B Does time dilation work in 1d space?

[jbriggs444]

Thanks for that! So comparing my initial reference frame, where I was standing still, to the frame of reference where I am moving, I need to do the calculations for time and distance transformations as these change according to the velocity I am traveling? As in, time and length are literally different due to the shift in spacetime? So in my own "rest" frame, no matter if I am still or running, it appears that nothing has changed with regard to time and space. But if we compared the two, time and distance has indeed changed because I've rotated through the "spacetime" plane. Please tell me I get it... because I feel like I am starting to grasp the twins paradox... but I don't want to think I get it, when I don't.

This sounds very much like you are starting to "get it".

It is not that space-time has changed really. It is that the coordinate system you've laid down over it has changed.

Still, it is not just a trick of coordinates. The underlying space-time geometry is such that all of these coordinate systems actually work. And work identically.

 

[Herman Trivilino]

If I am watching a car travel towards me and I run towards it our combined velocity will be equal to the velocity I experience from my frame of reference.

Yes, but the way you combine them is not by adding them. This is analogous to stacking a pair of wedges. You set the first wedge on a level table top and measure the angle that the upper surface of the wedge makes with the table top. You also measure the slope of that upper surface. Now you stack a second wedge on top of the first wedge in such a way that the upper surface of the second wedge forms a steeper slope. You can add the angles of the wedges to get the angle of the upper surface. But you cannot add the slopes to get the slope of the upper surface. However, if the slopes are small adding them gives a good approximation. Likewise, adding speeds gives a good approximation when the speeds are slow.

My movement in the spaceship towards the light does not add or subtract velocity from the speed of the light in the way my running towards a car increases the combined velocity.

We would use the same formula in both cases.

But from inside the spaceship the light would still appear to be traveling at the same speed, it would not appear to be traveling at the combined velocity of c plus the spaceship velocity.

Yes it would, provided that you combine them properly. Adding them is not the correct way to combine them.

The fact the light hits the spaceship more quickly is the strange part.

But it doesn't hit it more quickly. The flashes of light arrive more frequently because they are closer together.

If you have a continuous beam of light (instead of flashes) then you note that the crests of the light wave arrive more frequently because they are closer together, not because they move at a higher speed.

 

[YouAreAwesome]

Yes, but the way you combine them is not by adding them. This is analogous to stacking a pair of wedges. You set the first wedge on a level table top and measure the angle that the upper surface of the wedge makes with the table top. You also measure the slope of that upper surface. Now you stack a second wedge on top of the first wedge in such a way that the upper surface of the second wedge forms a steeper slope. You can add the angles of the wedges to get the angle of the upper surface. But you cannot add the slopes to get the slope of the upper surface. However, if the slopes are small adding them gives a good approximation. Likewise, adding speeds gives a good approximation when the speeds are slow.
View attachment 264738
We would use the same formula in both cases.
Yes it would, provided that you combine them properly. Adding them is not the correct way to combine them.
But it doesn't hit it more quickly. The flashes of light arrive more frequently because they are closer together.

If you have a continuous beam of light (instead of flashes) then you note that the crests of the light wave arrive more frequently because they are closer together, not because they move at a higher speed.

With the triangles you have shown, where is distance, time and velocity? What do the angles in the triangles represent? Is tan of the angle the velocity, the opposite side the distance and the adjacent side time?

 

[jbriggs444]

With the triangles you have shown, where is distance, time and velocity? What do the angles in the triangles represent? Is tan of the angle the velocity, the opposite side the distance and the adjacent side time?

It is a Euclidean analogy to a Minkowsky geometry.

An acceleration in Minkowski space corresponds to a rotation in Euclidean space.

The origin for the drawing it the common vertex on the right.

You have the original trajectory (the lower line) which is a time elapsed horizontally (horizontal = time) and a certain distance vertically (vertical = space).

You have the the first change in velocity. (the lower triangle). You have an additional distance moved (the vertical line on the left). This takes place over the original horizontal time line. This yields a new trajectory (the diagonal line on the lower triangle).

Now you reset your frame of reference so that this diagonal line is your new baseline trajectory. It is your new horizontal time line. Its "horizontal" length is elapsed time. The "vertical" distance moved is zero. This new baseline is at zero velocity.

The next short segment on the left is a displacement in this new frame. Over the duration of the trajectory some object moves that much distance. Its trajectory is the long upper diagonal line in the upper triangle. You can read this trajectory as distance in the original frame versus time in the original frame. Or you can rotate your point of view and read it as distance in the second frame versus time in the second frame.

 

[YouAreAwesome]

This sounds very much like you are starting to "get it".

It is not that space-time has changed really. It is that the coordinate system you've laid down over it has changed.

Still, it is not just a trick of coordinates. The underlying space-time geometry is such that all of these coordinate systems actually work. And work identically.

Awesome, thanks.

So back to the original idea that a spaceship is traveling back to Earth in view of a video projection from Earth of a clock, does the spaceship see the clock tick faster on the return journey? If yes, is it because the spacetime coordinate system "rotates"? I'm thinking it does tick faster, and it's quite a scary thought in practise. If one day we can travel at close to light speed, and we find ourselves far, far away from earth, then the only way back would be to arrive at an Earth that has aged many years. Thus we don't return to our loved ones. The faster we return, the less time we get to spend with them. A sad thought.

 

[jbriggs444]

So back to the original idea that a spaceship is traveling back to Earth in view of a video projection from Earth of a clock, does the spaceship see the clock tick faster on the return journey?

Yes. That's the Doppler effect. The space ship sees the clock tick fast. Not as fast as in a Newtonian universe, but still faster than the on board clock.

The faster you return, the more time you get to spend with them. You see the clock speed up, yes. But you see it run faster for a shorter time. Less total elapsed time shown on the video.

 

[Dale]

Remark: Yes, it is legal to simply add or subtract these speeds to obtain the relative speeds as viewed by B.

I thought this was illegal. I thought if we measure the speed of light in the positive x direction relative to the speed of light in the negative x direction we are not allowed to attain a velocity of 2c.

It is a valid mathematical operation. However, I would not call it “the relative speed as viewed by B” precisely to avoid the confusion you are feeling.

I would reserve the term “relative speed” of two objects for the speed of one object measured in the frame of the other. I would use the term “separation speed” to describe the difference in velocity in a frame where neither is at rest.

 

[Herman Trivilino]

With the triangles you have shown, where is distance, time and velocity?

Nowhere. It's a drawing of a pair of wedges. A wedge is something that you might use, for example, to keep a door open.

 

[robphy]

Here is the spacetime-diagram example of the Euclidean "slope composition" diagram given by @Mister T and elaborated on by @jbriggs444 .
(Time runs upwards.)

Below, \bot indicates that the two lines meeting there are Minkowski-perpendicular (that observer's spaceline (their x-axis) is Minkowski-perpendicular to that observer's timeline (their t-axis, their worldline)).

Choosing arithmetically-convenient values...

By counting diamonds, the velocity of Bob with respect to Alice [using Alice's diamonds for her time and space units]
v_{BA} =\frac{6}{10}, and so forth.

<br /> \begin{align*}v_{CA}<br /> &amp;=\frac{v_{CB}+v_{BA}}{1+v_{CB}v_{BA}}\\<br /> &amp;=\frac{(\frac{5}{13})+(\frac{6}{10})}{1+(\frac{5}{13})(\frac{6}{10})}=(\frac{4}{5})=\frac{16}{20}<br /> \end{align*}<br />

 

[YouAreAwesome]

Again thanks for all the replies. So if a person is traveling at close to the speed of light towards a clock, does it appear the person traveling that the clock is ticking faster?

 

[Nugatory]

Again thanks for all the replies. So if a person is traveling at close to the speed of light towards a clock, does it appear the person traveling that the clock is ticking faster?

If the traveller is watching through a telescope, yes.
However, when we allow for the light travel time (we're getting closer to the clock so with every tick the light reaches us sooner) we will calculate that the clock is ticking slower than ours. We see the ticks happening faster because of the Doppler effect, but not as much faster as we'd expect from a non-relativistic calculation.

And note that we will get the same result if we say that the "traveller" is not moving while the clock is rushing towards them at close to teh speed of light.

 

[Ibix]

Again thanks for all the replies. So if a person is traveling at close to the speed of light towards a clock, does it appear the person traveling that the clock is ticking faster?

@Nugatory already answered this, noting that there's a distinction between what you literally see (clocks apparently running faster or slower as you approach or recede from them) and what you calculate is happening once you correct for the changing light travel time (moving clocks run slowly). Just to add to that, you need to be very wary whenever you come across the word "see" in a discussion of relativity, because it is often used sloppily to mean the calculated effect that you don't literally see (!). For example, it's quite common to come across statements like "a moving observer sees my clock tick slowly just as I see hers tick slowly". That's a correct statement about time dilation if I take "see" to mean "measure the apparent clock rate and correct for the changing distance", but that's not what "see" normally means. Such usage is, therefore, potentially very confusing and unfortunately rather common. Most people here will try to be clear about the distinction, and say things like "you literally see..." and "once you've corrected for the changing distance you'll find...", because we've corrected the misunderstandings so many times that being careful about it is second nature, but many sources are nowhere near as careful.

 

[PeroK]

Again thanks for all the replies. So if a person is traveling at close to the speed of light towards a clock, does it appear the person traveling that the clock is ticking faster?

Yes, but that has nothing to do with special relativity. That's simply the varying travel time of the signal from a sequence of events to an observer. In fact, the first good estimate of the speed of light was made using this idea back in 1676 by measuring the amount by which the orbit of one of Jupiter's moons got out of sync as Jupiter got further or closer to the Earth. You can read about that here, for example:

https://en.wikipedia.org/wiki/Speed_of_light#First_measurement_attempts

Note that SR is not about the finiteness of the speed of light but about the invariance of the speed of light across all inertial frames of reference.