Understanding 4-Vector Momentum^2

[Ene Dene]

I'm having a problem understanding this:

$$P^2=P_{\mu}P^\mu=m^2$$

If we take c=1.

Here is what bothers me:

$$P(E, \vec{p})=E^2-(\vec{p})^2$$

Now, I assume that E=mc^2, and for c=1, E^2=m^2? Is that correct?

And I don't know what p^2 is, I look at it as:

$$(\vec{p})^2=m^2(\vec{v})^2$$

What am I doing wrong?

 

[nrqed]

I'm having a problem understanding this:

$$P^2=P_{\mu}P^\mu=m^2$$

If we take c=1.

Here is what bothers me:

$$P(E, \vec{p})=E^2-(\vec{p})^2$$

Now, I assume that E=mc^2, and for c=1, E^2=m^2? Is that correct?

And I don't know what p^2 is, I look at it as:

$$(\vec{p})^2=m^2(\vec{v})^2$$

What am I doing wrong?

I am not exactly sure what bothers you but one point: $$E = \gamma m c^2$$, not mc^2 (which is valid only in the rest frame of the particle). Also the momentum is the relativistic three-momentum so it's $\gamma m \vec{v}$

 

[Meir Achuz]

Your first equation shows m^2=E^2-p^2.
Your assumption E=m is wrong.

 

[dynamicsolo]

I'm having a problem understanding this:

$$P^2=P_{\mu}P^\mu=m^2$$

If we take c=1.

Here is what bothers me:

$$P(E, \vec{p})=E^2-(\vec{p})^2$$

Now, I assume that E=mc^2, and for c=1, E^2=m^2? Is that correct?[/tex]

The symbol 'E' in this equation is the total energy of the particle. The popularly-known equation E = mc^2 refers to the "rest mass-energy of the particle" and is really an incorrect use of the symbol.

'p' here is just the regular ol' 3-momentum, mass times the 3-dimensional velocity vector for the particle. The so-called "4-momentum" $$P^{\mu}$$ is a 4-dimensional vector whose components are

( iE, p_x, p_y, p_z ) [or flip signs depending on whose notational convention you use].

Your original relation then,

$$P^2=P_{\mu}P^{\mu}$$

is then just taking the "dot product" of P with itself to get the square of the magnitude,

$$P^2 = E^2 - p^2 = (mc^2)^2$$ , again with appropriate adjustments for local notational practice.

 

[CompuChip]

As already said, the popular formula E = mc^2 -- mostly misquoted -- is a) not generally applicable and b) not even a main result of special relativity, it's more like a small remark buried somewhere deep inside the text.

To stick with Ene Dene's approach: if you plug in the correct formula
$$E = \sqrt{ (\gamma m c^2)^2 + (m p^2)^2 }$$
you will get a consistent result.
In units where c = 1, it'd be
$$E^2 = \gamma m^2 + p^4$$
and
$$p = \gamma m v$$.
Also note that you can deduce where E = mc^2 is applicable; the general formula reduces to it in the rest frame (p = 0) at non-relativistic speeds ($\gamma \approx 1$).