E=mcÂ²

ERRORIST · 2004-03-23 20:24:24

If it has no inherint mass then how does it gain temperatue?

GCNRevenger · 2004-03-23 20:58:07

*blink* Didn't I just state in the last post that it has no temperature because it isn't an atom?

ERRORIST · 2004-03-23 22:27:08

Light is hot because it has mass. Mass generates heat because it is moving. Light generates heat because it is moving.

Euler · 2004-03-24 00:22:54

As GNCRevenger said, temperature and heat are not concepts that apply to subatomic particles. Light has no temperature. It is neither hot nor cold. The temperature of a substance affects the wavelengths of light that it can emit, but it does not mean that the emmited light is at that temperature.

C M Edwards · 2004-03-24 17:00:08

I feel obligated to point out that the actual equation is:

E = SQRT( ( a * m * c^2 )^2 + ( p * c )^2 ),

where E is energy, m is mass, c is the speed of light, p is momentum, and a is the relativistic correction factor:

a = 1 / SQRT( 1 - (v / c)^2 ),

where v is the velocity. (SQRT() is the square root operator that I can't make HTML give me...) (The whole universe can be thought of as being "off" by the relativistic correction factor. For example, m(v) = a * m is the formula for relativistic mass increase, L(v) = a * L is the formula for relativistic length contraction, E(v) = a * E is a good approximation for the energy formula above, etc.)

E = m * c^2 holds true for the special case of v=0. It's also about right when v is very small.

For a photon, E is not equal to m * c^2. E = p * c for a photon, because m = 0 for a photon in this convention.

That said, the argument still can't be put to rest just by saying "Aha! Photons have no mass! See!", because photons still have momentum. Without mass.

And you guys thought energy without mass was odd...

SBird · 2004-03-24 18:50:19

CM, are you sure those equations are correct? I believe that the standard convention is for m to equal the rest mass and M to equal relativistic mass. Therefore your equation should be E=SQRT(((M*a)^2*c^4)+(p^2*c^2)). Just to make sure we're on the same wavelength, these are the equations and variables that I'm using:

E=Mc^2
Where M is the relativistic or apparent mass.

E=SQRT((m^2*c^4)+(p^2*c^2))
Where m = the invariant or rest mass.

The relativistic correction term you mention (a=SQRT(1-v^2/c^2) is the conversion term between M and m. eg: M = m/a

I double checked these equations in both the Usenet Physics FAQ and Wikipedia. If they're in error, please let me know - my physics textboks are in storage so I can't refer to them right now.

According to my memory and the online references I've looked at, this is the relationship between M and m.

m is the rest/invariant mass and is what modern physics refers to when they say mass. It's used in the equation E=SQRT((m^2*c^4)+(p^2*c^2)). In this case, m represents the actual physical mass of a particle or the total energy of the particle after kinetic energy is removed. For example, if you combine an electron and positron, the energy released is equivalent to the rest masses of the two particles. The SQRT(m^2*c^4) is basically the rest mass of the particle and the SQRT(p^2*c^2) is the kinetic energy portion.

M is the relativistic/apparent mass and is an antiquated term. However, it is what most physicists refer to as mass when writing popular science texts like The Brief History of Time. This no doubt is the source of much of the confusion related to layperson comprehension of relativity. For M, E = Mc^2. In this case, M represents the total energy of the system, including the kinetic energy. Basically, M = E. The whole E=Mc^2 is just a unit conversion equation that really translates to E=E or M=M. M=E is why relativistic/apparent mass isn't used anymore since it's just another term for energy.

For a photon, m = zero since the physical photon has no actual, physical mass. However, M is non-zero. For a photon, M is the kinetic energy or momentum of the particle. Basically, whether a photon has mass depends on how you define mass. The actual photon doesn't have mass but its kinetic energy does - sort of. If you ask a physicist if the kinetic energy has mass, s/he will say no since mass is defined as m. However, the kinetic energy term which is all that's left in M can still convey momentum and is capable of bending spacetime just like regular old mass does.

So, it really depends on how you define mass. If you define it as the physical mass of particle, then mass = m and a photon has no mass. If you define mass as the ability to impart momentum and bend spacetime, mass = M and photons do have mass. Personally, I prefer the latter definition just because its easier to wrap my mind around but the overwhelming majority of physicists and textbooks will go with the former. It's purely a semantic argument, really. You just have to be careful about which 'mass' you are talking about.

Here's an example:

If you accelerate a particle to nearly the speed of light, does it collapse into a black hole?

No, as you approach the speed of light, M goes to infinity but m remains the same. m is the actual mass of the particle and it stays the same therefore the particle itself doesn't collapse since it hasn't gotten more massive. However, the kinetic energy term which is included in M goes to infinity and does gain gravitational energy, at least in theory - I doubt that this effect is measurable with modern instruments, even with high energy cosmic rays. However, the kinetic energy is not a physical object and therefore can't collapse into a black hole.

ERRORIST · 2004-03-24 19:25:28

Photons could be traveling around a smaller object. Simlar to the electron traveling around an atom but millions of times smaller, and traveling millions of times faster. This circle could be what we see in the photons wavelength as it travels along.

SBird · 2004-03-24 19:47:11

Although your photon 'atom' idea isn't terribly plausible, you have found one of the biggest outstanding problems of modern science - the particle/wave duality. All subatomic particles can be treated as either a wave or a particle. It's clear that the actual particle part of a photon has no measurable size. The wave, on the other hand is quite big. For visible light, the size of the wave is about 0.5 microns or about 2000 times the size of an average atom.

If you want to learn about the wave/particle duality, read a copy of In Search of Schrodinger's Cat by John Gribbin. You can get a copy from Amazon used for $3.00 or at a local used bookstore for about the same price. It's easy to ready and does a far better job of explaining the problem that I can in a message here.

The wave/particle duality is partially explained by a concept called quantum decoherence. Basically, a particle acts like a wave and quantum mechanically as long as it doesn't interact too much with the rest of the universe. As soon as it passes some sort of threshold of interaction, it starts acting like a particle governed by relativity. Some recent experiments have started to put some fairly well defined limits on this transition which should help to explain it.

There's also the Heisenberg uncertainty principle which says that you can't know a particle's momentum and position at the same time. I'm not well versed enough in quantum physics to know if there's a connection between the two but intuitively, it seems that there should be.

Additionally, string theory and quantum loop gravity theory predict a fundamental graniness to the universe which again might help to explain both quantum decoherence and the uncertainty principle.

However, at this time, the underlying cause is not known.

ERRORIST · 2004-03-24 20:00:28

The photon particle as they call it must have mass. It could be traveling in a circle that looks like the total peak to peak of a wavelength, which would be quite large because of the velocity the photon travels.

GCNRevenger · 2004-03-24 20:26:21

Nooo, there is no requirement that all particles have rest mass.

The wave's width doesn't have anything to do with the size of the particle persay, because you can't use the two principles at the same time... if you are dealing with waves, there is no mass concern. If you are dealing with particles, there is no wave nature.

ERRORIST · 2004-03-24 20:33:22

The only thing that has no mass is empty space. If anything exists in space it must have mass.

GCNRevenger · 2004-03-24 20:38:34

Noooo, because as has already been stated, energy exsists and does not have to have any rest mass at all.

GCNRevenger · 2004-03-24 20:42:13

Oh and I am not so sure that a nonzero rest mass particle pushed to a high enough speed won't "go black" and implode to a singularity Sbird, because there is no upper limit to the mass it can achieve. There simply isn't enough energy available anywhere to induce this fantastic quantity of momentum.

ERRORIST · 2004-03-24 20:43:25

Yeeeeees, Energy has mass. E=M THEREFORE M=E.
Math proves it.

GCNRevenger · 2004-03-24 22:14:18

No it does not. Otherwise, it would be M=M. This is physics, not algebra.

Edit: E=MC^2 is still a conversion from one thing to another, it does not mean the two things are the same thing.

ERRORIST · 2004-03-24 23:44:36

If M=E then it works both ways. Math proves it.

SBird · 2004-03-25 01:04:31

True, although one has to wonder about what the definition of a singularity is. An electron has 0 radius and a finite mass which would seem to imply that it should be a singularity itself. I suspect that we'll have to have a fully developed theory of quantum gravity first before little problems like that are resolved.

High energy cosmic rays (~10^20 eV) have momentums measured in Joules which should have an additional apparent mass of something like 10^10 AMU. When packed into an object as small or smaller than an atomic nucleus it should be well past the point of collapsing into a singularity. AFAIK, the particle fallout from those high energy collisions don't exhibit behavior consistent with black holes. Therefore, I'm fairly sure that the apparent mass is incapable of collapsing into a singularity. However, so few of those high energy events have been observed that it's difficult to say for certain.

GCNRevenger · 2004-03-25 07:43:51

Errorist, this does not work. Not in the way you want it to work. Plug in some units for E and M and come back to us.

SBird: I was thinking a "classical" particle... electrons and their annoying leptonic nature and photons with their "whatever the h*ll they are" massless background probably don't apply.

ERRORIST · 2004-03-25 08:02:24

Ok I'll plug in some numbers. M=2 and E=2 therefore 2=2

GCNRevenger · 2004-03-25 08:28:18

No, not numbers. Units. What are the UNITS for E and M?

ERRORIST · 2004-03-25 08:35:27

Algebra is math and is part of physics. Correct?

GraemeSkinner · 2004-03-25 08:53:59

Mathematics is a big part of physics yes but knowing how to use the equations as part of physics is a bigger part still. You say E=M so M=E why?

GCNRevenger · 2004-03-25 09:01:07

Without units, that is, algebraic concepts that link the pure math to reality, the pure math is meaningless. Algebra alone is not enough, physics demands the numbers be assigned to something... If you apply units for E and M, strictly speaking without synthetic alteration, it becomes quite clear that they do NOT reprisent the same thing (at rest mass).

C M Edwards · 2004-03-25 10:29:01

Maybe it's not energy after all, but angular momentum...

Or is angular momentum energy instead?

After all they both have the same units...

At any rate, GNC Revenger is right. The algebra of "E = m, therefore M = E", as derived from the mass-energy relation, is flawed because some steps are skipped.

E = m * c^2
E / c^2 = m
let E' = E / c^2

making the actual relation, as described by Errorist:

E' = m, therefore m = E'

which is a true relationship. A transform exists that allows derivation of E from m and vice versa. But that transform is not the actual energy all by itself. Nor, mathematically, is the mass.

The dimensional analysis is very important, too. Units have to be treated as variables during the solution of this formula in order to have any claim to mathematical rigor, and when you're trying to make a point based on math, rigorousness is vital.

For example:

Where this concept gets really cool is when apparently very different things are describable using the same units. Electron spin is often described as "having units of angular momentum", but that's hand waving, because energy and angular momentum both have the same units. In the mathematical sense, angular momentum = energy, with no need for a transform. Angular momentum just happens to be directional, a difference which isn't always relavent to the formula at hand.

Since the units are the same for angular momentum as for energy, the equation E = m c^2 mathematically allows the conversion of angular momentum to mass, too. This is actually observed in nature, too, as relativistic inertial frame dragging.

ERRORIST · 2004-03-25 10:40:30

Well said CM. A certain amount of mass has a certain amount of energy and a certain amount of energy has a certain amount of mass.

New Mars Forums

Announcement

#26 2004-03-23 20:24:24

Re: E=mcÂ²

#27 2004-03-23 20:58:07

Re: E=mcÂ²

#28 2004-03-23 22:27:08

Re: E=mcÂ²

#29 2004-03-24 00:22:54

Re: E=mcÂ²

#30 2004-03-24 17:00:08

Re: E=mcÂ²

#31 2004-03-24 18:50:19

Re: E=mcÂ²

#32 2004-03-24 19:25:28

Re: E=mcÂ²

#33 2004-03-24 19:47:11

Re: E=mcÂ²

#34 2004-03-24 20:00:28

Re: E=mcÂ²

#35 2004-03-24 20:26:21

Re: E=mcÂ²

#36 2004-03-24 20:33:22

Re: E=mcÂ²

#37 2004-03-24 20:38:34

Re: E=mcÂ²

#38 2004-03-24 20:42:13

Re: E=mcÂ²

#39 2004-03-24 20:43:25

Re: E=mcÂ²

#40 2004-03-24 22:14:18

Re: E=mcÂ²

#41 2004-03-24 23:44:36

Re: E=mcÂ²

#42 2004-03-25 01:04:31

Re: E=mcÂ²

#43 2004-03-25 07:43:51

Re: E=mcÂ²

#44 2004-03-25 08:02:24

Re: E=mcÂ²

#45 2004-03-25 08:28:18

Re: E=mcÂ²

#46 2004-03-25 08:35:27

Re: E=mcÂ²

#47 2004-03-25 08:53:59

Re: E=mcÂ²

#48 2004-03-25 09:01:07

Re: E=mcÂ²

#49 2004-03-25 10:29:01

Re: E=mcÂ²

#50 2004-03-25 10:40:30

Re: E=mcÂ²

Board footer