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A question on photons

by bucher
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bucher
#1
Aug5-09, 10:43 AM
P: 77
This isn't homework but I need to ask my question with an example.

Say you're out in space and you have a laser pointer. You point the laser pointer straight out in front of you and turn it on. You're able to leave the laser pointer on without holding onto it, so the laser pointer is just floating out in front of you. Now, say, you decide to push the laser pointer down. You push the laser pointer and it begins to float downwards (still pointing straight out in front of you).

My question is do the photons of the moving laser pointer change their trajectory by moving down as well as out, or do they still just move straight out? Likewise, if the laser pointer were to spin, would the photons spiral around the pointer while moving out?
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jtbell
#2
Aug5-09, 11:24 AM
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Each individual photon follows a straight-line path. but in the situations you describe, different photons follow different paths. If you take a "shapshot" of the positions of all the photons at a certain time, in your first case the photons lie along a straight line that does not line up with the pointer; in the second case the photons lie along a curved line.
bucher
#3
Aug5-09, 12:47 PM
P: 77
Thanks. I'm still trying to get the duality of light concept and wasn't sure how a photon would react in these situations.

humanino
#4
Aug5-09, 12:51 PM
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A question on photons

How is it different from playing with a water hose in your garden ?
(at least with enough pressure so that one can neglect Earth gravity over a significant length)
bucher
#5
Aug5-09, 01:00 PM
P: 77
Yes, though the affects are insignificant with the hose, the affects are still there (small, but there). I didn't know if the trajectory of the laser pointer affected the trajectory of the photon no matter how small.
Nabeshin
#6
Aug5-09, 01:43 PM
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You will get hit with the laser pointer. Better wear hockey pads.
bucher
#7
Aug5-09, 02:32 PM
P: 77
Let me rephrase my example.

I'm out in space with an emitter that fires one photon at a time. I get the emitter to point away from me and move downward at a constant speed of 10 m/s. When the emitter is going at this constant speed it fires a single photon. Will the photon have a downward velocity of 10 m/s (along with going straight out) like the emitter?
scupydog
#8
Aug5-09, 02:37 PM
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I like your question.
jtbell
#9
Aug5-09, 06:14 PM
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Quote Quote by bucher View Post
Will the photon have a downward velocity of 10 m/s (along with going straight out) like the emitter?
For all practical purposes, yes. If what you're after is the relativistic answer which keeps the speed of light constant, I applied the matrix form of the Lorentz transformation to the photon's energy-momentum four-vector, and ended up with a downward velocity component of 9.9999999999999944 m/s. The horizontal velocity component is very slightly less than c, by the right amount to make the magnitude exactly equal to c.
bucher
#10
Aug5-09, 08:44 PM
P: 77
Thanks, this helps me out a lot.
jtbell
#11
Aug5-09, 11:14 PM
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Oops, I'm sorry. I went back over my calculation again and realized that I'd made a small algebra mistake. The downward component of the velocity of the light from the moving laser pointer is exactly equal to the downward velocity of the pointer.

The horizontal and vertical components of the velocity of the light coming from the pointer:

[itex]u_x = (1 - 5.5556 \times 10^{-16}) c [/itex]

[itex]u_y = \beta c = (3.3333 \times 10^{-8}) c [/itex] = 10 m/s

which gives the magnitude (speed)

[tex]u = \sqrt {u_x^2 + u_y^2} = c [/tex]

apart from a smidgen of roundoff error in the last decimal place.
bucher
#12
Aug5-09, 11:30 PM
P: 77
thanks again, this makes a bit more sense to me.
jtbell
#13
Aug6-09, 08:37 AM
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Just to make sure, here are the key steps in my calculation, so you or someone else can check them if you like:

In the rest frame of the laser pointer, the photon travels horizontally. Assuming it has energy E, its four-momentum is

[tex]Pc = \left( \begin{array}{cccc} E \\ p_x c \\ p_y c \\ p_z c
\end{array} \right) = \left( \begin{array}{cccc} E \\ p c \\ 0 \\ 0 \end{array} \right) = \left( \begin{array}{cccc} E \\ E \\ 0 \\ 0 \end{array} \right)[/tex]

For a photon, E = pc. I'm used to dealing with energy and momentum in energy-units, so I use Pc instead of plain P for the four-vector.

The laser pointer is moving downwards in your rest frame, so the relative velocity v of your rest frame w.r.t. the laser pointer rest frame is upward.

[tex]\beta_x = 0[/itex]

[tex]\beta_y = \beta = v / c[/tex]

[tex]\beta_z = 0[/tex]

Substituting these into the Lorentz transformation matrix on the Wikipedia page that I linked to, I got the following:

[tex]\left( \begin{array}{cccc}
\gamma & 0 & - \beta \gamma & 0 \\
0 & 1 & 0 & 0 \\
- \beta \gamma & 0 & \gamma & 0 \\
0 & 0 & 0 & 1
\end{array} \right)[/tex]

Multiplying this matrix by the four-momentum of the photon in the laser pointer rest frame, I get the four-momentum in your rest frame:

[tex]P^{\prime} c = \left( \begin{array}{cccc}
E^{\prime} \\ p_x^{\prime} c \\ p_y^{\prime} c \\ p_z^{\prime} c
\end{array} \right) = \left( \begin{array}{cccc}
\gamma E \\ E \\ - \beta \gamma E \\ 0
\end{array} \right)[/tex]

The x-component of the photon's momentum is unchanged, and there is now a downwards y-component as we expect.

The photon's velocity (in your rest frame) [itex]{\vec u}^{\prime}[/itex] is in the same direction as its momentum [itex]{\vec p}^{\prime}[/itex], so for the x-components we have the following proportionality:

[tex]\frac {u_x^{\prime}} {u^{\prime}} = \frac {p_x^{\prime} c} {p^{\prime} c} [/tex]

The magnitude of the velocity, [itex]u^{\prime}[/itex], is of course c, and [itex]p^{\prime} c = E^{\prime}[/itex], so

[tex]\frac {u_x^{\prime}} {c} = \frac {p_x^{\prime} c} {p^{\prime} c} = \frac {p_x^{\prime} c} {E^{\prime}} = \frac {E} {\gamma E} = \frac {1} {\gamma}[/tex]

[tex]u_x = \frac {c} {\gamma} = c \sqrt {1 - v^2 / c^2}[/tex]

Similarly for the y-component of the photon's velocity in your frame:

[tex]u_y = - \beta c = v[/tex]

In your example v = 10 m/s, for which [itex]\beta = 3.3333 \times 10^{-8}[/tex] and [itex]\gamma = 1 + 5.5556 \times 10^{-16}[/tex]. The results are easier to work with numerically if you use a larger v, say v = 0.5c.
Austin0
#14
Aug9-09, 07:41 AM
P: 1,162
Quote Quote by jtbell View Post
For all practical purposes, yes. If what you're after is the relativistic answer which keeps the speed of light constant, I applied the matrix form of the Lorentz transformation to the photon's energy-momentum four-vector, and ended up with a downward velocity component of 9.9999999999999944 m/s. The horizontal velocity component is very slightly less than c, by the right amount to make the magnitude exactly equal to c.
Is this actually a downward velocity component or an angle deviation during emission??

Would this be due to the finite duration of emission time and the movement of the emitting electron during the process?

It is correct to assume isn't it, that movement of the lazer forward or backward along the line of emission would not have any effect on the photons four-vector??
WOuld this mean that to get a completely transverse emission and travel vector you would have to aim the lazer slightly backward in an inertial frame??
Thanks
Austin0
#15
Aug9-09, 08:09 AM
P: 1,162
Quote Quote by jtbell View Post
For all practical purposes, yes. If what you're after is the relativistic answer which keeps the speed of light constant, I applied the matrix form of the Lorentz transformation to the photon's energy-momentum four-vector, and ended up with a downward velocity component of 9.9999999999999944 m/s. The horizontal velocity component is very slightly less than c, by the right amount to make the magnitude exactly equal to c.
A very good question indeed. The more I have thought about it the more puzzling it seems. This would seem to conflict with the basic; light being independant of the motion of the source. It seems to imply a kind of Newtonian inheritance of momentum principle.

Also on reflection I have realized that with the emission time/angle hypothesis you would actually need to aim the lazer slightly forward, not backward , to get a perfectly transverse emission. And in the falling lazer situation the path would be slightly upward.
Thanks
jtbell
#16
Aug9-09, 09:23 AM
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Did you notice I corrected my answer two posts further on? The vertical component isn't 9.9999999999999944 m/s, it's 10 m/s exactly.

And the speed of the light is independent of the motion of the source in this example. Using [itex]u_x[/itex] and [itex]u_y[/itex] from post #13 you can easily show that

[tex]u = \sqrt {u_x^2 + u_y^2} = c[/tex]
Austin0
#17
Aug10-09, 08:32 AM
P: 1,162
Quote Quote by jtbell View Post
Did you notice I corrected my answer two posts further on? The vertical component isn't 9.9999999999999944 m/s, it's 10 m/s exactly.

And the speed of the light is independent of the motion of the source in this example. Using [itex]u_x[/itex] and [itex]u_y[/itex] from post #13 you can easily show that

[tex]u = \sqrt {u_x^2 + u_y^2} = c[/tex]
I understood that it wasn't speed. But am I correct in looking at it as a change of direction???
And that this had to have taken place during emission?

Thanks
jtbell
#18
Aug10-09, 09:07 AM
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The speed of a photon does not depend on the motion of the source. Its direction of motion, on the other hand, does depend on the motion of the source, in general.


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