What Are the Space-Time Coordinates of an Event in a Different Frame?

[forty]

Question 1:
An event occurs at x'=60m, t'= 8*10^-8s in frame S' (with y'=0 and z'=0) The frame S' has a velocity of 3c/5 in the x direction with respect to the frame S. The origins of S and S' coincide at t=0,t'=0. What are the space time coordinates of the event in S?

Just want to make sure I'm using the Lorentz formulas properly..

ɣ=(1 - v²/c²)^-1/2

x = ɣ(x'+vt) = (1-(9/25))^-1/2*(60+(3c/5)*(8*10^-8))=93m

t = ɣ(t' + (vx'/c^2)) = (1-(9/25))^-1/2*((8*10^-8)+(3c/5c^2)*(60))=45s (For some reason I think this is very wrong)

Question 2:
The space time coordinates of two events measured in the frame S are as follows:
Event 1: x₁=x_o, t₁=x_o/c
Event 2: x₂=2x_o, t₁=x_o/2c

(a) There exists a frame in which these events occur at the same time. Find the velocity of this frame with respect to S.
(b) What is the value of t at which both events occur in the new frame.

(a) Would doing the following work?

ɣ(t₁ + (vx₁/c²))=ɣ(t₂ + (vx₂/c²))

then solve for v which gives v=-c/2 (again I don't think this is right)

(b)

ɣ=(1 - v²/c²)^-1/2

Then using the value of v from a and the equation t'=ɣ(t - vx/c²) solve for t'



Any help on this would be greatly appreciated... This is a lot more confusing than I first anticipated.

 

[Doc Al]

Just want to make sure I'm using the Lorentz formulas properly..

ɣ=(1 - v²/c²)^-1/2

x = ɣ(x'+vt) = (1-(9/25))^-1/2*(60+(3c/5)*(8*10^-8))=93m

t = ɣ(t' + (vx'/c^2)) = (1-(9/25))^-1/2*((8*10^-8)+(3c/5c^2)*(60))=45s (For some reason I think this is very wrong)

You have the correct formulas, but I get a different value for t. Double check your arithmetic.

Question 2:
The space time coordinates of two events measured in the frame S are as follows:
Event 1: x₁=x_o, t₁=x_o/c
Event 2: x₂=2x_o, t₁=x_o/2c

(a) There exists a frame in which these events occur at the same time. Find the velocity of this frame with respect to S.
(b) What is the value of t at which both events occur in the new frame.

(a) Would doing the following work?

ɣ(t₁ + (vx₁/c²))=ɣ(t₂ + (vx₂/c²))

then solve for v which gives v=-c/2 (again I don't think this is right)

Looks OK to me.

(b)

ɣ=(1 - v²/c²)^-1/2

Then using the value of v from a and the equation t'=ɣ(t - vx/c²) solve for t'

That should work.

 

[forty]

Double checked my value of t and get 2.5*10^-7 which seems better...

While I'm on a roll...!

(a) In frame S', a stick makes an angle of θ' with the x' axis. What is the angle θ measured in the S frame. What is the length in the S frame.

I presume frame S' has velocity v in the x direction (Which mean the length of the stick is only changed in the x direction)

So the angle in the S frame is..

Cos(θ) = (Cos(θ') + (v/c))/(1 + Cos(θ')*(v/c))
θ = Cos^-1((Cos(θ') + (v/c))/(1 + Cos(θ')*(v/c)))

then

L_p = proper length = length in S frame

L_p = ɣL

lCos(θ') = The length of the stick in the x-direction

L_p = ɣlCos(θ')

Then the length of the stick in S = L_p/Cos(θ)


(b) Now we consider a particle moving with a velocity u' in frame S' at an angle θ' to the x' axis. Find the angle θ in frame S.

To do this I transformed the x and y components to the S frame then used tan(θ) = u_y/u_x

u'_y=u'Sin(θ')
u'_x=u'Cos(θ')

Then the transforms

u_x = (u'_x + v)/(1 + u'_xv/c²)

and

u_y = (u'_y)/(ɣ(1 + u'_yv/c²))

then..

tan(θ) = u_y/u_x
θ = tan^-1(u_y/u_x)


Again does this look right? I'm not sure if I'm trying to over complicate things...

 

[Doc Al]

I presume frame S' has velocity v in the x direction (Which mean the length of the stick is only changed in the x direction)

So the angle in the S frame is..

Cos(θ) = (Cos(θ') + (v/c))/(1 + Cos(θ')*(v/c))
θ = Cos^-1((Cos(θ') + (v/c))/(1 + Cos(θ')*(v/c)))

That should work, but even simpler is to consider the x & y components of the stick's length and compute how tanθ would transform.

then

L_p = proper length = length in S frame

L_p = ɣL

lCos(θ') = The length of the stick in the x-direction

L_p = ɣlCos(θ')

Then the length of the stick in S = L_p/Cos(θ)

Not sure what you did here. For one thing, only the x-dimension contracts. How does that affect the total length of the stick? (Again, consider the x & y components of the stick's length.) Also, presumably the stick is moving in the S' frame, so its length in that frame is its proper length.

 

[Doc Al]

(b) Now we consider a particle moving with a velocity u' in frame S' at an angle θ' to the x' axis. Find the angle θ in frame S.

To do this I transformed the x and y components to the S frame then used tan(θ) = u_y/u_x

u'_y=u'Sin(θ')
u'_x=u'Cos(θ')

OK.

Then the transforms

u_x = (u'_x + v)/(1 + u'_xv/c²)

and

u_y = (u'_y)/(ɣ(1 + u'_yv/c²))

Double check that second transform.

then..

tan(θ) = u_y/u_x
θ = tan^-1(u_y/u_x)

That should work.

 

[forty]

then

Lp = proper length = length in S frame

Lp = ɣL

lCos(θ') = The length of the stick in the x-direction

Lp = ɣlCos(θ')

Then the length of the stick in S = Lp/Cos(θ)

Not sure what you did here. For one thing, only the x-dimension contracts. How does that affect the total length of the stick? (Again, consider the x & y components of the stick's length.) Also, presumably the stick is moving in the S' frame, so its length in that frame is its proper length.

Ahhhh always confuse the proper length... So the proper length is the length as measured by an observer at rest with the object, so Lp is the length in S'

So If the stick makes an angle θ' with the x' axis in the S' frame then its x component of it's length is lcos(θ') = L_px

ɣ=(1 - v2/c2)^-1/2

so in the S frame this length becomes L_px/ɣ

So the length of the stick in the S frame is (L_px/ɣ)/Cos(θ)

Where θ is the angle the sticks makes with the x-axis in the S frame.

 

[forty]

uy = (u'y)/(ɣ(1 + u'yv/c2)) uy = (u'y)/(ɣ(1 + u'xv/c2))

fixed!

Thanks for all the help!

 

[Doc Al]

Ahhhh always confuse the proper length... So the proper length is the length as measured by an observer at rest with the object, so Lp is the length in S'

So If the stick makes an angle θ' with the x' axis in the S' frame then its x component of it's length is lcos(θ') = L_px

ɣ=(1 - v2/c2)^-1/2

so in the S frame this length becomes L_px/ɣ

So the length of the stick in the S frame is (L_px/ɣ)/Cos(θ)

Where θ is the angle the sticks makes with the x-axis in the S frame.

OK.

But generally you'd want to express the answer in terms of the original S' variables, something like L = Lp*f(θ', v). (At least that's what I would expect if I asked for the length measured in the S frame.)

 

[forty]

Yes I agree, was just trying to save writing it all out. If i knew how to do the fancy code I would ;)

Thanks Doc Al, always so helpful!

 

[maddawg]

hey doc how did you ge thte angle in

Cos(θ) = (Cos(θ') + (v/c))/(1 + Cos(θ')*(v/c))
θ = Cos-1((Cos(θ') + (v/c))/(1 + Cos(θ')*(v/c)))

?

 

[Doc Al]

hey doc how did you ge thte angle in

Cos(θ) = (Cos(θ') + (v/c))/(1 + Cos(θ')*(v/c))
θ = Cos-1((Cos(θ') + (v/c))/(1 + Cos(θ')*(v/c)))

?

It's a variant of the aberration formula. One way to derive it is to imagine a beam of light parallel to the stick, then use the addition of velocities formula to compute the horizontal component of the light velocity in the S frame. (The horizontal component is c*Cosθ)