# Bohr radius and length contraction

from wikipedia - "The Bohr radius is a physical constant, approximately equal to the most probable distance between the proton and electron in a hydrogen atom in its ground state."

in planck units bohr radius (a0) = (mp/me).(1/$\alpha$).(lp)
the 1st 2 terms of the RHS of the equation are dimensionless.

now , there are 2 different inertial observers one at rest and the other moving at some uniform velocity relative a hydrogen atom whose radius they are measuring....

would or wouldn't the 2 measurements differ ?
if they differ, wouldn't it mean a change in the term '(lp)' of the above equation ,since the other terms of the RHS of the equation are dimensionless?
if they don't, why? it has the dimensions of length after all? does this mean that an absolute length scale is possible?

bcrowell
Staff Emeritus
Gold Member
The Bohr radius only equals the size of a hydrogen atom if the hydrogen atom is at rest in your frame of reference.

We don't really know what significance the Planck length has, since we don't have a theory of quantum gravity. Some people have suggested that it might be a minimum distance scale, like a granularity of spacetime. An argument against this is that such a distance should suffer length contraction, which wouldn't make sense, because we can't define motion "relative to spacetime." Lorentz transformations do preserve area and volume. (A boost along the x axis conserves area in the x-t plane, and similarly for 3-volume and 4-volume.) Therefore it kind of makes sense that in loop quantum gravity, what's quantized is not length but area and volume.

the 2 observers are measuring the average distance between the electron and the proton of the H-atom relative to them, shouldn't they disagree in their answers (as they are measuring lengths)?

in general, any body in uniform relative motion must detect a decrease in length of its surrounding world......for instance, from the perspective of a very fast moving train, all the outside world appears contracted in length, and from the outside world's perspective, the train appears to be contracted in length parallel to its velocity.
so for the outside world the volume of the train has decreased compared to what it was measured while the train was at rest. And for the observers attached with the moving train, the outside world has shrinked in volume compared to what it was while measured at rest.
but as the number of particles (atoms, for instance) must be the same (while at rest or i motion) irrespective of the state of the motion of the observer, from the perspective of the outside world, shouldn't every atom making the train would have shrinked in volume? and so for the outside world relative to the train?
doesn't this mean that a uniformly moving observer would detect a greater force between the electrons and the protons constituting the atoms of something that is being measured compared to any other observer at rest relative to what is being measured? Won't it lead to different properties of atoms in the 2 different cases?

The Bohr radius is calculated in a frame where the atom is at rest. It is not a relativistically invariant number, and it would suffer a Lorentz contraction.

Here is a similar situation--If you were to look up the mass of a electon in some table of fundamental constants you would get a fixed number, but that does not mean the electron would have that same mass in all frames--the table is really only referring to the mass in the rest frame, the rest mass.

jtbell
Mentor
Physicists nowadays assume that those tables of particle masses refer to the invariant mass given by

$$mc^2 = \sqrt{E^2 - (pc)^2}$$

which gives the same result no matter how fast the particle is moving. This is of course the same number that we call the "rest mass", but the particle doesn't need to be at rest in order for us to calculate it.

"Rest mass" simply refers to what the mass would be *if* the particle *were* at rest. A moving particle thus could have a rest mass.

Our disagreement appears only to be semantical, though.

the 2 observers are measuring the average distance between the electron and the proton of the H-atom relative to themselves, not anything that is called the bohr's radius....!
would not they disagree in their answers?
wouldn't this mean that a uniformly moving observer would detect a greater force between the electrons and the protons constituting the atoms of something that is being measured compared to any other observer at rest relative to what is being measured? Won't it lead to different properties of atoms in the 2 different cases?

Physicists nowadays assume that those tables of particle masses refer to the invariant mass given by

$$mc^2 = \sqrt{E^2 - (pc)^2}$$

which gives the same result no matter how fast the particle is moving. This is of course the same number that we call the "rest mass", but the particle doesn't need to be at rest in order for us to calculate it.

but the 'p' in the above equation is a function of velocity? isn't it?

jtbell
Mentor
Yes, of course. Both E and p depend on velocity. Nevertheless, when you measure E and p for a given kind of particle, e.g. an electron, and then calculate that quantity, you always get the same result. This is one technique for identifying particles in an accelerator experiment (CERN, Fermilab, etc.).

"wouldn't this mean that a uniformly moving observer would detect a greater force between the electrons and the protons constituting the atoms of something that is being measured compared to any other observer at rest relative to what is being measured? Won't it lead to different properties of atoms in the 2 different cases?"

Suppose that in one frame the Lorentz contraction leads to there being a greater force. Well. since the force is gretaer the electron is pulled more strongly to the nucleus, leading physically to....a contraction! So the presense of a contraction (i.e. the Lorentz contraction) implies that there is a contraction! This is a logically consistent state of affairs. Things work out nicely.

One expects shrinkage should somehow be there because one knows of the Lorentz contaction, and the actual solved equations give the shrinkage expected.

"Suppose that in one frame the Lorentz contraction leads to there being a greater force. Well. since the force is gretaer the electron is pulled more strongly to the nucleus, leading physically to....a contraction! So the presense of a contraction (i.e. the Lorentz contraction) implies that there is a contraction! This is a logically consistent state of affairs. Things work out nicely.
One expects shrinkage should somehow be there because one knows of the Lorentz contaction, and the actual solved equations give the shrinkage expected.

but how to reconcile it with the fact that the position of electron inside an atom is quantized and the average separation between the electron and the nucleus cannot change arbitrarily (the radius of an atom cannot change continuously)?

ZapperZ
Staff Emeritus
but how to reconcile it with the fact that the position of electron inside an atom is quantized and the average separation between the electron and the nucleus cannot change arbitrarily (the radius of an atom cannot change continuously)?

There is a more fundamental issue at work here than just tackling this hydrogen atom. You need to figure out how we reconcile the different observations being made in different reference frame. How do you reconcile the fact that one frame observes one length while the other observes a different length. In other words, you really should have done this with a simpler situation than using the hydrogen atom.

The fact that something is moving in one frame is the way we reconcile things, because we now know Relativity! If we don't, then yes, we see a paradox that we can't explain. The fact that we know Relativity means that if we want to reconcile while such-and-such a frame sees what it sees, we simply do a Lorentz transformation! In other words, I look at a coin and I see Head, while you look at the same coin and you see Tail, how do we reconcile that difference? Easy. We both already know that a coin as two sides! To see what the other is seeing, we simply spin the coin 180 degrees, and voila, we see what the other saw!

The same with this situation. If you are in a frame in which the hydrogen atom is moving, you won't get the same result as a stationary frame. However, this should not create any problem, because you are aware that the hydrogen atom is moving, AND, you know Relativity! You know that your frame is not the rest frame of the hydrogen atom, and what you are trying to calculate/measure are not invariant quantities with respect to a Lorentz transformation. There's no reason why they should be the same!

It is why physics and physicists like to look for invariant parameters, because this type of issues do not appear for these quantities.

Zz.

<<but how to reconcile it with the fact that the position of electron inside an atom is quantized >>

I figured out what you meant by the above, but some people might be confused. In a hydrogen atom, energy, total angular momentum and angular momentum in the "z" direction are quantized, but not position.

<<of electron inside an atom is quantized and the average separation between the electron and the nucleus cannot change arbitrarily (the radius of an atom cannot change continuously)?>>

A "contracted" version of the standard version of the hydrogen wavefunction would have to be a solution to the appropriate wave equation. You can get whatever the equations allows you to get. And we know that the equations must allow you to get these weird wavefunctions because we trust that Lorentz contractions take place.

You might wonder why when one solves the wave equation you get the standard textbook "hydrogen wavefunctions" but not these new ones. For the moving atom, the potential is no longer a simple Q/r, and so the equation for the electron's wavefunction is a different equation, and thus will give different solutions.

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Me: <<For the moving atom, the potential is no longer a simple Q/r>>

I should also note that when the atom is moving, not only is the electric potential changed, but the four-vector of which the electric potential is the zeroth component, a four vector which in the rest frame of atom only had (ignoring the effects of nuclear spin) a non-vanishing zeroth component, now also has a non-vanishing component in the direction of motion. So that too modifies the wave equation.