Understanding Time Dilation: Explaining the Constant Speed of Light

[rabcarl]

I've read that time and space dilate under acceleration to ensure the constant speed of light. I'm having trouble visualizing a scenario that illustrates this concept. Could someone please explain this to me? Thanks.

 

[Drakkith]

Time dilation and other related affects are a result of different frames of reference. Because the speed of light is invariant, meaning that it's always measured the same by any observer no matter what their speed is relative to something, its a requirement.

Note that the effects are between frames of reference, so you can have a million different frames of reference all moving differently relative to each other and you would have a million different time dilations, length contractions, etc.

Note that by "frame of reference" I refer to an observer moving at a particular velocity relative to another object.

 

[DaveC426913]

Simplistically:

Imagine a guy in a rocket traveling at .99c. He has a flashlight, which he shines out the front window.

He sees the beam traveling at c ahead of his ship. In one second, his light beam has traveled one light second ahead of him.

On Earth, we see a rocket with a guy with a flashlight shining out a window. The light beam is traveling at c, but the rocket is following very close behind it at .99c. At this rate, it will take not 1 second, but 100 seconds for the light beam to extend 1 light second in front of the ship.

How is this possible?

Because, as seen from Earth, the guy on spaceship is time-dilated. He is moving very slowly according to Earth. (As are all clocks on his rocketship.) On Earth, we watch his clock tick by so slowly that is takes 100 seconds for the clock to tick over 1 second - just as the guy in the rocket watches one second tick by while the light beam reached out 1 light second.

 

[sammy4u]

Okay, whenever I have trouble visualizing a concept, I always try to look up some proofs with the aid of a diagram. Let us do the same.
Let's assume that we have a clock with a laser, that 'ticks' every time the beam hits the laser after reflecting off:
___________ (That is the top surface that it reflects off)
| (That is the beam of light that is going to hit the surface and reflect off


____()______ (That is the bottom surface, and the thing in between is the part where the beam was blasted)

Let's say that the distance between the shooting surface and the reflecting surface is L and the speed of light is C. So the time taken : T = 2L/C.

Now let's assume that the surfaces are moving at some velocity, let's call it V.

_______A_________ (Again our top surface, but in motion)
/ \
/ \ (Since, the ship is in motion, the beam blaster thingy will move, and due
/ \ to relativity, the beam will be like that too, on an angle)
/ \
/ \
___()______()___
B D C


So the distance between the midpoint at the bottom and the beginning is point where it reaches the top will be: BD = vt/2 and the length is still L. Using pythagorean theorem, we can say that BD^2 + L^2 = AB^2, so (BD^2 + L^2)^0.5 = AB. Since AB and CD are identical, we can say that : 2*(BD^2 + L^2)^0.5 = AB = Ct(1). C is still the speed of light, and t(1) will be the time it took for it to travel that distance. Since the speed of light is a constant, we can say the above.

So now we have the following:

Ct(1) = 2*(BD^2 + L^2)^0.5

With a some math bashing, which you should be able to do, like square both sides and get t(1) on one side, you should get this:

t(1) = 2L / (C (1 - V^2/c^2))

But remember we said already that 2L/C was T? So now:

t(1) = T / (1 - c^2/V^2)

where t(1) is the time indicated by the faster person's clock, T is the time indicated by the slower person's clock, and V is the velocity of the faster person with respect to the slower person. Still don't get it, well here's a story for you :D

There are two twin brothers. On their thirtieth birthday, one of the brothers goes on a space journey in a superfast rocket that travels at 99% of the speed of light. The space traveller stays on his journey for precisely one year, whereupon he returns to Earth on his 31st birthday. On Earth, however, seven years have elapsed, so his twin brother is 37 years old at the time of his arrival. This is due to the fact that time is stretched by factor 7 at approx. 99% of the speed of light, which means that in the space traveller’s reference frame, one year is equivalent to seven years on earth. Yet, time appears to have passed normally to both brothers, i.e. both still need five minutes to shave each morning in their respective reference frame.

Still not? Then go to youtube, and write Physics for Smarties: Time Dilation. He does pretty much what I did, but with better diagrams :S. Anyways, hope I helped. :D

 

[sammy4u]

Oops, my diagrams are messed, just go to the youtube link if you don't understand verbally :/

 

[Dale]

To get ASCII art to show correctly just use [ code ] tags.

 

[rabcarl]

Simplistically:

Imagine a guy in a rocket traveling at .99c. He has a flashlight, which he shines out the front window.

He sees the beam traveling at c ahead of his ship. In one second, his light beam has traveled one light second ahead of him.

On Earth, we see a rocket with a guy with a flashlight shining out a window. The light beam is traveling at c, but the rocket is following very close behind it at .99c. At this rate, it will take not 1 second, but 100 seconds for the light beam to extend 1 light second in front of the ship.

How is this possible?

Because, as seen from Earth, the guy on spaceship is time-dilated. He is moving very slowly according to Earth. (As are all clocks on his rocketship.) On Earth, we watch his clock tick by so slowly that is takes 100 seconds for the clock to tick over 1 second - just as the guy in the rocket watches one second tick by while the light beam reached out 1 light second.

But if we saw his craft moving at all, even if it was moving extremely slowly due to time dilation, wouldn't that be enough to put the light's net speed over 1.0c? Sorry if I sound dumb I've only taken an introductory course in physics in high school and the instructor's explanation didn't make much sense to me.

 

[Nugatory]

But if we saw his craft moving at all, even if it was moving extremely slowly due to time dilation, wouldn't that be enough to put the light's net speed over 1.0c?

What we see: A spaceship past us, from left to right, .99c and a beam of light moving ahead of the spaceship at speed c. Note that we see the difference between the speed of the light and the spaceship as .01c, but we see the light moving at c.

What he sees: He is in a spaceship floating at rest in empty space, while the Earth is zooming away from him in the sternwards direction at .99c. He also sees a beam of light moving forwards away from him at speed c. Note that he sees the light moving at c.

This all works (and we don't see the light moving at 1.99c as you'd expect of you just added the speeds to get a "net speed") because speed is defined as distance divided by time, and thanks to time dilation and length contraction, the two of us are using different notions of time and distance.

 

[DaveC426913]

But if we saw his craft moving at all, even if it was moving extremely slowly due to time dilation, wouldn't that be enough to put the light's net speed over 1.0c?

No, in the Earth's frame of reference the flashlight beam is measured as moving at c relative to Earth. Since the rocket is moving at .99c, on Earth they would see the flashlight beam receding relative to the rocket at only .01c. That's why it takes 100 seconds before the Earth observatory declares the light beam has reached 1 light second ahead of the ship.

 

[Dale]

Since the rocket is moving at .99c, on Earth they would see the flashlight beam receding relative to the rocket at only .01c.

I would say "on Earth they would see the flashlight beam moving with a separation speed from the rocket of 0.01 c." I usually use "relative speed" to mean the speed of the other in one of their rest frames, and "closing speed" or "separation speed" to mean the change in distance over the change in time in a frame where neither are at rest.

 

[rabcarl]

Ok, this makes a little more sense now. So in the Earth's frame of view we could theoretically see the light beam receding from his craft more and more slowly as his speed increased arbitrarily close to c. But for the man in the rocket the light would always be receding at c. But if the Earth observer says the light travels distance x in 100 seconds whereas the observer in the rocket says the light travels distance x in 1 second wouldn't there be a discrepancy in the speed of light between the two frames?

 

[DaveC426913]

Ok, this makes a little more sense now. So in the Earth's frame of view we could theoretically see the light beam receding from his craft more and more slowly as his speed increased arbitrarily close to c. But for the man in the rocket the light would always be receding at c. But if the Earth observer says the light travels distance x in 100 seconds whereas the observer in the rocket says the light travels distance x in 1 second wouldn't there be a discrepancy in the speed of light between the two frames?

No. On Earth, that beam of light is moving at c. (It doesn't matter what the rocket is doing; it could turn around and go home, the light beam is still moving at c.)

 

[harrylin]

But if we saw his craft moving at all, even if it was moving extremely slowly due to time dilation, wouldn't that be enough to put the light's net speed over 1.0c? Sorry if I sound dumb I've only taken an introductory course in physics in high school and the instructor's explanation didn't make much sense to me.

Just to make sure: if for you the craft is moving slowly, the clock inside the craft will tick at about the normal rate according to you. Only if the craft is going extremely fast, then you will measure that the clock in the craft is ticking slower than normally. And that is called "time dilation".
An animation may be helpful, here is one (in fact two, first simply and then combined) that DuckDuckGo found for me:
http://webphysics.davidson.edu/physlet_resources/special_relativity/illustration4.html

 

[DaveC426913]

Just to make sure: if for you the craft is moving slowly, the clock inside the craft will tick at about the normal rate according to you. Only if the craft is going extremely fast, then you will measure that the clock in the craft is ticking slower than normally. And that is called "time dilation".

Just to make even more sure, all of what harrylin said applies to the Earth observing the rocket, not the passenger of the rocket. i.e.:

if for you on Earth the craft is moving slowly, you will see the clock inside the craft will tick at about the normal rate according to you on Earth. Only if the craft is going extremely fast, then you will measure that the clock in the craft is ticking slower than normally. And that is called "time dilation".

No observer - on Earth or in a rocket - ever observes their own frame doing anything other than passing at normal time.

 

[rabcarl]

Would there be a discrepancy between the frames as to the velocity the light was receding from the spacecraft ?

 

[DaveC426913]

Would there be a discrepancy between the frames as to the velocity the light was receding from the spacecraft ?

Both rocket passenger and Earth observer measure the beam of light moving at c. (i.e. they both observe the beam traveling a distance of one light second over a duration of one second.)

Rocket passenger observes himself as stationary, and that light is receding from him at c - which makes perfect sense to the rocket passenger.

Earth passenger observes light moving at c, and rocket following closely behind light beam at .99c. So they see the light separating from the rocket at a net speed of .01c - which makes perfect sense to Earth.

 

[FieldTheorist]

I've read that time and space dilate under acceleration to ensure the constant speed of light. I'm having trouble visualizing a scenario that illustrates this concept. Could someone please explain this to me? Thanks.

Actually, not exactly. It's under relative velocity differences (say you're traveling at a constant 0.1 c from me) that time and space distort to keep the speed of light constant. Under acceleration, the speed of light isn't constant (although time still does not appear to pass from the photon's frame of reference).

The speed of light is, fundamentally, constant because time and space are distorted, relative to two viewers who're moving with respect to one another. Explicitly, the distortions are given by the Lorentz transformation:

x' = \gamma_v ( x - v t )
t' = \gamma_v \left( -\frac{v}{c^2}x + t \right)

Where gamma is defined as:

\gamma_v = \frac{1}{\sqrt{ 1 - \left( \frac{v}{c} \right)^2}},

such that u is the velocity that you are traveling relative to one another. Here we see that under this transformation, the speed of light is constant between two reference frames:

u = \frac{\Delta x}{\Delta t} = c \Longrightarrow u' = \frac{\Delta x'}{\Delta t'} = c

(i.e. if an object is traveling at the speed of light in one frame, it's traveling at the speed of light in any frame relative to it. Here's the proof:


u' = \frac{\Delta x'}{\Delta t'} = \frac{\gamma_v ( \Delta x - v \Delta t )}{\gamma_v \left( -\frac{v}{c^2} \Delta x + \Delta t \right)} = \frac{ \Delta x - v \Delta t }{ -\frac{v}{c^2} \Delta x + \Delta t } = \frac{ \Delta x - v \Delta t }{ -\frac{v}{c^2} \Delta x + \Delta t }\frac{1/ \Delta t}{1/ \Delta t} = \frac{ \frac{ \Delta x}{ \Delta t} - v}{ -\frac{v}{c^2} \frac{\Delta x}{\Delta t} + 1 }

Here we can recall that

u = \frac{\Delta x}{\Delta t} = c

Because it's moving at the speed of light in the other observer's frame of reference. Plugging this in, we see that:

\frac{ c - v}{ -\frac{v}{c^2} c + 1 } = \frac{ c - v}{ -\frac{v}{c^2} c + 1 } \frac{\frac{c}{c}}{1} = \frac{ 1 - \frac{v}{c}}{ -\frac{v}{c} + 1 } c = c

So the time dilation and spatial contractions --what the Lorentz transformations amount to physically-- conspire to cause the speed of light to remain constant between inertial observers.

 

[D H]

I've read that time and space dilate under acceleration to ensure the constant speed of light.

I prefer to look at it the other way around: Time dilation and length contraction are a necessary consequence of the constancy of the speed of light. Taking time dilation and length contraction as axiomatic is bit too ad hoc for my tastes.

Starting with some very basic concepts of continuity dictates that some speed must be the same to all observers. You get Newtonian mechanics if this universally agreed-upon speed is infinite. You get a Lorentz manifold if this universally agreed-upon speed is finite. Maxwell's equations and multiple experiments indicate that this universally agreed-upon speed is finite and is in fact the speed of light.