# Question on shifting field gradients by an angle

• Goldstone1
In summary, this conversation is discussing a possible way to have a complex energy in a field, and whether it is theoretically possible. It is not clear why a complex energy would be necessary for this to happen, and the relativistic energy-momentum relation is not given by what was written.
Goldstone1
If it is known that the energy of a field can be given as $$E=M\phi$$, then it must also be allowed that $$\phi$$ can undergo a shift. If we allow it to be shifted by $$\pi \in (\mathcal{R},\mathcal{C})$$ then surely there is a change in that field specified by using the following shifts?

$$\Delta E=M(e^{i(\theta + \pi)})$$

$$\Delta E=M(e^{-i(\theta + \pi)})$$

In principle yes, but how would a complex phase for a quantity like energy make sense?

Polyrhythmic said:
In principle yes, but how would a complex phase for a quantity like energy make sense?

This seems like an odd question because we deal with negative energies in the Dirac Equation all the time. The negative energy solution of the Dirac equation is basically a change in sign:

right moving waves are given as

$$\frac{i \phi_R}{\partial t} = -i \partial_x \psi_R +M\psi_L$$

and left movers are given as:

$$\frac{i \phi_L}{\partial t} = +i \partial_x \psi_L +M\psi_R$$

It must be noted, the beauty of \beta the matrix, interchanges the sign of the particle, so a coupled equation is given also:

$$i\frac{\psi}{\partial t} = -\alpha \partial_x \psi + M\beta$$

But we never question why the antiparticle in the sea has a negative energy... So why can a complex shift hold the same comparisons?

What you refer to as Dirac sea is an outdated concept. In Quantum Field theory, you get both particle and anti-particle solutions. But I still don't get what a complex energy would have to do with this.

Polyrhythmic said:
What you refer to as Dirac sea is an outdated concept. In Quantum Field theory, you get both particle and anti-particle solutions. But I still don't get what a complex energy would have to do with this.

It's not outdated as such: The principle of the Dirac Sea is still at the core understanding of particles at the quantized level. Virtual photons, for instance have a negative energy. Even the Casimir Effect has an increasingly small negative energy solution between the two plates. The vacuum is heathing with a bath of particles which have negative solutions in quantum field theory.

But the point of the Dirac Equation was that it predicted negative energy solutions. I was replying to you in this sense, because you brought up why negative energy should even be considered in a field. As I qoute your words:

''In principle yes, but how would a complex phase for a quantity like energy make sense? ''

Well, the reply is as before: The same way it makes sense of all the other negative energy appearances in physics. They require a field which satisfies $$\Phi=\phi_R+\phi_C$$ otherwise my post could not have declared that $$\pi \in (\mathcal{R},\mathcal{C})$$ range.

Polyrhythmic said:
What you refer to as Dirac sea is an outdated concept. In Quantum Field theory, you get both particle and anti-particle solutions. But I still don't get what a complex energy would have to do with this.

By the way, the equations:

right moving waves are given as

$$\frac{i \phi_R}{\partial t} = -i \partial_x \psi_R +M\psi_L$$

and left movers are given as:

$$\frac{i \phi_L}{\partial t} = +i \partial_x \psi_L +M\psi_R$$

are particles and antiparticles by the way :P

Now suppose I want to evolve my question. Is it correct in principle then, since $$E=M\phi$$ is true for a phase shift in the field, then the relativistic cases would simply be:

$$E=(M(e^{i(\theta + \pi)})+pc$$

$$E=(M(e^{-i(\theta + \pi)})+pc$$

Would make the field ''of unknown name'' $$\phi$$ naturally invariant yes?

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Goldstone1 said:
Now suppose I want to evolve my question. Is it correct in principle then, since $$E=M\phi$$ is true for a phase shift in the field, then the relativistic cases would simply be:

$$E=(M(e^{i(\theta + \pi)})+pc$$

$$E=(M(e^{-i(\theta + \pi)})+pc$$

Would make the field ''of unknown name'' $$\phi$$ naturally invariant yes?

I still don't see your point. Negative energies, alright. But you're obviously writing down complex energies, which is something I don't understand.

The relativistic energy-momentum relation is not given by what you wrote down here, it's given by

$E=\sqrt{m²c^4+p^2c^2}$.

Polyrhythmic said:
I still don't see your point. Negative energies, alright. But you're obviously writing down complex energies, which is something I don't understand.

The relativistic energy-momentum relation is not given by what you wrote down here, it's given by

$E=\sqrt{m²c^4+p^2c^2}$.

I know that that is the relativistic equation, is given by the above, but we don't have the Mc^2 term, we have the $$E=M\phi$$ which makes the energy.

I think I should have written

$E=\sqrt{m²\phi^4+p^2c^2}$

but I should still be able to do my shifts right?

Goldstone1 said:
It's not outdated as such: The principle of the Dirac Sea is still at the core understanding of particles at the quantized level. Virtual photons, for instance have a negative energy. Even the Casimir Effect has an increasingly small negative energy solution between the two plates. The vacuum is heathing with a bath of particles which have negative solutions in quantum field theory.

Not really. The Dirac Sea was a starting point for undestanding quantum field theory, but one should definitely look at the theory in a different way. Instead of that Sea, you have particle and antiparticle solutions, both with the same mass but the latter with oposite charge. As for virtual photons, in what way would they have negative energy? They violate the relativistic energy-momentum relation. I wouldn't explain the Casimir effect by virtual photons.

Polyrhythmic said:
Not really. The Dirac Sea was a starting point for undestanding quantum field theory, but one should definitely look at the theory in a different way. Instead of that Sea, you have particle and antiparticle solutions, both with the same mass but the latter with oposite charge. As for virtual photons, in what way would they have negative energy? They violate the relativistic energy-momentum relation. I wouldn't explain the Casimir effect by virtual photons.

I agree with looking at it a different way, but the general idea still holds, and that our vacuum must harbour an energy density which must be negative in mathematical standards.

Goldstone1 said:
I know that that is the relativistic equation, is given by the above, but we don't have the Mc^2 term, we have the $$E=M\phi$$ which makes the energy.

I think I should have written

$E=\sqrt{m²\phi^4+p^2c^2}$

but I should still be able to do my shifts right?

Relativistic dynamics doesn't work that way. you can't just replace c by some field $\phi$ and claim that the energy relation would still be valid. I initially thought you were talking about some classical field given by some mass parameter times some field. But now that we're talking about special relativity, it just doesn't make sense anymore.

As for the photons, they where an example. They may be seen as violations, but they don't last long anyway.

Goldstone1 said:
As for the photons, they where an example. They may be seen as violations, but they don't last long anyway.

They don't exist in the sense of a particle, but that's a different story we shouldn't discuss here.

Polyrhythmic said:
Relativistic dynamics doesn't work that way. you can't just replace c by some field $\phi$ and claim that the energy relation would still be valid. I initially thought you were talking about some classical field given by some mass parameter times some field. But now that we're talking about special relativity, it just doesn't make sense anymore.

So then I have a new question,

if $$E=M\phi=Mc^2$$ holds true, how do you make $$E=M\phi$$ relativistic?

Polyrhythmic said:
Relativistic dynamics doesn't work that way. you can't just replace c by some field $\phi$ and claim that the energy relation would still be valid. I initially thought you were talking about some classical field given by some mass parameter times some field. But now that we're talking about special relativity, it just doesn't make sense anymore.

Come to think of this post again, i don't think the highlighted was what I was heading at anyway. It is true there I was replacing c by $$\phi$$... but I did mention the shift mechanism in the field itself - what if that can compensate the energy relation problem?

Afterall, $$\phi$$ is at the very dynamics of the energy of the theory. Perhaps the shifts can even any creases out.

Goldstone1 said:
So then I have a new question,

if $$E=M\phi=Mc^2$$ holds true, how do you make $$E=M\phi$$ relativistic?

If we consider M to be the relativistic mass, i.e. the rest mass $M_0$ times some velocity-dependent prefactor $\gamma$, the only way I could make sense of your field $\phi$ would be by the definition

$\phi(v)=\gamma(v)*c^2$.

The energy would be then given by

$E=M_0*\phi$.

Your field would then only be some rescaling of the gamma-factor.

Goldstone1 said:
Come to think of this post again, i don't think the highlighted was what I was heading at anyway. It is true there I was replacing c by $$\phi$$... but I did mention the shift mechanism in the field itself - what if that can compensate the energy relation problem?

Afterall, $$\phi$$ is at the very dynamics of the energy of the theory. Perhaps the shifts can even any creases out.

I don't think that this would make sense. You'd still have lost the factor of c².

Polyrhythmic said:
If we consider M to be the relativistic mass, i.e. the rest mass $M_0$ times some velocity-dependent prefactor $\gamma$, the only way I could make sense of your field $\phi$ would be by the definition

$\phi(v)=\gamma(v)*c^2$.

The energy would be then given by

$E=M_0*\phi$.

Your field would then only be some rescaling of the gamma-factor.

You would not try to expand the notation?>

$E=M_0*\phi$

into the full relativistic case?

Goldstone1 said:
You would not try to expand the notation?>

$E=M_0*\phi$

into the full relativistic case?

What do you mean?

$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Polyrhythmic said:
What do you mean?

$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Yes, beside the mass term?

See this is where I am confused. We talk about relativistic energy, but there is a contribution of energy let us say, in each shift of our scalar $$\phi$$. If the equation $$E=M\phi$$ is true for gravitational energy, what about the momentum of the field, or particle if you will? It seems

$$E=M\phi + P_{\mu}P^{\mu}= \sqrt{M^2c^4 +p^2c^2}$$
iff
$$\phi \in (\mathcal{R},\mathcal{C}) \to \mathcal{H} \in (\mathcal{R},\mathcal{C}) \forall \psi(\theta+\pi)$$

Plus the logical statement $$\phi \in (\mathcal{R},\mathcal{C}) \to \mathcal{H} \in (\mathcal{R},\mathcal{C}) \forall \psi(\theta+\pi)$$ answers another one of your questions. You ask why have the complex energy? If the field $$\Phi=\phi_R+\phi_C$$ is in the complex and real range of some abstract field of spacetime which is a directly leads to a Hamiltonian of the system which is also in the real and complex range, we find is because it is because it exists for all phase contributions to the fields energy.

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I fact you could apply even further logic. If the shifts belonged to a set of elements call them $$(\mathcal{E}_1,\mathcal{E}_2)$$, then call $$\phi= \mathcal{A}$$ and $$\phi*= \mathcal{B}$$

using doxastic logic

$$\Box \mathcal{A} \vdash \forall \mathcal{E} \in \mathcal{B}$$

''it is necessary that'' $$\mathcal{A}$$ derives from all elements in contained in $$\mathcal{B}$$.

Call this a new Group $$\mathfrak{G}$$. This new group defined is simply a special case of the rotation symmetry group $$U(1)$$.

Sorry, I'm completely lost with what you're trying to tell me...
I can only say that this equation can't be true, since on the left hand side you have momentum squared, while on the right side you have the squareroot of momentum squared. And I don't see how that E*phi would fit in.

Goldstone1 said:
$$E=M\phi + P_{\mu}P^{\mu}= \sqrt{M^2c^4 +p^2c^2}$$

Sorry I was getting ahead of myself and certain terms. the $$P_{\mu}P^{\mu}$$ should just be pc. sorry.

It still doesn't make any sense to me. The squareroot of a sum is not the sum of the squareroots of the summed terms.

Polyrhythmic said:
It still doesn't make any sense to me. The squareroot of a sum is not the sum of the squareroots of the summed terms.

I'm lost. You have a variable $$A$$, and square it to make $$A^2$$, are you saying $$\sqrt{A^2}=A$$ is not true?

Goldstone1 said:
I'm lost. You have a variable $$A$$, and square it to make $$A^2$$, are you saying $$\sqrt{A^2}=A$$ is not true?

What I meant to say was

$\sqrt{A+B}\neq\sqrt{A}+\sqrt{B}$.

Ah right, yes... i missed that one.

Then let's not so hastily equalize the terms. Let us consider just E=M\phi. If it is Mc^2 then how do you make it relativistic? I saw your last answer and liked it, any others?

Goldstone1 said:
Then let's not so hastily equalize the terms. Let us consider just E=M\phi. If it is Mc^2 then how do you make it relativistic? I saw your last answer and liked it, any others?

No, I think that's all I can contribute so far.

Polyrhythmic said:
No, I think that's all I can contribute so far.

Well thank you. I may have some more questions. These questions as you know are based on that paper I sent here, but was denied. The contributions of the posters will not be disregarded in the paper.

Right I do have more questions:

In fact I do have one. Even though you are saying $$E=M\phi=Mc^2$$ is technically correct, we are to ignore any contribution to momentum, which seems like a strange and incorrect statement: In our normal notation, it would be:

$$E=M\Delta\phi(x)$$

since

$$\Delta \phi(x) =\phi(x) \rightarrow \phi*(x)$$

which through substitution, gives a more simplistic form

$$\Delta E=M\phi(\Lambda^{-1} x)$$

There is also now the question of what energy has been taken by the shift. If the change in the Hamiltonian of the system was purely gravitational and no other added energies, then the shift mathematically can be given as:

$$\phi(x) \rightarrow \phi*(x)=\phi(\Lambda^{-1} x)$$

where the inverse $$\Lambda^{-1}$$ states that the field has definitely been shifted. If $$\phi$$ is no longer arbitrary, we can state it is actually the gravitational field, then it's also an energy perturbation of the field as well:

$$\Delta E_g=M_g\phi(\Lambda^{-1} x)$$

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Goldstone1 said:
Even though you are saying $$E=M\phi=Mc^2$$ is technically correct, [...]

I didn't say that this was correct, I provided you with a modified version I could make sense of.

<h2>1. What is a "shifting field gradient"?</h2><p>A shifting field gradient refers to a change in the strength or direction of a magnetic field over a given distance or area.</p><h2>2. How does shifting field gradients affect particles in the field?</h2><p>Shifting field gradients can cause particles in the field to experience a force, which can cause them to move or rotate in a certain direction.</p><h2>3. What is the purpose of shifting field gradients in scientific research?</h2><p>Shifting field gradients are often used in scientific research to manipulate and control the movement of particles for various purposes, such as studying their behavior or separating them for analysis.</p><h2>4. How is the angle of the shifting field gradient determined?</h2><p>The angle of the shifting field gradient is typically determined by the design of the magnetic field source and the parameters of the experiment, such as the desired force or movement of the particles.</p><h2>5. Are there any potential applications for shifting field gradients outside of scientific research?</h2><p>Yes, shifting field gradients have potential applications in various industries such as medical imaging, where they are used in MRI machines to produce detailed images of the body's tissues and organs.</p>

## 1. What is a "shifting field gradient"?

A shifting field gradient refers to a change in the strength or direction of a magnetic field over a given distance or area.

## 2. How does shifting field gradients affect particles in the field?

Shifting field gradients can cause particles in the field to experience a force, which can cause them to move or rotate in a certain direction.

## 3. What is the purpose of shifting field gradients in scientific research?

Shifting field gradients are often used in scientific research to manipulate and control the movement of particles for various purposes, such as studying their behavior or separating them for analysis.

## 4. How is the angle of the shifting field gradient determined?

The angle of the shifting field gradient is typically determined by the design of the magnetic field source and the parameters of the experiment, such as the desired force or movement of the particles.

## 5. Are there any potential applications for shifting field gradients outside of scientific research?

Yes, shifting field gradients have potential applications in various industries such as medical imaging, where they are used in MRI machines to produce detailed images of the body's tissues and organs.

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