# Question on shifting field gradients by an angle

1. Jun 20, 2011

### 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?

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

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

2. Jun 20, 2011

### Polyrhythmic

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

3. Jun 20, 2011

### Goldstone1

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?

4. Jun 20, 2011

### Polyrhythmic

What you refer to as Dirac sea is an outdated concept.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.

5. Jun 20, 2011

### Goldstone1

It's not outdated as such: The principle of the Dirac Sea is still at the core understanding of particles at the quantized level.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.

6. Jun 20, 2011

### Goldstone1

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

7. Jun 20, 2011

### Goldstone1

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?

Last edited: Jun 20, 2011
8. Jun 20, 2011

### Polyrhythmic

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

$E=\sqrt{m²c^4+p^2c^2}$.

9. Jun 20, 2011

### Goldstone1

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?

10. Jun 20, 2011

### Polyrhythmic

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.

11. Jun 20, 2011

### Goldstone1

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.

12. Jun 20, 2011

### Polyrhythmic

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.

13. Jun 20, 2011

### Goldstone1

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

14. Jun 20, 2011

### Polyrhythmic

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

15. Jun 20, 2011

### Goldstone1

So then I have a new question,

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

16. Jun 20, 2011

### Goldstone1

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.

17. Jun 20, 2011

### Polyrhythmic

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.

18. Jun 20, 2011

### Polyrhythmic

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

19. Jun 20, 2011

### Goldstone1

You would not try to expand the notation?>

$E=M_0*\phi$

into the full relativistic case?

20. Jun 20, 2011

### Polyrhythmic

What do you mean?

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