Finding momentum of a massless particle

[Saitama]

Homework Statement

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)

Homework Equations

The Attempt at a Solution

I can find the force between the particles by differentiating V(r) with respect to r.
[tex]F=-\frac{dV(r)}{dr}=\frac{ae^{br}(1-2br)}{2r\sqrt{r}}[/tex]

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!

 

[Curious3141]

Homework Statement

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)

Homework Equations

The Attempt at a Solution

I can find the force between the particles by differentiating V(r) with respect to r.
[tex]F=-\frac{dV(r)}{dr}=\frac{ae^{br}(1-2br)}{2r\sqrt{r}}[/tex]

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!

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.

 

[Saitama]

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.

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?

 

[Curious3141]

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?

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.

 

[Saitama]

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.

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?

 

[Curious3141]

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?

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 into 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.

 

[Saitama]

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.

See the attachments.

 

[Curious3141]

See the attachments.

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.

 

[Saitama]

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.

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?

 

[Curious3141]

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?

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.

 

[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):

Need a little help here - https://www.physicsforums.com/showthread.php?p=4320848

Could any kind souls please shed some light on that problem? If I'm totally on the wrong track, please correct me. Thanks!

I get the same answer for the contribution due to the potential energy. I think he should also add the contribution due to the momentum of the particles. If you think about it, the momentum of the particles (in the reference frame where the centre is stationary), is the 'invariant mass' of the system. So it is not really 'rest mass', but it is the most similar concept, when there are multiple particles. And then adding the potential energy I sort of interpret as "well, there is energy in the field which causes them to be pulled together, so this is another contribution to the invariant mass of the system".

To calculate the momentum... Well, as Pranav said, you can't use the usual equation for the centripetal force, because the particles are massless. But you can re-write the equation as:
[tex]- \frac{dV_{(r)}}{dr} = P \omega [/tex]
(Where P is the 3-momentum and omega is just the usual speed, divided by distance from the centre of the circle).
So now, the equation doesn't involve mass, and the momentum can be calculated. Also, I realize I have not shown that the usual equation for centripetal force should work for relativistic particles. I hope it does... It would make sense, because the gradient in potential tells us about how 3-momentum is (or is not) conserved, so I would hope it relates to 3-momentum in the usual way. And Eric Weisstein's world of physics : http://scienceworld.wolfram.com/physics/CentripetalForce.html
Seems to agree that it works for relativistic particles.

Thanks Bruce.

But even if that were the right approach, wouldn't ##\displaystyle \frac{dV_{(r)}}{dr} = 0## at the "magic" distance ##\displaystyle r = \frac{1}{2\beta}##?

darn. I hadn't thought that through. That does make sense though. For example, a ground-state electron in a coulomb potential has zero orbital momentum. So it shouldn't be surprising that these particles have zero momentum when the interaction potential is minimised. This is difficult to conceptually explain, since now we have two particles with zero rest mass and zero momentum. I guess we get this nonsense result because we haven't done the quantum mechanics 'properly'. So most likely, the question does not want us to try to calculate the momentum of the particles (since it turns out to be nonsense). But it is strange that they include some detail about the equation for the momentum of massless particles (E=PC) in the question statement.

Anyway, I was reading through the question again, and it says "The rest mass of this 2-particle bound state is defined as the lowest possible energy/c^2" The total energy of the system has no minimum. So by 'energy', they must mean 'interaction energy', i.e. the potential. So I think you are right, they only want the potential energy. So I don't know why the answer comes out differently to what the book says...

If you minimize V(r), you don't have a force to keep the particles in their orbit.

##F=\frac{dp}{dt}=\frac{pc}{r}##
Putting this into the derived formula for the force,
$$pc=-\frac{a e^{br}(1-2br)}{2\sqrt{r}}$$
The total energy is
$$E(r)=V(r)+2pc = 2ab e^{br}\sqrt{r}$$
Unfortunately, this does not have a minimum.


The attachment is used for a test, just ignore it.
Edit: Test done, attachment removed.

The r in this force expression is the radius of the orbit, which is not the same as the r in V(r). See if correcting that will lead to an expression for total energy that has a minimum. I think it does.

Thanks for all the added help.

TSny: using your suggestion to modify mfb's method, I get an expression that can be minimised:

[tex]E(r) = \frac{1}{2}\alpha e^{\beta r} r^{-\frac{1}{2}} (2\beta r + 1)[/tex]

Setting the derivative of this to zero, I get:

[tex]r_{min} = \frac{\sqrt{2}-1}{2\beta}[/tex]

Plugging that back into the expression for ##\displaystyle \frac{E(r)}{c^2}## to calculate the "rest mass", I get:

[tex]m = 671.57ng[/tex]

(is that the correct way to get to the rest mass from here, or should I have used ##E^2 = m^2c^4 + p^2c^2##)?

To be frank, I'm still not quite sold on this approach, but I can post it in that thread, if you all don't mind. But I would be grateful if someone could check my calculation.

I got the same ##r_{min}## value as you and the same value for the rest mass of the system. I think you can also use ##E^2 = m^2c^4 + p^2c^2## for the system. For the system, p is zero and m would represent the rest mass of the system. So, you still get m = E/c² where E is the internal energy of the system.

I'm not real confident about the whole thing either.

Thanks for confirming the computation.

I'll post this up in that thread, and let Pranav decide if he agrees with it (I guess a lot depends on whether our answer is accepted by his marking system - the empirical approach! ).

 

[Saitama]

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)
[tex]E(r) = \frac{1}{2}\alpha e^{\beta r} r^{-\frac{1}{2}} (2\beta r + 1)[/tex]
I get the following equation which has no minimum.
[tex]E(r)=\frac{\alpha e^{\beta r} (3-2\beta r)}{2\sqrt{r}}[/tex]

 

[Saitama]

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,
[tex]\frac{pc}{r/2}=-\frac{dV(r)}{dr}[/tex]
[tex]2pc=\frac{\alpha e^{\beta r}(2\beta r-1)}{2\sqrt{r}}[/tex]

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

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

 

[Curious3141]

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

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?

 

[Saitama]

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

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. :)

 

[Curious3141]

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. :)

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?

 

[Saitama]

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

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. :(

 

[Curious3141]

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 657.71 but the system did not accept this too. :(

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.

 

[Saitama]

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.

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. :-)

 

[Saitama]

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!