Why is the speed of light such an important concept in physics and atomic structure?

Factual Questions
21

Uh, actually, I thought nuclear and thermonuclear weapons worked precisely because mass was converted to energy. The curve whose name I forget shows quite an energy drop for nucleons as you work your way from hydrogen to helium and so on, until you get to iron, roughly, and then it starts to go up again a bit more slowly.

Nuclear weapons release energy when a large atom (e.g., plutonium) splits into two smaller ones, and all the pieces have less mass than what you started with. Thermonuclear weapons, conversely, release energy when two tiny atoms (e.g., hydrogen) combine into a larger one (e.g., helium) and the result has less mass than the original pieces did.

The vault containing the weapon, if it existed afterwards, would have the same mass as before, but the weapon’s explosive contents would not!

Right?

22

Mass and energy are not things, they’re properties of a system.

m[sup]2[/sup] = E[sup]2[/sup] - p[sup]2[/sup]

m = mass
E = energy
p = momentum
c = 1

If the momentum of a system is zero before an event (detonation) it must be zero after he event. Therefore:

m[sup]2[/sup] = E[sup]2[/sup] - 0[sup]2[/sup] or m = E

If you are consistant in defining your system mass and energy cannot change. They are both conserved.

In a nuclear explosion you can have a local mass defect but this is a result of a local loss of energy not the cause.

23

Ring, you obviously know more about this than I do, and I don’t want to mislead people (or be mistaken in my undersanding!), so can you help me out with this?

I always thought, according to E=MC^2, that matter and energy really are equivalent, and the only way you can tell them apart is that matter has mass/inertia. It would seem that if you blow up a nuclear bomb, and some of that matter is converted to energy, that the portion of the bomb that has been converted no longer is massive. The sum of the matter and energy is the same, but some of the mass is gone. I would think that in a closed container, the container maybe would absorb energy released by the nuclear reactions, and thus the whole thing would have the same weight on a scale, but if you emptied the container of its contents and weighed all the matter that you found inside, you would find less mass than when you started. Photons have momentum, but no “inertial mass” right? Also, I understand (or did once) the basics of Lorentz transformations in terms of Minkowski space, so I think I know a little bit of your talking about; but when you say “energy is the time component”, I’m stumped. I mean, energy is in units kg(m/s)^2. How could energy just be the time component? I thought the direction and magnitude of the 4-vector in direction t is the “time component” of the 4-vector.

24

Think of it this way Loopydude. A single photon has no mass. A system of two photons with parallel velocity also has no mass, but a system of two photons with anti-parallel velocity does have mass. The reason this is so is the latter system has a center of momentum frame and thus the system’s momentum equals zero. m = E.

You see energy isn’t radiation and mass isn’t matter. Mass and energy are properties of a system and that’s all they are. A system composed entirely of radiation has mass if it has a center of momentum frame and it therefore has inertia.

When a nucleus fissions potential energy is converted to kinetic and radiant energy and because of this local loss of energy there’s a local mass defect. But the mass of the entire system is absolutely conserved. There is no conversion. Energy remains energy and mass remains mass.

25

Hey, you got me riled! :slight_smile: So I went to my references.

Well, that may or may not be true for the fission/fusion cases. I see from Britannica that the issue there is the “binding energy”, although I am sure the binding energy (like any other form of energy) can be treated as mass.

However, I have found what I think is a complete refutation of your statement about conservation of mass. It is what happens when matter and antimatter meet:

I knew I didn’t imagine hearing that there was no such thing as conservation of mass - you have conservation of mass-energy, i.e., the combined values of the two properties don’t change.

But since I’m not a physicist (and I don’t know whether you are), I thought I’d bring in some backup.

26

The only issue is one of semantics. “Mass” does not have a unique definition; that four letter word is used to refer to different things in different contexts.

A thought experiment:

Take a closed box with an electron and a positron at the middle, ready to annihilate one another.



--------------------
|                  |
|     (e+)(e-)     |
|                  |
--------------------

Everyone agrees that the entire system has mass M equal to the sum of the box’s mass and the electron’s mass and the positron’s mass.

Let them annihilate, and pretend the walls can perfectly reflect the two resulting photons:



--------------------
|                  |
|  <~~~~    ~~~~>  |
|                  |
--------------------

If you could not look inside the box, you would not be able to tell that the annihilation had occured, even by measuring its gravitation pull (which will not have changed.) Using Ring’s sense of the word “mass”, the mass has not changed.

Now pretend the box is not there, and repeat the annihilation.

While one can still refer to the entire system, one does not need to. Subsets of this system are obvious, whereas in the “sealed box” version, one only had the entire boring sealed box after the annihilation (meaning the only natural thing one could talk about was, in fact, the entire system.) In these cases, Ring’s definition of mass is not (in my opinion) the most commonplace definition. Indeed, the general public and many subfields of physics would not casually say that there is mass present. In such situations, one can use the phrase “invariant mass” to refer explicitly to the quantity defined in Ring’s posts. (The adjective “invariant” refers to the fact that this quantity of the system has the same value regardless of the observer’s reference frame.)

When talking about general relativity or cosmology, subsets of systems are not usually what one is interested in. (E.g., It is the entire star’s gravitational signature that is of interest.) Thus, folks that deal with such things will usually use “mass” to refer to invariant mass.

In particle physics, the subsets of systems are the meat and potatoes (e.g., there are two photons now, and each has zero mass.) A particle physicist will usually use the unadorned word “mass” in a subset sense, using the phrase “invariant mass” when (s)he is referring to the system-wide quantity. And when this rule of thumb fails, context usually makes it clear which “mass” (s)he was using.

In the general public, well… invariant mass, schminvariant mass! If the system has distinct subsets, they are treated independently. Thus, an isotope that gamma decays:

X —> Y + photon

goes from having mass M[sub]X[/sub] to having mass M[sub]Y[/sub] + M[sub]photon[/sub] = M[sub]Y[/sub] + 0 = M[sub]Y[/sub].

“Mass” is by no means the only overloaded word in English or in science. As long as everyone involved understands the possible uses, it just plain doesn’t matter.

Now, regarding the OP… :slight_smile:

Many others have given insightful explanations about the ubiquitousness of c, but I wanted to condense the ideas into a single chain, lest the big picture be lost:

  1. Our universe seems to have three observable dimensions that behave in one way (a way we have chosen to call “spacelike”) and a fourth observable dimension that behaves in another way (a way we have chosen to call “timelike”.)
  2. Relativity describes how these spacelike and timelike dimensions affect (and are affected by) objects in them. In particular, an object’s motion in a spatial dimension is constrained by its motion in the temporal dimension (or vice versa, if you prefer). This constraint involves a parameter which is often written as a velocity and labeled c.
  3. Thus, in relativity contexts, c shows up because this constraint is present. In particular, the c in E=mc[sup]2[/sup] is not there because of light; it is there because of relativity.
  4. Now, relativity requires that all massless particles (such as photons) move at a speed equal to c. Thus…
  5. In electromagnetic contexts, c shows up because photons travel at this speed.

(Sorry not to credit the above posters who originally said all of this.)

  • If we’re in a relativistic regime, c will appear.
  • If we’re not in a relativistic regime, electromagnetic phenomena are the only things going on (save Newtonian gravity), so we still get c’s everywhere!

Thus, lot’s of c’s.

27

What about the energy of the electron, and the positron?

28

Yes, you can really understand relativity even without understanding all the maths. But I meant you can’t answer the question “why does the universe work this way”.

29

:smack: Okay, I suppose what I meant to say was that the value of “matter-energy” is conserved, not “mass-energy”. Of course the mass is the same, that’s what E=mc2 means! :smack: Okay, enough dope-slaps!

[tech question]Say, I can’t see the “smack” smilie on my browser - can y’all?[/tech question]

30

I was taking them to be at rest. Yes, if they had any kinetic energy, the mass of the sealed box system could be higher (depending on the details of that kinetic energy and any kinetic energy of the box itself.)

I was actually taking your side, MyOld85MG, that mass is not necessarily the same, depending on your definition of mass.

31

Without having read every post, I will nevertheless attempt to answer the OP with this approach: When we ask the question “Why?” we usually are looking for an explanation, not an answer to the question. In other words, when we say, for example, Why is the sky blue? we are looking for something that explains the blue color. And we get answers that talk about the nature of reflected light with regard to the materials that make up the atmosphere, and wave length, and such. And when we get that answer, we usually say, “Oh, now I know why.” But, in truth, you could still ask, when you get to that level, Why? You have in the above posts, is conversation that shows us how that speed limit connects with all manner of observations, calculations, and inferences. However, I believe that the best answer to your question could be, “Well, that’s the way the universe seems to be made.” xo C.

32

Excellent post Pasta but I disagree that it’s just a difference in semantics. The loose usage of the term ”mass is converted to energy” leads to some very basic misunderstandings.

First, fission isn’t about converting mass to energy its about converting potential energy to kinetic and electromagnetic energy. In other words it’s about changing one type of energy into another. The remnant of the nucleus has less internal energy and as a result of this decrease in internal energy it has less mass, not vice versa.

Second, with respect to the annihilation of an electron and a positron the magnitude of the momentum four-vector certainly doesn’t change and therefore the mass of the system doesn’t either. In this case I think it’s more correct to say that matter is completely converted to radiation rather than to say mass is converted to energy.

True, but these people understand what they are, and aren’t, saying.

True, but the point is that everyone doesn’t understand all the possible uses, and this is what leads to confusion. IMHO the only use of the word mass without a modifier should be the magnitude of momentum four vector

m = ( E[sup]2[/sup] - p[sup]2[/sup] )[sup]1/2[/sup]

And thusly so. RM Mentock the man of few words, but no errors, strikes again.

33

My point is that if one does not care (or know) about the internal structure of the object, talking about the internal energy is not useful. (Hence, the sealed box example.) Since such situations are commonplace, “mass” can be (and is) often used to refer to a particular measureable quantity of a complex object such as a nucleus. And, that measuremeable quantity changes when a nucleus fissions. (Implicit in this usage is the idea that the “before” and “after” nuclei are the same object whose mass can be measured. It is, of course, not as simple as that, but that is how the word “mass” is being used.)

To the extent that there is misunderstanding, I agree that semantics is not the issue. However, fist-pounding insistence that mass does not change misses the point that the word “mass” has different meanings, and such insistence perhaps doesn’t help anyone who does have a misunderstanding.

If only it were so. I would also wish that people would use “data” and “media” as the plurals they once were, but such is the nature of language. The cold fact, though, is that the word mass without a modifier is used to mean things other that the magnitude of the momentum four-vector, in particular to mean the sum of the masses of the “obvious” constituent parts of a system. Clearly this is a very context dependent use and one that is not rigorously defined, but again, such is the language of nature. When rigor is needed, one can use equations (or more words).

34

What, exactly, is c[sup]2[/sup]? Is c a speed? How do you square a speed? 1 light year per year squared is still 1, yet 186,000 miles per second squared is significantly larger.

I’ve never understood c squared as a numeric term…does it have one, and if so, what is it?

35

Well, I guess we’re going to have to agree to disagree.

It’s fine to use whatever definition of mass you feel like when dealing with others that understand what you mean. But for general consumption the accepted definition of mass is the magnitude of momentum four-vector. If you don’t believe me I suggest you ask about it on sci.physics.

m = ( E[sup]2[/sup] - p[sup]2[/sup] )[sup]1/2[/sup]
m = the magnitude of momentum four-vector

36

It uses the same units that I do in measuring my the growth of my carpet business.

I lay carpet in square yards. That’s distance[sup]2[/sup], right? The amount I lay in a week in square yards per week. That’s distance[sup]2[/sup]/time. And the rate at which my business is growing or shrinking is square yards per week per year. That’s distance[sup]2[/sup]/time[sup]2[/sup].

Of course when you bring in the factor of mass, that indicates the inertia that all of my lazy-ass workers exhibit in abundance and represents the energy I must expend to get them to finish the job on time. :wink:

37

This statement, besides possibly being out of line, indicates that you miss my point. In some circles (perhaps sci.physics is one), “mass” is taken to mean invariant mass by default. In other circles, it is not automatic. And in still other circles, the invariant mass would never be called simply “mass”.

Thus, it is not that MyOld85MG’s reference is plain wrong when it says “…the mass (m) that disappears…” Instead, they are using the word “mass” to mean something other than the magnitude of the momentum four-vector! I do not believe you will disagree with this.

Whether you think an encyclopedia should be allowed to use “mass” in this way is akin to a “datum/data” debate and is neither here nor there; debating such semantics does not help readers of the reference understand what it being described.

I am reminded of a passage from No Ordinary Genius, a Feynman biography by Christopher Sykes. Feynman tells a story:

They go one to discuss why the birds are pecking their feathers, what the bugs in the feathers are doing, etc.

38

Yes, but it is pretty much universal in physics circles that, without qualification, mass = invariant mass.