Can't get time dilation to add up

[lauge]

Hi.

I recently bought A Brief History of time by Stephen Hawking and has since been very intrigued by the whole idea of time dilation. I've been searching on the internet and reading a lot about it and it is beginning to make sense :)

However I have a problem that I just don't seem to get. I've been trying to find an answer online for the last couple of days, hopefully someone here can help me out.

The example is as follows:

If I'm sitting in a spacecraft , traveling at half the speed of light towards the sun, which again is shining light in my direction at the speed of light (duh!), Newtonian logic says that the light should pass by me at the speed of 3/2 c, right?

Since we know that this isn't possible, time dilation must be occurring, since the speed of light is always the same no matter my velocity.

My problem is the following. In order to experience the light as approaching me at the speed of c, time must be slowed down by 33% for me, right?

But according to the good old 1/sqrd(1-(v^2/c^2)) formula, the Relativistic Change Factor is only 1.1547005383792517 ?!

Now, I'm sure that I'm missing something, it's just what?

Can someone please explain this for me.

Thanks :)

 

[Irrational]

i don't think you can use the forumula $$t' = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}}$$ here.

one of the clocks is not traveling on the beam of light.

 

[fatra2]

Hi there,

time must be slowed down by 33% for me, right?

Wrong. You 33% is thinking into a linear sense. But like you said it, with the good old $$\frac{1}{\sqrt{c^2-v^2}}$$ tells you that we are not living into a linear world. Therefore, you cannot say that time must dilate by $$\frac{2}{3}$$.

Cheers

 

[lauge]

Thanks for your replies, but I'm afraid I still haven't gained any clarity :)

How does it add up then?

I mean; if someone watching all this from the sideline sees the light from the sun moving at the speed of light, and I, moving toward it, also see the light as moving from the sun at the speed of light, mustn't that imply that I'm experiencing time as moving 33% slower than the spectator on the sideline?

 

[fatra2]

That's the beauty of relativity. Time, for you on the spaceship, slows down (compared to someone on Earth). But it does not slow down linearly to your speed. Your time would slow down by $$\approx 13\%$$ compared to your Earthly observer.

 

[Doc Al]

The example is as follows:

If I'm sitting in a spacecraft , traveling at half the speed of light towards the sun, which again is shining light in my direction at the speed of light (duh!), Newtonian logic says that the light should pass by me at the speed of 3/2 c, right?

That would certainly be true according to pre-Einstein Galilean relativity. If someone at rest with respect to the sun sees the light move at speed c, then from your frame (moving at c/2 with respect to his) it would be moving at 3/2 c.

Since we know that this isn't possible, time dilation must be occurring, since the speed of light is always the same no matter my velocity.

Not only time dilation, but length contraction and clock desynchronization.

My problem is the following. In order to experience the light as approaching me at the speed of c, time must be slowed down by 33% for me, right?


But according to the good old 1/sqrd(1-(v^2/c^2)) formula, the Relativistic Change Factor is only 1.1547005383792517 ?!

You're thinking about how moving clocks are measured to run slow. If you observe a clock rushing past you at speed c/2, then you would measure it to run slow by a factor of 1.15. (By the way, you always see your own clocks running normally... it's only the other frame's clocks that are affected by time dilation.)

Now, I'm sure that I'm missing something, it's just what?

To understand how speeds transform from one frame to another, you have to include more than just time dilation of moving clocks. Think of how you would measure that speed of light with respect to you. Here's one way. Imagine your spacecraft is many, many miles long. You measure an incoming beam of light as it passes by your ship. Using a clock at the front of your ship, you measure the time at which it passes the front; similarly, using a clock at the rear of your ship, you measure the time at which it passes the rear. Knowing that your clocks are synchronized and the distance between them, you use that data to find the speed of the light beam. Of course, as relativity tells us, you find that speed to be c.

How do things seem from the observers in the other frame, moving at c/2 with respect to you? They of course also measure the speed of the beam as it passes by your ship to be c with respect to them. But according to them, your measurements need to be reinterpreted: You view your clocks as synchronized, but the moving frame sees the clock in the front of your ship running behind the clock in the rear of your ship (the relativity of simultaneity); you measure a certain distance between the front and rear of your ship, but the moving frame measures a smaller distance (length contraction); you see your clocks running at their normal rate, but the moving frame measures them to run slow (time dilation). When you put all those factors together, it turns out that the speed of light is the same in any frame. (This should be no surprise, since the relativistic effects are derived by assuming that the speed of light is the same in any frame. )

 

[Doc Al]

Just to add to my previous post, here's how velocities add up (just for the record):

Under Galilean relativity (as you know):

$$V_{a/c} = V_{a/b} + V_{b/c}$$

For this case:
V = c + c/2 = 3/2 c

Under special relativity:

$$V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}$$

For this case:
V = (c + c/2)/(1 + (1/2)*1) = (3/2 c) / (3/2) = c

(By the way, the velocity addition formulas apply for any speeds.)

 

[George Jones]

i don't think you can use the forumula $$t' = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}}$$ here.

one of the clocks is not traveling on the beam of light.

Right, the above expression only works for special cases.

Suppose I explode two firecrackers at different times. In my frame, the spatial separation of the two events is $\Delta x = 0$ and the time separation is $\Delta t \neq 0$.

Suppose further that I am moving with constant direction and constant speed $v$ with respect to Shazia. Then, Shazia measures the time separation between the explosions to be $\Delta t' = \Delta t / \sqrt{1 - v^2/c^2}$.

Now suppose that Pareesa moves with constant speed $V \neq v$ with respect to Shazia. Pareesa measures the time separation between the explosions to be $\Delta t''$, but Shazia's time separation [tex]\Delta t'[/itex] and Pareesa's time separation $\Delta t''$ are not related by a factor of $\sqrt{1 - V^2/c^2}$, because the spatial separation bewteen the explosions is non-zero for both Shazia and Pareesa. Elapsed times are only related by this simple factor when the spatial separation between the events is zero for one of the observers.

Now back to the speed of light. Consider two events that a photon experiences. An observer calculates the speed of light by dividing the spatial separation between the two events by the time separation between the two events. Different observers will measure different times separations, but the time separations fro two observers will not be related by a factor of $\sqrt{1 - v^2/c^2}$ because the spatial separations between the events is non-zero for all observers.

A more general relationship, called a Lorentz transformation, is needed. When the general relationship for time and space tranformations between observers is used,

$$\frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}$$

can be derived as the relativistic way to combine speeds.

Try $v_1 = c/2$, and $v_2 = c$.

[EDIT]I see Doc Al gave good explanations while I was typing.[/EDIT]

 

[Nisse]

Correct me if I'm wrong, but I think what the OP meant was:

"How much slower am I moving through time if I'm moving through space at 0.5c?"

I would have thought the answer to be 0.5c. Thus he still experiences the light coming towards him at speed c.

 

[granpa]

you arent taking relativity of simultaneity into account. that's what confuses all beginners. that's what doc was talking about.

 

[tiny-tim]

Welcome to PF!

Hi lauge! Welcome to PF!

If I'm on Earth, and you're on the spaceship, then the √(1 - v²/c) is the time dilation for my clock as seen by you (or yours as seen by me) …

but it's also the "length dilation" (ie contraction) …

when you measure the sunlight going distance ct in time t,

I measure it as going distance ct/√(1 - v²/c) in time t/√(1 - v²/c)

 

[DB Katzin]

Just to add to my previous post, here's how velocities add up (just for the record):

Under Galilean relativity (as you know):

$$V_{a/c} = V_{a/b} + V_{b/c}$$

For this case:
V = c + c/2 = 3/2 c

Under special relativity:

$$V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}$$

For this case:
V = (c + c/2)/(1 + (1/2)*1) = (3/2 c) / (3/2) = c

(By the way, the velocity addition formulas apply for any speeds.)

Hi- I am a bit new to PF and have also been struggling with some of the more basic aspects of Special Relativity--see "is the behavior of light all that remarkable" for a few juicy missteps. I was referred by George Jones for an understanding of how velocities add in special relativity. In the formula above, if V{b,c} = c the formula reduces to an identity which holds for all cases where V{a,b} is any real and finite number including zero and V{b,c) is any real and finite number excluding zero. This is because (simplifying subscripts) the denominator V1xV2/c*2 +1 where V2=c reduces to V1xx/c*2 +1 which equals V1/c + 1, and the numerator
V1 + V2 = V1 + c becomes V1xc/c + c by multiplying V1 by c/c=1 for all real, finite numbers c. Factoring c makes the numerator c(V1/c +1) which is c times the denominator, canceling the terms (V1/c +1) gives the result V(a,c)=c for not just all velocities V1 but for all real, finite numbers V1 as well and for all real, finite non-zero numbers c. Do you have an online reference for the derivation of the formula? Thanks