Relativistic Work: Deriving Formula & Application to Electrons

[Alettix]

Hello,

I have tried to derive a relativistic formula for work using the E = ϒmc^2 equation and that work is equivalent to change in energy. But I have not gotten especially far and everything is messy.
However, when looking at solutions to some harder problems I encountered the formula:
W = pc (1)
which I thought decribes the total energy of a massless particle, but this time it was used in connection with an electron changing directions (velocity but not speed) in a magnetic field.
So my question is, do you know if equation (1) holds for work done on/by all kinds of particles or do you have any other suggestion?

Thank you! :)

 

[Joshua L]

I am no expert on this topic, yet I know of one beautiful mathematical relationship that you may notice some familiar arguments within in it.

$E^2 = (m_0 c^2)^2 + (p c)^2$

where,

$E = γ m_0 c^2$

$p = γ m_0 v$ (massive particles)

$p = \frac{h}{λ}$ (photons)

This equations applies to all particles, whether massive or massless.

Now for work on relativistic particle: the definition of mechanical work gives the following.

$W = \int_{\vec r_i}^{\vec r_f} \vec F \cdot d\vec r$

where,

$\vec F = \frac{d\vec p}{dt}$

One needs to find the derivative of relativistic momentum with respect to time to find the net relativistic force on a particle. Then, it would be plugged back in the work equation.

$W = \int_{\vec r_i}^{\vec r_f} (\frac{d}{dt}(γm_0 \vec v)) \cdot d\vec r$

 

[rude man]

Work done on a particle = change in k.e. of that particle.
K.E. = E - E₀ = mc² - m₀c²
E = total energy of particle
E₀ = rest energy of particle.
m = relativistic mass
m₀ = rest mass

So if a particle starts from rest, K.E. = work done on that particle. If it's moving you compute m, then do the subtraction.
You can combine this equation with the aforesaid post one, to wit,
E² = (pc)² + E₀².

NOTE: you are using the new convention of m = rest mass & relativistic mass = γm.
I and the previous poster are using the older convention where m = relativistic mass.
Why this change was made I never hope to know.

 

[Alettix]

$\vec F = \frac{d\vec p}{dt}$

One needs to find the derivative of relativistic momentum with respect to time to find the net relativistic force on a particle. Then, it would be plugged back in the work equation.

$W = \int_{\vec r_i}^{\vec r_f} (\frac{d}{dt}(γm_0 \vec v)) \cdot d\vec r$

Firstly, thank you for the good answer! :)
But then my question is: How can I find the derivative of the change in momentum with respect to time if the magnitude of the velocity is not changing?
Maybe I haven't learned that math yet...

 

[Alettix]

Work done on a particle = change in k.e. of that particle.
K.E. = E - E₀ = mc² - m₀c²
E = total energy of particle
E₀ = rest energy of particle.
m = relativistic mass
m₀ = rest mass

So if a particle starts from rest, K.E. = work done on that particle. If it's moving you compute m, then do the subtraction.
You can combine this equation with the aforesaid post one, to wit,
E² = (pc)² + E₀².

NOTE: you are using the new convention of m = rest mass & relativistic mass = γm.
I and the previous poster are using the older convention where m = relativistic mass.
Why this change was made I never hope to know.

Thank you for your answer Sir! However, I still wonder if you have any suggestion of a formula which gives the work done on a particle which doesn't start from rest? The electron in the initial case mentioned only changes the direction of its speed, and I don't understand if W = p * c is a general formula or only applies in this case. :)

 

[Joshua L]

That is a terrific question! There is this one mathematical relationship that expresses the relativistic net force of a particle that has a constant speed.
Here is what I got.

The proof (or most of one) below applies to the motion of particles with a constant speed.

Newton's Second Law implies the following: \vec F = \frac{d \vec p}{dt}
It is worthy to note that, \vec p = \| \vec p\| \hat p
We can now expand the second part of the equation using the product rule of derivatives: \vec F = \frac{d \| \vec p\|}{dt} \hat p+\| \vec p\| \frac{d \hat p}{dt}
Since the speed isn't changing, \frac{d \| \vec p\|}{dt}=0
and therefore, \vec F = \| \vec p\| \frac{d \hat p}{dt}
It is crucial to realize that this force is perpendicular to the velocity of the particle due to the fact that earlier we canceled out the term containing $\hat p$ from the equation.

It can be geometrically proven that: \displaystyle\| \frac{d \hat p}{dt} \displaystyle\| = \frac{\| \vec v\|}{R}
(If you would like to know the proof, I'd be happy to share. This thread is already long.)

Now we know that the direction of the net force (perpendicular to the velocity) and the relationship above, we can now say that... \| \vec F\| = \frac{\| \vec p\| \| \vec v\|}{R}
or \| \vec F \| = \frac{\gamma m_0 \| \vec v\|^2}{R}
where $R$ is the radius of the "kissing circle."

The equation above is the magnitude of the net force acting on a relativistic particle moving with a constant speed. This applies to any massive particle moving at any constant speed under the conditional postulates of Einstein's theory of special relativity.

I'm glad that I could help.

 

[Alettix]

That is a terrific question! There is this one mathematical relationship that expresses the relativistic net force of a particle that has a constant speed.
Here is what I got.

This looks really useful, Thank you so much Sir!