# Equivalence of mass and energy

1. Feb 25, 2010

### Rasalhague

Some naive questions about the meaning of this expression.

In what follows, I'll use the word mass for "rest mass", the magnitude of the energy-momentum 4-vector. (Answers in terms of "relativistic mass" are fine, just let me know what definitions you're using. Taylor & Wheeler in Spacetime Physics sometimes use the term rest mass, as in the quote below, to avoid ambiguity, even though they depricate the term relativistic mass.)

In what sense is this equivalence to be understood? Presumably not in the obvious sense that any $E$ in an equation can be replaced with $m$ or $mc^2$. Not the former because it leads to this not-generally true conclusion:

$$E = \gamma m$$

$$E = mc^2$$

$$\therefore \gamma = c^2$$

Or if

then treating $E$ as equivalent to $m$ would lead to the conclusion that $2E$ and $2m$ are conserved while $E$ and $m$ are not, and that therefore 2 is not conserved!

And a photon is said to have no mass, and therefore its energy, $E$, is equal to its 3-momentum, $p$ because

$$\left(m_{phot}\right)^2 = \left(E_{phot}\right)^2 - \left(p_{phot}\right)^2$$

$$0 = \left(E_{phot}\right)^2 - \left(p_{phot}\right)^2,$$

which argument wouldn't work if $m$ or $mc^2$ were notationally equivalent to $E$.

So is "energy and mass are equivalent" merely a statement of the fact that, once the appropriate factors of $c$ are included, energy and mass have the same units, in which case could we just as well say that mass and energy and 3-momentum are "equivalent" to each other, in this sense, although not, of course, equivalent in the obvious sense that a statement about mass would be true if and only if it's true also of energy and 3-momentum.

Or is the equivalence something to do with a change from mass to energy, or vice versa, in a physical interaction? In Spacetime Physics, in chapter 2, section 13 "Equivalence of energy and rest mass" (which a previous owner of my copy has corrected to "Equivalence of rest energy and mass"), Taylor and Wheeler give the example of an inelastic collision between two lumps of putty which merge as a result of the collision. Before the collision, the first lump is moving, the second at rest. They use T for kinetic energy, subscript 1 for properties of the first lump, subscript 2 for properties of the second, and subscripts "initial" and "final" for the total mass, energy and 3-momentum of the system before and after the collision. From the conservation of energy and 3-momentum, and from the fact that the energy of a body at rest is equal to its mass, they reason:

$$E_{initial} = E_{final} = E_1 + E_2 = E_1 + m_2$$

$$p_{initial} = p_{final} = p_1$$

$$\left(m_{final}\right)^2 = \left(E_{final}\right)^2-\left(p_{final}\right)^2$$

$$\left(m_{final}\right)^2 = \left(E_1 + m_2\right)^2-\left(p_1\right)^2$$

$$\left(m_{final}\right)^2 = \mathbf{\left(E_1\right)^2 - \left(p_1\right)^2} + \left(m_2\right)^2 + 2E_1m_2 = \left(m_{initial}\right)^2$$

$$\left(m_{final}\right)^2 = \mathbf{\left(m_1\right)^2} + \left(m_2\right)^2 + 2\left(m_1 + T_1\right)m_2 = \left(m_{initial}\right)^2$$

$$\left(m_{final}\right)^2 = \left(m_1\right)^2 + \left(m_2\right)^2 + 2m_1m_2 + 2T_1m_2 = \left(m_{initial}\right)^2$$

$$\left(m_{final}\right)^2 = \left(m_1+m_2\right)^2+2T_1m_2 = \left(m_{initial}\right)^2$$

(I've filled in some of the steps there.) Conclusion:

In connection with this, they say on p. 134 that "rest mass often changes in an inelastic encounter", contrasting the property of (frame-)invariance with conservation.

My question here: isn't the rest mass of the whole system always conserved? How can it not be, since it depends only on the energy of the whole system (which is conserved) and the momentum of the whole system (which is conserved)?

Perhaps I'm missing something, but the idea that rest mass "changes" in an inelastic collision, seems to rely on adopting an inconsistent convention whereby, before the collision, one uses the word "mass" for the sum of the individual masses of the seperate bodies, whereas after the collision one switches the referent of "mass" to the mass of the whole system.

The following quote deals with a similar, putty-based example:

I'm not sure what "effective mass" means in Taylor & Wheeler's terms. At first I thought it might be energy, give or take the relevant conversion factor, that some people call relativistic mass, the time component of the energy-momentum 4-vector, since this changes from frame to frame as $\gamma$ changes. But from the situation being described, this "effective mass" seems rather to play the role of Taylor & Wheeler's rest mass of the system. For them, the famous equation means, in natural units, "rest energy equals mass", i.e. in a frame where the total momentum of a system is zero, its rest mass is equal to the sum of the energies of its individual particles.

Or does the statement that mass and energy are equivalent relate to the fact that mass is the source of gravity in Newtonian mechanics, whereas in GR, the source of gravity is energy-stress, whose components include energy density and energy flux density, among other things, and therefore energy, in GR, is partly responsible for what mass is wholly responsible for in Newtonian mechanics, and therefore total relativistic energy is very roughly "equivalent" to Newtonian mass in some simplified special case?

Last edited: Feb 25, 2010
2. Feb 26, 2010

### Rasalhague

Well, I rambled a bit there, but in a nutshell:

In what sense are mass and energy equivalent?

(To Taylor and Wheeler, if I've understood them, the statement that mass and energy are equivalent only means "rest energy is equivalent to rest mass", but others sometimes seem to mean more by it.)

3. Feb 26, 2010

### Staff: Mentor

I prefer to think of them as distinct concepts. Specifically, in units where c=1 the total energy is the timelike component of the http://en.wikipedia.org/wiki/Four-momentum" [Broken] and mass is the norm of the four-momentum. The equivalence comes from the simple fact that if the momentum is 0 (the object is at rest and the spacelike components of the four-momentum are 0) then the timelike component is equal to the norm (the total energy of a body at rest is equal to the mass).

Last edited by a moderator: May 4, 2017
4. Feb 26, 2010

### Staff: Mentor

Yes. (provided of course that the system is "closed").

Note, however, that your preceding quote from Taylor and Wheeler says:

(I added the boldface.) In general, the rest mass of a system does not equal the sum of the rest masses of the objects that comprise the system. In this example, the rest mass of the total system of two objects before the collision is greater than the sum of the rest masses of the individual objects.

5. Feb 26, 2010

### Rasalhague

Thanks for the replies. This all agrees with what I've read in Taylor & Wheeler. I'm still puzzled as to how the equivalence of rest mass and rest energy can be used, as in that first quote, to reason about the gravitational influence exerted by a pulse of light which has no rest mass. Don't the components of the stress-energy tensor include only energy density, energy flux density, momentum density and momentum fluc density, but not explicitly mass? This also puzzles me:

"Mass and energy are equivalent. The energy of a beam of light is equivalent to a certain amount of mass, and the beam is therefore deflected by a gravitational field."
http://www.lightandmatter.com/html_books/0sn/ch07/ch07.html [Broken]

I thought the idea was that the light, and anything else not subject to a (non-gravitational) force, follows a geodesic, regardless of its composition.

"in relativity there are no separate laws of conservation of energy and conservation of mass. There is only a law of conservation of mass plus energy (referred to as mass-energy)" (same source).

Taylor and Wheeler say explicity that energy is conserved, and you've confirmed that (rest) mass is conserved. In Question 16, Benjamin Crowell gives the definition $E = m \gamma$, calling it "total mass-energy". So I'm thinking perhaps in this quote, and here:

"The total gravitational attraction between two objects is proportional not just to the product of their masses, m1m2, as in Newton's law of gravity, but to the quantity (m1+E1)(m2+E2). (Even this modification does not give a complete, self-consistent theory of gravity, which is only accomplished through the general theory of relativity.)"

"Energy" and "E" have been used for what Taylor & Wheeler call kinetic energy (i.e. the time component of the energy-momentum 4-vector minus rest mass), whereas elsewhere in this chapter "E" and "energy" or "mass-energy" are used for what Taylor & Wheeler call simply energy, i.e. the time component of the energy-momentum 4-vector.

Last edited by a moderator: May 4, 2017
6. Feb 26, 2010

### atyy

Mass is not a fundamental thing in general relativity. However, from special relativity, there is the equivalence of energy and relativistic mass, from which there is a heuristic that if mass produces gravity, then energy should too, hence energy is gravitational mass, and the stress-energy tensor is the source of spacetime curvature. In a number of cases in GR, the heuristic that energy=mass works out technically for photons in a box.

http://arxiv.org/abs/gr-qc/9909014
http://arxiv.org/abs/gr-qc/0510041

7. Feb 27, 2010

### Staff: Mentor

Yes, but you have to be careful with the heuristic. The thing that introductory GR students often miss, I believe, is that momentum, pressure, and stress also produce gravity. So the photons in a box is a good place to apply the heuristic because the system's momentum is the same (although there is pressure and stress to consider), but a highly relativistic baseball is not a good place to apply the heuristic because the system's momentum is dramatically different. This is where people get into trouble with erroneous ideas like things turning into black holes in one frame but not in another.

Last edited: Feb 27, 2010
8. Feb 27, 2010

### atyy

Yes, definitely.

9. Feb 28, 2010

### dimsun

The equation $$m_0c^2 = \sqrt{E_r^2 - p_x^2 - p_y^2 - p_z^2}$$ is a bit misleading.

It is more appropriate to write it as:
$$E_0 = \sqrt{E_r^2 - p_x^2 - p_y^2 - p_z^2}$$

This is the same as
$$E_r = \gamma E_0$$

In case the vector quantities are zero we have the following two equivalences:
$$E_r = m_r c^2$$

This is the same as:
$$E_0 = m_0 c^2$$

Mass is a different physical manifestation then energy is. For example the fission of U-235 is an exchange of mass for energy.

Dimsun

10. Mar 1, 2010

### Rasalhague

Thanks, all.

Dimsun, are you just saying that you prefer to think of the magnitude of the energy-momentum 4-vector as $E_0$, rest energy, rather than rest mass? In Taylor & Wheeler's terminology, energy is the time component of the energy-momentum 4-vector, and mass its magnitude, with the appropriate scaling factor.

What units are you using? Taylor and Wheeler give the equation in "natural units", where c=1, with no explicit factors of c, and in "conventional" (e.g. S.I. units) as

$$mc^2=\sqrt{E^2-p^2c^2},$$

where p is the magnitude of 3-momentum. This is what I'd expect, with energy having units $J=kg\cdot m^2\cdot s^{-2}$, and momentum $kg \cdot m \cdot s^{-1}$. You seem to have energy and momentum in the same units, but mass in different units.

I take it by "vector quantities", you mean when the 3-momentum is zero, i.e. in a frame in which the system has no net 3-momentum, the vector sum of the 3-momenta of its constituents being zero.

11. Mar 2, 2010

### dimsun

Rasalhaque, I was not using natural units, I just forgot adding c to momentum. You are right, equations must be dimensional correct.

With vector quantities I mean both the 3-momentum is zero and vector quantity $$\frac{m_{rel}v^2}{c^2}$$ is zero.
Restmass:
$$m_0 = \sqrt{\frac{E_{rel}^2}{c^4} - \frac{p^2_x}{c^2} - \frac{p^2_y}{c^2} - \frac{p^2_z}{c^2}}$$

Relativistic mass:
$$m_{rel} = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

$$\sqrt{m_{rel}^2 - \frac{m_{rel}v^2_x}{c^2} -\frac{m_{rel}v^2_y}{c^2} -\frac{m_{rel}v^2_z}{c^2}} = \sqrt{\frac{E_{rel}^2}{c^4} - \frac{p^2_x}{c^2} - \frac{p^2_y}{c^2} - \frac{p^2_z}{c^2}}$$

A lot of people disagree with me, saying that $$\frac{m_{rel}v^2}{c^2}$$ is the same as momentum, but I see this quantity as a different quantity then momentum, just like mass and energy are different physical manifestations.
It is mathematically correct.

Dimsun.

12. Mar 2, 2010

### Rasalhague

I can appreciate the difference between (rest) mass and energy, since they're only equal in a special case, but I don't quite understand the distinction you're making. Is it that the 3-momentum, which we could see as a 3-vector, or as a frame-dependent 4-vector having no time component, is not conceptually the same thing as its magnitude, even though its magnitude is zero if and only if the 3-momentum vector is zero?

13. Mar 3, 2010

### dimsun

Yes, I do not understand the concept of rest-mass and rest-energy. rest-energy is used in the following expression:

$$E_0 = \sqrt{E_{rel}^2 - p_x^2c^2 - p_y^2c^2 - p_z^2c^2}$$
How could this energy be rest energy when the 3-momentum is not zero?

And how could rest mass be incorporated into the equation when rest mass and rest energy are the same only when the 3-momentum is zero?
$$m_0c^2 = \sqrt{E_{rel}^2 - p_x^2c^2 - p_y^2c^2 - p_z^2c^2}$$

In the two equations above, I think it is not logical to use the concepts of rest energy and rest mass. Can someone explain it to me?

In the following two equation relativistic energy and relativistic mass are used:

$$E_0 = \sqrt{E_{rel}^2 - \frac{E_{rel}^2v_x^2}{c^2} - \frac{E_{rel}^2v_y^2}{c^2} - \frac{E_{rel}^2v_z^2}{c^2}$$

$$m_0 = \sqrt{m_{rel}^2 - \frac{m_{rel}^2v_x^2}{c^2} - \frac{m_{rel}^2v_y^2}{c^2} - \frac{m_{rel}^2v_z^2}{c^2}$$

In the two equations above relativistic mass is a different phenomena then relativistic energy, then also the 3-momentum in the first equation must be a different phenomena then the vector quantity in the last equation. I don't think it is appropriate to use the concept of rest-energy and rest-mass in the two equations above.

It is better to write it in the following manner:
$$\sqrt{E_{rel}^2 - \frac{E_{rel}^2v_x^2}{c^2} - \frac{E_{rel}^2v_y^2}{c^2} - \frac{E_{rel}^2v_z^2}{c^2}} = c^2\sqrt{m_{rel}^2 - \frac{m_{rel}^2v_x^2}{c^2} - \frac{m_{rel}^2v_y^2}{c^2} - \frac{m_{rel}^2v_z^2}{c^2}}$$

Only when the two vector quantities are zero:
$$\sqrt{E_{rel}^2} = c^2\sqrt{m_{rel}^2}$$
And that is the only appropriate situation in which we can use the concepts of rest-mass and rest-energy:
$$\sqrt{E_0^2} = c^2\sqrt{m_0^2}$$

14. Mar 3, 2010

### Rasalhague

Okay, I think I get it. The magnitude of this "(3-)vector quantity" you define, call it $q$,

$$q=\sqrt{E_{rel}^2v_x+E_{rel}^2v_y+E_{rel}^2v_z},$$

is related to the magnitude, $p$, of 3-momentum

$$p=\sqrt{m_{rel}^2v_x+m_{rel}^2v_y+m_{rel}^2v_z},$$

like this:

$$q=\sqrt{p^2+E_{rel}^2c^2-m_{rel}^2c^6}.$$

I just rearranged your definitions to solve for q there. And you say

$$E_{rel}=m_{rel}c^2.$$

So

$$q=\sqrt{p^2+c^2\left(E_{rel}^2-E_{rel}^2\right)},$$

which is to say,

$$q=p.$$

So in all situations, they're the same.

15. Mar 3, 2010

### dimsun

Agreed.

That is only true when the two vectors are zero.
In the situation you showed $$q = p = 0$$

Dimsun

16. Mar 3, 2010

### Rasalhague

Oh, right. I see. Sorry, I overlooked that condition in your definitions. On the other hand, it can be established that wherever relativistic mass is defined (that is, wherever the system has mass), it will have the same value as the energy of the system (see, for example, http://www.karlscalculus.org/pdf/einstein.pdf [Broken] derivation by Karl Hahn, which Altabeh pointed me towards in a recent thread), with the relavant scaling factor to balance the units. So the only significant difference between relativistic mass and energy would be that energy is a more general concept, also applying to a massless system. If I've understood your definitions right, and if we accept Karl Hahn's reasoning, q=p whenever p is defined (that is, for a system with mass). But it's common to talk about the momentum (i.e. 3-momentum) of a photon, which has no mass, this 3-momentum being equal to the energy of the photon. So, I could be mistaken, but the distinction you're making (between the momentum of a massive system and the "some other kind of (3-)vector quantity" that plays the equivalent role for a massless system) isn't a standard one, as far as I know.

Here's what Taylor and Wheeler do in Spacetime Physics. They define an energy-momentum 4-vector. That's a vector with four components, three space components (momentum), and one time component (energy). In natural units, where c=1,

$$\left ( E,p_x,p_y,p_z \right ).$$

Mass (i.e. rest mass), is the magnitude of this 4-vector,

$$m = \sqrt{E^2 - p_x - p_y - p_z}.$$

When the magnitude is zero (i.e. for a lightlike energy-momentum 4-vector), energy equals 3-momentum:

$$E=p=\sqrt{p_x + p_y + p_z}.$$

When the particle or system has mass, the components of the energy-momentum 4-vector can also be written:

$$\left(\gamma m,\gamma \beta_x m, \gamma \beta_y m, \gamma \beta_z m \right)$$

where

$$\gamma = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$$

and $\beta$ is $\frac{v}{c}$.

Equivalently:

$$\left ( m\frac{\mathrm{d} t}{\mathrm{d} \tau},m\frac{\mathrm{d} x}{\mathrm{d} \tau},m\frac{\mathrm{d} y}{\mathrm{d} \tau},m\frac{\mathrm{d} z}{\mathrm{d} \tau} \right ),$$

where $\tau$ is proper time. In the rest frame--I've also seen the term centre of momentum frame--of a (massive) particle or system, and only in such a reference frame,

$$m=E.$$

They deprecate the term "relativistic mass", and see it as a source of confusion, prefering to call the time component (of the energy-momentum 4-vector) energy.

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