# I Morin relativistic rocket example (chapter 12, page 606)

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1. May 11, 2016

### almarpa

Hello all.

In Morin's classical mechanics book, chapter 12 (relativistic dynamics), in the axample about the relativistic rocket, we have a rocket that propels itself converting mass into photons and firing them back. Here Morin takes dm as a negative quantity, so the instantaneous mass of the rocket goes from m to m+dm.

I have tried to solve the same example taking dm as a positive quantity, and assuming that the instantaneous mass goes from m to m-dm, but I do not get the same answer. The correct rocket motion equation should be:

(dm/m)+(dv/(1-v2))=0

(dm/m)-(dv/(1-v2))=0

As you see, the solution is the same, except for one minus sign.
What could be the problem? Any ideas?

Thank you so much.

2. May 11, 2016

### Vitro

If dm is a positive quantity in one equation and a negative quantity in the other then it is the same equation.

3. May 12, 2016

### almarpa

Yes, that is what I thought.

Nevertheless, in Morin example he integrates this equiation from M to m and from 0 to v to get:

m=M[(1-v)/(1+v)]1/2

But if I integrate my equation in the same terms (from M to m and from 0 to v ), what I get is:

m=M[(1+v)/(1-v)]1/2

As you see, the signs are exchanged in the quotient, so the result is nor the same.

What is the problem?

4. May 12, 2016

### robphy

This sounds like it might be a confusion between
the use and/or interpretation of an "infinitesimal change in mass" [whose sign depends on whether one adds or subtracts]
and as an "increment of the mass variable m" (used in doing an integral) [which is always positive].

I think it might analogous to this [my reply in an old thread]
Dot product in the Gravitational Potential Energy formula

5. May 12, 2016

### PAllen

Ah, then per robphy's point if the OP reverses the limits of integration, they will get the same answer as the other method.

6. May 12, 2016

### robphy

Or is that, when $dm$ changes sign (in the attempted change of variables)
then $v$ (or $dv$) changes sign?

7. May 16, 2016

### almarpa

Confusing, isn't it?

8. May 16, 2016

### robphy

It could be, if one isn't careful.

I took a closer look.
Let's clarify variable names.
m is the instantaneous rocket-mass function (which varies with time).
dm is the incremental change in that rocket-mass function (following what Morin is doing)
so that m+dm represents the new mass (which is smaller than the old mass since the increment is a negative amount).
M is the original mass.

Now if you wish change variables
so that you can (better?) display the fact that the rocket-mass decreases,
let $(d\mu)=-dm$;
this $d\mu$ is positive (which could be interpreted as the positive photon-energy ejected in the rocket-frame).

Following Morin,
the spatial-momentum of the rocket in the ground frame is
$(m\gamma v)_{new}=(m\gamma v)_{old}-\gamma(1-v)dm$
Using $(d\mu)$,
$(m\gamma v)_{new}=(m\gamma v)_{old}-\gamma(1-v)(-d\mu)$,
which could be written as
$d(m\gamma v)= -\gamma(1-v)(-d\mu)$.
When expanding out the left-hand side, you have $dm$, but you have $d\mu$ on the right-hand-side.
If you expand then try to combine terms (in favor of $d\mu$) to get the diff eq, you get
$\frac{-d\mu}{m}+\frac{dv}{1-v^2}=0$.
However, you can't continue to solve for m unless you use $d\mu=-dm$.

So, you are forced to same differential equation for $m$.

9. May 18, 2016

### almarpa

All right, I see.

I will take a deeper look to it, to check if I really get it.

Thank you so much, I spent a lot of time thinking about it.