Solving the Mystery: Exploring E << m Condition for Carbon 12 Ions

[dsilvas]

Homework Statement

What is the length L of the longest drift tube in a linac, which, operating at a frequency f = 20 MHz, is capable of accelerating Carbon-12 ions to a maximum energy of E = 100 MeV?

Relevant Equations

(1): E = ½mv^2
(2): L = v/2f

Clarification:
The statement in the title is actually from the solution to the homework question, as given by the textbook (you can see the whole thing below under "Textbook solution"). The solution doesn't explain everything, which is where my confusion comes from. Usually in my classes we ignore relativistic effects when we know v << c, but in this case it says to ignore relativistic effects because E << m where m is the rest mass of a carbon 12 ion. I don't understand why you can do that.

Numbers/Unit conversion for rest mass and energy:
The max energy in the problem is given to be 100 MeV. I looked up the mass of the Carbon 12 ion and it was m = 1.9927e-26 kg. The energy can be converted from 100 MeV to be
E = 1.6021773e-11 Joules = 1.6021773e-11 kg*m²/s²

What I'm confused about:
One of the problems I'm having is the energy in this case is clearly much GREATER than the rest mass, which is the exact opposite of the condition stated in the solution. The other problem I'm having is understanding where this condition came from in the first place.
____________________________________________________________________________________________________________________________________________________

How to get the solution:
It looks like we are supposed to end up using equation (1) (which is justified by saying E << m somehow), which ends up getting the desired result of v = 4.01e7, which we can then plug into equation (2) to get the length L of the drift tube. My final answer for L is 1.00251143 meters, and the textbook solution just rounds it to 1 meter which is fair. The final answer itself isn't the trouble I'm having.

Textbook solution:
"For constant acceleration, the ions must travel the length of the drift tube in half a cycle of the r.f. field. Thus, L = υ/2 f , where υ is the velocity of the ion. Since the energy is far less than the rest mass of the ion, we can use nonrelativistic kinematics to find υ, i.e. υ = 4.01 × 10⁷ m s−1 and finally L = 1 m. "

Final remark:
Sorry if my formatting isn't that great, this is my first time on this website and I'm sure using code makes things easier to read, but I don't really know how to use that yet. Hope this works!

 

[tnich]

Homework Statement: What is the length L of the longest drift tube in a linac, which, operating at a frequency f = 20 MHz, is capable of accelerating Carbon-12 ions to a maximum energy of E = 100 MeV?
Homework Equations: (1): E = ½mv^2
(2): L = v/2f

Clarification:
The statement in the title is actually from the solution to the homework question, as given by the textbook (you can see the whole thing below under "Textbook solution"). The solution doesn't explain everything, which is where my confusion comes from. Usually in my classes we ignore relativistic effects when we know v << c, but in this case it says to ignore relativistic effects because E << m where m is the rest mass of a carbon 12 ion. I don't understand why you can do that.

Numbers/Unit conversion for rest mass and energy:
The max energy in the problem is given to be 100 MeV. I looked up the mass of the Carbon 12 ion and it was m = 1.9927e-26 kg. The energy can be converted from 100 MeV to be
E = 1.6021773e-11 Joules = 1.6021773e-11 kg*m²/s²

What I'm confused about:
One of the problems I'm having is the energy in this case is clearly much GREATER than the rest mass, which is the exact opposite of the condition stated in the solution. The other problem I'm having is understanding where this condition came from in the first place.
____________________________________________________________________________________________________________________________________________________

How to get the solution:
It looks like we are supposed to end up using equation (1) (which is justified by saying E << m somehow), which ends up getting the desired result of v = 4.01e7, which we can then plug into equation (2) to get the length L of the drift tube. My final answer for L is 1.00251143 meters, and the textbook solution just rounds it to 1 meter which is fair. The final answer itself isn't the trouble I'm having.

Textbook solution:
"For constant acceleration, the ions must travel the length of the drift tube in half a cycle of the r.f. field. Thus, L = υ/2 f , where υ is the velocity of the ion. Since the energy is far less than the rest mass of the ion, we can use nonrelativistic kinematics to find υ, i.e. υ = 4.01 × 10⁷ m s−1 and finally L = 1 m. "

Final remark:
Sorry if my formatting isn't that great, this is my first time on this website and I'm sure using code makes things easier to read, but I don't really know how to use that yet. Hope this works!

You haven't yet put the two values in the same units for comparison. You have the kinetic energy in joules and the rest mass in kg.

 

[collinsmark]

Hello @dsilvas,

Welcome to PF!

Homework Statement: What is the length L of the longest drift tube in a linac, which, operating at a frequency f = 20 MHz, is capable of accelerating Carbon-12 ions to a maximum energy of E = 100 MeV?
Homework Equations: (1): E = ½mv^2
(2): L = v/2f

Clarification:
The statement in the title is actually from the solution to the homework question, as given by the textbook (you can see the whole thing below under "Textbook solution"). The solution doesn't explain everything, which is where my confusion comes from. Usually in my classes we ignore relativistic effects when we know v << c, but in this case it says to ignore relativistic effects because E << m where m is the rest mass of a carbon 12 ion. I don't understand why you can do that.

Numbers/Unit conversion for rest mass and energy:
The max energy in the problem is given to be 100 MeV. I looked up the mass of the Carbon 12 ion and it was m = 1.9927e-26 kg. The energy can be converted from 100 MeV to be
E = 1.6021773e-11 Joules = 1.6021773e-11 kg*m²/s²

What I'm confused about:
One of the problems I'm having is the energy in this case is clearly much GREATER than the rest mass, which is the exact opposite of the condition stated in the solution. The other problem I'm having is understanding where this condition came from in the first place.
____________________________________________________________________________________________________________________________________________________

How to get the solution:
It looks like we are supposed to end up using equation (1) (which is justified by saying E << m somehow), which ends up getting the desired result of v = 4.01e7, which we can then plug into equation (2) to get the length L of the drift tube. My final answer for L is 1.00251143 meters, and the textbook solution just rounds it to 1 meter which is fair. The final answer itself isn't the trouble I'm having.

Textbook solution:
"For constant acceleration, the ions must travel the length of the drift tube in half a cycle of the r.f. field. Thus, L = υ/2 f , where υ is the velocity of the ion. Since the energy is far less than the rest mass of the ion, we can use nonrelativistic kinematics to find υ, i.e. υ = 4.01 × 10⁷ m s−1 and finally L = 1 m. "

Final remark:
Sorry if my formatting isn't that great, this is my first time on this website and I'm sure using code makes things easier to read, but I don't really know how to use that yet. Hope this works!

When working with relativistic bodies it is very common to establish a system of units $c = 1$ (where $c$ is the speed of light).

In other words, it's common to establish a system of units where length and time are on the same footing. Light travels one light year per year, or one light second per second. $c = \frac{1 \mathrm{\ light \ second}}{1 \mathrm{\ second}} = 1$

In such system of units, you still have $E = mc^2$. But since $c = 1$, you can simplify that to $E = m$.

Don't over-think that though. All I'm saying is that just as it's possible to think of both time and distance in terms of light-seconds, it's also possible to express both mass and energy in units of electron-Volts.

I think the main problem you're having is you are comparing the mass of the $^{12}\mathrm{C}$ atom, in units of kilograms to the kinetic energy in units of Joules.

If you're going to compare the two, you need to make sure the units match. Either convert the mass of the carbon atom to units of joules or electron-Volts (by multiplying by $c^2$), or convert the kinetic energy to kilograms (by dividing by $c^2$).

 

[dsilvas]

Hello @dsilvas,

Welcome to PF!
When working with relativistic bodies it is very common to establish a system of units $c = 1$ (where $c$ is the speed of light).

In other words, it's common to establish a system of units where length and time are on the same footing. Light travels one light year per year, or one light second per second. $c = \frac{1 \mathrm{\ light \ second}}{1 \mathrm{\ second}} = 1$

In such system of units, you still have $E = mc^2$. But since $c = 1$, you can simplify that to $E = m$.

Don't over-think that though. All I'm saying is that just as it's possible to think of both time and distance in terms of light-seconds, it's also possible to express both mass and energy in units of electron-Volts.

I think the main problem you're having is you are comparing the mass of the $^{12}\mathrm{C}$ atom, in units of kilograms to the kinetic energy in units of Joules.

If you're going to compare the two, you need to make sure the units match. Either convert the mass of the carbon atom to units of joules or electron-Volts (by multiplying by $c^2$), or convert the kinetic energy to kilograms (by dividing by $c^2$).

Thank you! I understand that now. I got a value for E/$c^2$ less than the rest mass of $^{12}\mathrm{C}$, but I still don't understand why this means relativistic effects can be ignored when finding velocity.

 

[dsilvas]

You haven't yet put the two values in the same units for comparison. You have the kinetic energy in joules and the rest mass in kg.

Thank you, I divided Energy by $c^2$ and I got E/$c^2$ = 1.441959e-26. I think this is now less than the rest mass of carbon-12, but why does this mean we can exclude special relativity? Besides, It doesn't even seem to be that much less than the rest mass.

UPDATE:

I tried the other way and got m*$c^2$= 1.79343e-9, which shows the energy when in units of joules is clearly much less than the rest "mass". so E << m$c^2$. My final confusion with the question now is why this means we can use this non-relativistic formula: E = ½m$v^2$

 

[tnich]

Thank you, I divided Energy by c2 and I got E/c2 = 1.441959e-26. I think this is now less than the rest mass of carbon-12, but why does this mean we can exclude special relativity? Besides, It doesn't even seem to be that much less than the rest mass.

It looks like you calculated $E/c^2$ by multiplying 1.6 joules by $(3\times 10^{-8} m/s)^2$. You might want to take another look at that. Under what conditions would you normally use Newtonian assumptions?

 

[dsilvas]

It looks like you calculated $E/c^2$ by multiplying 1.6 joules by $(3\times 10^{-8} m/s)^2$. You might want to take another look at that. Under what conditions would you normally use Newtonian assumptions?

Thank you for spotting that. I was a little too fast with myself and didn't realize I was multiplying by 9 instead of dividing by it. Normally I use Newtonian assumptions when v << c. I suppose we are working with kinetic energy, not total (unless the total energy is just kinetic energy in this case, i.e: PE = 0), so I guess I can say ½mv² << mc², which implies that ¼v << c or essentially v << c?

 

[collinsmark]

Thank you, I understand that now. I got a value for E/$c^2$ less than the rest mass of $^{12}\mathrm{C}$, but I still don't understand why this means relativistic effects can be ignored when finding velocity.

The total energy of the $^{12} \mathrm{C}$ atom is

$E_{tot} = \gamma m c^2$,

where

$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$,

and $m$ is the "rest mass" of the $^{12} \mathrm{C}$ atom. That total includes both the particle's rest mass energy and its kinetic energy.

The kinetic energy (alone) of the $^{12} \mathrm{C}$ atom is

$E_{kinetic} = \gamma m c^2 - mc^2$.

Take the Taylor Series Expansion of that, near $v = 0$ and you'll find a very familiar result.

So what does that mean? Well, when the particle's velocity is small (i.e., $v \ll c$), the particles's total energy is pretty much the same as its rest-mass energy. The particle's kinetic energy is not a significant part of its total energy. Therefore, in that case, there's no need to consider relativistic effects.

Does that make sense?

 

[dsilvas]

The total energy of the $^{12} \mathrm{C}$ atom is

$E_{tot} = \gamma m c^2$,

where

$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$,

and $m$ is the "rest mass" of the $^{12} \mathrm{C}$ atom. That total includes both the particle's rest mass energy and its kinetic energy.

The kinetic energy (alone) of the $^{12} \mathrm{C}$ atom is

$E_{kinetic} = \gamma m c^2 - mc^2$.

Take the Taylor Series Expansion of that, near $v = 0$ and you'll find a very familiar result.

So what does that mean? Well, when the particle's velocity is small (i.e., $v \ll c$), the particles's total energy is pretty much the same as its rest-mass energy. The particle's kinetic energy is not a significant part of its total energy. Therefore, in that case, there's no need to consider relativistic effects.

Does that make sense?

This is very helpful, thank you. So from here can we just assume $E_{kinetic}$ = ½m$v^2$ for the carbon-12 ion? You said it was $\gamma m c^2 - mc^2$ just now, so how do you get from that to the other equation you can then use to find velocity ($E_{kinetic}$ = ½m$v^2$)? I see that you can move $mc^2$ to the left hand side to get $\gamma$ = 1 (as the KE is much less than mc^2), but I don't see how anything leads to $E_{kinetic}$ = ½m$v^2$

 

[collinsmark]

This is very helpful, thank you. So from here can we just assume $E_{kinetic}$ = ½m$v^2$ for the carbon-12 ion?

Right.

You said it was $\gamma m c^2 - mc^2$ just now, so how do you get from that to the other equation you can then use to find velocity ($E_{kinetic}$ = ½m$v^2$)? I see that you can move $mc^2$ to the left hand side to get $\gamma$ = 1 (as the KE is much less than mc^2), but I don't see how anything leads to $E_{kinetic}$ = ½m$v^2$

I'll leave it as an exercise for you (it's worth doing at least once in one's life). It involves the Taylor Series Expansion (link: https://en.wikipedia.org/wiki/Taylor_series).

Start with the kinetic energy $E$ as a function of velocity $v$,

$E(v) = \left( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1 \right) mc^2$

Then calculate a few derivatives of $E(v)$,
$E'(v) = \frac{d \ E(v)}{dv}$,

$E''(v) = \frac{d^2 \ E(v)}{dv^2}$

$E'''(v) = \frac{d^3 \ E(v)}{dv^3}$

and so on. (Three of them should be enough to get the idea.)

Then put it all together,

$E(v)_{v \mathrm{\ near \ 0}} = E(0) + \frac{E'(0)}{1!}v + \frac{E''(0)}{2!}v^2 + \frac{E'''(0)}{3!}v^3 + \ldots$