Relativistic rocket equation

In particular, the term "binomial expansion" is often used even when the exponent is not an integer. That's what I did in the post above.
  • #1
Kutuzov
8
0

Homework Statement


In this thread the author performs the following calculation under "method 1":
[tex]v+dv=\frac{v+dv'}{1+vdv'}=v+(1-v^2)dv'\implies dv'=\frac{dv}{1-v^2}[/tex]

He's set c=1 so the second expression is the relativistic velocity addition formula. What I don't understand is how he gets the third expression from the second one.

Homework Equations


Velocity addition formula
[tex]u=\frac{v+u'}{1+\frac{v u'}{c^2}}=\left[ c=1 \right]=\frac{v+u'}{1+vu'}[/tex]

The Attempt at a Solution


When I solve for dv' between the first and second expression, I get
[tex]dv'=\frac{dv}{1-v^2-vdv}[/tex]
Instead of what he gets
[tex]\frac{dv}{1-v^2}[/tex]
Perhaps vdv is negligible or something like that? I can't figure it out.
 
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  • #2
Kutuzov said:

Homework Statement


In this thread the author performs the following calculation under "method 1":
[tex]v+dv=\frac{v+dv'}{1+vdv'}=v+(1-v^2)dv'\implies dv'=\frac{dv}{1-v^2}[/tex]

He's set c=1 so the second expression is the relativistic velocity addition formula. What I don't understand is how he gets the third expression from the second one.

Homework Equations


Velocity addition formula
[tex]u=\frac{v+u'}{1+\frac{v u'}{c^2}}=\left[ c=1 \right]=\frac{v+u'}{1+vu'}[/tex]

The Attempt at a Solution


When I solve for dv' between the first and second expression, I get
[tex]dv'=\frac{dv}{1-v^2-vdv}[/tex]
Instead of what he gets
[tex]\frac{dv}{1-v^2}[/tex]
Perhaps vdv is negligible or something like that? I can't figure it out.

Yes, the ##dv##'s are infinitesimal, and so terms of order ##(dv')^2## or ##(dv)^2## are neglected. In the original derivation, the author used the binomial expansion on the denominator and dropped any terms ##(dv')^2## or higher.

If you use the binomial expansion on your derivation, you will similar find that you get the desired result plus terms of order ##(dv)^2## or higher, which you neglect. (You are not just neglecting ##v dv##; it's only because there is already a ##dv## in the numerator that the ##v dv## term ends up vanishing.)
 
  • #3
I need an example of said binomial expansion to see what it's about. I'm sure I will recognize it straight away, it's just what I think is binomial expansion is stuff like [itex](a+b)^3=a^3+3a^2b+3ab^2+b^3[/itex]. In this case ([itex]1+vdv'[/itex]) the exponent is one, so an expansion like this would leave the denominator unchanged...
 
  • #4
Kutuzov said:
I need an example of said binomial expansion to see what it's about. I'm sure I will recognize it straight away, it's just what I think is binomial expansion is stuff like [itex](a+b)^3=a^3+3a^2b+3ab^2+b^3[/itex]. In this case ([itex]1+vdv'[/itex]) the exponent is one, so an expansion like this would leave the denominator unchanged...

You want to expand ##(1-v^2 - vdv)^{-1}##. Actually, it's better if you pull out a factor of ##1-v^2##, so that you have

$$\frac{dv}{1-v^2}\left(1-\frac{vdv}{1-v^2}\right)^{-1}.$$

You want to expand the factor ##\left(1-\frac{vdv}{1-v^2}\right)^{-1}## now.
 
  • #5
Thank you. Works perfectly! I call that kind of expansion Taylor series expansion, and only say binomial expansion if I use the binomial theorem. But it might be the same thing.
 
  • #6
Kutuzov said:
Thank you. Works perfectly! I call that kind of expansion Taylor series expansion, and only say binomial expansion if I use the binomial theorem. But it might be the same thing.

That particular Taylor series is called the Binomial series, hence why I called it a binomial expansion. :) It reduces to the binomial theorem in the case of positive integer exponent, as well.
 
  • #7
Kutuzov said:
Thank you. Works perfectly! I call that kind of expansion Taylor series expansion, and only say binomial expansion if I use the binomial theorem. But it might be the same thing.
Like Mute said, binomial expansion is a special case of Taylor expansion, so you could say they're the same thing in that sense.

It's standard practice to say "binomial expansion" to mean any expansion of this sort, even if it doesn't use the binomial theorem for nonnegative integer exponents.
 

1. What is the relativistic rocket equation?

The relativistic rocket equation is an equation used in physics to calculate the velocity of a rocket in terms of the mass of its fuel and the velocity of its exhaust gases.

2. How is the relativistic rocket equation derived?

The equation is derived from the principles of conservation of momentum and energy, taking into account the effects of relativity on the motion of the rocket and its exhaust gases.

3. What is the significance of the relativistic rocket equation?

The equation is significant because it allows scientists to accurately calculate the velocity of a rocket traveling at relativistic speeds, which is important for space exploration and understanding the behavior of objects in motion at high speeds.

4. Can the relativistic rocket equation be used for all types of rockets?

No, the equation is specifically designed for rockets that use exhaust gases to propel themselves forward, and it does not take into account other types of propulsion systems such as ion thrusters or solar sails.

5. Are there any limitations to the relativistic rocket equation?

Yes, the equation assumes that the rocket is traveling in a vacuum and does not take into account external forces such as air resistance or gravitational pull from other objects. It also assumes that the rocket is traveling in a straight line and does not account for changes in direction or acceleration.

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