Time Elapsed between Light Emission and Absorption at Different Frames

[byerly100]

A flash of light is emitted at point O and is later reabsorbed at point P. In frame S, the line OP has a length l and makes an angle theta with the x axis. In a frame S' moving relative to S with a constant velocity v along the x axis:
How much time tau' elapses between emission and absorption of the light?

The answer is (1-betacos(theta))gamma*l/c, where beta=v/c

How does one get this answer?

 

[OlderDan]

It took some effort, but I managed to do this. I used length contraction in the x direction to figure out the coordinates of P in S' when the pulse is emitet. Since S' is moving relative to points O and P in S, the pulse receiver is moving through S' toward the y' axis a distance v*tau' as the light pulse is moving a distance c*tau'. It's just vector addition to find the distance the pulse travels to the receiver. You get a quadratic equation for tau' that is most easily handled in terms of the combination c*tau'/gamma. Solve the quadratic and simplify and you've got it.

 

[byerly100]

Could you give some more detail on "length contraction in the x direction to figure out the coordinates of P in S' when the pulse is" emitted?

[tex]l= l_p/gamma[/tex]

Also,
(c*tau')^2=(v*tau')^2+(x)^2 ?...

 

[OlderDan]

Could you give some more detail on "length contraction in the x direction to figure out the coordinates of P in S' when the pulse is" emitted?

If O is the origin, then the coordinates of P in S are
x = l*cos(theta)
y = l*sin(theta)

Let the origins of S and S' coincide when the pulse is emitted. Then the coordinates of P in S' are found by contracting in the direction of motion

x' = x/gamma
y' = y

You could use Pythagoras to find the distance from O to P in S', but the light pulse will not go that far because P is moving through S' with speed v. You need to figure out how far it is from the origin to where the light pulse intercepts P. That will be at
(x' - v*tau', y') = (l*cos(theta)/gamma - v*tau', l*sin(theta))
Use Pythagoras to find the distance from O this point and set it equal to c*tau'

 

[byerly100]

[(l*cos(theta)/gamma - v*tau')^2 + (l*sin(theta))^2 = d^2 = (c*tau')^2
[(l*cos(theta)/gamma - v*tau')^2 + (l*sin(theta))^2]^(1/2)/c=tau'

"most easily handled in terms of the combination c*tau'/gamma"

 

[OlderDan]

Do I need to solve this equation for something, like x', y'?

(x', y') = (l*cos(theta)/gamma - v*tau' + v*tau', l*sin(theta)) ?
(x', y') = (l*cos(theta)/gamma, l*sin(theta)) ?

(l*cos(theta)/gamma)^2+ (l*sin(theta))^2 = d^2 = (c*tau')^2 ?

[(l*cos(theta)/gamma)^2+ (l*sin(theta))^2]^(1/2)/[c]= tau'

That must not be right since v is in the answer.

"You get a quadratic equation for tau' that is most easily handled in terms of the combination c*tau'/gamma. Solve the quadratic and simplify and you've got it."

---

(l*cos(theta)/gamma - v*tau')^2 + (l*sin(theta))^2 = d^2 = (c*tau')^2 ?
[(l*cos(theta)/gamma - v*tau')^2 + (l*sin(theta))^2]^(1/2)/c=tau' ?

The blue is all you need, and then some algebra skills and one well known trig identity.

 

[OlderDan]

http://www.mathwizz.com/algebra/help/help32.htm
http://en.wikipedia.org/wiki/Trigonometric_identity

Somehow the sin is eliminated...

cos^2(a) + sin^2(a) = 1?

cos^2(a) = 1 - sin^2(a) ? (maybe not)

http://en.wikipedia.org/wiki/Trigonometric_identity#Power-reduction_formulae ?

(The answer is (1-(v/c)cos(theta))gamma*l/c).

What kind of algebra skills?

Should I FOIL (expand) (l*cos(theta)/gamma - v*tau')^2 ?
l^2cos^2(theta)/gamma^2 - 2v*tau'*l*cos(theta)/gamma + v^2tau'^2 ?

whole equation =
l^2cos^2(theta)/gamma^2 - 2v*tau'*l*cos(theta)/gamma + v^2tau'^2 +
l^2sin^2(theta) ?

"most easily handled in terms of the combination c*tau'/gamma"




*I need to have this finished tomorrow so I should finish it tonight.

The sin^2 = 1-cos^2 is the first step after squaring terms. Combine the cos^2 terms and you will have a factor of 1-1/gamma^2 which is beta^2. Replace any v with beta*c and you should have something in the form of a quadratic
0 = x^2 + bx + c where b = 2*l*beta*cos(theta), c = l^2[1-beta^2cos(theta)^2] and x = c*tau'/gamma.
Use the quadratic formula. The beta*cos(theta) contributions to the discriminant cancel leaving you with l*sqrt(1). Only the + root is kept. Solve for tau' and your done.

 

[OlderDan]

Did "squaring" include FOILing or just having a ^2?

Yes. FOIL it

 

[OlderDan]

[l^2cos^2(theta)/gamma^2 - 2v*tau'*l*cos(theta)/gamma + v^2tau'^2 + l^2sin^2(theta)]^(1/2)/c= tau'

sin^2 = 1-cos^2

square the whole thing to get rid of the ^(1/2) and combine the v^2tau'^2with the c^2tau'^2 into a term with a gamma^2
use the sin^2 = 1-cos^2 and combine the cos^2 terms

 

[OlderDan]

[l^2cos^2(theta)/gamma^2 - 2v*tau'*l*cos(theta)/gamma + v^2tau'^2 + l^2sin^2(theta)]/c^2=tau'^2

[l^2cos^2(theta)/gamma^2 - 2v*tau'*l*cos(theta)/gamma + v^2tau'^2 + l^2sin^2(theta)]=c^2tau'^2

Keep going :!)
[l^2cos^2(theta)/gamma^2 - 2v*tau'*l*cos(theta)/gamma + l^2sin^2(theta)]=c^2tau'^2 - v^2tau'^2 = c^2tau'^2/gamma^2

 

[OlderDan]

sin^2 = 1-cos^2

[l^2cos^2(theta)/gamma^2 - 2v*tau'*l*cos(theta)/gamma + l^2{1-cos^2(theta)}]= c^2tau'^2/gamma^2


I'm not sure how you said c^2tau'^2 - v^2tau'^2 = c^2tau'^2/gamma^2.

I think I'm supposed to combine the cos^2 terms now.

l^2cos^2(theta) - 2v*tau'*l*cos(theta)gamma + [l^2{1-cos^2(theta)}]gamma= c^2tau'^2

c^2tau'^2 - v^2tau'^2 = (c^2 - v^2)tau'^2 = c^2(1 - v^2/c^2)tau'^2= c^2tau'^2/gamma^2.

Replace that v in the linear term with beta*c

 

[byerly100]

l^2cos^2(theta) - 2beta*c*tau'*l*cos(theta)gamma + [l^2{1-cos^2(theta)}]gamma= c^2tau'^2

l^2cos^2(theta) - 2beta*c*tau'*l*cos(theta)gamma + [{l^2-l^2cos^2(theta)}]gamma= c^2tau'^2

What do I need to do to get the below?
At some point, I need to combine the cos^2 terms.

"quadratic
0 = x^2 + bx + c where b = 2*l*beta*cos(theta), c = l^2[1-beta^2cos(theta)^2] and x = c*tau'/gamma. (-> 0= (c*tau'/gamma)^2 + 2*l*beta*cos(theta)(c*tau'/gamma) + l^2[1-beta^2cos(theta)^2] )

a=1?

Use the quadratic formula."

*I still need to get the above but I'm going to insert things into the quadratic formula now.

http://mathworld.wolfram.com/images/equations/QuadraticEquation/equation5.gif

c*tau'/gamma= (-2*l*beta*cos(theta)) +/- [(2*l*beta*cos(theta))^2-4(1){l^2[1-beta^2cos(theta)^2]}]^(1/2)) / (2(1))

You said, "The beta*cos(theta) contributions to the discriminant cancel leaving you with l*sqrt(1). Only the + root is kept. Solve for tau' and your done." ...?

c*tau'/gamma = (-2*l*beta*cos(theta)) + l ?

Again, the answer is tau' = (1-betacos(theta))gamma*l/c

 

[OlderDan]

When the file appears, you can see it all. Check my work!

I can't get this LaTeX stuff to appear either

 

[OlderDan]

At some point, I need to combine the cos^2 terms.

Yes you do. They are both part of the "c" term

 

[byerly100]

I was wondering exactly where you started in the file.
It looks like you started around https://www.physicsforums.com/showpost.php?p=1129304&postcount=6.

 

[OlderDan]

I was wondering exactly where you started in the file.
It looks like you started around https://www.physicsforums.com/showpost.php?p=1129304&postcount=6.

I started by saying the distance from the origin to the position where the pulse is received is c*tau'. The unlabeled diagram (in S') shows this as the red arrow. The black arrow is v*tau'. The black slanted line is the separation of O and P when the pulse leaves O. Its components are l*cos(theta) and l*sin(theta). The radical is the sum of the squares of the components of c*tau'