Groups > sci.physics.relativity > #428840
Re: What Do Dingleberries Actually Mean When They Claim Moving Clocks Run Slow?

From	Odd Bodkin <bodkinodd@gmail.com>
Newsgroups	sci.physics.relativity
Subject	Re: What Do Dingleberries Actually Mean When They Claim Moving Clocks Run Slow?
Date	2017-08-15 15:02 +0000
Organization	Aioe.org NNTP Server
Message-ID	<omv2e6$etl$3@gioia.aioe.org> (permalink)
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HGW... <hgw@....> wrote:
> On 14/08/17 03:23, dancouriann@gmail.com wrote:
>> On Sunday, August 13, 2017 at 12:39:51 AM UTC-7, HGW... wrote:
>>> The second postulate is leads to the RoS on which the rest of SR 
>>> crucially depends and which is logically impossible. It is not
>>> easy to see why it is logically impossible, which is why I have
>>> written up a comprehensive paper about it. Read this and stop 
>>> complaining...
>> 
>> I read it, but the claimed explanation is invalid. It boils down to 
>> the "Paradox" toward the end, which simply notes that, if inertial 
>> coordinate systems are related by Lorentz transformations, then 
>> pulses of light traversing a moving rod in both directions will take 
>> equal times in terms of the inertial coordinates in which the rod is 
>> at rest, but will take different amounts of time in terms of the 
>> inertial coordinates in which the rod is moving.  You declare this
>> to be a logical contradiction, but all you've really done is
>> re-state the relativity of simultaneity, you haven't shown any
>> contradiction. Two events can be simultaneous in terms of one system,
>> and not simultaneous in terms of another.  There is no
>> contradiction.
> 
> You seem to have missed the most important point....a common dingleberry
> habit. IF THE LIGHT SOURCES ARE CONNECTED TO THE MOVING ROD, THE PULSES
> WILL OBVIOUSLY TAKE THE SAME TIME TO TRAVERSE THE ROD IN OPPOSITE
> DIRECTIONS.

That's not obvious at all. In fact that answer is frame dependent.
Obviously. 


> Get it now?
> 
> Frankly I cannot see much point in arguing with someone with
> insufficient intelligence to even discuss Einstein's funny theories.
> 
>> Look, you can obviously set the clock at the west end of your house 
>> to read one minute ahead of the clock at the east end.  This may be
>> a dumb thing to do, but it isn't logically impossible.  Likewise, we 
>> can obviously set clocks at various locations in different states of 
>> motion so they match the coordinate times related by Lorentz 
>> transformations.  This is not logically impossible.  Your complaint 
>> should really be, not that it's logically impossible to define a 
>> class of coordinate systems with skewed time coordinates, but rather 
>> that you don't understand why we would claim that clocks set in this 
>> way represent the time coordinates of inertial coordinate systems.
> 
> You are talking nonsense again.
> 
>> Remember, an inertial coordinate system is a system of space and
>> time coordinates in terms of which Newton's equations of motion are 
>> satisfied in the low speed limit, which suffices to establish a 
>> system of coordinates in any uniform state of motion.  These 
>> coordinate systems are defined using inertial behavior of objects, 
>> but if energy has inertia (something which Newton didn't know, 
>> because the constant of proportionality is 1/c^2), inertial 
>> coordinate systems in different states of motion will necessarily 
>> have skewed time coordinates.  This is all explained in detail in
>> any good book on relativity.  It isn't magic, and it isn't even hard
>> to understand.
> 
> I told you there is no such thing as a good book that teaches Einstein's
> logically impossible version of relativity

Well, that claim explains why you've read nothing about relativity. 

> 
>>> Einstein's SR has no unifying mechanism for light speeds from 
>>> different sources...
>> 
>> Not true.  Forget about light.  The inertia of energy for ballistic 
>> particles explains the invariant speed.
> 
> I don't know of any energy that resists change. It just happens
> spontaneously. Energy clearly has no inertia.
> 
> If you have two identical guns with identical bullets with muzzle speed
> u, one fired northward by a gun on the ground and one fired northward on
> a northbound train moving with speed v.  In terms of the ground inertial
> coordinates, the speed of the ground-fired bullet will be u, but what
> will be the speed of the train-fired bullet, and why?  (Hint:  The
> answer is not v+u, because you must account for the inertia of energy).
> And if u and v are very large, approaching c, you will find that their
> speeds are "unified" asymptotically toward c.  This is not due to an
> ether, it is due to the inertia of energy.
> 
> No, it is due to the silly theory of a crazed plagiarizing hoaxer.
> 
>>> Energy can neither be created or destroyed, but can be transformed 
>>> from one form to another.
>> 
>> Right.
>> 
>>> As for being movable from one place to another, I'm not sure what 
>>> you have in mind. Sure you can carry gasoline around in a car but 
>>> so what? It is its mass which resists such movement, not the 
>>> chemical energy it contains.
>> 
>> The rest mass of ordinary substances includes a contribution from
>> its chemical energy, and from the kinetic energy of the molecules (a
>> hot object has slightly more rest mass than a cold one), and the
>> nuclear binding energy, and the kinetic energy of the bound quarks,
>> etc.  All forms of bound energy contribute to the rest mass of an
>> object.
> 
> Yes we know that.
> 
> Also, unbound energy exhibits inertia as well.  For example, Laue use
> this to explain the momentum flows corresponding to electromagnetic
> fields, etc., as shown in the Trouton-Noble experiment.
> 
> Well all the energy I can think of has no inertia.
> 
>>> ...stopping one object will cause another to gain momentum and KE.
>> 
>> Sure, that's what conservation of energy and momentum means.  This 
>> doesn't support your claim that kinetic energy doesn't exist...
>> quite the contrary.  If kinetic energy didn't exist, then
>> conservation of energy would be violated.
> 
> An object in remote space has no kinetic energy. In reference to another
> moving object, it is associated with an amount of energy equal to
> 1/2mv^2, which is called kinetic energy for convenience. That term is
> used in calculating the amount of energy that can be obtained by the
> remote object in he send object's frame.
> 
>>> 1/2mv^2 is obviously frame dependent... but it is not a quantity 
>>> that should be considered separately as energy.
>> 
>> Again, if kinetic energy didn't exist, then conservation of energy 
>> would be violated.  Consider a mass-spring system.  The energy flows 
>> alternately between kinetic energy of the mass and potential energy 
>> of the spring.
> 
> Yes we physicists know all about these things. That's why we try to help
> non-physicists like you and little gawy harnagel to understand the finer
> points.
> 
>> If the kinetic energy isn't energy, then energy is not conserved. But
>> we've agreed that energy is conserved, i.e., energy is never created
>> or destroyed, so it can only move and change form.  One of the forms
>> it can take is kinetic energy.
> 
> Energy is conserved when an object is stopped by another.
> 
> Let a mass m be moving at v relative to a second mass M.
> Let is coalesce with M, in which case some heat will be generated. Let
> the combined mass move at u relative to the original position of M.
> In M's frame, the kinetic energy of m is:
> KE = is 1/2mv^2.
> The KE of M is zero.
> After the collision, the final kinetic energy KE'is:
> Initial KE = 1/2(M+m)u^2............................(1)
> 
> From the conservation of momentum,  mv = (M+m)u, so u = mv/(M+m)
> If M = km
> Substituting for u in (1), Final KE' = 1/2m^2v^2/(M+m)
> If M = km
> u = v/m(k+1)
> Final KE' = 1/2mv^2/(k+1)
> Loss of KE = 1/2mv^2[1-1/(k+1)]
> = 1/2mv^2(k/(k+1)
> 
> The bigger k, the more closely does the heat generated approach the 
> initial KE.
> 
> Now, if the situation is viewed from any other frame, it can be shown 
> that the total KE lost is always the same.
> In a third observer's frame, both m and M are moving towards O.
>> v+w.....M>w.......O
> Let M = km
> Initial momentum = m(v+w)+kmw.
> Final momentum = m(k+1)u
> From conservation of momentum:
> m(v+w)+kmw = m(k+1)u
> v+w(k+1)= (k+1)u
> v/(k+1)+w = u
> u^2 = [v^2/(k+1)^2]+[2vw/(k+1)]+w^2
> w = u-[v/(k+1)]
> 
> Initial KE = 1/2m[(v+w)^2+kw^2]
> Final KE = 1/2m(k+1)u^2
> 
> Difference = 1/2m[(v+w)^2+kw^2-(k+1)u^2]
> Substituting for u:
> Diff = 1/2m[(v+w)^2+kw^2]-[(k+1)(v^2/(k+1)^2]+[2vw/(k+1)]+w^2)
> = 1/2m [v^2 +2vw+w^2 +kw^2]-[(v^2/(k+1)]+[2vw]+(k+1)w^2)]
> = 1/2mv^2(k/(k+1)
> 
> .....which is independent of w.
> So any change in KE is frame independent.
> Which implies that any creation of KE is also frame independent.
> 
> 
> 
> 
> 
> 
> 
> 



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