Higgs Boson Discovered?!

[Perpetual Student]

It's instructive to review this discussion from its beginning (edited for clarity and continuity):

Farsight

Originally Posted by sol invictus
It's a matter of historical record that Einstein near the end of his life was trying to formulate a unified field theory that would explain both gravity and matter. One of his primary goals? Understanding the origin of mass, so as to account for the electron/proton mass ratio (the other particles were just beginning to be discovered in the 20s and 30s when he started on this). Did Einstein's own ideas about the origin of mass also violate E=mc^2, Farsight?

No. His ideas about the origin of mass gave us E=mc². As far as I know he never worked out the proton/electron mass ratio, which is c^½ / 3π with a small binding-energy adjustment:

c^½ = 17314.5158177
3π = 9.424778
c^½ / 3π = 17314.5158177 / 9.424778
r = 1837.12717877
Actual = 1836.15267245

OK, the debate begins:

That's some nice numerology there. Shame one side of the equation has units and the other doesn't, rendering the whole thing absolutely meaningless.

You've got to be kidding. You've heard of "units", right?

The first convoluted response:

Sure. The c^½ / 3π expression sits on top of another expression λ = 4π / n c^1½ metres where n is a dimensionality conversion factor n with a value of 1. It's all to do with harmonics and ratios and spin ½, and everything is based on the motion of light. If you change your definition of c everything else changes too, but the sense of E=mc² and E=p/c still holds. It's the same for these expressions. The thing we call c isn't so much a speed as a conversion factor between our units of distance and time. They're both defined using the motion of light. Everything relates back to the motion of light. Check out the watt balance section of the wikipedia Kilogram article and note the bit that says this: "The Planck constant defines the kilogram in terms of the second and the meter. By fixing the Planck constant, the definition of the kilogram would depend only on the definitions of the second and the meter." The article goes on to say "the definition of the second depends on a single defined physical constant: the ground state hyperfine splitting frequency of the caesium 133 atom". However there's a little flaw in that in that you can't define the second using a frequency, which is cycles per second. Anyway, SI is the kilogram-metre-second system, and will end up being more of a metre-second system where everything relates back to the motion of light. Interesting stuff I think. A bit off topic mind, but I think we've almost exhausted it anyway.

Some dialog:

Measure c in units of feet/second or Smoots/century, and your expression would give a different result for the electron-proton mass ratio. Therefore, it's manifest nonsense.

Huff puff. It would give a different result for E=mc² too. It isn't manifest nonsense, you just don't understand spin ½, or that c^½ and c^1½ equates to c², or that everything hangs off the motion of light. Again, that's new-thread territory.

No, it would not.

It's quite clear that you don't understand units, at all. Units are the first thing you study in an introductory course in any physical science.

The hammers fall:

Let's use natural units then...
c^½ = 1
3π = 9.424778
c^½ / 3π = 1 / 9.424778
r = 0.106
Actual = 1836.15267245

Do you see the problem now? You wrote a completely nonsensical formula down.

c^1/2 = 8.41529061292597326e+05 inches^1/2 minutes^-1/2
3π = 9.424778
Therefore the proton-electron mass ratio is 89000.

This works great!

Does anyone even know what the heck he was even trying to show with that "calculation"? I mean, besides his atrocious use (or non-use) of units...

More dialog:

No I didn't. You can't just set c to 1 without making provision elsewhere. It's the conversion factor between distance and time, and frequency is the reciprocal of time. Try reading this. Space is like the guitar string. When its length is x it vibrates with a first harmonic frequency of 1/x, not x. That's why the n is there in λ = 4π / n c^1½. The 4π is there because you're sweeping a sphere. The c^1½ is there because youre doing it like a moebius strip. You're going round the equator at c and over the pole at ½c. And there's only one size sphere where you can get the spherical harmonic. The c^½ and the 3π is something on top of that, and it's a bit more complicated. But hey, since you don't understand the mass of a body is a measure of its energy-content[/url], you aren't going to understand quantum harmonics. Next!

Oh dear. This looks frustratingly close to what we are trying to tell you. Just assume my distances are in light seconds and my times are in seconds. Is your formula still valid? If so why does it give a completely different answer? If not why does it care about some dead dudes in Paris slightly misestimating the size of the Earth?

Almost everything else you just said is gibberish.

A commitment to vanquish the opposition:

Oh dear edd. I give physics, you just sneer. You know, there was a time when I thought you had some sincerity. Not any more. Not when your response is gibberish! That's no counteragument, now is it? In order to bring this home, I will look out for what you say, and I will carefully offer a counterargument that isn't gibberish. In addition I will provide surgical evidence and logic and references. And when you then exclaim "gibberish!", everybody will see that my surgical evidence took your gibberish apart.

I am now anticipating a powerful response including some "surgical evidence."

 

[Farsight]

Being in your time zone I do agree it is bedtime. But first you do not see a teensy inconsistency above given energy is scalar?

That's why I said But wait a minute, that's OK, force is a vector quantity too. You missed the hint of sarcasm. People say momentum is something very different to energy instead of referring to energy-momentum. They point to the fact that momentum is a vector quantity whilst energy is a scalar. When you then refer them to force x time and force x distance, for momentum they take force as a vector quantity but for energy they don't. it's like I said to Robo, conservation of momentum is there because for every action there is a reaction. A force is only there if there's an interaction, and if I exert force x time on you, you exert a force x time on me. Kinetic energy isn't generally conserved because collisions aren't generally elastic. Some of it gets converted into heat etc. Which is a form of kinetic energy as it happens, but not kinetic energy of the overall body.

Maybe a vector displacement rather than scalar distance combined with a vector dot product might make a touch more sense.
That bit at least is straightforwardly corrected anyway.

You can't use displacement for energy because that's a vector quantity whilst distance is a scalar. You can have a negative displacement. You can't have a negative distance.

 

[Farsight]

Oh my... there are so many errors in this one post it reads like a failing exam in my high school physics class.

No there aren't.

No, it isn't. Vector momentum is not the same thing as Newton's Third Law of action/reaction. If you are referring to the impulse-momentum theorem (which is derived from Newton's Second Law), then you are doing it in a piss-poor manner.

I'm doing it really well. Momentum is conserved. In a collision between two bodies, the bodies exert the same force on each other for the same time. Body A doesn't exert a greater force on B than B exerts on A. And body A doesn't exert that force for a greater time than body B. One force we call the action, the other we call the reaction, and they are the two halves of an interaction. Simple.

What? Do you just pull these things out of your nether region? According to the impulse-momentum theorem, a change in momentum is identical to force integrated over a time interval

Yeah yeah, we all know about

..and work (a transfer of energy) is identical to force integrated over a displacement.

The usual expression is

Like I said to edd, energy (rather than the transfer of energy) shouldn't be associated with displacement, because that can be negative whilst distance can't.

..Wow. Talk about misunderstanding the basics...

I don't misunderstand the basics. I'm the one here who really undertands them.

Farsight, when discussing impulse (what you call force x time) you need to remember that an impulse is equal to a change in momentum; so if the object loses momentum, the impulse is negative (because the force acting on the object is in the opposite direction of the motion). It has nothing to do with the concept of "negative time"!

I know that, I pointed it out. And I also know that impulse is change in momentum, and has the same units as momentum. And do note that if a cannonball is just sitting there in front of you, it's got no momentum. But after you push it, now it does. And guess what, its momentum changed from zero! You're saying nothing Mattus.

You make a similar mistake with work and kinetic energy.

I'm not making mistakes.

According to the work-energy theorem, assuming no transfer of potential energy, the work done on an object is equal to the change in kinetic energy of the object.

And if the cannonball is just sitting there in front of you, it's got no kinetic energy. But after you push it, now it does. And guess what, it's kinetic energy changed from zero! Let's run that little exercise again shall we? But with a twist. Here's the cannonball just sitting there in front of you. Now try reducing its kinetic energy.

Therefore, if an object slows down (loses KE), then there is negative work done on it (i.e. energy is transferred away from the object); this is due to the fact that the work is defined as the scalar product of force and displacement, and if the force acting is in the opposite direction of the displacement then the work comes out negative. And, for the record, because displacement is a vector, it can be a negative quantity depending upon how it is oriented.

I've said displacement is a vector on plenty of previous occasions.

No, you are just digging the hole even deeper. You are doing nothing more than displaying your ignorance of not only cutting edge physics but high school physics as well.

LOL, in your dreams Mattus.

Again, the work done is equal to the change in kinetic energy in this case, not the kinetic energy. Therefore because the change in kinetic energy can be negative (i.e. the object loses KE), then the work can be negative.

Yawn. And the change in kinetic energy of the cannonball that was sitting there in front of you, the one you pushed, is the kinetic energy.

Saying the same wrong things over and over again doesn't make you any more correct, Farsight. Please, follow these steps:

1. Go to the library.

2. Get a book on basic physics.

3. Read the book, and work the problems.

4. Realize your errors.

Geddoutofit Mattus. You've got nothing. next!

 

[MattusMaximus]

Farsight,

As I pointed out before, work (a transfer of energy) is defined as the scalar product of force and displacement (technically, work is the scalar product of force integrated over displacement). A scalar product is a mathematical operation performed on two vectors which yields a scalar result. One would expect that a self-described expert on these topics would know what is a scalar product, wouldn't you? One would further expect that such an expert would also then realize the role that displacement (a vector) plays in work (a scalar), right?

But no, not you, Farsight. In your arrogance, you continue to ignore your basic mistakes on these and other points, yet you expect those of us trained (and those of us who teach the topic) in physics to bow to your self-declared superior knowledge?

Give me a break. I was willing to give you a fair shake, but now it is patently obvious to me that you have no idea what you're talking about, whether it be the Higgs mechanism or basic high school physics. Worse yet, you are not interested in learning anything - your loss, because there is much to learn here.

If I were handing out grades to posters on this thread, you would get an 'F'. Congratulations.

 

[edd]

Farsight,

As I pointed out before, work (a transfer of energy) is defined as the scalar product of force and displacement (technically, work is the scalar product of force integrated over displacement). A scalar product is a mathematical operation performed on two vectors which yields a scalar result. One would expect that a self-described expert on these topics would know what is a scalar product, wouldn't you? One would further expect that such an expert would also then realize the role that displacement (a vector) plays in work (a scalar), right?

But no, not you, Farsight. In your arrogance, you continue to ignore your basic mistakes on these and other points, yet you expect those of us trained (and those of us who teach the topic) in physics to bow to your self-declared superior knowledge?

Give me a break. I was willing to give you a fair shake, but now it is patently obvious to me that you have no idea what you're talking about, whether it be the Higgs mechanism or basic high school physics. Worse yet, you are not interested in learning anything - your loss, because there is much to learn here.

If I were handing out grades to posters on this thread, you would get an 'F'. Congratulations.

This. The scalar product is commonly called the dot product. The above is exactly what I meant, and is extremely basic physics. Thanks for putting it so clearly MattusMaximus.

 

[ben m]

You can't use displacement for energy because that's a vector quantity whilst distance is a scalar.

Seriously? I mean, really? Amazing.

 

[Farsight]

I think it'd be just peachy if you'd respond to the issue of your not using units half the time.

What, the quantum harmonics? As epitomised by ben's miles per hour and edd's grumblings and Perpetual Student's so-called review? Given that people here don't even understand the mass of a body is the measure of its energy-content let alone the Higgs mechanism, I'm not sure it's going to be productive. But OK, let's make a start. We'll take it one step at a time like we did when I explained why the Higgs mechanism contradicted E=mc².

First off look at what I said to sol in post 677 about the Watt balance. I included this quote: "by fixing the Planck constant, the definition of the kilogram would depend only on the definitions of the second and the meter". The dimensionality of energy is kg·m²/s², so if we're going to be defining the kilogram using the second and the metre, you need to understand how they're defined. The second is defined using the NIST caesium fountain clock. There's a reference to that on wikipedia:

"Since 1967, the second has been defined to be the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom..."

In the NIST caesium fountain clock, lasers and a microwave cavity are employed to cause hyperfine transitions, which are electron “spin flips”. These emit microwaves - light in the wider sense. The story goes that there’s a peak frequency which is found and measured by the detector. But note that frequency is measured in Hertz, which is defined as cycles per second, and the second isn't defined yet. When you haven't defined the second, you can't talk in terms of frequency! What's really happening here is that we count light waves coming at us in a wavetrain. When we get to 9,192,631,770 we say that's a second. Then the frequency is 9,192,631,770 Hertz by definition. Note that we used the motion of light to do this. You might challenge that and refer to the hyperfine transition, but that's an electromagnetic thing, so again that's light in the wider sense.

Now take a look at the definition of the metre:

Since 1983, it has been defined as "the length of the path travelled by light in vacuum during a time interval of 1/299,792,458 of a second."

So we define the metre using the motion of light. Along with the second, which we also defined using the motion of light. Note that regardless of how fast light actually moves, this means that you always measure the motion of light to be 299,792,458 m/s. Because you use the motion of light to define the second and the metre, which you then use to measure the motion of light. Also note that frequency is the inverse of time, so wavelengths and frequencies are all based upon the motion of light. The units are derived from it. We quote c, the speed of light, in terms of the units derived from light moving through space.

Are you happy with this? This is what lies at the heart of it. You have to understand this to understand the rest. And if you don't, well, it is off-topic, so nevermind.

 

[ben m]

(three paragraphs mysteriously NOT containing the three lines of math it ought to take to explain a unit-conversion)

Farsight: Miles are units of length. Hours are units of time. The speed of light in miles per hour is 670,616,629 miles/hour. Go ahead, do the five lines of algebra you did with meters and seconds.

 

[MattusMaximus]

You can't use displacement for energy because that's a vector quantity whilst distance is a scalar. You can have a negative displacement. You can't have a negative distance.

Nope, no mistakes there

Have fun with your delusions, Farsight.

ETA: Do you even know what a scalar/dot product is? Have you ever performed calculations with scalar/dot products?

 

[Tubbythin]

Hey, I've already tried. Farsight has shown that he is impervious to learning physics.

I said a good physi... only kidding

 

[MattusMaximus]

I said a good physi... only kidding

I see what you did there

 

[Farsight]

Seriously? I mean, really? Amazing.

Yep. Seen any -511KeV electrons recently? Nope. You've seen me talking about binding energy, which is said to be negative energy. But all it really is, is less positive energy. You can't exert a force for a negative distance. When you push you might think you exert a positive force → whereas when you pull you exert a negative force ←. But just turn yourself around and now your push ← is like your pull was. It isn't really negative force. You can't reduce the kinetic energy of that motionless cannonball. Like I said, people get confused about direction and sign. Just as they get confused about positive and negative charges and positive and negative energy, like Zig did here.

Farsight: Miles are units of length. Hours are units of time. The speed of light in miles per hour is 670,616,629 miles/hour. Go ahead, do the five lines of algebra you did with meters and seconds.

Pay attention and maybe you'll get to understand this. See my post 747. Do you you understand what I said, and do you agree with it?

 

[Tubbythin]

List of basic school-level physics Farsight doesn't understand:

• Scalar

• Vector

• Scalar product

• First law of thermodynamics

• Conservation of momentum

• Difference between energy and momentum

• When to use E=pc and when to use E²=(pc)²+(mc²)²

Any more?

 

[MattusMaximus]

From another thread, Farsight also seems to think that the potential energy associated with gravitational fields is positive. I see that he is also extending this misunderstanding to the concept of binding energy as well.

The hits just keep on coming.

 

[edd]

ETA: Do you even know what a scalar/dot product is? Have you ever performed calculations with scalar/dot products?

Scalar multiples of vectors (along with the associated basics you might expect for scalars and vectors of course) are typically in the mathematics GCSE syllabus at the higher end at least - GCSEs being exams taken at the age of ~16 in England and Wales by everyone, covering a wide range of abilities (so not everyone getting what is technically a pass at GCSE will be expected to know what a scalar product is, but those getting the better grades should).

Farsight says he has A-level mathematics and has tutored up to A-level mathematics, which is taught between the ages of 16 and 18, and only generally speaking to those that have done well at the GCSE level. A-level will certainly cover the vector product and harder problems, but probably not generally vector calculus (no grad, div, or curl), at least to any great extent. Then he has his CS degree.

On the other hand, we have the posts here.

[shrug]

 

[MattusMaximus]

I'm starting to think his supposed A-level performance in math is composed largely of two other letters: BS

 

[W.D.Clinger]

What, the quantum harmonics? As epitomised by ben's miles per hour and edd's grumblings and Perpetual Student's so-called review?

Perpetual Student's "so-called review" mostly consisted of quotations from what you and others had written in this thread. By following the links, anyone can verify that Perpetual Student's quotations were accurate.

edd's "grumblings" have been spot on. You have not formulated any coherent response.

ben m's calculation involved inches per minute, not miles per hour. Not that it matters, but your inability to get even irrelevant details right illustrates your larger problems.

Snipping a bunch of related nonsense:

....When you haven't defined the second, you can't talk in terms of frequency!....

phunk explained your mistake four days ago.

Most tellingly, you still have not acknowledged your sub-high-school-level misunderstanding of units.

Although we acknowledge your mastery of the Gish gallop, we aren't going to let you forget the hilarity of your numerological crackpottery.

 

[Perpetual Student]

It's instructive to review this discussion from its beginning (edited for clarity and continuity):

OK, the debate begins:

The first convoluted response:

Some dialog:

The hammers fall:

More dialog:

A commitment to vanquish the opposition:

I am now anticipating a powerful response including some "surgical evidence."

Still waiting for the surgical evidence. Maybe one more bag of popcorn...

 

[edd]

Wait, rewind. I misread the syllabuses I was looking at. A scalar product of two vectors should be an A-level topic (but a pretty basic one in it). GCSE may only cover the product of a scalar and a vector. Sorry. Still... most of the [shrug] is still there.

 

[Tubbythin]

I'm starting to think his supposed A-level performance in math is composed largely of two other letters: BS

Maybe we all misunderstood and the reference to A-level was the most recent letter of the alphabet that his tutees understood?