# Conservation of energy-mass

1. Jan 13, 2007

### pivoxa15

It is one of the beauties of relativity that there is no need to consider whether a collision is elastic or inelastic. Because any loss in energy after collision is accounted by the resultant mass(es) via E=mc^2. So goes into increasing the masses of the products.

What I like to ask is some reactions result in energy lost via heat, sound etc which does not go into increasing the mass of the resultant product. How does relativity account for that? Is the answer that it dosen't? It only tells you how much energy is absorbed after the collision which goes into increasing the mass of the product. This extra mass could then escape away from the product via friction, heat etc but the formuals in relativity doesn't tell you exactly how much goes where, it only tells you the (very) initial stage of increasing the mass of the product(s).

2. Jan 13, 2007

### Hootenanny

Staff Emeritus
Einstein's equation applies to a system, not a collision. That is, the total energy(including momentum and mass) of a system must be conserved; not just an isolated collision.

3. Jan 13, 2007

### pervect

Staff Emeritus
I'm not positive what pivoxa15 (henceforth - the OP) is trying to understand, but it's possible that he's still confused about the fact that that the answer to the question

"Can a photon -- that has no mass -- give mass to an absorber?" is yes.

Reference: Taylor & Wheeler, "Space-time Physics" in the section entitled "Dialog: use and abuse of the concept of mass".

Similarly, absorbers (or clouds of nuclear debris) that emit photons can lose mass.

If this is in fact what he is confused about, revewing the defintion of mass (invariant mass!) should explain why this is true.

I think that the following problems might be helpful for the OP.

Question 1:

Suppose we have a single photon, of frequency f.

1) What is the energy of the photon?
2) What is the momentum of the photon?
3) What is the magnitude of the momentum of the photon?
4) What is the mass of the photon, per the wikipedia formula for invariant mass

$$m = \frac {\sqrt{E^2 - (pc)^2}}{c^2}$$

See for instance http://en.wikipedia.org/wiki/Mass_in_special_relativity#The_mass_of_composite_systems
or a textbook

Question 2:

Suppose we have two photons, of frequency f, moving in opposite directions

1) What is the energy of the system of photons?
2) What is the momentum of the system of photons?
3) What is the magnitude of the momentum of the system of photons
4) Using the formula for invariant mass given earlier, what is the mass of the system of photons?

Question 3:
Was the mass of the composite system of two photons the sum of the masses of its constituents?

It may additionally be helpful for the OP to work the problems with the appropriate formulas for "relativistic mass" if he intends to use that concept.

4. Jan 18, 2007

### pivoxa15

My answers could all be wrong but I'll have a go. Please tell me where I have made my mistake.

I accept this. Why do you use the word 'still'. I don't remember being confused and discussing this matter on this forum?

1) E=hf
2) P=E/c = h/(lambda=wavelength)
3) p=E/c = h/lambda
4) m=0 (using invariant mass)

1) E=2hf
2) p=0
3) p=0
4) m=0

I haven't used the formula for relativistic mass because it recquries invariant mass which is always 0 with photons no matter how many there are in the system.

I haven't distinguished between momentum with and without direction because the formula dosen't allow it.

Last edited: Jan 18, 2007
5. Jan 18, 2007

### pivoxa15

You could include a collision as part of the system. I know from classical mechanics that one had to carefully distinguish between elastic and inelastic collisions because with the latter some energy is lost as heat, sound etc. But if one use relativity, everything (other energies like heat and sound) can be fitted in via (relativistic) mass. So my question is, are these wate energy like heat and sound from collisions first stored in as mass of the resultant product and then these tiny masses are converted into energy such as heat and sound. However, to calculate the exact porportions, one must get into the details and relativity alone cannot tell you that.

6. Jan 18, 2007

### Hootenanny

Staff Emeritus
Your thinking is along the right lines. Suppose to bodies X and Y combine into a single body XY as a result of an inelastic collision, then the mass of the combined body XY will be slightly less than the sum of the individual masses of X and Y. That is mY + mX > mXY; note that theses masses are invariant mas, not relativistic. This mass in then converted into energy and released in the form of heat and sound etc. as you say above.

Last edited: Jan 19, 2007
7. Jan 18, 2007

### Ich

This is not quite correct. Wait for pervect to explain the addition of invariant masses.

8. Jan 18, 2007

### Hootenanny

Staff Emeritus
I know that is not how it works; however, I was explaining the concept in lay terms, perhaps it was an error on my part to infer that invariant mass is subject to simple addition.

pivoxa15: In relativity, when two particles collide (of masses mx and my) to form a composite body (of mass mxy); then one must consider not only the particle's individual masses, but also their respective energies and momenta thus;

$$m_{xy} = \sqrt{m_x^2 + m_y^2 + 2\left(E_xE_y - \vec{p_x}\cdot\vec{p_y}\right)}$$

Last edited: Jan 18, 2007
9. Jan 18, 2007

### pervect

Staff Emeritus
OK, the answer to the first question is correct.

The answer to the second question is not correct.

are all correct, but

is incorrect. If you calculate m, and follow the formula, you should get 2hf/c, not zero.

Writing down the direction isn't essential to the problem - we are really only interested in the magnitude of the momentum, so this is just a notational issue (how to write down a vector). I don't think we need to get into it more unless you are curious.

Much more important is the illustration of the fact that a system of photons, each with a zero invariant mass, can have a non-zero invariant mass when regarded as a whole.

Last edited: Jan 18, 2007
10. Jan 19, 2007

### Ich

The mass of a particle is the norm of its momentum four vector.
The mass of many particles is the norm of the sum of their momentum vectors, not the sum of the norms.

11. Jan 19, 2007

### pivoxa15

There is a contradiction in your post. Is the inequality correct but your description not? Should it be the mass of the combined body XY be more than the sum of the individual masses of X and Y.

i.e. Consider two equal masses moving in opposite directions and towards each other and clashing with both obejcts stuck together and stationary. Relativistic mass is conserved. The final mass is M composed of the two masses with gamma factor so larger than the combined invariant mass of the two objects prior to collision. This increase in mass is not retained in the final object but some is dissapated away and the mass that is left could even be less than the two separate objects prior to collision.

12. Jan 19, 2007

### pivoxa15

What should have been the answer?

That is amazing. However, it may be an advanced/non intuitive illustration of my above post. The photon itself always have 0 invariant mass so the non zero invariant mass in the system could be due to energy converted into sound, heat etc?

13. Jan 19, 2007

### Hootenanny

Staff Emeritus
My apologies, a typo, my inequality is the wrong way round. My description, however, is correct. Take note on the additional comment regarding the summation of invariant mass. The correct inequality is as follows; mX + mY > mXY

14. Jan 20, 2007

### pervect

Staff Emeritus
I'm not getting feedback that sounds to me like you understand what I am trying to say. We have been given the formula for invariant mass.

$$m = \frac {\sqrt{E^2 - (pc)^2}}{c^2}$$

Now, you've already anwered correctly that p = 0 in the second case.

So basic algebra, substituting for the value of p in this expression with it's known value (0) tells us that

$$m = \frac {\sqrt{E^2 - (pc)^2}}{c^2} = \frac {\sqrt{E^2 }}{c^2} = \frac{E}{c}$$

Since E=2hf, (which you also got correct), m = 2hf/c.

There is no "energy being converted into sound (?) or heat" in the problem. There is just some routine algebra and a defintion for invariant mass.

15. Jan 20, 2007

### bernhard.rothenstein

is there some redundancy in the title of the thread?

16. Jan 22, 2007

### pivoxa15

Should that be m=E/c^2?

What about my original question? Relativity only seems to account for the masses and energy of reactants and products. What about other energy involved in the system or reaction like sound and heat?

17. Jan 22, 2007

### pivoxa15

I assume these masses are invariant.

Are you suggesting that the lower invariant product mass is due to some of the mass from the reactants converted into heat and sound etc?

But I provided an example that is the opposite of your inequality (same as your first inequality). Could you provide evidence for your claim.

18. Jan 22, 2007

### Hootenanny

Staff Emeritus
Reactants is a poor word, particles would be better; but yes, the 'lost' invariant mass manifests itself as heat, sound etc.

19. Jan 22, 2007

### pervect

Staff Emeritus
Try Taylor & Wheeler's book, "Spacetime physics" if you want a more formal treatment and a textbook references.

But it's basically just a matter of some VERY simple algebra plus the formal defintions of invariant mass you've seen from several sources.

I definitely would NOT want to discourage you from digging up a relativity textbook (while I personally like Taylor and Wheeler, I would encourage you to pick up ANY relativity textbook) and do some checking some of the formulas mentioned. Though I believe you were the one who intially posted the formula in question.

You will need to be able to do some very basic algebra to be able to follow special relativity, though. While not much math is needed, one DOES need to be able to perform high school algebra. For instance, in this thread, it's important to be able to substitute p=0 into an expression such as the one for invariant mass, get the right answer after doing the substitution, and furthermore it would be highly recommended to be skilled enough at this task to have confindence to be able to tell when you've got the correct answer without having to rely on some sort of "higher authority" to tell you when your algebra is right, and when it is wrong.

If you're not skilled enough at algebra yet to accomplish this, you need to polish up your algebra skills a bit. While it might be better to do this polishing before you study SR, it would be OK to do it while you study SR, as long as you do it.

Last edited: Jan 22, 2007
20. Jan 25, 2007

### pivoxa15

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