Mystery111

A

Ok, if you didn't someone did lol

Your question is inherently irrelevant. After all, it was the legendary Schrodinger who defined the first helix structure, or something similar.

the helix structure is a matter of all biological life.

The only thing that is clear is your propensity to pontificate on subjects about which you know absolutely nothing. In fact you know less than nothing as what you say is often wrong, but couched in buzz words designed to bamboozle neophytes.

 

Matter most certainly will not always radiate away as gravity waves. Were that the case there would be no such thing as a stable particle.

is that really true?

I mean, I did actually say I should have mentioned it was for accelerating objects, well before this post.

So if anyone is misrepresenting facts, it is you!

No, it wasn't clear to me what you meant (you don't get to decide if it was clear to me, only I can decide that), and there is absolutely a need for it.

Well, ... ok?

Consistent with what I said, I think. He appears to have applied physics to various processes in biology, but did not actually define life. AFAIK biologists can't even come up with a complete definition.

reading your post, would assume the question of life is not a physics question.

Now, taking into regards what he predicted, since DNA the helix stucture is completely inherent in all life, I think you have mistaken your facts. Physics has a lot to say about this subject! And has done!

Swansont downgraded my post! LOL

Swan, you are wrong, period!

"Matter radiates gravitational waves" is not the same as "matter will radiate away as gravitational waves" Where does the matter go? I have a bunch of quarks (in some form) and possibly leptons. And suddenly they aren't there?

stop being pedantic. It is clear what I meant!

This reminds me of the time you corrected someone on the difference between mass and matter.

There is no need for it.

Gravitational waves are not matter.

Given enough time, matter will radiate away, and all that will be left is graviational waves. Gravitational waves are non-trivial, they are a presence of curvature in the universe even when matter is not present.

The fact matter is predicted to radiate gravitational waves is an aspect predicted by dyson and his equations.

 

10^1500 years - iron, seriously ? we will be in a warm radiation dominated universe by then - ie heat death, the big boring, dullsville tennessee .

Actually I agree with this statement. This is what Dyson predicted.

stationary mass does not radiate wave's however. Only an acceletating mass will. I should have mentioned that.

Gravitational waves basically carry energy away from their sources.

''and in more detail in a 1918 paper, Einstein showed that when a mass accelerates – in other words, changes its state of motion – it can't help but give rise to time-varying gravitational fields that travel away from the source at light-speed as undulations in the surface of spacetime.''

http://www.daviddarling.info/encyclopedia/G/gravwave.html

Imaatfal

I think you have the concept of heat death wrong. It does not imply a warm radiated universe.

It means that the universe will cool down.

Null, for a photon for instance?

I don't quite understand what you mean.

I tink the physics view is that life is something that biologists define. You can analyze processes with the laws of physics and whether the item is alive doesn't change any of that; our body is a heat engine driven by combustion, we can do mechanical work, etc. Similar analysis can be applied to any other organism. Being alive is not a special state from a physics standpoint.

Not all cases.

Schrodinger had some very interesting thoughts. In fact, I believe his book was named ''what is life?''

http://en.wikipedia.org/wiki/What_Is_Life%3F

can you cut this down a bit? I loose concentration before the wall of words has finished. Please?

Did I say this wrong?

I always get confused with the initial setup to the final.

If you measure the position accurately enough - you cannot know the trajectory because the momentum is so uncertain. If you know the position of a particle with zero error then the error of the momentum is such that it could be heading in any direction at any speed - ie the first measurement is impossible

''Doing so will excite the particle, so the position of the electron becomes certain if the photon has a small wavelength - the momentum of the electron then becomes very uncertain,''

Is what I said. I don't think this is in dispute.

I can only just barely visualize four dimensions!!! lol

Don't be silly. No question is a dumb question if you don't know the answer

Of course matter will eventually radiate away in the form of gravitational waves. This is what Dyson worked out from his calculations. I will give you three examples of solutions to what will happen, there are more and will recite them all if you want.

Within [math]10^{19}[/math] the central region of the Galaxy may be expected to collapse into a Black Hole while stars in the outer region are detached from it.

After [math]10^{24}[/math] years, steller orbits around the Galaxy will decay into gravitational waves

After [math]10^{1500}[/math] years, all ordinary matter will have fissioned or fusioned to iron through radioactive processes.

Please don't contradict me, I know what I am talking about. I knew this about his equations many years before today.

That was to guenter, sorry^^

Just a thought. Could two universe's exsist in the same space? You know, over lap each other? And not even know it?

Short answer, this is how the multiple universe theory works.

They overlap each other in the act of superpositioning and fly apart when a measurement is made. Then overlap all over again.

no

Of course they can rocket. This is how a single particle is shared among universes in the multiverse theory. Everything exists as a superposition until something comes along and disturbs that wave function. Then universes fly apart, create universes equal to as many possibilities which is permissible.

Because logically, there should not be two universes. And according to law of science, there should not be two universes. But.... with the "space" outside space, it's possible. So if there is more then one, shouldn't what we know basically mean nothing or be changed?

Yes. Two universes cannot work. It turns out as an Oxymoron.

Someone did sort of the same thing once. It didn't end well:

 

https://en.wikipedia...natoli_Bugorski

My god. The man was irradiated beyond belief!!! Eeeek!

You know... you could technically know the exact position and momentum of an object.

If you make a calculation of it's position in the past, then made an accurate calculation of it's trajectory in the future, you can know with certainty the position and trajectory made in the present. Of course, you can only do this because you are not measuring the two simultaneously in the present moment.

Anyway, an easy way to reconcile this feature of reality is by assuming you want to observe an electron. You might observe it by hitting a photon of the electron. Doing so will excite the particle, so the position of the electron becomes certain if the photon has a small wavelength - the momentum of the electron then becomes very uncertain, but it's position would have been known with reasonable accuracy.

(I think I got that right)

Hi and thanks for the quick answer.

My whole question is: How to get from [math]\langle \phi|A| \Psi \rangle ^* [/math] to [math] \langle \Psi|A^\dagger| \phi \rangle[/math]. Is this only per definition that way (I already understood that this is the case for matrix elements A, because this is the way for creating the adjoint operator - transpose and conjugate-complex.). Because I tried to review the single steps to get from the left side to the right side.

 

 

Or maybe my question is:

If I transpose the scalar product, do I just change the bra and the ket?

Sorry if this sounds confused, but in fact, I am o.O

 

 

Thanks in advance, Patrick

You should check what I say in this following post. Prove to yourself for instance that the inner product of [math]<b|[/math] and [math]|a>[/math] is the complex conjugate of [math]<a|b>[/math]. Here is one that is less obvious. Take any matrix, hermitian or not hermitian as [math]<b|M|a>[/math]

Take M and multiply it onto the ket will give you a new vector. Then you take the inner product of b. This is related to

[math]<a| M^{\dagger}|b>^{*}[/math]

where the <a| is acting like a complex conjugate, in fact, this has been complex conjugated where all rows and columns have been interchanged. Incidently, M on (a) will give you a vector, but M and (a) after this onto (b) will give you a number. It won't give you a vector in this case, it's just a number.

So you get from [math]<b|M|a>[/math] to the expression [math]<a| M^{\dagger}|b>^{*}[/math] by complex conjugating it. In fact, if it is hermitian you can now state it as

[math]<b|M|a> = <a| M|b>^{*}[/math]

where we have just erased the conjugation dagger sign. This means it is equal to its own hermitian conjugate. That is just the definition of Hermitian. So the <b| and |a> matrix element of M is the same as the <a| and |b> matrix element of M.

Don't forget also you can have the case where a=b so

<a|M|a> = <a|M|a>*

Is real. It also has a special name. It is called the expectation value.

I will try and answer your questions... I little buisy at the mo, will be back on the comp soon.

Hi!

Since I forgot my PW... and I was too lazy to get a new one, I simply connected with my FB account, and here I am, posting some more questions...

I just reviewed linear operators. And found out that they can have an adjoint.

Well. I asked myself a question while looking at this equation:

[math]\langle \phi|A| \Psi \rangle ^* = \langle \Psi|A^\dagger| \phi \rangle[/math]

 

,where A is a linear operator, and [math]\phi[/math] and [math]\Psi[/math] are wave functions in the hilbert space. While trying to reproduce the steps I wondered if one is allowed to move bras and kets without forming there conjegate-complex.

Meaning:

[math]| \Psi \rangle \langle \Phi| = \langle \Phi| \Psi \rangle[/math]

 

Thanks for answering!

 

EDIT

 

I just realized that by definition [math] ( \Psi | \phi )= (\phi | \Psi )^* [/math].

Can this be applied to the bra-ket notation? And what happens if an operator is squeezed in between...?

Right... well, when you come across something like this [math]\langle \phi|A| \Psi \rangle ^* = \langle \Psi|A^\dagger| \phi \rangle[/math] the fact we can see [math]A^\dagger = A[/math] actually implies something important; if [math]A[/math] is a matrix, then this implies our matrix is hermitian, meaning all the diagonal elements in the matrix are real. Real things are associated to observables in quantum mechanics. This will take you into things like calculating matrices and the hermitian conjugate. You simply transponse your matrix but then complex conjugate it. I could write up some matrices, but I really can't be bothered unless you ask me to. Anyway... coming across something like [math]|\psi><\phi|[/math] is called an outer product, obviously the opposite name-meaning of the inner product. You may come across something like

[math]|\psi><\psi|\phi>[/math] this means that the eigenvector has been chosen, then this is an orthogonal vector [math]\phi[/math].

I would have written more, but am in a rush again!!! lol

How can someone measure the future in the present? Surely then it is no longer the future?

And if time is flowing, what is it flowing relative to?

Oh right cool, thanks!

If so, you can add me.

TY!!!

I am glad to have the nesxt couple of weeks here. I have missed the place.

I was wondering, like sound waves are vibrational waves of particles. How exactly are gravitational waves related. So are they vibrations of atoms ? and if so are they transverse or longitudinal vibrations ?

They are the distortions carried when matter will radiate. Needless to say, a large amount of time is required before matter will radiate away in the form of gravitational waves.

Curvature is caused by understanding the Christoffel Symbols.

A certain wave equation describing two kinds of waves,

[math]\frac{\partial^2 \phi}{\partial t} = c^2 \frac{\partial^2 \phi}{\partial x^2}[/math]

It describes basically the kind of waves we attribute to right movers and left movers. In three dimensions this is

[math]\frac{\partial^2 \phi}{\partial t^2} = \frac{\partial^2 \phi}{\partial x^2}+ \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2}[/math]

This can be written as

[math]\eta^{\mu \nu} \frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}} = 0[/math]

To make this into a tensor, we will invite now the Christoffel Symbols

[math]\nabla_{\mu} g^{\mu \nu} \frac{\partial \phi}{\partial X^{\nu}} + \Gamma^{\mu}_{\mu \alpha} g^{\nu \beta} \frac{\partial \phi}{\partial X^{\beta}} = 0[/math]

This equation describes the geodesic of a path taken by the likes a of a photon for an example.

Oh the Chronon... now there is a name I have not heard in a while. This number has possible massive implications for quantum mechanics. The Chronon is a wonderful idea.