# Dot product of 4-force and 4-velocity

1. Feb 15, 2010

### Rasalhague

In what follows, I'll use bold for 4-vectors and an arrow over a letter to denote a 3-vector. I'll use a dot both for the Minkowski dot product and the Euclidean one, and multiplication of real numbers; the meaning in each case should be clear from the symbols on either side. $\vec{v}$ and $\vec{a}$ are 3-veclocity and 3-acceleration. The same letters without the arrow are their magnitudes.

I've been working through the $E=mc^2$ derivation at the Shady Crypt Observatory site, which someone posted in a recent thread. I've got up to the detour establishing that $\mathbf{F} \cdot \mathbf{U} = 0$, equation 18. I'm with them for the first three lines of the derivation that follows this equation, but I don't yet see how

$$\frac{\mathrm{d} \gamma}{\mathrm{d} t} = \frac{\gamma^3}{c^2} \, \vec{a} \cdot \vec{v}.$$

I get

$$\frac{\mathrm{d} \gamma}{\mathrm{d} t} = \frac{\mathrm{d} }{\mathrm{d} t}\left ( \left [ 1-\left ( \frac{v}{c} \right )^2 \right ]^{1/2} \right )$$

$$=-\left [ 1-\left ( \frac{v}{c} \right )^2 \right ]^{-3/2}\cdot\left ( \frac{-2v}{c} \right )\cdot\frac{1}{c}\cdot a$$

$$=\frac{2av}{c^2\left [ 1-\left ( \frac{v}{c} \right )^2 \right ]^{3/2}} = \frac{2av \gamma^3}{c^2}=\frac{2av\gamma}{c^2-v^2}.$$

But $\vec{a} \cdot \vec{v} = a \, v\, \cos{\theta}$ and, whatever the angle, $\cos{\theta} \neq 2$.

2. Feb 15, 2010

### Altabeh

You forgot to put a 2 in the denominator of the derivative of $$\gamma$$.

AB

3. Feb 15, 2010

### Rasalhague

Ah, yes, so I did. Thanks, Altabeh. But how do we know that acceleration due to this arbitrary force will be in the same direction as the velocity of the object experiencing the force?

4. Feb 15, 2010

### DrGreg

In your argument you have taken

$$a = \frac{d|\vec{v}|}{dt}$$​

when in fact it is

$$a = \left|\frac{d\vec{v}}{dt}\right|$$​

To get the result you want, differentiate

$$v^2 = \vec{v} \cdot \vec{v}$$​

5. Feb 15, 2010

### Altabeh

We don't need to know how a and v vectors are directed if there is a force acting on an object! Why are you asking this?

AB

6. Feb 15, 2010

### Rasalhague

Got it! Thanks, DrGreg.

I see that now. I asked because I thought they'd just substituted $\vec{a}\cdot \vec{v}$ for $av$. But as DrGreg pointed out,

$$\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\vec{v}\cdot \vec{v}\right)^{1/2}=\frac{1}{2v} \left(\frac{\mathrm{d}\vec{v}}{\mathrm{d}t} \cdot \vec{v} + \vec{v} \cdot \frac{\mathrm{d}\vec{v}}{\mathrm{d}t} \right)$$

is not the same thing as

$$\vec{a} = \frac{\mathrm{d}\vec{v}}{\mathrm{d}t}.$$

So

$$\frac{\mathrm{d}\gamma}{\mathrm{d}t}=\frac{\vec{a}\cdot \vec{v}}{c^2}\gamma^3.$$

Last edited: Feb 15, 2010