MacM, I thinkl I see where there is divergence here. Are you saying that because you are stationary you will see the light blue shited and beacuse the train is moving the the detector will see the light red shifted relative to yourself, the stationary obvserver? In other words, the train detectors will see the light in the same frequency it was emitted, but relative to the staionary observer the light will be red shifted? Geistkiesel?
MacM, I think I see a problem here in terpretation. You are correct that relative to the stationary observer the light will appear red shifted as it approaches the out going train. This is because the stationary observer sees the light as blue shifted. The train detector will, however, see the light with the same frequency that is was emitted. This is so as the detector and source are moving at the same velocity. Geistkiesel
If you want to believe in magic then I guess that is your perogative. Please Register or Log in to view the hidden image! Vr = c + v (caboose) or Vr = c - v (engine) does not cancel out. Period. The only frame where doppler does not come into play is in the train frame itself.
That is correct. In the embankment frame you have two velocities. Light vl = c and train vt such that the relative velocity of the caboose is Vr = c + vt and for the engine Vr = c - vt. The only place doppler shift does not occur is in the train frame itself.
Correct because the detector functions in its frame which is the train. That is the point of this thread. The embankment frame is ineffective at participation in the event. The event is in the rest frame. The filters effective rule out detonation from the embankment frame. If the frame jphysics can be seperated then they are not part of the same event. This is geared to support my view that photons are generated,or observed, as an energy function and an observers relative velocity to a light source is causing the observer(s) to measure different photons. The invariance is an illusion by confusing what photons are being measured and is not by some magic of relativity so that a specific photon has the same velocity to multitudes of observers at different velocities simultaneously.
You don't? :bugeye: Tsk, mathematicians have known about it for centuries! It's called - elegance Please Register or Log in to view the hidden image! Of course not, but that's not the formula of interest. Relativistic Doppler shift is what you want here, plugging in the relative velocities v and -v respectively. That does cancel. Either way gives the same result.
MacM: Let's get this straight. Suppose you're standing on the track, watching the train move away with velocity v to the left, while light moves towards you with velocity c directed to the right. You say this means that the speed of light relative to the train is therefore v+c. But you're wrong. The phrase "speed of light relative to the train" means the speed of light as seen by somebody in the rest frame of the train, which is c, not c+v. In the track frame, there is a difference in speed between the light and the train of v+c, but that speed is not the speed of light relative to the train; it is a combination of the speed of light and the speed of the train relative to you, standing on the track. You're still not over your problem with reference frames, are you? You could have saved yourself years if you devoted an afternoon to understanding the concept, you know. Why don't you get yourself a good physics textbook and read it? It's telling that you level this charge at anybody who disagrees with you. Your attempt to redefine "physics" to mean "whatever MacM says" is juvenile. Yes. As I said above, "relative" is the wrong word to use here. You are using terms in a way which makes no sense. No "relative velocity" can be "the same as relativity of simultaneity", for the simple reason that relativity of simultaneity is a statement about times, not velocities. Done. I'm still correct.
Nope. Not so. The blue shift is to the caboose detector and the red shift is to the engine detector. Neither cancel :bugeye: You cannot take c+v and say it cancels the c-v. They are at opposite ends of the train. :bugeye:
Correction. Suppose you keep the scenario as it was in which case the train would be moving toward you as is the flash of light. To you the light is blue shifted but it is chasing the engine and has a relative velocity to the engine of c-v. That is red shifted to the detector. What a load of crap. Now you want to make a disclaimer for simultaneity which is based on the fact that light is chasing the engine and the caboose is moving toward the flash. Give it a break. You are starting to look really foolish here. It seems I am not the one with the problem here. Why don't you get off your high horse and stop pretending to be God All Mighty of physics. You clearly are not or at least you are ruining your record by playing stupid. Certainly what else have we learned to expect from you. Please Register or Log in to view the hidden image!
MacM: I was quite clear in making my point, but I can explain it again for you if you can gather together enough wit to express your own scenario clearly. Ok. Let me try to understand you. Is this right? I am standing on the tracks, with the train coming towards me. Light is emitted from the centre of the train and travels backwards towards the rear of the train. Relative to the train (engine, caboose, whatever), the light is travelling at c. Relative to me, standing on the tracks, the light is still travelling at c, since the speed of light is the same in all frames. However, relative to me, the light is travelling away from me at c, while the train is coming towards me at speed v. Therefore, the difference in the velocity of the light and the train, as measured by me, standing on the tracks, is c+v. Do you agree? Yes or no? Do you understand? Or do you want to change the scenario to something else, which you might be able to understand? I didn't mention simultaneity in my last post. Try again. You're just continuing. So, nyah! (Want to continue being childish?) You are in no position to judge such things.
Pardon me while I get personal. Screw you. Don't come on here and pretend to teach. The c+v is precisely what I wrote in the outset of this thread.
Not the formula of interest. Quantities are v and -v. Why? Observer/re-emitter is moving towards lightsource at v and away from engine detector at v. (Or lightsource is moving towards observer and detector is moving away from observer. Same thing.) We're only looking at the front half of the train. If you don't trust the Relativistic Doppler formula, it's fairly easy to derive from scratch (similar to normal Doppler but taking into account time dilation); but if you accept it there's no need to consider c+v etc.
Just as a side thought, please consider your experiment if you were to use sound pulses, with the air stationary w.r.t. the embankment: So long as v were below the speed of sound, why would the detector always hear the same frequency? Because, looking at the caboose side, the sound leaves a source which is running away from it, and 'red' shifts, but hits a detector that's moving towards it and 'blue' shifts back to its original frequency. Vice versa for the other side. While there are many differences between a classical and relativistic situation, what makes you think that something as basic as symmetry doesn't transfer? Why do you only consider the fact that the engine pulse chases the engine detector (according to the embankment frame), but not the fact that the light source chases the engine pulse?
Sorry but this statement is still absolutely correct.: "You cannot take c+v and say it cancels the c-v. They are at opposite ends of the train. " Nothing you have posted alters that.
It doesn't nor is light constant in color (frequency) That is the point. So what is your point.? Incorrect. The relationship of the light to the source has no impact on the relationship of the light to the detector. The light at the detector is and remains doppler shifted. What??? :bugeye: You only have the light flash and the engine detector. What is your third component?
The light source, aka the flashing thing, aka where do those light pulse things come from anyway? Where-ever it is, call this A. Now in the embankment frame, consider the velocity of both the engine-headed and the caboose-headed light pulses, w.r.t the light source, aka A. Forget the back end of the train for now. In the front half, what's of interest is the v with which A is moving towards an observer between A and the engine, and the v with which the engine is moving away from the observer. Or in 'towards' terms, v and -v.
But my point is that it does. Ask any classical physicist. So long as source and receiver are not moving relative to each other - even if they move relative to the air, so long as it's at constant speed lower than the speed of sound - although it will affect the speed at which the sound gets there, it doesn't change the frequency the receiver finally 'hears'.
You are mixing frames. The train frame is where the source and detector co-move (or are at relative rest to each other) and the light flash has a velocity of v = c to both detectors. It is the embankment frame which is being considered. In that frame the light still moves at v = c in both directions. The caboose is moving toward the flash or a relative velocity of c + v exists and the engine is moving away from the flash at v such that the relative velocity is c - v. PERIOD. The caboose dtector and the engine detector do not canel each other they both are doppler shifted and with a neutral optical filter the light flash will not be effective at detonation.
No, in my first post I was talking about the fact that Light is moving away from the source (A) at c. The source is moving towards the engine-bound light flash at v. Ignore v+c and the back half of the carriage. The only parts I'm talking about are (A) light source (B) light itself, heading from (A) towards (C) (C) light destination, aka engine-side detector. (My second post was talking about a similar situation using sound instead of light, to show that in classical physics there is still no nett shift. It was a reply to your reply to my previous post about a classical situation, so I thought you'd remember Please Register or Log in to view the hidden image! maybe I should've started a new thread about that instead...)
