Answer to high-school question

[mzambani]

Hello,

When I was studying physics in high-school, I often wondered what would be the meaning of sequence below in physics...

m + mv + mv² + mv³ + ...

This is equivalent to

$$\sum$$mvⁿ for n=0 to infinity

where m is mass, and v is velocity.

The though process was, that if mass (m) is conserved, momentum (mv) is conserved, and a form of energy (1/2 mv²) is conserved, then their sum probably should be conserved as well - resulting in above rudimentary sequence.

I never studied physics further, but I always wanted to see what this sequence - if any, would mean in physics. Can some one point me to it? Since velocity (v) is directional, I could never figure out how it can be summed in a sequence like this...

 

[Borek]

Units don't allow summation. Adding kg to kg*m/s to J and so on doesn't make sense.

 

[mzambani]

Thank you Borek. It is so obvious, and I did not see it...this sequence has no meaning in physics.

 

[HallsofIvy]

As a purely mathematical sum, it does have sense:
$$\sum mv^n= m \sum v^n$$
is a "geometric series". If -1< v< 1, its sum is
$$\frac{m}{1- v}$$

But yes, Borek is correct. If v has units of "distance/time" then the series, or even just "m+ mv+ mv^2" has impossible units.

 

[mzambani]

I remember I re-wrote it into this form:

m + mv + 1/2 mv² + 1/6 mv³ + ...
=
m + 1/1! mv + 1/2! mv² + 1/3! mv³ + ...
=
m ( 1 + v/1! + v²/2! + v³/3! + ...)

which is equal to

me^v (using exponential series)

But then I could never think of any meaning of e^v..what could possibly be e to the power of "velocity"?

 

[mzambani]

By taking it further...

m*e^v = some constant (k)
then
e^v = k/m
so
v = log(k/m)
where v is velocity and m is mass

 

[Redbelly98]

There is an issue with units again. Exponents must be unitless.

 

[mzambani]

HallsofIvy, Redbelly98 - thank you for taking time to answer my question. Yes, I realize there is issue with units. My curiosity is satisfied. Thanks.