# A Plankian Derivation

1. Apr 11, 2009

### ManyNames

1. The problem statement, all variables and given/known data

Just need to know if i have derived this correctly.

2. Relevant equations

3. The attempt at a solution

Taken the dimensions of the square of the mass-energy formula, and applying the equation to both sides (even though approximated) the equation of $$\frac{\lambda}{hf}$$ which in this case is also squared, which yields $$h^2c^2$$ Strictly using Natural Units, is then state:

$$h^2c^2(E^2) \approx \lambda^2 E^2(M^2c^4)$$

Now by simple derivation, divide both sides by the wavelength $$\lambda$$, so that

$$\frac{h^2}{\lambda}\frac{c^2}{\lambda}\frac{(E^2)}{\lambda} \approx \frac{\lambda^2}{\lambda}\frac{E^2}{\lambda}(\frac{M^2c^4}{\lambda})$$

which gives numerically:

$$E^2(\frac{M^2c^4}{\lambda})=\lambda E^2$$

since $$E=\frac{hc}{\lambda}$$ and also because $$\frac{\lambda^2}{\lambda}\frac{E^2}{\lambda}$$ reduces to $$\lambda E^2$$.

And finally, using the same equations, instead of dividing the derivation by the wavelength, i divided it by $$2\pi$$, and the result was in this expression:

$$\hbar^2c^2\frac{E}{2 \pi}$$

Last edited: Apr 11, 2009
2. Apr 12, 2009

### CompuChip

What have you derived? And from what?

3. Apr 13, 2009

### ManyNames

equation 1 in the OP requires to know that

$$hc=\lambda E$$

and to know we are using the sqaured value of $$E$$ in the generalized form $$E=Mc^2$$.

Knowing this, then equation 1 is simply the plugging in of these values. The rest of the commands of the derivation after this to equation 3, the equation yields $$\lambda E^2$$, simply a relation to sqaured value of $$E$$ and it's wavelength $$\lambda$$.

The final derivation took a different course. Using equation 1 again, the division of $$2\pi$$ gives:

$$\sqrt{\hbar^2 c^2(\frac{E}{2\pi})$$

(knowing that $$\frac{h^2}{2\pi}=\hbar$$)

which became an expression which leads to an equivalance between the kinetic energy

$$\hbar c(\frac{KE}{2\pi})=\sqrt{\hbar^2 c^2(\frac{E}{2\pi})}$$

If the equation was now squared on both sides, i came to the derivational expression of:

$$\hbar^2c^2 \frac{E}{2\pi}$$

Last edited: Apr 13, 2009
4. Apr 13, 2009

### ManyNames

I would also like approval of the following derivation, where i have derived the planck time and found relationships between the derivations to solve for the quantization of planck charge:

I define to begin with, the gravitational constant:

$$\frac{M - c(\hbar)}{M} = G$$

By rearrangement we can have:

$$M - c \hbar = GM$$ (1)

This is obviously quite a large value, if not quantized. Equation (1) can be rearranged also:

$$c \hbar = M(G+1)$$

We are simply returning the $$M$$ to the right hand side, but expressed in brackets that will distribute it. Therergo, we have:

$$c \hbar = GM^2$$

Now multiply $$c^2$$ to both sides:

$$c^2 \dot c \hbar = GMc^2$$

so that

If $$G(Mc^2)$$ is true, then it is the same as $$G(E)$$. Therefore, the following must also be true:

$$G(E)= \hbar c^3= GMc^2 = \sqrt{h^2 f^2 G^2}$$

because $$E=hf$$.

Knowing that $$f^2 \cdot G^2h^2 = h^2f^2 \cdot G^2$$ then now take away $$(f)$$ from both sides and rearrange:

$$G^2h^2 = h(G+1)$$

Now divide both sides by the quantization of $$c^5$$, and you have

$$\sqrt{G^2h^2/c^5}= t_{pl}$$

Which is exactly the Planck Time.

Now moving on to charge relationships, Knowing the previous work, one can derive this set of relationships:

$$\frac{M - \hbar c}{M} = \frac{M - q^2}{M} :\ [\sqrt{\hbar c} = q]$$

Where $$q$$ is Plancks Charge, we see that the square root of $$\hbar$$ is the precise value of the quantization of the charge. Since the fine structure constant is so much larger than the gravitational constant, $$\alpha > \alpha_g$$, where $$\alpha_g$$ here denotes the gravitational coupling constant, it seems interesting to note that:

$$\frac{G\sqrt{M^2}}{\hbar c} = \frac{GM}{\sqrt{M(G+1)}}=\frac{GM}{q}$$

which would be a relation to the charge again, and this leads to my final set of relations:

$$\frac{GM}{\sqrt{M(G+1)}}=\frac{G \sqrt{\alpha_g}}{q}$$

Does this all seem right?

(I had to edit for the expression of the planck time. I certainly didn't know in latex the expression P_t lead to... is it the number for silver... anyway, redited)

5. Apr 14, 2009

### ManyNames

No takers?

6. Apr 14, 2009

### ManyNames

I bolded this part becuse i've realized i've made some mistakes. So it's totallly invalid now. But the planck charge derivation seems correct.