I think I'm just confused, but for some reason I thought that light moving straight in one frame would have to move in the same direction in another frame. I know there are photons-but because I have not learned about them I was thinking purely from the wave model. If light has no mass, shouldn't a disturbance sent in one direction, regardless of perpendicular speed, move in that direction in all frames? In particular I had a homework problem that was asking how the Michaelson Morley experiment's null effects could be accounted for with Length contraction. They used the path of light in the vertical direction (Not a straight line.) Clearly the math works out this way, but I'm not sure why the vertically moving light would have a horizontal velocity just because the frame it was shot from had that velocity. Why does light have this inertia? Can someone explain it via wave theory?
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$\begingroup$ Ask yourself how vectors transform with changes in reference frame. $\endgroup$– WillCommented Jul 18, 2013 at 19:03
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1$\begingroup$ Isn't there something strange about this though? Why does light always go straight in a moving frame? While it would be weird if it didn't, I don't see what's allowing it to if it doesn't have some kind of 'inertia'. Light always moves as you would expect it to move from its source. $\endgroup$– user24082Commented Jul 19, 2013 at 2:11
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$\begingroup$ What exactly do you think a change in reference frame (from one inertial frame to another) does? You aren't touching the things in your spacetime. Lorentz transformations do no change the observed straightness of light's path. Read the answer by @joshphysics carefully and think about it. $\endgroup$– WillCommented Jul 19, 2013 at 3:31
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$\begingroup$ I just don't understand how, if I shoot a laser straight ahead of me, it could ever appear to be coming diagonally out of the laser to someone else. I feel like light would be more prone to some kind of equivalent drag. $\endgroup$– user24082Commented Jul 19, 2013 at 3:35
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$\begingroup$ What if I was to look in a rotated frame? Let's say you are pointing your laser ahead of you in your $x$ direction. But my frame is rotated 45 degrees w.r.t your frame, so I see it moving at 45 degrees w.r.t my $x$ axis. Does this help? Or are you only confused by frames moving w.r.t each other? $\endgroup$– WillCommented Jul 19, 2013 at 3:38
3 Answers
It's not that light has any inertia, it's simply that when you view the path of something from a different frame, the path can look different. Light's having no mass does not affect this qualitative fact. The masslessness simply means that its speed is frame-invariant, not its direction of travel.
In particular, if for example, a light ray is traveling in the $y$-direction in one frame, then it will look as though its path is angled towards the negative $x$-axis in a frame boosted along the positive $x$ axis.
Mathematically, consider a light ray along the $y$-axis described by the following path: \begin{align} t(\lambda) &= \lambda, \qquad x(\lambda) = 0, \qquad y(\lambda) = c\lambda \qquad z(\lambda) = 0 \end{align} The Lorentz transformation tells us that in a frame boosted in the $x$-direction, we have the following relationships: \begin{align} t' &= \gamma\left(t-\frac{v}{c^2} x\right), \qquad x' = \gamma(x-vt), \qquad y'=y, \qquad z'=z \end{align} so the path of the light ray in the new frame is \begin{align} t'(\lambda) = \gamma\lambda, \qquad x'(\lambda) = -v\gamma\lambda, \qquad y'(\lambda) = c\lambda, \qquad z'(\lambda) = 0 \end{align} Notice that the path of the ray has picked up a negative $x$ compoenent! This is purely because we are viewing the light ray in a different way. It's like if you were to view your computer monitor while sitting at your desk, it looks like its standing still, but you can make it look like its moving in any direction you please by yourself moving in an appropriate direction.
Notice, however, that in both frames, the path of the light ray satisfies the appropriate nullness condition; \begin{align} -c^2\dot t(\lambda)^2 +\dot{\mathbf x}(\lambda)^2 = 0 \end{align} where dots denote derivatives with respect to the parameter $\lambda$ which indicates that the speed of the light ray is invariant.
(source: virginia.edu)
In both reference frames, there is an event where the light strikes the center of the top mirror.
In the frame in which the mirrors are at rest, the light has a vertical path.
Is it not clear that, if the mirrors are moving in a reference frame, the light path cannot be vertical?
The confusion you have experienced appears to be common, judging by the number of related questions that have been posted on this site and others. The answer has nothing to do with light per se, and is simply the result of the fact that motion, and the direction of motion, are frame dependent.
If you stand at a corner of a crossing between two streets, one going north and south, the other east and west, you are motionless in your frame. Relative to anyone driving along the streets, you are moving. The direction in which you are moving depends upon the direction the driver is taking. To a driver heading north, you are moving south in their frame; to a driver heading east, you are moving west in their frame.
The effect becomes compounded if you move too. If you walk north up one road, while I walk west on the other, you will be moving north east relative to me, and I will be moving south west relative to you.
This effect applies generally- the speed and direction of all movement is relative.
If you roll a ball directly across the platform at a station, its path is at 90 degrees to the railway track. To someone sitting on a passing train, the path of the ball is not at 90 degrees but is angled in their frame as a result of their relative motion. The same is true of light- if you are on the platform and shine a laser beam at 90 degrees to the track, the light will move at a different angle in the frame of someone passing in the train. Since ordinary train speeds are negligible compared with the speed of light, the angle might be something like 89.9999999999999999999 degrees, say, but nonetheless it is still different from the angle in the frame of the platform.
