Unfortuantely this has nothing to do with the issue or this thread. This is about the relative velocity of the flash to the detectors in the embankment frame.
Mac, which frame is the detector in, the embankment frame or the train frame? Which frame is the flash in, the embankment frame or the train frame? Yes, both are in both frames, there is a flash/detector in the train frame and there is a flash/detector/OBSERVER in the embankment frame. The observer is the one that sees a Doppler shift of the flash, not the detector in the train frame. The flash is not 'chasing' the detector, the detector simply changes location while the flash is in flight, changing distance travelled but not Doppler shift in the train frame. The light from the sun takes about 8 1/2 minutes to reach us on Earth. We on Earth can change our location while the light is transiting the distance, but the light does not Doppler shift unless we are in motion the instant those photons reach us. The photons are not 'chasing' us when we are in the new location.
MacM: Your standard response these days to anyone who disagrees with you. Teaching requires two people: a teacher and a learner. So far, there's only me: the teacher. You're unwilling to take part in the process. You have to want to improve, MacM. So, you agree with me, then. Good.
You and I agree on most things but we do disagree on this. In the embankment frame the flash of light has a relative velocity to the engine detector of c - v. That is a different relative velocity than "c" of the train frame. c - v must be frequency shifted from a "c" frequency. That is from the embankment perspective fewer crests and valleys of the EM wave pass the detector per second.
Good gravy MacM, I agree with you that the relative velocity of the flash to the detectors in the embankment frame is important! The only difference is that I think the relative velocity of the flash to the source in the embankment frame is also important! If you can't agree with me on that, I'm afraid there's nothing more I can say on this subject.
I don't disagree that there are also issues associated with the relative velocity of the source to the flash but they don't impact the detonation question associated with the detectors.
Considering how this thread has evolved I tought I would add at least one outside Phd, Physicists view of the flywheel clock issue. ******************** e-mail Excerpt ********************** Dr Dowdye is a Phd, Physicist currently working at NASA. Here is his Bio. http://extinctionshift.com/author.htm Dr Dowdye's replies are in red. At 11:07 PM 2/6/2006, you wrote: Dr Dowdye, If you can, without getting to involved, either give a brief explanation or point me to a source which would explain the correlation (or affect) on a flywheel as viewed in SRT due to relative velocity between such rotating devices. Relativity does NOT treat the rotating frame at all!!! The stationary, rotator has an angular velocity that is, since it is stationary, is not considered to be shifted or so-called time-dilated. IN YOUR THEORY it would be interesting to see if one could put into space a spinning disc of high angular velocity where it is not interfered with, to see if it could serve as an accurate time reference. There would be NO ELECTRONICS in it. Nothing to be perturbed at all. Should be more constant than a time keeping device on earth. What do you think? I don't know of any that address the idea of using the Rpm as a clock standard. Nor do I see any direct relative velocity affect on the Rpm. The only affect I can envision would be to declare time dilation which would mean the apparent Rpm would increase? I don't think there would be any time shifts at all. EDowdye Dan
Hi everyone, while on a business trip I decided to draw out a spacetime diagram showing the Doppler effect. Here it is (click on the thumbnail to enlarge): Please Register or Log in to view the hidden image! This diagram is drawn in the embankment (unprimed) frame, which is represented by the black lines. The train (primed) frame, which is represented by the white lines, is also superimposed. The units of space and time are such that c = 1. The velocity of the train in this example is .6 c which corresponds to a gamma of 1.25. The blue and red lines represent the peaks of a steady light with unit proper frequency (one cycle per unit of proper time) broadcast in both directions from the train at x'=0. Due to the time dilation factor of 1.25, the blue and red lines leave the x'=0 line every t=1.25 which corresponds to every t'=1. Note that the blue lines cross the x=2 line every t=.5 which represents a blue shift for an observer stationary in the embankment frame on x=2 (until the train passes x=2). The red lines cross the x=-2 line every t=2, which represents a red shift for an observer stationary in the embankment frame on x=-2. In contrast, the blue lines cross the x'=2 line every t=1.25 (t'=1) which represents no Doppler shift. Finally, the red lines cross the x'=-2 line every t=1.25 (t'=1) which also represents no Doppler shift. The bottom line, once again, is that neither frame observes a Doppler shift between the flash and the detectors that are stationary in the train frame (e.g. at x'=±2). An embankment observer would think that the filters would stop the light only if the observer were egocentric or ignorant. -Dale
1 - For whatever reason your diagram does not show. 2 - I have read your presentation and frankly I am a bit miffed. It seems that anyone that doesn't realize that the light flash as observed by the embankment would be blue shifted as it approached but red shifted after it passed would be egocentric or ignorant. Unfortuantely that is not at issue. 3 - The issue is that in the embankment frame the light flash has a velocity of "c" and the train (according to your presentation) has a velocity of "0.6c". Omitting velocity addition for convience since it doesn't alter the arguement but complicates the presentation, the relative velocity between the light flash and the engine detector is Vr = c - v. That fact represents a doppler red shifted signal to the detector. PERIOD. I have already commented on the fact that in the train frame where the source and detectors are at relative (inertial) rest there is no doppler shift. Either try again or give a better description of the relationship between the light flash and the engine detector in the embankment frame. The above referring to approaching and receeding doppler of the light flash is misleading.
Does not show what? Do you mean it is too small? You should be able to click on the thumbnail and see the full version. If you have Internet Explorer's pop-up blocker enabled then you may need to hold down the Ctrl key while you click on the thumbnail. In any case, here it is as URL: http://img277.imageshack.us/my.php?image=lorentz5jj.png Correct, as I pointed out in my text and highlighted with the blue and red colors of the lines. Although technically they would just be ignorant if they didn't understand that. They would be egocentric if they thought that everyone else must also be seeing blue or red shifted light. That fact does not represent a Doppler red-shifted signal! EXCLAMATION POINT. Look at the diagram and count how often the light that leaves x'=0 hits x'=±2, it's exactly the same frequency. I know that math is not your strong point, but I had hoped you might be able to follow simple geometry. You do remember from geometry that opposite sides of a parallelogram are equal, right? QUESTION MARK. And we are all so proud of you for getting something right. OK, this is a reasonable request. I was envisioning a scenario where the emitter is at x'=0 the caboose detector is at x'=-2 and the engine detector is at x'=2. As I already mentioned the whole train is moving at .6c (gamma = 1.25) and c=1 in order to make everything easy to do geometrically, but it is just a matter of convenience and other numbers could certainly be used. The red and blue lines represent light leaving the emitter. They are drawn leaving from x'=0 at unit intervals of proper time representing the peak of a steady unit frequency radiation. The time between two peaks is the period and the distance between two peaks is the wavelength. Periods and wavelengths can therefore be calculated simply by counting the number of intersections, and any Doppler effects will show up clearly. In other words, counting the number of intersections on x'=0 shows the frequency of emission. Counting the number of intersections for x=2 shows a blue-shifted frequency, but the detector is not at x=2. It is at x'=2 which has no Doppler shift as demonstrated geometrically in the embankment frame. Counting the number of intersections for x=-2 shows a red-shifted frequency, but the detector is not at x=-2. It is at x'=-2 which has no Doppler shift as demonstrated geometrically in the embankment frame. -Dale
DaleSspam, OK, I can see your sketch in the link provided but NO thumbnail or link appears in your first post. I will be trying to follow your graph but frankly I find it absolutely unimaginable that c - v is not a red shifted signal and c - v is the absolute relative velocity between the light flash and the engine detector in the embankment frame. Do you dispute the relative velocity of the embankment frame?
Certainly opposite sides of a parallelogram are equal but unfortunately your graph is far from obvious. If the emitter and detectors are at t'=0 and t'=+/-2 in the embankment frame as stated, where are they in the train frame? t=0 and t=+/-2? Does the spacing difference therein represent length contraction? Graphically I measure a ratio of 1.27/1 (probably close enough). t'=0 intersects t=0 but t'=4 intersects t=2? That suggests a gamma=2 or 0.866c. You need more data showing how these angles are computed to represent the functions.
Just an observation. Dale's thumbnail appears on my screen. Clicking on the thumbnail takes me to the link. I have IE and a cable modem, pop-up blockers, firewall, spam blocker, etc. Must be something to do with your browser, Mac, but I don't know enough about software to guess what it could be.
Thanks, I have other ongoing problems with this SOB. i.e. - I get this Physics section but cannot get the Home page or other sections of SciF, no sound, no antimations, etc.
Sorry about the resulting confusion then. I didn't realize that some systems wouldn't display thumbnails. From now on I will post graphics as a thumbnail and as a URL. Have you not been able to see previous graphics that I have posted? If not, then that might explain why you seemed to ignore some of my best derivations. -Dale
Of course not, the relative velocity is clearly demonstrated in the graph. What is also demonstrated is that the relative velocity leads to no Doppler shift between the flash (x'=0) and the detector (x'=2). Your premise about the relative velocity is correct, your conclusion about the Doppler shift is incorrect. I know you don't like to hear it, but this is exactly why it is important to do math in physics and not just verbal logic. -Dale
That is correct but it doesn't seem to be a system problem. I've lost something in my setup because all these things quit at once. Camera, microphone, sound, some graphics and links, etc.
I am not quite sure what you are asking here. The emitter is located at x'=0, not t'=0. Similarly for the detectors. As far as the geometry in the train frame. The usual procedure is to establish which frame is the primed and which is the unprimed frame and then maintain that convention even when discussing things in the primed frame. So in the train frame I would still say that the emitter and detectors are at x'=0 and x'=±2. The only difference is that in the train frame the x' and t' lines would form squares instead of rhombuses. (rhombi?) In any case, the point of contention is the embankment frame, not the train frame. That is why I drew the diagram (http://img277.imageshack.us/my.php?image=lorentz5jj.png) in the embankment frame and I would like to limit the discussion to the embankment frame. That is correct. γ for v=.6c is 1.25 (1.27 is close enough for me). Actually, any t line intersects every t' line at some point in space (t=2 intersects t'=4, t'=2, and t'=0 on this small section alone). By looking at the point where t'=0 intersects t=0 and the point where t'=4 intersects t=2 you are considering a "clock" that is moving in both frames. What you want to do is consider clocks that are stationary in one frame, for example one located at x'=0. As you move from the origin along the x'=0 line you cross the t=2 line before you cross the t'=2 line. In fact, where the x'=0 line crosses the t'=2 line it is also crossing the (un-drawn) t=2.5 line. t/t'=2.5/2=1.25 gives the correct γ. Thus, the embankment frame sees that a clock moving at .6c runs slow by a factor of 1.25 Ok, this is another reasonable request. These lines were calculated from the Lorentz transform: t'=γ(t-vx/c²Please Register or Log in to view the hidden image! and x'=γ(x-vt). Since c=1 and v=.6=3/5 then γ=1.25=5/4, the slope of the x' lines is 5/3, and the slope of the t' lines is 3/5. For example, for the x'=0 line we have by the Lorentz transform: x'=γ(x-vt) 0=5/4(x-3/5t) t=5/3x which is what you would expect with the frame velocity of 3/5 c. The other lines are calculated similarly; I won't go through the details for each one, but please feel free to check my work if you wish. The light lines, of course, have a slope of ±1 since c=1, but when and where should each peak line start? If the first pair starts at x'=0 and t'=0 (x=0 and t=0) then the second pair starts at x'=0 and t'=1. That gives: 0=5/4(x-3/5t) 1=5/4(t-3/5x) Solving simultaneously for x and t yields x=3/4 and t=5/4. The other starting points are calculated similarly; again I won't go through the details for each one, but feel free to check. I hope that helps clear up the derivation sufficiently. For convenience I also put in "past" blue-shifted lines. The alternative was to make the graphic much bigger in the vertical direction. -Dale
In that case I sincerely apologize for my earlier uncivil comments on other threads about you ignoring my various derivations. I hope you can find the problem in your setup and I will be sure to include the URL in all future posts since that seems to work reliably. -Dale
