Recent content by astoria

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    Partial Derivatives: Solving T(x,t)=S(n) Chain Rule Mistake

    T(x,t)=S(n) where n is some given function of x and t. Why is (partial dT by partial dt)=dS/dn*(partial dn by partial dt) What happens to the extra (all partials) (dS/dx)*(dx/dt) I guess I'm misunderstanding the chain rule in partial derivatives but can someone point out my mistake...
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    How can a 4x4 matrix be a Lorentz transformation and what are its properties?

    I've found the formal definition of Lorentz transformation and proper and orthochronous, so ignore that bit.
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    How can a 4x4 matrix be a Lorentz transformation and what are its properties?

    Homework Statement What does it mean for a 4x4 matrix to be a Lorentz transformation? What does it mean for it to be proper and orthochronous? Show that for the standard Lorentz transformation, c^2t^2-x^2-y^2-z^2=c^2t'^2-x'^2-y'^2-z'^2 What is the significance of this statement when quantity...
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    How Is Total Energy Derived in a Special Relativity Collision?

    Having read more solutions of this form, am I right to assume that if V,W are the four velocities in ICS 1 and V' W' are the four velocities of two particles in ICS 2g(V,W)=g(V',W')?
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    How Is Total Energy Derived in a Special Relativity Collision?

    Essentially it just says: The four momenta of the particles are P=(E/c,p) and Q=m(c,0) in the given inertial frame. In centre of mass frame the four velocity V of the centre of mass decomposes as (c,0) and P+Q decomposes as (E'/c,0). Moreover g(V,V)=c^2 Therefore: V=c(P+Q)/sqrt(g(P+Q,P+Q))...
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    How Is Total Energy Derived in a Special Relativity Collision?

    Thanks, that came out fine :) I've had a look at the official solutions (which are only outline) and it seems to do it a completely different way, using g(P+Q,V) etc I've never seen this sort of problem done like that before, is it easier? unfortunately it's not comprehensively explained...
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    How Is Total Energy Derived in a Special Relativity Collision?

    I may be misunderstanding what you mean but I think what you are saying is that the energy is the sum of the Newtonian kinetic energy and mc^2 (i.e E=(mc^2+1/2mv^2)) This is only true if we take gamma as equal to 1+v^2/2c^2 and ignore the higher order terms.
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    How Is Total Energy Derived in a Special Relativity Collision?

    Homework Statement A particle of rest mass M and total energy E collides with a particle of rest m at rest. Show that the sum E' of the total energies of the two particles in the frame at which their centre of mass is at rest is given by (E')^2=(M^2+m^2)c^4+2Emc^2 3. The attempt at...
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