# Mass gained Infinitely?

1. Oct 26, 2006

### TDS

Mass gained Infinitely??

Would some of you learned folks outhere answer these questions?

It is said in GR and SR that an object will gain mass infinitely as it approaches the speed of light. How is this possible? Where does the mass come from? Does this apply to the Photon?

If you answer this, and you show a formula, please explain what the symbols in the formula represent. This will help me understand what you are saying.

Thanks!

2. Oct 26, 2006

### HallsofIvy

That isn't possible which is why that isn't said in either GR or SR! The mass is given by
$$m= \frac{m_0}{\sqrt{1- \frac{v^2}{c^2}}}$$
where m0 is the "rest mass", the mass as measured in a frame of reference in which the object is motionless, v is the speed relative to the frame of reference in which the mass is being measured. c is the speed of light. It is true that, as v increases toward c, the denominator "goes to 0" and m itself increases without bound. That may be what you meant by "gain mass infinitely" but I consider that very bad terminology. Since "conservation of mass" is replaced in relativity by "conservation of mass-energy", the mass "comes from" the increase in energy due to what ever forced is accelerating the object. In fact, it is more common now to refer to the rest-mass of an object as it's mass and include that "relativistic mass" in energy. It is precisely because it is impossible for that to "become infinite" that we say it is impossible for an object with non-zero rest mass to move at the speed of light.
Since a photon has 0 rest-mass that doesn't apply to photons except in the sense that 0 divided by anything is 0: the photon always has 0 mass.

3. Oct 26, 2006

### wxrocks

TDS -- this is the concept (in a sense) that particle accelerators use. Take a couple of atoms, smash them together at high speed, and you get a shower of particles with much more mass than the original 2 atoms -- this is because the energy becomes these particles (in a sense).

This actually happens in the atmosphere when high velocity/energy particles hit the atmosphere - the collision creates a shower of particles that certainly doesn't follow the conservation of mass, but the conservation of mass-energy.

4. Oct 26, 2006

### TDS

Thanks!

@HallsofIvy,

Thank you so much for this explaination!

While it is highly probable that I misunderstood GR and SR, I have posed the idea of traveling faster than the speed of light and it gets shot down by the statement: "you cannot travel faster than light because as you approach the speed of light, you gain mass infinitely and you could not carry enough fuel to allow you to travel that fast." This was told to me by a Physics Instructor at a school that I was attending. As for your statement concerning the Photon having 0 rest mass; are you saying that the Photon has no mass until it is accelerated to the speed of light? Is a Photon a Photon prior to it accelerating to "c"? And if it has 0 mass, how can it be affected by a gravitational source?

5. Oct 26, 2006

### pmb_phy

Although it was quite clear from the context that the OP meant as a body with finite rest mass approaches the speed of light the mass increases without bound. when he wrote gain mass infinitely it was an imprecise statement.
The correct definition of "mass" (aka relativistic mass/inertial mass) is m = p/v. Only when the particle has a finite rest mass will m = gamma*m_0 hold and then only under certain restrictions. But it is not a definition of mass.
Nothing about the conservation of mass was changed when going to and from Newton's theory to Einstein's theory. It is a simple matter to show that mass is conserved.
Or to be precise, the work done on the object will increase the kinetic energy of the object. This energy shows up as an increase in mass.
That is an incorrect statement. It assumes that E = mc2 in all cases, which it does not. It therefore "relativistic mass" cannot be replaced by "mass-energy". This assumption can lead a person to make statements which are very serious errors. E.g. people will almost always believe that mass density is identical to energy density. It is not. Rindler explains this in his 1982 intro to SR book.

Best wishes

Pete

6. Oct 26, 2006

### TDS

@wxrocks,

Perhaps I am having trouble distinguishing between "mass" and "weight".

Correct me if I am wrong; "Mass" is the measurement of the area that an object occupies, and "weight" is the measurement of the affect that gravity has on an object.

Having said that, if I take to atoms of gold and accelerate them to relativistic velocities and smash them together, the mass would increase because there would be the component particles that made up the two atoms of gold along with the energy produced by the collision and the weight of the two gold atoms would be reduced because part of the gold atoms was converted to energy??? Is this correct?

7. Oct 26, 2006

### pmb_phy

Have you ever heard of tachyons? These are particles which are already traveling faster than light when they are created. What you're refering to is a particle which is a first at rest (or moving such that v < c) and one attempts to accelerate it to the speed of light. These are two different cases.
I guess he never read that paper on tachyons in which all is explained quite well. As for your statement concerning the Photon having 0 rest mass; are you saying that the Photon has no mass until it is accelerated to the speed of light? Is a Photon a Photon prior to it accelerating to "c"? And if it has 0 mass, how can it be affected by a gravitational source?[/QUOTE]When someone uses the term "zero rest mass" they mean the m in the relation m^2 c^4 = E^2 - (pc)^2. With a photon E = pc so it follows that m = 0. Strictly speaking, the correct term to use is "proper mass" since "rest mass" means something else in GR. I.e. the "relativistic mass" is a function of the particle's gravitational potential as well as the particle's speed. So while the speed may be zero the rest mass may differ from the proper mass.

Best wishes

Pete

8. Oct 26, 2006

### ZapperZ

Staff Emeritus
Can you point out experimental evidence for this?

The problem I have with how you presented this "tachyons" is as if this is a done deal. Tachyons are still hypothetical particles and still have not been empirically verified. Someone who is not knowledgable about such a thing would read your post and would be mislead into thinking it is as well established as a neutrino. It is NOT!

So you cannot use it as evidence for something. A hypothetical evidence is even weaker than circumstantial evidence.

Zz.

Last edited: Oct 26, 2006
9. Oct 26, 2006

### TDS

@pmb_phy,

I am very familiar with the Tachyon, although I have not been able to find any materials on the subject. Would you providing me a source of information so that I can read about them?

As for the explaination for "zero rest mass" that you have provided, are you saying that:

m^2 = mass squared * speed of light to the 4th power equals Energy squared - (momentum*speed of light) squared?

When someone uses the term "zero rest mass" they mean the m in the relation m^2 c^4 = E^2 - (pc)^2. With a photon E = pc so it follows that m = 0. Strictly speaking, the correct term to use is "proper mass" since "rest mass" means something else in GR. I.e. the "relativistic mass" is a function of the particle's gravitational potential as well as the particle's speed. So while the speed may be zero the rest mass may differ from the proper mass.

Best wishes

Pete[/QUOTE]

10. Oct 26, 2006

### pmb_phy

Sure

Possibility of Faster-Than-Light Particles, G. Feinberg, Physical Review. Vol. 159, No. 5, July 1967
Yes, so long as you know that the "m" in the equation is proper mass.

Best wishes

Pete

11. Oct 26, 2006

### TDS

@pmb_phy,

Thanks for the article info and the answer to my question! Now I shall go and find the article.

12. Oct 26, 2006

### JesseM

That's not right, "the measurement of the area that an object occupies" would just be its volume. Mass is usually understood in terms of an object's inertia, or resistance to being accelerated. If you're familiar with the equation F=ma in Newtonian physics, you can see that the more mass m the object has, the greater the force F that must be applied to it in order to accelerate it by a fixed amount a. In relativity, the "rest mass" could be understood in terms of the amount of force (or energy) needed to accelerate an object by a small amount in its rest frame, while the "relativistic mass" could be understood in terms of the amount of force/energy needed to accelerate an object by a small amount in its direction of motion, as seen in a frame where it already has some sizeable velocity. The closer the object's speed is to the speed of light, the greater the energy needed to get its speed even closer to that of light, and it would take an infinite amount of energy to get it all the way there.

13. Oct 26, 2006

### TDS

At the risk of sounding stupid, would you mind giving me an example of this. Say for example that NASA wanted to launch a probe to Alpha Centauri and get it there in say 10 years.

14. Oct 26, 2006

### JesseM

I was actually thinking more along the lines of an incremental change in velocity, but if we assume the probe is "cruising" inertially throughout most of the trip, then to get there in 10 years it'll have to get up to a speed of around (4.35 light years/10 years) = 0.435c. In this case, we can use the equation $$E = mc^2/\sqrt{1 - v^2/c^2}$$ for the total energy of a moving object to figure out how much energy this takes. In this equation, m is the rest mass and v is the velocity, so the amount of energy needed to accelerate it would be (energy after acceleration, when v = 0.435c) - (energy before acceleration, when v = 0), which would equal $$mc^2/\sqrt{1 - 0.435^2} - mc^2$$ or $$1.11mc^2 - mc^2$$ or $$0.11mc^2$$. And if we want to define the rest mass in terms of the amount of energy $$E_a$$ it takes to accelerate an object to a given speed, we take the equation $$E_a = (mc^2/\sqrt{1 - v^2/c^2} - mc^2)$$, do a little algebra to turn this into $$E_a = (1 - \sqrt{1 - v^2/c^2})*mc^2/\sqrt{1 - v^2/c^2}$$ and solve for m to get $$m = E_a * \sqrt{1 - v^2/c^2} / (c^2*(1 - \sqrt{1 - v^2/c^2}))$$. In the example above, this would mean $$m = E_a /(c^2*0.11)$$, or $$m = 0.9E_a /c^2$$. So if we know the energy we had to use to accelerate the probe to 0.435c, this equation defines the probe's rest mass in terms of that energy.

Last edited: Oct 26, 2006
15. Oct 27, 2006

### TDS

@JesseM,

Thank you for explaining this to me. Now you have given me something to work with!

16. Oct 27, 2006

### pmb_phy

To be precise that is the increase in kinetic energy from a value of zero, i.e. the particle is accelerated from a position of rest.

Best wishes

Pete

17. Oct 31, 2006

### TDS

One more question.

@JesseM,

How would describe the amount on Energy used. Would it be in Ergs, Joules, Newtons, or Pounds of thrust?

18. Oct 31, 2006

### Staff: Mentor

Ergs and joules are units of energy. Newtons and pounds are units of force.

In the equations above, you can use ergs for energy provided that you also use grams for mass and cm for distance (cm/sec for velocity). You can use joules for energy provided that you also use kilograms for mass and meters for distance (m/sec for velocity). This is a just a matter of consistency of units (meter-kilogram-second units versus gram-centimeter-second units).

19. Oct 31, 2006

### JesseM

Ergs and Joules are units of energy, so you could use either of those. You could also use Newtons*meters to get the energy used in joules, since force*distance = work done to change the kinetic energy (with the distance in meters measured in the same direction of the force), and 1 pound of thrust = 4.45 Newtons so you could also do that conversion and then do 1 Newton*meter = 1 joule.

20. Oct 31, 2006

### TDS

Thanks!

@jtbell and JesseM,

My thanks to the both of you for your help! You have helped me more than you realize!