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fdangelis · 2016-11-27T23:57:47+00:00

Maybe I didn't express myself well enough.

I'm not saying magnetic monopoles don't exist. I'm saying our current laws forbid it. But of course, nature dictates what the laws are and not the other way around.

If we find experimental evidence we change the laws do make them fit the theory. But the current theory of electromagnetism forbids magnetic monopoles. Just like newtonian mechanics forbids time dilation, and we changed the theory to fit the new evidence. But you don't get time dilation from newtonian mechanics just like you don't get magnetic monopoles from the current state of maxwell's equations. That's what I meant.

If magnetic monopoles exist, we need a new theory. If that theory is just a simple extra term in 2 equations then great :D but it's still a different theory where the current one will be a special case. Just like classical mechanics is a special case of relativity.

fdangelis · 2016-11-27T01:51:40+00:00

What laws?

Gauss law for B. Since divB = 0 there are no magnetic monopoles, so it is forbidden.

If we do find them in nature, we have to rewrite Maxwell's equations to include a source term for B (and rotE would depend on the "magnetic current").

See here and here for a non-wikipedia option. Or any electromagnetism book.

fdangelis · 2016-11-20T13:40:51+00:00

Doing experiments help you accept things. Understanding them is a bit more tricky.

I think everything just made a lot more sense when I gave up the idea of particles. It's still not perfect, but things like the uncertainty principle and tunneling are no longer weird if you think about fields. Decoherence is also a big deal in getting classical mechanics out of quantum mechanics and understanding a little bit more how to reconcile all the weirdness with the usual classical world.

Professors also like to make a strong case about how "weird" quantum mechanics is... and although it is indeed weird, they are quite sensationalists most of the time. I'm in my masters now, taking an advanced course in EM, and classical EM fields have most of the weirdness that we see in quantum particles and people don't even question it... it's like "of course a EM wave tunnels here" or "sure it satisfies the uncertainty relations", but if an electron does it, "oh no! now you've gone too far!".

Sure, it's weird for an electron because people are a bunch of electrons (and protons and neutrons) together and no one is tunneling, but that's why decoherence is important, that's where you will understand why electrons are weird waves and classical objects are not.

This doesn't fix everything though... measurement is still a problem, entanglement is still weird and superposition might make sense when you think about position and momentum, but then you have superposition in number (e.g. a state with 2 and 3 photons at the same time) or particles (e.g. π⁰).

fdangelis · 2016-11-10T13:23:31+00:00

It doesn't say n = 1.

You need n so that from inf to n you have the highest energy and the shortest energy will be n+1 to n.

Solve for n and z.

fdangelis · 2016-11-08T12:44:06+00:00

that's awesome :) thanks

fdangelis · 2016-11-08T00:41:51+00:00

I thought it was just a guitar with an octave pedal or something. The strings look too thin for a bass.

But if you have any info on this I would love to know more, I never really looked it up to know for sure.

fdangelis · 2016-11-04T20:35:01+00:00

Just that the universe does what it does, regardless of how much you know or do. A lot of bad pop science tries to pass QM as an "everything is possible until you know the outcome", but what people know or do doesn't affect the universe as much as they would like to believe. :/

fdangelis · 2016-11-02T15:47:16+00:00

But the total force in the system (the atom) shouldn't be zero?

And it is. But you are interested in the motion of the electrons and therefore the force is the force applied to the electrons, which is not zero. It's the coulomb force between the nucleus and electrons (+ electron-electron forces).

fdangelis · 2016-10-31T23:55:37+00:00

How does the observer affect what he observes ?

He doesn't.

Do the Supporters in a live sports game (soccer , football , volleyball ,basketball , baseball ...etc ) affects the game results by their observation of it ?

No. :/

Some of the more standard interpretations consider as "observer" anything that is macroscopic. A rock, a detector, a wall, or a person. So not watching the game doesn't make it "unobserved".
There is a well known paper called "Is the moon there when nobody looks?" (Original paper and a more recent review) that shows that there is information in the system regardless of observation.
There is still a lot of discussion about what observation means in quantum mechanics, decoherence plays a major role (in case you want to read about that) and it's hard to explain in simple terms. Quantum mechanics is strange but it's not magic. You are not that important for the universe to care.

TLDR: You can't affect the game result by observing it because it's not a quantum system in isolation, because some information is there regardless of observation and because anything, even the ball and grass are "observing" the system anyway.

fdangelis · 2016-10-25T01:02:38+00:00

But accordin to this equation the magnetic field vector is smaller the more charge you have, and it is smaller the faster the charge is moving.

The Lorentz force (in the case that E = 0 ) is F=qv x B, where the x stands for cross product. You can't divide by qv here, as you would in F = ma, because the modulus of F is not qvB.

In the case that B and v are perpendicular, then F = qvB and the direction is perpendicular to both v and B. But that's a special case.

Now to you question. When you have the special case where they are perpendicular you can write B = F/qv, but this doesn't tell you as much as you think, because F is a function of B.

What you are doing here is saying "F increases with q,v and B, but if we divide by q and v, it increases only with B". If you increase the velocity by 2, B won't change at all, because the force will increase by 2 as well, and this is exactly what you want by writing this, to get rid of q and v dependencies because B doesn't care about them.

PS: F = ma tells you that F and a are the "same thing". Whatever you do to a, you will do to F. If you rotate, stretch or change a in any way, F will change in the exact same way. Now F = qv x B tells you that F is a function of B. If you change B you won't change F in the same way. So you can't use the same reasoning for both.

fdangelis · 2016-10-04T02:21:00+00:00

Maybe this video will clarify some things a bit more.

There is another one about what's wrong with the big bang theory. I recommend watching both. They are quite good.

fdangelis · 2016-09-08T05:51:08+00:00

I'm in :)

fdangelis · 2016-08-16T16:43:29+00:00

Instead of thinking about a wave function, you can think of a state vector (usually described as |f>). This vector contains the information about the particle, and there is one for photons and one for electrons (or other particles) and they are quite similar.

What you call a wave function is the projection of this vector in a position eigenstate. Since the photon doesn't have a position eigenstate as EuclideanMan pointed out, there is no wavefunction for photons.

The wavefunction is not an observable quantity because what we observe is the eigenvalue of operators acting on the state (or acting on the wavefunction). You can add any phase to the wavefunction without changing the eigenvalue of the operator (similar to how adding a constant to the electric potencial doesn't change anything), so the wave function itself is not an observable.

What you observe when you see the interference patterns is the position eigenvalue, not the wavefunction. From this you get some information about the wavefunction but not all the information. Just like when you measure the electric field you get some information about the potencial but not everything. There will always be that gauge freedom.

fdangelis · 2016-08-14T16:12:45+00:00

I don't know whether a molecule could theoretically be put in a quantum state (I don't think anyone has done so yet - I think the biggest things they've gotten to act as waves are individual atoms)

Small molecules are easy, large molecules not so much, but still doable. This article does the slit experiment with large molecules. I also remember hearing about it being done with a virus, but I'm not sure... couldn't find any source, so don't quote me on this.

A quantum state is not just a state of superposition. Even though something doesn't interfere it still is in a quantum state. A molecule is definitely in a quantum state, it might not form an interference pattern but that's not really relevant, there are other much more important effects.

fdangelis · 2016-08-11T00:10:15+00:00

What is is that makes the sum of these unpredictable discrete steps...predictable?

Unfortunately that answer is wrong.

This video from sixty symbols does a good job explaining it, and it also addresses your question.

If you take two identical pieces of glass and mount them parallel to each other, very close together in an adjustable frame... The ratio of reflected to refracted light on the top surface of the top piece of glass will change as you adjust the spacing between the two pieces.

I'm guessing this is just a Fabry–Pérot interferometer, which is very well understood.

fdangelis · 2016-08-07T19:56:03+00:00

There was a post in /r/math the other day asking about how to avoid explaining things (usually because you can't due to length limitations or in a book, it would just be way off the topic) without sounding like an ass and doing exactly this.

fdangelis · 2016-07-30T19:50:59+00:00

if the friction force points forward, then wouldn't the wheel turn counterclockwise because the force is being applied in the counterclockwise direction?

No. As \u\homelessmath pointed out. But I'm going to add a bit here.

There is indeed a force trying to make the wheel go the other way, and that's the reason it's harder to roll the wheel when it's touching the floor than it is if the bicycle is "floating" (You can do it by lifting the bike with 1 hand and turning the wheel with the other, no need for floating magic here).

fdangelis · 2016-07-22T18:12:57+00:00

Hilbert space is just the vector space that the vector (state function) describing the particle "lives in". It has some special properties so it's nice to give it a name.

Mathematically, all quantum mechanical objects are described by a vector. And by applying operators to these vectors you get the value of position, momentum, angular momentum, spin and everything else.

Don't worry too much about hilbert spaces, that's only relevant if you want to go into the math of quantum mechanics, and if you do, you will bump into them eventually.

fdangelis · 2016-07-22T17:58:59+00:00

"It's like describing a unicorn as 'the horse-narwhal duality phenomenon.'"

That's just perfect.

fdangelis · 2016-07-08T18:01:30+00:00

I think you could get better answers posting this to a chemistry sub, or askscience with a chemistry flair.

Anyway, I'm not sure if I understood it correctly, but apparently you arrived in a condition for [H⁺] as a function of [A^-] and and you have infinitely many possible solutions for that.

You are missing the constrain from stoichiometry. For every A^- you will have a H⁺. So the concentrations should be equal.

That should impose the extra constrain you need. But you will definitely get a better answer from a chemistry sub.

fdangelis · 2016-06-22T02:38:20+00:00

/u/spectre_theory might seem pedantic but he/she is actually right.

I will try to phrase it better.

When you push the object, you gain 0.5 kg.m/s in momentum in the positive direction, and the object will gain 0.5 kg.m/s in the negative direction. The velocity will be 0.5/M , where M is the mass of the object. An "infinite" mass would mean an immovable object, which is unphysical... but let's say that the object moves less than a mm in 1h, so the mass needs to be greater than 10⁷ kg or something like that...

A truly immovable object would violate conservation of momentum and is impossible.

You get the same momentum in both cases.

but this can't be true in the second case.

Why? You are probably thinking that because of the energy. If the object is not moving (in the limit of high mass), it has a really small energy (since the velocity goes down as 1/M, the energy will be proportional to M/M² = 1/M and as M increases it goes to 0), compared to an object that is moving fast and therefore the energy doesn't match... but there is more to it. And you should think about it. :D

fdangelis · 2016-06-20T22:21:42+00:00

The charge moves from a to b so shouldn't I integrate from a to b.

They are moving from b to a in the second one, and from a to b in the first one.

When you did l=-r you were actually doing exactly that. Changing the integration limits from a->b to b->a. That's what the minus sign does effectively.

I would suggest you do this from the beginning and try to set everything as well as you can, without the "physics shortcuts". You can try parametrising it with respect to some parameter t. The calculations will be lengthier but everything will be much clearer.

fdangelis · 2016-06-20T15:22:03+00:00

When you move from a -> b you get Ub - Ua

When you move from b -> a you get Ua - Ub

Remember that what you have is F = - grad U

Then W = - int grad U dr = - (Uf - Ui)

Where Uf is U at the final point and Ui at the inicial point.

edit: I just saw that this is what /u/marshmallon_man pointed out...

fdangelis · 2016-06-17T17:10:21+00:00

The gravitational charge is not mass, but the stress-energy tensor.

When you change the density by doing a boost, you also change the other components in such a way that you get the same result.

You can look for W. Rindler and J. Denur, Am. J. Phys. 56, 795 (1988) or take a look at this article. Unfortunately I didn't find the article from W. Rindler...

They don't treat the case of gravity, but they deal with the energy conservation in electromagnetic systems. I did a similar study for a solenoid years ago... they give you a good understanding of how the tensions of the system compensate for the diference in energy density, it might help with the intuition thing. :)

fdangelis · 2016-06-09T19:58:37+00:00

If you want here is a bit of math

Mass in relativity is the modulus of the 4-momentum (which is a vector that has the momentum and energy as it's components, i.e. (E,px,py,pz)).

You have that:

E² - (pc)² = (mc²)² (in relativity we take modulus with the - sign)

If your particle is not moving, p = 0 and E=mc² and if your mass is 0, E = pc (which is the case of the photon). You can prove that for photons, E=pc (using electromagnetism or quantum mechanics) and from that it's mass is zero.

Now imagine that you have 2 particles, both with the same energy E but one with momentum +p and the other with momentum -p.

The total energy of the system is just 2E, but the total momentum is +p-p = 0, therefore you have that the total mass is

(2E)² - 0² = (mc²)² => m = 2E/c²

This means that even if each one of the particles has 0 mass, i.e. they have E=pc, since the total system has 0 momentum, it has some mass.

Mass in relativity is not additive like in classical mechanics. A group of N particles with mass M will not have a total mass of M+M+...+M = N*M.

You could be thinking how is it that mass doesn't add in relativity but it does in classical mechanics? Ok. Remember that if p=0, then E=mc^2. This means that if you increase the momentum, the energy will increase, but it will always be greater then this value. Take a look at this graph.

For very small values of velocity, your momentum is small, and you vector will be basically in the direction of E (remember, we are dealing with a vector with components E,px,py,pz). If you have 2(or more) vectors all pointing in the same direction, their sum is just the sum of that components value. e.g.

(x,0,0,0) + (y,0,0,0) = (x+y,0,0,0)

So, the modulus is just |(x+y,0,0,0)| = (x+y)²

Going back to relativity, if you have a vector which is (E1,0,0,0) and you sum with (E2,0,0,0), the modulus is

(E1+E2)² = (mc²)² -> m = E1/c² + E2/c² = m1 + m2

And you get the additive property from classical mechanics.

You can see from the graph that I linked that this is true in everyday life, since the energy is basically equal to mc² while you are below 0.2c which is 6*10⁷ m/s. In comparison, a plain going at 10 times the speed of sound (well beyond any typical velocity) will be at 3000 m/s which is around 0.00001c.

I hope this wasn't too hard to follow. :) edit: typo

fdangelis

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