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Because of the equivalence principle, you cannot identify with an experiment your state of motion. That means, if the target moves together with the train, as well as the emitter, then everything what happens in the train does not allow to measure the velocity of the train. If the light ray would hit the target at rest, but not hit the target if the whole construction is moving, one would be able to find out if the train moves: try to hit the target with the light ray, if it is missed, the train is moving.
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05142016, 07:12 AM
(This post was last modified: 05142016, 09:29 AM by Schmelzer.)
ALT you misunderstood most of my post. I'll assume  this time  not deliberately.
I asked:
1. Do you believe SR says that the photon is affected by the transverse motion?
2. Do you believe SR is wrong?
#2 meant, of course, wrong about this particular point! I'll spell it out:
2. Do you believe SR is wrong about the photon being affected by the transverse motion?
I suppose your answer is "yes".
ALT: In your last paragraph you asked me 'why waste your time' ? Well, I don't think I am. I find this a fascinating field of enquiry, and great grey matter exercise. Better than doing crosswords, anyway :)
Of course I meant why waste your time talking to me, since we've already said what there is to say.
ALT: Evidence ? Experiment ? I have none.
I told you to find evidence on the net, not to generate it yourself!
[I have deleted here something which could be considered as becoming personal. I will do this very early, so that this is not even a reproach that you have written something inappropriate, more that it could be a starting point, which, after some possible escalation, could lead to something too personal later.]
Now, you answered a question I didn't mean to ask, about SR being wrong in general. You say it probably is, being provisional; someday it will be replaced by something more accurate. That's true. In fact I wouldn't be surprised if it's wrong already, in certain ways; that's what the ether theory (see that subforum) is about.
You mention Einstein's "enthronement": you're preaching to the choir, I couldn't agree more.
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In a certain sense, SR is already known to be wrong: It does not describe gravitational effects. That's why GR is used, and SR serves only as an approximation for situations where gravity does not matter. So, no ether necessary to show SR is wrong, this has been already done by Einstein himself.
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05142016, 12:50 PM
(This post was last modified: 05142016, 01:09 PM by ALT.)
Above, secur, you said;
ALT you misunderstood most of my post. I'll assume  this time  not deliberately.
I'm glad you so assumed. In fact, to say one misunderstands deliberately is rather a tautology .. or a contradiction .. or something like that. You see, if one 'misunderstands', implicit in that word is the fact that one really did not understand. A very different thing from one understanding but then being deceptive about it. And I'm sure you didn't mean to call me deceptive.
I will clarify. If it is clear to you that I have misunderstood something then please assume always that I have misunderstood it. The alternative being, that I am being deceptive. And if there is any suspicion in your mind that I AM being deceptive, ask yourself  how far would a (not even) layman get, by being deceptive to a group of professionals in any field ? I ain't THAT dumb :)
As to the red text in the above post, I'm not too sure what it's about or who made it.
No matter. I will respond to the above two posts (absent of personal assumptions or remarks) soon and will try very hard to stick to the OP.
(05122016, 05:11 PM)Schmelzer Wrote: Let's create the photon with a direction in the following way: Some explosion which creates light at the center of a sphere in all directions, and a small hole in the sphere. With this construction, it obviously matters if the sphere moves or not. Because the hole has moved some distance, together with the whole sphere (together with the train). You can invent something else which directs the light into some direction, but I see no reason to expect anything different. Anyway, we have the very simple equivalence principle argument, the one from the begin of my first answer, it will work also for quite complex constructions.
Schmelzer, having thought about this a little, I would say that the photon that gets through the hole, behaves like every other photon that doesn't. It seems to me that the sphere is an unnecessary addition. Let's remove the sphere for a moment. What do we have ? Something akin to a supernova maybe, whose light reaches us at c, independent of it's motion or ours.
So, putting the sphere back, we restrict all photons except one .. which behaves like all the others. If it is moving in a direction away from our line of sight, we would see nothing. If it WAS in our line of sight (or our photon detectors) we would say it was ravelling at c, independent of it's emitters motion or ours.
I'm not sure what this proves or disproves. But I will respond to your above (and later) comment about the equivalence principle soon.
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05142016, 01:28 PM
(This post was last modified: 05142016, 01:33 PM by ALT.)
(05142016, 06:48 AM)Schmelzer Wrote: Because of the equivalence principle, you cannot identify with an experiment your state of motion. That means, if the target moves together with the train, as well as the emitter, then everything what happens in the train does not allow to measure the velocity of the train. If the light ray would hit the target at rest, but not hit the target if the whole construction is moving, one would be able to find out if the train moves: try to hit the target with the light ray, if it is missed, the train is moving.
Thanks. I now understand what you meant. So it seems then, that I am also questioning (note, questioning) the equivalence principle.
It all hinges around whether the photon in my thought experiment is influenced by the transverse motion of the emitter, doesn't it ? It all comes back to that.
Note, in an earlier post, I said ..
Does it (get pushed sideways)? I don't know. This is the crux of my enquiry. I think as light travels at c independent of the speed of the source from the instant it is emitted and for it's history thereafter, the instant my photon is emitted, it will indeed travel at c in the direction it WAS emitted, independent of the motion of the emitter.
Put it this way  if my emitter had rectilinear motion at 1/2 c, we would not add this to c and say the photon is travelling at 150% c. So why then, add the transverse motion ? Why does it get pushed sideways but not forward ? After all, there would be much more forward push than sideways push in this case.
Does a photon leave the laser tip instantaneously and at full c and therefore unaffected by the emitters motion, or does it take some time to reach c (no matter how minute) and therefore defeat the c constant itself.
Have you given any thought to all that ? I would really appreciate your views.
Secur, re your post #22, I think I've covered your queries in the above posts. To avoid verbosity I will leave it there, but if you think I've missed something, please specifically point it out.
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05142016, 08:37 PM
(This post was last modified: 05142016, 08:39 PM by Schmelzer.)
(05142016, 01:28 PM)ALT Wrote: Put it this way  if my emitter had rectilinear motion at 1/2 c, we would not add this to c and say the photon is travelling at 150% c. So why then, add the transverse motion ? Why does it get pushed sideways but not forward ? After all, there would be much more forward push than sideways push in this case.
Does a photon leave the laser tip instantaneously and at full c and therefore unaffected by the emitters motion, or does it take some time to reach c (no matter how minute) and therefore defeat the c constant itself.
If the emitter would have a rectilinear motion \(v=\frac12 c\), we would nonetheless add this to c. Of course, as always, vector addition, thus, we have to add (0,c,0) and (v,0,0). Even this would not give 150% c, but (v,c,0), which for \(v=\frac12 c\) gives only \(\frac{\sqrt{5}}{2}c\) as the speed. And, of course, we would have to apply the relativistic formula for velocity addition, so that we get only \((v,c\sqrt{1v^2/c^2},0)\).
The same relativistic formula we would apply also to motion in the same direction. The general formula (see here) gives in the case for u and v in the same direction the result \[\frac{u+v}{1+\frac{uv}{c^2}},\] which, for \(u=c\) gives simply c.
We have a formula for velocity addition, and we apply this formula always. Nothing special if one of the velocities is c. So, the very question if the photon has always c (it has) or not does not change anything. The same formula for velocity addition would have to be applied anyway.
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(05142016, 08:37 PM)Schmelzer Wrote: (05142016, 01:28 PM)ALT Wrote: Put it this way  if my emitter had rectilinear motion at 1/2 c, we would not add this to c and say the photon is travelling at 150% c. So why then, add the transverse motion ? Why does it get pushed sideways but not forward ? After all, there would be much more forward push than sideways push in this case.
Does a photon leave the laser tip instantaneously and at full c and therefore unaffected by the emitters motion, or does it take some time to reach c (no matter how minute) and therefore defeat the c constant itself.
If the emitter would have a rectilinear motion \(v=\frac12 c\), we would nonetheless add this to c. Of course, as always, vector addition, thus, we have to add (0,c,0) and (v,0,0). Even this would not give 150% c, but (v,c,0), which for \(v=\frac12 c\) gives only \(\frac{\sqrt{5}}{2}c\) as the speed. And, of course, we would have to apply the relativistic formula for velocity addition, so that we get only \((v,c\sqrt{1v^2/c^2},0)\).
The same relativistic formula we would apply also to motion in the same direction. The general formula (see here) gives in the case for u and v in the same direction the result \[\frac{u+v}{1+\frac{uv}{c^2}},\] which, for \(u=c\) gives simply c.
We have a formula for velocity addition, and we apply this formula always. Nothing special if one of the velocities is c. So, the very question if the photon has always c (it has) or not does not change anything. The same formula for velocity addition would have to be applied anyway.
Thanks. The math is totally beyond me but it prompted me to search and I found this site which 'splains it rather well for one in my paygrade.
http://www.pitt.edu/~jdnorton/teaching/H...index.html
I am quite pleased to have found that site. You can see it reduces things nicely for complete dummies. I will study it a while and come back to this post.
One thing I would like to ask at this point  which I have before, but don't seem to have received any clarity.
At the instant of leaving the laser tip, does the photon do so at c or does it take some time to get to c ?
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Yes, Thanks.
I still feel as though my question has not been answered in a way so that I might understand.
I will try to sum it up soon.
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So here is where my curiosity lies.
The photon, before it is emitted, doesn't exist. Once it comes into existence (is emitted) it will travel at c instantly – no time taken to get to c, else, the speed of light constant is itself defeated.
So it has instantly left and is instantly independent of the emitter.
Thus it will neither know nor care that the emitter has moved sideways (or any way).
In fact, if I understand correctly, Einstein second postulate hinges on the fact that the photon, once emitted will be independent of the motion (if any) of its source.
So again, WHY does it move sideways with the emitter to hit the target, rather than travel directly vertical to miss it (given the target has moved away) ?
If there is good reason I have not yet seen it, or at least, understood it. What am I missing ? Where is my reasoning wrong ?
