# Radiation emitted by a decelerated particle

• physics_student123
In summary, the power radiated by an accelerated particle is equal to the product of the acceleration and the velocity squared, multiplied by the gamma factor.

#### physics_student123

Homework Statement
A particle with charge q and a relativistic velocity $$\vec{v}_0$$ penetrates a medium in which it is decelerated by a force proportional to its velocity, $$\vec{F} = -\alpha\vec{v}$$, with $$\alpha$$ constant. How much energy is emitted in the form of electromagnetic radiation until the particle stops? Assume the medium behaves like vacuum for the radiation.
Relevant Equations
relativistic Larmor formula: $$P = \frac{2e^2\gamma^6}{3c}[\dot{\beta}^2 - (\vec{\beta}\times\dot{\vec{\beta}})^2]$$ where $$\vec{\beta} = \vec{v}/c$$ and $$\gamma$$ is the usual Lorentz factor.
Honestly, folks, I don't even know how to start. I included in the Relevant Equations section the relativistic generalization of the Larmor formula according to Jackson, because that's the equation for the power emitted by an accelerated particle, but I don't see how that gets me very far.

The only thing I realized while thinking about this problem is that the second term inside the brackets in the equation, $$(\vec{\beta}\times\dot{\vec{\beta}})^2$$, is zero, because the velocity and the acceleration are collinear.

Any help or insight would be most appreciated!

There is the external force.
There will be a reaction force from the emitted radiation.
These together will decelerate the particle.
See if that gives you a place to start. ( I have not worked this out...)

physics_student123
physics_student123 said:
The only thing I realized while thinking about this problem is that the second term inside the brackets in the equation, $$(\vec{\beta}\times\dot{\vec{\beta}})^2$$, is zero, because the velocity and the acceleration are collinear.
Yes
physics_student123 said:
Any help or insight would be most appreciated!
You're given the force: ##F = -\alpha v ##. I think you are meant to assume that this relation holds for the entire deceleration of the electron as it comes to a stop and that ##F## is the net force. But I could be wrong. If you do make this assumption, then it's not too hard to determine the total energy radiated.

My hint to get you started: Can you derive an expression for the rate of change of relativistic momentum when the acceleration and velocity are colinear? How is this related to the net force ##F##?

physics_student123
Thank you for your replies, hutchphd and TSny.
TSny, the rate of change of relativistic momentum, which is equal to the net force F, is given by
$$\dot{\vec{p}} = \gamma m\vec{a} + \gamma^3m\left(\frac{va}{c^2}\right)\vec{v}$$
but I don't understand how to input the information that the acceleration and the velocity are collinear.

physics_student123 said:
the rate of change of relativistic momentum, which is equal to the net force F, is given by
$$\dot{\vec{p}} = \gamma m\vec{a} + \gamma^3m\left(\frac{va}{c^2}\right)\vec{v}$$
but I don't understand how to input the information that the acceleration and the velocity are collinear.
The particle moves along a straight line. Directions along the line can be distinguished as positive or negative. So, we don't really need to use vectors.

Thus, we can simply write ##p = \gamma mv = \gamma mc \beta##.

Take the time derivative and see if you can reduce it to $$\dot p = mc\frac{\dot \beta}{(1-\beta^2)^{3/2}}.$$.

physics_student123
Yes, I managed to do it, TSny. $$\dot{p} = \dot{\gamma}mc\beta + \gamma mc\dot{\beta}$$

$$\dot{p} = \frac{\beta^2\dot{\beta}mc}{(1-\beta^2)^{3/2}} + \gamma mc\dot{\beta}= \frac{\dot{\beta}mc}{\sqrt{1-\beta^2}}\left(\frac{\beta^2}{1-\beta^2}+1\right) = \frac{\dot{\beta}mc}{(1-\beta^2)^{3/2}}$$

This is equal to the net force F, right?

P.S. Sorry for taking so long to get back to you.

vanhees71
physics_student123 said:
Yes, I managed to do it, TSny. $$\dot{p} = \dot{\gamma}mc\beta + \gamma mc\dot{\beta}$$

$$\dot{p} = \frac{\beta^2\dot{\beta}mc}{(1-\beta^2)^{3/2}} + \gamma mc\dot{\beta}= \frac{\dot{\beta}mc}{\sqrt{1-\beta^2}}\left(\frac{\beta^2}{1-\beta^2}+1\right) = \frac{\dot{\beta}mc}{(1-\beta^2)^{3/2}}$$
Looks good.

physics_student123 said:
This is equal to the net force F, right?
Yes. This allows you to find ##\dot \beta## as a function of ##\beta## that you can use in the expression for the power radiated.

physics_student123 said:
P.S. Sorry for taking so long to get back to you.
No problem. That actually wasn't long. It's expected that other things can come up that we need to look after (not to mention sleep!).

vanhees71 and physics_student123
So we have $$\frac{\dot{\beta}mc}{(1-\beta^2)^{3/2}} = -\alpha c\beta$$ which implies $$\dot{\beta} = -\frac{\alpha\beta}{m}(1-\beta^2)^{3/2}$$
Using this in the relativistic Larmor formula, we get $$P = \frac{2e^2\gamma^6\alpha^2\beta^2}{3cm^2}(1-\beta^2)^3$$ but this is the power radiated at a given instant in time. I'm not seeing right now how to get to the total energy emitted in the form of electromagnetic radiation...

physics_student123 said:
$$P = \frac{2e^2\gamma^6\alpha^2\beta^2}{3cm^2}(1-\beta^2)^3$$
Yes. Express ##(1-\beta^2)^3## in terms of ##\gamma## and watch something neat happen. (It deserves at least one "hallelujah!")

physics_student123 said:
but this is the power radiated at a given instant in time. I'm not seeing right now how to get to the total energy emitted in the form of electromagnetic radiation...
How are power and energy related?

vanhees71 and physics_student123
$$P = \frac{2e^2\gamma^6\alpha^2\beta^2}{3cm^2}(1-\beta^2)^3 = \frac{2e^2\gamma^6\alpha^2\beta^2}{3cm^2}\frac{1}{\gamma^6} = \frac{2e^2\alpha^2\beta^2}{3cm^2}$$

Indeed, hallelujah!

Power is the time derivative of work, which is energy transferred, so in order to find the total energy emitted by the particle until it stops, I would have to integrate the expression above in time. So we have
$$\int Pdt = \int\frac{2e^2\alpha^2}{3cm^2}\frac{v^2}{c^2}dt$$
Using the fact that ##m\frac{dv}{dt} = -\alpha v## which implies ##-\frac{m}{\alpha}\frac{dv}{v} = dt##, we get
$$\int Pdt = -\frac{2e^2\alpha^2}{3cm^2}\frac{m}{\alpha}\frac{1}{c^2}\int_{v_0}^0v^2\frac{dv}{v} = -\frac{2e^2\alpha}{3c^3m}\int_{v_0}^0vdv = \frac{e^2\alpha v_0^2}{3c^3m}$$

physics_student123 said:
$$P = \frac{2e^2\gamma^6\alpha^2\beta^2}{3cm^2}(1-\beta^2)^3 = \frac{2e^2\gamma^6\alpha^2\beta^2}{3cm^2}\frac{1}{\gamma^6} = \frac{2e^2\alpha^2\beta^2}{3cm^2}$$

Indeed, hallelujah!
Good

physics_student123 said:
Power is the time derivative of work, which is energy transferred, so in order to find the total energy emitted by the particle until it stops, I would have to integrate the expression above in time. So we have
$$\int Pdt = \int\frac{2e^2\alpha^2}{3cm^2}\frac{v^2}{c^2}dt$$
OK. I would keep writing ##\frac{v^2}{c^2}## as ##\beta^2##.

Note that you can write $$dt = \frac{dt}{d\beta}d\beta = \frac{d \beta}{\dot{\beta}}$$ and you know how to express ##\dot{\beta}## in terms of ##\beta##.

physics_student123 said:
Using the fact that ##m\frac{dv}{dt} = -\alpha v##

##F \neq ma## in relativistic mechanics.

vanhees71 and physics_student123
Ok, so we have ##dt = -\frac{d\beta}{\frac{\alpha}{m}\beta(1-\beta^2)^{3/2}}## which implies
$$\int_{t=0}^{t=t_f}Pdt = -\frac{2e^2\alpha^2}{3cm^2}\int_{\beta_0 = \frac{v_0}{c}}^0\beta^2\frac{1}{\frac{\alpha}{m}\beta(1-\beta^2)^{3/2}}d\beta = -\frac{2e^2\alpha}{3cm}\int_{\beta_0}^0\frac{\beta}{(1-\beta^2)^{3/2}}d\beta$$
which gives us
$$\int_{t=0}^{t_f}Pdt = \frac{2e^2\alpha}{3cm}\left(\frac{1}{\sqrt{1-\beta_0^2}}-1\right) = \frac{2e^2\alpha}{3cm}(\gamma_0 - 1)$$
with ##\gamma_0 = \frac{1}{\sqrt{1-\beta_0^2}}##. Is that the answer, TSny?

physics_student123 said:
Ok, so we have ##dt = -\frac{d\beta}{\frac{\alpha}{m}\beta(1-\beta^2)^{3/2}}## which implies
$$\int_{t=0}^{t=t_f}Pdt = -\frac{2e^2\alpha^2}{3cm^2}\int_{\beta_0 = \frac{v_0}{c}}^0\beta^2\frac{1}{\frac{\alpha}{m}\beta(1-\beta^2)^{3/2}}d\beta = -\frac{2e^2\alpha}{3cm}\int_{\beta_0}^0\frac{\beta}{(1-\beta^2)^{3/2}}d\beta$$
which gives us
$$\int_{t=0}^{t_f}Pdt = \frac{2e^2\alpha}{3cm}\left(\frac{1}{\sqrt{1-\beta_0^2}}-1\right) = \frac{2e^2\alpha}{3cm}(\gamma_0 - 1)$$
with ##\gamma_0 = \frac{1}{\sqrt{1-\beta_0^2}}##.
I think that's right. (It's a good idea to check that the answer has the dimensions of energy.)

Good work.

vanhees71 and physics_student123
The units are indeed proving to be a problem, TSny. From the relation ##F = -\alpha v##, one arrives at the conclusion that the units of ##\alpha## are ##kilogram/second##. But from the expression that I derived above with your help, the units of ##\frac{2e^2\alpha}{3cm}(\gamma_0-1)## are ##[coulomb]^2/meter##, which are not, as far as I know, the units of energy. So there is something wrong somewhere.

I think the Larmor formula is in Gaussian units and the so $$q^2 \to \frac {q^2} {4 \pi \epsilon_0}$$ and maybe worse. Sure am glad its not my problem!

physics_student123 and TSny
Yes. In Gaussian units, (charge)2 divided by distance has dimensions of energy. For example, the potential energy of two point charges separated by distance ##r## is ##\frac{q_1q_2}{r}##.

vanhees71 and physics_student123
Ah, very well then. I'm not used to Gaussian units.

TSny, thank you SO MUCH for taking the time and effort to help me with this problem. I simply could not have done it without you. Your help was invaluable.