"Androcles" wrote in message
o.uk...
A diffraction grating spectrometer measures the velocity.

No . it lets you measure an angle which is in a known relationship to the
wavelength (independent of speed and frequency)

Paul B. Andersen wrote:
A diffraction grating spectrometer measures the wavelength.
lambda = d sin(phi) where d is the grating period.
The spectral lines only tell the wavelength shift.
It says nothing about the speed and the frequency.
It is experimentally proved that wavelength is Doppler shifted.

Right. Moreover, OTHER experiments show that the frequency is also
shifted (e.g. police radar guns). And surprising only to those who
refuse to understand SR, the two shifts are such that
speed = frequency * wavelength
remains constant.


Tom Roberts

On Feb 4, 12:50*am, "Jeckyl" wrote:
"snapdragon31" wrote in message
"snapdragon31" wrote in message
...
On Feb 1, 10:49 pm, "Jeckyl" wrote:
Let's look at this more clearly...
Let us assume our coordinates are such that at x=0,t=0 we also have
x'=0,
t'=0
Let us assume that we have a moving rod of length L travelling at
speed v
Let us assume the light is shining along the rod in the direction of
travel
Let us assume the light is emitted at t=t'=0, when the rear end of the
rod
is at x=x'=0

In the stationary frame of reference, the rod is moving while the
light
is
travelling at c
So the light gets to the other end of the rod, as seen by the
stationary
observer, at x, t where
t = L / (c - v)
x = c . t ... because x / t = c

That corresponds in the rods frame of reference to a point (x',t')

x' = gamma . ( x - v.t ) ... Lorentz
x' = gamma . ( c.t - v.t ) ... subs for x
x' = gamma . ( c - v ) . t ... factorise
x' = gamma . ( c - v ) . L / (c - v) ... subs for t
x' = gamma . L ... cancel

t' = gamma . ( t - v.x / c^2 ) ... Lorentz
t' = gamma . ( t - v.t / c ) ... subs for x
t' = gamma . (1 - v/c ) . t ... factorise
t' = gamma . (c - v) / c . L / (c - v) ... subs for t
t' = gamma . L / c ... cancel

Please try and come up something better. I believe you can because
you are smarter than the other guy.

Thanks for your patience .. you picked up my error in the above at the
line:

Let us assume that we have a moving rod of length L travelling at
speed v

It should have said that the L is the length of the moving rod measured
in
the stationary frame (ie the contracted length)

Your speed is incredible.

It helps when you know what you're talkin about .. even if one can stuff up
occasionaly on expressing it . *Indeed .. the equations would look better if
I used L' instead of L throughout .. not that it changes anything

I admit that I cannot think that fast.
The equations look good. *But I still need to plug in some numbers to
have a better feeling of what they are.

L' = 100m, Gamma = 1.34164
== L = 100/1.34164 = 74.54m

OK



Of course, one can do the analysis the other way .. by seeing that light
moving along the stationary rod at c transforms to light moving along the
moving rod at c

Lets assume the length of the stationary rod in the stationary frame is L
When light is emitted, we have at x=x'=0, t=t'=0

When light gets to the other end of the rod, we have x=L, t=L/c.
That point corresponds to (x',t') where

x = L = 74.54 m

No .. in this case, L = 100m

t = L/c = 2.48E-07 = 0.000000248 sec
Should t be L/(c-v)?

No .. but your L is wrong .. L is the length of the stationary rod in the
stationary frame

x' = gamma . ( x - v.t ) ... Lorentz
x' = 1.34 * (74.54 - 2E8 * 2.48E-07) = 33.33 m
(Sorry, it is not equal to 100m)

Because you stuffed up your L
Hi Jeckyl,

All calculations were based on the equations you provided. It showed
that your equations failed to work properly.


x' = gamma . ( L - v.L/c ) ... subs x=L, t=L/c
x' = gamma . (c - v) . L / c ... factorise

t' = gamma . ( t - v.x / c^2 ) ... Lorentz
t' = gamma . ( L/c - v.L / c^2 ) ... subs x=L, t=L/c
t' = gamma . ( c - v) L / c^2 ... factorise

x'/t' = ( gamma . (c - v) . L / c ) / ( gamma . ( c - v) L / c^2 ) = c

Again, the speed of light is c

Another way to prove x' / t' = c, given x = c * t is
x' = gamma . ( x - v.t ) ... Lorentz
t' = gamma . ( t - v.x / c^2 ) ... Lorentz
x'/t' = [gamma . (x - v.t)]/[gamma . (t - v.x / c^2)]
= (x - v.t)/(t - v.x / c^2)
= (c.t - v.t)/(t - v.c.t / c^2) * * *substitute x = c.t
= [t.(c-v)]/[t.(1-v/c)]
= (c-v)/[(c-v)/c]
= c

Indeed

In this case, there is no way to check what x, t, x' and t' are.

But if they are related by a Lorentz transformation, then the speed of light
is c.

But
one thing is sure that
if t t' due to time dilation then

You don't understand time dilation and simulatneity .. t is less that t'
because we are looking at different positions in space
Unfortunately, your equations were not correct. I could not show you
that the time dilation was not given by the formula t = gamma * t'
based on your formula.

Given v = 2E8 m/s, c = 3E8 m/s and x' = 100m
t' = x'/c = 3.33E-7 sec
Using Lorentz transformation to work backward, we get
x = gamma * (x' + v * t') = 223.61 m and
t = gamma * (t' + v * x' / c^2) = 7.45E-7 sec

t = 7.45E-7 sec t' = 3.33E-7 sec sorry your statement of t is less
than t' is not correct.

t/t' = 2.236 which is not equal to gamma (= 1.342)
2.236 = sqrt((c + v)/(c - v))
1.342 = gamma = 1 / sqrt(1 - v^2/c^2)

Interesting? The ratio between t and t' is not equal to gamma? It is
not a surprise if you understand Lorentz transformation thoroughly.


x must be greater than x', right?

At least you're thinking .. but you need to do some more learning first.
The idea of length elongation seems a bit difficult to you. Just
leave it until you fully understand the Lorentz transformation and
time dilation. One recommendation is to plug numbers into equations
to get a better feeling of what they are.

"snapdragon31" wrote in message
...
On Feb 4, 12:50 am, "Jeckyl" wrote:
All calculations were based on the equations you provided. It showed
that your equations failed to work properly.

As you were already told .. the equations are correct .. but you put the
wrong value into them . .you have to *understand* what they are for .. not
just plug in any old value and expect a results

You don't understand time dilation and simulatneity .. t is less that t'
because we are looking at different positions in space

Unfortunately, your equations were not correct.

No .. theey were correct

I could not show you
that the time dilation was not given by the formula t = gamma * t'
based on your formula.

unlikely

Given v = 2E8 m/s, c = 3E8 m/s and x' = 100m
t' = x'/c = 3.33E-7 sec
Using Lorentz transformation to work backward, we get
x = gamma * (x' + v * t') = 223.61 m and
t = gamma * (t' + v * x' / c^2) = 7.45E-7 sec
t = 7.45E-7 sec t' = 3.33E-7 sec sorry your statement of t is less
than t' is not correct.

You still fail to understand what the equations are telling you

t/t' = 2.236 which is not equal to gamma (= 1.342)
2.236 = sqrt((c + v)/(c - v))
1.342 = gamma = 1 / sqrt(1 - v^2/c^2)

Interesting? The ratio between t and t' is not equal to gamma? It is
not a surprise if you understand Lorentz transformation thoroughly.

I do .. you clearly do not .. you keep coming out with all this crap and
misapplying the equations because *you* do not know how to use them.

x must be greater than x', right?
At least you're thinking .. but you need to do some more learning first.
The idea of length elongation seems a bit difficult to you.

Because it is wrong

Just leave it until you fully understand the Lorentz transformation
and time dilation.

I do .. you do not

One recommendation is to plug numbers into equations
to get a better feeling of what they are.

I have .. many times .. your problem is you plug numbers in WITHOUT
UNDERSTANDING what you are plugging them in to, so you end up not
understanding the results. And worse.. you then publicise your errors to
the world

Koobee Wublee skrev:
On Feb 8, 3:46 am, "Paul B. Andersen" wrote:
Koobee Wublee wrote:

Thanks for bringing up what I want the Jackal to bring up.
These spectral lines only tell the frequency shift. It says nothing
about the speed and the wavelength. As you know, speed = frequency *
wavelength. Thus, the spectral shift does not invalidate Doppler/
Maxwell's model of classical Doppler shift --- with constant
wavelength but varying speed and frequency. However, to out-guess
what the happy professor might respond, you are going to toss in the
null results of the MMX. Well, does it really indicate invariance in
the speed? Could Voigt be wrong with all others plagiarizing from him
(without giving him any credit)?
Your out-guess is wrong.

Oh, boy. I love the unexpected. If the happy professor fails to
bring up the MMX to support his claim, you are walking on thin ice ---
once again.

My response is:
A diffraction grating spectrometer measures the wavelength.

lambda = d sin(phi) where d is the grating period.

Ah, yes. The diffraction grating spectrometer measures only the
wavelength at your observation. It says nothing about other important
parameters --- especially the ones at the source. shrug

The spectral lines only tell the wavelength shift.

Yes, at the moment of your observation.

It says nothing about the speed and the frequency.

That is correct.

It is experimentally proved that wavelength is Doppler shifted.

Well, I disagree very strongly.

I see.
So when different wavelengths are observed for
the H-alpha spectral line, it is not because the wavelengths
are Doppler shifted differently, but because the wavelengths of
the H-alpha spectral lines are different at the different sources?

And when we measure a cyclic shift in the H-alpha wavelength from a star,
we cannot conclude that it is a cyclic shift in the radial speed
of the star, but must conclude that the wavelength of the H-alpha
transition in all the Hydrogen atoms in that star by some mysterious
reason undergo a cyclic change?


I believe the Gregorian New Year is over. So, please stop drinking
too much unless the happy professor also celebrates the Chinese New
Year in Norway.

Quite.
A comment which reveals a lot of wisdom, of course.
Right to the point, and a very strong argument for why
the observed shifts of the wavelength of spectral lines
are caused by change of the laws of nature at the source,
rather than by Doppler shift.

--
Paul

http://home.c2i.net/pb_andersen/

"Paul B. Andersen" wrote in message
...

| So when different wavelengths are observed for
| the H-alpha spectral line, it is not because the wavelengths
| are Doppler shifted differently, but because the wavelengths of
| the H-alpha spectral lines are different at the different sources?

There are no silly questions ... just idiotic ones.



| And when we measure a cyclic shift in the H-alpha wavelength from a star,
| we cannot conclude that it is a cyclic shift in the radial speed
| of the star, but must conclude that the wavelength of the H-alpha
| transition in all the Hydrogen atoms in that star by some mysterious
| reason undergo a cyclic change?

Speed = distance * frequency.
Since the source has not changed frequency and the wavelength
is cyclic, it follows that the speed is cyclic.

Troll Dono trolled:


http://www.helinium.nl/trolltech.gif



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