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- Thread starter cragar
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In summary, if the cyclist sees the two beams go straight up, then an observer stationary on the ground will see them at an angle (but still parallel, of course). :wink:f

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From Special Relativity, if the cyclist sees the two beams go straight up, then an observer stationary on the ground will see them at an angle (but still parallel, of course).

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The machine gun moves at speed [tex] v=r\omega [/tex]. Now obviously the wheel is never stationary so even with the tiniest fraction of a second in between firing, the next bullet won't be fired exactly under the last, so in that sense "bends away". Also if we allow the column to conist of all bullets fired between r and [tex] r+\delta r [/tex], then we won't get a column of width [tex] \delta r [/tex] since the bullets fired at say, either side, of this interval will diverge out from each other, widening the column.

So a machine gun on such a trajectory would never really form a column of bullets exactly under each no matter what frequency the bullets are being fired at. But it may appear to form what we consider an acceptable column if the bullets are fired in fast enough succession.

Maybe a start would be, saying we observe the bullet/photon upto height h in our atmostphere, it takes time h/c to reach this height, then the next photon is emitted after time T=1/f, by which point the gun/laser has moved a linear distance of [tex] \delta s=r\omega \left(\frac{1}{f}\right) [/tex] or angular distance [tex] \delta \theta =\omega \left(\frac{1}{f}\right) [/tex]. Thus when this next photon is fired it starts off in a slightly different direction on the circle, inititally with the previous linear distance, but separation growing with time.

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From Special Relativity, if the cyclist sees the two beams go straight up, then an observer stationary on the ground will see them at an angle (but still parallel, of course).

I think he means just a guy sitting stationary in some frame spinning the wheel, not a guy moving at some linear speed v, seeing lines go straight up, vs a stationary observer on the side road. Alternatively you could imagine sitting on the laser in the non inertial frame, vs what someone would see in the inertial frame external to this motion (e.g. this could be a beacon shining up from the ground to the sky on earth, then what would somemone in the fixed reference frame of the stars observe, if the guy accelerating with the eath sees it go straight up)

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Non-inertial frames are pretty complicated in special relativity, so I would personally go about this by trying to understand it classically first (i.e. thinking of machine gun and bullets), and trying to work out how things look as you increase bullet frequency and and velocity, and increase the wheel's angular velocity.

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corresponding laser beam velocity component may be expected = V. But resulting beam velocity must remain = C, why the beam is inclining at angle ArcCos (V/C) to wheel plane.

I.e the emitted beam cluster should appear conical, seen from distance.

But I am no STR professional and may be wrong.

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Lets say we look at the point where the laser is moving away from us in the x-direction and treat that infinitesimal fragment of time in the usual SR Lorentz frame way (see diagram).

Then in [tex]\bar{O} [/tex] frame we have the four momentum of a photon in the laser beam given by:

[tex] \bar{P}=\left( h\bar{\nu}, h\bar{\nu}cos(\bar{\theta}), h\bar{\nu}sin(\bar{\nu}),0 \right) [/tex] for general angle.

Using the usual Lorentz transform matrix for x-direction boosts, we find components of the 4-momentum in the stationary frame (the frame moving -u in the x direction at the particular moment in time specifically)

This leads us to find the components as:

[tex] h\nu=\gamma h\bar{\nu}+\gamma u \left(h\bar{\nu} cos(\bar{\theta}) \right) [/tex]

[tex] h\nu cos(\theta)=\gamma u(h\bar{\nu})+\gamma(h\bar{\nu}cos(\bar{\theta)}) [/tex]

[tex] h\nu sin(\theta)=h\bar{\nu} sin(\bar{\theta}) [/tex]

[tex] P^z=P^{\bar{z}} [/tex]

The first equation here can be arranged as:

[tex] \frac{\nu}{\bar{\nu}}=\frac{1+ucos(\bar{\theta})}{1-u^2} [/tex]

This is the doppler shift, notice that even when [tex]\bar{\theta} =\frac{\pi}{2} [/tex], (which it actually does for our case, since the laser shoots straight up in the momentarily comoving reference frame), we still get a doppler shift of [tex] \frac{\nu}{\bar{\nu}}=\gamma [/tex]. This is what is known as the transverse Doppler shift, and is just an affect of time dilation, as you can see by using [tex] \nu=\frac{1}{T} [/tex] and [tex] \bar{\nu}=\frac{1}{\bar{T}} [/tex], leading to [tex] \frac{\bar{T}}{T}=\gamma [/tex]. So note also the colour of the beam will change too.

The next equation leads to [tex] \frac{\nu}{\bar{\nu}} cos(\theta)=\gamma (u+cos(\bar{\theta})) [/tex], and the third to: [tex] \frac{\nu}{\bar{\nu}} sin(\theta)=\sin(\bar{\theta}) [/tex]

you can sub the first into the second equation to obtain:

[tex] cos(\theta)=\frac{u+cos(\bar{\theta})}{1+ucos(\bar{\theta})} [/tex]

which looks a lot like the velocity addittion law (which of course it is):

[tex] v_x=\frac{u+v_{\bar{x}}}{1+uv_{\bar{x}}} [/tex]

[tex] \left|v_x\right|cos(\theta)=\frac{u+\left|v_{\bar{x}}\right|cos(\bar{\theta})}{1+u\left|v_{\bar{x}}\right|cos(\bar{\theta})} [/tex]

But since this is light [tex]\left|v_x\right|=\left|v_{\bar{x}}\right| =1 [/tex] to recover the above expression.

Now using this formula if the angle [tex] \bar{\theta}=\frac{\pi}{2} [/tex] we find that [tex] cos(\theta)=u [/tex] therefore [tex] \theta=arccos(u) [/tex].

u can range from 0->1 (the speed of light in these natural units). If the wheel is spinning at the speed of light 1 then, we find [tex] \theta=0 [/tex]. The light has bent so much that is no longer shooting upward but straight forward.

Then in [tex]\bar{O} [/tex] frame we have the four momentum of a photon in the laser beam given by:

[tex] \bar{P}=\left( h\bar{\nu}, h\bar{\nu}cos(\bar{\theta}), h\bar{\nu}sin(\bar{\nu}),0 \right) [/tex] for general angle.

Using the usual Lorentz transform matrix for x-direction boosts, we find components of the 4-momentum in the stationary frame (the frame moving -u in the x direction at the particular moment in time specifically)

This leads us to find the components as:

[tex] h\nu=\gamma h\bar{\nu}+\gamma u \left(h\bar{\nu} cos(\bar{\theta}) \right) [/tex]

[tex] h\nu cos(\theta)=\gamma u(h\bar{\nu})+\gamma(h\bar{\nu}cos(\bar{\theta)}) [/tex]

[tex] h\nu sin(\theta)=h\bar{\nu} sin(\bar{\theta}) [/tex]

[tex] P^z=P^{\bar{z}} [/tex]

The first equation here can be arranged as:

[tex] \frac{\nu}{\bar{\nu}}=\frac{1+ucos(\bar{\theta})}{1-u^2} [/tex]

This is the doppler shift, notice that even when [tex]\bar{\theta} =\frac{\pi}{2} [/tex], (which it actually does for our case, since the laser shoots straight up in the momentarily comoving reference frame), we still get a doppler shift of [tex] \frac{\nu}{\bar{\nu}}=\gamma [/tex]. This is what is known as the transverse Doppler shift, and is just an affect of time dilation, as you can see by using [tex] \nu=\frac{1}{T} [/tex] and [tex] \bar{\nu}=\frac{1}{\bar{T}} [/tex], leading to [tex] \frac{\bar{T}}{T}=\gamma [/tex]. So note also the colour of the beam will change too.

The next equation leads to [tex] \frac{\nu}{\bar{\nu}} cos(\theta)=\gamma (u+cos(\bar{\theta})) [/tex], and the third to: [tex] \frac{\nu}{\bar{\nu}} sin(\theta)=\sin(\bar{\theta}) [/tex]

you can sub the first into the second equation to obtain:

[tex] cos(\theta)=\frac{u+cos(\bar{\theta})}{1+ucos(\bar{\theta})} [/tex]

which looks a lot like the velocity addittion law (which of course it is):

[tex] v_x=\frac{u+v_{\bar{x}}}{1+uv_{\bar{x}}} [/tex]

[tex] \left|v_x\right|cos(\theta)=\frac{u+\left|v_{\bar{x}}\right|cos(\bar{\theta})}{1+u\left|v_{\bar{x}}\right|cos(\bar{\theta})} [/tex]

But since this is light [tex]\left|v_x\right|=\left|v_{\bar{x}}\right| =1 [/tex] to recover the above expression.

Now using this formula if the angle [tex] \bar{\theta}=\frac{\pi}{2} [/tex] we find that [tex] cos(\theta)=u [/tex] therefore [tex] \theta=arccos(u) [/tex].

u can range from 0->1 (the speed of light in these natural units). If the wheel is spinning at the speed of light 1 then, we find [tex] \theta=0 [/tex]. The light has bent so much that is no longer shooting upward but straight forward.

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I love the picture - the original 'back of a fag packet in a pub' ! But it gives the idea, which is what counts.yes i am talking about a stationary observer , maybe this picture will help.

I think this boils down to understanding what happens when you are traveling in a straight line and send out a light beam 'perpendicular' (from your point of view) to your path . Does another observer, at a different speed or 'stationary' see the beam going at right angles to your path or not? Once this has been sorted out, it may be possible to apply the result to the bicycle wheel experiment.

Assume you are flying through a thin ('stationary') mist (to make the beam visible). Ignore SR, initially.**

Imagine the beam is sent out horizontally and you are 'looking down' on it, sitting in your ship. The light returning from the beam (reflected by 'stationary' particles) will arrive later than it was sent and you will have moved forwards. Hence, it will appear to be coming from slightly behind the position it started from on the ship. The beam will appear as a line, tilted backwards from the perpendicular - if you are pointing the laser axis perpendicular to the ship's motion. The angle will be v/c, at low speeds, I think. By pointing the source forwards by this angle, to compensate, the beam will now appear to you as being perpendicular to the direction of motion.

For an observer, 'stationary' wrt the mist and vertically above, information takes the same amount of time to reach him from all parts of the ship and beam.

The observer will see the ship moving forwards and light scattered from mist particles will have taken time to get to them before they travel to the observer. Hence the beam will appear to lag behind also. But the difference in time lag will be half that of the light returning to the ship so I reckon the apparent angle will be halved.

SO, if the beam is pointed by the shipboard observer so as to appear to be perpendicular on the ship, the beam will appear to be pointing forwards for a remote observer.

Ignoring the extra problem introduced by the relativistic effects due to the acceleration associated with circular motion, this would transfer into the bicycle wheel question and it would imply that the beam would, indeed, appear to spread out as it would have a tangential component to its direction as far as the remote observer was concerned. I think I have also implied that this would also be true for the shipboard observer. Any ideas?

**Einstein proposed a not-too-dissimilar thought experiment involving cows, an electric fence and two observers. This did not involve SR but showed how our observation of an event is affected by our reference frame and the speed of light. It's described here: https://www.physicsforums.com/archive/index.php/t-79325.html

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corresponding laser beam velocity component may be expected = V. But resulting beam velocity must remain = C, why the beam is inclining at angle ArcCos (V/C) to wheel plane.

I.e the emitted beam cluster should appear conical, seen from distance.

But I am no STR professional and may be wrong.

Oh yeah, I just saw your post, after my typing out my lengthy beast, which I think is exactly the same as what you're saying, with my c set equal to 1. So the linear result is that beam bends away from the observer with an angle arccos(u), where in the case of the bike wheel [tex] u=r\omega [/tex].

How do you thus see this as conical? I am now picturing this as picture attached shows

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