# Constant velocity motion should be undetected

In special relativity, the constant velocity motion cannot be detected by any experiment. Even when light is used we end up with phenomenon of length contraction and time dilation but that is the case when we shine a beam of light in the direction of motion.

What happens if we were to shine a beam of light (a laser) within the spaceship perpendicular to the line of constant velocity motion, from wall to wall of the space ship?

If there was acceleration the light would bend BUT due to constant velocity motion, velocity being close to that of light, shouldn't the laser beam seem to emerge at an angle so that it is a straight line but deflected at an angle.

We could setup the lasers on one wall and align the detectors on opposite wall while the spaceship is at absolute rest. Then we can accelerate the ship close to speed of light and reach a steady speed. Then we shine the laser beams, would they still hit the same detectors on the opposite wall just like they did when the ship was at rest or will they be offset by an angle due to the motion of the ship close to speed of light?

If the laser should hit the same detectors as when it was stationary then how come we consider bending of light in accelerated motion.

By that analogy the light should seem to emerge at an angle in case of constant velocity motion.

And would therefore make it possible to detect constant velocity motions, with velocities closer to speed of light??

• "shouldn't the laser beam seem to emerge at an angle?" According to whom should light behave in that way? According to what theory? Experiments have shown that light actually does not behave in that way, and our our theories are designed to accommodate that observed behavior. Commented Aug 1 at 14:17
• It seems to me that the title of this question would better be: "Shouldn't constant velocity motion be detectable?" because this is what you actually ask. Commented Aug 1 at 14:41
• Thing is, if you do the experiment, you are never moving ;) Commented Aug 1 at 14:43
• The proposed experiment here is exactly the 1632 thought experiment known as Galileo's ship, only with a very high speed. This thought experiment is precisely meant to show that one cannot detect an uniform straight motion. Do you have any conceptual hint making you think that things should be different when the ship speed gets close to that of light? Commented Aug 1 at 15:09
• A reference frame is a system of coordinates, not a box. An experiment is performed in spacetime, not "in" a reference frame. It may be analyzed in any reference frame. Frequently it's most convenient to do this analysis in a frame (ie in coordinates) where some elements of the experiment are at rest. Commented Aug 1 at 17:47

Suppose you set up your spaceship and are floating in space. Another similar ship comes flying by at constant velocity. Which one of you is at absolute rest?

You are both free to consider yourselves at rest in your favorite frame of reference. You both consider the other guy to be moving at constant velocity. Both of you see light bouncing back and forth as you expect at rest. The other guy flying by does not change that.

Each of you sees his own laser bouncing straight sideways, just like a ball would. Each of you sees the other beam straight sideways too. Each of you sees each photon in the other beam bouncing in a zigzag path like a ball would.

The other ship is flying at velocity $$v$$ near the speed of light in your frame of reference. It is at rest in his own.

In his frame a pulse of light leaves the horizontal laser, flies horizontally across the ship, and hits a mirror. The pulse travels at the speed of light in his frame. The length of the path is $$w$$, the width of the ship. The time of flight is $$w/c$$.

In your frame the laser and mirror have a large velocity. The laser is pointed horizontally at the mirror. The pulse leaves the laser and stays even with it as it flies across the ship to hit the mirror. The pulse travels at the speed of light in your frame.

The width of the ship is the same in both frames. In your frame the pulse is traveling diagonally, so length of the path $$\sqrt{w^2 + (vt)^2}$$. The time of flight is $$\sqrt{w^2 + (vt)^2}/c$$.

So you and he measure two different path lengths and two different times of flight for the same trip. If your clocks read the same time at the start of the flight, the will read different times at the end. In your frame, his clock runs slower than yours. If you check the times on his clock at the start and end, you will see that he measures just the right time of flight for the speed of the pulse to be $$c$$ as it travels a distance he measures as $$w$$.

This is the single hardest part of special relativity. We are used to two people measuring two different distances for the same path. We are not at all used two people measuring two different time intervals as an object traverses this path.

All of the strange things about special relativity follow from this property of time. Time and space are much more alike than classical physics leads you to believe. This is the reason we combine space and time into spacetime. It is a lot to get used to.

• Comparing the spaceship at the start to the spaceship close to the speed of light is like comparing you to the other guy. Think of a ball bouncing back and forth in a tube. You don't need to point the tube at an angle to keep the ball bouncing correctly when the ship has reached its final velocity. Same for a laser. Commented Aug 1 at 15:24
• But that is because ball has a moment/inertia. Does light has that as well? Sorry for asking such a silly thing. This is way beyond my academics at the moment. Commented Aug 1 at 16:00
• Not a silly question. Yes, light does have momentum, even though it has no mass. How can a red light photon be different from a blue light photon? Commented Aug 1 at 16:15
• So is it like the spaceship imparts moment to laser photon (in the direction of motion of spaceship, just like in case of a ball)? and in turn the photon moves with reduced perpendicular velocity (because of limiting value of c) causing time dilation? More the speed of the spaceship more the time dilation there will be. Commented Aug 1 at 16:30
• Please do tell me if my interpretation is correct or not. You don't need to explain if it's horribly wrong. Commented Aug 1 at 17:06

Constant motion is relative, but acceleration is absolute. If you are in a airplane moving with constant velocity relative to the Earth, you feel no force on you, and you can move around and manipulate things just as though the plane is at rest. If the airplane is accelerating though, e.g during takeoff, you're pushed back in your seat -- you can feel the acceleration.

That's why the light is apparently curved in an accelerating rocket. As viewed from outside the rocket the light is going in a straight line, but the walls of the rocket are speeding up and so relative to those walls the light "appears" bent, i.e. appears subject to a force.

The distinction between accelerated motion and constant motion is fundamental not only in relativity but also in Newtonian mechanics.

• This explains the reason for bending of light relative to walls of spaceship. But why does the light go straight in uniform motion? Is it the momentum/inertia? Commented Aug 1 at 18:08
• Yes. Light has momentum, just like ordinary matter. It always travels at the maximum speed possible (c) but its direction is frame dependent. Commented Aug 1 at 19:04
• In uniform motion, the laser would appear diagonal to outside person. So why in case of accelerated frame should the light appear straight to the outside observer? The source of light is moving with the spaceship with some instantaneous velocity at any given moment. I understand it would appear bent from within the spaceship but to outside observer shouldn't it look somewhat diagonal (like in constant velocity motion) except probably with increasing slope I suppose as the instantaneous velocity increases? Commented Aug 2 at 5:34
• Once the light is emitted it is no longer connected to the frame of the space ship. So there are no longer any forces acting on it and in accordance with Newton's first law it keeps moving in a straight line at constant speed. From the outside observer's perspective this may be a diagonal straight line, yes. Commented Aug 2 at 7:26
• Ok I get it, it's a diagonal straight line not a straight straight line. Thanks a lot Commented Aug 2 at 7:33

Let's say Adam has a light clock. It has a light source on the floor and a small mirror on the ceiling. When the clock is switched on, he sees photons going straight up to the mirror and reflected straight back down. Eve also has a similar light clock and when Eve is moving horizontally relative to Adam, the photon in Eve's clock appears to be following a diagonal path between reflections, from Adam's point of view. However, from Eve's point of view the light in her clock is going straight up and down. Neither observer sees light in their own clock as moving diagonally in their own reference frame. Since both observers see the photons in their own clocks as moving straight up and down they see exactly the same thing and there is no way to determine which observer is 'really' moving.

How and What is making the light go diagonally in whoever's actually moving reference frame such that it appears vertical to them?

For convenience, I am going to swap horizontal and vertical for this next example. Imagine we has a toy battery powered car on the bottom of an elevator. it is designed to turn around every time it hit a elevator wall and head off to the opposite elevator wall at constant speed. At all times the toy car remains on the elevator floor and to a person inside the elevator the toy car is always moving in straight line horizontally. Now we film the whole thing from a video camera that is being lowered at constant velocity, so that when we play the video back, the elevator appears to be moving upward relative to the camera. In the video, the toy car appears to following a zig zag path, even though the motion of the video camera does not change anything physically inside the elevator. This zig zag path is an illusion created by the motion of the video camera and all this works the same using Newtonian physics. Now raise the elevator at a fixed velocity and place the video camera some distance away on a tripod fixed to the ground such that we film the motion of the elevator from ground to top without having to pan the camera. When we play back the recording of the moving elevator, the toy car appears to follow a zig zag path, but to the observer inside the elevator, the toy car remains on the floor of the elevator at all times and to that internal observer, the car is just moving horizontally.

Would light bend If the were to be shone into a elevator accelerating perpendicular to the path of light?? My whole question is based on this fact which I saw online. – user145522

Your whole question is based on a scenario that depends on the acceleration of the elevator, yet you constantly refer to the elevator moving with constant speed. Imagine we have gun mounted on one side of the elevator and it fires a constant stream of bullets to a target fixed on the opposite wall of the elevator. As long as the elevator has constant upward velocity, the bullets keep hitting the centre of the target, what the velocity without any adjustment to the angle of the gun. The bullets have an upward velocity component equal to the upward velocity component of the elevator, due to their initial momentum, so they never miss the target. Now imagine the elevator is initially stationary and we fir a single bullet at the target and after the bullet has left the gun, we suddenly accelerate the elevator upwards. The bullet maintains its horizontal velocity and is not aware of the change mid flight and as a consequence it misses the target and hits somewhere lower. As long as the target has the same upward velocity as the gun had the bullet was fired, the bullet hits the target, but if the target is accelerating upward it will miss. The difference between the accelerating and constant velocity elevator cases is very important. It is the acceleration of the elevator that makes the bullet's path appear to be curved downward. It works the same for photons, except the very high velocity of light makes the deflection hard to detect over short distances. The photon acquires the upward component of the emitting laser's velocity at the moment it leaves the laser similar to how a bullet acquires the vertical component of the gun at the time of leaving the gun barrel.

We could setup the lasers on one wall and align the detectors on opposite wall while the spaceship is at absolute rest. Then we can accelerate the ship close to speed of light and reach a steady speed. Then we shine the laser beams, would they still hit the same detectors on the opposite wall just like they did when the ship was at rest or will they be offset by an angle due to the motion of the ship close to speed of light?

The way you have set up your scenario, you wait until the elevator reaches a constant velocity and then you fire the laser. In that scenario the photon still hits the target because you have removed the acceleration during the flight of the photon. For bending to occur the velocity of the target has to to be different to the velocity of the laser at the time the photon was emitted. This only occurs if the photon is emitted while the elevator is still accelerating.

My question was why does that happen? Is that because the spaceship imparts momentum to the pulse of light/photon just like it would to a ball bouncing? – user145522

Here, you have reached essentially the right conclusion.

Strangely enough, we could still conclude the light path would bend even if we did not know about the existence of time dilation and length contraction. Ironically, Newton could have figured out a light path would curve in a accelerating elevator, because he considered photons to be "corpuscles" literally like little bullets. Imagine a child sitting on a train repeatedly throwing a ball straight up and catching it each time it comes back down. to the child it looks the like the ball is going straight up and straight down just as it would if he was doing the same thing sitting at home. To an observer filming the train going past, the path of the ball appears to following a curvy zig-zag path.

• How and What is making the light go diagonally in whoever's actually moving reference frame such that it appears vertical to them? Commented Aug 2 at 6:09
– KDP
Commented Aug 2 at 10:58
• Thankyou so much. Was momentum of light a known fact when Einstein published special theory of relatively? Commented Aug 2 at 12:11
• I think so. It is pretty much implicit in Einstein's light clock thought experiment. Momentum of a photon is also included in his energy momentum equation where $E = \sqrt{(pc)^2 + (mc^2)^2}$ where for a photon, (pc) is the momentum of the photon which in turn is a function of its frequency.
– KDP
Commented Aug 2 at 12:48

It would be better to do this thought experiment with a phased array antenna (maybe, I've worked with more microwave than laser recently).

You can make it a $$2+1$$ problem if you want, that is, two spatial dimensions: longitudinal and transverse, but I'll do $$1+1$$...just the longitudinal part.

The key is, you can't create this transverse beam with a point source, it has a longitudinal extend $$L$$, so in the initial rest frame, $$S$$, it has a phase:

$$\phi(x, t) = -\omega t = \phi(t)$$

for $$0 \le x \le L$$

(so this is from a cross section of a wave propagating in the $$y$$-direction:

$$A(x, y, t) = e^{i(ky-\omega t)}$$

which has no $$x$$ dependence..., I'm just looking at the transmitter (Tx) at $$y=0$$)

Now here's the thing about relativity thought experiments: we don't need to boost the apparatus, we can just boost the frame to $$S'$$, moving along $$x$$ at $$-\beta$$:

Here we have:

$$t'=\gamma(t+\beta x)$$

and the inverse transform:

$$t= \gamma(t' - \beta x')$$

so

$$\phi(t)=\phi( \gamma(t' - \beta x'))= \gamma\beta\omega -\gamma\omega t'$$

which means:

$$\phi'(x', t') = k'x'-\omega't$$

with

$$\omega' = k'c = \gamma\omega$$

From which we learn 2 things:

1) The frequency is time-dilated (as expected)
2) Moving frames see a phase ramp across the Tx.


Well how do you steer phased-array antennas? With a linear phase ramp.

So in moving frames, the beam is not perpendicular to the antenna, it's tilted to point at the Rx, which is a moving target in moving frames.

Another takeaway is that phase:

$$\phi(\vec x, t) = \vec k\cdot\vec x - \omega t$$

is a Lorentz scalar, and can be written as such manifestly:

$$\phi(x_{\mu}) = k^{\mu}x_{\mu}$$

I'm not sure why discussions of stellar aberrations (which are equivalent to the question) don't address this directly (afaik--it's been a while...the last time I saw this implemented was in flight software for Cassini's radar, and ofc, it was a $$3+1$$ mess in C)..but I digress.

Phase is a scalar because a zero crossing is a zero crossing in all reference frames.

The same is true for lasers, I just find it harder to visualize a phase ramp across the little opening...which make me think: what if you use a collimated Co-60 source? well, then you just Lorentz transform either 4-vector in:

$$p_{\mu}=\hbar k_{\mu}$$

and get the right answer, but I think that is less instructive.

Edit: one more thing I just thought of: I think the Rx gain pattern will also be phase-ramped to be staring at the retarded-time position of the Tx...I mean it has to work like that, right?

I asked a similar question a while back and then came across this: The special theory of relativity angle contradiction

Check out the answer @Hans de Vries posted there. The tilted beam in the reference frame outside of the rocket has to do with the relativity of simultaneity.

• This question accurately describes my doubt but just like In the comments the op asked how does the light know which direction and at what speed is the box moving. Because it seems like in phased array, the phase is finely tweaked to achieve the desired deflection. Is it because the single emitter itself acts as an array of emitters due to such high velocities (close to speed of light)? Commented Aug 2 at 5:07
• That somehow in my imagination just makes a single wave front by that logic and I don't know what that would look like. Commented Aug 2 at 5:16
• And how would that phased array reasoning hold up in accelerated frame? Commented Aug 2 at 5:19
• @user145522 I believe the basic situation is as follows for non-accelerated motion. In the space ship frame, every point on the laser beam is emitted simultaneously. In the external frame they are not. The different emission times result in a tilted wavefront, thus explaining the path of light in the external observer's frame and how the light "knows" which way to bend. Are you ok with that? Commented Aug 2 at 13:51

Consider diagram (a) carefully. In the reference frame of the laser source, the opposite wall is not moving. So in this frame, there is no reason to expect the beam to strike the opposite wall any closer to the floor.

Now consider diagram (b). Let us say the elevator is uniformly accelerated at the constant value of $$a$$ (in the reference frame of an outside observer). From the frame of the laser source (call it $$F1$$) at time zero when the pulse is emitted, the opposite wall is not moving, just as in scenario (a). Now after time elapsed time $$t$$, say the pulse has traveled half the distance to the opposite wall. With respect to the original frame of the laser source $$F1$$, the opposite wall and the laser source itself are moving with velocity $$v=at$$, and furthermore have moved a transverse distance $$at^2/2$$ from their original locations. After an additional time $$t$$, the light pulse strikes the opposite wall on a spot $$at^2$$ closer to the floor than the laser source, and the walls have gained a total of $$2at$$ velocity compared to frame $$F1$$ (the frame in which the pulse was emitted).

So the difference between (a) and (b) is that in (a), everything in the box stays in the same reference frame at all times, while in (b) the box is constantly changing reference frames, and processes that take a finite time to happen will begin and end with the box in different frames.

(Your light path in diagram (a) is what scientists expected to happen when they believed that all light traveled in the frame of the aether, regardless of the speed of the emmitter or detector. The failure to detect any "aether drag," e.g. due to the motion of the Earth, was partly the genesis of Einstein's description of relativity.)

• So this would imply that within the same reference frame (be it a stationary box or a box having uniform motion) everything including THE LIGHT would have the same momentum as that of the system/refrence frame under observation? Commented Aug 2 at 6:23
• No. In your example (a), the box and the light have the same upward velocity (zero) in the frame of the box. If we boost to a frame where the box has an upward velocity $v$, the light will follow a diagonal path, with upward velocity $v$, rightward velocity $\sqrt{c^2-v^2}$ and total speed $c$. Commented Aug 2 at 21:17

It does not depend of the magnitude of your speed, you never will be able to detect your velocity if you do not look outside to an other object you can't even decide in what direction you move. You can measure your velocity relative to an other object, so a spaceship would know its velocity relative to earth for example by the doppler effect of a known radio signal or light.

• The OP knows this, and is asking how to reconcile it with his proposed experiment. There is nothing helpful about repeating what he already knows. Commented Aug 1 at 17:38