Verifying $\gamma = \frac{E}{m}$ in QFT

[Jimmy Snyder]

Homework Statement

On page 41 of Ryder's QFT, just below eqn (2.84), it says: $$\gamma = E/m$$ I was unable to verify this, unless it is meant to be true only for small speeds.

Homework Equations

$$E = \pm(m^2c^4 + p^2c^2)^{1/2}$$ (2.24) page 29, but as suggested n the book, we let c = 1, so
$$E = \pm(m^2 + p^2)^{1/2}$$

The Attempt at a Solution

Well, I'm not sure this is legal, but I replaced p with mv as in the classical case, and then v with $$\beta$$.
$$E = \pm(m^2 + m^2v^2)^{1/2} = \pm m(1 + \beta^2)^{1/2} \approx \pm m\gamma$$
But that approximation is only good when v is small.

 

[meopemuk]

Homework Statement

On page 41 of Ryder's QFT, just below eqn (2.84), it says: $$\gamma = E/m$$ I was unable to verify this, unless it is meant to be true only for small speeds.

Homework Equations

$$E = \pm(m^2c^4 + p^2c^2)^{1/2}$$ (2.24) page 29, but as suggested n the book, we let c = 1, so
$$E = \pm(m^2 + p^2)^{1/2}$$

The Attempt at a Solution

Well, I'm not sure this is legal, but I replaced p with mv as in the classical case, and then v with $$\beta$$.
$$E = \pm(m^2 + m^2v^2)^{1/2} = \pm m(1 + \beta^2)^{1/2} \approx \pm m\gamma$$
But that approximation is only good when v is small.

In relativistic physics momentum is not given by classical formula $p = mv$. Try using

$$p = \frac{mv}{\sqrt{1-v^2/c^2}}$$

instead.
Eugene.

 

[Jimmy Snyder]

Got it. Thanks Eugene. Now I have:
$$E = \pm(m^2 + \frac{m^2v^2}{1 - v^2})^{1/2} = \pm m(\frac{1 - v^2 + v^2}{1 - v^2})^{1/2} = \pm m\gamma$$
Now my only problem is the matter of the $$\pm$$.

 

[meopemuk]

Got it. Thanks Eugene. Now I have:
$$E = \pm(m^2 + \frac{m^2v^2}{1 - v^2})^{1/2} = \pm m(\frac{1 - v^2 + v^2}{1 - v^2})^{1/2} = \pm m\gamma$$
Now my only problem is the matter of the $$\pm$$.

You should always take the positive sign. Energy is positive, by definition.

Eugene.

 

[Jimmy Snyder]

Thanks Eugene.