Why is the equation for a particle's wave function given by ##\nu = E/h##?

[Isaac0427]

I understand that E=mc^2 and E=hv can't be used to set mc^2 equal to hv, but why would the total equation be E=(mc^2)^2+(hv)^2 instead of E=mc^2+hv? I'm sorry if this question is stupid.

 

[Khashishi]

Why is the Pythagorean theorem $c^2 = a^2 + b^2$ instead of $c = a + b$? That's just how space works.

$E^2 = (mc^2)^2 + (pc)^2$ is just an application of the Pythagorean theorem in spacetime.

 

[Isaac0427]

Why is the Pythagorean theorem $c^2 = a^2 + b^2$ instead of $c = a + b$? That's just how space works.

$E^2 = (mc^2)^2 + (pc)^2$ is just an application of the Pythagorean theorem in spacetime.

Ok, thanks. By the way, what is the difference between pc and hv? I see them used interchangeably.

 

[Khashishi]

Ok, I gave a misleading answer and I think I need to address it. Space-time has a different metric than space, and Pythagorean theorem works differently. In familiar 3D, the metric is $dr^2 = dx^2 + dy^2 + dz^2$ so Pythagorean theorem is as above.
But in 4D space time, time has an opposite sign to space, so the corresponding metric is $ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$ or $ds^2 = dt^2 - dx^2 - dy^2 - dz^2$ depending on the convention you adopt.

So you can rewrite $E^2 = (mc^2)^2 + (pc)^2$ as
$(mc^2)^2 = E^2 - (pc)^2$
$(mc^2)^2 = E^2 - (p_xc)^2 - (p_yc)^2 - (p_zc)^2$

The invariant mass "m" is the 4D length of the energy and momentum 4-vector. But length is calculated using something like the Pythagorean theorem but with negative sign in front of the momentum. This is a result of the negative sign in the metric. This is clearer if you drop all the c constants from the equation.
$m^2 = E^2 - p_x^2 - p_y^2 - p_z^2$
You can drop c if you measure length and time in compatible units. If you measure time in seconds, you should measure length in light-seconds, and then you can set c = 1 light-second/second and then make it go away.

 

[Khashishi]

You shouldn't be using $h\nu$ in this equation. It isn't correct to interchange $h\nu$ with $pc$ except for massless particles.

 

[Isaac0427]

Thank you.

 

[jtbell]

why would the total equation be E=(mc^2)^2+(hv)^2

Who says it looks like that?

 

[Isaac0427]

Who says it looks like that?

My bad, I meant E^2

 

[my2cts]

E=hv also for a massive particle.

 

[nasu]

What is "v" here?

 

[jtbell]

My bad, I meant E^2

I was referring to the (hv)^2 part, not the E on the left which should indeed be E^2.

What is "v" here?

I'm pretty sure it's supposed to mean "nu" ($\nu$) for frequency, not "v" for velocity.

 

[my2cts]

$\nu$

 

[Isaac0427]

E=hv also for a massive particle.

What do you mean?

 

[jtbell]

A particle's wave function has a time-dependent part whose frequency is given by $\nu = E/h$.

For example, a free particle has $\Psi(x,t) = Ae^{i(px-Et)/\hbar} = Ae^{2 \pi i(px-Et)/h}$ so the time dependent part is $e^{-2 \pi i Et / h}$. One form of an oscillator in complex-number space is $e^{-i \omega t} = e^{-2 \pi i \nu t}$. Compare the two, and you get $\nu = E/h$.