Re: [Phys-l] E=mc^2 because E=mc^2?

["LaMontagne, Bob" <RLAMONT@providence.edu>]

All quite correct - but I think it misses the point of the original
question - How to present a plausible argument (derivation?) to a
conceptual physics student as to why we use E = mc2. Starting with the
invariant Energy-momentum expression is like answering a person who is
asking how to make an omelet by giving the details of how you first
raise a chicken.

Einstein gave a very simple derivation using a photon shuttling back and
forth by emission and absorption in a tube. There is no use of 4-space
and it is extremely intuitive. This would be more akin to telling the
prospective cook to first crack an egg.

Bob at PC

> -----Original Message-----
> From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
> bounces@carnot.physics.buffalo.edu] On Behalf Of John Denker
> Sent: Wednesday, April 25, 2007 2:24 PM
> To: Forum for Physics Educators
> Subject: Re: [Phys-l] E=mc^2 because E=mc^2?
>
> On 04/25/2007 12:33 PM, Fayngold, Moses wrote:
>
> >   In this respect I agree with Sciama, not with John.
>
> I don't know what Sciama said about this, but my usage agrees
> with Misner/Thorne/Wheeler, with Weinberger, ..... and with
> Einstein.
>
>    Reference:   Gary Oas, ``On the abuse and use of relativistic
mass''
>      http://arxiv.org/PS_cache/physics/pdf/0504/0504110.pdf
>
>
> > The photons
> > are massless - in the sense that they have zero rest mass;
> > however, they have a non-zero relativistic mass m.
>
> In a supposedly-introductory article, if they are going to
> use "mass" in a way contrary to decades-old convention, it
> might have been nice to explain that a little bit.
>
> >   Moreover, there are situations when a single photon in free space
> > has A NON-ZERO REST MASS! Such a situation can be created in a
quantum
> > optics experiement when a single photon is passed through a
> > beam-splitter and no attempt is made to locate path chosen by the
> > passed photon; in this case the meaningful characteristic of the
photon
> >  is its average momentum, which is, in this state, NOT equal to E/c.
>
> 1) I didn't say it was equal to E/c, and
> 2) This is not even remotely relevant to the article in question ...
>   ... so why bring it up?
>
> ======================
>
> In any case,
>    A) There are lots easier ways of getting to E=mc^2, and
>    B) That's not where you want to get to, anyway.
>
> The recommended modern (i.e. post-1908) approach is to write
>    m^2 c^4 = E^2 - ps^2 c^2                  [1]
> or simply
>    m^2 = E^2 - ps^2
>
> where m is "the" mass (which is an invariant Lorentz scalar)
> and where ps is the 3-momentum i.e. the spatial part of
> the 4-momentum.
>
> We see that -m^2 is just p*p i.e. the dot product of the
> 4-momentum with itself.  The 4-momentum is also known as
> the [energy,momentum] 4-vector.
>
> Equation [1] has the advantage that it includes E=mc^2 as a
> special case, and makes it clear that mc^2 is the /rest energy/
> (not the total energy).
>
> Another advantage is that it builds upon and reinforces what
> the students know about vectors.
>
> Next year is the 100th anniversary of spacetime.  All the
> students have heard of spacetime;  it's well established as
> part of pop culture.  Isn't it about time we taught them
> what spacetime really is, and how to use it?
>
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