This week, Amandeep from Toronto wrote in to ask:

According to Einstein’s theory of relativity time slows down as speed of the object increases. What is the rate of change of time? E.g. if time was being measured by a simple clock, can we see the hands of the clock slowing down at a certain rate as a result of increase in speed?

Thanks,

Amandeep

Amandeep,

Great question—and a surprisingly easy one to answer! There's a

*relatively*simple equation that relates dilated time to regular time, and it's something that you'll become very familiar with if you take courses involving relativity: the Lorentz factor.

Here, v is your velocity relative to the frame of reference for "normal time". |

For our readers who are unfamiliar with the concept of "time dilation", it's a prediction of special relativity which says, as Amandeep's question mentioned, that time passes slower for you the faster you're moving. It sounds more like science fiction than science fact, that you could "travel to the future" by making a round-trip at near lightspeed, but it's been tested repeatedly at small scales and proven real every time. (It is, however, used as a plot device in such sci-fi classics as

*The Left Hand of Darkness*and

*Ender's Game—*both of which I highly recommend.)

The prediction inevitably results from the principle that

*the speed of light is constant to all observers.*This is the cornerstone of special relativity, and it's a slight departure from our intuitive understanding of the world. When you throw a tennis ball forward at 20 mph while in a car moving 40 mph, the ball will appear to move 60 mph to someone on the side of the road. Light, however, will always appear to move at

*c*—2.99*10

^{8}meters per second—whether you're in the car or on the side of the road. The only difference is that the person on the side of the road would see light from the car as being higher in energy, or blueshifted. This is a lot like how the speed of sound is constant in air: the sound from an ambulance's siren

*doesn't*reach you faster if the ambulance is moving toward you, but it

*does*seem higher in pitch.

But time dilation is another story. Imagine you're on a spaceship passing by Earth—a futuristic one with a big bay window to let you look out at planets and stars as you whiz by—and you've got a mirror on both the floor and the ceiling, perfectly parallel to each other. (The classic thought-experiment used to demonstrate this idea involves mirrors on trains, but why not give it a bit of a 21st-century upgrade?)

Now let's say you shine a light upward, toward the mirror on the ceiling, and quickly pull your hand out of the way so that some of the photons are stuck bouncing between the two mirrors. If you measured the speed of this light, you'd find it to be the standard

*c*(with minor adjustments for the air in your ship).

But as your ship cruises by Earth at a high speed, someone on the ground looks up through a telescope and sees into your ship through the window. From their perspective, the light between those mirrors is not bouncing straight up and down, the way it is in your frame of reference—it's also moving with the ship, tracing out a zigzag path.

From the perspective of someone on the ship (left), the light is moving straight up and down. To an observer on the ground, though (right), it covers a greater distance.Image Credit: Sacamol, via Wikimedia Commons (CC BY-SA 4.0) |

Puzzling over this, Einstein realized there was only one way to resolve this paradox while maintaining the constancy of the light's speed: to warp

*time itself*for the person on the ship, so that light takes longer to move between the mirrors. The spacefarer will still measure the light as moving up and down at*c,*but this is because his clock is ticking slower than the clock of the person on the ground.
So, to finally answer Amandeep's question: how

*much*slower? If you take a look at the diagram above, you can see why we brought up Pythagoras earlier: the path taken by light aboard the ship is the hypotenuse of a right triangle. The distance between the mirrors forms one leg of the triangle, and the distance covered by the ship in the*time between bounces*forms the other leg. So, if we recall the Pythagorean theorem: A^{2}+B^{2}=C^{2}, the length of the hypotenuse*D*in the diagram above can be calculated with:
Since L and D both depend on the distance between your mirrors, finding a general equation to relate the spacefarer's time to the ground-observer's time means a little bit of substitution and rearrangement, so the equations only involve the time between bounces—represented by (∆t) and (∆t') in the diagrams above for the observer on the ship and ground, respectively.

Substituting, rearranging, and simplifying this equation gets us to the equation:

...which you'll recognize as the basis of the Lorentz factor we saw at the beginning! So if our ship is moving at 0.95c, doing the math tells us that one second of ship time (∆t) is equivalent to about 3.2 seconds of time on the ground (∆t')!

There's lots more fun weirdness in special relativity—reciprocity and the Twin "Paradox" are particularly interesting—but that's all we have time for today.

Thanks for reading, and remember to submit your questions on our "Ask a Physicist" page!

—

**Stephen Skolnick**
You can do it through trig too: just use cos(arcsin(v)) using speed units where c=1. E.g. 0.866c is the speed that gives you length contraction to half the rest length and makes clocks tick at half their rest rate. For a full understanding of this, see http://www.magicschoolbook.com/science/relativity.html

ReplyDelete"Time passes slower for you the faster you move" is not a correct statement because all inertial frames are equivalent. If you are in an inertial frame (not accelerating), everything in that frame, that is, moving at the same velocity as you, will experience the same time. Time as observed by you in any frame moving relative to yours will appear slower as you measure it. Your own inertial frame is always at zero velocity. All other frames will be moving away from you with velocities pointing outward around a sphere, and you will see the time dilation as described in this article for them, for any mirror placed along the perpendicular from the point of measurement in your frame to the relative velocity vector in the other frame. Only through acceleration of your own frame can you break the symmetry. A more correct statement would be: "You will observe more time dilation between your own frame and and any other frame if you accelerate away from it." Consequently you will observe less time dilation on the frames you accelerate towards, that is, frames for which you reduce your relative velocity.

ReplyDeleteThis is an explanation, but a graph at the end would have gone some way toward answering the question.

ReplyDeleteAlso, a small quibble; the train's pitch would "seem" higher because it would actually be higher.