# How can we be sure, that the dot-product of the four-vectors $U$ and $A$ in special relativity is zero?

I'am currently taking class in relativity and in the book that i have it is said that the four-vectors of velocity and acceleration is zero. I know that the invariance of the velocity is $$-c^2$$. Is it from this fact that the dot-product of four-acceleration and four-velocity must be orthogonal? It doesn't seem that calculating $$u^\alpha \cdot a_{\alpha}$$ is equal to zero. Maybe i did something wrong. Can someone please show me the calculations?

• Analogy: for uniform circular motion, the acceleration vector is perpendicular to the velocity vector. The velocity vector has time-independent magnitude (and thus time-independent dot product with itself). Dec 19 '20 at 15:58

Dot product of $$u$$ with itself is constant: $$u\cdot u=-c^2$$ Taking derivative with respect to proper time on both sides gives $$u\cdot \frac{du}{d\tau}=u \cdot a=0$$

• But why is it zero ? if $u=(c\gamma, c\gamma \beta^x) and a=(\gamma\gamma'c,\gamma'\gamma u+\gamma^2 a)$ Dec 19 '20 at 16:15
• Maybe try this link? Dec 19 '20 at 16:19
• Thank very much! Dec 19 '20 at 16:29
• Since the dot-product is zero, then your component expressions for $\tilde u$ and $\tilde a$ satisfy an identity. Dec 19 '20 at 19:52

One way to look at it is by multiplying by $$m$$:

$$mu^{\mu}=p^{\mu}=(E, \vec p)$$

with

$$p^{\mu}p_{\mu} = E^2-p^2 = m^2$$

If you add some momentum (and energy):

$$p^{\mu} +\delta p^{\mu} = (E+\delta E, \vec p + \delta\vec p)$$

Then:

$$(p^{\mu} +\delta p^{\mu})(p_{\mu} +\delta p_{\mu}) = E^2 + 2E\delta E - (p^2 +2\vec p \cdot \vec p) = m^2$$

Rearranging:

$$(E^2-p^2) + 2E\delta E - 2\vec p \cdot \vec p = m^2$$ $$2E\delta E = 2\vec p \cdot \vec p$$

$$\delta E = \frac{\vec p}E \cdot \vec p$$

Note that:

$$\frac{\vec p}E = \frac{\gamma m\vec v}{\gamma m} = \vec v$$

so

$$\delta E = \delta \vec p \cdot \vec v$$

Over the infinitesimal time interval $$\delta\tau$$:

$$\delta \vec p = \vec F \delta\tau$$

$$\vec v = \frac{\vec x}{\delta\tau}$$

so that:

$$\delta E = \vec F \cdot \vec x$$

When you apply a force to accelerate an object, the work done to change the energy exactly balances the force's change to momentum to keep the object on the mass shell, which is equivalent to keeping $$u^{\mu}u_{\mu}=c^2$$.

Thank very much. Both explanations are good. I would just like to say, that Yardan Sheffer's explanation just needed charlie's link to give the full explanation. So the answer in why they are orthogonal so $$a\dotv=0$$ lies in the fact that because of the invariance we can choose a rest system so $$\vec{p}=0$$, which implies that the time component in the acceleration is zero and the $$p$$ component in $$\vec{p}=0$$ so because $$U=(c,0)$$ and $$A=(0,\vec{a})$$ the dot production becomes zero.

• You probably mean $a\cdot v=0$ (written as a\cdot v=0) instead of a\dotv=0. Dec 21 '20 at 11:57
• i didn't notice that i missed the space . Thank you Dec 22 '20 at 12:23