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.

 

[paranormal]

Your approach seems correct so far. However, it would be helpful to explain the reasoning behind each step, especially for someone who may not be familiar with the equations and concepts involved. Also, it would be useful to state the units of each quantity involved, as this is important in physics. Additionally, it would be good to mention that this equation is known as the Planck-Einstein relation, which relates the energy of a photon (represented by E) to its frequency (represented by ω).