Same Charge, Same Mass? | Subatomic Particles

[a71cj34]

Homework Statement

Find the mass of a photon with charge 3.1eV, and an electron with charge 3.1eV.

Relevant Equations

E = mc^2

I converted 3.1eV into J, substituted into E = mc^2. Since the energy is the same, I got the same answer for both: 5.52*10^-36 kg. This doesn't seem quite right- I doubt that a photon and an electron have the same mass. So, when two particles have the same charge, does that mean they have the same mass?

 

[BvU]

a photon has no charge and an electron has charge 'one e' ...
eV is not the dimension of charge. Where does this exercise come from ?

 

[a71cj34]

a photon has no charge and an electron has charge 'one e' ...
eV is not the dimension of charge. Where does this exercise come from ?

Whoops, I meant energy. Wherever I say charge, I meant energy. Sorry for the confusion

 

[jbriggs444]

Homework Statement: Find the mass of a photon with charge 3.1eV, and an electron with charge 3.1eV.
Relevant Equations: E = mc^2

I converted 3.1eV into J, substituted into E = mc^2. Since the energy is the same, I got the same answer for both: 5.52*10^-36 kg. This doesn't seem quite right- I doubt that a photon and an electron have the same mass. So, when two particles have the same charge, does that mean they have the same mass?

The word "mass" normally means rest mass, also known as invariant mass. The term "invariant" means that it does not change depending on the frame of reference that you use. This, in turn, means that it does not depend on kinetic energy at all. Photons have zero invariant mass regardless of their kinetic energy.

Electrons, on the other hand, have a non-zero invariant mass.

https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

For any particle, invariant mass is given by the [invariant] magnitude of the energy-momentum four-vector. In simpler terms, you can compute it as $m^2c^4 = E^2 - (pc)^2$. As you may notice, when $p=0$, you obtain the popular equation: $E=mc^2$. However, for a photon, $p$ is never zero.

For a photon, $E=pc$ and it follows that $m=0$.

 

[a71cj34]

The word "mass" normally means rest mass, also known as invariant mass. The term "invariant" means that it does not change depending on the frame of reference that you use. This, in turn, means that it does not depend on kinetic energy at all. Photons have zero invariant mass regardless of their kinetic energy.

Electrons, on the other hand, have a non-zero invariant mass.

https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

I'm pretty sure the question is referring to rest mass. The question did not specify kinetic energy- just that it had energy. I'll have to look back at my course notes to see if the question is just bad writing or if zero is the intended answer for the photon. For the electron, I'll rephrase the question: Can energy affect the mass of an electron? The literature value that I've been using so far in physics is around 1*10^-5 away from the calculated value in this question.

 

[jbriggs444]

For the electron, I'll rephrase the question: Can energy affect the mass of an electron?

Kinetic energy cannot affect the mass of an electron. The term "invariant mass" makes this clear.

I suppose that it is conceivable that an electron in a bound state could have an energy deficit (or surplus?) so that its share of the total energy (and, hence, its share of the total mass) could deviate from the mass of a free electron.

For objects with non-zero invariant mass, "rest mass" is equal to "invariant mass" and is equal to the energy of the object in a frame of reference where it has zero momentum. Of course, this energy figure is divided by $c^2$.

 

[a71cj34]

Kinetic energy cannot affect the mass of an electron. The term "invariant mass" makes this clear.

I suppose that it is conceivable that an electron in a bound state could have an energy deficit (or surplus?) so that its share of the total energy (and, hence, its share of the total mass) could deviate from the mass of a free electron.

For objects with non-zero invariant mass, "rest mass" is equal to "invariant mass" and is equal to the energy of the object in a frame of reference where it has zero momentum. Of course, this energy figure is divided by $c^2$

Hmm, ok then. I'm just wondering because I don't remember learning this in my course. I'll take another look at the course notes.

Of course, this energy figure is divided by c^2

This looks like E = mc^2. Is that it? If so then my answer is correct.

 

[jbriggs444]

Hmm, ok then. I'm just wondering because I don't remember learning this in my course. I'll take another look at the course notes.

This looks like E = mc^2. Is that it? If so then my answer is correct.

Unfortunately, $E=mc^2$ in the context of a photon is incorrect. It would be correct if the $m$ meant "relativistic mass". But the concept of relativistic mass is deprecated for a reason.

 

[a71cj34]

Unfortunately, $E=mc^2$ in the context of a photon is incorrect. It would be correct if the $m$ meant "relativistic mass". But the concept of relativistic mass is deprecated for a reason.

Oh no, I'm talking about the electron's rest mass for the E = mc^2. About the photon, I found a small mention of it in my course notes:

Since a photon is a small bundle of energy with no mass [...]

So I do think zero is the intended answer for the photon. Thanks.

 

[jbriggs444]

Oh no, I'm talking about the electron's rest mass for the E = mc^2.

I am not sure what result you got for the rest mass of a 3.1 eV electron.

It would not be $\frac{3.1 \text{eV}}{c^2}$

 

[a71cj34]

I am not sure what result you got for the rest mass of a 3.1 eV electron.

It would not be $\frac{3.1 \text{eV}}{c^2}$

I got 5.525 x 10^-36 kg: which is off of the literature value of 9.1 x 10^-31 kg. Is that difference reasonable? Why is the mass different in the first place? I'm not sure what's going on here. At this point, I'm thinking I'll just put that as my answer as this is a high-school course: I might be overthinking it.

 

[kuruman]

I got 5.525 x 10^-36 kg: which is off of the literature value of 9.1 x 10^-31 kg. Is that difference reasonable? Why is the mass different in the first place? I'm not sure what's going on here. At this point, I'm thinking I'll just put that as my answer as this is a high-school course: I might be overthinking it.

Please show your calculation. This looks like a nonsense question to me for these reasons:
The photon is massless.
A 3.1 eV electron is non-relativistic. Asking for its mass, given its energy, is like asking for the mass of a car if it has an energy of 80,000 Joules.

It would make more sense if the problem asked for the speed of each particle.

 

[vela]

I'm pretty sure the question is referring to rest mass. The question did not specify kinetic energy- just that it had energy. I'll have to look back at my course notes to see if the question is just bad writing or if zero is the intended answer for the photon. For the electron, I'll rephrase the question: Can energy affect the mass of an electron? The literature value that I've been using so far in physics is around 1*10^-5 away from the calculated value in this question.

@jbriggs already answered your question. No, energy does not affect the mass of an electron. The mass is a constant. It's invariant. The mass doesn't change because the electron is moving.

You need to know what the symbols in an equation mean and, more importantly, how they are defined. In $E^2 - (pc)^2 = (mc^2)^2$, $E$ is the total relativistic energy; $p$ is the momentum; $c$ is the speed of light; and $m$ is the invariant or rest mass. For an electron, $m$ is $9.11\times 10^{-31}~\rm kg$.

If the electron isn't moving, it has zero momentum, and the relationship reduces to $E = mc^2$. That is, $E_0=mc^2$ is the energy an electron has when it's at rest, i.e., the rest energy. If the electron is moving, it has an energy $E$ that's greater than the rest energy $E_0$. The difference $K=E-E_0$ is the kinetic energy.

Hopefully, it's clear that $m=E_0/c^2$. That is, if you divide the rest energy by $c^2$, you get the particle's mass. You can't just plug in the kinetic energy or the total energy instead and hope to get an answer that makes sense.

 

[a71cj34]

Please show your calculation. This looks like a nonsense question to me for these reasons:
The photon is massless.
A 3.1 eV electron is non-relativistic. Asking for its mass, given its energy, is like asking for the mass of a car if it has an energy of 80,000 Joules.

It would make more sense if the problem asked for the speed of each particle.

This is the original question:

My calculation is:

\begin{align*}
E &= mc^2 \\
4.96675 \cdot 10^{-19} &= m(299792458^2) \\
m &= 5.5256 \cdot 10^{-36}
\end{align*}

 

[BvU]

This is the original question:

Better late than never... but you would have received much better help if you would have started with this !

$\$

 

[a71cj34]

@jbriggs already answered your question. No, energy does not affect the mass of an electron. The mass is a constant. It's invariant. The mass doesn't change because the electron is moving.

You need to know what the symbols in an equation mean and, more importantly, how they are defined. In $E^2 - (pc)^2 = (mc^2)^2$, $E$ is the total relativistic energy; $p$ is the momentum; $c$ is the speed of light; and $m$ is the invariant or rest mass. For an electron, $m$ is $9.11\times 10^{-31}~\rm kg$.

If the electron isn't moving, it has zero momentum, and the relationship reduces to $E = mc^2$. That is, $E_0=mc^2$ is the energy an electron has when it's at rest, i.e., the rest energy. If the electron is moving, it has an energy $E$ that's greater than the rest energy $E_0$. The difference $K=E-E_0$ is the kinetic energy.

Hopefully, it's clear that $m=E_0/c^2$. That is, if you divide the rest energy by $c^2$, you get the particle's mass. You can't just plug in the kinetic energy or the total energy instead and hope to get an answer that makes sense.

Alright, I'll do what you said. I'll assume the energy given is the rest energy. Thanks @jbriggs for the answer and everyone else who replied.

 

[vela]

I'll assume the energy given is the rest energy.

You already did that and found a mass that five orders of magnitude too small. That means your assumption that 3.1 eV is the rest energy is wrong. Why do you seem so unwilling to let go of this notion that 3.1 eV is the rest energy?

 

[kuruman]

Alright, I'll do what you said. I'll assume the energy given is the rest energy. Thanks @jbriggs for the answer and everyone else who replied.

Your mistake was that you used $E=mc^2$ which is the total energy which includes the rest mass energy. The 3.1 eV that you are given is the kinetic energy $K$ which for a free particle is the total energy minus the rest energy,
$$K=mc^2-m_0c^2=\gamma m_0c^2-m_0c^2=(\gamma-1)m_0c^2$$ where $\gamma = \dfrac 1 {\sqrt {1 - \frac {v^2} {c^2}}}.$ Note that when $K<<m_0c^2$ as is the case here, you will have to do a series expansion for $v<<c.$ The result for $K$ is something very familiar.

 

[topsquark]

This is the original question:
View attachment 329641
My calculation is:

\begin{align*}
E &= mc^2 \\
4.96675 \cdot 10^{-19} &= m(299792458^2) \\
m &= 5.5256 \cdot 10^{-36}
\end{align*}

The electron clearly has a total energy of 3.1 eV. So you have its energy (given), you have its rest mass, m (given in tables), and you can get its momentum from $E^2 = (pc)^2 + (mc^2)^2$. You can get the speed from the momentum, and you can get the wavelength from deBroglie.

The photon also has a total energy of 3.1 eV. It's rest mass is 0 eV, its speed should be obvious, and you can get its momentum from $E^2 = (pc)^2 + (mc^2)^2$. You can get wavelength from the Planck equation.

Why don't you fill in the whole table and post your reasoning? That will help us know what it is that you know.

-Dan

 

[kuruman]

The electron clearly has a total energy of 3.1 eV.

I disagree. The electron has total energy $E=0.511~\text{MeV}+3.1~\text{eV}.$ See post #18.

 

[a71cj34]

Your mistake was that you used $E=mc^2$ which is the total energy which includes the rest mass energy. The 3.1 eV that you are given is the kinetic energy $K$ which for a free particle is the total energy minus the rest energy,
$$K=mc^2-m_0c^2=\gamma m_0c^2-m_0c^2=(\gamma-1)m_0c^2$$ where $\gamma = \dfrac 1 {\sqrt {1 - \frac {v^2} {c^2}}}.$ Note that when $K<<m_0c^2$ as is the case here, you will have to do a series expansion for $v<<c.$ The result for $K$ is something very familiar.

I'm not entirely sure what this means. You lost me at the gamma and the series expansion. As I said, this is a Grade 12 Physics course, and I'm sorry, but I feel like this is getting a little too complex for my level. Is there any other way to solve the question without doing all of this? I get what you're saying about the total energy and the kinetic energy. Additionally, can I ask how you know the 3.1eV is the kinetic energy and not the total? Is it implied in the question?

 

[a71cj34]

The electron clearly has a total energy of 3.1 eV. So you have its energy (given), you have its rest mass, m (given in tables), and you can get its momentum from $E^2 = (pc)^2 + (mc^2)^2$. You can get the speed from the momentum, and you can get the wavelength from deBroglie.

The photon also has a total energy of 3.1 eV. It's rest mass is 0 eV, its speed should be obvious, and you can get its momentum from $E^2 = (pc)^2 + (mc^2)^2$. You can get wavelength from the Planck equation.

Why don't you fill in the whole table and post your reasoning? That will help us know what it is that you know.

-Dan

Ok, I understand. Thanks for laying it out like this. I wasn't taught that momentum equation but it looks very useful. When you say "given in tables" for the rest mass, can I ask what you mean? The way I see it, I have two possible answers: a literature value from online (but I'm pretty sure I'm supposed to derive it myself somehow) or the value I got from E = mc^2, (which doesn't match up with the literature value). I'm actually starting to lean towards just using a literature value. Does "given in tables" mean a literature value?

 

[kuruman]

Additionally, can I ask how you know the 3.1eV is the kinetic energy and not the total? Is it implied in the question?

Yes, it is implied in the question. A free electron has kinetic energy. It may have potential energy if it's in a field. Here no field is mentioned, kinetic energy is all that the 3.1 eV can be.

I think you should give up the idea of deriving the mass from the given energy of 3.1 eV and look it up. The idea is to use the kinetic energy and the mass to find the speed. The expression that I invited you to derive in post #18 is $$K=\frac{1}{2}m_0v^2$$ which you knew already. Knowing the speed, you can find the momentum and hence the wavelength.

Repeat for the photon. If you look up its mass, you will find that it is zero. You can also look up its speed, if you have not already memorized it. Fill up the photon entries as with the electron and you're done.

 

[a71cj34]

Yes, it is implied in the question. A free electron has kinetic energy. It may have potential energy if it's in a field. Here no field is mentioned, kinetic energy is all that the 3.1 eV can be.

I think you should give up the idea of deriving the mass from the given energy of 3.1 eV and look it up. The idea is to use the kinetic energy and the mass to find the speed. The expression that I invited you to derive in post #18 is $$K=\frac{1}{2}m_0v^2$$ which you knew already. Knowing the speed, you can find the momentum and hence the wavelength.

Repeat for the photon. If you look up its mass, you will find that it is zero. You can also look up its speed, if you have not already memorized it. Fill up the photon entries as with the electron and you're done.

Wow, that is more simple than I thought xD! I'm going to ask my teacher tomorrow about what she expects for the mass of the electron- I suspect that we're just supposed to use a literature value. I think this question is pretty much solved, unless I'm supposed to actually calculate the mass of the electron (but still, I think you've got a very good, if not a bit complex, explanation for how to do it). Thanks so much to everyone who replied and contributed to the question.

 

[kuruman]

Wow, that is more simple than I thought xD! I'm going to ask my teacher tomorrow about what she expects for the mass of the electron- I suspect that we're just supposed to use a literature value. I think this question is pretty much solved, unless I'm supposed to actually calculate the mass of the electron (but still, I think you've got a very good, if not a bit complex, explanation for how to do it). Thanks so much to everyone who replied and contributed to the question.

I am pretty sure that your teacher will confirm your suspicion. It's either look up the mass or not be able to finish the problem.

 

[Steve4Physics]

Wow, that is more simple than I thought xD! I'm going to ask my teacher tomorrow about what she expects for the mass of the electron

For a high school course, you may be given a data-booklet or data-sheet with relevant values and formulae. If so, the electron's rest mass is almost certainly in it and that's what you are expected to use.

Note that the formulae provided may not be complete. You are generally expected to memorise 'essential' formulae such as $KE = \frac 12 mv^2$.

 

[a71cj34]

I am pretty sure that your teacher will confirm your suspicion. It's either look up the mass or not be able to finish the problem.

Yup, we were supposed to Google it. Thanks to everyone who replied.

 

[kuruman]

You lost me at the gamma and the series expansion.

Now that the mass issue is resolved, I think it is worth your while to see what follows because it shows how Newtonian mechanics is an approximation of relativistic mechanics. As you know, the zero of any energy is an arbitrary choice. We normally take kinetic energy to be zero when an object is "at rest" but what does that mean? Is a baby strapped in her seat in the back of a car at rest? Relative to her father driving the car she is; relative to her grandparents standing on the driveway waving bye-bye she is moving therefore has non-zero kinetic energy.

In relativistic mechanics when a particle is at rest in an inertial frame, its energy is not zero. It is the rest mass energy $E_0=m_0~c^2$. If it starts moving relative to that frame, its energy becomes $E=\gamma~m_0~c^2$. Therefore the object's kinetic energy is $$K=E-E_0=(\gamma-1)m_0~c^2~~~\left(\gamma = \frac {1} {\sqrt {1 - \frac {v^2} {c^2}}}\right). $$The question is how to use this expression to calculate the kinetic energy when the speed of the object $v$ is of order 10 miles per hour and the speed of light $c$ is of order 100,000,000 miles per hour. Putting the numbers in a calculator won't work because $\gamma$ is extremely close but not equal to $1$. An approximate expression can be found using nothing more than simple high-school algebra. Note that $$(\gamma-1)=\frac{(\gamma-1)(\gamma+1)}{(\gamma+1)}=\frac{\gamma^2-1}{\gamma+1}.$$The numerator of this expression is $$N=\gamma^2-1=\frac {1} {1 - \frac {v^2} {c^2}}-1=\frac{1-\left(1-v^2/c^2\right)}{1 - v^2 /c^2}=\frac{v^2/c^2}{1-v^2/c^2}.$$Up to this point the expression for the numerator is exact. When $v/c$ is of order $10^{-8}$, it is an extremely good approximation to write $$N\approx \frac{v^2/c^2}{1}.$$ For the denominator, $$D=\gamma+1\approx 1+1=2.$$Then $$\gamma-1=\frac{N}{D}=\frac{v^2/c^2}{2}=\frac{1}{2}\frac{v^2}{c^2}.$$This gives the (extremely good) Newtonian approximation to the kinetic energy $$K=(\gamma-1)~m_0~c^2\approx\frac{1}{2}\frac{v^2}{c^2}~m_0~c^2=\frac{1}{2}~m_0~v^2.$$Example: Find $\gamma$ for a car that is moving at 30 m/s (67 mi/hr). $$\gamma -1=\frac{1}{2}\left(\frac{v}{c}\right)^2=\frac{1}{2}\left(\frac{30}{3\times 10^8}\right)^2 =5\times 10^{-15}\implies \gamma = 1+5\times 10^{-15}.$$ Now you see why a calculator is of limited use here.

 

[kuruman]

So to answer the original question in the title, two particles with the same rest mass (e.g. an electron and a positron) will always have the same relativistic mass if they have the same energy.

If one adopts the premise that "mass is energy and energy is mass", two particles that do not have the same rest mass could have the same relativistic mass if their energies are right. For example, the rest mass of the proton is about 1800 times the rest mass of the electron, $m_p=1800~m_e.$ Since $E=\gamma~m_0~c^2$, one could argue that the two particles have the same mass if the condition $~\gamma_e=1800~\gamma_p~$ is met.

In this example, just to match the rest mass of the proton ($\gamma_p=1$), the electron must have $\gamma_e=1800.$ If one does the math, one will find that the electron must be accelerated to speed $$v_e=\sqrt{\frac{1800^2-1}{1800^2}}~c=0.9999998~c.$$ If the proton is moving, ($\gamma_p>1$) the electron must be moving even faster than that.