Is My Derivation of the Planck Time and Charge Relationships Correct?

[ManyNames]

Homework Statement

Just need to know if i have derived this correctly.

Homework Equations

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}

Cheers in advance!

 

[CompuChip]

What have you derived? And from what?

 

[ManyNames]

What have you derived? And from what?

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}

 

[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}

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)

 

[ManyNames]

No takers?

 

[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)

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.