Energy and Mass: What Happens to Energy in Relativity Theory?

[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?

 

[bcrowell]

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.

 

[Dale]

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".

 

[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.

 

[DrGreg]

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.

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.

 

[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.

 

[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.

 

[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.

 

[jtbell]

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.

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:

https://www.physicsforums.com/showthread.php?p=2585116#post2585116

 

[PhizzicsPhan]

Thanks JTBell, let me noodle this some more.

 

[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-!

 

[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

 

[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.

 

[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.

 

[stevmg]

Still waiting...

 

[jtbell]

We've had some discussion here in the past about derivations of the relativistic mass and relativistic momentum formulas:

http://www.google.com/#hl=en&source...n+of+relativistic+mass+site:physicsforums.com

Maybe one of those threads has what you're looking for.

 

[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.

Thanks for your time...

 

[stevmg]

how do I send a file to you?

 

[Fredrik]

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.

 

[Dale]

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.

 

[stevmg]

The note by Fredrik is beyond meaningless as the square root as presented comes from where? It comes from Maxwell et al electromagnetic theories which are tautological in this as they are downstream from the relativistic mass equation.

Citing the Taylor series expansion is actually an expansion of
E = m(c^2)[1 - (v^2)/(c^2)]^(-1/2)
which already assumes the veracity of the relativistic mass equation. Again, tautological reasoning.

You cannot prove something is true by assuming it is true and then basing the proof on that "fact." You can only use that sort of technique and prove that the negative or contradiction of a given assumption false reductio ad absurdum.

 

[Dale]

Obviously it is tautological. Any attempt to "prove" a definition is asking for a tautology. Neither Frederik nor I were suggesting otherwise. I was just pointing out that it has the correct form in the classical limit, nothing more.

 

[stevmg]

I do have an article by a T. PLakhotnik from 2005 which is a direct mathematical proof using only SR and Lorentz which mathematically establishes the so-called relativistic mass formulas. It is tedious but looks correct.

I just found it. The reference is: Eur. J. Phys. 27 (2006) 103–107.

My e-mail is stevmg@yahoo.com. If you want it send me an e-mail and I will respond with the .pdf document attached.

Now, is that good enough for you?

 

[DrGreg]

The simplest derivations involve photons, but you don't like that.

Most books I've seen define momentum as p = γmv and/or energy as E = γmc² and go on to show that this definition reduces to the Newtonian values for small velocities and that these definitions transform correctly when you switch to a different frame, and then postulate that these quantities are conserved in collisions. The justification is that relativity built on these assumptions correctly predicts the results of experiments.

If you actually want to work out these formulas without knowing in advance what the answer is, it seems to be a difficult exercise. I had a go myself in the thread "Derivation of momentum & energy formulas".

 

[Dale]

Now, is that good enough for you?

The question is if it is good enough for you. I am perfectly comfortable without a derivation at all.

In my opinion the important things about the relativistic energy formula are not the derivation but the following:
1) It reduces to the classical formula for energy in the appropriate limit
2) It transforms as the timelike component of a four-vector
3) It is conserved

I would consider anything with properties 1) and 3) to be relativistic energy, but 2) is a really nice bonus.

 

[stevmg]

Sports Fans -

Try that Plakhotnik article... uses a first order diff eq but is much more intuitive.

Nobody on this planet could possibly follow that weird thread "Derivation of momentum & energy formulas" even though it is probably mathematically correct - I'll take your word for it.

Again, if you want the Plakhotnik article send an e-mail to me at stevmg@yahoo.com and I will attach the .pdf file as a reply. I don't know how to send a .pdf file to this forum directly or I would.

 

[stevmg]

Dalespam -

I bellieve the relativistic mass formula as well as velocity formulas and all the Lorentz formulas without knowing how they were derived - just as I do all the theorems of plane and solid geometry which I have forgotten.

The relativistic mass formula was just something I was curious about. I knew that it wasn't scrolled on tablets of stone by God, that's all so I wanted to know how it was derived - a step or two down from God.

 

[stevmg]

Hey, I got it...

Attached is the .pdf file I spoke of

Steve G

 

[Fredrik]

The note by Fredrik is beyond meaningless as the square root as presented comes from where? It comes from Maxwell et al electromagnetic theories which are tautological in this as they are downstream from the relativistic mass equation.

Citing the Taylor series expansion is actually an expansion of
E = m(c^2)[1 - (v^2)/(c^2)]^(-1/2)
which already assumes the veracity of the relativistic mass equation. Again, tautological reasoning.

You cannot prove something is true by assuming it is true and then basing the proof on that "fact." You can only use that sort of technique and prove that the negative or contradiction of a given assumption false reductio ad absurdum.

Steve, you have some very strange ideas (very wrong ideas) about these things, and you really need to think things through. I suggest that you read my post again, slowly, and think about what I'm saying this time. Nothing I said there is wrong, and I certainly didn't "prove something is true by assuming it is true".

You should also think about equation (2) in the article you uploaded. That's a definition too. It can't be derived from Newtonian mechanics. The best you can hope for is to find an argument for why it's not crazy to guess that the definition is appropriate.

SR is defined by Minkowski spacetime and a few axioms about how to interpret the mathematics as predictions about results of experiments. Terms like "energy" and "momentum" must be defined. They can't be derived. The proper way to justify the definitions is to perform experiments to test the accuracy of the theory's predictions.

 

[PhilDSP]

The derivation using only Maxwell's Equations (in Maxwell's original nomenclature) was done by J. J. Thompson for the electron only. Though that may be dependent on the de Broglie relations. One can then make the assumption that all charged particles function likewise. It's important to note that the E=mc² equivalence is a first order approximation only.

A proper exposition of that in modern terms would require a paper in itself. I plan to do that but have a bit too much in the works at the moment.

 

[stevmg]

Here is the tautology in that derivation posted - It is a very good derivation indeed, though.

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

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

Also, in Plakhotnik's article, equation 2 is derived eventually and NOT accepted as fact.

The loose part of his "intuitive article" is the approximation in equation (3) on page 105 as there is approximation he uses and then integrates THAT. That may not be so "kosher."

Since the equation (2) as cited is correct it will always work in any proof.

In my own history, for example, I remember a calculus test in which a cone was being filled up with water. Given was the angle of the cone, the current height of the water (so the area at the surface could be calculated), and the flow of water into the cone. The question was "what is the rate that the height was rising at that particular time"

My roommate answered the question and was marked wrong because he he made the assumption, which was correct, that the rising height rate was inversely proportional to the area of the surface. But, at that point in time, we had never proven that his assumption (and the assumption was correct) was true, so the instructor marked him wrong. Now, the instructor was a good guy and told him that had that one question made the difference between a B or a C or a D and an F he would have given him credit for it, but it did knock him down from A to a B, so the instructor felt comfortable "zinging" him to make his point. Mike (my roommate who is now a retired PhD professort at the University of Oregon in Biochemistry) was wrong, though.

P = mu is true but Plakhotnik doesn't accept it as true until shown
SQRT(1 - u^2/c^2) to be true

I am out of my league, now but attach another simpler proof from Karl Calculus Tutor.

I'm happy and I do know what a tautology is:

""The Bible says ...(thus and so)


(thus and so is true because it says so in the Bible"

I'm sure you've heard that before.

Stephen Garramone, M.D. (Col, Ret), USAF, MC

 

[Fredrik]

Here is the tautology in that derivation posted - It is a very good derivation indeed, though.

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

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

It's not a tautology. The first step just uses the definition of p (the same definition as in the article you posted), and some of the other steps use the definition of gamma.

Also, in Plakhotnik's article, equation 2 is derived eventually and NOT accepted as fact.

I noticed that they claimed to be doing that, but I don't even have to read the "derivation" to know that it's not really a derivation. You can't derive special relativistic stuff from the Newtonian/Galilean theory. So what they're doing is something else entirely. See my previous two posts for an explanation of what.

 

[stevmg]

To Fredrik -

I really do appreciate all the time you have spent trying to wrap my brain around this. I am not a physicist, nor do I work with physics, and my masters in math is also old . I took differential equations in the Spring of 1960, so we are going back a long way. Ooooh, do I forget that crap, although I can still do it if I have the texts in front of me. I am an M.D. and believe me, the only thing MDs and DO (osteopaths) do with math is ratio, at most for drug dosing.

Now, I forgot something about logic (I never had a course in logic and it was taught to me in a computer class much later) which your examples brutally reminded me of (yes, you can end a sentence in preposition as English is a Germanic language, not a Latin one.)

One CAN start off with a supposition such as momentum = gamma*mv

Then, taking that, and, by a series of logically EQUIVALENT statements one can work to something that has been previously proven or known to be true (but not on the basis of the original supposition:)

statement A: momentum = gamma*mv
A <--> B
B <--> C
C <--> D
.
.
.
Y <--> Z
Z "2 + 2 = 4" [for argument sake]

Then you have proven A. A is NOT a tautology in that case - something you have been tyrying to tell me for the past few days but, as stated, my old brain has had a hard time absorbing. My problem was that it has been so damn long since I did diff eq that I cannot follow what you did very well and I don't really need to. The two articles that I posted are of sufficient rigor that that's all I need to understand it at my level.

You do have to agree, though that this so-called relative mass equation (which really is not an increase in mass but an increase in the effect of the momentum at high velocity due to relativity - I use the word "velocity" as it is a vector that has to be turned) is not "proven" in physics texts but is assumed, while relativistic time, velocity and distance are mathematically proven in these same texts. I also know that particles with "increased mass" due to high velocity do NOT have increased gravitational effect commensurate with their "new" mass.

Muchas gracias, vielen Dank, whatever, for all your help!

Stephen M. Garramone, M.D,
Melbourne, FL

 

[stevmg]

Attached is a .pdf file from a lecture series which fully explains the derivation of p= gamma*u where u is the velocity in the moving time frame, m is the original mass at rest. It is lecture 4 that he goes into it.

The major contributor to this time dilation

Once momentum in relativity is explained, the so-called "relativistic mass" equations are explained:
m = m₀*gamma

Actually, when this "new" mass is created, it really has no true "mass" effects. Gravity is not affected nor is there more of a bending of light. This equation is to make the Newtonian equation p = m*v be more applicable.

 

[stevmg]

Great lectures on all aspects of simple relativity

http://www.youtube.com/watch?v=pHfFSQ6pLGU&feature=SeriesPlayList&p=FE3074A4CB751B2B

start with lecture 12 and go through lecture 15