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Science Forum Index » Physics Forum » Problem with a formula in special relativity
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| Eoghan |
 Posted: Thu May 01, 2008 12:46 am |
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Guest
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Hi there!
I'm studying the equation of motion for relativistic systems with a
variable rest mass. It's supposed that there aren't any external
forces.
The text says: according to the laws of conservation we can write:
My(V)=(M+dM)y(V+dV)+dmy(W) (1)
where y(x)=1/sqrt(1-x^2/c^2)
M is the initial mass
dM is the mass lost by the system
V is the speed of the system according to an inertial frame S
dV is the infinitesimal change of the speed in an
infinitesimal interval dT
dm is the expelled mass
W is the speed of the expelled mass according to the inertial
frame S
The equation (1) states that the total energy at a time T equals the
total energy at a time T+dT after a dm mass is expelled.
Moreover we can write the speed WR that is the speed of the expelled
mass relative to the system (W is relative to the inertial frame S)
W=(V-WR)/(1-WR*V/c^2)
where the minus sign means that WR and V have opposite directions.
Moreover: y(W)=y(WR)y(V)(1-WR*V/c^2)
According to Taylor's theorem we can also write:
y(V+dV)=y(V)+[y(V)^3]*VdV/c^2
Then the text says: after long calculus we get:
dM+y(WR)dm=0
The problem is that after long calculus I can't get this equation.
Where does it come from?
Please help |
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| Dirk Van de moortel |
| Posted: Thu May 01, 2008 10:53 am |
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"Eoghan" <lucagalbu@gmail.com> wrote in message news:393a66d3-69fe-49b0-91f2-babc270c6241@56g2000hsm.googlegroups.com...
Quote: Hi there!
I'm studying the equation of motion for relativistic systems with a
variable rest mass. It's supposed that there aren't any external
forces.
The text says: according to the laws of conservation we can write:
My(V)=(M+dM)y(V+dV)+dmy(W) (1)
where y(x)=1/sqrt(1-x^2/c^2)
M is the initial mass
dM is the mass lost by the system
V is the speed of the system according to an inertial frame S
dV is the infinitesimal change of the speed in an
infinitesimal interval dT
dm is the expelled mass
W is the speed of the expelled mass according to the inertial
frame S
The equation (1) states that the total energy at a time T equals the
total energy at a time T+dT after a dm mass is expelled.
Moreover we can write the speed WR that is the speed of the expelled
mass relative to the system (W is relative to the inertial frame S)
W=(V-WR)/(1-WR*V/c^2)
where the minus sign means that WR and V have opposite directions.
Moreover: y(W)=y(WR)y(V)(1-WR*V/c^2)
According to Taylor's theorem we can also write:
y(V+dV)=y(V)+[y(V)^3]*VdV/c^2
Then the text says: after long calculus we get:
dM+y(WR)dm=0
The problem is that after long calculus I can't get this equation.
Where does it come from?
from (1) you have (written with spaces):
M y(V) = (M+dM) y(V+dV) + dm y(W)
or, with the taylor expression:
M y(V) = (M+dM) ( y(V) + y(V)^3 V dV/c^2 ) + dm y(W)
giving, ignoring dM dV
0 = M y(V)^3 V dV/c^2 + dM y(V) + dm y(W)
and using the expression for y(W):
0 = M y(V)^3 V dV/c^2 + dM y(V) + dm y(WR) y(V) (1-WR V/c^2)
giving after dividing by y(V):
0 = M y(V)^2 V dV/c^2 + dM + dm y(WR) (1-WR V/c^2) [2]
Now you'll need conservation of momentum as well:
M y(V) V = (M+dM) y(V+dV) (V+dV) + dm y(W) W
or, with the taylor expression:
M y(V) V = (M+dM) ( y(V) + y(V)^3 V dV/c^2 ) (V+dV) + dm y(W) W
giving, ignoring all products of infinitesimals
0 = M y(V) dV + M y(V)^3 V^2 dV/c^2 + dM y(V) V + dm y(W) W
and using the expression for y(W):
0 = M y(V) dV + M y(V)^3 V^2 dV/c^2 + dM y(V) V + dm y(WR) y(V) (1-WR V/c^2) W
giving after dividing by y(V):
0 = M dV + M y(V)^2 V^2 dV/c^2 + dM V + dm y(WR) (1-WR V/c^2) W
and with the expression for W:
0 = M dV + M y(V)^2 V^2 dV/c^2 + dM V + dm y(WR) (1-WR V/c^2) (V-WR) / (1-WR V/c^2)
giving
0 = M dV + M y(V)^2 V^2 dV/c^2 + dM V + dm y(WR) (V-WR) [3]
Now, solve the system [2], [3] for dM and dm and calculate dM/dm.
You should get y(WR).
Dirk Vdm |
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| Eoghan |
| Posted: Thu May 01, 2008 10:57 am |
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Guest
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But the text says that I should get
dM + y(WR) dm = 0
M y(v)^2 dv - WR y(WR) dm = 0 |
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| Dirk Van de moortel |
| Posted: Thu May 01, 2008 4:08 pm |
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"Eoghan" <lucagalbu@gmail.com> wrote in message news:115d1e9f-0546-40c2-92a6-b15fab2e045e@25g2000hsx.googlegroups.com...
Quote: But the text says that I should get
dM + y(WR) dm = 0
Yes, I meant that yu should get dM/dm = - y(WR)
By the way, it might be easier to eliminate M from [2] and [3] and
then solve for dM, then you get
dM = - y(WR) dm
Quote: M y(v)^2 dv - WR y(WR) dm = 0
that should be part of the elimination process after re-inserting dM as - y(WR) dm.
There's no calculus here. It's all tedious but simple algebra.
Dirk Vdm |
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| mL |
| Posted: Thu May 01, 2008 5:39 pm |
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Guest
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Eoghan skrev:
Quote: Hi there!
I'm studying the equation of motion for relativistic systems with a
variable rest mass. It's supposed that there aren't any external
forces.
The text says: according to the laws of conservation we can write:
My(V)=(M+dM)y(V+dV)+dmy(W) (1)
where y(x)=1/sqrt(1-x^2/c^2)
M is the initial mass
Shouldn't M be "current" mass? - I guess you are aiming at the
relativistic rocket eqn.
Quote: dM is the mass lost by the system
V is the speed of the system according to an inertial frame S
dV is the infinitesimal change of the speed in an
infinitesimal interval dT
dm is the expelled mass
W is the speed of the expelled mass according to the inertial
frame S
The equation (1) states that the total energy at a time T equals the
total energy at a time T+dT after a dm mass is expelled.
Moreover we can write the speed WR that is the speed of the expelled
mass relative to the system (W is relative to the inertial frame S)
W=(V-WR)/(1-WR*V/c^2)
where the minus sign means that WR and V have opposite directions.
Moreover: y(W)=y(WR)y(V)(1-WR*V/c^2)
According to Taylor's theorem we can also write:
y(V+dV)=y(V)+[y(V)^3]*VdV/c^2
Then the text says: after long calculus we get:
dM+y(WR)dm=0
The problem is that after long calculus I can't get this equation.
Where does it come from?
Forget about frame S. Use the current *rest frame* instead!
Energy conservation:
Mc^2 = (M + dM)c^2 + y(WR) dm c^2,
0 = dM + y(WR) dm.
/mel |
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