studiot

I thought about this simpler version of just saying that EM radiation is a form of energy, and swans cross posted it whilst I was thinking.

Yes EM waves have energy. In the widest sense (of PE) it is a form of two potential energies which are periodically interchanging- hence the frequency of vibration. Potential energy can be taken in the narrow mechanical energy sense, which you are probably thinking of, or the wider sense of energy stored in a field such as (in this case a magnetic field and an electric field).

Yes I agree that the OP restricted his discussion to light which enters the eye. Since that is the only light he discusses, that must be the light he states quite clearly, although you did not underline it, that we never see. So the issue arise of does the light that enters the eye and forms some sort of image on the retinal constitute seeing or was he intending the creation of a mental model to be included. He was not clear and may even have meant something else entirely that neither you nor I have thought of. I have already demonstrated the some complications with the mental model interpretation the most important being the colour blindness tests where the hidden symbol is there and seen at the retinal stage but invisible in the mental model. As a matter of interest, yes this is an old thread and the correct proceedure would have been for the hijacker to have started a new thread, perhaps stating that it was inspired by this one. By coincidence I was asked another technical question in another old thread last night and I am collecting information to do just this and start a new thread in answer.

I really don't understand how you can refute what I have been saying consistently throughout this thread which is in short: It's complicated. Whilst at the same time agreeing with Swansont ( C "Exactly") and introducing see1, see2 and now see3 (A). Let me remind you of the OP. I have maintained consistently that the word see has multiple meanings and that the OP has not introduced sufficiently clarity or limitation. That's your (A) and (C), As to your (B) I do not see the tube because I added the context that I am not looking at it, but I can still see the light when standing under the streetlamp and see that it is yellow. That would be two of your different 'sees'. As far as I can make out (see) the process of seeing depends upon a chain of events and the wrord see is used to indicate both the process reaching stage partway through the chain and the stage at the very end. The process is further complicated by the fact that additional factors may be introduced at several of the stages along the way, including between the stage of formation of a retinal image and the final stage. Throughout this thread I have attempted to draw out the complications arising from both the multiple inputs and the multiple meanings. Your see1 etc was a valiant effort to try to bring some order to this jumbled up situation. But the bottom line is that unless the person using the word somehow indicates what's in and what's out, confusion will reign.

A bit of algebra, especially useful in exams. [math]1 - {\left( {\frac{5}{6}} \right)^{10}} = 1 - \frac{{{{\left( 5 \right)}^{10}}}}{{{{\left( 6 \right)}^{10}}}} = \frac{{{{\left( 6 \right)}^{10}} - {{\left( 5 \right)}^{10}}}}{{{{\left( 6 \right)}^{10}}}} = \frac{{60466176 - 9765625}}{{60466176}} = 0.84[/math]

Just because you can't see it does not mean there is no problem. If the OP premise that light itself cannot be seen is true, then how can you see the yellow light? How do you know there is yellow light and that it is yellow, if you can't see it? Of course if you think the OP premise is untrue then that's fine, but what makes your version any better than any one else's?

I think you still haven't appreciated the full import of John's question. The title and OP of this thread and the thesis of the suspended member is that you cannot see light itself. In particular furyan asserted that you can only see the source. John's point is: If a blue source is emitting yellow light how can you see the yellow light itself if you can't see light?

+1 Two good replies.

John asked a very good question about sodium, although this situation also applies to other materials. Sodium is a greyish/white/silver metal as a solid, but in the gaseous phase it is blue if illuminated with ordinary light. Under these conditions it does not give off any yellow light. If the blue gas is then stimulated with electricity it then also gives of yellow light. We no longer see the blue (although it must still be there) only the yellow. But the sodium gas itself remains blue. So we are seeing yellow light emitted by blue objects as a result of some electrical process.

Neat +1

Thank you John for telling me something new I didn't know +1 So I investigated. Eise, you didn't address John's very good comment. But you clearly agree that 'see' has multiple meanings. As a result I contend that it is encumbent upon the promoter to establish which one is to be used either by sufficient context or by a specific definiton at the outset.

Neither, there are other possibilities. Mordred is fond of saying that space is nothing more than volume for things to exist in. In similar manner time may be considered as nothing more than the arena for change to exist in.

Because it' conforms to both the rules of this forum and good practice and encourages helpful discussion.

Yes it's a very basic start, but Raider clearly doesn't care about the subject any more. 30lbs lifting is a very small amount for an electromagnet. They were lifting tons, well over 50 years ago.

The first use of Shannon in this thread occurs in my post here: I said that you should lay out your stall properly and warned others that there is more than one definition of the crucial word 'entropy' I think you are saying that there is experimental evidence that consciousness is emergent from the complexity of the human neurological system, but I am not sure enough to comment. The simplest emergent phenomenon (non sentient) I can think of is the action of the arch.

I don't feel anything about the use of Shannon Entropy (here) one way or the other, and nothing I have posted could remotely lead to that conclusion. I have not read your papers, since you have not introduced sufficient summary to understand what is going on, in contravention of the rules of this forum. Can you not provide a succinct summary of your point which I have yet to divine?

Since you wished to preach about the subject, surely you understand what entropy is and the difference between Shannon Entropy, Classical Thermodynamic Entropy and Statistical Mechanical Entropy?

I you wish to mention my post, then address my words. Don't quote another responder at me.

I've noticed my tex stuff is overcramped when displayed, these days so I'm finding out how to make it bigger. [math]{P_{n = 3}} = \left\{ {1 - {{\left( {\frac{5}{6}} \right)}^3}} \right\} = \left\{ {1 - \left( {\frac{{125}}{{216}}} \right)} \right\} = \frac{{91}}{{216}}[/math] It comes out bigger on my TEX editor, but is cut back here Any suggestions?

You mean he's off he premises?

Beware there are some incorrect formulae rattling around the net. This is the time to introduce a couple of new concepts. Firstly the stopping or termination criterion. I assume you will roll the die until you get a 1 and then stop (after n throws) Getting a 1 is the stopping criterion. This is equivalent to getting at least one 1 in n throws. Which brings us to the second concept. The probability of something happening and not happening equals 1 So if the probability of getting a 1 in one throw is 1/6 then the probability of not getting a 1 is (1 - 1/6) = 5/6 I am not going to produce a formal proof of the formula (you can have it if you really want it) I am just going to show that it works. The probability of getting a 1 in n throws is [math]{P_{1in\,n}} = \left\{ {1 - {{\left( {{P_{not\,n}}} \right)}^n}} \right\} = \left\{ {1 - {{\left( {\frac{5}{6}} \right)}^n}} \right\}[/math] So for n = 0 (no throws) the simplest possible there is zero probability of getting a 1 [math]{P_{n = 0}} = \left\{ {1 - {{\left( {\frac{5}{6}} \right)}^0}} \right\} = \left\{ {1 - 1} \right\} = 0[/math] For n = 1 we have what we already know that the probability is 1/6 [math]{P_{n = 1}} = \left\{ {1 - {{\left( {\frac{5}{6}} \right)}^1}} \right\} = \left\{ {1 - \left( {\frac{5}{6}} \right)} \right\} = \frac{1}{6}[/math] For n = 2 we have what we worked out above that the probability is 11/36 [math]{P_{n = 2}} = \left\{ {1 - {{\left( {\frac{5}{6}} \right)}^2}} \right\} = \left\{ {1 - \left( {\frac{{25}}{{36}}} \right)} \right\} = \frac{{11}}{{36}}[/math] and for n = 3 we have [math]{P_{n = 3}} = \left\{ {1 - {{\left( {\frac{5}{6}} \right)}^3}} \right\} = \left\{ {1 - \left( {\frac{{125}}{{216}}} \right)} \right\} = \frac{{91}}{{216}}[/math] I will leave you to do some work now for n = 10.

Note I have been careful to indicate my frame of reference (point of view) for the rotation. Your discussion elsewhere may be being conducted from diferent points of view.

How have we wandered so far off topic?

It is a great shame that two different branches of Science should employ the same word for quite different meanings, just because the numerical pat has the same mathematical form. Readers should be aware of which type of entropy is intended.

No problem. The easy way to approach this is to realise that rolling one die twice or two dice together is the same in probability theory. Please note the singular and plural of the word die. So it we roll two dice together and list all the possibilities we can see that they are 1 plus 1 , 2 , 3 , 4 , 5 or 6 that is 6 possibilities 2 plus 1 , 2 , 3 , 4 , 5 or 6 that is 6 possibilities 3 plus 1 , 2 , 3 , 4 , 5 or 6 that is 6 possibilities 4 plus 1 , 2 , 3 , 4 , 5 or 6 that is 6 possibilities 5 plus 1 , 2 , 3 , 4 , 5 or 6 that is 6 possibilities 6 plus 1 , 2 , 3 , 4 , 5 or 6 that is 6 possibilities Adding these up we see that there are a grand total of 36 or 6x6 possibilities (or outcomes) though not all include a 1. Now this is where we need to become more precise in our wording. Because there are two situations where a 1 is rolled. The roll can produce at least one 1 (two 1s in this scenario) or The roll can produce exactly one 1 In the first of the above lines each of the six possibilities produces at least one 1, making 6 ways to get a 1 In the last five of the above lines one of the 6 possibilities is produces at least one 1, adding a further 5 ways to get at least one 1 So there are a total of (6 + 5) = 11 ways to get at least one 1 out of a grand total of 36 possibilities. So the probability of getting at least on 1 is 11/36. In the first of the above lines, five of the six ways will produce exactly one 1 and the sixth will produce two 1s, making 5 ways to get exactly one 1. In the last five of the above lines one of the 6 possibilities is produces exactly one 1; there are no ways to produce more than one 1. So this adds a further 5 ways to get exactly one 1. So there are a total of (5 + 5) = 10 ways to get exactly one 1 out of a grand total of 36 possibilities. So the probability of getting exactly one 1 is 10/36 This analysis is meant to help understanding and provide something to fall back on. You can always use formula, which is quicker, once you have the understanding. Does this help?