How do you derive the units for momentum for photons

[TheCelt]

Homework Statement:

Units for photon momentum confuse me

Relevant Equations:

p=hf/c = E/c

So if i have a photon of some energy and i want to find the magnitude of the momentum, i can get the right answer but the units don't make sense.

So i derive p = E/c since i know the energy of the photon and i used f=E/h and substituted this into p=hf/c

This means for units of the equation p = E/c i get:

kg m/s = J / (m/s) = J s/m

This is confusing me since I've always been told to check my units but these equations don't have the same units, so I don't know if i have done something wrong, or momentum has different units applied for photons? Hope some one can explain what's going on here with the units.

Thanks

 

[DaveE]

OK, so can you express the Energy units in terms of the basic units kg, m, sec instead of Joules? Then you can simplify the left side of your last equation.

Hint: I thought it was useful to think of Energy=Force*Distance, but there are other ways too. Perhaps the formula for kinetic energy?

 

[TheCelt]

Oh true i can use kinetic energy to match them! Thanks! :)

 

[DaveE]

Resolving units like this is a great application for all of those simple formulas that you've memorized. This problem had nothing to do with kinetic energy, but that formula told you the equivalence of some of the units you did have.

So, for example, how can you express Volts in basic SI units? Well, I know that Volts*Amps=Watts=Energy/sec, so Volts=Energy/(Amp*sec)=(kg*m²)/(Coulomb*sec²).

 

[haruspex]

these equations don't have the same units

Yes they do.
The set of standard units in any system is richer than it needs to be. While it is usual to express pressure in Pascals, there's nothing to stop you using the same number of kg m^-1 s^-2 or J m^-3 or any other equivalent.