1
$\begingroup$

I might be misunderstanding a basic concept here, so forgive me. I know that the faster an object gets, the more it's dimensions will contract according to the following equation:

$${1\over D} = 1-{V^2\over C^2} $$

For point-like particles, how does this relationship behave? We can surely halt the motion of an electron, and quickly speed it up to near the speed of light, but does it's dimensions or any other of it's attributes change at all?

$\endgroup$
2
$\begingroup$

The proper formula for length contraction is: $$l=\frac{l_0}{\sqrt{1-v^2/c^2}}$$ Where $l$ is the measured length in the moving frame and $l_0$ is the rest length of the object. If $l_0=0$ you can see immediately that $l=0$. There is no length contraction for objects with no length.

$\endgroup$
1
$\begingroup$

The electron is in reality not a point particle, but is spread out in space. The function we use to describe it is also spread out in space, and upon changing reference frames, the shape of this function can stretch as well. You can imagine it a bit like a cloud which changes shape depending on the reference frame you are in. However, mathematically, it is more $\mathbb{C}$omplex than that, and I have only written an analogy to physicists' description of a particle.

If you are curious to learn what exactly is spread out in space: roughly, it is a probability density. More accurately, in quantum mechanics, the object we use to predict experimental outcomes on particles is a function $\psi(x)$ which assigns a complex number to each point in space.

electron orbitals

Above is a picture of an electron cloud in different configurations, within a hydrogen atom. Even though the cloud extends throughout ALL space, this picture highlights the area with the highest probability.

$\endgroup$
0
$\begingroup$

Electron is not a point. It does have dimensions, though very tiny.

Length contraction is experimentally observed in case of subatomic particles called muons/mesons, where their density increases due to flattening of their shape.

$\endgroup$
  • 1
    $\begingroup$ Why was this downvoted? $\endgroup$ – doublefelix Nov 15 '18 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.