How Does a Photon Rocket's Velocity Depend on Its Mass Reduction?

In summary, the spacecraft's velocity relative to its initial rest-frame is given by v= gamma*Mf*c^2.
  • #1
ajclarke
35
1

Homework Statement



A spacecraft begins a journey with rest mass Mi. Its method of propulsion involves converting matter entirely to photons, which are emitted in the direction opposite to the direction of motion. After a period of acceleration the rest mass has been reduced to Mf. Show that the velocity v of the spacecraft , relative to its initial rest-frame, is then given by

[tex]\frac{v}{c}[/tex]=[tex]\frac{Mi^2-Mf^2}{Mi^2+Mf^2}[/tex]

Homework Equations



Conservation of Momentum
Conservation of Energy


The Attempt at a Solution



Well I think I have at least got the Physics right to this.
Thining in two stages.

Before:


Rocket is stationary. Therefore

P(before)=Mi*v=0
E(before)=Mi*c^2

After:


E(photons)=pc
p(photons)=E/c

E(after)=Gamma*Mf*c^2
[(after)=Gamma*Mf*v

Calculations

1. Conservation of E

E(before)=E(photons)+E(after)
Mi*c^2=E{photons)+Gamma*Mf*c^2

2. Conservation of P

P(before)=P(after)-P(photons) [Negative for photons since ejected in -ve x direction]
0=Gamma*Mf*v - E(photons)/c

Rearranging this: E(photons)=Gamma*Mf*v*c

I then subbed this in for consv. of E equation above. Multiplied by the denomintor of Gamma, squared out to get rid of the root etc, and at a suggestion from the lecturer let Beta=v/c

However when I square all the terms, I end up with a factor of 2 in it. Bascially I end up with:


Mi^2*Mf^2=Beta(Mf^2*Beta+2Mf^2+Mi^2*Beta)

Which clearly will not rearrange to the answer. I have worked through this about 5 or six times now all to no avail. Anyone able to see a mistake or omething I have missed out. I checked the Physics at the beginning with the lecturer and he said it was fine, so i think its some mathmatical treatment which is causing me issues here.

Once again, thank you for your help.
Adam
 
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  • #2
Welcome to PF!

Hi Adam! Welcome to PF! :smile:

(have a beta: β and a gamma: γ and try using the X2 tag just above the Reply box :wink:)

I don't see what differential equation you've used. :confused:

And I think you may have the wrong formula for the energy of the photons …

find ∆(photon energy) in the rocket frame, then use the redshift formula for photons to convert that to the rest frame, then covert that to momentum. :smile:
 
  • #3
Thanks for the welcome.

I didn't use differential equations at all, just simple convservation ones with the lorentz factor thrown in for anything relativistic. At this section of the coure we hadn't covered redshift so he won't expect us to use that to solve it, though I'm not doubting it would work.

The E=pc for massless particles came straight out of the textbook (French, Special Relativity)

Hummm :/

P.S Thanks, I only noticedafter this post that i could use LATEX in here, as i did in the Slow Neutron Capture one. :D
 
  • #4
Hi Alan! :smile:

I don't understand which frame you're working in. :confused:

If it's the Earth frame, how are you converting the momentum of the photons?

(the usual β/γ formula won't work, it's for converting form a frame in which the body is stationary)
 
  • #5
Adam * :P :D :D

Hmmm, I hadn't really considered frames I have to admit. The lecturer said it was treatable as a classic conservation of momentum and energy in basis, with γ placed in for everything moving relativistically.

I thought E=pc for massless particles already took into consideration relativistic properties, though perhaps this is where I am going majorly wrong. :/
 
  • #6
Hi Adam! :smile:

(oops! … I don't know you from Alan! :biggrin:)
ajclarke said:
I thought E=pc for massless particles already took into consideration relativistic properties …

Yes it does, but if you use it in the Earth frame to find p, first you need to find E in the Earth frame …

I can't for the moment think of a way of doing that with a bare γ factor :confused:

(btw, it may help you to notice what (1+v)/(1-v) equals :wink:)​
 
  • #7
Don't worry about it =D

Hmm I had indeed thought it looked familiar, but i can't find where from. I will sleep on it, and have a gander through my notes to try and consider what you said.

Thanks for the input
 
  • #8
Solved. Thankd for your input. I was on the right lines. Some simple maths I happened to skip over in the algebraic manipulation stage was causing me to not be able to rearrange it properly.
 

FAQ: How Does a Photon Rocket's Velocity Depend on Its Mass Reduction?

What is a photon rocket?

A photon rocket is a hypothetical spacecraft that uses the momentum of photons (particles of light) to propel itself through space.

How does a photon rocket work?

A photon rocket works by releasing photons in one direction at high speeds, which in turn creates a force in the opposite direction, propelling the spacecraft forward.

What is special relativity and how does it relate to photon rockets?

Special relativity is a theory developed by Albert Einstein that explains the relationship between space and time. It is relevant to photon rockets because it shows that as an object approaches the speed of light, its mass increases, making it more difficult to accelerate. This means that a photon rocket would require a tremendous amount of energy to reach the speed of light.

What are the challenges of building a photon rocket?

One of the main challenges of building a photon rocket is the amount of energy needed to accelerate the spacecraft to near the speed of light. Additionally, the spacecraft would also need a way to efficiently convert energy into photons and direct them in a specific direction.

Could a photon rocket travel faster than the speed of light?

No, according to special relativity, nothing can travel faster than the speed of light. As an object approaches the speed of light, its mass increases, making it more and more difficult to accelerate. This means that even a photon rocket, which uses the speed of light to propel itself, would not be able to exceed the speed of light.

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