Is the kinetic energy of an electron the same in all frames reference?

[LoadedAnvils]

Homework Statement

Is the kinetic energy of an electron the same in all frames of reference?

Homework Equations

I think that the kinetic energy of an electron is E = hc / λ. However, I am not sure: I got this from searching Google rather than learning it myself.

The Attempt at a Solution

I think that, due to length contraction, the wavelength is different in both frames, and hence it is not the same in all frames of reference. However, I am not sure about this.

 

[Mute]

What is the kinetic energy of an electron if you are in a co-moving frame with it? i.e., if you are "running" alongside an electron at the same speed as it, what speed is it going relative to you? Does it have any kinetic energy?

 

[LoadedAnvils]

if you are "running" alongside an electron at the same speed as it, what speed is it going relative to you?

Zero.

If I considered classical kinetic energy then I would say that the kinetic energy is 0. However, I know that this is not the case.

Can you be more clear because I am not sure to which conclusion you are leading me towards.

 

[Mute]

Zero.

If I considered classical kinetic energy then I would say that the kinetic energy is 0. However, I know that this is not the case.

Can you be more clear because I am not sure to which conclusion you are leading me towards.

But it is the case. There's no such thing as a privileged reference frame. This means that if you are in the same frame as an electron its kinetic energy is zero. However, from another frame of reference its kinetic energy will not be zero. This tells you that the kinetic energy of an object is not frame-invariant. This is true even in special relativity. In special relativity, it's the magnitude of the four-momentum that is invariant: $(E/c)^2 - p^2 = (mc)^2$, where $E$ is the total energy.

So what about the matter-wave relation, then? The matter-wave relation (the de Broglie relation) is $\lambda = h/p = (hc)/(pc)$. Now, we know $pc = \sqrt{E^2 - (mc^2)^2}$, where E is again the total energy, from relativity. We can also show that $E = KE + mc^2$. With some manipulations you can then show

$$pc = KE \sqrt{1-2\frac{mc^2}{KE}}$$.

This means that your assumption that $KE = hc/\lambda$ isn't quite true. It's only (approximately) true if the particle's kinetic energy (in the lab frame) is much greater than its rest energy.

But, let's make that approximation anyways - basically, let's say we're in the non-relativistic limit. The problem you are having, then, is that if kinetic energy is not invariant, then it would seem like wavelength is not invariant also. Should wavelength be invariant? Since you posted this in the homework forum, I'll have to try and guide you to the answer.

Consider, say, a free (massive) particle moving in one dimension. The corresponding Schrödinger equation is

$$\hat{p}^2 \psi(x,t) = -\frac{\hbar^2}{2m} \psi(x,t) = i\hbar \frac{\partial \psi(x,t)}{\partial t}$$.

What's the solution to this equation? What sets the particle's energy - and hence, wavelength? What would happen if we put the particle in some potential?