Why did Einstein considered it “evident”?

  • #1
At the end of the (fourth), 1905’s E=mc² paper Einstein claimed: “The fact that the energy withdrawn from the body becomes energy of radiation evidently makes no difference”. This sounds odd to me, given that the change L(γ-1) in kinetic energy resulted from calculating radiation energy. Light emitted in the two opposite directions has the same speed for any reference system, and that is a special property which holds for light only. If you think of the symmetric heat release without a change in the rest state of the body in the rest reference system, then no change in velocity should be observed from the moving inertial system too. But the same would not be true respect to the particles in thermal motion. A different Doppler effect (corresponding to different temperatures and thermal propagation rate) should be calculated from the moving system for the total amount of thermal energy emission from the two sides (as resulted for the light). But I would expect a kinetic energy change different from L(γ-1). Would somebody like to help me understand the basis of Einstein’s implicit deduction?
 
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  • #2
phinds
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I'm not answering your question, just alerting you to an incorrect statement, to wit:
Light emitted in the two opposite directions has the same velocity for any reference system, and that is a special property which holds for light only.
should say "the same SPEED". Obviously things moving in different directions cannot have the same velocity (which is a vector), just the same speed (which is a scalar).
 
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  • #3
Ibix
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I think Einstein is merely pointing out that (edit: ...it turns out that...) he hasn't actually assumed that the things carrying away energy are packets of radiation (edit: unlike On the Electrodynamics of Moving Bodies, in which he does assume that it's radiation). He's simply said that energy is carried away somehow. So whether it's radiation or something else makes no difference.

Edit 2: link to the paper under discussion http://www.fourmilab.ch/etexts/einstein/E_mc2/www/
 
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  • #4
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Here is some context and a slightly different translation (closer to the original):

"The kinetic energy of the body with respect to ##(\xi, \eta, \zeta)## [my edit: which is the moving coordinate system] decreases as a result of the emission of light by an amount independent of the qualities of the body. The difference ##K_0 - K_1## depends on the velocity as well as the kinetic energy of the electron [my edit: ... depends on the velocity].

Neglecting fourth and higher order terms, we can set:
$$
K_0-K_1 = \dfrac{L}{V^2} \dfrac{v^2}{2}
$$
From this equation follows immediately:

If a body releases the energy ##L## in the form of radiation, its mass decreases by ##L / V^2##. Obviously, it is not essential that the energy withdrawn from the body transitions into the energy of radiation, leading us to the more general conclusion:

The mass of a body is a measure of its energy content; If the energy changes by ##L##, the mass changes by ##L / 9\cdot 10^{20}##, when the energy is measured in erg and the mass in grams.

It is not excluded that in bodies whose energy content is highly variable (for example in the case of the radium salts), an examination of the theory will succeed.

When the theory is factual, the radiation transmits inertia between the emitting and absorbing bodies."
 
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  • #5
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At the end of the (fourth) 1905’s E=mc² paper Einstein claimed: “The fact that the energy withdrawn from the body becomes energy of radiation evidently makes no difference”. This sounds odd to me, given that the change L(γ-1) in kinetic energy resulted from calculating radiation energy. Light emitted in the two opposite directions has the same velocity for any reference system, and that is a special property which holds for light only. If you think to a symmetric heat release without a change in the rest state of the body in the rest reference system, than no change in velocity should be observed from the moving system. But the same would not be true respect to the particles in thermal motion. A different Doppler effect (corresponding to different temperatures and thermal propagation rate should be calculated from the moving system for the two sides. I would expect a kinetic energy change different from L(γ-1). Would somebody like to help me understand the basis of Einstein’s implicit deduction?

It would be very odd if a block of copper with initial mass of 1kg that has cooled from 500K to 400K by radiative cooling was in any way distinguishable from a block of copper with initial mass of 1 kg that has cooled from 500K to 400K by conductive or convective cooling.

Addition:

If we choose temperatures on both sides of the melting point of copper, then we can first show that a block of copper loses mass when it solidifies by radiating, which means loss of potential energy, then we can say that evidently some other way of losing the same potential energy would result in the same mass loss.
 
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  • #6
I'm not answering your question, just alerting you to an incorrect statement, to wit:
should say "the same SPEED". Obviously things moving in different directions cannot have the same velocity (which is a vector), just the same speed (which is a scalar).
Thank you. I did correct the mistake and re-phrase some parts hoping it is clearer now.
 
  • #7
Here is some context and a slightly different translation (closer to the original):
If a body releases the energy ##L## in the form of radiation, its mass decreases by ##L / V^2##. Obviously, it is not essential that the energy withdrawn from the body transitions into the energy of radiation, leading us to the more general conclusion:

The mass of a body is a measure of its energy content; If the energy changes by ##L##, the mass changes by ##L / 9\cdot 10^{20}##, when the energy is measured in erg and the mass in grams..."
Thank you very much Fresh_42 for your in-depth exam. Along with this translation, we could reason that Einstein's calculation would hold for any kind of transition from any kind of emitted energy (included thermal) even though it is not eventually converted in radiant energy.
So, E = mc² is demonstrated in the case that the energy change is thermal (kinetic energy transmitted to the molecules of an external medium) and remains thermal-kinetic.
It is that correct?
Anyway, in "Einstein's Miraculous Year", Five papers that changed the face of Physics, the editor (for Princeton University Press) John Stachel translates, at page 164: "Here it is obviously inessential that the energy taken from the body turns into radiant energy, so we are led to the more general conclusion..."
This lead us to the same conclusion. Apparently, Einstein thought the amount of energy taken was all that matter, although its calculation was based on radiant energy and equal light speed from both directions and both reference systems. Energy should be conserved in each intertial reference system, so the kinetic energy change must be the same even "before" heat get transformed into radiant energy.
I also attempted a translation with Google translator from the original German: "Hierbei ist es offenbar unwesentlich, das die der Körper entzogene Energie gerade in Energie der Strahlung ubergeht" that corresponds to your translation: "Obviously, it is immaterial that the energy taken from the body just transits into the energy of the radiation".
In the sense that "if it could transit, it would have been converted in the same amount of radiant energy, but the loss in inertial mass had already occurred, in the same amount L/c²."
Am I right?
 
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  • #8
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As I understood the paper, he dealt with light emission in a previous paper and uses this result on measured energy. He considers different measurements of energy in a frame at rest and in motion.
"The kinetic energy of the body decreases due to light emission by an amount independent of the properties of the body."
so I don't see any problems with the direction of emission. His calculations are all scalar.

I also used Google translate, but with some adjustments: e.g. essential instead of immaterial, but this way I kept it close to the original, which the sometimes strange constructions show.
 
  • #9
Mister T
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So, E = mc² is demonstrated in the case that the energy change is thermal (kinetic energy transmitted to the molecules of an external medium) and remains thermal-kinetic.
Think of it this way. The mass of the object changes by ##\Delta m## in a way that's path-independent. It doesn't matter how the change occurs. If he shows that in one case that it's equal to ##E\c^2## then he's shown that ##\Delta m=E\c^2##. Thus any change in mass ##\Delta m## must be equivalent to ##E\c^2##.
 
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  • #10
A.T.
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I also attempted a translation with Google translator from the original German: "Hierbei ist es offenbar unwesentlich, das die der Körper entzogene Energie gerade in Energie der Strahlung ubergeht" that corresponds to your translation: "Obviously, it is immaterial that the energy taken from the body just transits into the energy of the radiation".
"Offenbar" is closer to "it turns out that", rather than "obviously" (which would be "offensichtlich").
 
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  • #11
Think of it this way. The mass of the object changes by ##\Delta m## in a way that's path-independent. It doesn't matter how the change occurs. If he shows that in one case that it's equal to ##E\c^2## then he's shown that ##\Delta m=E\c^2##. Thus any change in mass ##\Delta m## must be equivalent to ##E\c^2##.
This is true, of course, because further reflections would convict us that mass and internal energy should be invariant (substantially the same invariant).
The reader of Einstein's paper only understands what he wrote: that the observer in motion calculates a decrease in kinetic energy not due to a velocity decrease, thus only imputable to an inertial mass loss. The reader imagines that the evaluation of the rest observer is "no kinetic energy change, then no mass change either: it is only from the point of view of the moving observer that mass changes, like other stuff, as time intervals and distances. Moreover, that observer is a foul pretending to evaluate the mass of an object while moving".

There are a lot of unsaid-unwritten claims behind the (correct) generalisation by Einstein. First: both observers "E" and "H" agree on the kinetic energy of a body as related to the relative motion between the body and observer "H". This is necessary for both the observers to agree on (second inference) the same value for internal energy (never mentioned by Einstein) of the body. That is internal energy is invariant. Therefore, when any energy amount is emitted, both observers continue to agree on the same ΔEinternal = -L; so they are forced to ascribe the variation in kinetic energy L(γ-1) to a change in an invariant (but not conserved) mass.
In this context, let me evaluate the principle of energy conservation from the POV of the moving observer. For the moving observer "H" the initial total energy was Eint+K whereas, after the light emission is [E'int + L] + [K - L(γ-1)], where E'int is the new (agreed and invariant) internal energy E'int = Eint-L and [K - L(γ-1)] is K', the new kinetic energy. The difference is nonzero, ΔH = -L(γ-1) ~ -L/c², that is the moving observer does not calculate a conserved total energy, as long as he considers kinetic energy as a component of the total energy of the body. Only mass/c² plus internal energy is conserved from H's POV, whereas the rest observer "E" calculates a conserved total energy ΔE = E'int + L - Eint = 0.
What do you think?
 
  • #12
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"Offenbar" is closer to "it turns out that", rather than "obviously" (which would be "offensichtlich").
I do not agree in this case. "It turns out that" or "offenbar" leaves open the question: "How does it turn out?". As he didn't answer this, nor even mentioned it, obviously fits better in this case. There is generally rarely a one by one translation possible as natural languages are highly context sensitive.

In any case, the distinction between "offenbar" (=literally "quasi open") and "offensichtlich" (=literally "open to see") is nitpicking.

And to put evidence to my argument, here is what the Duden has to say:

upload_2018-9-12_13-14-24.png
 

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  • #13
A.T.
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In any case, the distinction between "offenbar" (=literally "quasi open") and "offensichtlich" (=literally "open to see") is nitpicking.
"Ist offensichtlich" usually implies that it is trivially obvious, and doesn't require further explanation.

"Offenbar" doesn't imply that it is trivialy obvious, but rather that it was revealed to be the case (Offenbarung = revelation).
 
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  • #14
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"Ist offensichtlich" usually implies that it is trivially obvious, and doesn't require further explanation.

"Offenbar" doesn't imply that it is triviality obvious, but rather that it was revealed to be the case (Offenbarung = revelation).
The Duden lists them as synonym, that's all I need. At least more than this ridiculous discussion here. Any bet that Einstein hasn't thought about such nuances. Obviously is correct. Full stop.
 
  • #16
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The thing about that paper is that Einstein introduces the invariant rest energy without ever saying so explicitly. It's so subtle that you wonder if he even realized how revolutionary it was!

The meat of the paper is "between the lines," IMO. He focuses on kinetic energy (now defined as "total energy minus rest energy"), but hidden in there is the derivation of the key relationship between total energy ##E## and rest energy ##E_0##, namely ##E = \gamma E_0##. His thought experiment that leads us there involves a body emitting electromagnetic radiation without changing velocity, but the relation holds for any system with non-zero rest energy precisely because "energy is energy," so to speak (topic of the thread), and also I think because of the clock hypothesis (i.e., ##E = \gamma E_0## holds for an accelerating body, too).

Of course it follows that mass and rest energy are entirely equivalent concepts (##E_0 = mc^2##), and that's a beautiful unifying result. But once established it's no more "useful" than the fact that 1 inch = 2.54 cm. In practice it's that hidden ##E = \gamma E_0## relation that's so important. In tandem with the momentum equation ##\vec p c = E \vec \beta## that was derived soon after, it anchors the relativistic dynamics.
 
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  • #17
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Merriam-Webster lists "force", "energy" and "power" as synonyms:
https://www.merriam-webster.com/thesaurus/force
That doesn't mean they are synonymous in the context of physics.
This is even more ridiculous, as offenbar has absolute nothing to do with mathematics. The question is absolutely about linguistic. And as long as you aren't a Deutschlehrer, this discussion is obsolete - and off topic, but I don't want to moderate discussions I'm involved in. And I do not understand why you started this obviously unnecessary discussion at all.

Let me reformulate it: Sie haben recht und ich meine Ruhe.
 
  • #18
PeroK
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This has become a linguistic issue. From the context, "evidently" or "apparently" would appear to be better translations. But, English is such a subtle language, that "obviously" would also do, as that can be used as a near synonym of "evidently".

I doubt you can ever know exactly what was meant without asking the man himself.

In fact, if I were translating it, I would use "evidently" as that ambiguously covers both possibilities!
 
  • #19
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In fact, if I were translating it, I would use "evidently"
Which is also the original translation in the OP, and better than Google-Translate.
 
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