Groups > sci.physics.relativity > #428792
Re: What Do Dingleberries Actually Mean When They Claim Moving Clocks Run Slow?

From	"HGW..." <hgw@....>
Newsgroups	sci.physics.relativity
Subject	Re: What Do Dingleberries Actually Mean When They Claim Moving Clocks Run Slow?
Date	2017-08-15 10:56 +1000
Organization	Aioe.org NNTP Server
Message-ID	<omtgs6$6om$1@gioia.aioe.org> (permalink)
References	(14 earlier) <76184cfa-1e5e-4f73-8fa8-b673588e68a1@googlegroups.com> <omo3nc$dqf$1@gioia.aioe.org> <feebd636-3253-464f-89bb-db4fcd50ad9d@googlegroups.com> <omovo3$1ff2$1@gioia.aioe.org> <7f17ad2a-69ed-4608-8e0a-632d5e071635@googlegroups.com>
Show all headers | View raw
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.
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

>> 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.
m>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.








--
Thread

csiph-web