1. Nov 26, 2005

### FeynmanMH42

Light always travels at c.
No object with mass can travel at c.
E = mc^2, all energy has mass.
Light is a form of energy.
How can light travel at c??

2. Nov 26, 2005

### Staff: Mentor

In this statement, "mass" means what is variously named "proper mass" or "invariant mass" or "rest mass", not "relativistic mass".

In this statement, "mass" means "relativistic mass".

Photons have a "proper mass" of zero.

Last edited: Nov 26, 2005
3. Nov 26, 2005

### JesseM

And if you want to use rest mass in the energy equation, the equation must be E^2 = m^2 * c^4 + p^2 * c^2 , where p is the relativistic momentum $$p = mv/\sqrt{1 - v^2/c^2}$$. You can see that if p=0 then this reduces to E=mc^2.

4. Dec 1, 2005

### zoobyshoe

Quote:
E = mc^2, all energy has mass.
I'm confused now. I understood E=MC2 to be a formula for determining how much energy was contained in any given rest mass. In other words, I've been thinking it would be read as: "E (the energy contained in a given amount of rest mass) is equal to that mass times the speed of light squared." I've been under the impression a person could take, say, an apple, determine its mass, and then apply the formula to find the total energy (expressed in joules, I suppose) that was locked up in the apple. You're saying the "m" in mc2 stands for "relativistic mass"?

5. Dec 1, 2005

### DrGreg

If your apple is stationary relative to the observer, then its rest mass m and relativistic mass M are the same thing.

If your apple is moving relative to the observer, then the difference between Mc2 and mc2 is the apple's kinetic energy.

So E = mc2 is correct for rest mass m provided you understand that in this case E excludes the apple's kinetic energy. (But it does include other sorts of energy such as thermal and chemical.)

6. Dec 1, 2005

### Staff: Mentor

If M is the "rest mass", then your statement is correct. If M is the relativistic mass, the equation gives you the total energy of the object: the "rest energy" associated with the "rest mass", plus the kinetic energy associated with the object's motion.

7. Dec 1, 2005

### DaveC426913

This is not true. Energy can be converted to mass. The formula shows the equivalency between the amount of energy and the amount of mass.

8. Dec 1, 2005

### zoobyshoe

OK. I see where relativistic mass figures in.

Just to make sure I'm thinking forward from this properly, is the following correct:

In the case of a literal apple I suspect the difference between the result of figuring the energy in its rest mass and the result of figuring the energy in its relativistic mass were I to throw it or shoot it out of a canon is pretty neligible. I could hardly add much kinetic energy to an actual apple compared to what's locked up in its rest mass. In the case of a particle, though, let's say a molecule, I'm sure the difference between the energy in its rest mass, and what you might add by accelerating it, becomes important. Correct?

9. Dec 1, 2005

### JesseM

Actually, the ratio between the kinetic energy and the rest mass energy should be the same for all objects at a given velocity regardless of each object's rest mass, because kinetic energy is given by $$(\gamma - 1) mc^2$$ and rest mass energy is given by $$mc^2$$, so if you take the ratio the m cancels out and you get kinetic energy:rest mass energy = $$(\gamma - 1)$$. If you want to know how fast an object must go for its kinetic energy to become equal to its rest mass energy, just solve $$(\gamma - 1) = 1$$ -> $$\gamma = 2$$ -> $$\sqrt{1 - v^2/c^2} = 0.5$$ -> $$v = (\sqrt{3}/2)*c$$, or about 0.866c. At any velocity higher than that, an object's kinetic energy will be greater than its rest mass energy.

There is still a sense in which we need to worry more about kinetic energy of small particles than of large objects, for the simple reason that it's a lot easier to accelerate small particles up to relativistic velocities than it is to do so with large objects like apples.

By the way, if it helps, relativistic mass $$m_R$$ is equal to $$\gamma m$$, where m is the rest mass (and $$\gamma = 1/\sqrt{1 - v^2/c^2}$$). The total energy of an object is $$E = m_R c^2 = \gamma mc^2$$, and since the kinetic energy is total energy - rest mass energy, you can see why this means the kinetic energy is $$(\gamma - 1)mc^2$$. Also, a little algebra will show that the formula $$E = \gamma mc^2$$ is equivalent to the formula $$E^2 = m^2 c^4 + p^2 c^2$$ I gave earlier, with $$p = mv/\sqrt{1 - v^2/c^2}$$.

Last edited: Dec 1, 2005
10. Dec 2, 2005

### zoobyshoe

Thanks, JesseM, That clicked things into place for me.

11. Dec 4, 2005

### FeynmanMH42

Now that's just strange.
I used to think that... until I was told it was wrong...
...what happened????

12. Dec 6, 2005

### Longstreet

I hope this doesn't make things more confusing.

Say you have some massless boxes, which are stationary in your frame of reference. If you put the light into a box, the box now has a mass related to the energy of the light: E = mc^2. Now you can do fun stuff. The box cannot reach the speed of light, but you can change the relativistic mass of the box by getting it *close* to the speed of light. If you then open the box again, the light comes back out, but it has more energy. Say you put in ultraviolet light, and you get out x-rays. If you put the x-rays into another box, the new box is at rest but has the same mass as the box that was close to the speed of light. This is just showing how mass and energy are related. The key is the boxes can't reach the speed of light, even though it is the light inside which is determining its mass.

Now you are probably completely confused, heh.

13. Dec 7, 2005

### Enuma_Elish

Is it
or "No object with mass can be gradually accelerated to c over a finite period of time"?

I am not a physicist, so don't take this as a rhetorical question.

14. Dec 7, 2005

### JesseM

Both. An object with nonzero rest mass travelling at c would have infinite energy, since $$E = \gamma m c^2$$ where m is the rest mass and $$\gamma = 1/\sqrt{1 - v^2/c^2}$$.