Finding momentum of a massless particle

1. Mar 25, 2013

Pranav-Arora

1. The problem statement, all variables and given/known data
This question is a part of a problem I am doing. I am stuck at finding momentum of the massless particles.

The two particles move around a stationary point (evenly spaced from the particles). The potential energy between the particles is a function of distance r between the massless particles. $V(r)=ae^{br}r^{-1/2}$ where a and b are constants. I have to find the momentum of the particles.

(Actually I have to find the minimum possible energy of the system but I guess I need to find the momentum before that. The question asks for the rest mass of this 2-particle bound state)

2. Relevant equations

3. The attempt at a solution
I can find the force between the particles by differentiating V(r) with respect to r.
$$F=-\frac{dV(r)}{dr}=\frac{ae^{br}(1-2br)}{2r\sqrt{r}}$$

Since the particles move in circles around the point, some force (probably a component of the force I found above) is balanced by the centrifugal force. But the mass of the particles is zero and the centrifugal force would be zero then.

The energy of the system , $E=V(r)+2pc$ where p is the momentum of the particles.

Any help is appreciated. Thanks!

2. Mar 25, 2013

Curious3141

In a bit of a rush, so I might have to make a more detailed post later if you still need help, but what exactly are you asked to find?

If it's just a matter of finding the minimum energy of the system, why not just find the derivative and set it to zero? Sketching the curve will help you verify it's a global minimum.

Also, is this given to be a massless system? Why would they then ask for a "rest mass"?

If it's truly a massless system, then you shouldn't be using Newtonian mechanics at all.

And the momentum is easy to find, since for massless particles, E = pc holds. You're already given E(r) (you called it V), so determing p(r) is trivial.

3. Mar 25, 2013

Pranav-Arora

Sorry, I should have been more clear. The problem states that the system acquires mass through interaction of massless particles.

Yes, I could find the minimum energy by setting the derivative equal to zero but I am not sure if the momentum of particles is a function of r or not. Is my equation for energy of system correct?

4. Mar 25, 2013

Curious3141

Isn't the expression for energy what you're given? It's not something you came up with, right? To be dimensionally consistent, a has to have the dimension of Joule.metre^0.5, so we have to assume it does.

If that's the energy of the system, and it's divided equally between the two particles, then I think you can find the momentum of a single particle just by taking half that and dividing by c (speed of light). In this case, the momentum should be dependent on r.

5. Mar 25, 2013

Pranav-Arora

Yes, a and b have proper dimensions. Their values are given in the problem with proper units. a is given to be 1 Nm^(3/2) and b is 10^(15) m^(-1).

I was talking about the last equation in the first post I mentioned for the energy of system.
$E=V(r)+2pc$. Is this right?

6. Mar 25, 2013

Curious3141

I don't think so, I would've thought V(r) = 2pc is an adequate description of the system. Else, I'm completely misinterpreting the problem, in which case someone else should step in to help.

But in the meantime, perhaps it's better you just post the exact entire problem as it's written, so that we're all clear.

EDIT: I can see why you wrote what you did: Total energy = PE + KE. I'm just not sure if this is the way to approach this. I still think you should post the entire problem including all givens and figures.

7. Mar 25, 2013

Pranav-Arora

See the attachments.

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8. Mar 25, 2013

Curious3141

They're apparently trying to model the Higgs boson, something that I'm not intimately familiar with. The model is very simplistic, so you should just go with the most direct approach.

From what's given, this is how I would approach it:

First use calculus to find the minimum E(r) as I suggested in my first post. The implication is that this is the distance at which the particles will be orbiting the centre, since it minimises the PE of the system.

Use that minimal E(r) in E = mc^2 to calculate the effective rest mass of the system. You don't have to divide by 2, since you're considering the two particles as components of a single system.

Plug in the values and see what you get.

I think the E = pc part is unnecessary here.

9. Mar 25, 2013

Pranav-Arora

I think I have tried the same before.
The minimum E(r) is when r=1/2b. (I replaced alpha with a and beta with b)
At r=1/(2b),
$E(r)=ae^{1/2}\sqrt{2b}$.
$m=E(r)/c^2$
Plugging in the values, I get m=819.25 ng. But this is incorrect. :(

Why is 2pc part unnecessary? What I think is the total energy of system comprises of the energy of individual particles plus the energy due to their interaction. Where am I wrong with my reasoning?

Last edited: Mar 25, 2013
10. Mar 25, 2013

Curious3141

I get 819.26ng. In the same ballpark.

I think the 2pc part is unnecessary because the momentum (and the mass) arises from the very interaction between the particles. This is not like a system with mass where you can say $E^2 = m^2c^4 + p^2c^2$ (i.e. there's "rest" mass/energy and "kinetic" energy). Beyond that, I can't explain it.

Never mind, let me ask other HH to see if they can shed some light on the matter.

11. Mar 25, 2013

Curious3141

Pranav, sorry it took a while, but I've asked some of the other homework helpers, and here's where we're at (I'm quoting the posts verbatim to ensure they get due credit):

12. Mar 25, 2013

Pranav-Arora

Thanks a lot all of you! I never thought that this amount of discussion was going on at the back! :)

I read all the replies. It looks to me that TSny has considered the massless particles to be just opposite to each other whereas the question doesn't state this (unless I am misinterpreting the question).

I tried BruceW's method before of rewriting the centripetal force as Pω but the function I obtained had no minimum. Setting the derivative equal to zero gave me a quadratic with no real roots. :(

@Curious3141: Can you explain how did you reach the equation for E(r)
$$E(r) = \frac{1}{2}\alpha e^{\beta r} r^{-\frac{1}{2}} (2\beta r + 1)$$
I get the following equation which has no minimum.
$$E(r)=\frac{\alpha e^{\beta r} (3-2\beta r)}{2\sqrt{r}}$$

Last edited: Mar 25, 2013
13. Mar 26, 2013

Pranav-Arora

I redid the algebra and this time I did obtain the same expression as Curious3141.

If r is the distance between the particles (assuming that they are on the opposite ends of the stationary point), the radius of orbit is r/2. Hence,
$$\frac{pc}{r/2}=-\frac{dV(r)}{dr}$$
$$2pc=\frac{\alpha e^{\beta r}(2\beta r-1)}{2\sqrt{r}}$$

$$E(r)=V(r)+2pc$$
$$E(r)=\frac{\alpha e^{\beta r}}{\sqrt{r}}+\frac{\alpha e^{\beta r}(2\beta r-1)}{2\sqrt{r}}$$
$$\Rightarrow E(r)=\frac{\alpha e^{\beta r}(1+2\beta r)}{\sqrt{r}}$$

But is it correct to assume that particles revolve as shown in the attachment?

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• particles.png
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14. Mar 26, 2013

Curious3141

We're not given a good description of the model, but the circular model of motion best fits what we are given. The particles have to stay a constant distance of r from each other, while being in motion at a fixed speed c. That's a circular orbit with the particles diametrically opposed while moving in the same sense (either both clockwise or both anticlockwise).. If it were anything other than a circle, the distance would not be constant. If the particles were not diametrically opposed, the distance would be constant, but not r. If they were not moving in the same sense, the distance wouldn't be constant.

I guess from your lingering doubt, that answer also wasn't accepted by the system?

15. Mar 26, 2013

Pranav-Arora

I still haven't tried your answer because I only have my last try left so I would like to confirm everything before submitting the answer. :)

16. Mar 26, 2013

Curious3141

I am by no means confident of that answer. I can't speak for the other HHs who chipped in, but at least some have reservations.

Is there a lot riding on this test? Would it be possible to seek a clarification from your tutor or TA before committing yourself?

17. Mar 26, 2013

Pranav-Arora

This isn't a test. :)

And the problem isn't by my tutor. :P

EDIT: When I plug in the values, I don't get the same answer as you.

EDIT2: Got the same answer as you, missed a factor of 1/2.

EDIT3: I tried 671.57 but the system did not accept this too. :(

Last edited: Mar 26, 2013
18. Mar 26, 2013

Curious3141

I'm sorry to hear that. Where's the problem from? Can you find out how it's meant to be done? I'm definitely interested, and I'm sure some of the HHs are too.

19. Mar 26, 2013

Pranav-Arora

This is from a website I came across a month before. The solution is posted after a week. So for this problem, it will be out by next Wednesday or Thursday. Four problems are posted for Physics per week, I solved the rest three but got stuck on this. Discussing the problem here was fun. Thanks to everyone who participated. :-)

20. Mar 30, 2013

Pranav-Arora

I got an e-mail from the website about my submitted answer. It addressed that due to some mistake from their side, my answer was marked wrong. The answer 671.57 ng is correct. Thanks everyone!