Exploring the Impact of M = E/c^2 in Modern Physics

[TrillAeon]

TL;DR Summary

E = mc^2
means
m = E/c^2

<illegible image removed, please post equations using LaTeX and type text>

 

[Dale]

For a system with no momentum, yes. The equation does not apply for a photon which always has momentum.

 

[Nugatory]

TL;DR Summary: E = mc^2
means
m = E/c^2

Yes, but do remember that $E=mc^2$ is the $p=0$ special case of the more general relationship $E^2=(m_0c^2)^2+(pc)^2$ where $p$ is the momentum and $m_0$ is the rest mass. It doesn’t apply when $p$ is non-zero.

 

[sha1000]

Yes, but do remember that $E=mc^2$ is the $p=0$ special case of the more general relationship $E^2=(m_0c^2)^2+(pc)^2$ where $p$ is the momentum and $m_0$ is the rest mass. It doesn’t apply when $p$ is non-zero.

Why people always underline that the general equation is different from the classical E=mc2?
These are the exact some equations, we just simply put terms in different way.
Its not like E=mc2 is a special (or simplified) case of the general equation.

E=mc2 is one of the most elegant equations. Shouldn't we take it as it is and conclude that if something has energy then the rest mass must be non-zero? Why we try to play with it just to move ourselves away from the obvious conclusion?

 

[weirdoguy]

Its not like E=mc2 is a special (or simplified) case of the general equation

Um, yes, it is. Just as the post you quoted says.

 

[jbriggs444]

Why people always underline that the general equation is different from the classical E=mc2?
These are the exact some equations, we just simply put terms in different way.

In the two equations, the $m$ term means two different things.

If you take $E=mc^2$ as a general statement that applies whether or not the object is at rest then the $m$ must be "relativistic mass". Relativistic mass is a quantity that varies with velocity relative to your chosen reference frame.

By contrast in the general equation, $E^2 = (m_0c^2)^2 + (pc)^2$, $m_0$ is the invariant mass. Invariant mass is a quantity that does not change with velocity. More precisely, it is a quantity that is the same, no matter what reference frame you select.

 

[Dale]

Why people always underline that the general equation is different from the classical E=mc2?
These are the exact some equations, we just simply put terms in different way.
Its not like E=mc2 is a special (or simplified) case of the general equation.

It absolutely is exactly a special case of the general equation. $E=mc^2$ is the special case of $m^2 c^2 = E^2/c^2-p^2$ where $p=0$.

The reason we always "underline" that is that people always misapply the usual $E=mc^2$ in a way that is resolved by using the general equation.

Shouldn't we take it as it is and conclude that if something has energy then the rest mass must be non-zero? Why we try to play with it just to move ourselves away from the obvious conclusion?

Because the universe doesn't behave according to the "obvious conclusion". It behaves according to the general equation in general.

E=mc2 is one of the most elegant equations.

It is even more elegant if you spend a couple of seconds to write it with LaTeX.

 

[sha1000]

In the two equations, the $m$ term means two different things.

If you take $E=mc^2$ as a general statement that applies whether or not the object is at rest then the $m$ must be "relativistic mass". Relativistic mass is a quantity that varies with velocity relative to your chosen reference frame.

By contrast in the general equation, $E^2 = (m_0c^2)^2 + (pc)^2$, $m_0$ is the invariant mass. Invariant mass is a quantity that does not change with velocity. More precisely, it is a quantity that is the same, no matter what reference frame you select.

Let's imagine that we are in the begging of the theory of relativity. After some derivations the first equation we have obtained is :
E^2= (m_0c^2)^2 + (pc)^2

And now we are wondering if the rest mass of the photon is zero or non-zero. Wouldn't it be more reasonable to play with this equation and get E = mc^2 (with m= m0/sqrt(1 -v/c^2) relativistic mass), and conclude that the rest-mass of the photon must be non-zero to satisfy this equation?

And now we are doing the exact opposite.

 

[Ibix]

Shouldn't we take it as it is and conclude that if something has energy then the rest mass must be non-zero?

No, because that isn't true. You could conclude that the relativistic mass is non-zero. This is trivially true since relativistic mass is just energy divided by $c^2$. But rest mass may be zero in things with positive energy - notably, light.

 

[Dale]

E^2= (m_0c^2)^2 + (pc)^2
...
E = mc^2

Seriously, use LaTeX.

 

[Ibix]

Wouldn't it be more reasonable to play with this equation and get E = mc^2 (with m= relativistic mass), and conclude that the rest-mass of the photon must be non-zero to satisfy this equation?

No, because $E=pc$ for something traveling at light speed, so your conclusion would obviously contradict your starting point.

 

[Ibix]

Seriously, use LaTeX.

Indeed - @sha1000, you literally only need to put a pair of # characters before each equation and after each one. You're already using the syntax.

 

[sha1000]

Indeed - @sha1000, you literally only need to put a pair of # characters before each equation and after each one. You're already using the syntax.

Sorry for that.

But do we really need to assume that photons travel with the "speed of light" to derive our equations? If I'm not mistaken we can just assume that there is some limit speed in the universe (not necessarily the speed of light).

 

[Ibix]

If I'm not mistaken we can assume that there is some limit speed in the universe (not necessarily the speed of light).

You can assume that. Relativity still says that $E=mc^2$ is a special case of $E^2=m^2c^4+p^2c^2$ and things traveling at $c$ are massless - it's just that light is no longer such a thing, by assumption.

The experimental upper bound on photon mass is about $10^{-50}\mathrm{kg}$ if memory serves.

 

[Nugatory]

Why people always underline that the general equation is different from the classical E=mc2?

The general equation is every bit as classical as its $E=mc^2$ special case, is it not? Both are relativistic, neither is quantum mechanical.

And now we are wondering if the rest mass of the photon is zero or non-zero. Wouldn't it be more reasonable to play with this equation and get E = mc^2 (with m= m0/sqrt(1 -v/c^2) relativistic mass), and conclude that the rest-mass of the photon must be non-zero to satisfy this equation?

When I try that I don’t conclude that the photon has non-zero rest mass, I conclude that I’m dividing by zero and therefore have messed something up and cannot take any results after that seriously.

But do we really need to assume that photons travel with "speed of light" to derive the our equations? If I'm not mistaken we can assume that there is some limit speed in the universe (not necessarily the speed of light).

You are right that relativity implies an invariant speed, not necessarily the speed of light. However you are mistaken when you suggest that we are “assuming” that photons travel at the invariant speed. It’s the Lorentz invariance of Maxwell’s equations that tells us that that light (not photons! We don’t know about them yet) travels at the invariant speed. Photons moving at the speed of light (to the extent that photon speed is even a meaningful concept) appear after we develop quantum mechanics and then a relativistic quantum mechanical treatment of electromagnetism.

Long before then we are using $E^2=(m_0c^2)^2+(pc)^2$ instead of $E=m_0/\sqrt{1-v^2/c^2}$ because in most physically reasonable relativistic problems we can measure $p$ but not $v$. That’s a much stronger motivation than the need to explain to confused laypeople that they’ve misapplied $E=mc^2$.

 

[PeterDonis]

Shouldn't we take it as it is and conclude that if something has energy then the rest mass must be non-zero?

Obviously not since photons have energy but have zero rest mass. More precisely, the upper bound on their rest mass from various observational tests is so tiny that, in the context of this thread, and indeed for all practical purposes, it can be taken as zero. It certainly is much, much smaller than the photon's energy divided by $c^2$.

 

[sha1000]

The general equation is every bit as classical as its $E=mc^2$ special case, is it not? Both are relativistic, neither is quantum mechanical.

When I try that I don’t conclude that the photon has non-zero rest mass, I conclude that I’m dividing by zero and therefore have messed something up and cannot take any results after that seriously.

As I mentioned before we don't need to assume that photons travel at the speed of "light" to derive $E=mc^2$. For that we only need to assume that there is some critical speed in the universe (c = critical speed).

In this case you don't have to divide your expression by zero. Since the speed of the photons don't has to be equal to c. Thus, non-zero rest-mass.

 

[PeterDonis]

we don't need to assume that photons travel at the speed of "light" to derive $E=mc^2$. For that we only need to assume that there is some critical speed in the universe (c = critical speed).

Yes, but then, as @Nugatory pointed out, we know from the Lorentz invariance of Maxwell's Equations that light travels at whatever the critical speed is. So while we don't assume that photons travel at that speed, we do deduce that they do, so we end up at the same place.

the speed of the photons don't has to be equal to c.

Yes, they do. See above.

 

[sha1000]

Long before then we are using $E^2=(m_0c^2)+(pc)^2$ instead of $E=m_0/\sqrt{1-v^2/c^2}$ because in most physically reasonable relativistic problems we can measure $p$ but not $v$. That’s a much stronger motivation than the need to explain to confused laypeople that they’ve misapplied $E=mc^2$.

I'm really not against the idea of using $E^2=(m_0c^2)+(pc)^2$ .
I just can't understand why this expression is considered as general one? It is exactly the same as $E=m_0/\sqrt{1-v^2/c^2}$ . We just played with the terms in order to get $p$ there, for practical purposes.

My question is simple. Imagine that we just have derived: $E^2=(m_0c^2)+(pc)^2$. For that we made an assumption about the existence of critical speed $c$ in the universe (nothing about the speed of the photons).
And now we are wondering if the mass of the photons is zero or non-zero. The logical process would be to play with this expression and get the $E=m_0/\sqrt{1-v^2/c^2}$ form. From here we can conclude that any particle with some energy must have non-zero rest-mass and the velocity $v << c$ (critical speed).

 

[PeterDonis]

I'm really not against the idea of using $E^2=(m_0c^2)+(pc)^2$ .
I just can't understand why this expression is considered as general one? It is exactly the same as $E=m_0/\sqrt{1-v^2/c^2}$ .

No, it isn't, because the latter expression is only valid for objects with $m_0 > 0$. Whereas the former expression is valid for all objects.

We just played with the terms in order to get $p$ there, for practical purposes.

No, we didn't. The expression with $p$ there comes first. Then, for the case $m_0 > 0$, you derive the expression for $E$ with the square root in it. You can't do it the other way around.

My question is simple. Imagine that we just have derived $E=m_0/\sqrt{1-v^2/c^2}$.

You've already gone off the rails here, because we can't derive this expression without first deriving the more general expression involving $p$. See above.

For that we made an assumption about the existence of critical speed $c$ in the universe (nothing about the speed of the photons).

For deriving the general equation with $p$ in it, yes. Then, to get the equation with the square root in it, we added the assumption $m_0 > 0$.

And now we are wondering if the mass of the photons is zero or non-zero. The logical process would be to play with this expression and get the $E=m_0/\sqrt{1-v^2/c^2}$ form.

No, the logical process is to first look at the equations that tell us the dynamics of light, namely, Maxwell's Equations. And, as has already been stated several times, we find that those equations tell us that light moves at whatever the invariant speed is.

From here we can conclude that any particle with some energy must have non-zero rest-mass

Wrong. You have it backwards: we have to assume $m_0 > 0$ in order to derive the equation with the square root in it. See above.

 

[sha1000]

Yes, but then, as @Nugatory pointed out, we know from the Lorentz invariance of Maxwell's Equations that light travels at whatever the critical speed is. So while we don't assume that photons travel at that speed, we do deduce that they do, so we end up at the same place.Yes, they do. See above.

Does Lorentz invariance of Maxwell equations tell us that speed of photons must be exactly Equal to $c$ (critical speed)? Or it can be "very" close to it in order to get things to work?

Once we have the assumption about the invariant critical speed. All other velocities become invariant right? The speed of the bullets that I fire will always be the same in my reference frame etc.

 

[PeterDonis]

Does Lorentz invariance of Maxwell equations tell us that speed of photons must be exactly Equal to $c$ (critical speed)?

Yes.

Or it can be "very" close to it in order to get things to work?

No.

Once we have the assumption about the invariant critical speed. All other velocities become invariant right?

No.

The speed of the bullets that I fire will always be the same in my reference frame etc.

Assuming you always use the same gun with the same properties, yes. But the bolded qualifier that I have put in the quote above means the speed of the bullets is not invariant--it will be different in different frames. Whereas the critical speed is invariant--it's the same in all frames.

 

[sha1000]

No, it isn't, because the latter expression is only valid for objects with $m_0 > 0$. Whereas the former expression is valid for all objects.No, we didn't. The expression with $p$ there comes first. Then, for the case $m_0 > 0$, you derive the expression for $E$ with the square root in it. You can't do it the other way around.You've already gone off the rails here, because we can't derive this expression without first deriving the more general expression involving $p$. See above.For deriving the general equation with $p$ in it, yes. Then, to get the equation with the square root in it, we added the assumption $m_0 > 0$.No, the logical process is to first look at the equations that tell us the dynamics of light, namely, Maxwell's Equations. And, as has already been stated several times, we find that those equations tell us that light moves at whatever the invariant speed is.Wrong. You have it backwards: we have to assume $m_0 > 0$ in order to derive the equation with the square root in it. See above.

Sorry. My mistake.

I intended to say that first we derive $E^2=(m_0c^2)+(pc)^2$

From this point we just play with some terms and we can get the $E=m_0/\sqrt{1-v^2/c^2}$. We don't need to make any assumptions about $m_0$ to do that, right? So the non-zero $m_0$ is not an assumption but the consequence of this equation.

Once again here we only talk about the relativity. In order to derive this expression we don't need Maxwell or Lorentz. Just basic assumption about the critical speed and some trigonometry.

 

[PeterDonis]

I intended to say that first we derive $E^2=(m_0c^2)+(pc)^2$

Ok.

From this point we just play with some terms and we can get the $E=m_0/\sqrt{1-v^2/c^2}$. We don't need to make any assumptions about $m_0$ to do that, right?

Wrong. We need to assume $m_0 > 0$. Otherwise you get a mathematically invalid equation during your attempted derivation.

 

[sha1000]

Yes.No.No.Assuming you always use the same gun with the same properties, yes. But the bolded qualifier that I have put in the quote above means the speed of the bullets is not invariant--it will be different in different frames. Whereas the critical speed is invariant--it's the same in all frames.

Ok I see. Thank you for your explanations.

 

[Nugatory]

Does Lorentz invariance of Maxwell equations tell us that speed of photons must be exactly Equal to c (critical speed)?

Yes. The experiments which have established that astoundingly small upper bound on the possible mass of the photon are looking for violations of Lorentz invariance, and finding none to within the limits of experimental accuracy.

 

[Mister T]

E=mc2 is one of the most elegant equations.

But the correct equation is $E_o=mc^2$.

Shouldn't we take it as it is and conclude that if something has energy then the rest mass must be non-zero?

A photon is an obvious counterexample. It has energy but zero mass.

 

[Dale]

We don't need to make any assumptions about m0 to do that, right? So the non-zero m0 is not an assumption but the consequence of this equation.

Well, clearly that equation assumes that $v\ne c$. So that is an assumption even if it is not an assumption about $m$. The consequence of the equation is that no massive particle can travel at $v=c$. But since we did assume $v\ne c$ in the derivation we cannot use this equation to rule out massless particles at $v =c$.

 

[vanhees71]

In the two equations, the $m$ term means two different things.

If you take $E=mc^2$ as a general statement that applies whether or not the object is at rest then the $m$ must be "relativistic mass". Relativistic mass is a quantity that varies with velocity relative to your chosen reference frame.

By contrast in the general equation, $E^2 = (m_0c^2)^2 + (pc)^2$, $m_0$ is the invariant mass. Invariant mass is a quantity that does not change with velocity. More precisely, it is a quantity that is the same, no matter what reference frame you select.

One should also once more state that the use of "relativistic mass" is out of fashion since 1907. It was introduced by Einstein at a time (1905), before the correct relativistic version of point-particle mechanics has been understood (Planck 1906). Einstein also clearly stated that it's preferrable not to use the notion of any relativistic mass to avoid misunderstandings and misconceptions. For the history, see

https://physicstoday.scitation.org/doi/abs/10.1063/1.881171?journalCode=pto

 

[jbriggs444]

One should also once more state that the use of "relativistic mass" is out of fashion since 1907.

Indeed. Which is why I chose to link https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/ when I used the phrase.

 

[Dale]

"relativistic mass" is out of fashion since 1907.

Oh, that is a very specific date. What happened in 1907?

 

[PeroK]

Oh, that is a very specific date. What happened in 1907?

https://en.wikipedia.org/wiki/1907

Nothing about relativistic mass, though.

 

[vanhees71]

Oh, that is a very specific date. What happened in 1907?

Einstein learnt about the consistent description of classical point-particle mechanics within SRT from Planck.

 

[martinbn]

Oh, that is a very specific date. What happened in 1907?

It may be unralated, but at that time he was asked to write a review article about relativity. Then he decided that gravity needs to be treated relativistically and it is a problem that can be worked on. May be he also realized that relativistic mass is more confusing than usefull.

 

[Sagittarius A-Star]

One should also once more state that the use of "relativistic mass" is out of fashion since 1907.

Einstein learnt about the consistent description of classical point-particle mechanics within SRT from Planck.

The statement 'the use of "relativistic mass" is out of fashion since 1907' is true, if you focus on Einstein. However, other physicists used "relativistic mass" after 1907:

Planck (1906a) defined the relativistic momentum and gave the correct values for the longitudinal and transverse mass by correcting a slight mistake of the expression given by Einstein in 1905. Planck's expressions were in principle equivalent to those used by Lorentz in 1899.[80] Based on the work of Planck, the concept of relativistic mass was developed by Gilbert Newton Lewis and Richard C. Tolman (1908, 1909) by defining mass as the ratio of momentum to velocity. So the older definition of longitudinal and transverse mass, in which mass was defined as the ratio of force to acceleration, became superfluous. Finally, Tolman (1912) interpreted relativistic mass simply as the mass of the body.[81] However, many modern textbooks on relativity do not use the concept of relativistic mass anymore, and mass in special relativity is considered as an invariant quantity.

Source:
https://en.wikipedia.org/wiki/History_of_special_relativity#Relativistic_momentum_and_mass