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March 9[edit]

Feynman Lectures. Lecture 15 (34). Special relativity [1][2][edit]

We have the Lorentz transformation:
${\begin{aligned}x'&={\frac {x-ut}{\sqrt {1-u^{2}/c^{2}}}},\\y'&=y,\\[2ex]z'&=z,\\t'&={\frac {t-ux/c^{2}}{\sqrt {1-u^{2}/c^{2}}}},\end{aligned}}$
Suppose a man Joe in $S$ reference frame measures coordinates of some point $A$ at $t=0$ . In $S$ the point $A$ has coordinates:
$A(x,y,z)$ .

From Joe's point of view a moving (in positive $x$ direction) man Moe in his reference frame $S'$ at $t=0$ has coordinates:
$A(x',y',z')=A({\frac {x}{\sqrt {1-u^{2}/c^{2}}}},y,z)$ .

But from Moe's point of view $A$ is moving with speed $-u$ . So his coordinate $x'$ contracts to $x{\sqrt {1-u^{2}/c^{2}}}$ .

Moe can highlight a position of $A$ on his $x'$ -axis. What will see Joe ? Will he see 2 highlighted marks ${\frac {x}{\sqrt {1-u^{2}/c^{2}}}}$ and $x{\sqrt {1-u^{2}/c^{2}}}$ ?

Username160611000000 (talk) 14:04, 3 March 2019 (UTC)[reply]

I'm having a hella time following that, but I think what you're saying is that if you look at a Lorentz-contracted object and imagine how far that object thinks it has gone past you, you suppose that because it is contracted it thinks you are expanded. But in truth the object sees you as contracted as much as it is, so the distance is contracted in both frames! The reason is that you and it don't see the same times as simultaneous. You could highlight the same point at different times, but that shouldn't be sufficient to actually expand the distance to x*gamma factor; when two points are at relative rest, and the distance between them is x in that frame, it won't be longer than that in any frame, I don't think. ??? Wnt (talk) 17:12, 3 March 2019 (UTC)[reply]

@Wnt:The reason is that you and it don't see the same times as simultaneous. Both men stay at coordinates

x=x'=0;t=t'=0

. The only difference is that they move relative to each other, and point

A

is at rest in system

S

. Username160611000000 (talk) 18:24, 3 March 2019 (UTC)[reply]

The time coordinates are:

in

S

:

A(x,0)

;

S'

as seen from

S

:

A(x',t')=A({\frac {x}{\sqrt {1-u^{2}/c^{2}}}},{\frac {-ux/c^{2}}{\sqrt {1-u^{2}/c^{2}}}})

;

in

S'

(as seen from

S'

):

A(x'',0)=A(x{\sqrt {1-u^{2}/c^{2}}},0)

;

S

as seen from

S'

:

A(x,t)=A(x,{\frac {+ux/c^{2}}{\sqrt {1-u^{2}/c^{2}}}})

.

t=t' *only* where x=x'=0. Wherever x != x', now one of the times is nonzero relative to the other. Wnt (talk) 12:34, 4 March 2019 (UTC)[reply]

In the Lecture 34 Feynman says

...

We shall now give two more derivations of this same interesting and important result. Suppose, now, that the source is standing still and is emitting waves at frequency $\omega _{0}$ , while the observer is moving with speed $v$ toward the source. After a certain period of time t the observer will have moved to a new position, a distance $vt$ from where he was at $t=0$ . How many radians of phase will he have seen go by? A certain number, $\omega _{0}t$ , went past any fixed point, and in addition the observer has swept past some more by his own motion, namely a number $vtk_{0}$ (the number of radians per meter times the distance). So the total number of radians in the time t, or the observed frequency, would be $\omega _{1}=\omega _{0}+k_{0}v$ . We have made this analysis from the point of view of a man at rest; we would like to know how it would look to the man who is moving. Here we have to worry again about the difference in clock rate for the two observers, and this time that means that we have to divide by ${\sqrt {1-v^{2}/c^{2}}}$

— Feynman • Leighton • Sands , http://www.feynmanlectures.caltech.edu/I_34.html#mjx-eqn-EqI3413

Why have we to divide by ${\sqrt {1-v^{2}/c^{2}}}$ ? Is it because a moving person thinks that time $t$ (or better $\Delta t$ ) has passed at a slow clock rate? The moving man takes $\Delta t$ from $S$ , converts it to his time $\Delta t'=\Delta t/{\sqrt {1-v^{2}/c^{2}}}$ and then does he take the same radians and divide? Username160611000000 (talk) 09:52, 8 March 2019 (UTC)[reply]

Mute swan wings[edit]

I've noticed there's a particular way mute swans and some black swans swim, with their wings partially raised off of their backs. This is in contrast to the trumpeter swan, which holds its wings completely folded while swimming. Is there a specific name for what these birds are doing?

Mute swan
Black swan
Tundra and trumpeter swans

BlueSkinnyJeans (talk) 02:45, 9 March 2019 (UTC)[reply]

(Reformatted your image list.) --76.69.46.228 (talk) 05:46, 9 March 2019 (UTC)[reply]

By raising the profile of theirs wings the Mute swan and the Black swan are exhibiting threatening or assertive behaviour. Swans under non-threatening circumstances will swim with the wings flat against the body but in response to a perceived threat, particularly when protecting young or a nest, will raise their wings which increases the apparent size of the bird. Wing raising as a threat in Trumpeter swans seems much rarer or possibly non existent. I was unable to find any image to demonstrate it. Richard Avery (talk) 09:21, 9 March 2019 (UTC)[reply]

It's called "busking". [3] Alansplodge (talk) 14:50, 9 March 2019 (UTC)[reply]

Then I suggest a short article busk (birds) or busking (birds). --76.69.46.228 (talk) 06:40, 10 March 2019 (UTC)[reply]

Or, since it only seems to be associated with swans, maybe just a section in the swan article? Or in Display (zoology)? —107.15.157.44 (talk) 07:43, 10 March 2019 (UTC)[reply]

National Geographic suggests they may also use this posture to "windsurf" across water: [4] —107.15.157.44 (talk) 08:07, 10 March 2019 (UTC)[reply]