Re: Einstein's Doppler equation wrong?
From: Bilge (dubious_at_radioactivex.lebesque-al.net)
Date: 11/02/04
- Next message: Bilge: "Re: Scientifically valid definitions."
- Previous message: John C. Polasek: "Re: Is Einstein's Principle of Equivalence true?"
- In reply to: Androcles: "Re: Einstein's Doppler equation wrong?"
- Next in thread: V ertner Vergon: "Re: Einstein's Doppler equation wrong?"
- Messages sorted by: [ date ] [ thread ]
Date: Tue, 02 Nov 2004 05:25:12 -0000
Androcles:
>
>I do not need to go quote mining on the internet.
>I = E/R. --- Ohm.
>
>To suggest that Maxwell did not know this is an affront to Jimmy
>Clerk.
Maxwell did not show that. In fact, maxwell's equations fail to
explain it. Ohm's law is a completely empirical result obtained by
fitting data points with a line. Until drude proposed a very basic
theory of conductors around 1900, there was no explanation for
ohm's law. Drude's theory was also only a partial explanation. One
needs quantum mechanics to explain ohm's law properly.
>I can't very well change your stubborn mind, of course, but staring at
>equations you do not fully understand is no comparison to direct hands
>on experience.
>Where did you not learn physics?
If you think maxwell explained ohm's law, you are free to attempt
a derivation using maxwell's equations. Since coulombs law explicitly
states that charges accelerate in an electric field, I have a great
deal of difficulty seeing how you'll turn that into the constant
velocity in J = \sigma E, which is the more fundamental form of
V = IR.
[...]
>:
>: Well, sure -- a resistive load would melt! :-)
>
>Engineers know how to prevent that.
>Ever heard of wire wound resistors? Tungsten has a high melting point.
They know an even better way: Match impedances so that the antenna
radiates the power rather than having it dumped into a resistor.
You match the source impedance to the antenna if you want want
the maximum power transfer. If you have an ethernet card or a scsi
bus, you have a cable impedance of 50 to a few hundred ohms and devices
with input impedances on the order of at least k\Ohms if not M\Ohms,
which optimizes the voltage transfer, but still requires proper termination.
[...]
>
>
>You can quite effectively terminate the coax with a suitable LC
>and lose nothing as heat in the resistor. However, because the
>frequencies of digital signals are irregular, a resistor is a
>compromise to dampen the ringing, just a so-called "shock absorber"
>fitted to a car
No, it isn't. The transmission line doesn't ring unless you mismatch
impedances somewhere. Transmission lines have a characteristic impedance
that is independent of frequency:
Z = sqrt(L/C)
An LC circuit at the end of the line will make it ring like a telephone,
precisely because an LC circuit has a resonant frequency. In fact, if
you want to build a simple oscillator, just take a piece of coax with
a BNC on one end, attach a BNC ``T'' with some noise source at the
middle of the ``T'' and at the other end of the ``T'', use a BNC
connector that shorts the coax. You'll have a 1/4 wave resonant cavity
which resonates at a frequency determined by th length of the coax.
Just pick off the signal somewhere along the line, like the far end
of the coax.
>is a compromise. We have no idea what the frequency of bumps
>in the road will be, but I COULD arrange resonance as an experiment.
>The main shock absorbers are the tyres, backed up by the springs. The
>dampers are the resisitive "shock absorbers".
In case this hasn't occured to you, the idea here is to drop the
voltage across the resistor, not create a resonant circuit that acts
as dead short and burns up your cables.
>: goes out into the aether ... ER oops I mean onto the air ... anyway
>it
>: goes away from the antenna.
>
>When the current is running up the antenna it is charging a capacitor,
>the "plates" of which are the top and bottom of the rod. dE/dt is at a
>maximum
Uh, no. Antennas radiate for the simple reason that accelerated
charges radiate and changing E and B fields accelerate charges.
Your description of how an electric dipole works is incorrect, and
there is nothing special about electric dipole radiation. A loop
antenna is a magnetic dipole, for example.
- Next message: Bilge: "Re: Scientifically valid definitions."
- Previous message: John C. Polasek: "Re: Is Einstein's Principle of Equivalence true?"
- In reply to: Androcles: "Re: Einstein's Doppler equation wrong?"
- Next in thread: V ertner Vergon: "Re: Einstein's Doppler equation wrong?"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
