Question about units, if the units are given in the equation or as

  • Thread starter Thread starter rwooduk
  • Start date Start date
  • Tags Tags
    Units
rwooduk
Messages
757
Reaction score
59
... part of the value?

first example:

v (MHz) = 110 B(nT) E^2 (GeV)

could someone confirm that when an equation is like this you simply change E into GeV before squaring it and making the calculation? slightly confused.

second example:

rest mass of a Pion is stated by Wiki as 139.57018(35) MeV/c^2, why is this? are there units of c^2? what if you want to use the mass in a calculation, do you divide 139MeV by the speed of light squared? not sure what to do with this.

Thanks for any help.
 
Last edited:
Physics news on Phys.org
If E is given in "GeV" then squaring will give units of "GeV squared". Of course, they should then combine with whatever units you have for "B(nT)" to give the correct units for v.

MeV is a measure of a type of energy which has basic MKS units "kilogram meters per second squared" so that dividing by a speed, squared, like c^2, leaves mass units: kilograms.
 
  • Like
Likes 1 person
HallsofIvy said:
If E is given in "GeV" then squaring will give units of "GeV squared". Of course, they should then combine with whatever units you have for "B(nT)" to give the correct units for v.

MeV is a measure of a type of energy which has basic MKS units "kilogram meters per second squared" so that dividing by a speed, squared, like c^2, leaves mass units: kilograms.

thanks for that!

so why is it given over c^2 and not in kilograms? and also if there is a calculation where the pion mass is multiplied by c^2, do the c^2's simply cancel?

thanks again.
 
HallsofIvy said:
MeV is a measure of a type of energy which has basic MKS units "kilogram meters per second squared" so that dividing by a speed, squared, like c^2, leaves mass units: kilograms.
Correction: "kilogram meters per second squared" -- that's the Newton (i.e., units of force). Energy has dimensionality mass*velocity2, or units of kilogram*meters squared / seconds squared.

rwooduk said:
so why is it given over c^2 and not in kilograms?
Metric units can be rather inconvenient in particle physics. Expressing mass in terms of energy/c2 tells a much better story. For example, how much energy is released in an annihilation event? Expressing rest mass in energy/c2 is much more informative than expressing it in kilograms. In fact, in mass is expressed in electron-volts rather than electron-volts/c2; that missing division by c2 is implied.
 
  • Like
Likes 1 person
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
Back
Top