# Mass gain at high velocities

1. Dec 24, 2012

### DiracPool

Every schoolboy knows that a principal reason that nobody or no thing can go faster than the speed of light is that we gain mass at relativistic speeds approaching c. However, I have yet to find an explanation as to how that mass gain manifests itself in the accelerating body. Say we have a mass of 1000 gold atoms approaching c. How is this mass gain manifested to slow down its acceleration or to increase its intertia? Does the 1000 atoms all of a sudden build to 1500 atoms (or so)? Does the mass-energy of each atom increase? If so, how? Stronger vibrations between the subatomic particles? Or is it between the particles? More violent modes of oscillation between the gold atoms as we approach c? Or is it something else?

2. Dec 24, 2012

### Staff: Mentor

Actually, the concept of relativistic mass has been generally discarded by the scientific community. The usual concept used now is the invariant mass, which does not increase.

Instead, it is recognized that energy and momentum are nonlinear functions of velocity which go to infinity as the velocity goes to c.

3. Dec 24, 2012

### s0ft

Interesting. So, what they generally accept today is that the only reason for the slow down in acceleration of a body in relativistic speeds is simply the fact that speed of light is unachievable? And the fact that massless particles must travel at c to exist is rather empirical than theoretical?

4. Dec 24, 2012

### DiracPool

Wow, that's news to me. Somebody ought to tell those "pop" physicists on the science channel. I submitted this thread cuase I just saw a show on it last night and started thinking about it. Just when you thought there was something you could count on...

Ok..., I'm not sure if I understand what the difference is here. Sounds kind of like the same thing. What energy and momentum goes to infinity as the velocity goes to c? Where does it come from? Where is it? Inside the massive body? Around it? Is it a mathematical thing and we're supposed to just shut up and calculate? Is it something that just suddenly appears about the point we reach c, and then shoots rapidly to infinity to put up a wall between the massive body and the speed limit? Or is it something like time dilation where there is a graded adjustment the entire relativistic way through. The relativistic mass is still calculated using the Lorentz transform, isn't it? I'm still a bit confused?

5. Dec 24, 2012

### Staff: Mentor

No. The first is circular and the second is false. How did you get either of those from my comments?

6. Dec 24, 2012

### Staff: Mentor

Agreed. It is rather annoying and as a result we have to cover it on a pretty regular basis.

OK, that is a lot of questions, I cannot answer them all, but if you can pick the most important one then I will try to answer.

The energy and momentum equations come out of the Lorentz transform, so it is very similar to the time dilation formula.

The easiest way to understand this is through the use of four-vectors (http://en.wikipedia.org/wiki/Four-vector) especially the four-momentum (http://en.wikipedia.org/wiki/Four-momentum). From those you can easily derive the expression for relativistic kinetic energy and see its behavior.

7. Dec 24, 2012

### s0ft

So what changes in the kinetic energy expression for relativistic speeds? Mass or velocity?

8. Dec 24, 2012

### Staff: Mentor

The expression is:
$$E=\frac{mc^2}{\sqrt{1-v^2/c^2}}=mc^2+\frac{1}{2}mv^2+\frac{3}{8c^2}mv^4+O(v^6)$$

m is the ordinary (invariant) mass and v is the ordinary velocity. Neither mass nor velocity changes, what changes is the expression for energy. The kinetic energy is the total energy above minus the rest energy, mc^2.

Last edited: Dec 25, 2012
9. Dec 25, 2012

### Naty1

Hello again Diracpool!

In a way you can say that, although Dalespam did not imply such. In a sense it's an inexact explanation. In another discussion, someone provided this related explanation which is better:

A related insight is this:

[This is not so far from one of your suggestions:"Or is it something like time dilation where there is a graded adjustment the entire relativistic way through.]

This gets at the heart of the issue!! This means velocities do not add in the simple linear fashion [u + v] we all use before relativity. And this directly relates to Lorentz factor identity with cosh [rapidity].

http://en.wikipedia.org/wiki/Rapidity

The expression posted by Dalespam is probably THE source of common explanations like "You can't accelerate anything to lightspeed because it would take infinite energy".....meaning E becomes unbounded as v approaches c. Some people seem not to like that explanation either. But I can't see why it is 'wrong'.

[This stuff takes time,be patient and pick posters here to study that you can trust [like Dalespam, advisors, mentors,etc] and compare replies to get a better perspective on issues when you can. The only way an old dude like me can 'remember' all this stuff is by taking notes. So I often post 'expert' answers that 'rang a bell with me'.]

Last edited: Dec 25, 2012
10. Dec 26, 2012

### DiracPool

Hello yourself Naty1, and thanks again for the thoughtful comments. You didn't really address relativistic mass though, and what I am taking from all of this is that it is really a "shut up and calculate" kind of problem. I thought at least someone would bring up the colliding protons at the LHC and mention how the physicists model what actually happens physically to the protons as they are accelerated to near c velocities. Maybe they don't even know. Perhaps they just model it as an amorphous growing ball of energy that collides with another one in the Atlas measuring center.

You would think, though, that the "pop" physicists in particular would take a minute to think about this issue before they start talking about their time machines and warp drives. I mean, it makes a difference how this energy-mass gain is accumulated in a traveling body, or if it is accumulated. That is, if the energy is manifested as the non-linear but real hyper-agitation of molecular bonds as someone in a spaceship approaches c, then those intrepid intergalactic explorers are gonna die long before they hit warp speed. If the relativistic mass is manifested in another way or no way at all, it would be a good idea to explain how if you want your Star Wars fantasy to have teeth.

11. Dec 26, 2012

### Staff: Mentor

It isn't any kind of problem. What problem do you think there is?

12. Dec 26, 2012

### DiracPool

I'm posting this in this thread and in the "why does space contract" thread because they are related.

Well, now I’m thinking that it may help to approach the problem from a different angle. I just started thinking that as Bob starts approaching the speed of light by firing his rocket pack at full blast, Alice sitting at home will see Bob get shorter (length contraction) and heavier (relativistic mass-energy increase). However, Bob, being in his own reference frame, will see none of this change.

This seeming inconsistency leads us to somewhat of an epistemological question which is, “Is Bob actually getting heavier and shorter, or isn’t he?” The easy answer would be to say it depends on your reference frame and that’s all we can say about it, but let’s think about this a little deeper with a practical example. Let’s shrink Bob down to the size of a proton and have Alice watch him race around the LHC while she is safely stationary in her own reference frame at the Atlas building. To Alice, Bob started out his “jog around the track” at 1 GeV, but now he’s put on a few pounds and is riding about 3 TeV. However, Bob still thinks he weighs 1 GeV because he’s in his own reference frame. Well, he thinks so, until he runs into George coming at him from the other direction and instead of them just bumping chests at 1+1=2GeV, they explode upon impact at 3+3=6 TeV.

So how do we explain this? What actually happens at the collision point in space between Bob and George? It doesn’t seem possible that, ostensibly in the same position in space, we could have one interaction at 2GeV and another at 6TeV, and that it all depends on who is looking at it? Do have something wrong here?

13. Dec 26, 2012

### Staff: Mentor

Please don't do that. If you believe that it is related, then simply post a link in one thread. Do not duplicate the question, it just leads to confusion.

Shorter, yes, heavier I would say no. Again, the concept of relativistic mass is deprecated in modern physics. Of course, he is gaining energy, he is firing his rocket pack at full blast!

The easy answer is a correct answer. I find this to be a rather odd psychological phenomenon that I have seen before. People find an easy and correct answer but insist on rejecting it anyway. It seems like some mental form of masochism to me.

However, if you want a "better" answer I would say that the question itself is incomplete as stated. Since we recognize that length is a relative quantity depending on the reference frame then the word "shorter" is a meaningless word without the identification of the reference frame to which is pertains. A correct statement of the question would be: “Is Bob actually getting shorter in Alice's frame, or isn’t he?”. And the correct answer is obviously: "Yes, Bob is actually getting shorter in Alice's frame."

I find that if you write correct questions the answers often become clear.

No, nothing is wrong here (other than your persistent use of the concept of relativistic mass). However, working problems with γ = 3000 (v = 0.99999994444444 c) can be a pain, so let's use γ = 1.25 (v = 0.6 c) and units where c = 1.

In Alice's frame Bob has an energy of 1.25 and a momentum of -.75 while George has an energy of 1.25 and a momentum of .75. So the Bob/George system has a total energy of 2.5 and a total momentum of 0. The square of the invariant mass of the Bob/George system is the square of the energy minus the square of the momentum, so the system has an invariant mass of 2.5 meaning that the collision can produce a particle of mass 2.5. Since the system has a momentum of 0 that particle would be at rest.

In Bob's frame Bob has an energy of 1 and a momentum of 0 while George (v=0.882 c) has an energy of 2.125 and a momentum of 1.875. So the Bob/George system has a total energy of 3.125 and a total momentum of 1.875. The system again has an invariant mass of 2.5 meaning that the collision can produce a particle of mass 2.5. Since the system has a momentum of 1.875 that particle would be travelling at 0.6 c, as expected from the results in Alice's frame.

So the energy and the momentum do depend on the reference frame, but the results of the collision are the same in terms of what particles can be produced. In reference frames where the collision has a higher energy the system also has a higher momentum. Since both energy and momentum must be conserved the "extra" energy in the high energy and momentum frame must remain as kinetic energy and cannot be converted into additional particles.

Last edited: Dec 26, 2012
14. Dec 26, 2012

### Naty1

We are 'amassing' [sorry to use that term, Dalespam!!] enough discussion points in this thread for recalling a closely related idea.

The last quote says it all....that's 'reality'....

Let's start with Dalespam's energy equation. In the frame of the mass, as you know,
velocity is zero...so there is NO additional momentum, no additional kinetic energy, for example. That is the essential reason relativistic mass is no longer popular: The 'mass' is unaffected by it's speed in its own frame. Electrons continue in the normal orbitals. A closely related aspect is the source of gravity: the stress energy tensor, which is also unaffected by high velocity.

It MASS [and associated gravity] did change, the object could turn into a black hole via high speed...yet we never observe that,so MASS cannot increase at high velocities!!!! So the way I think about this now, after a lot of tutoring from Dalespam, bcrowell, DocAl and others is in the reverse order from that I just used....

Just as different observers pass different relative TIME in SR, there is no 'real' time except for the clock you carry with you; Others 'see' things differently. And in GR not only does relative velocity affect the apparent passage of time, so does gravitational potential: Time for an observer on earth passes more slowly from that of an orbiting astronaut!!

Here is an interesting example from TomStoer:

How do we interpret this?? A test particle outside the collapse would be unaffected!!!
So here we have what might be an apparent gain in energy, but again, not in the frame of the collapsing star.

A contra example: Heat a mass and it's rest energy DOES increase.....electrons speed up in their orbitals, molecular bonds become more active, lattice structure may change, and there is no frame in which all this movement is at rest. So a hot mass will follow a slightly different trajectory than when cold.

15. Dec 27, 2012

### DiracPool

Ok, thank you to everyone who has commented so far. Unfortunately, all the numbers and equations are not helping me understand what is going on here. So, I’ll strip my quandary down to the barest minimum conceptual sense I can and see if I get anywhere.

Again, at the LHC, we take a proton of roughly 1 GeV, and then send it through the loops where it gains an energy of upwards of 3 TeV (or so) right before it collides with the other proton moving in the opposite direction. Call me crazy, but a simple man would look at that and say, “it looks to me as though that proton gained some energy by accelerating through those loops.” Now, if we are to assume that e=mc^2, then it looks to the simple man as though that proton must have gained some mass, or mass-energy, or however you want to characterize it. I mean, this is why they spent 2 billion dollars on the LHC, right, in order to increase the mass-energy of the protons?

This argument is not in trying to eschew Dalespam’s assertion of the concept of invariant mass, it is just trying to square it with the above-mentioned anomaly, or perceived anomaly a layperson may experience. I guess to put it succinctly, the question is what are we getting for our 2 billion dollars if we’re not getting more mass? We must at least be getting more energy, right? So are we to treat this energy as we treat the energy of a photon, having energy but just not “rest mass?” Is this the equivalent of how invariant mass remains invariant under acceleration? That is, is it the rest mass that is invariant?

16. Dec 27, 2012

### Staff: Mentor

Because mass and energy are in some sense equivalent, we can choose to think in terms of getting more mass OR more energy. When relativity was first discovered, it seemed most natural to think in terms of mass increasing with velocity. Since then, we have discovered mathematical formulations of relativity that make it much more natural to think in terms of a constant rest mass.

But as long as mass and energy are equivalent, we can always shift from one way of speaking to the other if it's convenient. It just so happens that we encounter very few problems today that are more easily solved using the language of mass increase, while there are many ways that the language of mass increase can be misleading and confusing. so it's not very popular these days.

In any case, there's no question about what the LHC does: It pumps enormous amounts of energy into particles and then smashes them together. It's up to you whether you want to use the equations of relativistic mass increase or invariant mass to calculate what happens next; but if you apply the equations correctly you'll get the same results either way.

17. Dec 27, 2012

### Staff: Mentor

No problem so far.

Here is the problem. In the current language used by most physicists, E=mc2 is a statement about the "rest-energy" of an object, that is, the energy an object has when it is at rest. The "m" is what popularizations and some/many serious introductory treatments of relativity call the object's "rest mass."

When the object is in motion, we include the effect of the motion on the energy either by way of the velocity:

$$E = \frac{mc^2}{\sqrt{1 - v^2/c^2}}$$

or by way of the momentum:

$$E = \sqrt{(mc^2)^2 + (pc)^2}$$

Yes, we get more energy by accelerating the particles. Before a collision, this extra energy is "motion energy", that is, kinetic energy. After the collision, some of that energy has become the rest-energies of the newly created particles, corresponding to their masses. The rest of the energy of course remains as kinetic energy, in the motion of the newly-created particles as they fly away from the collision point.

The total energy before equals the total energy afterwards. The sum of the masses of the incoming particles generally does not equal the sum of the masses of the outgoing particles. That is, the total rest-energy of the incoming particles does not equal the total rest-energy of the outgoing particles. Neither does the total kinetic energy of the incoming particles equal the total kinetic energy of the outgoing particles. But the total (rest + kinetic) energy is the same before and after.

18. Dec 27, 2012

### DiracPool

Alright, now things are making a little more sense. Thanks Nugatory.

19. Dec 27, 2012

### Staff: Mentor

That is why I put a nice little no-math conceptual summary at the end. Please read it and ask any specific questions you have:
It did gain energy, but the E=mc^2 applies only for objects at rest. The formula for moving objects is E^2/c^2 = m^2 c^2 + p^2. Increasing v increases p and therefore E, but not m. For p=0 it reduces to the familiar formula.

Energy, yes.

Yes, the rest mass is invariant, that is what makes it a useful quantity. As far as what we are getting for the money, that is more energetic collisions which can produce larger particles.

Also, note from my example above, that the mass of a system of two particles is greater than the sum of the masses of the individual particles.

20. Dec 28, 2012

### DiracPool

Ok, I think I got it, rest-mass bearing particle invariant mass is conserved in all cases and all reference frames. The only thing that changes depending on your reference frame is the apparent kinetic energy of the system as a whole. Is that right? But isn't that why we are giving the protons more energy in the LHC? So they can form heavier particles such as the Higgs boson which is 125 times the rest mass of the proton? The Higgs is a rest mass particle at 125 GeV, isn't it? If the relativistic energy stayed as kinetic, how are we to produce these heavier particles with just protons?