Re: Nobel Prize for David Thomson?!

From: David Thomson (news5_at_volantis.org)
Date: 01/27/05 

Date: Thu, 27 Jan 2005 13:26:51 -0600

"Richard Schultz" <schultr@mail.biu.ack.il> wrote in message
news:cta0ci$9m5$7@news.iucc.ac.il...
> In sci.physics.particle David Thomson <news5@volantis.org> wrote:
> : "Richard Schultz" <schultr@mail.biu.ack.il> wrote in message
> : news:ct8bo9$k3i$1@news.iucc.ac.il...
> :> In sci.physics.particle David Thomson <news5@volantis.org> wrote:
>
> :> :> :> : When c is taken for 1 then the two sides don't match units.
>
> :> : c = 1 x 10^0 ly/y
> :>
> :> How do you reconcile your value of c = 1 ly/y with your previous
> statement
> :> that when c = 1 the units don't match on the two sides?
> :
> : It should be obvious to you. The other side doesn't have units of ly/y.
>
> The other side has units of energy (that's what that letter "E" stands
> for). If m is in kilograms, then E is in units of kg * (ly)^2/y^2. With
> the conversion factors you had previously calculated, one could turn that
> into whatever other units (e.g. joules) that one wanted.

You still don't get it. We're talking about an equality here. E is equal
to kg * ly^2 / y^2 therefore

kg * ly^2 / y^2 = kg * ly^2 / y^2

So if you are going to change the unit of c on one side of the equation, you
need to do it on the otherside, even if the variable looks like the letter
"E". When you do, you end up with kg = kg, not E = kg.

> : You can change the units of c to whatever you want. But c is not a unit
> of
> : itself (at least not in the MKS, SI, or cgs systems), it is a constant
> which
> : has units (dimensions, actually). The equality sign means that on the
> other
> : side of the equation there is something over there that is equal to c.
>
> No, the equality sign on the other side of the equation says that there
> is something equal to the product of m times c^2.

They both have the same value and dimensions. That's what the two parallel
lines in between the expressions means.

> : So if you change c on the right side, you need to find and change
> : whatever is equal to c on the left side as well.
>
> No one has "changed c on the right side," only expressed the quantity in
> different units. Thus, if one changes the units to those in which c = 1,
> one comes up with equivalent energy units.

Expressing in different units is the same thing as "changing". But have it
your way, if you express c differently on the right side, you need to
express c differently on the left side. Even if c is now a part of E, you
need to factor out c to make the change (excuse me, "express it in different
units").

> Is this something you disagree with?
>
> Or to put it another way -- suppose there is a sign on a highway that says
> "Speed limit 55 miles per hour." A person is stopped for speeding and
> says
> to the policeman "I wasn't going 62 mph -- I was going 100 kilometers per
> hour, a unit which isn't mentioned on the sign." Does the person caught
> for speeding have a reasonable argument?

No, according to your argument, the speeder could say that mph is the same
thing as seconds per hour and claim he was going under the speed limit.

> : Is this something you disagree with?
>
> Yes, because I can do eighth-grade algebra.

Apparently, you cannot.

Dave

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