# B Relationship between mass and energy

#### fog37

Great. I will be moving into general relativity next. Before that, I will touch on one more topic: energy and mass, a topic of special relativity, and the formula $E=m c^2$.

There is rest mass $m_0$ (the amount of stuff an object is made off, an invariant) and the inertial mass $m_{inertial}$.

As an object'a speed $v$ increases, its $KE$ increases, and the object gains mass $m$" in the amount $$m=\frac {KE}{c^2}$$, correct? Is that mass increase m the inertia mass? Or do we call the entire mass (including both rest mass and extra gained mass) the inertial relativistic mass?

The same goes if an object gains gravitational potential energy $PE$: its mass increases by an amount $$m=\frac{PE}{c^2}$$

correct?

I hear that many don't like to use the term "relativistic mass". Why? Is the relativistic mass the sum of the rest mass and the gained or lost by adding energy (either potential or gravitational) to the system?

Thanks!

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#### jbriggs444

Homework Helper
Great. I will be moving into general relativity next. Before that, I will touch on one more topic: energy and mass, a topic of special relativity, and the formula $E=m c^2$.

There is rest mass $m_0$ (the amount of stuff an object is made off, an invariant) and the inertial mass $m_{inertial}$.
If by "inertial mass" you mean the m in F=ma then there is no such thing in special relativity. Erase it from your vocabulary.

The correct formulation is that $F=\frac{dp}{dt}$ and $p=m\gamma v$. If you differentiate, you do not get $F=m \frac{dv}{dt} = ma$. There is no way to redefine m to get F=ma to work. The force required to produce a given acceleration depends on the direction of the acceleration relative to the body's current motion. You cannot possibly have the same mass work for all directions.
As an object'a speed $v$ increases, its $KE$ increases, and the object gains mass $m$" in the amount $$m=\frac {KE}{c^2}$$, correct?
Its invariant mass, m, does not vary. It gains energy but not mass.
Is that mass increase m the inertia mass? Or do we call the entire mass (including both rest mass and extra gained mass) the inertial relativistic mass?
Since relativistic mass (computed as $\frac{E}{c^2}$) does not work with F=ma, calling it "inertial relativistic mass" would be a misnomer.

• nasu, vanhees71 and Ibix

#### Ibix

I hear that many don't like to use the term "relativistic mass". Why? Is the relativistic mass the sum of the rest mass and the gained or lost by adding energy (either potential or gravitational) to the system?
The term "relativistic mass" is fine, although the whole concept has basically been dropped in modern physics. Note, however, that you should not use the word (as you have done several times) "mass" to mean relativistic mass. Modern usage is that "mass" means the invariant mass, and if you want to mean relativistic mass you need to say relativistic mass (and probably which of the several quantities that go loosely under that heading you actually mean).

The concept has largely been dropped because it's not an invariant, and invariant quantities are the ones that work best with relativity. And calling it a mass is misleading as @jbriggs444 has pointed out - it does not behave at all like a Newtonian mass.

I think people introduced relativistic mass because they hoped they could make relativity more friendly to students familiar with Newtonian physics. But to my mind it's a mistake, because it's an attempt to hide the fundamental differences between relativistic and Newtonian physics, rather than drawing out and highlighting the differences.
The same goes if an object gains gravitational potential energy PE: its mass increases by an amount
$$m=\frac{PE}{c^2}$$
correct?
I've never seen that, and I don't think it's correct. Localising the energy in a gravitational field is notoriously difficult - even if you want to think of it as a mass term I do not think you'll find it "in" an object in the field.

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• PeroK

#### pervect

Staff Emeritus
Great. I will be moving into general relativity next. Before that, I will touch on one more topic: energy and mass, a topic of special relativity, and the formula $E=m c^2$.

There is rest mass $m_0$ (the amount of stuff an object is made off, an invariant) and the inertial mass $m_{inertial}$.
Unfortunately, especially if you wish to go onto general relativity, you'll need to revise this.

Where you really want to end up to deal with General Relativity is saying that there is a stress-energy tensor $T^{ab}$ that describes the distribution and density of energy and momentum of a medium. Getting there may be a bit of a journey, though.

If you don't have the background to go to the stress energy tensor yet - for instance you are likely not familiar with tensors- there are intermediate steps you can take.

In special relativity, a point particle with zero volume, or an isolated system with a finite volume, has an energy-momentum 4-vector, that gives the amount of energy and momentum the object has in a given frame. This 4-vector transforms the same way as any other 4-vector in special relativity, via the Lorentz transform. You'll still have to learn about 4-vectors - a good text for this is Taylor & Wheeler "Space-time physics".

Then mass, which you call rest mass, is the "length" of this 4-vector. In units where c=1, there's a simple expression:
$$m^2 = E^2 - p^2$$

where E is the energy and p is the momentum. In more standard units, where c is not necessarily one, this becomes

$$(mc^2)^2 = E^2 - p^2c^2$$

Relativistic mass is just a synonym for energy with a scaling factor of c^2, and is generally not used much.

It's not stressed much, but systems with a finite volume that are not isolated doesn't have an invariant mass. You can still compute $E^2$ and $p^2$, but $E^2 - p^2$ is no longer invariant for such a system.

This can be regarded a a consequence of the relativity of simultaneity. Energy is flowing into and out of a non-isolated system. How much energy is in the system at a given time winds up depending on the definition of simultaneity used. The standard way of dealing with this (and the only way I know of that works) is to reformulate physics in terms of the stress-energy tensor.

Understanding special relativity with the 4-vector approach is a good first step towards the background needed for General realtivity. 4-vectors can be regarded as a rank 1 tensor, though they are often introduced without the full tensor machinery.

• DEvens

#### Nugatory

Mentor
...the formula $E=mc^2$
That formula is correct only for the special case of an object that is at rest. The $m$ that appears in this formula is therefore going to be the invariant rest mass and the energy calculated from it will also be invariant; in fact you will occasionally hear it referred to as the "rest energy". Because they’re both invariant and related by the constant $c^2$, you should even think of rest mass and rest energy as two ways of talking about the same thing. The $c^2$ is just the conversion factor between the units that we most often use for energy and the units that we most often use for mass. (This is why the particle physicists will say that the mass of a proton is $938 MeV/c^2$ - MeV is a unit of energy).

But that’s all for the invariant and frame-independent rest mass. The frame-dependent notion of kinetic energy isn't involved so if you're going to include it you need a different formula: $E^2=(mc^2)^2+(pc)^2$ where $p$ is the momentum. This gives you the total energy including the kinetic energy. Because $p$ is frame-dependent the $E$ in this formula is also frame-dependent which is to be expected because kinetic energy is frame-dependent.
(Start with this formula and grind through the algebra and you can calculate that for a massive particle traveling at subluminal speeds the total energy is $\gamma{m}c^2$, the kinetic energy is $(\gamma-1)mc^2$, and the rest energy is $mc^2$. Note that we’ve gotten all the physics in without ever introducing the dubious and unhelpful concept of relativistic mass).

Now we have enough background to take on your questions.
As an object's speed v increases, its KE increases....
Yes, but do remember that that speed is frame-dependent so the kinetic energy will also be frame-dependent
....and the object gains mass m" in the amount $m=\frac{KE}{c^2}$, correct?
Not correct. The kinetic energy is $(\gamma-1)mc^2$ and it’s $\gamma$ that’s increasing, not the invariant mass $m$.
The same goes if an object gains gravitational potential energy PE: its mass increases by an amount $m=PE/c^2$
correct?
Not correct. The rest energy of the system consisting of the object and whatever is creating the gravitational field will increase by that amount, because that’s the amount of energy we had to add to the system to further separate the two to increase the potential energy. However, the mass of the object is doesn’t change. (If the energy had come from inside the system - perhaps we dug up some coal, burned it to generate electricity, then used the electricity to run a motor that lifted the object to increase its potential energy - then the rest energy of the system won’t change either because we’re just moving energy around inside the system).

• vanhees71 and overmancer

#### fog37

Thank you. I am processing all this great information...

Very simplistically, is it ok to look at the (rest) mass of an object as the sum of all the internal energy + the energy trapped in the mass (atoms, molecules, the electrons, protons, neutrons composing them) composing the object? After all, quantum chromodynamic theory is a field theory which helps (me) think of this thing called mass as energy since it is truly a field..The mass of protons is the energy fluctuations in the gluon fields...

#### Nugatory

Mentor
Very simplistically, is it ok to look at the (rest) mass of an object as the sum of all the internal energy + the energy trapped in the mass (atoms, molecules, the electrons, protons, neutrons composing them) composing the object?
Pretty much yes, although "represented by the rest mass of all the particles" would be a bit more accurate than "trapped in the mass".

#### DEvens

Gold Member
Thank you. I am processing all this great information...

Very simplistically, is it ok to look at the (rest) mass of an object as the sum of all the internal energy + the energy trapped in the mass (atoms, molecules, the electrons, protons, neutrons composing them) composing the object? After all, quantum chromodynamic theory is a field theory which helps (me) think of this thing called mass as energy since it is truly a field..The mass of protons is the energy fluctuations in the gluon fields...
For a static object, yes that's fine.

Think of a trivial example. I'm going to use the $c=1$ units because I'm very lazy. Consider a mass at rest with mass $M$. Its four-vector momentum is just $(M,0,0,0)$.

Now suppose it splits up into two equal portions of $m$, and $2m < M$, so there is some energy to play with. The two parts go flying off in the +x and -x direction. Recall that $\gamma = \sqrt{\frac{1}{1-v^2} }$. We must then have $\gamma$ such that $2 \gamma m = M$. So we then get two four-vector momenta as follows. The one to the right gets $( \gamma m, \gamma m v, 0 ,0)$ and the one to the left gets the same 0-component but negative of the x-component, or $( \gamma m, - \gamma m v, 0 ,0)$. And for each of these, the three-vector has magnitude $\gamma m v$ but the four-vector has magnitude $m$.

Recall that the magnitude of a four-vector $(x_0,x_1,x_2,x_3 )$ is $\sqrt{ -x_0^2 + x_1^2 + x_2^2 + x_3^2 }$.

It's a beautiful thing. The four-vector is a four dimensional vector with an invariant magnitude under Lorentz transforms, and the three-vector part has a magnitude that is invariant under rotations.

But now we look at the total four-momentum after the split. We just add the components and we get
$( \gamma m, \gamma m v, 0 ,0) + ( \gamma m, - \gamma mv, 0 ,0) =(2 \gamma m, 0, 0 ,0) = (M,0,0,0)$. So gathering up all the bits-and-pieces we see a very cool thing indeed. Four-momentum is conserved. And that's what mass is about. Notice that the two parts of the split-up mass, if considered a single system, still have mass $M$. That is, it should be called the "invariant mass" not the rest mass. For one particle the invariant mass is $m$. For the system, with each part flying away from the other, the invariant mass is $M$, even though the parts are flying away from each other.

It even works for photons. Then the mass would split into two back-to-back photons, each with energy $k = M/2$ and the four-momenta of the photons would be $(k,k,0,0)$ and $(k,-k,0,0)$. Each photon has invariant mass of zero. But the invariant mass of the total system is still $M$ even when the photons are flying away from each other.

That's also a contributing factor to deprecating "relativistic mass," though jbriggs444 has a good reason also. Invariant objects in the theory, and invariant relationships, are what are important. It makes the name somewhat ironic. • nasu

#### DrStupid

There is no way to redefine m to get F=ma to work. The force required to produce a given acceleration depends on the direction of the acceleration relative to the body's current motion. You cannot possibly have the same mass work for all directions.
You can, but that that "mass" would be different from the concepts of mass as used in relativity or classical mechanics. It would not even be a scalar (and with scalar I mean scalar and not Lorentz scalar).

• jbriggs444

#### Mister T

Gold Member
There is rest mass $m_0$ (the amount of stuff an object is made off, an invariant) [...]
No. It's true that it's an invariant, but the rest mass (I just call it the mass because it's the only "kind" of mass you need to consider in special relativity) is not the amount of stuff an object is made of. Any internal energy that the object has also contributes to the mass.

That is the true meaning of the Einstein mass-energy equivalence. I prefer to write it as $E_o=mc^2$ where $E_o$ is called the rest energy and $m$ is the mass.

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