Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Does a particle and antiparticle have the same velocity and acceleration in an electric field?

+ 0 like - 2 dislike
1109 views

As I understand it, according to the CPT theorem an antiparticle can be thought of as having either the opposite charge (C) or the opposite spacetime parity (PT). For example in Feynman diagrams a positron going forwards in spacetime is equivalent to an electron going backwards in spacetime.

Let us assume that we have a particle of mass $m$ and charge $q$ moving in the x-direction under the influence of an electric field $\vec{E}$ directed along the x-axis.

The velocity $\vec{v}=d\vec{x}/dt$ and acceleration $\vec{a}=d^2\vec{x}/dt^2$ of the particle are given by
\begin{eqnarray}
\frac{d\vec{x}}{dt} &=& \frac{\vec{p}}{m},\tag{1}\\
\frac{d^2\vec{x}}{dt^2} &=& \frac{q\vec{E}}{m}.\tag{2}
\end{eqnarray}

Now consider the particle's antiparticle. Instead of assuming it has the opposite charge we can assume that it has the opposite spacetime parity. This means that it has negative momentum and negative mass implying that it is traveling backwards in both space and time.

When we apply the following transformations to Eqn.$(1)$ and Eqn.$(2)$ they remain the same:
\begin{eqnarray}
\vec{p} &\rightarrow& -\vec{p},\\
m &\rightarrow& -m,\\
d\vec{x} &\rightarrow& -d\vec{x},\\
dt &\rightarrow& -dt.
\end{eqnarray}

Thus the antiparticle has the same velocity $\vec{v}$ and acceleration $\vec{a}$ as the particle.

Is this correct?

asked Aug 19, 2022 in Theoretical Physics by John [ revision history ]
edited Aug 20, 2022

m -->-m ? Particles and antiparticles have the same mass .

Dirac considered that antiparticles could have a negative energy. This would explain why an electron annihilates with a positron giving out two photons. In a sense the electron falls into the negative energy well which is the positron.

A hole in the negative energy spectrum (positron) and a patch (electron) are figurative notions, not real ones, because all particles and antiparticles have positive energies. The resulting photons have positive energies!

Perhaps I should say that antiparticles can be considered to have either the opposite charge or negative rest-mass. Antiphotons have zero charge and zero rest-mass so that they are identical to photons.

Stick to the opposite charge - this is proven to be the right description.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
$\varnothing\hbar$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...