Solving the Equation: \hbar c = Gm^2

[coolSmith]

Homework Statement

Homework Equations

The Attempt at a Solution

Seems pretty strightforward:
$$\hbar c = Gm^2$$

multiply both sides by $$c^4$$

$$\hbar c^5 = Gm^2c^4$$

knowing that $$m^2c^4$$ is $$E^2=(\hbar \omega)^2$$ then

$$\hbar c^5 = G\hbar^2 \omega^2$$

dividing $$\hbar^2$$ off both sides

$$\frac{c^5}{\hbar}=G\omega^2$$

then this is the same as

$$\frac{c^5}{\hbar}=G\frac{k}{m}$$

because $$\omega^2 = \frac{k}{m}$$ so multiplying $$m$$ on both sides gives

$$\frac{c^5}{\hbar}m=Gk$$

Then finally multiplying $$\hbar$$ on both sides gives you

$$\frac{c^5}{\hbar}m \hbar=Gk\hbar$$

Has there been anything wrong in this so far?

 

[coolSmith]

Basically it was suggested I had my dimensions wrong, but the derivation is so simple, I couldn't see anything wrong with. My question was simple enough I think.. is there anything wrong with it? Either it is wrong and I am doing this with the wrong dimensions, or it is right and there was no problem to begin with..

which is it, please answer!

 

[dextercioby]

What does this derivation do ? What's it good for ?

 

[coolSmith]

I haven't fully decided on this. There are two main features from the final equation written here in the OP.

$$\frac{c^5}{\hbar}M\hbar = G(k\hbar)$$

which in another form is simply

$$\frac{c^5}{\hbar}S = G(k\hbar)$$

This equation has the features that the left handside is in fact just a rescalling of the spin $$S$$ and the right handside has the feature of momentum of any quanta given as $$(\hbar k)$$. So for a specific wave equation, the right hand side may have an implication for

$$\psi(x,t) = Aexp(ikx - i \hbar k2t / 2m)$$

because this has a definite momentum, p = ħk. But because this has the form of spin on the left handside, it can be futher written as

$$\frac{c^5}{\hbar} \frac{\hbar}{2} \sigma_i = G(\hbar k)$$

Here $$\sigma_i$$ is the Pauli Spin Matrices, where the subscript represents any of the three dimensions of space $$\sigma_i = (\sigma_x, \sigma_y, \sigma_z)$$ and of course, if you use all three dimensions, this directly effects the momentum component to make $$(\vec{\hbar k})$$.

Of course, I have thought about other applications.