]

Language is not the problem, your imbecility is: gamma is not a constant, you need to have it under the derivative, so you cannot blindly divide F by it, as you keep doing.

[RLH]

I should re-read what I say. :-(

A differential is applied over a range of point values in a field to produce a rate of change.

An Integral is applied over a range of point values in a field to produce a sum.

Damn fingers.

[RLH]

I had failed to realise that we had switched from F as describing a point value to F as describing a range of point values whilst still using the same word, Force.

I apologised.

F = ma

and

F = dp / dt

are not the same F.

[Gary Harnagel]

On Monday, October 2, 2017 at 12:16:42 PM UTC-6, RLH wrote:
>
> F = ma
>
> and
>
> F = dp / dt
>
> are not the same F.

Well, that's true if m isn't a constant, as in
the rocket equation:

p = m*v (in Newtonian)

dp/dt = m*dv/dt + v*dm/dt

dv/dt = a, so

dp/dt = m*a + v*dm/dt

If m = constant, then F = m*a.

In Mewtonian mechanics, of course.

Which isn't as exact as relativistic mechanics.

[JanPB]

On Saturday, September 30, 2017 at 4:14:26 AM UTC-7, RLH wrote:
> The changes required by Lorentz, Einstein and others to Newton revolve around the observations, experiments, etc. using applications of the Second Law
>
> F = ma
>
> The other 2 are left unchanged.
>
> The change is to
>
> F = gamma ma

No, that's incorrect.

--
Jan

[RLH]

One way of putting it is that

F = gamma ma

Happier now?

[RLH]

Doesn't give correct results as v approaches c. I know.

The required adjustment involves gamma. SR requires it to be used for Force, Mass, Length and Time in order to be exact at relativistic speeds.

Apparent Mass, Apparent Length and Apparent Time as observed from another frame vary as described overall by

F = gamma ma

Force is considered to be velocity independent even though its propagation is limited to c in all frames.

My contention is that Apparent Force (I called it Effective Force previously elsewhere but you get the point I hope) changes not the others as v approaches c.

Thus a force generated in one frame will have less and less effect in another frame as their relative velocity approaches c.

Of course the active/reactive forces will have to diminish in the same way as each other in order to satisfy symmetry and obey the Third Law.

[Dono,]

No. Imbecile. Stubborn.

[Tom Roberts]

On 10/2/17 10/2/17 1:07 PM, RLH wrote:
> A differential is applied over a range of point values in a field to produce a rate of change.

Still wrong. You need to learn what these words ACTUALLY MEAN. Your GUESSES are
wrong.

> An Integral is applied over a range of point values in a field to produce a sum.

Also wrong. You need to learn what these words ACTUALLY MEAN. Your GUESSES are
wrong.

Tom Roberts

[Tom Roberts]

On 10/2/17 10/2/17 9:52 AM, danco...@gmail.com wrote:
> On Sunday, October 1, 2017 at 8:47:55 PM UTC-7, tjrob137 wrote:
>> Moreover, Newton's third law is not valid in SR and classical
>> electrodynamics.
>
> Not true.

Yes, true.

> People who say this are overlooking the momentum of the
> interaction fields.

No! You misunderstand. It is precisely because the EM field can carry momentum
that the 3rd law is not valid.

The 3rd law says that if particle A exerts a force on particle B, then B must
exert a force on A that is equal in magnitude and opposite in direction [#]. But
for charged particles in classical electrodynamics is it easy to construct a
situation in which this is violated, because radiation is generated. But when
one totals the energy and momentum for particles and fields (including
radiation), the total 4-momentum is conserved. Moreover, the difference in those
two forces is related to the net momentum carried by the fields.

[#] Newton was explicitly thinking of both contact forces and
action at a distance (gravitation). He didn't know about
electrodynamics; if he had, he surely would have significantly
re-phrased the 3rd law.

The problem is not in the physics, or the basic conservation laws, it is merely
related to the way Newton phrased his third law. One cannot expect him to have
anticipated developments in physics hundreds of years in the future!

> [... lecture that misses the point.]
If you attempt to apply the 3rd law to the force on particle A exerted by the EM
field at its location, you'll have great difficulty defining how a "force" is
"exerted" on the field.

IOW: Radiation reaction exerts a force on charged particles, but is not itself a
"force" applied to the EM field.

The modern approach is to not attempt to discuss "forces" at all, and instead
consider conservation of 4-momentum [@]. After all, the primary, most useful
consequence of the 3rd law is conservation of 3-momentum (long before and
independent of Noether's theorems).

[@] This is forced on us in particle physics, as the QFTs we use
have no forces, nor any objects on which forces could be exerted.

Yes, in relativity all interactions are local. But they cannot all be considered
"forces", and one cannot apply the 3rd law to anything but forces.

Today when we discuss the "strong and weak forces", there is a
PUN on the word "force", as it does not mean what it did to
Newton. "Strong and weak interactions" is better phrased, and
is also in common usage.

Tom Robert

[Tom Roberts]

On 10/2/17 10/2/17 4:27 AM, RLH wrote:

> On Monday, October 2, 2017 at 4:47:55 AM UTC+1, tjrob137 wrote:
>> Moreover, Newton's third law is not valid in SR and classical
>> electrodynamics.
>

> The 3 laws exist and apply across all physics. Suspending one in order to get
> the maths/algebra to get the equations to work is a definite sign that all is
> not well.

What GOD whispered in your ear and told you this?

Moreover, your GOD is wrong, because Newton's laws are a TERRIBLE model of the
world we inhabit, in many different regimes. But they happen to work very well
in our everyday lives -- you have been deceived by thinking "common sense"
applies in regimes far removed from your everyday life where it was acquired. It
doesn't.

> 3d space requires magnitude and direction. F = gamma ma only gives magnitude.
> Relative position gives direction. That is unchanged by using F /gamma = ma.

That's just word salad combined with some blatantly false claims. You REALLY
need to learn about the subject before attempting to write about it.

For instance, in the case to which it applies (force perpendicular
to velocity), F = gamma m a is a 3-vector equation.

You are clearly just making stuff up and pretending it is true. That's USELESS.

> The change from a a 1d space to a 3d space is what creates the 3 vector.

More word salad. There never was a "1d space", the world has always been 3d, so
there never was any such "change" (except, apparently, in your mind).

> These are not guesses as you put it.

The stuff you make up is just plain wrong. "Guesses" is me being charitable.

> It is an observation that you can only determine the overall relationship
> described by an equal sign. You cannot decide which side of an equation a
> term sits, only that if it swaps sides it must follow the rules as regards
> add/subtract and multiply/divide.

But you have to apply equations to situations in which they apply. You consider
just one special case out of an infinity of similar cases and think you have
discovered something "profound". You have merely discovered the limits of your
understanding, which are VASTLY smaller than the subject at hand. Go study SR,
don't try to re-write it -- you have no hope of doing that until you actually
understand it.

> I know what SR says,

CLEARLY NOT. Apparently you are too ignorant to be able to recognize the limits
of what you think is understanding.

Tom Roberts

[JanPB]

On Monday, October 2, 2017 at 6:33:09 PM UTC-7, tjrob137 wrote:
> On 10/2/17 10/2/17 9:52 AM, danco...@gmail.com wrote:
> > On Sunday, October 1, 2017 at 8:47:55 PM UTC-7, tjrob137 wrote:
> >> Moreover, Newton's third law is not valid in SR and classical
> >> electrodynamics.
> >
> > Not true.
>
> Yes, true.
>
> > People who say this are overlooking the momentum of the
> > interaction fields.
>
> No! You misunderstand. It is precisely because the EM field can carry momentum
> that the 3rd law is not valid.
>
> The 3rd law says that if particle A exerts a force on particle B, then B must
> exert a force on A that is equal in magnitude and opposite in direction [#]. But
> for charged particles in classical electrodynamics is it easy to construct a
> situation in which this is violated, because radiation is generated.

Not even radiation, just the fact that the Lorentz force acts "sideways".

--
Jan

[RLH]

Thank you for your consideration.

[RLH]

Your observations are noted.

[danco...@gmail.com]

On Monday, October 2, 2017 at 6:33:09 PM UTC-7, tjrob137 wrote:
> > On Sunday, October 1, 2017 at 8:47:55 PM UTC-7, tjrob137 wrote:
> >> Moreover, Newton's third law is not valid in SR and classical
> >> electrodynamics.
> >
> > Not true.
>
> Yes, true.
>
> > People who say this are overlooking the momentum of the
> > interaction fields.
>
> No! You misunderstand. It is precisely because the EM field can carry momentum that the 3rd law is not valid.

No, Newton's third law is "To any action there is always an opposite and equal reaction". Action applies to any entity that carries momentum, which includes the electromagnetic and other fields. Granted, Newton didn't know about the electromagnetic or other fields, let alone that the energy of those fields carries momentum, but the third law (equal and opposite action and reaction) is valid provided we take into account all the carriers of momentum.

You can even find the frequently asked question on many web sites. For example, at the University of Illinois physics site:

=======
Question: Some people say relativity breaks Newton's third law, while others say it doesn't, but it appears to all come down to disagreement on the details of Newton's third law. - Sam (age 17) Springfield, IL

Answer: Newton's third law simply says, in its modern form, that the momentum of any isolated collection of things is conserved. Someone might have gotten the impression that for certain long-range forces (electromagnetism) momentum was not conserved, but that's only because they forgot to include the momentum of the electromagnetic field itself. Relativity is not only compatible with the third law, it's very closely related. The overall symmetry of special relativity, in which all places are fundamentally the same and any uniformly moving object can be treated as being at rest, actually implies Newton's third law.
=======

> The 3rd law says that if particle A exerts a force on particle B,
> then B must exert a force on A that is equal in magnitude and
> opposite in direction [#].

There is no action at a distance in modern physics, because all interactions are local. Again, you're overlooking the momentum of the fields that are interacting wit5h the material particles. The third law (equal and opposite action and reaction) applies to everything that carries momentum, including e.g., photons. Of course, Newton didn't know about photons, etc., but that doesn't invalidate the third law.

[danco...@gmail.com]

On Monday, October 2, 2017 at 7:44:53 PM UTC-7, JanPB wrote:
> Not even radiation, just the fact that the Lorentz force acts "sideways".

You're overlooking the momentum carried by the electromagnetic field. Study the "right angle lever paradox" and Laue's explanation of the Trouton-Noble experiment.

[Odd Bodkin]

<danco...@gmail.com> wrote:
> On Monday, October 2, 2017 at 6:33:09 PM UTC-7, tjrob137 wrote:
>>> On Sunday, October 1, 2017 at 8:47:55 PM UTC-7, tjrob137 wrote:
>>>> Moreover, Newton's third law is not valid in SR and classical
>>>> electrodynamics.
>>>
>>> Not true.
>>
>> Yes, true.
>>
>>> People who say this are overlooking the momentum of the
>>> interaction fields.
>>
>> No! You misunderstand. It is precisely because the EM field can carry
>> momentum that the 3rd law is not valid.
>
> No, Newton's third law is "To any action there is always an opposite and
> equal reaction". Action applies to any entity that carries momentum,
> which includes the electromagnetic and other fields. Granted, Newton
> didn't know about the electromagnetic or other fields, let alone that the
> energy of those fields carries momentum, but the third law (equal and
> opposite action and reaction) is valid provided we take into account all
> the carriers of momentum.

I’ve heard it described that, despite phrasing interpretations, the third
law is really the statement that interactions are pairwise, always
involving a momentum exchange between two things. In the conversation you
two have been having, there seems to be a counting problem about what the
“things” are. So if, for example there are two wire lops with current in
it, it’s possible to find cases where the third law does not work for the
two things being the loops, because the field is a third thing. I’ve always
tried to reconcile this by reverting to a Feynman diagram way of thinking,
where the counting is done at the vertex level. Momentum exchange is still
clean but there are creation operators in play.

>
> You can even find the frequently asked question on many web sites. For
> example, at the University of Illinois physics site:
>

> ======Question: Some people say relativity breaks Newton's third law,

> while others say it doesn't, but it appears to all come down to
> disagreement on the details of Newton's third law. - Sam (age 17) Springfield, IL
>
> Answer: Newton's third law simply says, in its modern form, that the
> momentum of any isolated collection of things is conserved. Someone might
> have gotten the impression that for certain long-range forces
> (electromagnetism) momentum was not conserved, but that's only because
> they forgot to include the momentum of the electromagnetic field itself.
> Relativity is not only compatible with the third law, it's very closely
> related. The overall symmetry of special relativity, in which all places
> are fundamentally the same and any uniformly moving object can be treated
> as being at rest, actually implies Newton's third law.
> ======

>> The 3rd law says that if particle A exerts a force on particle B,
>> then B must exert a force on A that is equal in magnitude and
>> opposite in direction [#].
>
> There is no action at a distance in modern physics, because all
> interactions are local. Again, you're overlooking the momentum of the
> fields that are interacting wit5h the material particles. The third law
> (equal and opposite action and reaction) applies to everything that
> carries momentum, including e.g., photons. Of course, Newton didn't know
> about photons, etc., but that doesn't invalidate the third law.
>

--
Odd Bodkin -- maker of fine toys, tools, tables

[danco...@gmail.com]

On Tuesday, October 3, 2017 at 10:40:28 AM UTC-7, Odd Bodkin wrote:
> I’ve heard it described that... the third law is really the

> statement that interactions are pairwise, always involving a
> momentum exchange between two things.

That would be a strange interpretation, since (for example) Newton's Principia included many calculations involving the simultaneous interactions between the sun, moon, and earth. An "action" need not be due to a single elementary entity. Two or more entities could act on two or more other entities. Regardless of how we partition them, the net actions and reactions are always equal and opposite.

I think Sam (age 17) had it right, i.e., it depends on how we define "the third law". For those who define it as referring to action at a distance between material particles, then clearly it is invalid in special relativity. In fact, action at a distance itself is invalid under special relativity. On the other hand, for those who take the third law to mean "equal and opposite action and reaction", including all forms of (necessarily local) action, the third law remains valid. This was all discussed at length by Poincare and Lorentz, even before 1905. Poincare noted that Lorentz's theory violated the third law, at least if we limit it to just the material particles. Then he had the brilliant realization that the third law would be rescued if we assume that the energy flux given by Poynting's equations is accompanied by a momentum flow E/c.

[Tom Roberts]

That's not sufficient.

For the case of two equal-mass and equal-charge particles, Newton's 3rd law
holds in the center-of-mass frame, even though the force between them is not
directed along the line between them, and even though radiation is emitted. This
is true in this case for all initial conditions. Think about the symmetry....

It doesn't hold in other frames, because the change in
simultaneity breaks the symmetry (speaking loosely, one is not
comparing the forces on the two particles "at the same time").

But for non-equal mass, non-equal charge, or other frames there is no such
symmetry, and the statement of the 3rd law above does not hold (except possibly
for certain very special situations).

But as dancouriann points out, if one interprets the 3rd law as LOCAL
conservation of (3-)momentum, then the 3rd law does hold. It cannot be globally
applicable, because of the frame-dependence alluded to above.

Tom Roberts

[danco...@gmail.com]

On Wednesday, October 4, 2017 at 6:25:26 AM UTC-7, tjrob137 wrote:
> ... the 3rd law ... cannot be globally applicable, because of the
> frame-dependence alluded to above.

Not true. The third law is globally applicable (at least in flat spacetime), provided we account for the momentum of the fields, not just the material particles. It isn't a matter of global vs local, it's a matter of correctly taking into account all the carriers of momentum (action) vs not.

[Tom Roberts]

On 10/3/17 10/3/17 11:02 AM, danco...@gmail.com wrote:
> No, Newton's third law is "To any action there is always an opposite and

> equal reaction". Action applies to any entity that carries momentum, [...]

Basically you want to interpret Newton's words as LOCAL conservation of
(3-)momentum. I agree that with that interpretation it is valid in SR and
classical electrodynamics.

Newton himself would be surprised at that interpretation, as he certainly
applied his laws to gravity, which to him was quintessentially NON-local.

So we must re-interpret his words in a MODERN way, and that implies ambiguities.
My re-statement of it as a balance of forces is certainly not unusual today. Nor
is your interpretation as local conservation of (3-)momentum. So this comes down
to a disagreement about how to interpret ancient, rather ambiguous words.

One certainly cannot fault Newton for not anticipating developments
in the field that were hundreds of years in the future when he wrote.

> There is no action at a distance in modern physics, because all interactions are local. [...]

Sure. But there certainly was back when Newton wrote.

There is ambiguity in how to interpret his ancient words today, and I feel no
compelling reason to argue about that.

Tom Roberts

[RLH]

On Tuesday, October 3, 2017 at 6:40:28 PM UTC+1, Odd Bodkin wrote:

> > ======Question: Some people say relativity breaks Newton's third law...

Or all/none of them.

Let me give you a model.

Michelson/Morley showed that we can model light (and thence variations in EMF and gravity) as a circular pond with waves travelling from the centre to the edge and, if you like, reflected back to the centre again.

This will hold true for observers at the centre and one passing by directly overhead from edge to edge.

Of course, Doppler says that the number of waves counted from edge to centre will differ as the direction to the source changes but the overall count will still be the same. Thus if we were to count the number of outgoing waves and the number of incoming waves from edge to centre or centre to edge and divide by 2 they would be the same.

[Tom Roberts]

On 10/4/17 10/4/17 9:55 AM, RLH wrote:
> Michelson/Morley showed that we can model light (and thence variations in EMF
> and gravity) as a circular pond with waves travelling from the centre to the
> edge and, if you like, reflected back to the centre again.

They did no such thing, as they most definitely did not have such a "circular
pond".

It's not clear at all what you are talking about, but it sure isn't the MMX.

Tom Roberts