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

Physics Asked by JohanL on January 28, 2021

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?

3 Answers

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):

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

Then:

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

Rearranging:

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

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

Note that:

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

so

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

Over the infinitesimal time interval $deltatau$:

$$ delta vec p = vec F deltatau$$

$$ vec v = frac{vec x}{deltatau}$$

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 $u^{mu}u_{mu}=c^2$.

Correct answer by JEB on January 28, 2021

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

Answered by Yarden Sheffer on January 28, 2021

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 $adotv=0$ 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.

Answered by JohanL on January 28, 2021

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