# B Special Relativity & Relativistic Mass

1. Nov 23, 2017

### Jimmy87

Hi,

If a particle is within 2 m/s of the speed of light (like at CERN) and then more energy is transferred to the particle, where does this energy go? Some sources I have read talk about relativistic mass getting bigger and approaching infinity as a particle approaches 'c'. However, this source which seems reliable (as he has worked all his life in particle accelerators) says that there is no such thing as relativistic mass - it is just a useful concept to help make equations easier to work with:

So if a particle really only has one single mass and there is no such thing as relativistic mass then where does the energy go if you keep transferring it to a particle that is within 1 or 2 m/s of 'c'?

Thanks

2. Nov 23, 2017

### Ibix

The added energy increases the kinetic energy of the particle. The total energy of the particle (including both the contributions of its rest mass and its kinetic energy) is $\gamma mc^2$, where $\gamma=1/\sqrt {1-v^2/c^2}$. That can be as large as you like without ever exceeding the speed of light.

Some sources call the quantity $\gamma m$ the relativistic mass. That has largely dropped out of favour because it tends to lead to incorrect guesses (e.g. "do I become a black hole if I go really fast?") and lots of confusion between "rest mass" and "relativistic mass". So now we use "mass" to mean "rest mass" and don't use "relativistic mass". Relativistic mass is just total energy divided by $c^2$, anyway, so you can always talk about total energy if you need the concept.

3. Nov 23, 2017

### Jimmy87

Thanks that helps! If the mass stays the same though I still don't get where the energy physically is when you transfer more to it. If the kinetic energy goes up but it can't go faster or get more massive then the energy must be doing something?

4. Nov 23, 2017

### Ibix

Energy isn't anywhere. It's just an accounting mechanism, really.

For example, you are presumably sitting somewhere. Your kinetic energy is zero in your local frame of reference. But the Earth is doing 20km/s around the Sun. So your kinetic energy in the solar system frame of reference is about ten billion Joules (if my mental arithmetic is reliable). Both answers are correct; your kinetic energy depends on who is doing the measuring. So it can't "be" anywhere.

5. Nov 23, 2017

### PeroK

This is quite a good question, I think. There is, in fact, a similar conundrum in good old classical physics, where KE is $\frac{1}{2}mv^2$.

If, in your reference frame, an object of unit mass accelerates from $0$ to $1m/s$ then it gains $0.5J$.

But, in a reference frame where the object accelerates from $10$ to $11m/s$ it gains $10.5J$.

How can the energy be frame dependent? How can you get different answers for the energy gained depending on how you are moving relative to the object?

If the object is a car, could you not simply measure the fuel consumption to see who is right?

6. Nov 23, 2017

### Jimmy87

Ok, well let's say we are in the frame of a particle physicist measuring the speed of the particle. He/she increases the kinetic energy (and speed) of the particle until it gets to within 2 m/s of 'c' from their frame of reference. They keep injecting more and more energy into the particle but it can't get any faster (at least not much) or any more massive, so if the physicist is transferring more energy to the particle what is happening to the particle as a result. It like dropping a tennis ball on the floor and asking "where did the energy go?" - answer - into heat and sound. So "where did the energy I transferred to the particle that was already going within 2 m/s of 'c'?". Into sound? Heat? Obviously not - but where?

7. Nov 23, 2017

### Jimmy87

How could that situation exist? If I am in a car going at 30m/s and increase to 31m/s surely both me and stationary observer agree I have gone from 30 to 31m/s. In my car I could use the time it takes to go between road signs to deduce my speed couldn't I? Which would still give me the same as the stationary observer?

8. Nov 23, 2017

### Ibix

Into the kinetic energy of the particle. Do you have any program that can plot a graph? Excel for example? Or just pen and paper. You might like to plot the kinetic energy as velocity increases towards c (I gave you the total energy in #2 - kinetic energy is just the total energy, $\gamma mc^2$, minus the rest mass energy, $mc^2$, so $K.E.=(\gamma-1)mc^2$).

9. Nov 23, 2017

### Ibix

According to the roadside observer the road signs are stationary and you are moving north (picking a direction out of a hat). According to you, you are stationary and the road signs are moving south. So, no. You determine that the road signs have the same speed in the opposite direction as the roadside observer says you have.

This was the point behind Einstein's famous "when does Oxford stop at this train" comment. Conventionally, we regard the train as moving. But there's nothing wrong with thinking of Oxford as moving and the train as stationary. Either view is fine, and it is a key principle of relativity (actually called "the principle of relativity") that neither is wrong.

10. Nov 23, 2017

### Jimmy87

I think I get what your saying. You are saying that the gamma factor gets bigger since the particles mass and velocity can't change (if it is nearly at c). So the kinetic energy is not increasing because the mass is increasing but because the gamma factor increases?

11. Nov 23, 2017

### nitsuj

I think there is a refinement in the wording, that is invariant mass From the exception of netting momentum of massless particles to zero. i.e. things not at rest, but invariant.

12. Nov 23, 2017

### Mister T

The speed continues to increase, it's just that at low speeds most of the energy goes into making the speed increase, but at high speeds most of it goes into making $\gamma$ bigger. The speed continues to increase as you transfer energy, getting closer and closer to $c$, but $\gamma$ increaes beyond all bounds, there is no limit to how large it can get.

Last edited: Nov 23, 2017
13. Nov 23, 2017

### Mister T

I think he's saying just the opposite. There is such a thing as relativistic mass, but it's a rather useless quantity. Particle physicists have never used it, as far as I know, and if there were cases where they did use it, they would have been quite rare. Certainly its use among authors of physics books has fallen off fairly rapidly in the last few decades.

14. Nov 23, 2017

### PeroK

That's how fast the car is going relative to the ground. But, of course, the Earth is rotating and orbiting the Sun, so your car is going much faster relative to a "stationary" observer in space.

To them, if your car accelerates by 1m/s that could represent a huge increase in kinetic energy.

15. Nov 24, 2017

### Ibix

The velocity changes, but it changes less for each additional injection of energy. This is so even at low speeds - 1J added to a stationary 1kg mass accelerates it to about 1.4m/s. Adding another 1J accelerates it to 2m/s, less than double the speed. This diminishing return effect gets stronger as you get closer to c.

Of course, an object with 2J of kinetic energy will hit twice as hard as one with 1J, whether they're 1kg masses moving at walking pace or atoms moving a tiny fraction under the speed of light.

16. Nov 24, 2017

### Ibix

I agree that invariant mass is a better term than rest mass. The point I was making was that you don't have to use either - you just use "mass", and use "total energy" for what used to be called "relativistic mass".

17. Nov 24, 2017

### nitsuj

oh, I totally missed that point but agree.

18. Nov 24, 2017

### nitsuj

Over a long(er) period of time, the car is only 100 horse power.

19. Nov 24, 2017

### Jimmy87

So is gamma technically an energy store. Let me explain. We are taught that when energy is transferred it goes from one store to another - this helps show conservation of energy. So burning a fuel for example you may start off with 200J in the chemical store (fuel) and if you burn it all you will end up with 200J in the thermal store. So is gamma an energy store in this sense? If a particle is going at 99.99999999999% speed of light and transfer 1000000000J to it then would this be contained in the gamma store? Is this kosher?

20. Nov 24, 2017

### Jimmy87

Thanks for all the excellent info guys btw

21. Nov 24, 2017

### Mister T

In a particle accelerator energy is being transferred to the particle, and it's accounted for in the kinetic energy of the particle. That is, you transfer energy $E$ to the particle and the particle's kinetic energy increases by $E$.

The only issue you're having is that the relationship between speed and kinetic energy is not what you think it is. You think that as the kinetic energy increases so does the speed. And you're absolutely correct about that. It's just that it doesn't increase in the way that doesn't seem to make sense to you.

$\gamma$ is simply a mathematical function of the speed $v$. The only reason it has a physical implication is that it's used in the formula that relates the kinetic energy to $v$.

22. Nov 24, 2017

### SiennaTheGr8

Momentum: $\mathbf{p} = \gamma m \mathbf{v}$
Energy: $E = \gamma m c^2$

It is true that if something is moving near light-speed and you transfer momentum to it in the direction it's already going, the increase in its magnitude-of-momentum that you measure will be almost entirely in its $\gamma$, and hardly at all in its $v$ (I'm assuming its mass stays constant). At very low speeds, on the other hand, the "roles" of $\gamma$ and $v$ here would be reversed: the increase in $\gamma$ would be negligible, while the increase in $p$ would be almost entirely accounted for by the increase in $v$.

You'd also measure a corresponding increase in the object's energy, and that increase would be entirely accounted for by the increase in $\gamma$, regardless of the object's initial speed.

23. Nov 24, 2017

### jbriggs444

The rate of kinetic energy gain by a car powered by a 100 horse power motor can greatly exceed 100 horse power if the highway is moving at multiple miles per second. This does not violate energy conservation once you consider the force the tires apply to the highway.

24. Nov 24, 2017

### PeroK

You're getting a bit confused by the mathematics. There is only a change in kinetic energy here. There is no gamma store. In fact if you expand the expression for gamma, you get:

$\gamma mc^2 = mc^2 + \frac{1}{2}mv^2 + \dots$

Where the remaining terms in the expansion have increasing orders of $1/c^2$.

From this you can see that the relativistic expression for kinetic energy reduces to the classical expression where $v$ is small compared to $c$.

In any case, there is no need to see relativistic kinetic energy as fundamentally different from classical kinetic energy. They just have different formulas, where one is an approximation of the other.

Last edited: Nov 25, 2017
25. Nov 25, 2017

### nitsuj

Are you able to reword that? I don't understand what you're saying.