Momentum 4 Vector Notation: Differences Explained

[PAK108]

I have noticed three ways to write the momentum 4 vectors
i. P = (E/c, p)
ii. P = (E, p)
iii. P = (E, pc)

The three ways to write the momentum 4 vectors is really confusing since I know how to derive the
way it is written in the form as (i). But the two other forms (ii and iii) does not make any sense at all.
Could someone please "enlighten" me in this matter?

 

[vela]

Physicists quite often use units where $c=1$, in which case (ii) is the four-momentum. The last one, to me, is wrong because the units don't really work out unless $c=1$ but then why bother writing the factor of $c$?

 

[PAK108]

Physicists quite often use units where $c=1$, in which case (ii) is the four-momentum. The last one, to me, is wrong because the units don't really work out unless $c=1$ but then why bother writing the factor of $c$?

Thanks for answering :)
Regarding equation 3, it is actually from my homework in modern physics. I had to use this equation to solve a comptom scattering problem made by my professor. But it is just like you said, it also does not make any sense to me...

 

[Orodruin]

Well, the last option is also built from the components of a 4-vector which differs from the 4-momentum only by a constant. Some might say it is convenient to use it as its magnitude is $mc^2$, even if it has the dimensions of energy instead of momentum.

Why you would use units where c is not one is a mystery though.

 

[PAK108]

Well, the last option is also built from the components of a 4-vector which differs from the 4-momentum only by a constant. Some might say it is convenient to use it as its magnitude is $mc^2$, even if it has the dimensions of energy instead of momentum.

Why you would use units where c is not one is a mystery though.

Thanks :)

 

[vanhees71]

Well, in introductory lectures it's good practice to keep $c \neq 1$ (it's bad practice in these days to use the SI in the theory lecture about electromagnetism, but that's another topic). Then, it's a matter of convention, how you define the four vectors. It's pretty common to have the space-time four-vectors with dimension length, i.e., $(x^{\mu})=(c t,\vec{x})$. It's not so clear to me, what's preferred for the four-momentum vector. I think it's a bit more common to use the dimensions of the spatial piece, i.e., $(p^{\mu})=(E/c,\vec{p})$, of course since $c$ is a scalar, also $c p^{\mu}$ are four-vector components. In electromagnetism the charge-current-density vector usually one chooses the dimension as that of a current density, i.e., $j^{\mu}=(c \rho,\vec{j})$.

In short, you have to check your textbook/paper carefully to figure out which convention is used.

 

[PAK108]

In short, you have to check your textbook/paper carefully to figure out which convention is used.

Thanks for answering. My homework exercises starts with P = (E/c, p), but suddenly they started to use P = (E, pc). Not only that, they use a different convention than my textbook uses for many of the formulas. So I cannot hep it but get confused time to time

 

[Orodruin]

Thanks for answering. My homework exercises starts with P = (E/c, p), but suddenly they started to use P = (E, pc). Not only that, they use a different convention than my textbook uses for many of the formulas. So I cannot hep it but get confused time to time

I suggest using units where $c = 1$ and all your confusions regarding powers of $c$ will disappear. If you want to insist on having the appropriate powers of $c$ in your answers, you can always reinsert them using dimensional analysis.

 

[PAK108]

I suggest using units where $c = 1$ and all your confusions regarding powers of $c$ will disappear. If you want to insist on having the appropriate powers of $c$ in your answers, you can always reinsert them using dimensional analysis.

Thanks! I shall do that from now on :)

 

[jtbell]

even if it has the dimensions of energy instead of momentum

At least back when I was in grad school, most of us experimentalists did say things like, "the momentum of this particle is xxx MeV."

 

[Orodruin]

At least back when I was in grad school, most of us experimentalists did say things like, "the momentum of this particle is xxx MeV."

Most of us theorists do this too, because we like to set $c = 1$. In addition, masses are often quoted in GeV instead of GeV/c².

 

[PeterDonis]

The last one, to me, is wrong because the units don't really work out

Yes, they do; the units of E, energy, are the units of p, momentum, times the units of c, velocity. So both components have units of energy.

 

[vela]

I was thinking if you're not using $c=1$, then the four-momentum should have units of momentum, not energy. That's all I meant.

 

[PeterDonis]

I was thinking if you're not using $c=1$, then the four-momentum should have units of momentum, not energy.

It's more a matter of preference than anything else. The most common units in particle physics are energy units, and with that convention form (iii) in the OP would be the most natural if you weren't using $c = 1$. The term "four-momentum" should not be interpreted too literally; for one thing, it's not the only name for that 4-vector, other names I've seen are "energy-momentum 4-vector" and "momenergy". And the invariant length of this vector is usually referred to as "invariant mass" or "rest mass", implying mass units, which would imply components of (E/c^2, p/c) if we're not using $c = 1$.