1

$p^{\nu} = (E/c,\textbf{p})^T$

After time reversal Lorenz transformation ( $t'=-t,r'=r$) it becomes:

$p'^{\nu} = (E/c,\textbf{-p})^T$

But if we multiply transformation matrix by old coordinates, we obtain the opposite result:

$(E'/c,\textbf{p}')^T =diag(-1,1,1,1)*(E/c,\textbf{p})^T = (-E/c,\textbf{p})^T $

So, 4-momentum is not a vector...

Qmechanic
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Михаил Игоревич Краснов
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2 Answers2

2

It's because the time reversal operator $T$ is anti-unitary. You need to understand exactly how the transformation of the momentum is derived. The unitary operator that generates translations $U(a)$ satisfies $$ T^{-1} U(a)T = U ( {\cal T}a) , \qquad {\cal T} = diag(-1,+1,+1,+1). \tag{1} $$ Now, we recall that $U(a) = \exp ( i a^\mu P_\mu )$ where $P_\mu$ is the translation generator. Expanding equation (1) to first order in $a^\mu$, we find $$ T^{-1} ( i P^\mu )T = {\cal T}^\mu{}_\nu ( i P^\nu ) $$ But, now we recall that $T$ is anti-unitary so $T i T^{-1} = - i$. Using this, we find $$ T^{-1} P^\mu T = - {\cal T}^\mu{}_\nu P^\nu $$ The rest of the argument is as follows. Let $| p \rangle$ be a momentum eigenstate $$ P^\mu | p \rangle = p^\mu | p \rangle $$ Then, consider the state $T | p \rangle$. This satisfies $$ P^\mu ( T | p \rangle ) = T T^{-1} P^\mu T | p \rangle = T ( - {\cal T}^\mu{}_\nu P^\nu | p \rangle = - {\cal T}^\mu{}_\nu p^\nu ( T | p \rangle ) $$ It follows that the state $T | p \rangle$ has momentum $- {\cal T}^\mu{}_\nu p^\nu$. This is often written concisely by stating that under time reversals $$ p^\mu \to - {\cal T}^\mu{}_\nu p^\nu \quad \implies \quad ( E , \vec{p} ) \to ( E , - \vec{p} ). $$

Prahar
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1

This is only a proof for massive particles as: $$p^\nu = m\frac{\text{d}x^\nu}{\text{d}\tau}$$ where $\tau$ is the proper time, $m$ the mass and $x^\nu$ the position of the particle in spacetime.

Under the transformation $t'=-t$ and $\vec{r}\,'=\vec r$ the proper time is unchanged! Thus, you only get a change in sign of the $\nu=0$ component. Therefore, your second expression holds true:

$$(E'/c,\textbf{p}')^T =diag(-1,1,1,1)*(E/c,\textbf{p})^T = (-E/c,\textbf{p})^T$$

The error in your first expression is you use the derivative with respect to coordinate time not the proper time for your definition of the 4-momentum.

Chris Long
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  • Would be interested to see someone extend this proof to massless particles (other than just stating that the expression should be continuous as $m$ and $\text{d}\tau$ tend to zero). – Chris Long Jul 01 '21 at 11:19
  • Prahar Mitra's answer does just that! – Chris Long Jul 01 '21 at 11:59
  • You are not right. If you do the time reversal transformation, parameterization of the curve x(t) change and $d\tau$ get a negativ sign – Михаил Игоревич Краснов Jul 04 '21 at 12:40
  • No $\tau$ is the time in the rest frame of the particle. We reverse time by transforming to another frame where the time coordinate is the negative of the initial frame of intrest which may or may not be the rest frame. In either case we are not altering the rest frame but specificing a new coordinate system. This is why my proof holds. – Chris Long Jul 04 '21 at 13:15
  • Expression $\frac{d\textbf r}{d\tau}$ got negative sign when we do the time reversal transformation, because the particle moves in opposite direction in new coordinates – Михаил Игоревич Краснов Jul 04 '21 at 17:12
  • It becomes clear if we write this expression in another way: $\frac{d\textbf r}{d\tau} = \gamma \frac{d\textbf r}{dt} = \gamma \textbf v$ – Михаил Игоревич Краснов Jul 04 '21 at 17:17
  • If what you are agruing is the case then how do you produce the correct result? I think in this case $\frac{\text{d}t}{\text{d}\tau}=-\gamma$ as when deriving the Lorentz factor you have to take the square root and so actually there is a positive and negative solution. See this post for derivations: https://physics.stackexchange.com/questions/173268/deriving-the-lorentz-factor-gamma – Chris Long Jul 04 '21 at 19:11