Recent content by romanski007

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    Doppler effect in Special relativity -- Solution verification

    I am assuming that B is a stationary observer here. For the first part of the trip, using the formula, rocket A is approaching B at velocity $$v_A$$ =0.6 c. The length that A travels is $$L_A = v_A t_1 $$ where $$t_1 = $$ 6 hrs. For the first part of the trip, B is receiving signals at a rate...
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    Two light sources are a distance D apart

    I did it in the first case so that $$x_2'-x_1' = \gamma (x_2 - x_1 - v (t_2 - t_1) = \gamma D \because t_2 - t_1 = 0 $$ but the final answer impliesthat $$t_2-t_1 = \frac{D}{c} $$
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    Two light sources are a distance D apart

    Yes, the way I worked it out was by letting ##x_2 = x_0 + D + ct_2 ## and ##x_1 = x_0 + ct_1## so that ##x_2 -x_1 = D + c(t_2 - t_1) ##
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    Two light sources are a distance D apart

    In relation to the attached diagram, I described the position of ##x_1, x_2 ## and hence took their difference, and transformed it to the primed reference frame where ##t' = ## constant
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    Two light sources are a distance D apart

    Oh so actually it's ## t'_1 = t'_2 ##, so in that case, ##x_2-x_1 = D + c(t_2 - t_1 ) ## as both sources would emit light simultaneously and hence $$x_2'-x_1' = \gamma (D + c(t_2 - t_1) - v(t_2 - t_1)$$ and since in O', the event is simultaneous, I got $$t_2'-t_1' = 0 = \gamma \[ t_2 - t_1 -...
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    Two light sources are a distance D apart

    Why do I need ##t'## coordinates if I am only considering the length difference in O' ? Furthermore in O, how does the time difference of ##\frac{D}{c}## emerge if both events are simultaneous in O.
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    Two light sources are a distance D apart

    I let E1 be the event where source 1 emits the photon and E2 for the second source with the respective coordinates in O as $(x_1, t_1$) and $(x_2,t_2)$ such that $t_2=t_1 \because$ simultaneous and $x_2-x_1 =D$. Using Lorentz transformation I obtained that in O', $$x'_2-x'_1 = \gamma...
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    Show that Tv^(R/c_v) = constant

    The solutions start from the fact that c_v= (dT/dV)_s = -[(du/dv)_T + P], however I cannot reason where did that come from. Any help will be appreciated.
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    A substance has an isothermal compressibility kappa = (aT^3)/(P^2)...

    I proceeded, then differentiated wrt P and got that f'(P) = (T^3/P^2)(b/3 -a ). Hence I proved that b/3 -a must be zero by contradiction as otherwise f would be a function of T and P. Hence a/b = 1/3, and f'(T) = 0 so that f'(T) is come constant. Initial conditions would be V_0 P_0 and T_0...
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    A substance has an isothermal compressibility kappa = (aT^3)/(P^2)...

    Starting from v(P,T), dv=(dv/dp)_T dp + (dv/dT)_P dt i implemented conditions when T and P are constant and ended up with ln V = aT^3/P + constant and ln V = bT^3 /3P + constant If i assume that the constant is 0, i can say that a/b = 1/3 but how do i justify this assumption?
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    Calculate the Temperature and Pressure of a melting point

    Should I assume that the partial derivative of P wrt T remains constant throughout the process or that the specific volume of ice remains constant to work out the graph of P vs T for solid phase? Thanks.
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    Calculate the Temperature and Pressure of a melting point

    Adiabatic increase in pressure implies Tds=0, can someone tell me how to proceed?
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