Radiation emitted by a decelerated particle

[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!

 

[hutchphd]

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...)

 

[TSny]

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

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.

 

[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.

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.

 

[TSny]

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.

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.

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!).

 

[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...

 

[TSny]

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

Yes. Express $(1-\beta^2)^3$ in terms of $\gamma$ and watch something neat happen. (It deserves at least one "hallelujah!")

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?

 

[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
$$\int Pdt = -\frac{2e^2\alpha^2}{3cm^2}\frac{m}{\alpha}\frac{1}{c^2}\int_{v_0}^0v^2\frac{dv}{v} = -\frac{2e^2\alpha}{3c^3m}\int_{v_0}^0vdv = \frac{e^2\alpha v_0^2}{3c^3m}$$
Is that the answer?

 

[TSny]

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

Indeed, hallelujah!

Good

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 $\beta$.

Using the fact that $m\frac{dv}{dt} = -\alpha v$

$F \neq ma$ in relativistic mechanics.

 

[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?

 

[TSny]

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.

 

[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.

 

[hutchphd]

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!

 

[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 $\frac{q_1q_2}{r}$.

 

[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.

Also, thank you hutchphd for your answers.