How Do Mass and Length Correspond in Physics?

[latentcorpse]

The electron mass m_e = 10^6eV translates to a length scale \lambda_e = 2 \times 10^{-12}m

and the Planck scale has M_p=10^19GeV which translates to a length l_p \tilde 10^{-33}cm

i don't know how to show these masses correspond to these length scales. I'm sure it's fairly straightforward...can anyone help out?

thanks.

 

[dextercioby]

Use the formula

\lambda=\frac{\hbar}{mc}

You should be getting the 2 numerical values you mention.

 

[latentcorpse]

yes but we are working in natural units so the formula would become \lambda=\frac{1}{m}, right?

then \lambda_e = \frac{1}{10^6}=10^{-6} which isn't what we want - so I assume I need to change the units of mass around or something?

 

[vela]

You can't use \hbar=c=1 if your energies are still in units of eV.

A useful combination you should know is \hbar c = 197~\mathrm{MeV~fm} = 197~\mathrm{eV~nm}.

 

[latentcorpse]

You can't use \hbar=c=1 if your energies are still in units of eV.

A useful combination you should know is \hbar c = 197~\mathrm{MeV~fm} = 197~\mathrm{eV~nm}.

i thought that when we were in natural units, energy and mass were both measured in eV?
also how will \hbar c help when we have \frac{\hbar}{c}?

 

[vela]

Maybe I'm misremembering, but my recollection is in natural units, you measure everything in terms of length. Maybe it's a theorist vs. experimentalist thing.

In any case, you can't just ignore the units you're given. In natural units, you have λ=1/m, so if λ is in units of length, then m is in units of 1/length, which is obviously not the same as an eV. You can't just plug the numbers in blindly. You have to convert the quantities to the proper units first.

\hbar c is a more natural combination than \hbar/c. Take the formula bigubau gave. What is mc? Nothing readily identifiable. On the other hand, mc² you know is the rest energy of the particle. In terms of this quantity, you'd have

\lambda = \frac{\hbar c}{mc^2}

Note how the combination \hbar c appears in the numerator. This isn't an isolated instance. The same combination pops up all over the place.

You can also think of \hbar c as a conversion factor since

1 = \hbar c = 197~\mathrm{MeV~fm}

or equivalently

197~\mathrm{MeV} = 1~\mathrm{fm}^{-1}

 

[latentcorpse]

Maybe I'm misremembering, but my recollection is in natural units, you measure everything in terms of length. Maybe it's a theorist vs. experimentalist thing.

In any case, you can't just ignore the units you're given. In natural units, you have λ=1/m, so if λ is in units of length, then m is in units of 1/length, which is obviously not the same as an eV. You can't just plug the numbers in blindly. You have to convert the quantities to the proper units first.

\hbar c is a more natural combination than \hbar/c. Take the formula bigubau gave. What is mc? Nothing readily identifiable. On the other hand, mc² you know is the rest energy of the particle. In terms of this quantity, you'd have

\lambda = \frac{\hbar c}{mc^2}

Note how the combination \hbar c appears in the numerator. This isn't an isolated instance. The same combination pops up all over the place.

You can also think of \hbar c as a conversion factor since

1 = \hbar c = 197~\mathrm{MeV~fm}

or equivalently

197~\mathrm{MeV} = 1~\mathrm{fm}^{-1}

thanks. i still seem to be a little bit out

\lambda = \frac{\hbar c}{E} \frac{197 \times 10^6 eVfm}{10^6 eV} \simeq 200 fm = 200 \times 10^{-15} m = 2 \times 10^{-13}m

and for the other one

\lambda = \frac{\hbar c}{10^{19} \times 10^9 eV} = \frac{197 \times 10^6 eVfm}{10^{22} \times 10^6 eV} = \frac{197}{10^{22}} \times 10^{-15}m \simeq 200 \times 10^{-37} m = 2 \times 10^{-35}m

 

[vela]

Those are right. Note the Planck length you were given was in centimeters, not meters. I'd gather the other answer was just a typo.