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There are lots of different ways of arriving at the relativistic relations involving mass, energy, and momentum such as $E=mc^2$ and $m^2=E^2-p^2$ (the latter with $c=1$). One that I've seen in some textbooks is to start with the equation $W=\int F dx$ for mechanical work. Here is a particularly careful and rigorous version, which makes it explicit that this is a nontrivial assumption, along with $F=dp/dt$: http://www.physicsforums.com/showthread.php?p=2416765 Historically, Einstein made use of $W=\int F dx$ in section 10 of his 1905 paper on SR http://www.fourmilab.ch/etexts/einstein/specrel/www/ . This was before the "Does the inertia of a body..." paper, in which he derived $E=mc^2$ and gave it its full relativistic interpretation. Einstein later decided that force wasn't a very useful concept in relativity and stopped appealing to it.

What I have never seen in any of these treatments (neither the careful ones above nor the sloppy ones in some freshman physics texts) is any argument as to why the nonrelativistic relation $W=\int F dx$ should be expected a priori to hold without modification in SR. If one has already established the form and conservation of the energy-momentum four-vector, then I don't think it's particularly difficult to show that $W=\int F dx$. But what justification is there for assuming $W=\int F dx$ before any of that has been established? Is there any coherent justification for it?

(A secondary problem with some of these treatments based on $W=\int F dx$ is that they need to establish the constant of integration.)

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I do not see where the problem is. $W=\int F dx$ follows from the relativistic formula $F=\frac{dp}{dt}$ which has been proved experimentally. –  Martin Gales Aug 18 '11 at 7:44
    
@MartinGales: You may be right, but that would depend on spelling out what your assumptions are and what reasoning you have in mind when you say "follows from." If you don't consider $F=dp/dt$ to be a definition, then what do you consider to be the relativistic definition of force? –  Ben Crowell Sep 22 '13 at 1:47

2 Answers 2

There is no reason for the formula to obviously work, as you say, just because there are at least two different generalizations of force, the proper-time relativistic curvature of the trajectory ($d^2x\over d\tau^2$ not the right notion) and the derivative of the relativistic energy times the velocity with respect to time (${d\over dt} (m{dx\over d\tau})$ the one that works). Which one is the proper generalization?

Einstein came from a thermodynamic background, and he always accorded the laws of thermodynamics an absolute status, being deduced as they were so inexorably from such elementary postulates. It was the deductive form of thermodynamics that led him to write the special relativity in such a deductive style, building up from simple axioms.

Potential energies, and energies in general, have a thermodynamic meaning mostly independent of their mechanical meaning, which is unchanged by relativity (if you like to think of this statistically, the Boltzmann distribution is still $\exp(-\beta E)$ in either theory). The electrostatic potential is the energy per unit charge in a region of space. The electrostatic energy of a charged ball moving with velocity v in an electrostatic potential $\phi$ can't depend on the velocity, because then I could transfer the charge from a fast moving object to a stationary object as the fast object comes whooshing by, and make a perpetual motion machine. So the potential energy of the charge is still $q\phi$ even in relativity, for thermodynamic reasons.

The electric field is the gradient of the static potential, and this is true in Maxwell's equations, which are already relativistic. The integral of the gradient of the static potential along the path, times the charge, then gives the work done along the path. So the correct notion of force to use is whatever mechanical thing is equal to $qE$. It is physically all but certain that different forces, no matter what their origin, do the same work if they have the same magnitude, so you can conclude that the work formula is generally true for this notion of force, which is the notion of force in the equation of motion. For Einstein this is so clear, he doesn't bother to explain.

This assumption seems unnatural to many people nowadays, since we are comfortable with symmetries much more than thermodynamics. But Einstein's exposition, survives in textbooks to this day.

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You're a gold mine on these historical topics! It seems that you're saying that the four-force $ma^j$, where $a^j$ is the acceleration four-vector and $m$ the invariant mass, is "not the right notion" of force. Why not? If I'm generalizing the three-force, I'd really dearly love to have the generalization be a valid four-vector. The quantity $d(E v^j)/dt$, where $E$ is the mass-energy, is not a four vector, so I react to it with horror. Why do you consider it "the one that works?" –  Ben Crowell Aug 18 '11 at 1:53
    
I don't think that the four-force is "wrong", just wrong for generalizing the horribly non-covariant expression $\int F dx$. I personally am more comfortable with four-force, which is why I asked myself the exact same question "why does Einstein consider this formulat obvious?" many years ago. –  Ron Maimon Aug 18 '11 at 2:01
    
The relationship between the second half of your answer and the first half confuses me. The first half says we have multiple possibilities for generalizing the notion of force to SR. The second half argues that $W=\int F dx$ holds in SR, without ever having to invoke either of the two definitions of F. Surely its validity should depend on which definition one uses...? –  Ben Crowell Aug 18 '11 at 2:02
    
What I am saying is that the notion of force which can be seen to have the property that $W = \int F dx$ is whatever mechanical quantity that appears on the left hand side of "F=qE" in relativity (where the right hand side is just the spatial gradient of the electrostatic potential, without any velocity factors). Einstein figures out this quantity in his special relativity paper by transforming from a frame moving along with the particle. –  Ron Maimon Aug 18 '11 at 2:11
    
Also, in skimming over the paper, he does say that this quantity must be the work done, because it must equal the gain in electrostatic potential energy along the integration path. This is the thermodynamic a-priori argument I was talking about. –  Ron Maimon Aug 18 '11 at 2:12

I suspect there's nothing here that isn't already in Ron Maimon's answer, but maybe I've improved the sequencing...

Recall that Einstein, in that famous 1905 paper you reference, is considering E&M; indeed, the title is "On the Electrodynamics of Moving Bodies". Prior to the section 10 considered here, he has demonstrated the covariance of Maxwell's equations.

I'd expect him to exploit that covariance, modifying Newtonian dynamics to "fit" relativistically correct E&M.

Accordingly, Einstein simplifies to a one-dimensional electrostatic case, with a charge $q$ moving from point 1 to point 2 in response to an electric field $X$. Then, by conservation of energy, the increase in the charge's kinetic energy $W$ (the functional form of which Einstein determines in a separate calculation) must just be the drop in electrical potential energy $P$, which is just the line integral of the electric force:

$$ W = P = q \int_1^2 X dx = \int_1^2 F dx $$

Voila.

[I can't help noting that Einstein specifies that the charge acceleration be small, so that the charge "not give off any energy in the form of radiation". (related) ]

When time-varying magnetic fields are included, there is no longer a potential function, but the increase in kinetic energy is again equal to the line integral of force, evaluated over the actual path of the particle: $$ W = \int_1^2 \boldsymbol{F \cdot dl} $$

The force $\boldsymbol{F}$ here is obviously not the four-force; it's just the standard Lorentz vector force: $$ \boldsymbol{F} = \frac{d \boldsymbol{p}}{dt} =q (\boldsymbol{E} + \boldsymbol{v \times B} ) $$

You can see this formulation in Feynman's "Classical Physics" Table 18-1 in Volume 2 of The Lectures.

Well, I've convinced myself anyway. When I first saw this question I was mystified by the "horribly non-covariant expression", as Ron put it. It's still non-covariant, but now it's obvious.  

[Trivial observation: Einstein uses $\beta$ for what we now call $\gamma$. What's up with that?]

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