"Light is frame-dependent in PF, but constant in SR"

To me what you said above is so obvious as to be an axiom, I don't need you to explain something that obvious but you continue to think I do. In constantly making the same statements trying to explain what I knew and would agree with, it's clear you don't understand what I'm saying; I know it may not be your fault, it might be I just haven't been able to express my implications clearly enough, or perhaps some people are too quick to make a conclusion without really trying to understand?

That's what I’ve been doing for a while, I learn than calculate the equations for hypothetical scenarios such as the muon/ground frame and many more, in order to understand the equations translation into a physical form where I can more precisely understand their meaning and relationships. I examine those outcomes and my thoughts against evidence such as, atomic clocks being flown on plane and compared to a ground frame clock, how length contraction explains Coulomb law in accelerators, empirical evidence and function of time, curvature and focusing of light around a strong gravitational field, etc.

Because you jump to conclusions and don't try to understand or ask me to clarify what I may of meant, you wind up making posts like this, repeating things I know and I would have told you I agreed with, and making implications which are not true, i.e. "you need to learn the theoretical side to derive the predictions and the experimental side to test the predictions" which I have done and continue to do. We could have legitimately discussed if I yet know the theoretical side well enough, or what experimental evidence I do, or do not ,know of, however you post the same incorrect general repetitive innuendo over and over and almost nothing important is learned or accomplished.

If you look closely the nature of most of my posts they are not about being right, impressing others, or ego (I said most), they are about trying to make constructive progress and being open to analyze constructive criticisms whether talking about my ideas or others.

Maxila

P.S. While I'm new to this forum you have seen my post for many years on physforum so you know my history. After rpenner left it deteriorated beyond what even I could tolerate, and I am patient with a lot of nonsense.

 

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Then you should probably take the time to address the specific issue with your post that Bruce, and I, mentioned. Namely "muon sees 1/5 c". It is this statement of yours that casts the most doubt on your understanding, so why not address it?

You went so far as to erroneously criticize the knowledge of others to come off as "being right", all the while having yet to answer for what you have been criticized for several times now. If you can justify your "muon sees 1/5 c" then do so already. Otherwise you are coming off as evasive, which is definitely not conducive to constructive criticism. The initial criticisms of this statement were constructive, but instead of fostering an open discussion of your reasoning you chose to go on the offensive.

 

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Try again: Are you saying "the speed of light is variant (is a variable) in every frame that is not equivalent to a local coordinate and/or a local proper frame"?

I understand how you can measure the speed of light in a local coordinate or a local proper frame but how do you measure it a remote frame? Is it even possible?

 

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This is Chapter 2, Curving, from Taylor and Wheeler's text Exploring Black Holes.
http://www.eftaylor.com/pub/chapter2.pdf

Page 2-34 Starting where it says: Bookkeeper coordinates r, φ, and t

This is an example of a measurement from remote coordinates [Schwarzschild coordinates].
http://hubblesite.org/newscenter/archive/releases/exotic/black-hole/2001/03/text/

This is a reenactment of the HST observation.
http://imgsrc.hubblesite.org/hu/db/videos/hs-2001-03-a-low_mpeg.mpg

 

Thanks BruceP Is there any explanation what we are looking at?
http://imgsrc.hubblesite.org/hu/db/v...a-low_mpeg.mpg

 

I could not open the third link, apparently a mpg(?). Still, I am at a loss as to how the first two links, address Robittybob1's question, (paraphrazed) "...how do you measure it, the speed of light, in a remote frame?".

Tossing out a reference to 32 out of 44 pages, in the first link, without any clarifying or defining comment, is really no better than saying figure it out for yourself! Similarly, the second link did not seem to, at least directly, or within the context of the information provided, address the question with greater clarity.

So how do you measure the speed of light in a remote frame? Or was there some other point you were attempting to raise?

 

It is really simple, locally the speed of light is c. This allows us to write the solution to the EFE in, say, the Schwarzschild form:

\(ds^2=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2(d \theta^2+sin^2 \theta d \phi^2)\)

Light follows null geodesics \(ds^2=0\) so:

\(0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)-r^2(d \theta^2+sin^2 \theta d \phi^2)\)

Now, if we consider radial motion: \(d \theta=d \phi =0\), we get:

\(0=(1-r_s/r)(cdt)^2-dr^2/(1-r_s/r)\)

The above results into:

\(\frac{dr}{dt}=c(1-r_s/r)\)

In the above \(\frac{dr}{dt}\) represents the "coordinate speed of light". You can see that it has the following properties:

1. it is dependent on the radial coordinate (r), hence the name "coordinate speed of light".
2. It is smaller (or equal ) than c
3. It is equal to c only locally, at infinity.
4. It is zero at the Schwarzschild radius \(r=r_s\).
5. It isn't a measurable quantity, it is a mathematically derived quantity.

For circular orbits, we can do the same exercise but this time we need to set \(dr=d \phi =0\) and , if you do the calculations correctly, you will get:

\(r \frac{d \theta}{dt}=c\sqrt{1-r_s/r}\)

 

The invariant value of c is its ratio of distance per time (x/t), that ratio is constant and preserved in every frame of reference. In the example I posted where the ground saw lights physical change of 10km/33.33µs, (@.98c relative to the ground) the muon frame saw lights physical change of 2km/6.66µs or 1/5 less. Both preserve the constant ratio of c, i.e. as in .3km/µs, however the physical observation of light’s travel distance and time as c=x/t for the muon was 1/5 that of the ground frame; i.e. ground c=10km/33.33, muon c=2km.6.66, both see the constant ratio of c=.3km/µs, however the muon observation was 1/5 that of the ground.

That is why I put "most" in parentheses, specifically meaning greater than 50%. I am not immune to, wanting to be right, impress others, or having an ego; however I am very aware how such traits can thwart constructive progress and I try very hard to be aware of, and suppress them, in pursuit of constructive progress; I am not always successful.

I did not respond to several additional posts where you merely made personal insults and there was nothing constructive to discuss, and I digressed a few times also. The scientific value is to acknowledge them, learn from them, and move forward by trying not to repeat them; anything else is egotistical BS that gets in the way of a constructive discussion.

Maxila

 

O.K. So you can calculate it.., now how do you measure it?

 

Did you notice this:

5. It isn't a measurable quantity, it is a mathematically derived quantity

 

Yes, however my point was that the bulk of your post is calculation... And it is a better clarification that you point out the limiting statement!

Tach, I know you know the difference. It is not as clear that everyone following and involved in the discussion does.

 

Let's assume that l'/t' does not equal l/t, and your velocity is calculated with v = l/t or v = l'/t' in other words the distance traveled is the length measured by an observer. So then find the value of l' and t', for a given velocity. Then with those values find the velocity given by l'/t'. Oh, but then they are not the same velocity that we started with! So then you should plug the new velocity back into the spactime dialation equations, to find their real values and then determine what velocity they observe then. Oh, but then they are not the same as the new velocity that we found! So then we must need to plug this new velocity back into the spacetimes dialation equations. Oh, but then this is not the same velocity that we have found again. So on etc etc, do you see a pattern here?. I will let you plug in the numbers yourself since you claim to be really good at math. There is no reason why the relative velocity should be different for each observer. They can both claim they are at rest and the difference between their velocities is the same. To say otherwise would mean that any velocity determined by l' and t' is not the true relative velocity and leads to an incorrect value for velocity! You could never get a correct value for velocities after using a known velocity in the spacetime dialation equations and then using those values to determine velocity if 'l/t' does not equal l/t. I already showed you how to correctly derive time dialation and length contraction in other threads and showed how that gives you the same velocity in another thread. You claim what mathmatics says means more than anything, but clearly this just went in one ear and out the other. If your finding another relativistic velocity between the same two objects than what you started with, than that means you are doing something wrong. Both observers should always agree an the relative velocity between them, spacetime dialation is dependent on velocity, it cannot create a different velocity. You would end up having two different velocities describing the same situation that cannot happen. A ship doesn't travel at 1m/s and 2m/s relative to the same observer that is at rest, and he can't pick and choose which one he uses because that is what he feels like using and still be a good physicist. One of them has to be the correct value and the other has to be the wrong value!

 

You prove AlphaNumeric's point! It does to appear that you understood what he was saying. I am not even sure you understand what you said, in reply to what you thought he was saying.

Go back and re-read his post, paying special attention to what superficial and flawed mean in context. The simple math as presented is a superficial example of the transforms and the flawed aspect is what it is interpreted to mean comceptually, by many lay readers, relying on wordy descriptions.

And yes, this has been a superficial and to some extent flawed interpretation...

 

It is clear you didn't understand what I was saying. The only difference between a primed frame and a frame that is not prime is that it is the frame that you are in or the one at rest. Every frame is equally valid in saying that his frame is the correct frame or the one that is really at rest. So then if the prime frame is choosen arbitrarily, there would be no way to know what frame had the true velocity if they where not equal. You might as well flip a coin in saying what velocity is the true velocity, it will be just as random as picking what frame to assinged the prime frame too.

I agree that the simple inference of relativity is flawed, and I have given the corrected version of it on these forums that leads to the correct equation of tau. There would be no reason to think that the mathmatics of manipulating an equation would lead to incorrect answers if you started with the correct equation and then did correct manipulations. If you worked backwards from tau then everything you worked backwards to should be correct.

What is superficial and flawed is the statment that the prime velocity would not be the same as the velocity that is not primed. That two observers could measure different difference in velocity between each other. I think this hasn't been known because no one has showed a correct simple derivation of the proper time. But, with that derivation it is clear to see in the mathmatics that anything traveling at a constant speed would measure the same relative velocity between each other, no matter what frame is given the prime value. I didn't work much on showing it for acceleration, but acceleration is different. You can prove what object in what frame is accelerating. But, considering the average velocity of an object that has been accelerating still gives the same average velocity that I also did show. His only comment on this was that the whole thing was for a object in constant motion, but it was actually the average velocity of an accelerating frame.

http://www.sciforums.com/showthread...and-Its-Relation-to-Velocity-and-Acceleration

 

Prof.Layman, read what I actually said. Then read a book on Lorentz transforms. Then read it again. Then work through a bunch of mathematical examples to test your understanding. I hate to break it to you but you DON'T grasp Lorentz transforms or special relativity or anything of this. Did you not learn from last time we 'discussed' things and I demonstrated time and time and time again you don't understand any of this? If you think that your trivial rearranging of Pythagoras' theorem to get \(t' = \gamma t\) amounts to doing Lorentz transforms in a proper way than you're wrong. And it has been explained to you at length. I've also explained it to Maxilla. \(dt' = \gamma dt\) and \(dx' = \gamma dx\) are NOT Lorentz transforms. If they were then they would mean \(-dt^{2} + dx^{2} = -(dt')^{2} + (dx')^{2}\), ie the preservation of the space-time line element which is the DEFINITION of a Lorentz transform. The line element is formed by the generalisation of Pythagoras' theorem to non-Euclidean geometries. It must apply to ALL \(ds^{2} = -dt^{2} + dx^{2}\).

As for me thinking I'm good at maths, I am provably competent at special relativity. Shit, I've taught it at university level. You, on the other hand, clearly don't grasp Lorentz transforms, metrics, space-time intervals and all the rest of the stuff which makes up special relativity. I get it, you have so little experience with real physics you think what you did is non-trivial but reshuffling expressions like that is something I'd be unimpressed by from a 1st year, not the almost PhD level you thought it to be. It is funny you complain that supposedly I didn't listen to you when you didn't listen to me and have now repeated your demonstration you don't grasp special relativity. Special relativity is one of those areas hacks think they grasp because notions like \(E=mc^{2}\) and applying Pythagoras' theorem is just about within their (your) grasp. It is also therefore great at exposing those people who haven't actually done any relativity because if they had they would know that \(E=mc^{2}\) and \(dt' = \gamma dt\) are NOT consistent with Lorentz invariance. Instead, as I've explained at length, it should be \(E^{2} = (mc^{2})^{2} + |\mathbf{p}c|^{2}\) and \(dt' = \gamma (dt - v dx)\) (along with \(dx' = \gamma (dx - v dt)\), where c=1). Those expressions satisfy Lorentz and 4-vector norm invariance.

But hey, let's not let little things by accuracy and facts get in the way of you deluding yourself. Please carry on, it'll give us something to point and laugh at.

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It is a mathematically derived quantity but it's also a measurable quantity. It's a global measurement. The example I included is a measurement confirming the theoretical prediction. Read my post. Specifically "Bookkeeper coordinates r, φ, and t" from the text.

 

It appears the main flaw was how I presented the information, as it was clearly understood by some differently than I Intended. For example if I did imply "two observers could measure different difference in velocity between each other", I had no intention, nor would I knowingly make or imply such a statement. There were other assumptions that were at odds with my intent, so I must assume some fault in my presentation. Rest assured nothing I discussed (as intended) should have disagreed with experimental evidence of, time dilation, the invariant value of c seen in experiments such as Michelson–Morley , length contraction as seen in accelerators being consistent with Coulomb's law. etc.

Maxila

 

I have a better grasp of special relativity than anyone since Einstein. That is how I found the error in the light clock example and then corrected it. It is normally the little mistakes and the common ones that lead to big errors in math and science. So then if you don't mind it having these small common errors, like misassigning a variable, then go ahead and just forget about it. And while we are at it, we don't need a geometrical based relativity, we have gone without one for this long why not. Why even bother with one that leads to the same equations that have been proven by experiment. And who cares that there is actually a correct way to arrive at this equations using a correct method. Scientist have been doing this way for over a hundred years now and saying it is not correct, how could there possibly be a correct way of doing it when they where all so smart.

The shocker here is that the derivation I gave hasn't even been considered yet until I started posting it on the net. If the scientist that came up with the idea that the light clock can't be solved correctly this way didn't even consider that derivation then it really makes you wonder if they where even able to even come up with it themselves. It is really easy to say, oh the observer at rest sees the beam go straight up and down and that is how they measure time. It isn't as easy to see that the observer in constant motion does the same thing. The common sense approuch to assigning the variables that way was wrong. It is a common mistake and it is no surprise that it has popped up in modern physics.

It is simply an incorrect method that leads to incorrect answers, and just because everyone did it incorrectly before doesn't mean that it can't be done right. I know my derivation is correct because the observer on the ship would then use his time to measure a beam that he sees go straight up and down perpendicular to the direction of motion of the ship. So then if you plug in the proper time into the equation d' = c t', then you will get the correct speed of light and the correct distance. The prime distance in that equation wouldn't actually be smaller because from this frame the beam of light is actually seen to travel a shorter distance. That shorter distance is then equal to the distance of the observer at rest. That is why distance only contracts in the direction of motion. The beam isn't seen to travel a greater distance for the observer traveling on the ship in the other directions. In this case, d' = d in that direction. The distance is actually shorter so it doesn't need to be considered as being contracted.

 

How about the first sentence of Prof nonsense reply. That is so 'delusional intellectually dishonest' there should be some rule 'if you go there' you'll find yourself at local coordinates outside the forum back door.

 

Let me know when an outside observer measures your velocity differently and makes the door hit you on your way out.