Einstein's Starting Point

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19 years 1 month ago #14336 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by rodschmidt</i>
<br />If SR is wrong and LR is right, then where in this chain of reasoning did Einstein go wrong?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">In LR, the local gravitational potential field is a "preferred frame", and nothing happens to time. So the Lorentz transformations do not apply both ways between two frames. That makes following a light wave analogous to following a water wave, say, the way a surfer might. The wave does look "static" from that perspective. -|Tom|-

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19 years 1 month ago #13628 by wisp
Replied by wisp on topic Reply from Kevin Harkess
If the light were generated in your frame, you would see it travelling at speed c (Lorentz invariance law).
If the light were generated in an absolute external reference frame, you would see it travelling towards you at greater than c (optical time dilation illusion). Or away from you at less that c (optical time dilation illusion), assuming both SR and LR are incorrect and that motion have relativistic properties relative to an absolute frame.

wisp

- particles of nothingness

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19 years 1 month ago #13526 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by wisp</i>
<br />If the light were generated in your frame, you would see it travelling at speed c (Lorentz invariance law).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">If you are using electromagnetic-based clocks, the observed speed depends on how those clocks are synchronized. Unless you are in the local gravity field or use Einstein synchronization, the measured speed will not be c.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If the light were generated in an absolute external reference frame, you would see it travelling towards you at greater than c (optical time dilation illusion). Or away from you at less that c (optical time dilation illusion), assuming both SR and LR are incorrect and that motion have relativistic properties relative to an absolute frame.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The result still depends on how you synchronize clocks unless you use a near-infinite-speed synchronization device such as gravitational force signals. -|Tom|-

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19 years 1 month ago #13570 by rodschmidt
Replied by rodschmidt on topic Reply from Rod Schmidt
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">
The result still depends on how you synchronize clocks unless you use a near-infinite-speed synchronization device such as gravitational force signals. -|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
How is that statement compatible with the following one (also made on Aug 14)? How can you send a "signal" without "disturbing" the field?

What is a signal, besides a disturbance (i.e. a change) in the field?
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">
Gravitation and electrodynamics do have similar behavior. Their forces propagate at FTL speeds. <b>And when something disturbs the associated potential fields, those waves travel at speed c in both cases.</b>
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

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19 years 1 month ago #14114 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by rodschmidt</i>
<br />How can you send a "signal" without "disturbing" the field?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I referred to a "Gravitational force signal". In principle, any oscillation of a source mass would send such a signal. But binary pulsars also accomplish this, and would make a nice gravitational clock.

Naturally, any change in gravitational force affects the local gravitational potential field too, just as it affects material bodies within its range. But changes in potential fields (gravitational waves) are so weak that they are so far undetectable by any existing instrument.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">What is a signal, besides a disturbance (i.e. a change) in the field?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Gravitational force and gravitational potential are separate physical entities. One can change the potential without changing the force. For example, consider the potential inside a uniform spherical shell. The force in the interior is everywhere zero. And it remains zero as mass is added to or taken from the uniform shell, changing the potential.

So changes in potential do not imply force changes. While force changes are accompanied by potential changes, the latter are normally negligible and undetectable.

Note that force is proportional to acceleration, and potential is proportional to velocity squared. Yet acceleration and velocity squared remain independent physical entities in the sense that one is a vector and the other a scalar, and either can become indefinitely large or small in the presence of any (instantaneous) value for the other.

It might (or might not) also help to think of the gravitational potential field as synonymous with "elysium", the modern aether (in the sense used by Einstein) whose local density is determined by its proximity to masses. -|Tom|-

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19 years 1 month ago #11159 by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The force in the interior is everywhere zero. And it remains zero as mass is added to or taken from the uniform shell, changing the potential.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Potential is, by definition, relative to some arbitrary reference potential. Adding mass to a spherical shell changes both the the gravitational force and the potential relative to some arbitrary point outside the sphere. Potential relative to the center of the sphere, or any other point inside the sphere, would not change. The gravitational potential is constant within the sphere, and the force of gravity is zero inside the sphere. From a geometric perspective, the question of cause and effect purely semantic. From a field perspective, Tom is correct in pointing out that the force causes the potential. This is starting to sound like a religious debate, in which the two sides think their disagreement is substantive, when it is actually semantic.

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