# Questions about Electrons and Photons **URL:** \ **Category:** Factual Questions\ **Created:** [October 9, 2011, 6:43pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095 "2011-10-09T18:43:39Z")\ **Posts on this page:** 12\ **Page:** 1
**Author:** ![drewtwo99](https://avatars.discourse-cdn.com/v4/letter/d/d07c76/32.png) [@drewtwo99](https://boards.straightdope.com/u/drewtwo99)\ **Post date:** [October 9, 2011, 6:43pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095/1 "2011-10-09T18:43:39Z")
Hi everyone… I have a couple of physics questions. I did my best to search the board and I got some pretty good information, but nothing that I could find quite answered my questions. Question 1: When an electron of an atom absorbs a photon, does it’s mass increase, in any sense? I know that there is a lot of discrepancy on what the word mass means, but if you could break it down for me I’d appreciate it. I was a physics major so I can probably handle a fairly technical explanation. Question 2: Photons have no rest mass, but can you attribute the word mass in _any_ sense to a photon? I have always been taught that photons don’t have any mass, period, whatsoever. They have MOMENTUM, but not mass, and photons having any sort of mass just makes no sense whatsoever. Am I mistaken on this one? Question 3: All things considered equal, does a hot brick of copper have more mass than a cold brick of copper? Why is this the case, if true? If true, is this related to a higher energy electron having more mass than a lower energy electron? You’d think as a physics major I’d know the answer to these questions, but yeah… we didn’t really go over this stuff in undergrad. Thanks so much for the help. ---
**Author:** ![drewtwo99](https://avatars.discourse-cdn.com/v4/letter/d/d07c76/32.png) [@drewtwo99](https://boards.straightdope.com/u/drewtwo99)\ **Post date:** [October 9, 2011, 7:36pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095/2 "2011-10-09T19:36:48Z")
I’ve continued to do some research on my questions, which have lead me to even more questions. “This mass decrease is also equivalent to the energy required to break up the nucleus into individual protons and neutrons (in this case, work and mass would need to be supplied). Similarly, the mass of the solar system is slightly less than the masses of sun and planets individually.” - From Wikipedia article on mass-energy equivalence. Question 4: Why is the mass of the solar system slightly less than the sum of the masses of its constituent parts? I understand the necessity to supply work in order to get stuff done, but what does it mean that “mass would need to be supplied”? ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [October 9, 2011, 7:39pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095/3 "2011-10-09T19:39:51Z")
1. The electron does not absorb the photon. The atom absorbs the photon, with the electron changing from a low energy state to a higher energy state. For example, the electron in a Hydrogen atom may move from the S0 orbital to P1. The mass of the atom in that state will be slightly higher than in the lower energy state. 2. No, you’re correct. Individual photons do not have mass. A system of two or more photons can have mass. 3. Yes, it has more mass, because it has more energy. The difference will be very small. 2 and 3 both assume the definition of mass as the energy of a system that can’t be transformed away by changing to a different reference frame. This gets covered here every month or so, it seems, and written better than I’d be able to do. Searching for “energy” and “rest mass” should get you multiple threads to wade through. ETA: 4) The gravitational binding energy is negative. To eject Jupiter from the Solar System, for example, you’d need to supply a lot of energy to move it away from the Sun, but the energy of Jupiter itself wouldn’t change. ---
**Author:** ![drewtwo99](https://avatars.discourse-cdn.com/v4/letter/d/d07c76/32.png) [@drewtwo99](https://boards.straightdope.com/u/drewtwo99)\ **Post date:** [October 9, 2011, 7:50pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095/4 "2011-10-09T19:50:54Z")
Thanks ZenBeam. After reading through some of the articles here, it seems that the knowledgeable people are saying that mass and energy are completely different properties of a system, and both are conserved independently in a closed system. So why does increasing energy necessarily also increase mass (in the example of hot copper vs cold copper). Is it possible to increase the energy of a system without increasing the mass? Can a free electron interact (absorb?) a photon and increase its energy that way? When they accelerate electrons in particle accelerators, can you think of them as absorbing photons (or virtual photons)? Does a free electron with more energy necessarily have more mass? I guess what I am asking is: Does more energy in a system always require there to be more mass? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [October 9, 2011, 8:17pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095/5 "2011-10-09T20:17:29Z")
Mass and energy are not completely different properties: In fact, mass is one portion of the total energy. And if you considered the system of atom plus photon heading towards the atom, before the absorbtion, that system does in fact have the same mass as the system of energized atom after it’s absorbed the photon. The biggest distinction between mass and any other form of energy is that the total energy of a system depends on what reference frame you’re looking at it from, while the mass does not. In the reference frame of Pauline tied to the railroad tracks, the oncoming train has a great deal of kinetic energy, but in the reference frame of a passenger on the train, the train is at rest, and therefore has zero kinetic energy. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [October 9, 2011, 8:21pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095/6 "2011-10-09T20:21:20Z")
We had a discussion just the other day on a closely related topic. [http://boards.straightdope.com/sdmb/showthread.php?p=14292397](http://boards.straightdope.com/sdmb/showthread.php?p=14292397) ---
**Author:** ![drewtwo99](https://avatars.discourse-cdn.com/v4/letter/d/d07c76/32.png) [@drewtwo99](https://boards.straightdope.com/u/drewtwo99)\ **Post date:** [October 9, 2011, 8:26pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095/7 "2011-10-09T20:26:44Z")
OK Chronos, I gotcha. That helps explain why there isn’t any mass to energy conversion going on. Mass is a type of energy, but by definition, it cannot change and is invariant from reference frame to reference frame. So how is the mass of a closed system calculated? I think I understand what you are saying when you say that the mass of a system doesn’t change from one point in time to the next, but how can a system of two or more photons have a calculable mass if when individual photons are considered, they do not have mass. I am assuming this has something to do with mass being a part of (property of?) momentum but I’m not sure. Andy: Yeah I read through that one all the way but was still confused. I realize I’m asking questions that have been more or less answered, but I am still unclear on some of the details. ---
**Author:** ![Asympotically\_fat](https://avatars.discourse-cdn.com/v4/letter/a/e47c2d/32.png) [@Asympotically\_fat](https://boards.straightdope.com/u/Asympotically_fat)\ **Post date:** [October 9, 2011, 8:30pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095/8 "2011-10-09T20:30:58Z")
> [@drewtwo99](#): > > OK Chronos, I gotcha. That helps explain why there isn’t any mass to energy conversion going on. Mass is a type of energy, but by definition, it cannot change and is invariant from reference frame to reference frame. > > So how is the mass of a closed system calculated? I think I understand what you are saying when you say that the mass of a system doesn’t change from one point in time to the next, but how can a system of two or more photons have a calculable mass if when individual photons are considered, they do not have mass. I am assuming this has something to do with mass being a part of (property of?) momentum but I’m not sure. > [@Why is mass conserved in ordinary chemical reactions but not in nuclear reactions?](https://boards.straightdope.com/t/why-is-mass-conserved-in-ordinary-chemical-reactions-but-not-in-nuclear-reactions/597507/11): > > Whether mass is conserved (in special relativity) just depends on how you look at it. Mass is the norm (i.e. the ‘length’) of the four momentum vector **P**. For an isolated system **P** [sub]system[/sub] is constant and hence it’s norm | **P** [sub]system[/sub]| is constant, i.e. the rest mass of the system considered as a whole is conserved > > Let’s say the system is made up of a number of n particles with four momenta **P** [sub]1[/sub], **P** [sub]2[/sub], …, **P** [sub]n[/sub], then > > **P** [sub]system[/sub] = **P** [sub]1[/sub] + **P** [sub]2[/sub] + … + **P** [sub]n[/sub] > > however the sum (which I just call M for convenience,not to imply that it’s more proper to think of it as the mass of the system) > > M[sub]system[/sub] = | **P** [sub]1[/sub]| + | **P** [sub]2[/sub]| + … + | **P** [sub]n[/sub]| > > (where **P** [sub]1[/sub], etc, are the rest masses of the individual particles) will not neccesarily be equal to | **P** [sub]system[/sub]| and is not necessarily constant, i.e. it is not conserved. > > In many ways most prefer to talk about the conservation of| **P** [sub]system[/sub]| in terms of the conservation of energy (it’s just the statement that the system’s energy is conserved in it’s rest frame) rather than describe at as the conservation of mass. On the other hand it’s a much of a muchness: | **P** [sub]system[/sub]| is conserved and M isn’t conserved so if you choose to talk about the conservation of mass in terms of the former then mass is conserved and if you choose the latter it isn’t conserved. (slightly edited to correct a typo) [http://boards.straightdope.com/sdmb/showthread.php?p=14292397](http://boards.straightdope.com/sdmb/showthread.php?p=14292397) ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [October 9, 2011, 8:31pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095/9 "2011-10-09T20:31:03Z")
An (admittedly) imperfect analogy: Consider happiness. A single person can have happiness, and a system of two or more people can have happiness, but the happiness of a pair of people is not necessarily just the sum of the happinesses of the two individuals. ---
**Author:** ![drewtwo99](https://avatars.discourse-cdn.com/v4/letter/d/d07c76/32.png) [@drewtwo99](https://boards.straightdope.com/u/drewtwo99)\ **Post date:** [October 9, 2011, 8:43pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095/10 "2011-10-09T20:43:56Z")
These are my own pants: Thanks for reposting your explanation, and I’ve read through it several times but I feel like I missing some crucial understanding in order to make sense of it. If mass is the norm of a momentum vector P, why do we say photons are massless when they certainly have momentum? How is the norm of a photon’s P vector equal to 0? Where is the breakdown in my understanding of what you mean when you represent the vector P and it’s norm as mass. Chronos: Nice analogy. But if one person has no happiness, and another person in the room has no happiness, couldn’t one say that the room contains no happiness? This seems to be what’s happening with the photons. I do understand that adding up the mass of each constituent within a system with mass is NOT going to give you the mass of the system (like an atom, or the solar system), but I don’t understand how two massless photons taken together as a closed system could contain mass. Especially considering the explanation that what we call mass is the norm of a momentum vector. Also, I just want to say thank you for trying to help clear this up for me. I’m not trying to be argumentative or picky or anything, just genuinely trying to grasp the concepts. ---
**Author:** ![Asympotically\_fat](https://avatars.discourse-cdn.com/v4/letter/a/e47c2d/32.png) [@Asympotically\_fat](https://boards.straightdope.com/u/Asympotically_fat)\ **Post date:** [October 9, 2011, 9:12pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095/11 "2011-10-09T21:12:23Z")
> [@drewtwo99](#): > > These are my own pants: Thanks for reposting your explanation, and I’ve read through it several times but I feel like I missing some crucial understanding in order to make sense of it. If mass is the norm of a momentum vector P, why do we say photons are massless when they certainly have momentum? How is the norm of a photon’s P vector equal to 0? Where is the breakdown in my understanding of what you mean when you represent the vector P and it’s norm as mass. Sorry, maybe I should make this clear, it’s [four momentum](http://en.wikipedia.org/wiki/Four-momentum)I’m talking about, the 4-D spacetime analog of momentum. In spacetime, due to the different geometry, four-vectors, like a four-momentum vector, can have non-zero components (in some basis) but have norms (lengths) of zero. Such vectors are called null vectors (and are distinct from zero vectors which also have norms of zero, but all components equal to zero too). So we say a photon has null four momentum, which instantly implies it must have zero mass. ---
**Author:** ![drewtwo99](https://avatars.discourse-cdn.com/v4/letter/d/d07c76/32.png) [@drewtwo99](https://boards.straightdope.com/u/drewtwo99)\ **Post date:** [October 9, 2011, 9:39pm UTC](https://boards.straightdope.com/t/questions-about-electrons-and-photons/599095/12 "2011-10-09T21:39:02Z")
> [@These are my own pants](#): > > Sorry, maybe I should make this clear, it’s [four momentum](http://en.wikipedia.org/wiki/Four-momentum)I’m talking about, the 4-D spacetime analog of momentum. > > In spacetime, due to the different geometry, four-vectors, like a four-momentum vector, can have non-zero components (in some basis) but have norms (lengths) of zero. Such vectors are called null vectors (and are distinct from zero vectors which also have norms of zero, but all components equal to zero too). So we say a photon has null four momentum, which instantly implies it must have zero mass. Ah alright! This stuff is so cool. Makes me want to go back and get a graduate degree in quantum mechanics or atomic physics, or whatever this stuff would fall under, haha! I just want to make sure I am understanding this all correctly, because I want to be able to intelligently pass on the information to my co-worker who doubted me when I was telling her that an atom’s mass increases when it absorbs a photon, but I couldn’t really explain why. (From here on out I’m just going to use the term momentum synonymously with four-momentum since that’s what I’m concerned with) A given photon has a momentum vector P, which is null, which means it is massless. Calculating the norm of any individual photon’s P will always yield 0. When you are calculating the P of an entire closed system, however, the P’s of each individual photon, when combined, may make a new system P vector that is not null, meaning that the system as a whole has mass. So when you have a photon coming in toward an atom, the P vector of the photon combines with the P of the atom, and this slightly increases the norm of the original P somehow. In this sense, the “mass” of the system is not conserved, because the sum of the norms of each momentum vector is less than the norm of the new combined momentum vector. Do I have it about right, or am I still missing something?