• Support PF! Buy your school textbooks, materials and every day products Here!

A Plankian Derivation

  • Thread starter ManyNames
  • Start date
  • #1
136
0

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 [tex]\frac{\lambda}{hf}[/tex] which in this case is also squared, which yields [tex]h^2c^2[/tex] Strictly using Natural Units, is then state:

[tex]h^2c^2(E^2) \approx \lambda^2 E^2(M^2c^4)[/tex]

Now by simple derivation, divide both sides by the wavelength [tex]\lambda[/tex], so that

[tex]\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})[/tex]

which gives numerically:

[tex]E^2(\frac{M^2c^4}{\lambda})=\lambda E^2[/tex]

since [tex]E=\frac{hc}{\lambda}[/tex] and also because [tex]\frac{\lambda^2}{\lambda}\frac{E^2}{\lambda}[/tex] reduces to [tex]\lambda E^2[/tex].

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

[tex]\hbar^2c^2\frac{E}{2 \pi}[/tex]

Cheers in advance!
 
Last edited:

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
4,302
47
What have you derived? And from what?
 
  • #3
136
0
What have you derived? And from what?
equation 1 in the OP requires to know that

[tex]hc=\lambda E[/tex]

and to know we are using the sqaured value of [tex]E[/tex] in the generalized form [tex]E=Mc^2[/tex].

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 [tex]\lambda E^2[/tex], simply a relation to sqaured value of [tex]E[/tex] and it's wavelength [tex]\lambda[/tex].

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

[tex]\sqrt{\hbar^2 c^2(\frac{E}{2\pi})[/tex]

(knowing that [tex]\frac{h^2}{2\pi}=\hbar[/tex])

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

[tex]\hbar c(\frac{KE}{2\pi})=\sqrt{\hbar^2 c^2(\frac{E}{2\pi})}[/tex]

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

[tex]\hbar^2c^2 \frac{E}{2\pi}[/tex]
 
Last edited:
  • #4
136
0
equation 1 in the OP requires to know that

[tex]hc=\lambda E[/tex]

and to know we are using the sqaured value of [tex]E[/tex] in the generalized form [tex]E=Mc^2[/tex].

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 [tex]\lambda E^2[/tex], simply a relation to sqaured value of [tex]E[/tex] and it's wavelength [tex]\lambda[/tex].

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

[tex]\sqrt{\hbar^2 c^2(\frac{E}{2\pi})[/tex]

(knowing that [tex]\frac{h^2}{2\pi}=\hbar[/tex])

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

[tex]\hbar c(\frac{KE}{2\pi})=\sqrt{\hbar^2 c^2(\frac{E}{2\pi})}[/tex]

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

[tex]\hbar^2c^2 \frac{E}{2\pi}[/tex]
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:

[tex]\frac{M - c(\hbar)}{M} = G[/tex]

By rearrangement we can have:

[tex]M - c \hbar = GM[/tex] (1)

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

[tex]c \hbar = M(G+1)[/tex]

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

[tex]c \hbar = GM^2[/tex]

Now multiply [tex]c^2[/tex] to both sides:

[tex]c^2 \dot c \hbar = GMc^2[/tex]

so that

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

[tex]G(E)= \hbar c^3= GMc^2 = \sqrt{h^2 f^2 G^2}[/tex]

because [tex]E=hf[/tex].

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

[tex]G^2h^2 = h(G+1)[/tex]

Now divide both sides by the quantization of [tex]c^5[/tex], and you have

[tex]\sqrt{G^2h^2/c^5}= t_{pl}[/tex]

Which is exactly the Planck Time.

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

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

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

[tex]\frac{G\sqrt{M^2}}{\hbar c} = \frac{GM}{\sqrt{M(G+1)}}=\frac{GM}{q}[/tex]

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

[tex]\frac{GM}{\sqrt{M(G+1)}}=\frac{G \sqrt{\alpha_g}}{q}[/tex]

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
136
0
No takers?
 
  • #6
136
0
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:

[tex]\frac{M - c(\hbar)}{M} = G[/tex]

By rearrangement we can have:

[tex]M - c \hbar = GM[/tex] (1)

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

[tex]c \hbar = M(G+1)[/tex]

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

[tex]c \hbar = GM^2[/tex]

Now multiply [tex]c^2[/tex] to both sides:

[tex]c^2 \dot c \hbar = GMc^2[/tex]

so that

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

[tex]G(E)= \hbar c^3= GMc^2 = \sqrt{h^2 f^2 G^2}[/tex]

because [tex]E=hf[/tex].

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

[tex]G^2h^2 = h(G+1)[/tex]

Now divide both sides by the quantization of [tex]c^5[/tex], and you have

[tex]\sqrt{G^2h^2/c^5}= t_{pl}[/tex]


Which is exactly the Planck Time.

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

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

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

[tex]\frac{G\sqrt{M^2}}{\hbar c} = \frac{GM}{\sqrt{M(G+1)}}=\frac{GM}{q}[/tex]

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

[tex]\frac{GM}{\sqrt{M(G+1)}}=\frac{G \sqrt{\alpha_g}}{q}[/tex]

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.
 
Top