Particle in an electric field- formulae suggestions.

[Alpha&Omega]

1. A particle of charge q and rest mass m_0 is accelerated from rest at t=0 in a uniform electric field of magnitude E. All other forces are negligible in comparison to the electric field force. Show that after a time t the relativistic expression for the speed of the particle is:

v=\frac{c}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}}

Homework Equations

I think the equations relevant here have got to be E_k=(\gamma-1)m_0c^2 and E=\frac{F}{q}.

I've tried rearranging the above equations and substituting them into each other but I can't get anything good. Does anyone know any other useful equations that I might be able to use?

 

[tiny-tim]

Hi Alpha&Omega!

(how come you have two different names? )

v=\frac{c}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}}
…
Does anyone know any other useful equations that I might be able to use?

Hint: with that v, what is (v/c)/√(1 - v²/c²) ?

 

[Alpha&Omega]

Hi Alpha&Omega!

(how come you have two different names? )

I felt insecure. XD

Hint: with that v, what is (v/c)/√(1 - v²/c²) ?

Do you mean as a physical quantity?

I would just do this:

\frac{\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}}

\sqrt{\frac{\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}

\sqrt{\frac{1}{1-\frac{v^2}{c^2}}-1

Is this what you meant? =S

 

[tiny-tim]

No , I meant what is (v/c)/√(1 - v²/c²) if v=\frac{c}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}} ?

 

[Alpha&Omega]

No , I meant what is (v/c)/√(1 - v²/c²) if v=\frac{c}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}} ?

Ah, darn. Let me see:

\frac{{\frac{c}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}}}{c}}{\sqrt{1-\frac{(\frac{c}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}})^2}{c^2}}}

\frac{{\frac{c^2}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}}}}{\sqrt{1-\frac{(\frac{c^2}{{1+\frac{m_0^2c^2}{q^2E^2t^2}}})}{c^2}}}

\frac{{\frac{c^2}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}}}}{\sqrt{-\frac{m_0^2c62}{q^2E^2t^2}}}

Ummm, am I doing it correctly? I've got a complex number as the denominator! =S

 

[tiny-tim]

uhhh? whadya do?

should be qEt/m₀c ​

 

[Alpha&Omega]

magic!

This may sound weird, but how did you know to use \frac{\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}}?

 

[tiny-tim]

Because I knew that m₀(v/c)/√(1 - v²/c²) is the … ?

(btw, it would have save a lot of paper and time if you'd written X, say, instead of that fraction! )

 

[Alpha&Omega]

The momentum divided by c?

So if momentum is MeV/c then in this equation we have MeV/c² which is the mass!

(I know it would have saved trees, but for some reason I decided to do it this long and complicated way =D).

So I have to work from this equation to the equation posted in the thread.

I just tried it and I noticed that the question doesn't mention whether the applied force is parallel to the x-axis or perpendicular to it. Which expression for relativistic force should I use?

 

[tiny-tim]

The momentum divided by c?

Yup!

So the question is asking you to prove that after time t the momentum is qEt …

what principle of physics would you use to do that? ​

I noticed that the question doesn't mention whether the applied force is parallel to the x-axis or perpendicular to it. Which expression for relativistic force should I use?

uhh? the x-axis is wherever you want it to be

 

[Alpha&Omega]

Yup!

So the question is asking you to prove that after time t the momentum is qEt …

what principle of physics would you use to do that? ​

Would it be an integral of an expression for the force with respect to time?

uhh? the x-axis is wherever you want it to be

In my book it has two expressions for relativistic force. One is F=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}ma if it's perpendicular to the x-axis and the other is F=\frac{1}{(1-\frac{v^2}{c^2})^\frac{3}{2}}ma if the force is in the same direction as the field.

 

[tiny-tim]

Would it be an integral of an expression for the force with respect to time?

I was thinking more along the lines of "Force = …"

(work backwards from the answer! )

In my book it has two expressions for relativistic force. One is F=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}ma if it's perpendicular to the x-axis and the other is F=\frac{1}{(1-\frac{v^2}{c^2})^\frac{3}{2}}ma if the force is in the same direction as the field.

ah …

A particle of charge q and rest mass m_0 is accelerated from rest at t=0 in a uniform electric field of magnitude E.

so the velocity will always be in the same direction as E

btw, I don't normally approve of "relativistic mass", but in this case I find it easier to remember the equations as

F_perp = m_rela

F_para = m_rela/(1 - v²/c²)

so I remember the perpendicular force equation as being the same as Newton's