Accelerating an electron from speed u1 to u2

[Alan Ezra]

Hi!

I am trying to find the time for an electron to accelerate in a uniform electric field from u1 to u2.

In the textbook, I found that time is found by integrating d(u/(1-u^2/c^2)^(1/2))=(qE/m_0)⋅dt
and they get u/(1-u^2/c^2)^(1/2) = qEt/m_0 by integrating from t=0 and u=0 to t=t and u=u
I don't quite get the left hand side of the equation, what is it integrating with respect to? what if the particle starts at t=0 and u=u1 and end at t=t1 and u=u2?

Thanks a lot

 

[PeroK]

I've suggested that this is moved to the homework forum.

The assumption here is that the three-force of the electric field is constant. How is the three force defined?

I assume you are trying to calculate time, in terms of the coordinate time in the IRF in which the electron accelerates from $u_1$ to $u_2$.

 

[Alan Ezra]

I've suggested that this is moved to the homework forum.

The assumption here is that the three-force of the electric field is constant. How is the three force defined?

I assume you are trying to calculate time, in terms of the coordinate time in the IRF in which the electron accelerates from $u_1$ to $u_2$.

Thank you for the reply! Yes I am trying to calculate how energy and velocity change wrt time. I am actually following the reasoning of a book, and I cannot quite understand how it gets from eq 2.26 to 2.27. Isn't u a function of time too? why after the integration the left hand side stays the same? Thank you a lot!

Alan

 

[PeroK]

The left-hand side doesn't stay the same - it loses the time derivative. The integral of the derivative of a function is the function.

 

[Alan Ezra]

The left-hand side doesn't stay the same - it loses the time derivative. The integral of the derivative of a function is the function.

Thank you a lot. That makes sense.