Energy and mass in relativity

1. Feb 20, 2010

PhizzicsPhan

I've come across an objection to relativity theory a few times, for which I've not seen a good answer, so I'm looking for clarification. A key result of relativity theory is that all speed is relative. So while we can measure a distant quasar moving in relative terms away from us at almost the speed of light, to an alien rotating the quasar, the quasar could be considered stationary, just as we can consider our own sun stationary. However, in relativity theory, E=mc^2, so energy increases dramatically as relative speed increases toward the speed of light. The energy of a spaceship moving at .99 the speed of light is far higher, to an outside observer, than a stationary spaceship. But the same spaceship can of course be considered moving at .99 the speed of light or stationary, depending on one's perspective. If energy is a real thing, however, which is inter-convertible with matter (as Einstein's famous equation establishes), then where does this energy "go"? Merely by flipping perspectives, we can create or eliminate massive amounts of energy, and thus of mass. What gives?

2. Feb 20, 2010

bcrowell

Staff Emeritus
This is not so much a relativity question as an issue relating to Newtonian mechanics. In Newtonian mechanics, energy is conserved, but it's frame-dependent. Pick a frame, and you have some amount of energy in that frame, which is conserved. Pick some other frame, and you have some other amount of energy in that frame, and it's also conserved.

If you pick some specific process, such as electron-positron annihilation, that converts between mass and energy, it will obey the conservation of mass-energy in all frames.

3. Feb 21, 2010

Staff: Mentor

In addition to bcrowell's explanation about the difference between invariance and conservation you should also know that there are at least two definitions of mass. One is known as "relativistic mass", it is the kind of mass that changes based on relative velocity, you cannot think of it as a property of the object, but a relationship between the object and the observer. The other kind is the "invariant mass" or "rest mass" or just plain "mass". That mass can be thought of as a property of the object because all observers will agree on its value. The invariant mass is the kind of mass that most physicists mean when they say "mass".

4. Feb 22, 2010

PhizzicsPhan

Thanks for the responses. Doesn't a distinction between relativistic mass and "invariant" mass lead to problems with, for example, nuclear fission? To clarify, under the basic equation relating energy to mass, if relativistic mass has any meaning at all, it must have a role in how much energy is released in nuclear fission. But if relativistic mass is completely dependent on the frame of reference chosen, the amount of relativistic mass becomes completely arbitrary in terms of its real world effects. An atomic bomb could be considered moving at .9999 c, in which case it would have huge amounts of kinetic energy and thus increased relativistic mass, which it seems should be a factor in how much energy is released in the fission explosion. But we could also consider the bomb stationary, thus eliminating by fiat this extra relativistic mass/energy. So for the stationary observer the bomb would have x energy, but for the observer who considers the bomb moving at .9999 c the explosion would be much larger in terms of its actual effects in all frames of reference. Perhaps I'm missing the point here, but this distinction seems to be difficult to work with.

5. Feb 22, 2010

DrGreg

If an atomic bomb exploded as it hit a planet at 0.9999c it would cause a lot more damage than a stationary bomb. But if an inert lump of metal hit a planet at 0.9999c it would cause almost as much damage. The extra energy is the object's kinetic energy and there's no need to count this energy as "mass" to explain what's happening.

6. Feb 22, 2010

PhizzicsPhan

Dr. Greg, I should have been more clear: I was comparing the explosion of a bomb on Earth, considered stationary to someone on Earth, to the same bomb considered by an observer traveling at .9999 c relative to the Earth. The bomb's surroundings and content haven't changed, but according to the equivalency of mass and energy the bomb will have, from the perspective of the observer moving at .9999 c, far more mass/energy than it does for the observer on Earth. If this is the case and if relativistic mass/energy mean anything, it seems that there is something screwy going on with the appearance/disappearance of mass/energy simply by shifting frames.

7. Feb 22, 2010

Ich

Just calculate the energy of earth itself in that frame. Isn't that a fantastic amount?
Energy is frame-dependent, and has always been. Same with change of energy: consider the earth hitting an object with 1 km/s relative speed. The damage done is the same in all frames, but earth's change in energy can be anything from negative to positive infinity - even in newtonian physics. Try to get a grip on the classical concept of energy, the relativistic one doesn't seem weird after that.

8. Feb 22, 2010

PhizzicsPhan

Ich, there seems to be a difference between Earth's relativistic mass and the bomb's relativistic mass insofar as the Earth's relativistic mass is not in my thought experiment being converted into energy, which is the case for the bomb.

We can, however, easily extend the idea to the Earth in toto. If we could initiate a chain reaction that led to fission of the entire Earth, converting it all into energy, this energy would range from zero relativistic mass/energy up to almost infinite relativistic mass/energy merely by frame shifting between stationary and very fast frames of reference in relation to Earth. Again, this seems problematic.

9. Feb 22, 2010

Staff: Mentor

The energy released is the same in all inertial reference frames. Someone asked a very similar question just last week, and I worked out a numerical example:

10. Feb 26, 2010

PhizzicsPhan

Thanks JTBell, let me noodle this some more.

11. Feb 28, 2010

stevmg

I understand the Lorentz equations for x, and t

The equation e = mc2 uses the basic concept:

m = m0
SQRT(1 - v2/c2) (the so-called gamma factor.)

Can someone show me how this relationship is derived. The gamma factor is a consequence of space-time relativity but how does one get to this point with mass?

I can see it intuitively by assuming the conservation of momentum: m*v

since distance is contracted by the gamma factor when jumping across frames of reference with different speeds, one would have to increase by the inverse of the gamma factor to preserve the product "m" * v resulting in the familiar equation given avove to offset the loss of velocity by the contraction of distance in the second inertial frame but there has to be more rigorous proof than that and I would like to know what it is. Please, no discussions of photons, etc. as the derivation of the Lorentz equations does not require jumping outside the box to get there.

I can't find it anywhere on line but I am not a great web surfer.

H-E-L-P-!

12. Mar 1, 2010

PhilDSP

You can always try Google or Wikipedia with something like "energy mass equivalence"

On Wikipedia the results turned up a nice article with lots of information. At the bottom there is a list of web pages including a "high school" level explanation of the derivation:

http://en.wikipedia.org/wiki/Mass–energy_equivalence

13. Mar 1, 2010

stevmg

To PhilDSP:

The Wikipedia article is almost tautological - starting with the assumption of what is trying to be proven and using that as the basis for proving what you are seeking to prove.

What has to shown is that (using Lorentz is OK because these are derived "from scratch") m = m_0/[1 - v^2/c^2] from scratch... no photons, no electromagnetic waves, ... "no nuthin" - you can't go outside the box.

The intuitive approach would be to establish that given a particle of mass m_0 is accelerated up to v from v = 0.

Using the that same particle as the reference point and using the reference frame as X', , Y', Z and Z' coordinates in which the particle sits at 0, 0, 0 and this second frame of reference is moving at velocity v (in the direction of the X and X' axis - we will drop references to the Y, Y', Z and Z' axes from here on out.)

In the X' axis space contraction occurs by the gamma factor (Lorentz.) But the "real" momentum is still the original m_0*v but in the X' axis the shortening of distance would reduce the momentum by the gamma factor. Hence, one must "factor up" by the inverse of the Lorentz gamma factor which "juices" the m_0 to m up to compensate and maintain the conservation of momentum.

Now, that is intuitive, not a proof. Can somebody take this and prove it???

Also, I looked at the list of references at the bottom of that Wikipedia article you cited and could not make heads or tails of the which of the cited references was the "high school explanation" of which you mentioned. Maybe just indicating which one of the cited references (by article number) may help. Sherlock Holmes I am not.

14. Mar 1, 2010

stevmg

To PhilDSP:

I found the external link to which you referred bit IT USES PHOTONS. No fair. Have to use Lorentz and laws of physics (non electrical) as we know them to dervive that equivalence.

15. Mar 2, 2010

stevmg

Still waiting...

16. Mar 2, 2010

Staff: Mentor

17. Mar 2, 2010

stevmg

Dear jtbell:

Those sites are just Google pages. I've looked at them before and none of them do the trick. I am sending you by e-mail an article which supposedly explains the derivation but it, too is total Martian to me.

It appears, after reviewing Einstein's book "Relativity" that his energy equation which has the gamma factor in and would by backwards steps come up with the relativistic mass equation, thta Albert got it from Maxwell.

I bet you no one has ever derived it except Maxwell in relation to elecxtromagnetic phenomena and Einstein just generalized it to all phenomena.

Again, my intuitive feeling is that that faster an object goes, the velocity is relatively less in the inertial frame than in the moving frame by the gamma factor and so, in order to maintain the equality of momentum and energy, one must jack up the mass with the reciprocal of gamma to maintain equality of energy and momentum.

Am I barking up the right tree here or at least the right frame of reference? Look in your e-mail for that weird article I found. I didn't know how to post it on this message.

18. Mar 2, 2010

stevmg

how do I send a file to you?

19. Mar 2, 2010

Fredrik

Staff Emeritus
Why don't you just post a link to the place where you found the article?

Any derivation of E=mc2 will be based on the SR definitions of some other terms, so the value of such "derivations" is questionable. You might as well start by defining energy to be

$$E=\sqrt{\vec p^2c^2+m^2c^4}$$

and use that to motivate some of the other definitions. Many of the "derivations" in traditional presentations of SR are actually pretty meaningless for this very reason. The thought process that Einstein had to go through in order to use his knowledge of non-relativistic physics and electrodynamics to guess the appropriate definitions of observables in SR if of course of interest to historians, but you don't have to study those things to understand SR. What you need to know are the definitions and how to use them to make predictions about results of experiments.

Nevertheless, some of these "meaningless" calculations can be good exercises. As an example of that, here's the calculation of the work required to accelerate a particle with mass m from speed 0 to speed v in proper time t:

$$c=1$$ (This is just a choice of units).

A dot above a symbol denotes the derivative with respect to proper time.

$$\gamma=\frac{1}{\sqrt{1-v^2}}=(1-\dot x^2)^{-\frac{1}{2}}$$

$$\dot\gamma=-\frac 1 2(1-\dot x^2)^{-\frac 3 2}(-2\dot x\ddot x)=\gamma^3\dot x\ddot x$$

$$W=\int_0^t F(x'(\tau))x'(\tau)d\tau=\int_0^t \dot p\dot x d\tau$$

$$\dot p=\frac{d}{d\tau}(\gamma m \dot x)=\dot\gamma m\dot x+\gamma m\ddot x=m\gamma^3\dot x^2\ddot x+\gamma m\ddot x=\gamma m\ddot x(\gamma^2\dot x^2+1)$$

$$=\gamma m\ddot x\left(\frac{\dot x^2}{1-\dot x^2}+1\right)=\gamma m\ddot x\left(\frac{\dot x^2+1-\dot x^2}{1-\dot x^2}\right)=\gamma^3 m\ddot x$$

$$W=\int_0^t\gamma^3 m\ddot x\dot x d\tau=\int_0^t m\dot\gamma d\tau=m\gamma(t)-m\gamma(0)=m\gamma-m$$

If we restore factors of c, this becomes

$$W=\gamma mc^2-mc^2$$

This "derivation" assumes that we have already accepted the SR definitions of four-velocity, four-momentum, force and work.

Last edited: Mar 2, 2010
20. Mar 2, 2010

Staff: Mentor

Another good calculation is to do a Taylor series expansion in powers of v. You recover mc^2 as the first term and 1/2 m v^2 as the second term. So it definitely has the right form in the classical limit.